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Kinetic Theory Of Gases And Thermodynamics

Kinetic Theory Of Gases:

The kinetic theory of gases is based on the following basic assumptions.

1. A gas consists of a very large number of molecules. These molecules are identical, perfectly elastic, and hard spheres. They are so small that the volume of molecules is negligible compared with the volume of the gas.
2. Molecules do not have any preferred direction of motion, motion is completely random.
3. These molecules travel in straight lines and free motion most of the time. The time of the collision between any two molecules is very small.
4. The collision between molecules and the wall of the container is perfectly elastic. It means kinetic energy is conserved in each collision.
5. The path traveled by a molecule between two collisions is called the free path and the mean of this distance traveled by a molecule is called the mean free path.
6. The motion of molecules is governed by Newton’s law of motion
7. The effect of gravity on the motion of molecules is negligible.

Expression For The Pressure Of A Gas

Let us suppose that a gas is enclosed in a cubical box having length l. Let there be ‘ N ‘ identical molecules, each having mass ‘ m’ since the molecules are of the same mass and perfectly elastic, so their mutual collisions result in the interchange of velocities only.

Only collisions with the walls of the container contribute to the pressure of the gas molecules. Let us focus on a molecule having velocity v₁ and components of velocity v_x1, v_y1, and v_z1 along the x, y, and z-axis as shown in the figure.

The change in momentum of the molecule after one collision with wall BCHE

⇒ \(m v_{x_1}-\left(-m v_{x_1}\right)=2 m v_{x_1}\)

The time taken between the successive impacts on the face BCHE = \(\frac{\text { distance }}{\text { velocity }}\)

⇒ \(\frac{2 \ell}{v_{\mathrm{x}_1}}\)

Time rate of change of momentum due to collision = \(\frac{\text { change in momentum }}{\text { time taken }}\)

⇒ \(\frac{2 m v_{\mathrm{x}_1}}{2 \ell / \mathrm{v}_{\mathrm{x}_1}}=\frac{\mathrm{mv}_{\mathrm{x}_1}^2}{\ell}\)

Hence the net force on the wall BCHE due to the impact of n molecules of the gas is :

⇒ \(\mathrm{F}_{\mathrm{x}}=\frac{\mathrm{mv}_{\mathrm{x}_1}^2}{\ell}+\frac{\mathrm{mv}_{\mathrm{x}_2}^2}{\ell}+\frac{\mathrm{mv}_{\mathrm{x} 3}^2}{\ell}+\ldots \ldots \ldots .+\frac{\mathrm{mv}_{\mathrm{x}_{\mathrm{n}}}^2}{\ell}\)

⇒ \(\frac{m}{\ell}\left(v_{x_1}^2+v_{x_2}^2+v_{x_3}^2+\ldots \ldots \ldots \ldots+v_{x_n}^2\right)\)

⇒ \(\frac{\mathrm{mN}}{\ell}<\mathrm{v}_{\mathrm{x}}^2>\)

where < V²_x > = mean square velocity in x-direction. Since molecules do not favor any particular direction therefore

⇒\(\left\langle\mathrm{v}_{\mathrm{x}}^2\right\rangle=\left\langle\mathrm{v}_{\mathrm{y}}^2\right\rangle=\left\langle\mathrm{v}_{\mathrm{z}}^2\right\rangle\)

But

⇒\(\left\langle\mathrm{v}^2\right\rangle=\left\langle\mathrm{v}_{\mathrm{x}}^2\right\rangle+\left\langle\mathrm{v}_{\mathrm{y}}^2\right\rangle+\left\langle\mathrm{v}_{\mathrm{z}}^2\right\rangle\)

⇒ \(\left\langle\mathrm{v}_{\mathrm{x}}^2\right\rangle\)

⇒ \(\frac{\left\langle\mathrm{v}^2\right\rangle}{3}\) Pressure is equal to force divided by area.

P = \(\frac{\mathrm{F}_{\mathrm{x}}}{\ell^2}=\frac{\mathrm{M}}{3 \ell^3}\left\langle\mathrm{v}^2\right\rangle=\frac{\mathrm{M}}{3 \mathrm{~V}}\left\langle\mathrm{v}^2\right\rangle\).

Pressure is independent of x, y, and z directions.

Where l³ = volume of the container = V

M = total mass of the gas, <c² > = mean square velocity of molecules

⇒ \(P=\frac{1}{3} \rho\left\langle v^2\right\rangle\)

As PV = nRT , then total translational K.E. of gas = \(\frac{1}{2} M<v^2>=\frac{3}{2} P V=\frac{3}{2} n R T\)

Translational kinetic energy of 1 molecule = \(\frac{3}{2} \mathrm{kT}\)(it is independent of nature of gas) kT

⇒ \(\left\langle v^2\right\rangle=\frac{3 P}{\rho}\)

or \(v_{rm s}=\sqrt{\frac{3 P}{\rho}}=\sqrt{\frac{3 R T}{M_{\text {mole }}}}=\sqrt{\frac{3 k T}{m}}\)

Where v_rms is the root mean square velocity of the gas.

Pressure exerted by the gas is P = \(\frac{1}{3} \rho\left\langle v^2\right\rangle=\frac{2}{3} \times \frac{1}{2} \rho<v^2>\)

or \(P=\frac{2}{3} E, E=\frac{3}{2} P\)

Thus, the total translational kinetic energy per unit volume (it is called energy density) of the gas is numerically equal to \(\frac{3}{2} k T\) times the pressure exerted by the gas.

Important Points

1. \(\mathrm{v}_{\mathrm{rms}} \propto \sqrt{\mathrm{T}} \text { and } \mathrm{v}_{\mathrm{rms}} \propto \frac{1}{\sqrt{\mathrm{M}_{\text {mole }}}}\)
2. At absolute zero, the motion of all molecules of the gas stops.
3. At higher temperatures and low pressure or higher temperatures and low density, a real gas behaves as an ideal gas.

Maxwell’s Distribution Law

Distribution Curve – A plot of \(\frac{\mathrm{dN}(\mathrm{v})}{\mathrm{dv}}\)(number of molecules per unit speed interval) against c is known as Maxwell’s distribution curve. The total area under the curve is given by the integral.

⇒ \(\int_0^{\infty} \frac{d N(v)}{d v} d v=\int_0^{\infty} d N(v)=N\)

• Figure shows the distribution curves for two different temperatures. At any temperature, the number of molecules in a given speed interval dv is given by the area under the curve in that interval (shown shaded).
• This number increases, as the speed increases, upto a maximum and then decreases asymptotically toward zero.
• Thus, a maximum number of the molecules have speed lying within a small range centered about the speed corresponding to the peak (A) of the curve. This speed is called the ‘most probable speed’ v_P or v_mp. dN(v)

The distribution curve is asymmetrical about its peak (the most probable speed v_P) because the lowest possible speed is zero, whereas there is no limit to the upper speed a molecule can attain. Therefore, the average speed v is slightly larger than the most probable speed v_P. The root-mean-square speed, v_rms, is still larger

⇒ \(\left(v_{\text {rms }}>\bar{V}>v_p\right)\)

Average (or Mean) Speed:

⇒ \(\bar{v}=\sqrt{\frac{8}{\pi} \frac{k T}{m}}=1.59 \sqrt{\mathrm{kT} / \mathrm{m}}\) (derivation is not in the course)

RMS Speed:

⇒ \(v_{\mathrm{rms}}=\sqrt{\left\langle v^2\right\rangle}=\sqrt{\frac{3 \mathrm{kT}}{\mathrm{m}}}=1.73 \sqrt{\frac{\mathrm{kT}}{\mathrm{m}}}\)

Most Probable Speed:

The most probable speed v_P or v_mp is the speed possessed by the maximum number of molecules and corresponds to the maximum (peak) of the distribution curve. Mathematically, it is obtained by the condition.

⇒ \(\frac{d N(v)}{d v}=0\)[by substitution of formula of dN(v) (which is not in the course)]

Hence the most probable speed is

⇒ \(v_p=\sqrt{\frac{2 k T}{m}}=1.41 \sqrt{\mathrm{kT} / \mathrm{m}}\)

From the above expression, we can see that

⇒\(\mathrm{v}_{\mathrm{rms}}>\overline{\mathrm{v}}>\mathrm{v}_{\mathrm{p}}\)

Degree Of Freedom

A total number of independent coordinates that must be known completely specify the position and configuration of a dynamical system completely is known as “degree of freedom f”. The maximum possible translational degrees of freedom are three i.e.

⇒ \(\left(\frac{1}{2} m V_x^2+\frac{1}{2} m V_y^2+\frac{1}{2} m V_z^2\right)\)

The maximum possible rotational degrees of freedom are three i.e.

⇒ \(\left(\frac{1}{2} I_x \omega_x^2+\frac{1}{2} I_y \omega_y^2+\frac{1}{2} I_z \omega_z^2\right)\)

Vibrational degrees of freedom are two i.e. (Kinetic energy. of vibration and Potential energy of vibration)

Mono Atomic: (all inert gases, He, Ar, etc.) f = 3 (translational)

Diatomic: (gases like H₂, N₂, O₂ etc.) f = 5 (3 translational + 2 rotational)

If temp < 70 K for diatomic molecules, then f = 3

If the temp is between 250 K to 5000 K, then f = 5

If temp > 5000 K f = 7 [ 3 translational.+ 2 rotational + 2 vibrational ]

Maxwell’s Law Of Equpartition Of Energy

Energy associated with each degree of freedom = \(\frac{1}{2} \mathrm{kT}\). If the degree of freedom of a molecule is f, then the total kinetic energy of that molecule U = \(\frac{1}{2} \mathrm{fkT}\)

Internal Energy

The internal energy of a system is the sum of the kinetic and potential energies of the molecules of the system. It is denoted by U. Internal energy (U) of the system is the function of its absolute temperature (T) and its volume (V). i.e. U = f (T, V)

• In the case of an ideal gas, the intermolecular force is zero. Hence its potential energy is also zero.
• In this case, the internal energy is only due to kinetic energy, which depends on the absolute temperature of the gas. i.e. U = f (T).

For an ideal gas internal energy U = \(\frac{f}{2} n R T\)

Question 1. A light container having a diatomic gas enclosed within is moving with velocity v. The Mass of the gas is M and the number of moles is n.

1. What is the kinetic energy of gas w.r.t. center of mass of the system?
2. What is K.E. of gas w.r.t. ground?

Answer:

1. K.E. = \(\frac{5}{2} n R T\)

2. Kinetic energy of gas w.r.t. ground = Kinetic energy of center of mass w.r.t. ground + Kinetic energy of gas w.r.t. center of mass.

K.E. = \(\frac{1}{2} M V^2+\frac{5}{2} n R T\)

Question 2. Two nonconducting containers having volumes V₁ and V₂ contain monoatomic and diatomic gases respectively. They are connected as shown in the figure. Pressure and temperature in the two containers are P₁, T_1, and P₂, T₂ respectively. Initially stop cock is closed, if the stop cock is opened find the final pressure and temperature.

Answer:

⇒ \(n_1=\frac{P_1 V_1}{R T_1}\)

⇒ \(\mathrm{n}_2=\frac{\mathrm{P}_2 \mathrm{~V}_2}{R \mathrm{RT}_2}\)

n = n₁+ n₂(number of moles are conserved)

Finally, the pressure in both parts and the temperature of both gases will become equal.

⇒ \(\frac{P\left(V_1+V_2\right)}{R T}=\frac{P_1 V_1}{R T_1}+\frac{P_2 V_2}{R T_2}\)

From energy conservation

⇒ \(\frac{3}{2} n_1 R T_1+\frac{5}{2} n_2 R T_2\)

⇒ \(\frac{3}{2} n_1 R T+\frac{5}{2} n_2 R T\)

T = \(\frac{\left(3 P_1 V_1+5 P_2 V_2\right) T_1 T_2}{3 P_1 V_1 T_2+5 P_2 V_2 T_1}\)

P = \(\left(\frac{3 P_1 V_1+5 P_2 V_2}{3 P_1 V_1 T_2+5 P_2 V_2 T_1}\right)\left(\frac{P_1 V_1 T_2+P_2 V_2 T_2}{V_1+V_2}\right)\)

Indicator Diagram

A graph representing the variation of pressure or variation of temperature or variation of volume with each other is called or indicator diagram.

1. Every point of the Indicator diagram represents a unique state (P, V, T) of gases.
2. Every curve on the Indicator diagram represents a unique process.

Thermodynamics

Thermodynamics is mainly the study of the exchange of heat energy between bodies and the conversion of the same into mechanical energy and vice versa.

Thermodynamic System

The collection of an extremely large number of atoms or molecules confined within certain boundaries such that it has a certain value of pressure (P), volume (V), and temperature (T) is called a thermodynamic system.

Anything outside the thermodynamic system to which energy or matter is exchanged is called its surroundings. Taking into consideration the interaction between a system and its surroundings thermodynamic system is divided into three classes :

1. Open system: A system is said to be an open system if it can exchange both energy and matter with its surroundings.
2. Closed system: A system is said to be a closed system if it can exchange only energy (not matter with its surroundings).
3. Isolated system: A system is said to be isolated if it can neither exchange nor matter with its surroundings.

Zeroth Law Of Thermodynamics

If two systems (B and C) are separately in thermal equilibrium with a third one (A), then they are in thermal equilibrium with each other.

Equation Of State ( For Ideal Gases)

The relation between the thermodynamic variables (P, V, T) of the system is called an equation of state. The equation of state for an ideal gas of n moles is given by PV = nRT,

Work Done By A Gas

Let P and V be the pressure and volume of the gas. If A is the area of the piston, then the force exerted by a gas on the piston is, F = P × A.

Let the piston move through a small distance dx during the expansion of the gas. Work done for a small displacement dx is dW = F dx = PA dx

Since A dx = dV, the increase in the volume of the gas is dV

⇒ dW = P dV

mg

or \(W=\int d W=\int P d V\)

The area enclosed under the P-V curve gives work done during the process.

Different Types Of Processes

1. Isothermal Process:

T = constant [Boyle’s law applicable] PV = constant

There is an exchange of heat between the system and its surroundings. The system should be compressed or expanded

very slowly so that there is sufficient time for the exchange of heat to keep the temperature constant.

The slope of the P−V curve in the isothermal process:

PV = constant = C

⇒ \(\frac{d P}{d V}=-\frac{P}{V}\)

Work done in the isothermal process:

W = \(n R T \quad \ell n \frac{V_f}{V_i}\)

[If v_f > v_i then W is positive

If v_f < v_i then W is negative]

W = \(\left[2.303 n R T \log _{10} \frac{V_f}{V_i}\right]\)

Internal energy in the isothermal process:

U = f (T) ⇒ ΔU = 0

2. Iso-Choric Process (Isometric Process):

V = constant

⇒ change in volume is zero

⇒ \(\frac{P}{T}\) is constant

⇒ \(\frac{P}{T}\)= const.(Galussac-law)

Work done in the isochoric process:

Since the change in volume is zero therefore dW = p dV = 0

Indicator diagram of the isochoric process:

Change in internal energy in isochoric process: ΔU = \(n \frac{f}{2} R \Delta T\)

Heat given in isochoric process: ΔQ = ΔU = \(n \frac{f}{2} R \Delta T\)

3. Isobaric Process: Pressure remains constant in the isobaric process

∴ P = constant

⇒ \(\frac{\mathrm{V}}{\mathrm{T}}\) = constant

Indicator diagram of the isobaric process:

Work done in the isobaric process:

ΔW = P ΔV = P (V_final – V_initial) = nR (T_final – T_initial)

Change in internal energy in the isobaric process: ΔU = n C_VΔT

Heat given in the isobaric process:

ΔQ = ΔU + ΔW

ΔQ = \(n \frac{f}{2} R \Delta T+P\left[V_f-V_i\right]\)

⇒ \(n \frac{f}{2} R \Delta T+n R \Delta T\)

The above expression gives the idea that to increase the temperature by ΔT in the isobaric process heat required is more than in the isochoric process.

4. Cyclic Process: In the cyclic process initial and final states are the same therefore initial state = final state

Work done = Area enclosed under P-V diagram.

Change in internal Energy ΔU = 0

ΔQ = ΔU + ΔW

∴ ΔQ = ΔW

If the process on the P-V curve is clockwise, then the net work done is (+ve) and vice-versa. The graphs shown below explain when work is positive and when it is negative

Question 1. The cylinder shown in the figure has conducting walls and the temperature of the surroundings is T, the position is initially in equilibrium, and the cylinder contains n moles of a gas. Now the piston is displaced slowly by an external agent to make the volume double the initial. Find work done by an external agent in terms of n, R, T

Answer:

1st Method:

Work done by external agents is positive because Fext and displacement are in the same direction. Since walls are conducting therefore temperature remains constant.

Applying the equilibrium condition when the pressure of the gas is P

⇒ \(P A+F_{\text {ext }}=P_{a t m} A\)

⇒ \(F_{\text {ext }}=P_{a t m} A-P A\)

⇒ \(W_{e x t}=\int_0^d F_{e x t} d x=\int_0^d P_{a t m} A d x-\int_0^d P A d x \)

⇒ \(P_{a t m} A \int_0^d d x-\int_V^{2 V} \frac{n R T}{V} d V=P_{a t m} A d-n R T \ln 2\)

⇒ \(P_{a t m} \cdot V_0^2-n R T \ln 2=n R T(1-\ln 2)\)

2nd Method

Applying the work-energy theorem on the piston

Δk = 0

Wall = Δk

W_gas+ W_atm + W_ext = 0

⇒ \(n R T \ln \frac{V_f}{V_i}-n R T+W_{\text {ext }}=0\)

W_ext = nRT (1 – ln2)

Question 2. Find out the work done in the given graph. Also, draw the corresponding T-V curve and P-T curve.

Answer:

Since in P-V curves area under the cycle is equal to work done therefore work done by the gas is equal to P₀ V₀. Lines A B and CD are isochoric lines, and lines BC and DA are isobaric lines.

∴ The T-V curve and P-T curve are drawn as shown.

Question 3. The t-V curve of the cyclic process is shown below, number of moles of the gas is n to find the total work done during the cycle.

Answer: Since path AB and CD are isochoric therefore work done is zero during path AB and CD. Process BC and DA are isothermal, therefore

⇒ \(W_{B C} =n R 2 T_0 \ln \frac{V_C}{V_B}=2 n R T_0 \ln 2\)

⇒ \(W_{D A} =n R T_0 \ln \frac{V_A}{V_D}=-n R T_0 \ln 2\)

Total work done = \(W_{B C}+W_{D A}=2 n R T_0 \ln 2-n R T_0 \ln 2\)

⇒ \(n R T_0 \ln 2\)

Question 4. The P-T curve of a cyclic process is shown. Find out the work done by the gas in the given process if several moles of the gas are n.

Answer: Since path AB and CD are isochoric therefore work done during AB and CD is zero. Path BC and DA are isobaric.

Hence W_BC = nRΔT = nR(T₃– T₂)

W_DA = nR(T₁– T₄)

Total work done = W_BC + W_DA = nR(T₁+ T₃–T₄– T₂)

Question 5. Consider the cyclic process ABCA on a sample of 2.0 mol of an ideal gas as shown in the figure. The temperatures of the gas at A and B are 300 K and 500 K respectively. A total of 1200 J heat is withdrawn from the sample in the process. Find the work done by the gas in part BC. Take R = 8.3J/mol–K.
Answer:

The change in internal energy during the cyclic process is zero. Hence, the heat supplied to the gas is equal to the work done by it. Hence,

W_AB + W_BC + W_CA = –1200 J. …….(1)

The work done during the process AB is

W_AB = P_A(V_B– V_A)

= nR(T_B– T_A)

= (2.0 mol) (8.3 J/mol–K) (200 K)

= 3320 J

The work done by the gas during the process CA is zero as the volume remains constant. From (1),

3320 J + W_BC = –1200 J

or W_BC = –4520 J.

= –4520 J.

First Law Of Thermodynamics

The first law of thermodynamics is the law of conservation of energy. It states that if a system absorbs heat dQ and as a result the internal energy of the system changes by dU and the system does a work dW, then dQ = dU + dW.

But, do = P dV dQ = dU + P dV

which is the mathematical statement of the first law of thermodynamics.

Heat gained by a system, work done by a system, and an increase in internal energy is taken as positive.

Heat lost by a system, work done on a system, and a decrease in internal energy are taken as negative.

Question 1. 1 gm water at 100ºC is heated to convert into steam at 100ºC at 1 atm. Find out the change in the internal energy of water. It is given that a volume of 1 gm water at 100ºC = 1 cc. volume of 1 gm steam at 100ºC = 1671 cc. Latent heat of vaporization = 540 cal/g. (Mechanical equivalent of heat J = 4.2J/cal.)
Answer:

From the first law of thermodynamic ΔQ = Δu + Δw

ΔQ = mL = 1 × 540 cal. = 540 cal.

ΔW = PΔV = \(\frac{10^5(1671-1) \times 10^{-6}}{4.2}\)

⇒ \(\frac{\left.10^5 \times 1670\right) \times 10^{-6}}{4.2}\) = 40 cal.

Δu = 540 – 40 = 500 cal.

Question 2. Two moles of diatomic gas at 300 K are kept in a non-conducting container enclosed by a piston. Gas is now compressed to increase the temperature from 300 K to 400 K. Find work done by the gas

Answer:

ΔQ = Δu + Δw

Since the container is non-conducting therefore

ΔQ = 0 = Δu + Δw

⇒ ΔW = – Δu = \(-\Delta u=-n \frac{f}{2} R \Delta T\)

⇒ \(-2 \times \frac{5}{2} R(400-300)\)

= –5 × 8.314 × 100 J

= – 5 × 831.4 J

= –4157 J

Question 3. A sample of an ideal gas is taken through the cyclic process of abaca (figure. It absorbs 50 J of heat during part ab, no heat during bc, and rejects 70 J of heat during ca. 40 J of work done on the gas during part bc.

1. Find the internal energy of the gas at b and c if it is 1500 J at a.
2. Calculate the work done by the gas during the part ca.

Answer:

1. In part ab the volume remains constant. Thus, the work done by the gas is zero. The heat absorbed by the gas is 50 J. The increase in internal energy from a to b is

ΔU = ΔQ = 50J.

As the internal energy is 1500 J at a, it will be 1550 J at b. In part bc, the work done by the gas is ΔW = –40J and no heat is given to the system. The increase in internal energy from b to c is

ΔU = –ΔW = 40 J.

As the internal energy is 1550 J at b, it will be 1590 J at c.

2. The change in internal energy from c to a is

ΔU = 1500 J – 1590 J = – 90 J.

The heat given to the system is ΔQ = – 70J.

Using ΔQ = ΔU + ΔW,

ΔW_ca= ΔQ – ΔU

= – 70 J + 90 J

= 20 J.

Question 4. The internal energy of a monatomic ideal gas is 1.5 nRT. One mole of helium is kept in a cylinder of cross-section 8.5 cm². The cylinder is closed by a light frictionless piston. The gas is heated slowly in a process during which a total of 42 J heat is given to the gas. If the temperature rises through 2ºC, find the distance moved by the piston. Atmospheric pressure = 100 kPa.
Answer:

The change in internal energy of the gas is

ΔU = 1.5 nR (ΔT)

= 1.5 (1 mol) (8.3 J/mol-K) (2K)

= 24.9 J.

The heat is given to the gas = 42 J

The work done by the gas is

ΔW = ΔQ – ΔU

= 42 J – 24.9 J = 17.1 J.

If the distance moved by the piston is x, the work done is

ΔW = (100 kPa) (8.5 cm²) x.

Thus, (10⁵ N/m²) (8.5 × 10^-4 m²) x = 17.1 J

or, x = 0.2m = 20 cm.

Question 5. A sample of ideal gas (f =5) is heated at constant pressure. If an amount of 140 J of heat is supplied to the gas, find

1. The change in internal energy of the gas
2. The work done by the gas.

Answer:

Suppose the sample contains n moles. Also, suppose the volume changes from V₁ to V₂ and the temperature changes from T₁ to T₂.

The heat supplied is

ΔQ = ΔU + PΔV = ΔU + nRΔT = \(\Delta U+\frac{2 \Delta U}{f}\)

1. The change in internal energy is

ΔU = \(n \frac{f}{2} R\left(T_2-T_1\right)=n \frac{f}{2} R\left(T_2-T_1\right)\)

⇒ \(\frac{\mathrm{f}}{2+\mathrm{f}} \Delta Q=\frac{140 \mathrm{~J}}{1.4}\)

= 100 J.

2. The work done by the gas is

ΔW = ΔQ – ΔU

= 140 J – 100 J

= 40 J.

Efficiency Of A Cycle (η):

⇒ \(\eta=\frac{\text { total Mechanical work done by the gas in the whole process }}{\text { Heat absorbed by the gas (only }+ \text { ve) }}\)

⇒ \(=\frac{\text { area under the cycle in } \mathrm{P}-\mathrm{V} \text { curve }}{\text { Heat injected into the system }}\)

⇒ \(\eta=\left(1-\frac{Q_2}{Q_1}\right)\) for Heat Engine,

⇒ \(\eta=\left(1-\frac{T_2}{T_1}\right)\) for Carnot cycle

Question 6. n moles of a diatomic gas have undergone a cyclic process ABC as shown in the figure. The temperature at a is T₀. Find

1. Volume at C?
2. Maximum temperature?
3. Total heat given to gas?
4. Is heat rejected by the gas, if yes how much heat is rejected?
5. Find out the efficiency

Answer:

1. Since triangles OA V₀ and OC V are similar therefore

⇒ \(\frac{2 P_0}{V}=\frac{P_0}{V_0}\)

⇒ \(V=2 V_0\)

2. Since process AB is isochoric hence

⇒ \(\frac{P_A}{T_A}=\frac{P_B}{T_B}\)

⇒ \(T_B=2 T_0\)

Since process BC is isobaric therefore \(\frac{T_B}{V_B}=\frac{T_C}{V_C}\)

⇒ \(T_{\mathrm{C}}=2 \mathrm{~T}_{\mathrm{B}}=4 \mathrm{~T}_0\)

3. Since the process is cyclic therefore

∴ ΔQ = ΔW = area under the cycle = \(\frac{1}{2} P_0 V_0\)

4. Since Δu and ΔW both are negative in process CA

∴ ΔQ is negative in process CA and heat is rejected in process CA

ΔQ_CA = Δw_CA + Δu_CA

⇒ \(-\frac{1}{2}\left[P_0+2 P_0\right] V_0-\frac{5}{2} n R\left(T_c-T_a\right)\)

⇒ \(-\frac{1}{2}\left[P_0+2 P_0\right] V_0-\frac{5}{2} n R\left(\frac{4 P_0 V_0}{n R}-\frac{P_0 V_0}{n R}\right)\)

= –9P₀V₀

= Heat injected.

4. η = efficiency of the cycle = \(=\frac{\text { work done by the gas }}{\text { heat injected }}\)

⇒ \(\eta=\frac{P_0 V_0 / 2}{Q_{\text {injected }}} \times 100\)

ΔQ_inj = ΔQ_AB + ΔQ_BC

⇒ \(\left[\frac{5}{2} n R\left(2 T_0-T_0\right)\right]+\left[\frac{5}{2} n R\left(2 T_0\right)+2 P_0\left(2 V_0-V_0\right)\right]\)

⇒ \(\frac{19}{2} P_0 V_0\)

⇒ \(\eta=\frac{100}{19} \%\)

Specific Heat

The specific heat capacity of a substance is defined as the heat supplied per unit mass of the substance per unit rise in the temperature. If an amount ΔQ of heat is given to a mass m of the substance and its temperature rises by ΔT, the specific heat capacity s is given by the equation

⇒ \(s=\frac{\Delta Q}{m \Delta T}\)

The molar heat capacities of a gas are defined as the heat given per mole of the gas per unit rise in the temperature. The molar heat capacity at constant volume, denoted by C_v, is :

⇒ \(C_v=\left(\frac{\Delta Q}{n \Delta T}\right)_{\text {constant volume }}\)

⇒ \(\frac{f}{2} R\)

and the molar heat capacity at constant pressure, denoted by C_P is,

⇒ \(C_{\mathrm{P}}=\left(\frac{\Delta Q}{\mathrm{n} \Delta T}\right)_{\text {constant volume }}\)

⇒ \(\left(\frac{f}{2}+1\right) R\)

• where n is the amount of the gas in several moles and f is a degree of freedom. Quite often, the term specific heat capacity or specific heat is used for molar heat capacity.
• It is advised that the unit be carefully noted to determine the actual meaning. The unit of specific heat capacity is J/kg-K whereas that of molar heat capacity is J/mol–K.

Molar Heat Capacity Of Ideal Gas In Terms Of R:

1. For a monoatomic gas f = 3

⇒ \(C_v=\frac{3}{2} R\)

⇒ \(C_p=\frac{5}{2} R\)

⇒ \(\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=\frac{5}{3}\)

= 1.67

2. For a diatomic gas f = 5

⇒ \(\mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R},\)

⇒ \(\mathrm{C}_{\mathrm{P}}=\frac{7}{2} \mathrm{R}\)

⇒ \(\gamma=\frac{C_P}{C_V}\)

= 1.4

3. For a Triatomic gas f = 6

⇒ \(C_V=3 R, C_P=4 R\)

⇒ \(\gamma=\frac{C_P}{C_V}=\frac{4}{3}\) = 1.33 [Note for CO₂; f = 5, it is linear]

In general, if f is the degree of freedom of a molecule, then,

⇒ \(C_V=\frac{f}{2} R, \quad C_P=\left(\frac{f}{2}+1\right) R\)

⇒ \(\gamma=\frac{C_P}{C_V}=\left[1+\frac{2}{f}\right]\)

Question 1. In a thermodynamic process, the pressure of a certain mass of gas is changed in such a way that 20 Joule heat is released from it and 8 Joule work is done on the gas. If the initial internal energy of the system is 30 joule then the final internal energy will be –
Answer:

dQ = dU + dW ⇒ dQ = U_final– U_initial + dW

U_final = dQ – dW + U_initial or U_final = –20 + 8 + 30 of U_final = 18 Joule

Question 2. A gas is contained in a vessel fitted with a movable piston. The container is placed on a hot stove. A total of 100 cal of heat is given to the gas and the gas does 40 J of work in the expansion resulting from heating. Calculate the increase in internal energy in the process.
Answer:

Heat given to the gas is ΔQ = 100 cal = 418 J.

Work done by the gas is ΔW = 40 J

The increase in internal energy is

ΔU = ΔQ – ΔW

= 418J – 40 J = 378 J

Question 3. A gas is compressed from volume 10 m³ to 4 m³ at a constant pressure of 50N/m². Gas is given 100 J energy by heating then its internal energy.
Answer:

P = 50 N/m²

dV = 10 – 4 = 6 m³

δW = PdV = 6 × 50 = 300 J (Volume is decreasing, δQ = 100 J)

W = – 300 J

δQ = δW + dU

100 + 300 = dU

dU = increased by 400 J

Question 4. Two moles of a diatomic gas at 300 K are enclosed in a cylinder as shown in the figure. The piston is light. Find out the heat given if the gas is slowly heated to 400 K in the following three cases.

1. The piston is free to move
2. If the piston does not move
3. If the piston is heavy and movable.

Answer:

1. Since pressure is constant

∴ ΔQ = nC_P ΔT = 2 × \(\frac{7}{2}\) 2× R × (400 – 300) = 700 R

2. Since volume is constant

∴ ΔW = 0 and ΔQ = Δu (from first law)

ΔQ = Δu = nC_vΔT = 2× \(\frac{5}{2}\) × R × (400 – 300) = 500 R

3. Since pressure is constant

∴ ΔQ = nC_P ΔT = 2 × \(\frac{7}{2}\) × R × (400 – 300) = 700 R

Question 5. The p-V curve of a diatomic gas is shown in the figure. Find the total heat given to the gas in the process AB and BC

Answer:

From the first law of thermodynamics

ΔQ_ABC = Δu_ABC + ΔW_ABC

⇒ \(\Delta W_{A B C}=\Delta W_{A B}+\Delta W_{B C}=0+n R T_B \ln \frac{V_C}{V_B}=n R T_B \ln \frac{2 V_0}{V_0}\)

⇒ \(n R T_B \ln 2=2 P_0 V_0 \ln 2\)

Δu = \(n C_v \Delta T=\frac{5}{2}\left(2 P_0 V_0-P_0 V_0\right)\)

⇒ ΔQ_ABC = \(\frac{5}{2}\) 2P₀V₀+ 2P₀V₀ln 2.

Question 6. Calculate the value of the mechanical equivalent of heat from the following data. The specific heat capacity of air at constant volume = 170 cal/kg-K, γ=C_P/C_v= 1.4, and the density of air at STP is 1.29 kg/m³. Gas consant R = 8.3 J/mol-K.
Answer:

Using pV = nRT, the volume of 1 mole of air at STP is

V = \(\frac{\mathrm{nRT}}{\mathrm{p}}=\frac{(1 \mathrm{~mol}) \times(8.3 \mathrm{~J} / \mathrm{mol}-\mathrm{K}) \times(273 \mathrm{~K})}{1.0 \times 10^5 \mathrm{~N} / \mathrm{m}^2}\)

= 0.0224m³

The mass of 1 mole is, therefore,

(1.29 kg/m³) × (0.0224 m³) = 0.029 kg.

The number of moles in 1 kg is \(\frac{1}{0.029}\). The molar heat capacity at constant volume is

⇒ \(C_v=\frac{170 \mathrm{cal}}{(1 / 0.029) \mathrm{mol}-\mathrm{K}}\)= 4.93 cal/mol-K.

Hence, C_P = γC_v= 1.4 × 4.93 cal/mol-K

or, C_P – C_v= 0.4 × 4.93 cal/mol-K

= 1.97 cal/mol-K.

Also, C_P– C_v= R = 8.3 J/mol-K.

Thus, 8.3 J = 1.97 cal.

The mechanical equivalent of heat is

⇒ \(\frac{8.3 \mathrm{~J}}{1.97 \mathrm{cal}}\) =4.2 J/cal.

Average Molar Specific Heat of Metals:

[Dulong and Petit law]

At room temperature average molar specific heat of all metals are same and is nearly equal to 3R ( 6 cal. mol^-1 K^-1).

[Note: Temp. above which the metals have constant C_V is called Debye temp.]

Mayer’s Equation: C_P− C_V= R (for ideal gases only)

Adiabatic Process:

When no heat is supplied or extracted from the system the process is called adiabatic. The process is sudden so there is no time for the exchange of heat. If the walls of a container are thermally insulated no heat can cross the boundary of the system and the process is adibatic.

The equation of the adiabatic process is given by

PV^γ = constant [Poission Law]

Tγ P^1-γ = constant

T V^γ-1 = constant

The slope of P−V−curve in the adiabatic process:

Since PV^γ is a constant

∴ \(\frac{\mathrm{dP}}{\mathrm{dV}}=-\gamma\left(\frac{\mathrm{P}}{\mathrm{V}}\right)\)

The slope of P−T−curve in the adiabatic process: Since T^γ P^1-γ is a constant

∴ \(\frac{d P}{d T}=-\frac{\gamma}{(1-\gamma)} \frac{P}{T}=\frac{(\gamma)}{(\gamma-1)} \frac{P}{T}\)

Slope of T−V−curve:

⇒ \(\frac{d V}{d T}=-\frac{1}{(\gamma-1)} \frac{V}{T}\)

Work Done In Adiabatic Process:

\(\Delta W=-\Delta U=n C_v\left(T_i-T_f\right)=\frac{P_i V_i-P_f V_f}{(\gamma-1)}=\frac{n R\left(T_i-T_f\right)}{\gamma-1}\)

work done by the system is (+ve), if T_i> T_f (hence expansion)

work done on the system is (−ve) if T_i< T_f (hence compression)

Question 7. A quantity of air is kept in a container having walls that are slightly conducting. The initial temperature and volume are 27ºC (equal to the temperature of the surroundings) and 800cm³ respectively. Find the rise in the temperature if the gas is compressed to 200cm³

1. In a short time
2. In a long time. Take γ = 1.4.

Answer:

1. When the gas is compressed in a short time, the process is adiabatic. Thus,

⇒ \(\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1}=\mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}\)

or \(\mathrm{T}_2=\mathrm{T}_1\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1}\)

⇒ \((300 \mathrm{~K}) \times\left[\frac{800}{200}\right]^{0.4}\)

= 522K

Rise in temperature = T₂– T₁= 222 K.

2. When the gas is compressed for a long time, the process is isothermal. Thus, the temperature remains equal to the temperature of the surroundings which is 27ºC. The rise in temperature = 0.

Question 8. A monoatomic gas is enclosed in a nonconducting cylinder having a piston that can move freely. Suddenly gas is compressed to 1/8 of its initial volume. Find the final pressure and temperature if the initial pressure and temperature are P₀ and T₀ respectively.

Answer:

Since the process is adiabatic therefore

⇒ \(P_0 V^{\frac{5}{3}}=P_{\text {final }}\left(\frac{V}{8}\right)^{5 / 3}\)

⇒ \(\gamma=\frac{C_P}{C_V}=\frac{5 R}{2} / \frac{3 R}{2}=\frac{5}{3}\)

⇒ \(P_{\text {final }}=32 P_0\)

Since the process is adiabatic therefore

⇒ \(\mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1}\)

⇒ \(\mathrm{T}_0 \mathrm{~V}_0^{2 / 3}=\mathrm{T}_{\text {frall }}\left(\frac{\mathrm{V}_0}{8}\right)^{2 / 3}\)

⇒ \(\mathrm{T}=4 \mathrm{~T}_0\)

Question 9. A cylindrical container having nonconducting walls is partitioned into two equal parts such that the volume of each part is equal to V₀. A movable nonconducting piston is kept between the two parts. The gas on the left is slowly heated so that the gas on the right is compressed upto volume \(\frac{V_0}{8}\). Find pressure and temperature on both sides if the initial pressure and temperature were P₀ and T₀ respectively. Also, find heat given by the heater to the gas. (number of moles in each part is n)

Answer:

Since the process on the right is adiabatic therefore PV^γ = constant

⇒ P₀ V^γ₀ = P_final (V₀/ 8)^γ ⇒ P_final = 32

P₀ T₀ V^γ-1₀ = T_final (V₀/8)^γ-1 ⇒ T_final = 4T₀

Let the volume of the left part be V₁

⇒ \(2 V_0=V_1+\frac{V_0}{8}\)

⇒ \(V_1=\frac{15 V_0}{8}\)

Since several moles on the left parts remain constant therefore for the left part

⇒ \(\frac{\mathrm{PV}}{\mathrm{T}}\) = constant.

The final pressure on both sides will be the same

⇒ \(\frac{P_0 V_0}{T_0}=\frac{P_{\text {final }} V_1}{T_{\text {final }}}\)

⇒ \(\mathrm{T}_{\text {final }}=60 \mathrm{~T}_0\)

⇒ \(\Delta Q=\Delta u+\Delta w\)

⇒ \(\Delta Q=n \frac{5 R}{2}\left(60 T_0-T_0\right)+n \frac{3 R}{2}\left(4 T_0-T_0\right)\)

⇒ \(\Delta Q=\frac{5 n R}{2} \times 59 T_0+\frac{3 n R}{2} \times 3 T_0\)

Free Expansion

If a system, say a gas expands in such a way that no heat enters or leaves the system and also no work is done by or on the system, then the expansion is called the “free expansion”.

ΔQ = 0 , ΔU = 0 and ΔW = 0. Temperature in the free expansion remains constant.

Question 1. A nonconducting cylinder having volume 2V₀ is partitioned by a fixed nonconducting wall in two equal parts. Partition is attached with a valve. The right side of the partition is a vacuum and the left part is filled with a gas having pressure and temperature P₀ and T₀ respectively. If the valve is opened the final pressure and temperature of the two parts are.

Answer: From the first law of thermodynamics ΔQ = Δu + ΔW

Since gas expands freely, therefore, ΔW = 0 since no heat is given to gas ΔQ = 0

⇒ Δu = 0 and temperature remains constant.

T_final = T₀

Since the process is isothermal therefore P₀× V₀= P_final × 2V₀

⇒ Pfinal = P₀/2

Reversible And Irreversible Process

A process is said to be reversible when the various stages of an operation in which it is subjected can be traversed back in the opposite direction in such a way that the substance passes through the same conditions at every step in the reverse process as in the direct process.

A process in which any one of the conditions stated for the reversible process is not fulfilled is called an irreversible process.

Comparison Of Slopes Of Iso-Thermal And Adiabatic Curve

In compression up to the same final volume: | W_adia |> | W_isothermal |

In Expansion up to the same final volume: W_isothermal > W_adia

Limitations Of Ist Law Of Thermodynamics:

The first law of thermodynamics tells us that heat and mechanical work are interconvertible. However, this law fails to explain the following points:

1. It does not tell us about the direction of the transfer of heat.
2. It does not tell us about the conditions under which heat energy is converted into work.
3. It does not tell us whether some process is possible or not.

Mixture Of Non-Reacting Gases:

1. Molecular weight = \(\frac{\mathrm{n}_1 \mathrm{M}_1+\mathrm{n}_2 \mathrm{M}_2}{\mathrm{n}_1+\mathrm{n}_2}\)

M₁ and M₂ are molar masses.

2. Specific heat CV = \(\frac{\mathrm{n}_1 \mathrm{C}_{\mathrm{V}_1}+\mathrm{n}_2 \mathrm{C}_{\mathrm{V}_2}}{\mathrm{n}_1+\mathrm{n}_2}\)

⇒ \(C_p=\frac{n_1 C_{P_1}+n_2 C_{P_2}}{n_1+n_2}\)

3. For mixture, γ = \(\gamma=\frac{\mathrm{C}_{{mix}}}{\mathrm{C}_{\mathrm{v}_{{mix}}}}\)

⇒ \(\frac{n_1 C_{p_1}+n_2 C_{p_2}+\ldots \ldots}{n_1 C_{v_1}+n_2 C_{v_2}+\ldots \ldots \ldots}\)

Question 1. 5 gm air is heated from 4ºC to 6ºC. If the specific heat of air at constant volume is 0.172 cal/gm/ºC, then the increase in the internal energy of air will be –
Answer:

dU = mC_vdT

dU = 5 × 0.172 × 2

dU = 1.72 calorie

Question 2. In the following indicator diagram, the net amount of work done will be –

Answer:

The cyclic process 1 is clockwise and the process 2 is anti clockwise. Therefore W₁ will be positive and W₂ will be negative area 2 > area 1, Hence the network will be negative.

Question 3. Two gram-moles of gas, which are kept at a constant temperature of 0ºC, are compressed from 4 liter to 1 liter. The work done will be
Answer:

W = 2.303 μ RT \(\log _{10} \frac{V_2}{V_1}\)

W = 2.303 × 2 × 8.4 × 273 \(\log _{10} \frac{1}{4}\)

W = 2.303 × 2 × 8.7 × 273 × (log₁₀ – log⁴₁₀)

∵ log⁴₁₀ = 0.6021

∴ W = –6359 Joule

Question 4. Air is filled in a motor car tube at 27ºC temperature and 2-atmosphere pressure. If the tube suddenly bursts then the final temperature will be \(\left[\left(\frac{1}{2}\right)^{2 / 7}=0.82\right]\)
Answer:

⇒ \(\mathrm{T}_2=\mathrm{T}_1\left(\frac{\mathrm{P}_2}{\mathrm{P}_1}\right)^{\frac{\gamma-1}{\gamma}}\)

⇒ \(\mathrm{T}_2=300\left(\frac{1}{2}\right)^{\frac{0.4}{1.4}}\)

⇒ \(300\left(\frac{1}{2}\right)^{2 / 7}\)

= 300×8.2

T₂ = 246K

Question 5. One liter of air at NTP is suddenly compressed to 1 c.c. the final pressure will be.
Answer:

⇒ \(P_2=P_1\left(\frac{V_1}{V_2}\right)^\gamma\)

⇒ \(P_2=10^5\left(10^3\right)^{5 / 3}=10^5 \times 10^5\)

⇒ \(P_2=10^{10} \text { Pascal }\)

Question 6. In the following fig. the work done by the system in the closed path ABCA is Answer: Work done in closed path ABCA

W_ABCA = Area of ΔABC = \(\frac{1}{2}\) AB × BC

W_ABCA = –\(\frac{1}{2}\)(P₂– P₁) (V₂– V₁)

Question 7. According to the fig. if one mole of an ideal gas is in a cyclic process the work done by the gas in the process will be
Answer:

Work done W = area of PV curve

⇒ \(\frac{1}{2}\)[3P₀– P₀][2V₀– V₀]

= P₀V₀

Question 8. In the above question, heat given by the gas is
Answer:

δQ = μC_pdT, μ = 1, dT = T_A– T_C, and for monoatomic ideal gas C_P= 5/2 R

∴ (δQ)_CA = \(\frac{5}{2}\) R [T_A– T_C] = \(\frac{5}{2}\)[P_AV_A– P_CV_C]

But P_A= P₀, V_A= V0, V_C= 2V0, P_C= P₀

and \(\frac{PV}{T}\)

∴ (δQ)C_A = \(\frac{5}{2}\)[P₀V₀– P₀2V₀] = – \(\frac{5}{2}\)P₀V₀

Question 9. In the above question, the heat taken by gas in the path AB will be Answered:

(δQ)_AB = μC_VdT (the process is on constant volume)

C_V= \(\frac{3}{2}\)R, μ = 1

⇒ \((\delta Q)_{A B}=\frac{3}{2} R\left[T_B-T_A\right]\)

⇒ \(\frac{3}{2}\left[3 P_0 V_0-P_0 V_0\right]\)

⇒ \(3 \mathrm{P}_0 \mathrm{~V}_0\)

Question 10. In the above question, the absorbed heat by gas in path BC will be Answered:

If the heat given for the complete process is δQ then

(δQ) = (δQ)_AB + (δQ)_BC + (δQ)_CA

dU = 0 in a cyclic process, thus by the first law of thermodynamics

δQ = δW

∴ (δQ)_AB + (δQ)_BC + (δQ)_AC = δW

(δQ)_BC = δW – (δQ)_AB – (δQ)_BC

⇒ \(P_0 V_0+\frac{5}{2} P_0 V_0-3 P_0 V_0\)

⇒ \(\frac{P_0 V_0}{2}\)

Question 11. For a given cyclic process as shown in fig. the magnitude of absorbed energy for the system is
Answer:

In cyclic process

Q = W (∵ dU = 0)

Q = area of closed loop

Q = 10²π Joule

Second Law Of Thermodynamics

This law gives the direction of heat flow.

According To Classius: It is impossible to make any such machine that can transfer heat from an object with low temperature to an object with high temperature without an external source.

According To Kelvin: It is impossible to obtain work continuously by cooling an object below the temperature of its surroundings.

Statement Of Kelvin-Planck: It is impossible to construct any such machine that works on a cyclic process and absorbs heat from a source, converts all that heat into work, and rejects no heat to sink.

Heat Engine:

The device, used to convert heat energy into mechanical energy, is called a heat engine.

• For the conversion of heat into work with the help of a heat engine, the following conditions have to be met. There should be a body at a higher temperature ‘T₁’ from which heat is extracted. It is called the source.
• The body of the engine contains the working substance. There should be a body at a lower temperature ‘T₂’ to which heat can be rejected. This is called the sink.

Working Of Heat Engine:

Schematic diagram of heat engine

Engine derives an amount ‘Q₁’ of heat from the source.

A part of this heat is converted into work ‘W’.

The remaining heat ‘Q₂’ is rejected to the sink.

Thus Q₁= W + Q₂

or the work done by the engine is given by W = Q₁– Q₂

Efficiency Of Heat Engine:

The efficiency of the heat engine (η) is defined as the fraction of total heat, supplied to the engine which is converted into work.

Mathematically

∴ \(\eta=\frac{W}{Q_1}\)

or \(\eta=\frac{Q_1-Q_2}{Q_1}=1-\frac{Q_2}{Q_1}\)

Carnot Engine And Carnot Cycle

Carnot Engine: The Carnot engine is an ideal heat engine. It consists of the following parts.

Schematic Diagram:

Source: It is a reservoir of heat energy with a conducting top maintained at a constant temperature T₁K. The source is so big that extraction of any amount of heat from it does not change its temperature.

Body Of Heat Engine: It is a barrel having perfectly insulating walls and a conducting bottom. It is fitted with an air-tight piston capable of sliding within the barrel without friction. The barrel contains some quantity of an ideal gas.

Sink: It is a huge body at a lower temperature T₂ having a perfectly conducting top. The size of the sink is so large that any amount of heat rejected to it does not increase its temperature.

Insulating Stand: It is a stand made up of perfectly insulating material such that the barrel when placed over it becomes thoroughly insulated from the surroundings.

Carnot Cycle: As the engine works, the working substance of the engine undergoes a cycle known as the Carnot cycle. The Carnot cycle consists of the following four strokes.

Graphical representation of the Carnot cycle:

First Stroke (Isothermal expansion): In this stroke, the barrel is placed over the source. The piston is gradually pushed back as the gas expands.

The fall of temperature, due to expansion, is compensated by the supply of heat from the source and consequently, temperature remains constant. The conditions of the gas change from A(P₁, V₁) to B(P₂, V₂). If W₁ is the work done during this process, then heat Q₁ derived from the source is given by

⇒ \(\mathrm{Q}_1=\mathrm{W}_1=\text { Area } \mathrm{ABGE}=\mathrm{RT} \log _e\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)\)

Second Stroke (Adiabatic expansion):

The barrel is removed from the source and placed over the insulating stand. The piston is pushed back so that the gas expands adiabatically resulting in a fall of temperature from T₁ to T₂. The conditions of the gas change from B(P₂, V₂) to C(P₃, V₃). If W₂ is the work done in this case then

⇒ \(\mathrm{W}_2=\text { Area } \mathrm{BCHG}=\frac{\mathrm{R}}{\gamma-1}\left(\mathrm{~T}_1-\mathrm{T}_2\right)\)

Third Stroke (isothermal compression): The barrel is placed over the sink. The piston is pushed down there by compressing the gas. The heat generated due to compression flows to the sink maintaining the temperature of the barrel constant.

The state of the gas changes from C(P₃, V₃) to D(P₄, V₄). If W₃ is the work done in this process and Q₂ is the heat rejected to the sink, then

⇒ \(\mathrm{Q}_2=\mathrm{W}_3=\text { Area CDFH }=\mathrm{RT}_2 \log _{\mathrm{e}}\left(\frac{\mathrm{V}_3}{\mathrm{~V}_4}\right)\)

Fourth Stroke (Adiabatic compression): The barrel is placed over the insulating stand. The piston is moved down thereby compressing the gas adiabatically till the temperature of the gas increases from T₂ to T₁.

The state of gas changes from D(P₄, V₄) to A(P₁, V₁). If W₄ is the work done in this process, then

⇒ \(\mathrm{W}_4=\text { Area ADFE }=\frac{\mathrm{R}}{\gamma-1}\left(\mathrm{~T}_1-\mathrm{T}_2\right)\)

Heat Converted Into Work In Carnot Cycle:

During the four strokes, W₁ and W₂ are the work done by the gas, and W₃ and W₄ are the work done on the gas. Therefore the net, work performed by the engine

W = W₁+ W₂– W₃– W₄= Area ABGE + Area BCHG – Area CDFH – Area ADEF = Area ABCD

Thus net work done by the engine during one cycle is equal to the area enclosed by the indicator diagram of the cycle. Analytically

⇒ \(\mathrm{W}=\mathrm{R}\left(\mathrm{T}_1-\mathrm{T}_2\right) \log _e\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)\)

Efficiency Of Carnot Engine:

The efficiency (η) of an engine is defined as the ratio of useful heat (heat converted into work) to the total heat supplied to the engine. Thus.

⇒ \(\eta=\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1}\)

or \(\eta=\frac{R\left(T_1-T_2\right) \log _e\left(\frac{V_2}{V_1}\right)}{R T_1 \log _e\left(\frac{V_2}{V_1}\right)}\)

⇒ \(\frac{T_1-T_2}{T_1}\)

or \(\eta=1-\frac{Q_2}{Q_1}=1-\frac{T_2}{T_1}\)

Some Important Points Regarding Carnot Engine

The efficiency of an engine depends upon the temperatures between which it operates.

η is independent of the nature of the working substance.

η is one only if T₂= 0. Since absolute zero is not attainable, hence even an ideal engine cannot be 100 % efficient.

η is one only if Q₂= 0. But η = 1 is never possible even for an ideal engine. Hence Q₂≠ 0. Thus it is impossible to extract heat from a single body and convert the whole of it into work.

If T₂= T₁, then η = 0

In actual heat engines, there are many losses due to friction, etc. and various processes during each cycle are not quasistatic, so the efficiency of actual engines is much less than that of an ideal engine.

Question 1. A Carnot engine has the same efficiency between

1. 100 K and 500K and
2. Tk and 900 K. The value of T is

Answer:

Efficiency η = \(1-\frac{T_2}{T_1}\)

or η = \(1-\frac{100}{500}\)

⇒ \(1-\frac{T}{900}\)

or \(\frac{100}{500}=\frac{T}{900}\)

∴ T = 180K

Question 2. A Carnot engine takes 10³ kilocalories of heat from a reservoir at 627ºC and exhausts it to a sink at 27ºC. The efficiency of the engine will be.
Answer:

Efficiency of Carnot engine

η = \(1-\frac{T_2}{T_1}\)

⇒ \(1-\frac{300}{900}=\frac{2}{3}\)

or η = 66.6 %

Question 3. In the above problem, the work performed by the engine will be
Answer:

Work performed by the engine

W = \(\eta Q_1=\frac{2}{3} \times 10^6 \times 4.2\)

or W = 2.8 × 10⁶ Joule

Question 4. A Carnot engine has an efficiency of 40% when the sink temperature is 27ºC. The source temperature will be
Answer:

⇒ \(\eta_{\text {efficiency }}=1-\frac{T_2}{T_1}\)

or \(\frac{2}{5}=1-\frac{300}{\mathrm{~T}_1}\)

∴ T₁= 500K

Question 5. A reversible engine takes heat from a reservoir at 527ºC and gives out to the sink at 127ºC. The engine is required to perform useful mechanical work at the rate of 750 watts. The efficiency of the engine is
Answer:

Efficiency η = \(1-\frac{T_2}{T_1}\)

or η = \(1-\frac{400}{800}=\frac{1}{2}\)

η = 50%

Question 6. The efficiency of Carnot’s engine is 50%. The temperature of its sink is 7ºC. To increase its efficiency to 70%. The increase in heat of the source will be
Answer:

Efficiency in first state η = 50% = 1/2

T₂= 273 + 7 = 280 K

Formula η = \(1-\frac{T_2}{T_1}\)

⇒ \(\frac{1}{2}=1-\frac{280}{\mathrm{~T}_1} \Leftrightarrow \frac{280}{\mathrm{~T}_1}=\frac{1}{2}\)

or T₁ = 560ºK (temperature of source)

In the second state (1) \(\frac{70}{100}=1-\frac{280}{\mathrm{~T}_1}\)

∴ \(\mathrm{T}_1=\frac{2800}{3}=933.3 \mathrm{~K}\)

∴ Increase in source temperature = (933.3 – 560) = 373.3 K

Question 7. A Carnot’s engine works at 200ºC and 0ºC and another at 0ºC and –200ºC. The ratio of efficiency of the two is
Answer:

⇒ \(\eta=\frac{\left(T_1-T_2\right)}{T_1}\)

⇒ \(\eta_1=\frac{(473-273)}{473}=\frac{200}{473}\)

and \(\eta_2=\frac{(273-73)}{273}=\frac{200}{273}\)

⇒ \(\frac{\eta_1}{\eta_2}=\frac{273}{473}\)

= 0.577

Question 8. A Carnot engine works as a refrigerator in between 0ºC and 27ºC. How much energy is needed to freeze 10 kg of ice at 0ºC?
Answer:

Heat absorbed by the sink

Q₂= 10 × 10² × 80 = 800 k.cal

Now \(\frac{Q_1}{Q_2}=\frac{T_1}{T_2}, Q_1=Q_2 \cdot \frac{T_1}{T_2}\)

∴ \(Q_1=800 \times \frac{300}{273} \mathrm{k} . \mathrm{cal}\)

= 879 kcal

Question 9. The work efficiency coefficient in the above question
Answer:

Work efficiency coefficient (coefficient of performance)

⇒ \(\beta=\frac{Q_2}{Q_1-Q_2}\)

⇒ \(\frac{800 \times 10^3}{(879-800) \times 10^3}\)

= 10.13

Question 10. A Carnot engine works as a refrigerator in between 250K and 300K. If it acquires 750 calories from a heat source at a low temperature, then the heat is generated at a higher temperature. (in calories) will be.
Solution:

⇒ \(\eta=\frac{Q_2}{Q_1-Q_2}=\frac{T_2}{T_1-T_2}\)

⇒ \(\frac{750}{Q_1-750}=\frac{250}{300-250}\)

Q₁= 900 Calories

Question 11. A vessel of volume 2 x 10^-2 m³ contains a mixture of hydrogen and helium at 47º C temperature and 4.15 x 10⁵ N/m² pressure. The mass of the mixture is 10^-2 kg. Calculate the masses of hydrogen and helium in the given mixture.
Solution:

Let the mass of H₂ is m₁ and He is m₂

∴ m₁+ m₂= 10^-2 kg = 10 × 10^-3 ….(1)

Let P₁, P₂ are partial pressure of H₂ and He

P₁+ P₂= 4.15 × 10⁵ N/m²

for the mixture

⇒ \(\left(P_1+P_2\right) V=\left(\frac{m_1}{n_1}+\frac{m_2}{n_2}\right) R T\)

⇒ \(4.15 \times 10^5 \times 2 \times 10^{-2}\)

⇒ \(\left(\frac{m_1}{2 \times 10^{-3}}+\frac{m_2}{4 \times 10^{-3}}\right) 8.31 \times 320\)

⇒ \(\frac{m_1}{2}+\frac{m_2}{4}=\frac{4.15 \times 2}{8.31 \times 320}=0.00312=3.12 \times 10^{-3}\)

⇒ 2m₁+ m₂= 12.48 × 10^-3 kg …..(2)

Solving (1) and (2)

m₁= 2.48 × 10^-3 kg ≅2.5 × 10^-3 kg

and m = 7.5 × 10^-3 kg.

Question 12. The pressure in a monoatomic gas increases linearly from 4 x 10⁵ N m^-2 to 8 x 10⁵ N m^-2 when its volume increases from 0.2 m³ to 0.5 m³. Calculate the following:

1. Work done by the gas.
2. Increase in the internal energy.

Answer:

1. As here pressure varies linearly with volume, work done by the gas

ΔW = ∫PdV = area under P-V curve

which in the light of Figure 1 becomes:

ΔW = P_I(V_F– V_I) (P_F– P_I) × (V_F– V_I)

i.e., ΔW = \(P_I\left(V_F-V_I\right)+\frac{1}{2}\left(P_F-P_I\right) \times\left(V_F-V_I\right)\)

i.e., ΔW = \(\frac{1}{2}\)(0.5 – 0.2) (8 + 4) × 10⁵

i.e., ΔW = 1.8 × 10⁵ J

2. The change in internal energy of a gas is given by

ΔU = \(\mu C_V \Delta T=\frac{\mu R \Delta T}{(\gamma-1)}\)

⇒ \(\frac{\left(P_F V_F-P_I V_I\right)}{(\gamma-1)}\)

As the gas is monatomic γ = (5/3)

So, ΔU = \(\frac{10^5(8 \times 0.5-4 \times 0.2)}{[(5 / 3)-1]}\)

⇒ \(\frac{3}{2} \times 10^5(4-0.8)\)

i.e., ΔU = 4.8 × 10⁵ J

Refrigerator Or Heat Pump

A refrigerator or heat pump is a heat engine that runs in revenue one direction.
It essentially consists of three parts

1. Source: At higher temperature T₁.
2. Working substance: It is called refrigerant liquid ammonia and freon works as a working substance
3. Sink: At lower temperature T₂.

• The working substance takes heat Q₂ from a sink (contents of the refrigerator) at a lower temperature, has a net amount of work done W in it by an external agent (usually the compressor of the refrigerator), and gives out a larger amount of heat Q₁ to a hot body at temperature T₁(usually atmosphere)
• Thus, it transfers heat from a cold to a hot body at the expense of mechanical energy supplied to it by an external agent. The cold is thus cooled more and more.
• The performance of a refrigerator is expressed using the “coefficient of performance” β which is defined as the ratio of the heat extracted from the cold body to the needed to transfer it to the hot body.

i.e \(\beta=\frac{\text { Heat extracted }}{\text { Work done }}\)

⇒ \(\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2}\)

A perfect refrigerator transfers heat from a cold to a hot body without doing work e. W = 0 so that Q₁=Q₂ hence β = ∞

Carnot Refrigerator:

For Carnot refrigerator \(\frac{Q_1}{Q_2}=\frac{T_1}{T_2}\)

⇒ \(\frac{Q_1-Q_2}{Q_2}=\frac{T_1-T_2}{T_2}\)

or \(\frac{Q_2}{Q_1-Q_2}=\frac{T_2}{T_1-T_2}\)

So Coefficient of performance β = \(\frac{T_2}{\mathrm{~T}_1-\mathrm{T}_2}\)

⇒ \(\beta=\frac{T_2}{T_1-T_2}\)

here T₁= temperature of surrounding T₂ = temperature of cold body. It is clear that β = 0 when T₂= 0 i.e. the coefficient of performance will be zero if the cold body is at a temperature equal to absolute zero.

Relation between the coefficient of performance and efficiency of a refrigerator

We know β = \(\frac{Q_2}{Q_1-Q_2}=\frac{Q_2 / Q_1}{1-Q_2-Q_1}\)…………(1)

But the efficiency η = \(\frac{Q_2}{Q_1} \text { or } \frac{Q_2}{Q_1}=1-\eta\)…………(2)

Form (1) and (2) we get, \(\frac{1-\eta}{\eta}\)

Entropy

Entropy is a measure of the disorder of molecular motion of a system. The greater the disorder, the greater the entropy.

The change in entropy i.e

dS = \(=\frac{\text { Heat absorbed by system }}{\text { Absolute temperature }}\)

or \(\mathrm{dS}=\frac{\mathrm{dQ}}{\mathrm{T}}\)

The relation is called the mathematical form of the Second Law of Thermodynamics.

For solids and liquids

1. When heat is given to a substance changes its state at a constant temperature, then changes in entropy

⇒ \(d S=\frac{d Q}{T}= \pm \frac{m L}{T}\)

where the positive sign refers to heat absorption and the negative sign to heat evolution.

2. When heat given to a substance raises its temperature from T₁ to T₂ then changes in entropy

⇒ \(d S=\int \frac{d Q}{T}=\int_{T_1}^{T_2} m c \frac{d T}{T}=m c \log _e\left(\frac{T_2}{T_1}\right)\)

⇒ \(\Delta S=2.303 \mathrm{mc} \log _{10}\left(\frac{T_2}{T_1}\right)\)

For a perfect gas: The perfect gas equation for n moles is PV = nRT

ΔS = \(\int \frac{d Q}{T}=\int \frac{\mu C_V d T+P d V}{T}\) [As dQ = dU+dW]

ΔS = \(\int \frac{\mu C_V d T+\frac{\mu R T}{V} d V}{T}\)

⇒ \(\mu C_V \int_{T_1}^{T_2} \frac{d T}{T}+\mu R \int_{V_1}^{V_2} \frac{d V}{V}\) [ As PV = μ RT]

∴ ΔS = \(\mu C_V \text { long }_e\left(\frac{T_2}{T_1}\right)+\mu \log _e\left(\frac{V_2}{V_1}\right)\)

In terms of T and P, ΔS = \(m C_p{lon}_g\left(\frac{T_2}{T_1}\right)-\mu R \log _e\left(\frac{P_2}{P_1}\right)\)

and in terms of P and V ΔS = \(\mu C_V \log _e\left(\frac{P_2}{P_1}\right)+\mu C_p \log _e\left(\frac{V_2}{V_1}\right)\)

 

Gravitation Introduction

The motion of celestial bodies such as the sun, the moon, the earth the planets etc. has been a subject of fascination since time immemorial.

• Indian astronomers of ancient times have done brilliant work in this field, the most notable among them being Arya Bhatt the first person to assert that all planets including the Earth revolve around the sun.
• A millennium later the Danish astronomer Tycobrahe (1546-1601) conducted a detailed study of planetary motion which was interpreted by his pupil Johnaase Kepler (1571-1630), ironically after the master himself had passed away.
• Kepler formulated his important findings in three laws of planetary motion

Universal Law Of Gravitation: Newton’s Law

According to this law, “Each particle attracts every other particle. The force of attraction between them is directly proportional to the product of their masses and inversely proportional to the square of the distance between them”.

⇒ \(\mathrm{F} \propto \frac{m_1 m_2}{r^2}\)

or \(\mathrm{F}=G \frac{m_1 m_2}{r^2}\)

where G = 6.67 × 10^-11 Nm² kg^-2 is the universal gravitational constant. This law holds good irrespective of the nature of two objects (size, shape, mass etc.) at all places and at all times. That is why it is known as a universal law of gravitation.

Dimensional Formula Of G:

G = \(\frac{F r^2}{m_1 m_2}\)

⇒ \(\frac{\left[M L T^{-2}\right]\left[L^2\right]}{\left[M^2\right]}\)

= [M^-1 L³ T^-2]

Newton’s Law Of Gravitation In Vector Form:

⇒ \(\overrightarrow{\mathrm{F}}_{12}=\frac{G m_1 m_2}{r^2} \hat{r}_{12}\)

⇒ \(\overrightarrow{\mathrm{F}}_{21}=\frac{G m_1 m_2}{r^2}\)

Where \(\overrightarrow{\mathrm{F}}_{12}\) is the force on mass m₁ exerted by mass m₂ and vice-versa.

⇒ \(\overrightarrow{\mathrm{F}}_{12}=\frac{G m_1 m_2}{r^2} \hat{r}_{12}\)

⇒ \(\overrightarrow{\mathrm{F}}_{21}=\frac{G m_1 m_2}{r^2} \hat{\mathrm{r}}_{21}\)

Now \(\hat{r}_{12}=-\hat{r}_{21}\)

Thus \(\vec{F}_{21}=\frac{-G m_1 m_2}{r^2} \hat{r}_{12}\) Comparing above, we get \(\vec{F}_{12}=-\vec{F}_{21}\)

Important Characteristics Of Gravitational Force

1. Gravitational force between two bodies forms an action and reaction pair i.e. the forces are equal in magnitude but opposite in direction.
2. Gravitational force is a central force i.e. it acts along the line joining the centres of the two interacting bodies.
3. Gravitational force between two bodies is independent of the nature of the medium, in which they lie.
4. The gravitational force between two bodies does not depend upon the presence of other bodies.
5. Gravitational force is negligible in the case of light bodies but becomes appreciable in the case of massive bodies like stars and planets.
6. Gravitational force is long range-force i.e., gravitational force between two bodies is effective even if their separation is very large. For example, the gravitational force between the sun and the earth is of the order of 10²⁷ N although the distance between them is 1.5 × 10⁷ km

Question 1. The centres of two identical spheres are at a distance of 1.0 m apart. If the gravitational force between them is 1.0 N, then find the mass of each sphere. (G = 6.67 × 10^-11 m³ kg^-1 sec^-1)
Answer:

Gravitational force F = \(\frac{G m \cdot m}{r^2}\)

on substituting F = 1.0 N , r = 1.0 m and G = 6.67 × 10^-11 m³ kg^-1 sec^-1

we get m = 1.225 × 10⁵ kg

Principle Of Superposition

The force exerted by a particle or other particle remains unaffected by the presence of other nearby particles in space.

The total force acting on a particle is the vector sum of all the forces acted upon by the individual masses when they are taken alone.

⇒ \(\vec{F}_1=\vec{F}_1+\vec{F}_2+\vec{F}_3+\ldots \ldots\)

Question 2.

Four point masses each of mass ‘m’ are placed on the corner of a square of side ‘a’. Calculate the magnitude of gravitational force experienced by each particle.

Answer:

F_r= resultant force on each particle = 2F cos 45º + F₁

⇒ \(\frac{2 G \cdot m^2}{a^2} \cdot \frac{1}{\sqrt{2}}+\frac{G m^2}{(\sqrt{2} a)^2}=\frac{G \cdot m^2}{2 a^2}(2 \sqrt{2}+1)\)

Gravitational Field

The space surrounding the body within which its gravitational force of attraction is experienced by other bodies is called the gravitational field.

• A gravitational field is very similar to an electric field in electrostatics where charge ‘q’ is replaced by mass ‘m’ and electric constant ‘K’ is replaced by gravitational constant ‘G’.
• The intensity of the gravitational field at a point is defined as the force experienced by a unit mass placed at that point.

⇒ \(\vec{E}=\frac{\vec{F}}{m}\)

The unit of the intensity of the gravitational field is N kg^-1.

Intensity of gravitational field due to point mass:

The force due to mass m on test mass m0 placed at point P is given by :

⇒ \(\mathrm{F}=\frac{\mathrm{GMm}_0}{\mathrm{r}^2}\)

Hence \(E=\frac{F}{m_0}\)

⇒ \(E=\frac{G m}{r^2}\)

In vector form \(\vec{E}=-\frac{G M}{r^2} \hat{r}\)

Dimensional formula of intensity of gravitational field = \(\frac{F}{m}=\frac{\left[M L T^{-2}\right]}{[M]}\)

⇒ \(\left[\begin{array}{ll}M^0 & L T^{-2}\end{array}\right]\)

Question 1. Find the relation between the gravitational field on the surface of two planets A and B of masses m_A, mB and radius R_A and R_B respectively if

1. They have equal mass
2. They have equal (uniform) density

Answer: Let EA and EB be the gravitational field intensities on the surface of planets A and B.

then, \(\mathrm{E}_{\mathrm{A}}=\frac{G m_A}{R_A^2}=\frac{G \frac{4}{3} \pi R_A^3 \rho_A}{R_A^2}\)

⇒ \(\frac{4 G \pi}{3} \rho_A R_A\)

Similarly, \(E_B=\frac{G m_B}{R_{B^2}}=\frac{4 G}{3} \pi \rho_{\mathrm{B}} \mathrm{R}_{\mathrm{B}}\)

For \(m_A=m_B\)

⇒ \(\frac{E_A}{E_B}=\frac{R_B^2}{R_A^2}\)

For \(\rho_{\mathrm{A}}=\rho_{\mathrm{B}}\)

⇒ \(\frac{E_A}{E_B}=\frac{R_A}{R_B}\)

Gravitational Potential

The gravitational potential at a point in the gravitational field of a body is defined as the amount of work done by an external agent in bringing a body of unit mass from infinity to that point, slowly (no change in kinetic energy). Gravitational potential is very similar to electric potential in electrostatics.

Gravitational Potential Due To A Point Mass :

Let the unit mass be displaced through a distance dr towards mass M, then work done is given by

dW = F dr = \(\frac{\mathrm{Gm}}{\mathrm{r}^2} \mathrm{dr}\)

The total work done in displacing the particle from infinity to point P is −

⇒ \(\mathrm{W}=\int d W=\int_{\infty}^r \frac{G M}{r^2} d r=\frac{-G M}{r}\)

Thus gravitational potential, \(V=-\frac{G M}{r}\)

The unit of gravitational potential is J kg^-1. Dimensional Formula of gravitational potential

⇒ \(\frac{\text { work }}{\text { mass }}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{[\mathrm{M}]}\)

= [M°L² T^-2].

Question 1. Find out the potential at P and Q due to the two-point mass system. Find out work done by external agents in bringing unit mass from P to Q. Also find work done by gravitational force.

Answer: V_P1 = potential at P due to mass ‘m’ at ‘l’ = –\(\frac{G m}{\ell}\)

⇒ \(\mathrm{V}_{\mathrm{P} 2}=-\frac{G m}{\ell}\)

∴ \(\mathrm{V}_{\mathrm{p}}=\mathrm{V}_{\mathrm{P} 1}+\mathrm{V}_{\mathrm{P} 2}=-\frac{2 G m}{\ell}\)

⇒ \(V_{Q 1}=-\frac{G M}{\ell / 2}\)

⇒ \(V_{Q 2}=-\frac{G m}{\ell / 2}\)

∴ \(\mathrm{v}_{\mathrm{Q}}=\mathrm{v}_{\mathrm{Q} 1}+\mathrm{v}_{\mathrm{Q} 2}=-\frac{G m}{\ell / 2}-\frac{G m}{\ell / 2}=-\frac{4 G m}{\ell}\)

Force at point Q = 0

Work done by external agent = \(\left(\mathrm{V}_{\mathrm{Q}}-\mathrm{V}_{\mathrm{P}}\right) \times 1=-\frac{2 G M}{\ell}\)

Work done by gravitational force = V_P– V_Q = \(\frac{2 G M}{\ell}\)

Question 2. Find the potential at a point ‘P’ at a distance ‘x’ on the axis away from the centre of a uniform ring of mass M and radius R.
Answer:

The ring can be considered to be made of a large number of point masses (m₁, m₂ ………. etc) Ring

⇒ \(\mathrm{V}_{\mathrm{p}}=-\frac{G m_1}{\sqrt{R^2+x^2}}-\frac{G m_2}{\sqrt{R^2+x^2}}-\ldots \ldots .\)

⇒ \(-\frac{G}{\sqrt{R^2+x^2}}\left(m_1+m_2 \ldots . .\right)=-\frac{G M}{\sqrt{R^2+x^2}}\) where, M = m₁+m₂+m₃+……..

Potential at centre of ring = \(-\frac{G \cdot M}{R}\)

Relation Between Gravitational Field And Potential

The work done by an external agent to move unit mass from one point to another point in the direction of the field E, slowly through an infinitesimal distance dr = Force by external agent × distance moved = – Edr.

Thus dV = – Edr

⇒ E = – \(\frac{d V}{d r}\)

Therefore, the gravitational field at any point is equal to the negative gradient at that point.

Uniform Solid Sphere

Point P inside the shell. r ≤ a, then

V = \(-\frac{\mathrm{Gm}}{2 \mathrm{a}^3}\left(3 \mathrm{a}^2-\mathrm{r}^2\right)\)

and E = \(-\frac{G M r}{a^3}\) ,and at the centre V = \(-\frac{3 G M}{2 a}\) and E = 0

Point P outside the shell. r > a, then V = \(-\frac{G M}{r}\) and E = \(-\frac{G M}{r^2}\)

Uniform Thin Spherical Shell

Point P Inside the shell. r ≤ a , then V = \(-\frac{G M}{a}\) and E = 0

Point P outside shell. r ≥ a, then V = \(-\frac{G M}{r}\) and E = \(-\frac{G M}{r^2}\)

Gravitational Potential Energy

The gravitational potential energy of two mass systems is equal to the work done by an external agent in assembling them, while their initial separation was infinity.

Consider a body of mass m placed at a distance r from another body of mass M. The gravitational force of attraction between them is given by

⇒ \(\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{r}^2}\)

Now, Let the body of mass m is displaced from a point. C to B through a distance ‘dr’ towards the mass M, then work done by internal conservative force (gravitational) is given by,

⇒ \(\mathrm{dW}=\mathrm{F} \mathrm{dr}=\frac{G M m}{r^2} \mathrm{dr}\)

⇒ \(\int d W=\int_{\infty}^r \frac{G M m}{r^2} \mathrm{dr}\)

∴ Gravitational potential energy, \(\mathrm{U}=-\frac{\mathrm{GMm}}{r}\)

Increase In Gravitational Potential Energy:

Suppose a block of mass m on the surface of the earth. We want to lift this block by ‘h’ height. Work required in this process = increase in P.E. = U_f– U_i= m(V_f– V_i)

⇒ \(\mathrm{W}_{\mathrm{ext}}=\Delta \mathrm{U}=(\mathrm{m})\left[-\left(\frac{G M_c}{R_e+h}\right)-\left(-\frac{G M_c}{R_e}\right)\right]\)

W_ext = ΔU = \(\mathrm{GM}_e \mathrm{~m}\left(\frac{1}{R_e}-\frac{1}{R_e+h}\right)=\frac{G M_e m}{R_e}\left(1-\left(1+\frac{h}{R_e}\right)^{-1}\right)\)

(as h << R_e, we can apply the Binomial theorem)

W_ext = ΔU = \(\frac{G M_e m}{R_e}\left(1-\left(1-\frac{h}{R_e}\right)\right)=(\mathrm{m})\left(\frac{G M_e}{R_e^2}\right) \mathrm{h}\)

W_ext = ΔU = mgh

This formula is valid only when h << R_e

Question 1. A body of mass m is placed on the surface of the earth. Find the work required to lift this body by a height

1. h = \(\frac{R_c}{1000}\)
2. h = R_e

Answer: h = \(\frac{R_c}{1000}\) as h << R_e , so

we can apply

W_ext = \(\mathrm{W}_{\mathrm{ext}}=\mathrm{U} \uparrow=\mathrm{mgh}\)

⇒ \(\mathrm{W}_{\mathrm{ext}}=\left(\frac{G M_e}{R_e{ }^2}\right)\left(\frac{R_e}{1000}\right)(\mathrm{m})=\frac{G M_e m}{1000 R_e}\)

h = R_e, in this case, h is not much less than R_e, so we cannot apply ΔU = mgh so we cannot apply ΔU = mgh

W_ext = U↑ = U_f– U_i= m(V_f– V_i)

⇒ \(\mathrm{W}_{\mathrm{ext}}=\mathrm{m}\left[\left(-\frac{G M_e}{R_e+R_e}\right)-\left(-\frac{G M_e}{R_e}\right)\right]\)

⇒ \(\mathrm{W}_{\mathrm{ost}}=-\frac{G M_e m}{2 R_e}\)

Acceleration Due To Gravity

It is the acceleration, a freely falling body near the earth’s surface acquires due to the earth’s gravitational pull.

• The property by virtue of which a body experiences or exerts a gravitational pull on another body is called gravitational mass mG, and the property by virtue of which a body opposes any change in its state of rest or uniform motion is called its inertial mass
• m₁ thus if \(\overrightarrow{\mathrm{E}}\) is the gravitational field intensity due to the earth at a point P, and g is acceleration due to gravity at the same point, then \(m_1 \vec{g}=m_G \vec{E}\)
• Now the value of inertial and gravitational mass happens to be exactly the same to a great degree of accuracy for all bodies. Hence,\(\vec{g}=\vec{E}\)

The gravitational field intensity on the surface of the earth is therefore numerically equal to the acceleration due to gravity (g), there. Thus we get,

⇒ \(g=\frac{G M_e}{R_e^2}\)

where, M_e= Mass of earth

R_e = Radius of earth

Note: Here the distribution of mass in the earth is taken to be spherical and symmetrical so that its entire mass can be assumed to be concentrated at its centre for the purpose of calculation of g.

Variation Of Acceleration Due To Gravity

Effect of Altitude

Acceleration due to gravity on the surface of the earth is given by,

g = \(\frac{G M_e}{R_e^2}\)

Now, consider the body at a height ‘h’ above the surface of the earth, then the acceleration due to gravity at height ‘h’ given by

⇒ \(\mathrm{g}_{\mathrm{h}}=\frac{G M_e}{\left(R_e+h\right)^2}=\mathrm{g}\left(1+\frac{h}{R_e}\right)^{-2} \simeq \mathrm{g}\left(1-\frac{2 h}{R_e}\right)\)

when h << R.

The decrease in the value of ‘g’ with height h = g – g_h= \(\frac{2 g h}{R_e}\)

The percentage decrease in the value of \(‘ \mathrm{~g}^{\prime}=\frac{g-g_h}{g} \times 100=\frac{2 h}{R_e} \times 100 \%\)

Effect Of Depth

The gravitational pull on the surface is equal to its weight i.e. mg = \(\frac{G M_e m}{R_e^2}\)

∴ \(\mathrm{mg}=\frac{G \times \frac{4}{3} \pi R_e^3 \rho m}{R_e^2}\)

or \(\mathrm{g}=\frac{4}{3} \pi \mathrm{GR}_{\mathrm{e}} \rho\) ………(1)

When the body is taken to a depth d, the mass of the sphere of radius (R_e – d) will only be effective for the gravitational pull and the outward shall will have no resultant effect on the mass. If the acceleration due to gravity on the surface of the solid sphere is gd, then

⇒ \(g_d=\frac{4}{3} \pi G\left(R_e-d\right) \rho\) ……………(2)

By dividing equation (2) by equation (1)

⇒ \(\mathrm{g}_{\mathrm{d}}=\mathrm{g}\left(1-\frac{d}{R_e}\right)\)

Important Points

At the center of the earth, d = R_e, so g_centre = \(g\left(1-\frac{R_e}{R_e}\right)\) = 0.

Thus, the weight (mg) of the body at the centre of the earth is zero.

Percentage decrease in the value of ‘g’ with the depth

⇒ \(\left(\frac{g-g_d}{g}\right) \times 100=\frac{d}{R_e} \times 100\)

Effect Of The Surface Of Earth:

The equatorial radius is about 21 km longer than its polar radius.

We know, g = \(\frac{G M_e}{R_e^2}\) Hence g_pole > g_equator.

The weight of the body increases as the body is taken from the equator to the pole.

Effect of rotation of the Earth: The earth rotates around its axis with angular velocity ω. Consider a particle of mass m at latitude θ. The angular velocity of the particle is also ω.

⇒ \(\mathrm{g}^{\prime}=\mathrm{g}\left[1-\frac{R_e \omega^2}{g} \cos ^2 \theta\right]\)

At pole θ = 90° ⇒ g_pole = g,

At equator θ = 0 ⇒ g_equator = g \(\left[1-\frac{R_e \omega^2}{g}\right]\)

Hence g _pole > g _equator

If the body is taken from the pole to the equator, then g′ = g\(\left(1-\frac{R_e \omega^2}{g}\right)\)

Hence % change in weight = \(\frac{m g-m g\left(1-\frac{R_e \omega^2}{g}\right)}{m g} \times 100\)

⇒ \(\frac{m R_e \omega^2}{m g} \times 100\)

⇒ \(\frac{R_e \omega^2}{g} \times 100\)

Escape Speed

The minimum speed required to send a body out of the gravity field of a planet (send it to r → ∞)

Escape Speed At Earth’s Surface:

Suppose a particle of mass m is on the earth’s surface

We project it with a velocity V from the earth’s surface so that it just reaches r → ∞ (at r → ∞, its velocity becomes zero)

Applying energy conservation between the initial position (when the particle was at the earth’s surface) and finding positions (when the particle just reaches r → ∞)

K_i+ U_i= K_f+ U_f

⇒ \(\frac{1}{2} m v^2+m_0\left(-\frac{G M_e}{R}\right)\)

⇒ \(0+\mathrm{m}_0\left(-\frac{G M_e}{(r \rightarrow \infty)}\right)\)

⇒ \(\mathrm{v}=\sqrt{\frac{2 G M_0}{R}}\)

Escape speed from earth is surface \(V_e=\sqrt{\frac{2 G M_e}{R}}\)

If we put the values of G, M_e, and R we get

V_e = 11.2 km/s.

Escape Speed Depends On :

1. Mass (Me) and size (R) of the planet
2. Position from where the particle is projected.

Escape Speed Does Not Depend On :

1. Mass of the body which is projected (m0)
2. The angle of projection.

If a body is thrown from the Earth’s surface with escape speed, it goes out of the earth’s gravitational field and never returns to the Earth’s surface. But it starts revolving around the sun.

Kepler’s Law For Planetary Motion

Suppose a planet is revolving around the sun, or a satellite is revolving around the earth, then the planetary motion can be studied with the help of Kepler’s three laws.

Kepler’s Law Of Orbit

Each planet moves around the sun in a circular path or elliptical path with the sun at its focus. (In fact circular path is a subset of an elliptical path)

Law Of Areal Velocity:

To understand this law, let us understand the angular momentum conservation for the planet.

• If a planet moves in an elliptical orbit, the gravitation force acting on it always passes through the centre of the sun. So torque of this gravitation force about the centre of the sun will be zero.
• Hence we can say that the angular momentum of the planet about the centre of the sun will remain conserved (constant) τ about the sun = 0

⇒ \(\frac{d J}{d t}\) = 0

J_planet/ sun = constant

⇒ mvr sinθ = constant

Now we can easily study Kapler’s law of areal velocity.

If a planet moves around the sun, the radius vector \((\vec{r})\) also rotates are sweeps area as shown in the figure. Now let’s find a rate of area swept by the radius vector \((\vec{r})\).

Suppose a planet is revolving around the sun and at any instant its velocity is v, and the angle between radius vector \((\vec{r})\) and velocity \((\vec{v})\). In dt time, it moves by a distance vdt, during this dt time, the area swept by the radius vector will be OAB which can be assumed to be a triangle

dA = 1/2 (Base) (Perpendicular height)

dA = 1/2 (r) (vdtsinθ)

so rate of area swept \(\frac{d A}{d t}=\frac{1}{2} \mathrm{vr} \sin \theta\)

we can write \(\frac{d A}{d t}=\frac{1}{2} \frac{m v r \sin \theta}{m}\)

where mvr sinθ = angular momentum of the planet about the sun, which remains conserved (constant)

⇒ \(\frac{d A}{d t}=\frac{L_{\text {planet } / \text { sun }}}{2 m}\)= constant

so the Rate of area swept by the radius vector is constant

Kepler’s Law Of Time Period: Suppose a planet is revolving around the sun in a circular orbit

then \(\frac{m_0 v^2}{r}=\frac{G M_s m_0}{r^2}\)

v = \(\sqrt{\frac{G M_s}{r}}\)

Time period of the revolution is

⇒ \(\frac{2 \pi r}{v}=2 \pi \mathrm{r} \sqrt{\frac{r}{G M_s}}\)

⇒ \(\mathrm{T}^2=\left(\frac{4 \pi^2}{G M_s}\right) \mathrm{r}^3\)

For all the planets of a sun, T² ∝ r³

Circular Motion Of A Satellite Around A Planet

Suppose a satellite of mass m0 is at a distance r from a planet. If the satellite does not revolve, then due to the gravitational attraction, it may collide with the planet.

To avoid a collision, the satellite revolves around the planet, in a circular motion of a satellite.

⇒ \(\frac{G M_e m_0}{r^2}=\frac{m_0 v^2}{r}\) ….

⇒ v = \(\sqrt{\frac{G M_e}{r}}\) this velocity is called orbital velocity (v0)

⇒ \(\mathrm{v}_0=\sqrt{\frac{G M_e}{r}}\)

Total Energy Of The Satellite Moving In Circular Orbit:

KE = \(\frac{1}{2} m_0 v^2\) and from equation …..(1)

⇒ \(\frac{m_0 v^2}{r}=\frac{G M_e m_0}{r^2}\)

⇒ \(\mathrm{m}_0 \mathrm{v}^2=\frac{G M_e m_0}{r}\)

⇒ \(\mathrm{KE}=\frac{1}{2} m_0 v^2=\frac{G M_e m_0}{2 r}\)

Potential energy

⇒ \(\mathrm{U}=-\frac{G M_e m_0}{r}\)

Total energy = KE + PE = \(\left(\frac{G M_e m_0}{2 r}\right)+\left(\frac{-G M_e m_0}{r}\right)\)

⇒ \(\mathrm{TE}=-\frac{G M_e m_0}{2 r}\)

The total energy is –ve. It shows that the satellite is still bounded by the planet.

Geo – Stationery Satelite

We know that the earth rotates about its axis with angular velocity ω_earth and time period T_earth = 24 hours.

Suppose a satellite is set in an orbit which is in the plane of the equator, whose ω is equal to ω_earth, (or its T is equal to T_earth = 24 hours) and whose direction is also the same as that of earth. Then as seen from Earth, it will appear to be stationary. This type of satellite is called geostationary satellite. For a geo-stationery satellite,

w_satelite = w_earth

⇒ T_satelite = T_earth= 24 hr.

So time period of a geo-stationery satellite must be 24 hours. To achieve T = 24 hours, the orbital radius geo-stationery satellite :

⇒ \(\mathrm{T}^2=\left(\frac{4 \pi^2}{G M_e}\right) \mathrm{r}^3\)

Putting the values, we get an orbital radius of geo stationary satellite r = 6.6 Re(here Re = radius of the earth) height from the surface h = 5.6 Re.

Path Of A Satellite According To Different Speed Of Projection

Suppose a satellite is at a distance r from the centre of the earth. If we give different velocities (v) to the satellite, its path will be different

If v < v₀ \(\left(\text { or } v<\sqrt{\frac{G M_e}{r}}\right)\) then the satellite will move in an elliptical path and strike the earth’s surface.

But if the size of the earth were small, the satellite would complete the elliptical orbit, and the centre of the earth would be at its farther focus.

If v = v₀ \(\left(\text { or } \quad v=\sqrt{\frac{G M_e}{r}}\right)\), then the satellite will revolve in a circular orbit.

If v₀ > V > v₀ \(\left(\text { or } \sqrt{\frac{2 G M_e}{r}}>v>\sqrt{\frac{G M_e}{r}}\right)\), then the satellite will revolve in an elliptical orbital, and the centre of the earth will be at its nearer focus.

If v = v_e \(\left(\begin{array}{ll}\text { or } and v=\sqrt{\frac{2 G M_e}{r}}
\end{array}\right)\), then the satellite will just escape with a parabolic path.

Question 1. Suppose a planet is revolving around the sun in an elliptical path given by \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). Find a period of revolution. The angular momentum of the planet about the sun is L.

Answer: Rate of area swept = \(\frac{d A}{d t}=\frac{L}{2 m}\) constant

⇒ \(\mathrm{dA}=\frac{L}{2 m} d t\)

⇒ \(\int_{A=0}^{A=\pi a b} d A=\int_{t=0}^{t=T} \frac{L}{2 m} d t\)

⇒ \(\pi \mathrm{ab}=\frac{L}{2 m} \mathrm{~T}\)

⇒ \(\mathrm{T}=\frac{2 \pi m a b}{L}\)

Question 2. The Earth and Jupiter are two planets of the sun. The orbital radius of the earth is 10⁷ m and that of Jupiter is 4 × 10⁷ m. If the time period of the revolution of earth is T = 365 days, find the time period of revolution of the Jupiter.

Answer:

For both the planets T² ∝ r³

⇒ \(\left(\frac{T_{\text {jupitar }}}{T_{\text {earth }}}\right)^2=\left(\frac{T_{\text {jupiter }}}{r_{\text {earhh }}}\right)^3\)

⇒ \(\left(\frac{T_{\text {juppicr }}}{365 \text { days }}\right)^2=\left(\frac{4 \times 10^7}{10^7}\right)^3\)

T_jupiter = 8 × 365 days

Question 3.

Suppose earth has radius R and mass M. A point mass m₀ is at a distance r from the centre. Find the gravitational potential energy of the mass due to the earth.
Answer:

U_g= (m₀) (V_earth)

⇒ \(\mathrm{U}_{\mathrm{g}}=\left(\mathrm{m}_0\right)\left(-\frac{G M_e}{r}\right)=\left(-\frac{G M_e m_0}{r}\right)\)

Question 4.

Suppose the earth has mass and radius R. A small groove in made and point mass m0is placed at the centre of the sphere. With what minimum velocity should we project the particle so that it may escape out of the gravity field (reaches to r → ∞)
Answer:

Suppose the particle is projected with speed v, and to send it to infinity, its velocity should be zero at r → ∞. Applying energy conservation between its initial position (centre) and final position (r → ∞) K_i+ U_i= k_f+ U_f

⇒ \(\frac{1}{2} m_0 v^2+\left(m_0\right)\left(v_{\text {earth }}\right)\)

⇒ \(\frac{1}{2} \mathrm{~m}_0 \mathrm{v}^2+\left(\mathrm{m}_0\right)\left(-\frac{3 G M_e}{2 R}\right)\)

⇒ 0+m₀(0)

⇒ \(\mathrm{v}=\left(\sqrt{\frac{3 G M_e}{R}}\right)\)

Summary

Newton’s Law Of Gravitation:

Gravitational attraction force between two point masses

⇒ \(\mathrm{F}_{\mathrm{g}}=\frac{G m_1 m_2}{r^2}\) and its direction will be attractive.

Gravitational force on (1) due to (2) in vector form

⇒ \(\vec{F}_{12}=\frac{G m_1 m_2}{r^2}\)

Gravitational Field: Gravitational force acting on unit mass.

⇒ \(\mathbf{g}=\frac{F}{m}\)

Gravitational Potential:

⇒ \(\mathrm{v}_{\mathrm{g}}=\frac{U}{m}\)

Gravitation potential energy of unit mass

⇒ \(\mathrm{g}=-\frac{d V_g}{d r}\)

and \(\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=-\int_A^B \vec{g} \cdot d \vec{r}\)

For point mass: GM

⇒ \(\mathrm{g}=\frac{G M}{r^2}, \mathrm{~V}=-\frac{G M}{r}\)

For circular ring g = \(\frac{G M x}{\left(R^2+x^2\right)^{3 / 2}}\)

⇒ \(\mathrm{v}=-\frac{G M}{\sqrt{R^2+x^2}}\)

For thin circular disc

⇒ \(\mathrm{g}=\frac{2 G M}{R^2}\left(1-\frac{1}{\sqrt{1+\left(\frac{R}{x}\right)^2}}\right)\)

⇒ \(\mathrm{v}=\frac{-2 G M}{R^2}\left(\sqrt{R^2+x^2}-x\right)\)

Uniform thin spherical shell: 

⇒ \(\mathrm{g}_{\text {out }}=\frac{G M}{r^2}\)

⇒ \(\mathrm{g}_{\text {surface }}=\frac{G M}{R^2}\)

g_in = 0

Potential:

⇒ \(\mathrm{v}_{\text {out }}=-\frac{G M}{r}\)

⇒ \(\mathrm{V}_{\text {surtace }}=-\frac{G M}{R}\)

⇒ \(\mathrm{v}_{\mathrm{in}}=-\frac{G M}{R}\)

Uniform solid sphere: (Most Important)

⇒ \(\mathrm{g}_{\text {out }}=\frac{G M}{r^2}\)

⇒ \(\mathrm{g}_{\text {surface }}=\frac{G M}{R^2}\)

⇒ \(\mathrm{g}_{\mathrm{in}}=\frac{G M}{R^3} r\)

⇒ \(\mathrm{g}_{\text {centre }}=0\)

Potential: \(\mathrm{V}_{\text {out }}=-\frac{G M}{r}\)

⇒ \(\mathrm{V}_{\mathrm{in}}=-\frac{G M}{2 R^3}\left(3 R^2-r^2\right)\)

⇒ \(\mathrm{V}_{\text {surtace }}=-\frac{G M}{R}\)

⇒ \(\mathrm{V}_{\text {centre }}=-\frac{3}{2} \frac{G M}{R}\)

Self Energy:

Surface = \(\mathrm{U}_{\text {self }}=-\frac{1}{2} \frac{G M^2}{R}0\)

Gravitational Self energy of a Uniform Sphere = U_self= \(-\frac{3}{5} \frac{G M^2}{R}\)

Escape speed from earth’s surface

⇒ \(\mathrm{V}_{\mathrm{e}}=\sqrt{\frac{2 G M_e}{R}}\)

= 11.2km/sec.

If a satellite is moving around the Earth in a circular orbit, then its orbital speed is

⇒ \(\mathrm{V}_0=\sqrt{\frac{G M_e}{r}}\)

where r is the distance of the satellite from the centre of the earth.

PE . of the satellite = – \(\frac{G M_e m}{r}\)

KE of the satellite = \(\frac{1}{2} m v_0^2=\frac{G M_e m}{2 r}\)

TE of the satellite = \(-\frac{G M_e m}{2 r}\)

Time Period of Geo-stationary satellite = 24 hours

Kapler’s laws: 

1. Law of Orbit: If a planet is revolving around a sun, its path is either elliptical (or circular)
2. Law of Area :

View (1) If a planet is revolving around a sun, the angular momentum of the planet about the sun remains conserved

View (2) The radius vector from the sum to the planet sweeps the area at a constant rate

Areal velocity = \(\frac{d A}{d t}=\frac{L}{2 m}\) = constant

(3) For all the planets of a sun T² ∝ R³

⇒ \(\mathrm{T}^2=\left(\frac{4 \pi^2}{G M_s}\right) \mathrm{R}^3\)

Factors Affecting Acceleration Due to Gravity

1. Effect Of Altitude: \(\mathrm{g}_{\mathrm{n}}=\frac{G M_e}{\left(R_e+h\right)^2}\)

\(\mathrm{g}\left(1+\frac{h}{R_e}\right)^{-2} \simeq \mathrm{g}\left(1-\frac{2 h}{R_e}\right)\) when h << R.

2. Effect Of Depth: \(\mathrm{g}_{\mathrm{d}}=\mathrm{g}\left(1-\frac{d}{R_e}\right)\)

3. Effect Of The Surface Of Earth

The equatorial radius is about 21 km longer than its polar radius.

We know, g = \(\frac{G M_e}{R_e^2}\)

Hence g_pole > g_equator.

4. Effect Of Rotation Of The Earth

Consider a particle of mass m at latitude θ. g′ = g – ω²R_e cos²θ

At pole θ = 90°

⇒ g_pole = g , At equator θ = 0

⇒ g_equator = g – ω²Re.

Hence g_pole > g_equator

 

Friction Contact Force

When two bodies are kept in contact, electromagnetic forces act between the charged particles (molecules) at the surfaces of the bodies.

Thus, each body exerts a contact force on the other. The magnitudes of the contact forces acting on the two bodies are equal but their directions are opposite and therefore the contact forces obey Newton’s third law.

• The direction of the contact force acting on a particular body is not necessarily perpendicular to the contact surface.
• We can resolve this contact force into two components, one perpendicular to the contact surface and the other parallel to it.
• In the figure, the perpendicular component to the contact surface is called the normal contact force or normal force (generally written as N) and the parallel component is called friction (generally written as f). Therefore if R is the contact force then

⇒ \(\mathrm{R}=\sqrt{f^2+N^2}\)

Reasons For Friction

1. Ιnter-locking of extended parts of one object into the extended parts of the other object.
2. Bonding between the molecules of the two surfaces or objects in contact.

Friction Force Is Of Two Types.

1. Kinetic
2. Static

1. Kinetic Friction Force

Kinetic friction exists between two contact surfaces only when there is relative motion between the two contact surfaces. It stops acting when relative motion between two surfaces ceases.

Direction Of Kinetic Friction On An Object

• Ιt is opposite to the relative velocity of the object considered with respect to the other object in contact.
• Note that its direction is not opposite to the force applied it is opposite to the relative motion of the body considered which is in contact with the other surface.

Magnitude Of Kinetic Friction

The magnitude of the kinetic friction is proportional to the normal force acting between the two bodies. We can write

f_k= μ_k N

where N is the normal force. The proportionality constant μ_k is called the coefficient of kinetic friction and its value depends on the nature of the two surfaces in contact.

Question 1. Find the direction of the kinetic friction force

1. On the block, exerted by the ground.
2. On the ground, exerted by the block.

Answer:

where f₁ and f₂ are the friction forces on the block and ground respectively.

Question 2. In the above example, the correct relation between the magnitude of f₁ and f₂ is

1. f₁> f₂
2. f₂> f₁
3. f₁= f₂
4. It is not possible to decide due to insufficient data.

Answer: By Newton‘s third law the above friction forces are action-reaction pairs, equal but opposite to each other in direction. Hence (3).

Also, note that the direction of kinetic friction has nothing to do with applied force F.

Question 3. All surfaces as shown in the figure are rough. Draw the friction force on A and B

Answer:

Kinetic friction acts in such a way so as to reduce relative motion.

Question 4. Find out the distance travelled by the blocks shown in the figure before it stops.

Answer:

N – 10 g = 0

N = 100 N

f_k= µ_kN

f_k= 0.5 × 100 = 50 N

f_k= ma

50 = 10 a

⇒ a = 5

∴ v² = u² + 2as

0² = 10² + 2 (–5) (S)

∴ S = 10 m

Question 5. Find out the distance travelled by the block on the incline before it stops. The initial velocity of the block is 10 m/s and the coefficient of friction between the block and incline is μ = 0.5.

Answer:

N = mg cos37°

∴ mg sin 37° + µN = ma

a = 10 m/s² down the incline

Now v² = u² + 2as

0 = 10² + 2(–10) S

∴ S = 5 m

Question 6. Find the time taken in the above example by the block before it stops.

Answer:

a = g sin 37° + µg cos 37°

∴ a = 10 m/s² down the incline

∴ S = \(u t+\frac{1}{2} at^2\)

5 = \(\frac{1}{2} \times 10 \times \mathrm{t}^2\)

∴ t = 1sec.

Question 7. A block is given a velocity of 10 m/s and a force of 100 N in addition to friction force is also acting on the block. Find the retardation of the block?

Answer:

As there is relative motion

∴ kinetic friction will act to reduce this relative motion.

f_k = µN = 0.1 × 10 × 10 = 10 N

100 + 10 = 10a

a = 11 m/s²

Static Friction: Ιt exists between the two surfaces when there is a tendency of relative motion but no relative motion occurs along the two contact surfaces.

• For Question consider a bed inside a room; when we gently push the bed with a finger, the bed does not move.
• This means that the bed has a tendency to move in the direction of the applied force but does not move as there exists static friction force acting in the opposite direction of the applied force.

Question 8. What is the value of static friction force on the block?

Answer:

In the horizontal direction acceleration is zero.

Therefore Σ F = 0.

∴ ƒ = 0

Direction Of Static Friction Force: The static friction force on an object is opposite to its impending motion relative to the surface.

The following steps should be followed in determining the direction of static friction force on an object.

1. Draw the free body diagram with respect to the other object on which it is kept.
2. Include pseudo force also if the contact surface is accelerating.
3. Decide the resultant force and the component parallel to the surface of this resultant force.
4. The direction of static friction is opposite to the above component of the resultant force.

Note: Here once again the static friction is involved when there is no relative motion between two surfaces.

Question 9. In the following figure, an object of mass M is kept on a rough table as seen from above. Forces are applied to it as shown. Find the direction of static friction if the object does not move.

Answer:

In the above problem, we first draw the free-body diagram to find the resultant force.

As the object does not move this is not a case of limiting friction. The direction of static friction is opposite to the direction of the resultant force F_R as shown in the figure by f_s. Its magnitude is equal to 25 N.

Magnitude Of Kinetic And Static Friction

Kinetic Friction:

The magnitude of the kinetic friction is proportional to the normal force acting between the two bodies. We can write

f_k= μ_k N

where N is the normal force. The proportionality constant μ_k is called the coefficient of kinetic friction and its value depends on the nature of the two surfaces in contact.

If the surfaces are smooth μ_k will be small, if the surfaces are rough μ_k will be large. It also depends on the materials of the two bodies in contact.

Static Friction:

The magnitude of static friction is equal and opposite to the external force exerted, till the object at which force is exerted is at rest. This means it is a variable and self-adjusting force. However, it has a maximum value called limiting friction.

f_max = μ_sN

The actual force of static friction may be smaller than μ_sN and its value depends on other forces acting on the body. The magnitude of frictional force is equal to that required to keep the body at relative rest.

0, f_s, f_smax

Here μsand μk are proportionality constants. μ_s is called the coefficient of static friction and μ_k is called the coefficient of kinetic friction. They are dimensionless quantities independent of shape and area of contact. It is a property of the two contact surfaces.

Note: μ_s> μ_k for a given pair of surfaces. If not mentioned then μ_s= μ_k can be taken. The value of μ can be from 0 to ∞.

The following table gives a rough estimate of the values of the coefficient of static friction between certain pairs of materials.

The actual value depends on the degree of smoothness and other environmental factors. For example, wood may be prepared at various degrees of smoothness and the friction coefficient will vary.

table

Rolling Friction: When a body (say wheel) rolls on a surface the resistance offered by the surface is called rolling friction.

• Rolling friction forces arise as, for example, a rubber tyre rolls on pavement, primarily because the tyre deforms as the wheel rolls. The sliding of molecules against each other within the rubber causes energy to be lost.
• The velocity of the point of contact with respect to the surface remains zero.
• The rolling friction is negligible in comparison to static or kinetic friction which may be present simultaneously i.e., µ_R< µ_k< µ_S

Angle Of Friction

The angle of friction is the angle which the resultant of limiting friction F_S and normal reaction N makes with the normal reaction. It is represented by λ, Thus from the figure.

⇒ \(\tan \lambda=\frac{F_S}{N}\) (∵ F_s= µ N) ortan

or \(\frac{F_S}{N}\) = θtan λ = µ

For smooth surfaces, λ = 0 (zero)

Angle Of Repose (θ)

If a body is placed on an inclined plane and if its angle of inclination is gradually increased, then at some angle of inclination θ the body will just begin to slide down this angle is called the angle of repose (θ).

F_S = mg sinθ and N = mg cosθ

So, or µ = tanθ

Relation between angle of friction (λ) and angle of repose (θ)

We know that tan λ = µ and µ = tan θ

hence tan λ = tan θ or θ = λ

Thus, angle of repose = angle of friction

Question 1. Find the acceleration of the block. Initially, the block is at rest.

Answer: Zero

Question 2. Find out the acceleration of the block. Initially, the block is at rest.

Answer:

N + 24 – 100 = 0 for vertical direction

∴ N = 76 N

Now 0 ≤ f_s ≤ µ_s N

0 ≤ f_s ≤ 76 × 0.5

0 ≤ f_s≤ 38 N

∴ 32 < 38 Hence f = 32

∴ acceleration of the block is zero.

Question 3. Find out the acceleration of the block for different ranges of F.

Answer:

0 ≤ f ≤ µ_sN

⇒ 0 ≤ f ≤ µ_smg

a = 0 if F ≤ µ_smg

⇒ a = \(\frac{F-\mu M g}{M}\) if F > µMg M

Question 4. Find out the acceleration of the block. Initially, the block is at rest.

Answer:

0 ≤ f_s ≤ µ_sN

⇒ 0 ≤ f_s≤ 50

Now 51 > 50

∴ The block will move but if the block starts moving then kinetic friction is involved.

f_k= µ_k N = 0.3 × 100 = 30 N

∴ 51 – 30 = 10 a

∴ a = 2.1 m/s²

Question 5. Find out the minimum force that must be applied on the block vertically downwards so that the block doesn’t move.

Answer:

100-f_s = 0

∴ f_s= 100 ……..

F + 10 g = N ⇒ N = 100 + F ……….. (2)

Now 0 ≤ f_s≤ μN

100 ≤ 0.5 N

100 ≤ 0.5 [100 + F]

200 ≤ 100 + F

F ≥ 100 N

∴ Minimum F = 100 N

Question 6. A particle of mass 5 kg is moving on a rough fixed inclined plane with a constant velocity of 5 m/s as shown in the figure. Find the friction force acting on a body by plane.

Answer:

f_k= μ_kN = μ_k mg cos 37° = mg sin30° = 5 (10)\(\left(\frac{1}{2}\right)\)

⇒ f_k= 25 N

Question 7. The angle of inclination is slowly increased. Find out the angle at which the block starts moving.

Answer:

0 ≤ f ≤ µ_s N

mg sinθ > f_smax

mg sin θ > µN

mg sinθ > µ mg cos θ

∴ tan θ > µ

θ = tan^-1 µ

for tan θ ≤ µ no sliding on an inclined plane.

This method is used to find out the value of µ practically.

Question 8. Find out the acceleration of the block. If the block is initially at rest.

Answer: (FBD of the block excluding friction) N = 10 g cos 37° = 80 N

Now 0 ≤ f_s≤ µN

0 ≤ f_s≤ 0.5 × 80

∴ f_s≤ 40 N

We will put a value of f in the last i.e. in the direction opposite to the resultant of other forces. f acts down the incline and its value is of = 75 – 60 = 15 N

So acceleration is zero

Question 9. In the above problem, how much force should be added to 75 N force so that the block starts to move up the incline?

Answer:

∴ 60 + 40 = 75 + f_extra

∴ f_s= 25 N

Question 10. In the above problem, what is the minimum force by which 75 N force should be replaced so that the block does not move?
Answer:

In this case, the block has a tendency to move downwards.

Hence friction acts upwards.

∴ F + 40 = 60

∴ F = 20 N

Question 11. The top view of a block on a table is shown (g = 10 m/s²). Find out the acceleration of the block.

Answer:

Now f_s≤ µN

∴ f_s≤ 50

Hence the block will move.

a = \(\frac{40 \sqrt{2}-50}{10}\)

⇒ \((4 \sqrt{2}-5) \mathrm{m} / \mathrm{s}^2\)

Question 12. Find minimum µ so that the blocks remain stationary.

Answer: T = 100 g = 1000 N

∴ f = 1000 to keep the block stationary

Now f_max= 1000

µN = 1000

µ = 2

Can µ be greater than 1?

Yes 0 < µ ≤ ∝

Question 13. Find out the minimum acceleration of block A so that the 10 kg block doesn’t fall.

Answer:

Applying NL in a horizontal direction

N = 10 a ………(1)

Applying NL in a vertical direction

10 g = µ N …….(2)

10 g = µ 10 a from (1) and (2)

∴ a = 20 m/s²

Question 14. In the following figure force F is gradually increased from zero. Draw the graph between applied force F and tension T in the string. The coefficient of static friction between the block and the ground is μ_s.

Answer:

As the external force F is gradually increased from zero it is compensated by the friction and the string bears no tension. When limiting friction is achieved by increasing force F to a value till μ_smg, the further increase in F is transferred to the string.

Question 15. Find the acceleration of the two blocks. The system is initially at rest and the friction coefficient is as shown in the figure.

Answer:

f_max= 50 N

∴ f ≤ 50 N

If they move together a = \(\frac{101}{20}\)

= 5.05 m/s²

Check friction on B

f = 10 × 5.05 = 50.5 (required)

50.5 > 50 (therefore required > available)

Hence they will not move together.

Hence they move separately so kinetic friction is involved.

∴ for \(\mathrm{a}_{\mathrm{B}}=\frac{50}{10}=5.1 \mathrm{~m} / \mathrm{s}^2\)

⇒ \(a_A=\frac{101-50}{10}=5 \mathrm{~m} / \mathrm{s}^2\)

Also, a_A > a_B as force is applied on A.

Question 16. Find the acceleration of the two blocks. The system is initially at rest and the friction coefficient is as shown in the figure.

Answer:

Move Together Move Separately No need to calculate.

a = 2 m/s²

Check friction on 20 kg.

f = 20 × 2

f = 40 (which is required)

40 < 50 (therefore required < available)

∴ will move together.

Question 17. In the above example, find the maximum F for which two blocks will move together.
Answer:

Observing the critical situation where friction becomes limiting.

∴ F – f_max= 10 a ………(1)

f_max = 20 a ……….(2)

∴ F = 75 N

Question 18. Initially, the system is at rest. find out the minimum value of F for which sliding starts between the two blocks.

Answer:

At just sliding conditions limiting friction is acting.

F – 50 = 20 a ………..(1)

f = 10 a ………………(2)

50 = 10 a

∴ a = 5 m/s²

Hence F = 50 + 20 × 5 = 150 N

∴ F_min = 150 N

It is Easier to Pull Than to Push a Body. Why –

Let a force P be applied to pull a body of weight Mg.

The applied force is resolved into two components : P cosθ and P sinθ The normal reaction, R = (Mg – P sinθ)

Now, the kinetic force of friction is given by

F₁= μ_k R = μ_k(Mg – P sinθ) …….(1)

On the other hand, when the same force is applied to push a body of weight Mg, then a normal reaction,

R = (Mg + P sinθ) …….(2)

∴ Kinetic force of friction is F₂= μ_k R = μ_k(mg + P sinθ)

From eqn (1) and (2), it is clear and F₂> F₁

That is, the force of friction in the case of push is more than that in the case of pull.

Hence, it is easier to pull than to push the body.

Friction is a Necessary Evil :

Friction is a necessary evil. It means it has advantages as well as disadvantages. In other words, friction is not desirable but without friction, we cannot think of survival.

Disadvantages :

1. A significant amount of energy of a moving object is wasted in the form of heat energy to overcome the force of friction.
2. The force of friction restricts the speed of moving vehicles like buses, trains, aeroplanes, rockets etc.
3. The efficiency of machines decreases due to the presence of force of friction.
4. The force of friction causes a lot of wear and tear in the moving parts of a machine.
5. Sometimes, a machine gets burnt due to the friction force between different moving parts.

Advantages :

1. The force of friction helps us to move on the surface of the earth. In the absence of friction, we cannot think of walking on the surface. That is why, we fall down while moving on a smooth surface.
2. The force of friction between the tip of a pen and the surface of paper helps us to write on the paper. It is not possible to write on the glazed paper as there is no force of friction.
3. The force of friction between the tyres of a vehicle and the road helps the vehicle to stop when the brake is applied. In the absence of friction, the vehicle skid off the road when the brake is applied.
4. Moving belts remain on the rim of a wheel because of friction.
5. The force of friction between a chalk and the blackboard helps us to write on the board. Thus, we observe that irrespective of the various disadvantages of friction, it is very difficult to part with it. So, friction is a necessary evil.

Methods Of Reducing Friction

As friction causes the wastage of energy it becomes necessary to reduce the friction. Friction can be reduced by the following methods.

1. Polishing the surface. We know, that friction between rough surfaces is much greater than between the polished surfaces. So we polish the surface to reduce the friction. The irregularities on the surface are filled with polish and hence the friction decreases.
2. Lubrication. To reduce friction, lubricants like oil or grease are used. When the oil or grease is put in between the two surfaces, the irregularities remain apart and do not interlock tightly. Thus, the surfaces can move over each other with less friction between them.
3. By providing a streamlined shape. When a body (e.g. bus, train, aeroplane etc.) moves with high speed, air resistance (friction) opposes its motion. The effect of air resistance on the motion of the objects (stated above) is decreased by providing them with a streamlined shape.
4. Converting sliding friction into rolling friction. Since rolling friction is much less than sliding friction, we convert the sliding friction into rolling friction. This is done by using a ball bearings arrangement. Ball bearings are placed in between the axle (A) and hub (B) of the wheel as shown in the figure. The ball bearing tends to roll around the axle as the wheel turns and as such the frictional force is diminished.

Friction Key Concept

Part of the contact force that is tangential to the surface is called friction force. Microscopically friction force because of attraction between molecules of the two surfaces.

The friction force is of two types:

1. Kinetic friction
2. Static friction

Kinetic Friction Force:

Kinetic friction exists between two surfaces (in the case of a block), two points (in the case of a sphere), or two lines (in the case of a cylinder) when there is relative motion. It stops acting when relative motion ceases to exist.

Direction Of Kinetic Friction:

It is opposite to the relative velocity of contact surfaces.

Note: Its direction is not opposite to the force applied it is opposite to the motion of the body considered which is in contact with the other surface.

Kinetic friction f_k= μ_kN

The proportionality constant μ_k is called the coefficient of kinetic friction and its value depends on the nature of the two surfaces in contact.

Static Friction:

When two surfaces in contact have a relative velocity of zero, but there is a tendency of relative motion then the friction force acting will be static.

Direction Of Static Friction: If there is a tendency to slide between the contact surfaces, it will act in such a direction to prevent sliding.

Static friction is a variable and self-adjusting force. It can adjust its value upto a limit which is called limiting friction force (f_smax).

f_smax= μ_sN

Here μ_s is the coefficient of static friction which depends on the nature of two contact surfaces. The actual force of static friction may be smaller than μ_sN and its value depends on other forces acting on the body. The magnitude of frictional force is equal to that required to keep the body at relative rest.

0 f_s f_smax

The following steps should be followed in determining the direction of static friction force on an object.

1. Draw the free-body diagram to the other object on which it is kept.
2. Include pseudo force also if the contact surface is accelerating.
3. Decide the resultant force and the component parallel to the surface of this resultant force.
4. The direction of static friction is opposite to the above component of the resultant force.

Here μ_s and μ_k are dimensionless quantities independent of shape and area of contact. It is a property of the two contact surfaces. In general μ_s > μ_k for a given pair of surfaces. If it is not mentioned separately the μ_s = μ_k can be taken.

μ_s and μ_k can also be represented as angles. If θ_s and θ_k are angles of static friction and kinetic friction respectively, then

θ_s= tan^-1 μ_s

θ_k= tan^-1 μ_k

θ_s is also called the angle of repose.

Rolling Friction:

When a body rolls on a surface the resistance offered by the surface is called rolling friction.

Rolling Friction Is Less Than Sliding Friction

In sliding motion, elevation collides. This introduces friction. In rolling motion, elevations are crossed over. This avoids friction.

Rolling Friction Zero In Ideal Case:

• In ideal rolling the contact with the cylindrical surface of the body and lower surface must be along a straight line. Elevations must be crossed over and no friction be present.
• But no rolling is ideal. Due to the deformation of the moving cylindrical surface (wheel of a loaded truck), or the deformation of the lower surface (mud street), contact becomes over a flat surface. This introduces sliding, which causes friction.

 

Fluid Mechanics Definition Of Fluid

The term fluid refers to a substance that can flow and does not have a shape of its own. For example liquids and gases.

Fluid includes properties → (1) Density (2) Viscosity (3) Bulk modulus of elasticity (4) pressure (5) specific gravity

Pressure In A Fluid

The pressure p is defined as the magnitude of the normal force acting on a unit surface area.

P = \(\frac{\Delta F}{\Delta A}\)

ΔF = normal force on a surface area ΔA.

The pressure is a scalar quantity. This is because hydrostatic pressure is transmitted equally in all directions when force is applied, which shows that a definite direction is not associated with pressure.

Thrust. The total force exerted by a liquid on any surface in contact with it is called the thrust of the liquid.

Note:
The normal force exerted by liquid at rest on a given surface in contact with it is called the thrust of liquid on that surface.

The normal force (or thrust) exerted by liquid at rest per unit area of the surface in contact with it, is called pressure of liquid or hydrostatic pressure.

If F is the normal force acting on a surface of area A in contact with liquid, then the pressure exerted by the liquid on this surface is P = F/A

1. Units : N / m² or Pascal (S.I) and Dyne/cm²(C.G.S)
2. Dimension : (P) = \(\frac{[\mathrm{F}]}{[\mathrm{A}]}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^2\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\)
3. At a point pressure acts in all directions and a definite direction is not associated with it. So pressure is a tensor quantity.
4. Atmospheric pressure: The gaseous envelope surrounding the earth is called the earth’s atmosphere and the pressure exerted by the atmosphere is called atmospheric pressure its value on the surface of the earth at sea level is nearly 1.013 × 10⁵ N/m² or Pascal in S.I. Other practical units of pressure are atmosphere, bar and torr (mm of Hg) 1 atm = 1.01 × 10⁵ Pa = 1.01 bar = 760 torr.
   1. The atmospheric pressure is maximum at the surface of the earth and goes on decreasing as we move up into the earth’s
5. If P₀ is the atmospheric pressure then for a point at depth h below the surface of a liquid of density ρ. hydrostatic pressure P is given by P = P₀+ hρg.

6. Hydrostatic pressure depends on the depth of the point below the surface (h). nature of liquid (ρ) and acceleration due to gravity (g) while it is independent of the amount of liquid, the shape of the container, or the cross-sectional area considered.

• So if a given liquid is filled in vessels of different shapes to the same height, the pressure at the base in each vessel will be the same, though the volume or weight of the liquid in different vessels will be different.

7. In a liquid at the same level, the pressure will be the same at all points, if not, due to the pressure difference the liquid cannot be at rest. This is why the height of the liquid is the same in vessels of different shapes containing different amounts of the same liquid at rest when they are in communication with each other.

Gauge pressure: The pressure difference between hydrostatic pressure P and atmospheric pressure P0 is called gauge pressure.

P – P₀= hρg

Consequences Of Pressure

Railway tracks are laid on large-sized wooden or iron sleepers. This is because the weight (force) of the train is spread over a large area of the sleeper.

1. This reduces the pressure acting on the ground and hence prevents the yielding of ground under the weight of the train.
2. A sharp knife is more effective in cutting objects than a blunt knife. The pressure exerted = Force/area. The sharp knife transmits force over a small area as compared to the blunt knife. Hence the pressure exerted in the case of a sharp knife is more than in the case of a blunt knife.
3. A camel walks easily on sand but a man cannot in spite of the fact that a camel is much heavier than a man.

This is because the area of camel’s feet is large as compared to man’s feet. So the pressure exerted by the camel on the sand is very small as compared to the pressure exerted by man. Due to large pressure, sand under the feet of man yields, and hence he cannot walk easily on sand.

Variation Of Pressure With Height

Assumptions:

1. Unaccelerated liquid
2. Uniform density of liquid
3. Uniform gravity

The weight of the small element dh is balanced by the excess pressure. It means

⇒ \(\frac{d p}{d h}=\rho g\)

⇒ \(\int_{P_a}^P d p=\rho g \int_0^h d h\)

⇒ P = Pa + ρgh

Pascal’S Law

if the pressure in a liquid is changed at a particular, point the change is transmitted to the entire liquid without being diminished in magnitude. In the above case if P_a is increased by some amount then P must increase to maintain the difference (P – P_a) = hρg. This is Pascal’s Law which states that Hydraulic lift is a common application of Pascal’s Law.

1. Hydraulic press.

⇒ \(p=\frac{f}{a}=\frac{W}{A} \text { or } f=\frac{W}{A} \times a\)

as A >> a then f<<W…

This can be used to lift a heavy load placed on the platform of a larger piston or to press the things placed between the piston and the heavy platform. The work done by applied force is equal to the change in the potential energy of the weight in a hydraulic press.

Density: In a fluid, at a point, density ρ is defined as :

⇒ \(\rho=\lim _{\Delta V \rightarrow 0} \frac{\Delta m}{\Delta V}=\frac{d m}{d V}\)

1. In the case of a homogenous isotropic substance, it has no directional properties, so is a scalar.
2. It has dimensions (ML–and S.I. unit kg/m³ while C.G.S. unit g/cc with 1g /cc = 10³ kg/m³
3. The density of a substance means the ratio of the mass of a substance to the means the ratio of mass of a body to the volume of the body. So for a solid body. Density of body = Density of substance While for a hollow body, the density of the body is lesser than that of substance [As V_body > V_sub.]
4. When immiscible liquids of different densities are poured into a container, the liquid of the highest density will be at the bottom while that of the lowest density at the top, and interfaces will be plane.
5. Sometimes instead of density, we use the term relative density or specific gravity which is defined as:

⇒ \(\mathrm{RD}=\frac{\text { Density of body }}{\text { Density of water }}\)

6. If m₁ mass of liquid of density ρ₁ and m₂ mass of density ρ₂ are mixed. then as
m = m₁+ m₂ and V = (m₁/ ρ+ (m₂/ ρ₂)

[As V = m/ρ]

7. If  V₁ volume of liquid of density ρ₁ and V₂ volume of liquid of density ρ2are mixed, then as m = ρ₁V₁+ ρ₂V₂ and V = V₁+ V₂[As ρ = m/V]

If V₁= V₂= V ρ₁(ρ₁ + ρ₂)/2 = Arithmetic Mean

Question 1. A body of one kg is placed on two objects of negligible mass. Calculate pressure due to force on its bottom.

Answer:

⇒ \(P=\frac{F}{A}=\frac{m g}{A}\)

⇒ \(P_1=\frac{1 \times 10 \mathrm{~N}}{10 \times 10^{-2} \mathrm{~m}^2}=10^4 \mathrm{~N} / \mathrm{m}^2\)

⇒ \(P_2=\frac{1 \times 10 \mathrm{~N}}{2 \times 10^{-4}}=5 \times 10^4\left(=5 p_1\right)\)

Question 2. For a hydraulic system, A car of mass 2000 kg standing on the platform of Area 10m² while the area other side platform 10 cm² finds the mass required to balance the car
Answer:

According to the Pascal’s Law

P₁ = P₂ \(\frac{m_{\text {car }} g}{A_{\text {car }}}=\frac{\mathrm{mg}}{\mathrm{A}}\)

⇒ \(\mathrm{m}=\left(\frac{\mathrm{A}}{\mathrm{A}_{\text {car }}}\right) \times \mathrm{m}_{\text {car }}=\frac{10 \mathrm{~cm}^2}{10 \mathrm{~m}^2} \times 2000 \mathrm{~kg}\)

= 0.2 kg

= 200 gm

Question 3. If two liquids of the same masses but densities of P₁ and P₂ respectively are mixed, then the density of the mixture is given by

1. \(\rho=\frac{\rho_1+\rho_2}{2}\)
2. \(\rho=\frac{\rho_1+\rho_2}{2 \rho_1 \rho_2}\)
3. \(\rho=\frac{2 \rho_1 \rho_2}{\rho_1+\rho_2}\)
4. \(\rho=\frac{\rho_1 \rho_2}{\rho_1+\rho_2}\)

Answer:

ρ = \(\frac{\text { Total mass }}{\text { Totalvolume }}\)

⇒ \(\frac{2 m}{V_1+V_2}=\frac{2 m}{m\left(\frac{1}{\rho_1+\rho_2}\right)}\)

∴ \(\rho=\frac{2 \rho_1 \rho_2}{\rho_1+\rho_2}\)

Question 4. If two liquids of the same masses but different densities ρ₁ and ρ₂ are mixed, then the density of the mixture is given by

1. \(\rho=\frac{\rho_1+\rho_2}{2}\)
2. \(\rho=\frac{\rho_1+\rho_2}{2 \rho_1 \rho_2}\)
3. \(\rho=\frac{2 \rho_1 \rho_2}{\rho_1+\rho_2}\)
4. \(\rho=\frac{\rho_1 \rho_2}{\rho_1+\rho_2}\)

Answer:

ρ = \(\frac{\text { Total mass }}{\text { Total volume }}\)

⇒ \(\frac{m_1+m_2}{2 V} \frac{V\left(\rho_1+\rho_2\right)}{2 V}\)

⇒ \(\frac{\rho_1+\rho_2}{2}\)

Question 5. If pressure at the half depth of a lake is equal to 2/3 of pressure at the bottom of the lake, then the depth of the lake [ρ_water = 10³ kg m^-3, P₁= 10⁵ N/m²]

1. 10 m
2. 20 m
3. 60 m
4. 30 m

Answer: Pressure at the bottom of the lake = P₀+hpg

Pressure at half the depth of a lake = \(P_0+h \rho g\)

According to given condition \(P_0+\frac{1}{2} h \rho g=\frac{2}{3}\left(P_0+h \rho g\right)\)

⇒ \(\frac{1}{3} P_0=\frac{1}{6} h \rho g\)

⇒ \(h=\frac{2 P_o}{\rho g}=\frac{2 \times 10^5}{10^3 \times 10}=20 \mathrm{~m}\)

Question 6. A uniform tapering vessel is filled with a liquid of uniform density 900 kg/m³. The force that acts on the base of the vessel due to the liquid is (g = 10 ms^-2)

1. 3.6 N
2. 7.2 N
3. 9.0 N
4. 14.4 N

Answer: Force acting on the base

F = P × A = hdgA = 0.4 × 900 × 10 × 2 × 10^-3 = 7.2N

Question 7. The area of the cross-section of the two arms of a hydraulic press is 1 cm² and 10 cm² respectively (figure). A force of 5 N is applied to the water in the thinner arm. What force should be applied to the water in the thicker arms so that the water may remain in equilibrium?

Answer:

In equilibrium, the pressures at the two surfaces should be equal as they lie in the same horizontal level. If the atmospheric pressure is P and a force F is applied to maintain the equilibrium, the pressures are

⇒ \(P_0+\frac{5 \mathrm{~N}}{1 \mathrm{~cm}^2}\) and \([P_0+\frac{F}{10 \mathrm{~cm}^2}\) respectively. This gives F = 50 N.

Hydraulic Brake: Hydraulic brake system is used in automobiles to retard the motion.

Hydrostatic Paradox

Pressure is directly proportional to depth and by applying Pascal’s law it can be seen that pressure is independent of the size and shape of the containing vessel. (In all three cases the heights are the same).

P_A= P_B= P_C

Atmospheric Pressure

Atmospheric Pressure Definition: The atmospheric pressure at any point is numerically equal to the weight of a column of air of a unit cross-sectional area extending from that point to the top of the atmosphere.

At 0ºC, the density of mercury = 13.595 g cm^-3, and at sea level, g = 980.66 cm s^-2

Now P = hρg.

Atmospheric pressure = 76 × 13.595 × 980.66 dyne cm^-2 = 1.013 × 10^-5 N-m²(p_a)

Height of Atmosphere

The standard atmospheric pressure is 1.013 × 10⁵ Pa (N m^-2). If the atmosphere of the earth has a uniform density ρ = 1.30 kg m^-3, then the height h of the air column which exerts the standard atmospheric pressure is given by

⇒ hρg = 1.013 × 10⁵

h = \(\frac{1.013 \times 10^5}{\rho g}\)

⇒ \(\frac{1.013 \times 10^5}{1.13 \times 9.8}\mathrm{~m}\)

m = 7.95 × 10³ m ~ 8 km. 1.13 9.8

In fact, the density of air is not constant but decreases with height. The density becomes half at about 6 km high,\(\frac{1}{4}\)th at about 12 km, and so on.

Therefore, we can not draw a clear-cut line above which there is no atmosphere. Anyhow the atmosphere extends upto 1200 km. This limit is considered for all practical purposes.

Measurement Of Atmospheric Pressure

1. Mercury Barometer.

To measure the atmospheric pressure experimentally, Torricelli invented a mercury barometer in 1643.

p_a= hρg

The pressure exerted by a mercury column of 1mm

high is called 1 Torr.

1 Torr = 1 mm of mercury column

2. Open tube Manometer

An open-tube manometer is used to measure the pressure gauge. When equilibrium is reached, the pressure at the bottom of a left limb is equal to the pressure at the bottom of the right limb.

i.e. p + y₁ ρg = p_a+ y₂ ρg

p – p_a= ρg (y₂– y₁)= ρgy

p – p_a= ρg (y₂– y₁)= ρgy

p = absolute pressure, p – p_a= gauge pressure.

Thus, knowing y and ρ (density of liquid), we can measure the gauge pressure.

Question 1. A barometer tube reads 76 cm of mercury. If the tube is gradually inclined at an angle of 60° vertically, keeping the open end immersed in the mercury reservoir, the length of the mercury column will be

1. 152 cm
2. 76 cm
3. 38 cm
4. \(38 \sqrt{3} \mathrm{~cm}\)

Answer: cos 60º = \(\frac{\mathrm{h}}{\ell}\)

⇒ \(\ell=\frac{\mathrm{h}}{\cos 60^{\circ}}=\frac{76}{1 / 2}\)

∴ l = 152 cm

Question 2. When a large bubble rises from the bottom of a lake to the surface. Its radius doubles. If atmospheric pressure is equal to that of a column of water height H, then the depth of the lake is

1. H
2. 2H
3. 7H
4. 8H

Answer:

P₁V₁= P₂V₂

⇒ \(\left(P_{\circ}+h \rho g\right) \times \frac{4}{3} \pi r^2=P_0 \times \frac{4}{3} \pi(2 r)^3\)

Question 3. A beaker containing liquid is kept inside a big closed jar If the air inside the jar is continuously pumped out, the pressure in the liquid near the bottom of the liquid will

1. Increase
2. Decreases
3. Remain constant
4. First decrease and then increase

Answer: Total pressure at (near) bottom of the liquid

P = P₀+ hρg

As air is continuously pumped out from the jar (container), P0 decreases and hence P decreases.

Question 4. Write the pressure inside the tube

Answer:

⇒ \(P_A=P_0+\rho g \frac{8}{100}=P_B\)…(i)

⇒ \(P_{\text {tube }}=P_B-\rho g \frac{6}{100}\)…

(1) and (2)

⇒ \(P_{\text {tube }}=\left(P_0+\rho g \frac{8}{100}\right)-\rho g \frac{6}{100}\)

⇒ \(P_{\text {tube }}=\left(P_0+\frac{\rho g}{50}\right)\)

⇒ \(P_0+\frac{\rho g}{50}=P\)

Question 5. Find the pressure inside the tube

Answer:

⇒ \(P_A=\left(P_0+\rho g h\right)+3 \rho g(2 h)=P_B\)

⇒ \(P_{\text {tube }}=P_B-8 \rho g\left(\frac{2 h}{3}-\frac{h}{4}\right)\)

⇒ \(P_0+\rho g h+6 \rho g h-8 \rho g\left(\frac{5}{12} h\right)\)

⇒ \(P_0+\rho g h+6 \rho g h-\frac{10}{3} \rho g h\)

⇒ \(P_0+\frac{(21-10) g g h}{3}=P_0+\frac{11}{3} \rho g h\)

⇒ \(P_0+\frac{11}{3} \rho g h\)

Question 6. The manometer shown below is used to measure the difference in water level between the two tanks. Calculate this difference for the conditions indicated.

Answer:

p_a+ h₁ρg – 40ρ₁g + 40ρg = p_a+ h₂ρg

h₂ρg – h₁ρg = 40 ρg – 40 ρ₁g

as ρ₁= 0.9ρ

(h₂– h₁)ρg = 40ρg – 36ρg

h₂– h₁= 4 cm

3. Water Barometer.

Let us suppose water is used in the barometer instead of mercury.

hρg = 1.013 × 10⁵ or h = \(\mathrm{h}=\frac{1.013 \times 10^5}{\rho \mathrm{g}}\)

The height of the water column in the tube will be 10.3 m. Such a long tube cannot be managed easily, thus, a water barometer is not feasible.

Question 7. In a given U-tube (open at one end) find out the relation between p and p_a. Given d₂= 2 × 13.6 gm/cm³ d₁= 13.6 gm/cm³

Answer: Pressure in a liquid at the same level is the same i.e. at A – A–, p_a+d₂yg+xd₁g = p

In C.G.S.

p_a+ 13.6 × 2 × 25 × g + 13.6 × 26 × g = p

p_a+ 13.6 × g [50 + 26] = p

2p_a= p   [p_a= 13.6 × g × 76]

Question 8. The truck starts from rest with an acceleration of 2.5 ms^-2 then the angle (acute) between the vertical and surface at the liquid, In equilibrium (assume that liquid is at with respect to the truck)

1. sinθ = \(\frac{4}{\sqrt{17}}\)
2. cosθ = \(\frac{1}{\sqrt{17}}\)
3. tanθ = 4
4. None of these

Answer :

Consider a particle on the liquid surface

mg cos θ = ma cos θ

gcosθ = a sinθ

tanθ = \(\frac{\mathrm{g}}{\mathrm{a}}\)

tanθ = \(\frac{10}{2.5}\) = 4

ABC

Question 9. In the previous question pressure at points A, B, and C

1. P_A = P_B = P_C
2. P_A > P_B > P_C
3. P_A < P_B < P_C
4. None of these

Answer: 2. P_A > P_B > P_C

Question 10. In the previous question, three different points, above the points A, B, and C of an accelerated liquid surface in equilibrium called A’ B’, C’ then the pressure at the points A’ B’ and C’

1. P_A‘ = P_B‘ = P_C‘
2. P_A‘ > P_B‘ > P_C‘
3. P_A‘ < P_B‘ < P_C‘
4. Po = atmosphere pressure

Answer: 2. P_A‘ > P_B‘ > P_C‘

Question 11. Highest pressure at the point inside the liquid :

1. A
2. C
3. Pressure at A, B, and C are equal, and the highest
4. None of these

Answer: 1. A

Question 12. The slope of the line on which pressure is the same considers the direction of acceleration of the truck as the X-axis

1. –4
2. –0.25
3. –2.5
4. \(-\frac{1}{4}\)

Answer: 4. \(-\frac{1}{4}\)

Archimedes’ Principle

According to this principle, when a body is immersed wholly or partially in a fluid, it loses its weight which is equal to the weight of the fluid displaced by the body.

Up thrust = buoyancy = Vρ_lg

V = volume submerged

ρ_l= density of liquid.

Relation between the density of solid and liquid

weight of the floating solid = weight of the liquid displaced

V₁ρ₁g = V₂ρ₂ g

⇒ \(\frac{\rho_1}{\rho_2}=\frac{V_2}{V_1}\)

or \(\frac{\text { Density of solid }}{\text { Density of liquid }}=\frac{\text { Volume of the immersed portion of the solid }}{\text { Total Volume of the solid }}\)

This relationship is valid in accelerating fluid also. Thus, the forces acting on the body are :

1. Its weight is Mg which acts downward and
2. Net upward thrust on the body or the buoyant force (mg)

Hence the apparent weight of the body = Mg – mg = weight of the body – weight of the displaced liquid. Or Actual Weight of body – Apparent weight of body = weight of the liquid displaced.

• The point through which the upward thrust or the buoyant force acts when the body is immersed in the liquid is called its center of buoyancy. This will coincide with the center of gravity if the solid body is homogeneous.
• On the other hand, if the body is not homogeneous, then the center of gravity may not lie on the line of the upward thrust and hence there may be a torque that causes rotation in the body.
• If the center of gravity of the body and the center of buoyancy lie on the same straight line, the body is in equilibrium.
• If the center of gravity of the body does not coincide with the center of buoyancy (i.e., the line of upthrust), then torque acts on the body. This torque causes the rotational motion of the body.

Floatation

1. Translatory equilibrium: When a body of density p and volume V is immersed in a liquid of density σ, the forces acting on the body are

Weight of body W = mg = Vρg, acting vertically downwards through the center of gravity of the body. Upthrust force = Vσg acting vertically upwards through the center of gravity of the displaced liquid i.e., the center of buoyancy.

If the density is greater than that of liquid p>σ

Weight will be more than upthrust so the body will sink

If density is equal to that of liquid p>σ

Weight will be equal to upthrust so the body will float fully submerged in neutral equilibrium with its top surface just at the top of a liquid

If the density of the body is lesser than that of liquid p>σ

Weight will be less than upthrust so the body will, move upwards and in equilibrium will float and be partially immersed in the liquid such that,

W = V_in σg

⇒ Vρg = V_in σg

Vρ = V_in σ

Where V_in is the volume of a body in the liquid

1. A body will float in liquid only and only if ρ ≤ σ
2. In the case of floating the weight of the body = upthrust
3. So W_App = Actual weight – upthrust = 0
4. In the case of floating Vρg = V_inσg

So the equilibrium of floating bodies is unaffected by variations in g though both thrust and weight depend on g.

Rotatory Equilibrium: When a floating body is slightly tied from the equilibrium position, the center of buoyancy B shifts. The vertical line passing through the new center of buoyancy B’ and the initial vertical line meet at a point M called meta-center.

If the meta-center M is above the center of gravity the couple due to forces at G (weight of body W) and at B’ (upthrust) tends to bring the body back body the meta the center must always be higher than the center of gravity of the body.

• However, if the meta-center goes CG, the couple due to forces at G and B’ tends to topple the floating body.
• That is why a wooden log cannot be made to float vertically in water or a boat is likely to capsize if the sitting passengers stand on it. In these situations, CG becomes higher than MG and so the body will topple if slightly tilted.

Question 1. A concrete sphere of radius R has a cavity of radius r which is packed with sawdust. The specific gravities of concrete and sawdust are respectively 2.4 and 0.3 for this sphere to float with its entire volume submerged under water. A ratio of the mass of concrete to the mass of sawdust will be

1. 8
2. 4
3. 3
4. Zero

Answer: Let specific gravities of concrete and sawdust be ρ₁ and ρ₂ respectively.

According to the principle of floatation weight of the whole sphere = upthrust on the sphere

⇒ \(\frac{4}{5} \pi\left(R^3-r^3\right) \rho_1 g+\frac{4}{3} \pi r^3 \rho_2 g=\frac{4}{3} \pi R^3 \times 1 \times g\)

⇒ \(R^3 \rho_1-r^3 \rho_1+r^3 \rho_2=R^3\)

⇒ \(R^3\left(\rho_1-1\right)=r^3\left(\rho_1-\rho_2\right)=\frac{R^3}{r^3}=\frac{\rho_1-\rho_2}{\rho_1-1}\)

⇒ \(\frac{R^3-r^3}{r^3}=\frac{\rho_1-\rho_2-\rho_1+1}{\rho_1-1}\)

⇒ \(\frac{\left(R^3-r^3\right) \rho_1}{r^3 \rho_2}=\left(\frac{1-\rho_2}{\rho_1-1}\right) \frac{\rho_1}{\rho_2}\)

⇒ \(\frac{\text { Mass of concrete }}{\text { Mass of saw dust }}\)

⇒ \(\left(\frac{1-0.3}{2.4-1}\right) \times \frac{2.4}{0.3}=4\)

Question 2. A metallic block of density 5 gm cm^-3 and having dimensions 5 cm × 5 cm × 5cm is weighed in water. Its apparent weight will be

1. 5 × 5× 5 × 5 gf
2. 4 × 4 × 4 × 4 gf
3. 5 × 4× 4 × 4 gf
4. 4 × 5× 5 × 5 gf

Answer: Apparent weight

= V (ρ – σ) g = 1 × b × h × (5 –1) × g

= 5 × 5 × 5 × 4 × g

Dyne = 4 × 5 × 5 × 5 gf

Question 3. A cubical block is floating in a liquid with half of its volume immersed in the liquid. When the whole system accelerates upwards with an acceleration of g/3, the fraction of volume immersed in the liquid will be

1. \(\frac{1}{2}\)
2. \(\frac{3}{8}\)
3. \(\frac{2}{3}\)
4. \(\frac{3}{4}\)

Answer: Fraction of volume immersed in the liquid \(V_{\text {in }}=\left(\frac{\rho}{\sigma}\right) V\)

i.e. it depends upon the densities of the block and liquid. So there will be no change in it if the system moves upward or downward with constant velocity or some acceleration.

Question 4. A silver ingot weighing 2.1 kg is held by a string so as to be completely immersed in a liquid of relative density 0.8. The relative density of silver is 10.5. The tension in the string in kg-wt is

1. 1.6
2. 1.94
3. 3.1
4. 5.25

Answer: T = Apparent weight = V(ρ – σ) g = \(\frac{M}{\rho}(\rho-\sigma) g\)

T = \(M\left(1-\frac{\sigma}{\rho}\right) g=2.1\left(1-\frac{0.8}{10.5}\right) g=1.94 \mathrm{gN}\)

T = 1.94 Kg-wt

Question 5. A sample of metal weighs 210 gm in air, 180 gm in water and 120 gm in liquid. The density (RD) of

1. Metal is 3
2. Metal is 7
3. Liquid is 3
4. Liquid is \(\frac{1}{3}\)

Answer: Density of metal = ρ. Density of liquid = σ

If V is the volume of the sample then according to the problem

210 = Vρg ……(1)

180 = V (ρ – 1)g ……(2)

120 = V (ρ – σ)g ……(3)

By solving (1),(2) and (3) we get ρ = 7 and σ = 3.

Question 6. A cubical block of wood of edge 3 cm floats in water. The lower surface of the cube just touches the free end of a vertical spring fixed at the bottom of the pot. Find the maximum weight that can be put on the block without wetting it. The density of wood = 800 kg/m³ and the spring constant of the spring = 50 N/m. Take g = 10 m/s².

Answer:

The specific gravity of the block = 0.8. Hence the height inside water = 3 cm × 0.8 = 2.4 cm. The height outside after = 3 cm – 2.4 = 0.6 cm. Suppose the maximum weight that can be put without wetting it is W. The block in this case is completely immersed in the water. The volume of the displaced water

= volume of the block = 27 × 10^-6 m³.

Hence, the force of buoyancy

= (27 × 10^-6 m³)× 1(1000 kg/m³)× (10 m/s²)= 0.27 N.

The spring is compressed by 0.6 cm and hence the upward force exerted by the spring = 50 N/m × 0.6 cm = 0.3 N.

The force of buoyancy and the spring force taken together balance the weight of the block plus the weight W put on the block. The weight of the block is

W′ = (27 × 10^-6 m) × (800 kg/m× (10 m/s= 0.22 N.

Thus, W = 0.27 N + 0.3 N – 0.22 N

= 0.35 N.

Pressure In Case Of Accelerating Fluid

Liquid Placed In Elevator:

When the elevator accelerates upward with acceleration a0 then the pressure in the fluid, at depth ‘h’ may be given by,

p = hρ [g + a₀]

and force of buoyancy, B = m (g + a₀)

The free surface of a liquid in horizontal acceleration :

tan θ = \(\frac{a_0}{\mathrm{~g}}\)

p₁– p₂= lρ a₀ where p₁ and p₂ are pressures at point 1 and 2. Then h₁– h₂ 

= \(\frac{\ell \mathrm{a}_0}{\mathrm{~g}}\)

Question 1. An open rectangular tank 1.5 m wide 2m deep and 2m long is half filled with water. It is accelerated horizontally at 3.27 m/sec² in the direction of its length. Determine the depth of water at each end of the tank. [g = 9.81 m/sec²]

Answer: tan θ = \(\frac{\mathrm{a}}{\mathrm{g}}=\frac{1}{3}\)

Depth at the corner ‘A’

= 1 – 1.5 tanθ

= 0.5 m

Depth at corner ‘B’

= 1 + 1.5 tan θ = 1.5 m

The free surface of a liquid in the case of a rotating cylinder.

h = \(\frac{v^2}{2 g}=\frac{\omega^2 r^2}{2 g}\)

Streamline Flow

The path taken by a particle in flowing fluid is called its line of flow. In the case of steady flow, all the particles passing through a given point follow the same path and hence we have a unique line of flow passing through a given point which is also called streamline.

Characteristics Of Streamline

1. A tangent at any point on the streamline gives the direction of the velocity of the fluid particle at that point.

2. Two streamlines never intersect each other.

1. Laminar Flow: If the liquid flows over a horizontal surface in the form of layers of different velocities, then the flow of liquid is called Laminar flow. The particles of one layer do not go to another layer. In general, Laminar flow is a streamlined flow.
2. Turbulent Flow: The flow of fluid in which the velocity of all particles crossing a given point is not the same and the motion of the fluid becomes disorderly or irregular is called turbulent flow.

Reynold’S Number

According to Reynold, the critical velocity (vc) of a liquid flowing through a long narrow tube is

1. Directly proportional to the coefficient of viscosity (η) of the liquid.
2. Inversely proportional to the density ρ of the liquid and
3. Inversely proportional to the diameter (D) of the tube.

That is \(v_c \propto \frac{\eta}{\rho D} \quad \text { or } \quad v_c=\frac{R \eta}{\rho D} \quad \text { or }=\frac{v_c \rho D}{\eta}\) ……………(1)

where R is the Reynold number.

If R < 2000, the flow of liquid is streamlined or laminar. If R > 3000, the flow is turbulent. If R lies between 2000 and 3000, the flow is unstable and may change from streamlined flow to turbulent flow.

Equation Of Continuity

The equation of continuity expresses the law of conservation of mass in fluid dynamics.

a₁v₁ = a₂v₂

In general av = constant. This is called an equation of continuity and states that as the area of the cross-section of the tube of flow becomes larger, the liquid’s (fluid) speed becomes smaller and vice-versa.

Illustrations –

1. The velocity of the liquid is greater in the narrow tube as compared to the velocity of the liquid in a broader tube.
2. Deep waters run slowly can be explained by the equation of continuity i.e., av = constant. Where water is deep the area of cross-section increases hence velocity decreases.

Energy Of A Liquid

A liquid can possess three types of energies :

Kinetic Energy: The energy possessed by a liquid due to its motion is called kinetic energy. The kinetic energy of a liquid of mass m moving with speed v is \(\frac{1}{2} m v^2\)

∴ K.E. per unit mass = \(\frac{\frac{1}{2} m v^2}{m}=\frac{1}{2} v^2\)

Potential Energy: The potential energy of a liquid of mass m at a height h is m g h.

∴ P.E. per unit mass = \(\frac{\mathrm{mgh}}{\mathrm{m}}=\mathrm{gh}\)

Pressure Energy: The energy possessed by a liquid by virtue of its pressure is called pressure energy. Consider a vessel fitted with a piston at one side (figure).

• Let this vessel is filled with a liquid. Let ‘A’ be the area of the cross-section of the piston and P be the pressure experienced by the liquid. The force acting on the piston = PA
• If dx is the distance moved by the piston, then work done by the force = PA dx = PdV where dV = Adx, the volume of the liquid swept.

This work is equal to the pressure energy of the liquid.

∴ Pressure energy of liquid in volume dV = PdV.

The mass of the liquid having volume dV = ρdV,

ρ is the density of the liquid.

∴ Pressure energy per unit mass of the liquid = \(\frac{P d V}{\rho d V}=\frac{P}{\rho}\)

Bernoulli’s Theorem

It states that the sum of pressure energy, kinetic energy, and potential energy per unit mass or per unit volume or per unit weight is always constant for an ideal (i.e. incompressible and non-viscous) fluid having stream-line flow.

i.e. \(\frac{P}{\rho}+\frac{1}{2} v^2+g h\)= constant.

Question 1. An incompressible liquid flows through a horizontal tube as shown in the following fig. Then the velocity υ of the fluid is

1. v = 2v₁ + v₂
2. v = v₁+ v₂
3. \(v=\frac{v_1 v_2}{v_1+v_2}\)
4. \(\mathrm{v}=\sqrt{\mathrm{v}_1^2+\mathrm{v}_2^2}\)

Answer: 1. v = 2v₁+ v₂

m = m₁+ m₂

ρV = ρV₁+ ρV₂

ρAv = ρ2Av₁+ ρAv₂

v = 2v₁+ v₂

Question 2. Water enters through end A with speed υ₁ and leaves through end B with speed υ₂ of a cylindrical tube AB. The tube is always completely filled with water. In case I it is horizontal and in case II it is vertical with end A upwards and in case III it is vertical with end B upwards. We have υ₁= υ₂

1. Case 1
2. Case 2
3. Case 3
4. Each case

Answer: This happens in accordance with an equation of continuity and this equation was derived on the principle of conservation of mass and it is true in every case, either tube remains horizontal or vertical.

Question 3. Water flows in a horizontal tube as shown in the figure. The pressure of water changes by 600 N/m² between A and B where the areas of cross-section are 30cm² and 15cm² respectively. Find the rate of flow of water through the tube.

Answer:

Let the velocity at A = v_A and that at B = v_B.

By the equation of continuity, \(\frac{v_B}{v_A}=\frac{30 \mathrm{~cm}^2}{15 \mathrm{~cm}^2}=2\)

By Bernoulli’s equation,

⇒ \(P_A+\frac{1}{2} \rho v_A^2=P_B+\frac{1}{2} \rho v_B^2\)

or \(P_A-P_B=\frac{1}{2} \rho\left(2 v_A\right)^2-\frac{1}{2} \rho v_A^2=\frac{3}{2} \rho v_A^2\)

or \(600 \frac{\mathrm{N}}{\mathrm{m}^2}=\frac{3}{2}\left(1000 \frac{\mathrm{kg}}{\mathrm{m}^3}\right) \mathrm{v}_{\mathrm{A}}^2\)

or \(v_{\mathrm{A}}=\sqrt{0.4 \mathrm{~m}^2 / \mathrm{s}^2}\)

=0.63 m/s

The rate of flow = (30 cm²)(0.63 m/s) = 1800 cm³/s.

Application Of Bernoulli’s Theorem

1. Bunsen burner
2. Lift of an airfoil.
3. Spinning of a ball (Magnus effect)
4. The sprayer.
5. A ping-pong ball in an air jet
6. Torricelli’s theorem (speed of efflux)

At point A, P₁= P, v₁= 0 and h₁= h

At point B, P₂= P, v₂= v (speed of efflux) and h = 0

Using Bernoulli’s theorem \(\frac{P_1}{\rho}+g h_1+\frac{1}{2} v_1^2=\frac{P_2}{\rho}+g h_2=\frac{1}{2} v_2^2\)

we have \(\frac{P}{\rho}+g h+0=\frac{P}{\rho}+0+\frac{1}{2} v^2\)

⇒ \(\frac{1}{2} v^2=g h \text { or } v=\sqrt{2 g h}\)

Venturi meter: 

It is a gauge put on a flow pipe to measure the flow speed of a liquid. Let the liquid of density ρ be flowing through a pipe of area of cross-section A₁.

Let A₂ be the area of the cross section at the throat and a manometer is attached as shown in the figure. Let v₁ and P₁ be the velocity of the flow and pressure at point A, and v₂ and P₂ be the corresponding quantities at point B.

Using Bernoulli’s Theorem:

⇒ \(\frac{P_1}{\rho}+g h_1+\frac{1}{2} v_1^2=\frac{P_2}{\rho}+g h_2+\frac{1}{2} v_2^2\), we get

⇒ \(\frac{P_1}{\rho}+g h+\frac{1}{2} v_1^2=\frac{P_2}{\rho}+g h+\frac{1}{2} v_2^2\) (Since h₁=h₂=h)

or \(\left(P_1-P_2\right)=\frac{1}{2} \rho\left(v_2^2-v_1^2\right)\) ….

According to the continuity equation, A₁v₁= A₂v₂

or \(v_2=\left(\frac{A_1}{A_2}\right) v_1\)

Substituting the value of v₂ in the equation we have

\(\left(P_1-P_2\right)=\frac{1}{2} \rho\left[\left(\frac{A_1}{A_2}\right)^2 v_1^2-v_1^2\right] \frac{1}{2} \rho v_1{ }^2\left[\left(\frac{A_1}{A_2}\right)^2-1\right]\)

Since A₁> A₂, therefore, P₁> P₂

or \(v_1^2=\frac{2\left(P_1-P_2\right)}{\rho\left[\left(\frac{A_1}{A_2}\right)^2-1\right]}\)

⇒ \(\frac{2 A_2^2\left(P_1-P_2\right)}{\rho\left(A_1^2-A_2^2\right)}\)

where (P₁– P₂)= ρ_mgh and h is the difference in heights of the liquid levels in the two tubes.

⇒ \(v_1=\sqrt{\frac{2 \rho_{\mathrm{m}} g h}{\rho\left[\left(\frac{A_1}{A_2}\right)^2-1\right]}}\)

The flow rate (R) i.e., the volume of the liquid flowing per second is given by R = v₁A₁.

During A wind storm: The velocity of air just above the roof is large so according to Bernoulli’s theorem, the pressure just above the roof is less than the pressure below the roof. Due to this pressure difference an upward force acts on the roof which is blown off without damaging other parts of the house.

When a fast-moving train crosses a person standing near a railway track, the person has a tendency to fall towards the train.

This is because a fast-moving train produces a large velocity in the air between a person and the train and hence pressure decreases according to Bernoulli’s theorem. Thus the excess pressure on the other side pushes the person towards the train.

Question 1. Water flows through a horizontal tube of variable cross-section (figure). The area of a cross-section at A and B are 4 mm² and 2 mm² respectively. If 1 cc of water enters per second through A, find

1. The speed of the water at A,
2. The speed of the water at B and
3. The pressure difference P_A– P_B.

Answer: A₁v₁= A₂v₂

⇒ \(\rho_1+\frac{1}{2} \rho v_1^2+0\)

⇒ \(r_2+\frac{1}{2} \rho v_2^2+\rho g h\)

1. 25 cm/s,
2. 50 cm/s
3. 94 N/m²

Question 2. The velocity of the liquid coming out of a small hole of a large vessel containing two different liquids of densities 2ρ and ρ as shown in the figure is

1. \(\sqrt{6 g h}\)
2. \(2 \sqrt{g h}\)
3. \(2 \sqrt{2 \mathrm{gh}}\)
4. \(\sqrt{g h}\)

Answer: Pressure at : P₁= Patm + ρ g (2h)

Applying Bernoulli’s theory between points and

⇒ \(\left[P_{a t m}+2 \rho g h\right]+\rho g(2 h)+\frac{1}{2}(2 \rho)(0)^2\)

⇒ \(P_{\mathrm{atm}}+(2 \rho) g(0)+\frac{1}{2}(2 \rho) v^2\)

⇒ \(v=2 \sqrt{g h}\)

Question 3. A horizontal pipeline carries water in a streamlined flow. At a point along the pipe where the cross-sectional area is 10 cm², the water velocity is 1 ms^-1 and the pressure is 2000 Pa. The pressure of water at another point where the cross-sectional area is 5 cm² will be:

[Density of water = 103 kg. m^-3 ]

Answer:

From continuity equation

A₁v₁= A₂v₂

∴ \(v_2=\left(\frac{A_1}{A_2}\right) v_1=\left(\frac{10}{5}\right)(1)=2 \mathrm{~m} / \mathrm{s}\)

Applying Bernoulli’s theorem at 1 and 2

⇒ \(P_2+\frac{1}{2} \rho v_2{ }^2=P_1+\frac{1}{2} \rho v_1{ }^2\)

⇒ \(P_2=P_1+\frac{1}{2} \rho\left(v_1^2-v_2^2\right)\)

⇒ \(\left(2000+\frac{1}{2} \times 10^3(1-4)\right)\)

500 Pa

Question 4. Equal volumes of two immiscible liquids of densities ρ and 2ρ are filled in a vessel as shown in the figure. Two small holes are punched at depths h/2 and 3h/2 from the surface of a lighter liquid. If v₁ and v₂ are the velocities of efflux at these two holes, then v₁/v₂ is:

1. \(\frac{1}{2 \sqrt{2}}\)
2. 0.5
3. 0.25
4. \(\frac{1}{\sqrt{2}}\)

Answer: for hole (1)

⇒ \(P_0+\rho \frac{V_1^2}{2}=P_0+\rho g \frac{h}{2}\)

⇒ \(V_1=\sqrt{g h}\)

for hole (2)

⇒ \(P_0+\rho \frac{V_2^2}{2}\)

⇒ \(P_0+\rho g h+2 \rho h\left(\frac{h}{2}\right)\)

⇒ \(\frac{\rho V_2^2}{2}=2 \rho g h\)

⇒ \(V_2=2 \sqrt{g h}\)

⇒ \(\frac{V_1}{V_2}=\frac{1}{2}\)

= 0.5

 

Centre Of Mass

Every physical system has associated with it a certain point whose motion characterizes the motion of the whole system.

• When the system moves under some external forces, then this point moves as if the entire mass of the system is concentrated at this point and also the external force is applied at this point for translational motion.
• This point is called the center of mass of the system.

Centre Of Mass Of A System Of ‘N’ Discrete Particles

Consider a system of N point masses m₁, m₂, m₃, ……………. mn whose position vectors from origin O are given by \(\vec{r}_1, \vec{r}_2, \vec{r}_3, \ldots \ldots \ldots \ldots. \vec{r}_n\) respectively. Then the position vector of the center of mass C of the system is given by.

⇒ \(\vec{r}_{c m}=\frac{m_1 \vec{r}_1+m_2 \vec{r}_2+\ldots \ldots . .+m_n \vec{r}_n}{m_1+m_2+\ldots \ldots . .+m_n}\)

⇒ \(\vec{r}_{c m}=\frac{\sum_{i=1}^n m_i \vec{r}_i}{\sum_{i=1}^n m_i}\)

⇒ \(\vec{r}_{\mathrm{cm}}=\frac{1}{M} \sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{m}_{\mathrm{i}} \vec{r}_{\mathrm{i}}\)

where, \(\mathrm{m}_{\mathrm{i}} \vec{r}_{\mathrm{i}}\) iis called the moment of mass of the particle w.r.t O.

⇒ \(M=\left(\sum_{i=1}^n m_i\right)\) is the total mass of the system.

Note: If the origin is taken at the center of mass then \(\sum_{i=1}^n m_i \vec{r}_i=0\). Hence, the COM is the point about which the sum of the “mass moments” of the system is zero.

Position Of Com Of Two Particles

The Centre of mass of two particles of masses m₁ and m₂ separated by a distance r lies in between the two particles. The distance of the center of mass from any of the particles (r) is inversely proportional to the mass of the particle (m)

i.e. \(r \propto 1 / m\)

⇒ \(\frac{r_1}{r_2}=\frac{m_2}{m_1}\)

⇒ \(m_1 r_1=m_2 r_2\)

⇒ \(r_1=\left(\frac{m_2}{m_2+m_1}\right) r \text { and } r_2=\left(\frac{m_1}{m_1+m_2}\right) r\)

Here, r₁= distance of COM from m₁

and r₂= distance of COM from m₂

From the above discussion, we see that

r₁= r₂= 1/2 if m1= m₂, i.e., COM lies midway between the two particles of equal masses.

Similarly, r₁> r₂ if  m₁< m₂ and r₁< r₂if m₂<₁m1, i.e., COM is nearer to the particle having a larger mass.

Question 1. Two particles of mass 1 kg and 2 kg are located at x = 0 and x = 3 m. Find the position of their center of mass.
Answer:

Since both the particles lie on the x-axis, the COM will also lie on the x-axis. Let the COM be located at x = x, then

r₁= distance of COM from the particle of mass 1 kg = x

and r₂ = distance of COM from the particle of mass 2 kg = (3 – x)

Using \(\frac{r_1}{r_2}=\frac{m_2}{m_1}\)

or \(\frac{x}{3-x}=\frac{2}{1} \text { or } x=2 m\)

Thus, the COM of the two particles is located at x = 2 m.

Question 2. The position vector of three particles of masses m1= 1 kg, m2= 2 kg and m3= 3 kg are

⇒ \(\vec{r}_1=(\hat{i}+4 \hat{j}+\hat{k}) m, \vec{r}_2=(\hat{i}+\hat{j}+\hat{k}) m\) and \(\vec{r}_3=(2 \hat{i}-\hat{j}-2 \hat{k}) m\) respectively.

Find the position vector of their center of mass.

Answer:

The position vector of COM of the three particles will be given by \(\vec{r}_{\text {COM }}=\frac{m_1 \vec{r}_1+m_2 \vec{r}_2+m_3 \vec{r}_3}{m_1+m_2+m_3}\)

Substituting the values, we get

⇒ \(\overrightarrow{\mathrm{r}}_{\text {COM }}=\frac{(1)(\hat{i}+4 \hat{j}+\hat{k})+(2)(\hat{i}+\hat{j}+\hat{k})+(3)(2 \hat{i}-\hat{j}-2 \hat{k})}{1+2+3}\)

⇒ \(\frac{1}{2}(3 \hat{i}+\hat{j}-\hat{k}) m\)

Question 3. Four particles of mass 1 kg, 2 kg, 3 kg, and 4 kg are placed at the four vertices A, B, C, and D of a square of side 1 m. Find the position of the center of mass of the particles.

Answer:

Assuming D as the origin, DC as the x-axis, and DA as the y-axis, we have

m₁= 1 kg, (x₁, y₁) = (0, 1m)

m₂= 2 kg, (x₂, y₂) = (1m, 1m)

m₃= 3 kg, (x₃, y₃) = (1m, 0)

and m₄= 4 kg, (x₄, y₄) = (0, 0)

Co-ordinates of their COM are

⇒ \(x_{\text {com }}=\frac{m_1 x_1+m_2 x_2+m_3 m_3+m_4 x_4}{m_1+m_2+m_3+m_4}\)

⇒ \(\frac{(1)(0)+2(1)+3(1)+4(0)}{1+2+3+4}\)

⇒ \(\frac{5}{10}\)

⇒ \(\frac{1}{2}\)

= 0.5m

Similarly, \(\mathrm{y}_{\text {com }}=\frac{\mathrm{m}_1 \mathrm{y}_1+\mathrm{m}_2 \mathrm{y}_2+\mathrm{m}_3 \mathrm{y}_3+\mathrm{m}_4 \mathrm{y}_4}{\mathrm{~m}_1+\mathrm{m}_2+\mathrm{m}_3+\mathrm{m}_4}\)

⇒ \(\frac{(1)(1)+2(1)+3(0)+4(0)}{1+2+3+4}\)

⇒ \(\frac{3}{10}\)

0.3

∴ (x_COM, y_COM) = (0.5 m, 0.3 m)

Thus, the position of COM of the four particles is as shown in the figure.

Question 4. Consider a two-particle system with the particles having masses m₁ and m₁. If the first particle is pushed towards the center of mass through a distance d, by what distance should the second particle be moved to keep the center of mass at the same position?
Answer:

Consider figure. Suppose the distance of m1 from the center of mass C is x₁ and that of m₂ from C is x₂. Suppose the mass m₂ is moved through a distance d′ towards C to keep the center of mass at C.

Then, m₁x₁= m₂x₂………(1)

and m₁(x₁– d) = m₂(x₂– d′). ……… (2)

Subtracting from

m₁d = m₂d′

or, d′ = \(\frac{m_1}{m_2} d\)

Centre Of Mass Of A Continuous Mass Distribution

For continuous mass distribution, the center of mass can be located by replacing the summation sign with an integral sign. Proper limits for the integral are chosen according to the situation

⇒ \(x_{c m}=\frac{\int x d m}{\int d m}, y_{c m}=\frac{\int y d m}{\int d m}, z_{c m}=\frac{\int z d m}{\int d m}\)

∫dm= M (mass of the body)

⇒ \(\overrightarrow{\mathrm{r}}_{\mathrm{cm}}=\frac{1}{\mathrm{M}} \int \overrightarrow{\mathrm{r}} \mathrm{dm}\)

Note: If an object has symmetric mass distribution about the axis then the y coordinate of COM is zero and vice versa

Centre Of Mass Of A Uniform Rod

Suppose a rod of mass M and length L is lying along the x-axis with its one end at x = 0 and the other at x = L. Mass per unit length of the rod = \(\frac{M}{L}\)

Hence, dm, (the mass of the element dx situated at x = x is) = \(\frac{M}{L}\) dx

The coordinates of the element dx are (x, 0, 0). Therefore, the x-coordinate of COM of the rod will be

⇒ \(x_{\text {com }}=\frac{\int_0^L x d m}{\int d m}=\frac{\int_0^L(x)\left(\frac{M}{L} d x\right)}{M}\)

⇒ \(\frac{1}{L} \int_0^L x d x=\frac{L}{2}\)

The y-coordinate of COM is

⇒ \(y_{\text {com }}=\frac{\int y d m}{\int d m}=0\)

Similarly, Z_COM = 0

i.e., the coordinates of COM of the rod are \(\left(\frac{\mathrm{L}}{2}, 0,0\right)\) i.e. it lies at the center of the rod.

Question 1. A rod of length L is placed along the x-axis between x = 0 and x = L. The linear density (mass/length) λ of the rod varies with the distance x from the origin as λ = Rx. Here, R is a positive constant. Find the position of the center of mass of this rod.
Answer:

The mass of element dx situated at x = x is

dm = λ dx = Rx dx

The COM of the element has coordinates (x, 0, 0).

Therefore, the x-coordinate of COM of the rod will be

⇒ \(x_{\text {com }}=\frac{\int_0^L x d m}{\int d m}\)

⇒ \(\frac{\int_0^L(x)(R x) d x}{\int_0^L(R x) d x}\)

⇒ \(\frac{R \int_0^L x^2 d x}{R \int_0^L x d x}\)

⇒ \(\frac{\left[\frac{x^3}{3}\right]_0^L}{\left[\frac{x^2}{2}\right]_0^L}\)

⇒ \(\frac{2 L}{3}\)

The y-coordinate of COM of the rod is \(\mathrm{y}_{\mathrm{com}}=\frac{\int \mathrm{ydm}}{\int \mathrm{dm}}\) = 0 (as y = 0)

Similarly, Z_COM = 0

Hence, the center of mass of the rod lies at \(\left[\frac{2 L}{3}, 0,0\right]\)

1. The center of mass of a uniform rectangular, square, or circular plate lies at its center. Axis of symmetry plane of symmetry.

2. For a laminar type (2-dimensional) body with uniform negligible thickness the formulae for finding the position of the center of mass are as follows:

⇒ \(\vec{r}_{\text {COM }}=\frac{m_1 \vec{r}_1+m_2 \vec{r}_2+\ldots}{m_1+m_2+\ldots}=\frac{\rho A_1 t \vec{r}_1+\rho A_2 t_r+\ldots}{\rho A_1 t+\rho A_2 t+\ldots}\)

(∵ m = ρAt)

⇒ \(\vec{r}_{\text {COM }}=\frac{A_1 \vec{r}_1+A_2 \vec{r}_2+\ldots}{A_1+A_2+\ldots}\)

Here, A stands for the area,

3. If some mass of area is removed from a rigid body, then the position of the center of mass of the remaining portion is obtained from the following formulae:

⇒ \(\vec{r}_{\text {COM }}=\frac{m_1 \vec{r}_1-m_2 \vec{r}_2}{m_1-m_2}\)

or \(\overrightarrow{\mathrm{r}}_{\text {COM }}=\frac{\mathrm{A}_1 \overrightarrow{\mathrm{r}}_1-\mathrm{A}_2 \overrightarrow{\mathrm{r}}_2}{\mathrm{~A}_1-\mathrm{A}_2}\)

⇒ \(x_{\text {COM }}=\frac{m_1 x_1-m_2 x_2}{m_1-m_2}\)

or \(\mathrm{x}_{\text {COM }}=\frac{\mathrm{A}_1 \mathrm{x}_1-\mathrm{A}_2 \mathrm{x}_2}{\mathrm{~A}_1-\mathrm{A}_2}\)

⇒ \(\mathrm{y}_{\text {COM }}=\frac{\mathrm{m}_1 \mathrm{y}_1-\mathrm{m}_2 \mathrm{y}_2}{\mathrm{~m}_1-\mathrm{m}_2}\)

or \(\mathrm{y}_{\text {COM }}=\frac{\mathrm{A}_1 \mathrm{y}_1-\mathrm{A}_2 \mathrm{y}_2}{\mathrm{~A}_1-\mathrm{A}_2}\)

and \(z_{\text {COM }}=\frac{m_1 z_1-m_2 z_2}{m_1-m_2}\)

or \(z_{\text {COM }}=\frac{A_1 z_1-A_2 z_2}{A_1-A_2}\)

Here, m₁, A₁, x₁, y₁ and z₁ are the values for the whole mass while m₂, A₂, \(\vec{r}_2, \vec{x}_2\) y₂ and z₂are the values for the mass which has been removed. Let us see two Questions in support of the above theory.

Question 2. Find the position of the center of mass of the uniform lamina shown in the figure.

Answer:

Here,

A₁ = area of complete circle = πa²

A₂= area of small circle = \(\pi\left(\frac{a}{2}\right)^2=\frac{\pi a^2}{4}\)

(x₁, y₁) = coordinates of centre of mass of large circle = (0, 0)

and (x₂, y₂) = coordinates of centre of mass of small circle = \(\left(\frac{\mathrm{a}}{2}, 0\right)\)

Using \(x_{\text {COM }}=\frac{\mathrm{A}_1 \mathrm{x}_1-\mathrm{A}_2 \mathrm{x}_2}{\mathrm{~A}_1-\mathrm{A}_2}\)

we get \(\mathrm{x}_{\text {com }}=\frac{-\frac{\pi \mathrm{a}^2}{4}\left(\frac{\mathrm{a}}{2}\right)}{\pi \mathrm{a}^2-\frac{\pi \mathrm{a}^2}{4}}=\frac{-\left(\frac{1}{8}\right)}{\left(\frac{3}{4}\right)} \mathrm{a}=-\frac{\mathrm{a}}{6}\)

and y_COM = 0 as y₁ and y₂ both are zero.

Therefore, the coordinates of COM of the lamina shown in the figure are \(\left(-\frac{a}{6}, 0\right)\)

Centre Of Mass Of Some Common Systems

A system of two point masses m₁ r₁= m₂ r₂ The center of mass lies closer to the heavier mass.

Rectangular plate (By symmetry)

⇒ \(\mathrm{x}_{\mathrm{c}}=\frac{\mathrm{b}}{2}\)

⇒ \(y_c=\frac{L}{2}\)

A triangular plate (By qualitative argument)

at the centroid : \(y_c=\frac{h}{3}\)

A semi-circular ring \(y_c=\frac{2 R}{\pi}\)

xc = 0

A semi-circular disc

⇒ \(y_c=\frac{4 R}{3 \pi}\)

xc = 0

A hemispherical shell

⇒ \(y_c=\frac{R}{2}\)

xc = 0

A solid hemisphere

⇒ \(y_c=\frac{3 R}{8}\)

xc= 0

A circular cone (solid)

⇒ \(y_c=\frac{h}{4}\)

A circular cone (hollow)

⇒ \(y_c=\frac{h}{3}\)

Question 1. A uniform thin rod is bent in the form of closed loop ABCDEFA as shown in the figure. The y-coordinate of the center of mass of the system is

1. \(\frac{2 r}{\pi}\)
2. \(-\frac{6 r}{3 \pi+2}\)
3. \(-\frac{2 r}{\pi}\)
4. Zero

Answer:

The center of mass of a semicircular ring is at a distance from its center.

(Let λ = mass/length)

∴ \(Y_{c m}=\frac{\lambda \pi r \times \frac{2 r}{\pi}-\lambda \times 2 \pi r \times \frac{4 r}{\pi}}{\lambda \pi r+\lambda r+\lambda r+\lambda \times 2 \pi r}=-\frac{6 r}{3 \pi+2}\)

Motion Of Centre Of Mass And Conservation Of Momentum:

Velocity Of Centre Of Mass Of System

⇒ \(\vec{v}_{c m}=\frac{m_1 \frac{d \vec{r}_1}{d t}+m_2 \frac{d\vec{r}_2}{d t}+m_3 \frac{d \vec{r}_3}{d t} \ldots \ldots \ldots \ldots . .+m_n \frac{d \vec{r}_n}{d t}}{M}\)

⇒ \(\frac{m_1 \vec{v}_1+m_2 \vec{v}_2+m_3 \vec{v}_3 \ldots \ldots \ldots .+m_n \vec{v}_n}{M}\)

Here, the numerator of the right-hand side term is the total momentum of the system i.e., the summation of momentum of the individual component (particle) of the system

Hence, the velocity of the center of mass of the system is the ratio of the momentum of the system to the mass of the system.

∴ \(\overrightarrow{\mathrm{P}}_{\text {System }}=\mathrm{M} \overrightarrow{\mathrm{v}}_{\mathrm{cm}}\)

Acceleration Of Centre Of Mass Of System

⇒ \(\vec{a}_{c m}=\frac{m_1 \frac{d \overrightarrow{v_1}}{d t}+m_2 \frac{d \overrightarrow{v_2}}{d t}+m_3 \frac{d \overrightarrow{v_3}}{d t} \ldots \ldots \ldots \ldots . .+m_n \frac{d \overrightarrow{v_n}}{d t}}{M}\)

⇒ \(\frac{m_1 \vec{a}_1+m_2 \vec{a}_2+m_3 \vec{a}_3 \ldots \ldots \ldots . .+m_n \vec{a}_n}{M}\)

⇒ \(\frac{\text { Net forceonsystem }}{\mathrm{M}}\)

⇒ \(\frac{\text { Net External Force }+ \text { Net internal Force }}{\mathrm{M}}\)

⇒ \(\frac{\text { Net External Force }}{\mathrm{M}}\)

(∵ action and reaction both of an internal force must be within the system. Vector summation will cancel all internal forces and hence net internal force on the system is zero)

∴ \(\overrightarrow{\mathrm{F}}_{\mathrm{ext}}=\mathrm{M} \overrightarrow{\mathrm{a}}_{\mathrm{cm}}\)

Where \(\overrightarrow{\mathrm{F}}_{\text {ext }}\) is the sum of the ‘external’ forces acting on the system. The internal forces which the

particles exert on one another play absolutely no role in the motion of the center of mass.

If no external force is acting on a system of particles, the acceleration of center of mass of the system will be zero. If a_c= 0, it implies that v_c must be a constant and if v_cm is a constant, it implies that the total momentum of the system must remain constant. It leads to the principle of conservation of momentum in the absence of external forces.

If \(\overrightarrow{\mathrm{F}}_{\text {ext }}\) = 0 ext = then \(\overrightarrow{\mathrm{v}}_{\mathrm{cm}}\) = constant

“If a resultant external force is zero on the system, then the net momentum of the system must remain constant”.

Motion Of COM In A Moving System Of Particles:

COM at rest :

If F_ext = 0 and V_cm= 0, then COM remains at rest. Individual components of the system may move and have non-zero momentum due to mutual forces (internal), but the net momentum of the system remains zero.

All the particles of the system are at rest.

Particles are moving such that their net momentum is zero.

A bomb at rest suddenly explodes into various smaller fragments, all moving in different directions then, since the explosive forces are internal and there is no external force on the system for explosion therefore, the COM of the bomb will remain at the original position and the fragment fly such that their net momentum remains zero.

• Two men standing on a frictionless platform, push each other, then also their net momentum remains zero because the push forces are internal for the two-men system.
• A boat floating in a lake also has a net momentum of zero if the people on it change their position, because the friction force required to move the people is internal to the boat system.
• Objects initially at rest, if moving under mutual forces (electrostatic or gravitation)also have net momentum zero.
• A light spring of spring constant k is kept compressed between two blocks of masses m₁ and m₂ on a smooth horizontal surface. When released, the blocks acquire velocities in opposite directions, such that the net momentum is zero.

(In a fan, all particles are moving but COM is at rest

COM Moving With Uniform Velocity:

If F_ext = 0, then Vcm remains constant therefore, the net momentum of the system also remains conserved. Individual components of the system may have variable velocity and momentum due to mutual forces (internal), but the net momentum of the system remains constant and COM continues to move with the initial velocity.

1. All the particles of the system are moving with the same velocity.
2. For example: A car moving with uniform speed on a straight road, has its COM moving with a constant velocity.
3. Internal explosions/breaking does not change the motion of COM and net momentum remains conserved.
4. A bomb moving in a straight line suddenly explodes into various smaller fragments, all moving in different directions then, since the explosive forces are internal and there is no external force on the system for explosion therefore, the COM of the bomb will continue the original motion and the fragment fly such that their net momentum remains conserved.
5. Man jumping from a cart or buggy also exerts internal forces therefore net momentum of the system and hence, the Motion of COM remains conserved.
6. Two moving blocks connected by a light spring on a smooth horizontal surface. If the acting forces are only due to spring then COM will remain in its motion and momentum will remain conserved.
7. Particles colliding in the absence of external impulsive forces also have their momentum conserved.

COM Moving With Acceleration:

If an external force is present then COM continues its original motion as if the external force is acting on it, irrespective of internal forces.

COM Moving With Acceleration Example:

Projectile motion: An axe thrown in the air at an angle θ with the horizontal will perform a complicated motion of rotation as well as a parabolic motion under the effect of gravitation

⇒ \(\mathrm{H}_{\mathrm{com}}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}\)

⇒ \(R_{c o m}=\frac{u^2 \sin 2 \theta}{g}\)

⇒ \(T=\frac{2 u \sin \theta}{g}\)

Circular Motion: A rod hinged at an end, rotates, then its COM performs circular motion. The centripetal force (Fc) required in the circular motion is assumed to be acting on the COM.

F_c = ω²RCOM

Question 1. A projectile is fired at a speed of 100 m/s at an angle of 37º above the horizontal. At the highest point, the projectile breaks into two parts of mass ratio 1 : 3, the lighter piece coming to rest. Find the distance from the launching point to the point where the heavier piece lands.
Answer:

Internal force does not affect the motion of the center of mass, the center of mass hits the ground at the position where the original projectile would have landed. The range of the original projectile is,

⇒ \(\mathrm{x}_{\text {com }}=\frac{2 \mathrm{u}^2 \sin \theta \cos \theta}{\mathrm{g}}\)

⇒ \(\frac{2 \times 10^4 \times \frac{3}{5} \times \frac{4}{5}}{10} \mathrm{~m}\)

= 960 m

The center of mass will hit the ground at this position. As the smaller block comes to rest after breaking, it falls down vertically and hits the ground at half of the range, i.e., at x = 480 m. If the heavier block hits the ground at x2, then

⇒ \(x_{\text {COM }}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}\)

960 = \(\frac{(m)(480)+(3 m)\left(x_2\right)}{(m+3 m)}\)

x₂= 1120

Momentum Conservation:

The total linear momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its center of mass.

⇒ \(\overrightarrow{\mathrm{P}}=\mathrm{M} \overrightarrow{\mathrm{v}}_{\mathrm{cm}}\)

⇒ \(\vec{F}_{\text {ext }}=\frac{\overrightarrow{d P}}{\mathrm{dt}}\)

If \(\vec{F}_{\text {ext }}=0 \frac{\mathrm{dP}}{\mathrm{dt}} \Rightarrow=0\)

⇒ \(\vec{p}\) = constant

When the vector sum of the external forces acting on a system is zero, the total linear momentum of the system remains constant.

⇒ \(\vec{P}_1+\vec{P}_2+\vec{P}_3+\ldots \ldots \ldots \ldots \ldots+=\vec{P}_n\) constant

Question 2. A shell is fired from a cannon with a speed of 100 m/s at an angle 60º with the horizontal (positive x-direction). At the highest point of its trajectory, the shell explodes into two equal fragments. One of the fragments moves along the negative x-direction with a speed of 50 m/s. What is the speed of the other fragment at the time of the explosion?
Answer:

As we know in the absence of external force the motion of the center of mass of a body remains unaffected. Thus, here the center of mass of the two fragments will continue to follow the original projectile path. The velocity of the shell at the highest point of trajectory is

v_M= ucosθ = 100 ×cos60º = 50 m/s.

Let v₁ be the speed of the fragment that moves along the negative x-direction and the other fragment has speed v₂, which must be along the positive x-direction. Now from momentum conservation, we have

⇒ \(m v=\frac{-m}{2} v_1+\frac{m}{2} v_2\)

or 2v = v₂-v₁

or v₂ = 2v+v₁

= (2×50) + 50 = 150m/s

Question 3. A man of mass m is standing on a platform of mass M kept on smooth ice. If the man starts moving on the platform with a speed v relative to the platform, with what velocity relative to the ice does the platform recoil?
Answer:

Consider the situation shown in the figure. Suppose the man moves at a speed w towards the right and the platform recoils at a speed of V towards the left, both relative to the ice. Hence, the speed of the man relative to the platform is V + w. By the question,

V + w = v, or w = v – V ………….(1)

Taking the platform and the man to be the system, there is no external horizontal force on the system. The linear momentum of the system remains constant. Initially, both the man and the platform were at rest. Thus,

0 = MV – mw or, MV = m (v – V) [Using (1)]

or, V = \(\frac{\mathrm{mv}}{\mathrm{M}+\mathrm{m}}\)

Question 4. In a free space, a rifle of mass M shoots a bullet of mass m at a stationary block of mass M distance D away from it. When the bullet has moved through a distance d towards the block the center of mass of the bullet-block system is at a distance of :

1. \(\frac{(D-d) m}{M+m}\) from the block
2. \(\frac{m d+M D}{M+m}\) from the rifle
3. \(\frac{2 \mathrm{dm}+\mathrm{DM}}{\mathrm{M}+\mathrm{m}}\) from the rifle
4. \((D-d) \frac{M}{M+m}\) from the bullet

Answer: (1,2,4)

As; Mx = m(D – d – x) x = \(\frac{m(D-d)}{M+m}\) from the block

and x’ = D – d – x = \(\frac{(D-d) M}{M+m}\) from the bullet.

Question 5. The center of mass of two masses m & m′ moves by distance \(\frac{x}{5}\) when mass m is moved by distance x and m′ is kept fixed. The ratio \(\frac{m’}{m}\) is

1. 2
2. 4
3. 1/4
4. None of these

Answer: 2. 4

(m + m′)\(\frac{x}{5}\) = mx + m′O

∴ m + m′ = 5 m ; m’ = 4m

⇒ \(\frac{m’}{m}\) = 4

Question 6. A person P of mass 50 kg stands in the middle of a boat of mass 100 kg moving at a constant velocity of 10 m/s with no friction between water and boat and also the engine of the boat is shut off. With what velocity (relative to the boat’s surface) should the person move so that the boat comes to rest? Neglect friction between water and boat.

1. 30 m/s towards right
2. 20 m/s towards right
3. 30 m/s towards left
4. 20 m/s towards left

Answer: 1. 30 m/s towards right

The momentum of the system remains conserved as no external force is acting on the system in a horizontal direction.

∴ (50 + 100) 10 = 50 × V + 100 × 0 ⇒ V = 30 m/s towards right, as boat is at rest.

V = 30 m/s Pboat

Question 7. Two men of masses 80 kg and 60 kg are standing on a wood plank of mass 100 kg, that has been placed over a smooth surface. If both the men start moving toward each other with speeds 1 m/s and 2 m/s respectively then find the velocity of the plank by which it starts moving.

Answer:

Applying momentum conservation ;

(80) 1 + 60 (– 2) = (80 + 60 + 100) v

v = \(\frac{-40}{240}\)

⇒ \(-\frac{1}{6}\) m/sec.

Question 8. Each of the blocks shown in the figure has a mass of 1 kg. The rear block moves with a speed of 2 m/s towards the front block kept at rest. The spring attached to the front block is light and has a spring constant of 50 N/m. Find the maximum compression of the spring. Assume, on a frictionless surface

Answer:

Maximum compression will take place when the blocks move with equal velocity. As no net external horizontal force acts on the system of the two blocks, the total linear momentum will remain constant. If V is the common speed at maximum compression, we have,

(1 kg) (2 m/s) = (1 kg)V + (1 kg)V or, V = 1 m/s.

Initial kinetic energy = \(\frac{1}{2}\)(1 kg) (2 m/s)2 = 2 J.

Final kinetic energy = \(\frac{1}{2}\)(1 kg) (1m/s)2 + \(\frac{1}{2}\) (1 kg) (1 m/s)2 = 1 J

The kinetic energy lost is stored as the elastic energy in the spring.

Hence, \(\frac{1}{2}\)(50 N/m) x2 = 2J – 1J = 1 J

or x = 0.2 m.

Question 9. The figure shows two blocks of masses 5 kg and 2 kg placed on a frictionless surface and connected with a spring. An external kick gives a velocity of 14 m/s to the heavier block towards the lighter one. Find the velocity gained by the center of mass

Answer:

The velocity of the center of mass is

⇒ \(v_{c m}=\frac{5 \times 14+2 \times 0}{5+2}\)

= 10m/s

Question 10. The two blocks A and B of the same mass are connected to a spring and placed on a smooth surface. They are given velocities (as shown in the figure) when the spring is at its natural length:

1. The maximum velocity of B will be 10 m/s
2. The maximum velocity of B will be greater than 10 m/s
3. The spring will have maximum extension when A and B both stop
4. The spring will have maximum extension when both move toward the left.

Answer:

• Suppose B moves with a velocity more than 10 m/s a should move at a velocity greater than 5 m/s and increase the overall energy which is not possible since there is no external force acting on the system.
• Hence B should move with a maximum velocity of 10 m/s. Also, both A and B can never stop so as to keep the momentum constant.
• Also, both A and B can never move towards the left simultaneously for momentum to remain conserved. Hence only (A) is correct.

Impulse

Impulse of a force \(\overrightarrow{\mathrm{F}}\) acting on a body for the time interval t = t₁ to t = t₂ is defined as :- dv

⇒ \(\vec{I}=\int_{t_1}^{t_2} \vec{F} d t\)

⇒ \(\overrightarrow{\mathrm{I}}=\int \overrightarrow{\mathrm{F}} \mathrm{dt}=\int \mathrm{m} \frac{\mathrm{d} \overrightarrow{\mathrm{v}}}{\mathrm{dt}} \mathrm{dt}=\int \mathrm{md} \overrightarrow{\mathrm{v}}\)

⇒ \(\overrightarrow{\mathrm{I}}=\mathrm{m}\left(\overrightarrow{\mathrm{v}}_2-\overrightarrow{\mathrm{v}}_1\right)\)

ΔP= change in momentum due to force F

Also, \(\overrightarrow{\mathrm{I}}_{\text {Res }}=\int_{\mathrm{t}_1}^{\mathrm{t}_2} \overrightarrow{\mathrm{F}}_{\text {Res }} d t=\Delta \overrightarrow{\mathrm{P}}\)

(impulse-momentum theorem)

Note: Impulse applied to an object in a given time interval can also be calculated from the area under force time (F-t) graph in the same time interval.

Instantaneous Impulse:

There are many cases when a force acts for such a short time that the effect is instantaneous, for example., a bat striking a ball.

In such cases, although the magnitude of the force and the time for which it acts may each be unknown the value of their product (i.e., impulse) can be known by measuring the initial and final moment. Thus, we can write.

⇒ \(\overrightarrow{\mathrm{I}}=\int \overrightarrow{\mathrm{F}} \mathrm{dt}=\Delta \overrightarrow{\mathrm{P}}=\overrightarrow{\mathrm{P}}_{\mathrm{f}}-\overrightarrow{\mathrm{P}}_{\mathrm{i}}\)

Important Points:

1. It is a vector quantity.
2. Dimensions = [MLT^-1]
3. SΙ unit = kg m/s
4. Direction is along a change in momentum.
5. Magnitude is equal to the area under the F-t. graph.
6. \(\overrightarrow{\mathrm{I}}=\int \overrightarrow{\mathrm{F}}_{\mathrm{dt}}=\overrightarrow{\mathrm{F}}_{\mathrm{av}} \int \mathrm{dt}=\overrightarrow{\mathrm{F}}_{\mathrm{av}} \Delta \mathrm{t}\)
7. It is not a property of a particle, but it is a measure of the degree to which an external force changes the momentum of the particle.

Question 1. The hero of a stunt film fires 50 g bullets from a machine gun, each at a speed of 1.0 km/s. If he fires 20 bullets in 4 seconds, what average force does he exert against the machine gun during this period?
Answer:

The momentum of each bullet

= (0.050 kg) (1000 m/s) = 50 kg-m/s.

The gun has imparted this much amount of momentum with each bullet fired. Thus, the rate of change of momentum of the gun

⇒ \(\frac{(50 \mathrm{~kg}-\mathrm{m} / \mathrm{s}) \times 20}{4 \mathrm{~s}}\)

= 250 N

In order to hold the gun, the hero must exert a force of 250 N against the gun.

Collision Or Impact

A collision is an event in which an impulsive force acts between two or more bodies for a short time, which results in a change in their velocities.

Note:

1. In a collision, particles may or may not come in physical contact.
2. The duration of collision, Δt is negligible as compared to the usual time intervals of observation of motion.

The collision is in fact a redistribution of the total momentum of the particles. Thus, the law of conservation of linear momentum is indispensable in dealing with the phenomenon of collision between particles.

Line Of Impact

The line passing through the common normal to the surfaces in contact during impact is called a line of impact. The force during collision acts along this line on both bodies.

The direction of the Line of impact can be determined by:

1. The geometry of colliding objects like spheres, discs, wedges,s, etc.
2. Direction of change of momentum.

If one particle is stationary before the collision then the line of impact will be along its motion after collision.

Classification Of Collisions:

On The Basis Of The Line Of Impact

1. Head-on collision: If the velocities of the colliding particles are along the same line before and after the collision.
2. Oblique collision: If the velocities of the colliding particles are along different lines before and after the collision.

On The Basis Of Energy:

1. Elastic collision: In an elastic collision, the colliding particles regain their shape and size completely after the collision. i.e., no fraction of mechanical energy remains stored as deformation potential energy in the bodies. Thus, the kinetic energy of the system after collision is equal to the kinetic energy of the system before collision. Thus in addition to the linear momentum, kinetic energy also remains conserved before and after collision.
2. Inelastic collision: In an inelastic collision, the colliding particles do not regain their shape and size completely after collision. Some fraction of mechanical energy is retained by the colliding particles in the form of deformation potential energy. Thus, the kinetic energy of the particles after collision is not equal to that of before collision. However, in the absence of external forces, law of conservation of linear momentum still holds good.
3. Perfectly inelastic: If the velocity of separation along the line of impact just after collision becomes zero then the collision is perfectly inelastic. Collision is said to be perfectly inelastic if both the particles stick together after collision and move with the same velocity,

Note: Actually collisions between all real objects are neither perfectly elastic nor perfectly inelastic, it’s inelastic in nature.

Examples Of Line Of Impact And Collisions Based On Line Of Impact

Two balls A and B are approaching each other such that their centres are moving along line CD.

Head on Collision

Two balls A and B are approaching each other such that their center is moving along dotted lines as shown in the figure.

Oblique Collision

Ball is falling on a stationary wedge.

Oblique Collision

Coefficient Of Restitution (e)

e = \(\frac{\text { Velocity of separation along line of impact }}{\text { Velocity of approach along line of impact }}\)

The most general expression for the coefficient of restitution is

e = \(\frac{\text { velocity of separation of points of contact along line of impact }}{\text { velocity of approach of point of contact along line of impact }}\)

Example For Calculation Of e:

Two smooth balls A and B approach each other such that their centers are moving along line CD in the absence of external impulsive force. The velocities of A and B just before collision are u₁ and u₂ respectively. The velocities of A and B just after collision are v₁ and v₂ respectively.

∵ F_ext = 0 momentum is conserved for the system.

⇒ m₁u₁+ m₂u₂= (m₁+ m₂)v = m₁v₁+ m₂v₂

⇒ v = \(\frac{m_1 u_1+m_2 u_2}{m_1+m_2}=\frac{m_1 v_1+m_2 v_2}{m_1+m_2}\)…..(1)

Impulse of Deformation:

JD = change in momentum of any one body during deformation.

= m₂(v – u₂) for m₂

= m₁(–v + u₁) for m₁

Impulse Of Reformation:

JR = change in momentum of any one body during Reformation.

= m₂(v₂– v) for m₂

= m₁(v – v₁) for m₁

e = \(=\frac{\text { Impulse of Reformation }\left(\overrightarrow{\mathrm{J}}_{\mathrm{R}}\right)}{\text { Impulse of Deformation }\left(\overrightarrow{\mathrm{J}}_{\mathrm{D}}\right)}\)

⇒ \(\frac{v_2-v_1}{u_1-u_2}\)

⇒ \(\frac{\text { Velocity of separation along line of impact }}{\text { Velocity of approach along line of impact }}\)

Note: e is independent of the shape and mass of the object but depends on the material. The coefficient of restitution is constant for a pair of materials.

1. e = 1 Velocity of separation along the LOI = Velocity of approach along the LOI Kinetic energy of particles after collision may be equal to that of before collision. Collision is elastic.
2. e = 0 Velocity of separation along the LOI = 0 Kinetic energy of particles after collision is not equal to that of before collision. Collision Is Perfectly Inelastic.
3. 0 < e < 1 Velocity of separation along the LOI < Velocity of approach along the LOI Kinetic energy of particles after collision is not equal to that of before collision. Collision is Inelastic.

Note: In case of contact collisions e is always less than unity.

∴ 0 ≤ e ≤ 1

Collision In One Dimension (Head on)

⇒ \(u_1>u_2\)

⇒ \(v_2>v_1\)

⇒ \(e=\frac{v_2-v_1}{u_1-u_2}\)

⇒ \(\left(u_1-u_2\right) e=\left(v_2-v_1\right)\)

By momentum conservation,

m₁u₁+ m₂u₂= m₁v₁+ m₂v₂

v₂= v₁+ e(u₁– u₂)

and \(v_1=\frac{m_1 u_1+m_2 u_2-m_2 e\left(u_1-u_2\right)}{m_1+m_2}\)

⇒ \(v_2=\frac{m_1 u_1+m_2 u_2+m_1 e\left(u_1-u_2\right)}{m_1+m_2}\)

Special Case:

e = 0

v₁= v₂

For perfectly inelastic collision, both the bodies, move with the same vel. after collision.

e = 1

and m₁= m₂= m,

we get v₁= u₂ and v₂= u₁

i.e., when two particles of equal mass collide elastically and the collision is head-on, they exchange their velocities., for example.,

m₁>> m₂

⇒ \(\mathrm{m}_1+\mathrm{m}_2 \approx \mathrm{m}_1 \text { and } \frac{\mathrm{m}_2}{\mathrm{~m}_1} \approx 0\)

v₁ = u₁ No change

and v₂= u₁+ e(u₁– u₂)

Now If e = 1

v₂= 2u₁– u₂

Question 1. Two identical balls are approaching each other on a straight line with velocities of 2 m/s and 4 m/s respectively. Find the final velocities, after elastic collision between them.

Answer:

The two velocities will be exchanged and the final motion is the reverse of the initial motion for both.

Question 2. Three balls A, B, and C of the same mass ‘m’ are placed on a frictionless horizontal plane in a straight line as shown. Ball A is moved with velocity u towards the middle ball B. If all the collisions are elastic then, find the final velocities of all the balls.

Answer:

A collides elastically with B and comes to rest but B starts moving with velocity u

After a while B collides elastically with C and comes to rest but C starts moving with velocity u

∴ Final velocities VA = 0; VB= 0 and VC= u

Question 3. Four identical balls A, B, C, and D are placed in a line on a frictionless horizontal surface. A and D are moved with the same speed ‘u’ towards the middle as shown. Assuming elastic collisions, find the final velocities.

Answer:

A and D collide elastically with B and C respectively and come to rest but B and C start moving with velocity u towards each other as shown

B and C collide elastically and exchange their velocities to move in opposite directions

Now, B and C collide elastically with A and D respectively and come to rest but A and D start moving with velocity u away from each other as shown

∴ Final velocities V_A = u (←); V₂= 0; V_C= 0 and V_D = u (→)

Question 4. Two particles of mass m and 2m moving in opposite directions on a frictionless surface collide elastically with velocity v and 2v respectively. Find their velocities after a collision, also find the fraction of kinetic energy lost by the colliding particles.

Answer:

Let the final velocities of m and 2m be v₁ and v₂ respectively as shown in the figure:

By conservation of momentum:

m(2v) + 2m(–v) = m(v₁) + 2m (v₂)

or 0 = mv₁+ 2mv₂ or v₁+ 2v₂= 0 ………(1)

and since the collision is elastic:

v₂ – v₁ = 2v –(–v) or v₂ – v₁ = 3v ………(2)

Solving the above two equations, we get,

v₂= v and v₁= –2v

i.e., the mass 2m returns with velocity v while the mass m returns with velocity 2v in the direction shown in figure:

The collision was elastic therefore, no kinetic energy is lost, KE loss = KE_i– KE_f

or \(\left(\frac{1}{2} m(2 v)^2+\frac{1}{2}(2 m)(-v)^2\right)-\left(\frac{1}{2} m(-2 v)^2+\frac{1}{2}(2 m) v^2\right)\) = 0

Question 5. On a frictionless surface, a ball of mass m moving at a speed v makes a head-on collision with an identical ball at rest. The kinetic energy of the balls after the collision is 3/4th of the original. Find the coefficient of restitution.
Answer:

As we have seen in the above discussion, that under the given conditions :

By using conservation of linear momentum and equation of e, we get,

⇒ \(v_1^{\prime}=\left(\frac{1+e}{2}\right) v\)

and \(v_2^{\prime}=\left(\frac{1-e}{2}\right) v\)

Given that \(\mathrm{K}_{\mathrm{f}}=\frac{3}{4} \mathrm{~K}_{\mathrm{i}}\)

or \(\frac{1}{2} m v_1^{\prime 2}+\frac{1}{2} m v_2^{\prime 2}=\frac{3}{4}\left(\frac{1}{2} m v^2\right)\)

Substituting the value, we get

⇒ \(\left(\frac{1+e}{2}\right)^2+\left(\frac{1-e}{2}\right)^2\)

⇒ \(\frac{3}{4}\)

or \(e=\frac{1}{\sqrt{2}}\)

Question 6. A block of mass 2 kg is pushed toward a very heavy object moving with 2 m/s closer to the block (as shown). Assuming elastic collision and frictionless surfaces, find the final velocities of the blocks.

Answer:

Let v₁ and v₂ be the final velocities of the 2kg block and heavy object respectively then,

v₁= u₁+ 1 (u₁– u₂) = 2u₁– u₂ = –14 m/s

v₂= –2m/s

Question 7. A ball is moving with velocity of 2 m/s towards a heavy wall moving towards the ball with a speed 1m/s as shown in fig. Assuming collision to be elastic, find the velocity of the ball immediately after the collision.

Answer:

The speed of wall will not change after the collision. So, let v be the velocity of the ball after collision in the direction shown in figure. Since collision is elastic (e = 1),

separation speed = approach speed

or v – 1 = 2 + 1 or v = 4 m/s

Collision In Two Dimensions (oblique)

Question 8. A ball of mass m hits a floor with a speed v0 making an angle of incidence a with the normal. The coefficient of restitution is e. Find the speed of the reflected ball and the angle of reflection of the ball.
Answer:

The component of velocity v₀ along common tangential direction v₀ sin α will remain unchanged. Let v be the component along a common normal direction after collision. Applying, Relative speed of separation = e (Relative speed of approach) along a common normal direction, we get v = ev₀cos α

Thus, after collision components of velocity v’ are v₀ sin α and ev₀ cos α

⇒ \(v^{\prime}=\sqrt{\left(v_0 \sin \alpha\right)^2+\left(e v_0 \cos \alpha\right)^2}\)

and tan β = \(\frac{v_0 \sin \alpha}{e v_0 \cos \alpha}\)

α or tan β = \(\frac{\tan \alpha}{e}\)

Note: For elastic collision, e = 1

∴ v’ = v₀ and β = α

Question 9. A ball of mass m makes an elastic collision with another identical ball at rest. Show that if the collision is oblique, the bodies go at right angles to each other after a collision.
Answer:

In a head-on elastic collision between two particles, they exchange their velocities. In this case, the component of ball 1 along the common normal direction, v cos θ becomes zero after a collision, while that of 2 becomes v cos θ.

While the components along the common tangent direction of both the particles remain unchanged. Thus, the components along the common tangent and common normal direction of both the balls in tabular form are given below.

From the above table and figure, we see that both the balls move at the right angle after collision with velocities v sin θ and v cos θ.

Note: When two identical bodies have an oblique elastic collision, with one body at rest before the collision, then the two bodies will go in ⊥different directions.

Variable Mass System

If a mass is added or ejected from a system, at rate μ kg/s and relative velocity \(\overrightarrow{\mathrm{v}}_{\mathrm{rel}}\)(w.r.t. the system), then the force exerted by this mass on the system has magnitude \(\)

Thrust Force \(\left(\vec{F}_t\right)\)

⇒\(\overrightarrow{\mathrm{F}}_{\mathrm{t}}=\overrightarrow{\mathrm{v}}_{\mathrm{rel}}\left(\frac{\mathrm{dm}}{\mathrm{dt}}\right)\)

Suppose at some moment t = t mass of a body is m and its velocity is \(\vec{v}_r\). After some time at t = t + dt its mass becomes (m – dm) and velocity becomes.

The mass dm is ejected with relative velocity \(\mu\left|\vec{v}_{\text {rel }}\right|\)

Absolute velocity of mass ‘dm’ is therefore \(\left(\vec{v}+\vec{v}_r\right)\). If no external forces are acting on the system, the r linear momentum of the system will remain conserved, or

⇒ \(\vec{P}_{\mathrm{i}}=\vec{P}_{\mathrm{f}}\)

or \(m \vec{v}=(m-d m)(\vec{v}+d \vec{v})+d m\left(\vec{v}+\vec{v}_r\right)\)

or \(m \vec{v}=m \vec{v}+m d \vec{v}-(d m) \vec{v}-(d m)(d \vec{v})+(d m) \vec{v}+\vec{v} d m\)

The term (dm)\((\mathrm{d} \overrightarrow{\mathrm{v}} \text { ) }\) is too small and can be neglected.

∴ \(\mathrm{m}\left(\frac{\mathrm{d} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}\right)=\overrightarrow{\mathrm{v}}_{\mathrm{r}}\left(-\frac{\mathrm{dm}}{\mathrm{dt}}\right)\)

Here, \(\mathrm{m}\left(-\frac{\mathrm{d} \vec{v}}{d \mathrm{dt}}\right)=\text { thrust force }\left(\vec{F}_{\mathrm{t}}\right)\)

and \(-\frac{\mathrm{dm}}{\mathrm{dt}}\) = rate at which mass is ejecting or \(\vec{F}_t=\vec{v}_r\left(\frac{d m}{d t}\right)\)

Rocket Propulsion:

Let m₀ be the mass of the rocket at time t = 0. m its mass at any time t and v its velocity at that moment. Initially, let us suppose that the velocity of the rocket is u.

Further, let \(\left(\frac{-\mathrm{dm}}{\mathrm{dt}}\right)\) be the mass of the gas ejected per unit time and vrthe exhaust velocity of the gases with respect to rocket.

Usually \(\left(\frac{-\mathrm{dm}}{\mathrm{dt}}\right)\) and vr are kept constant throughout the journey of the dt rocket. Now, let us write few equations which can be used in the problems of rocket propulsion. At time t = t,

1. Thrust force on the rocket Ft= vrdm \(F_t=v_r\left(\frac{-d m}{d t}\right)\)(upwards)

2. Weight of the rocket W = mg (downwards)

3. Net force on the rocket Fnet = Ft– W (upwards)

or \(F_{\text {net }}=v_r\left(\frac{-d m}{d t}\right)-m g\)

4. Net acceleration of the rocket a = \(\frac{F}{m}\)

⇒ \(\frac{d v}{d t}=\frac{v_r}{m}\left(\frac{-d m}{d t}\right)-g\)

or \(\mathrm{dv}=\frac{v_r}{m}(-d m)-g d t\)

or \(\int_u^v d v=v_r \int_{m_0}^m \frac{-d m}{m}-g \int_0^t d t\)

Thus, v = u – gt + vr ln \(\left(\frac{m_0}{m}\right)\)…(i) m

Note:

1. Ft= vr \(\left(-\frac{\mathrm{dm}}{\mathrm{dt}}\right)\) is upwards, as v_r is downwards and \(\frac{\mathrm{dm}}{\mathrm{dt}}\) is negative.
2. If gravity is ignored and initial velocity of the rocket u = 0, Eq. (1) reduces to v = v_r ln \(\left(\frac{m_0}{m}\right)\)

Question 1. A rocket, with an initial mass of 1000 kg, is launched vertically upwards from rest under gravity. The rocket burns fuel at the rate of 10 kg per second. The burnt matter is ejected vertically downwards with a speed of 2000 ms^-2 relative to the rocket. If the burning stops after one minute. Find the maximum velocity of the rocket. (Take g as at 10 ms^-2)
Answer:

Using the velocity equation

v = u – gt + v_r ln\(\left(\frac{m_0}{m}\right)\)

Here u = 0, t = 60s, g = 10 m/s², vr= 2000 m/s, m₀= 1000 kg

and m = 1000 – 10 × 60 = 400 kg

We get v = 0 – 600 + 2000 ln \(\left(\frac{1000}{400}\right)\)

or v = 2000 ln 2.5 – 600

The maximum velocity of the rocket is 200(10 ln 2.5 – 3) = 1232.6 ms^-1

Linear Momentum Conservation In The Presence Of External Force.

⇒ \(\overrightarrow{\mathrm{F}}_{\mathrm{ext}}=\frac{\mathrm{d} \overrightarrow{\mathrm{P}}}{\mathrm{dt}}\)

⇒ \(\vec{F}_{\text {ext }} d t=d \vec{P}\)

⇒ \(\left.\mathrm{d} \overrightarrow{\mathrm{P}}=\overrightarrow{\mathrm{F}}_{\text {ext }}\right)_{\text {mpulsive }} \mathrm{dt}\)

∴ If \(\left.\vec{F}_{\text {ext }}\right)_{\text {mpuulise }}=0\)

⇒ \(\mathrm{d} \overrightarrow{\mathrm{P}}=0\)

or \(\vec{P}\) is constant 94

Note: Momentum is conserved if the external force present is non-impulsive. eg. gravitation or spring force

Question 1. Two balls are moving toward each other on a vertical line that collides with each other as shown. Find their velocities just after the collision.

Answer:

Let the final velocity of 4 kg ball just after collision be v. Since, an external force is gravitational which is non – impulsive, hence, linear momentum will be conserved.

Applying linear momentum conservation:

2(–3) + 4= 2+ 4(v) or v = \(\frac {1}{2}\)m/s

Question 2. Three particles of masses 0.5 kg, 1.0 kg, and 1.5 kg are placed at the three corners of a right-angled triangle of sides 3.0 cm, 4.0 cm, and 5.0 cm as shown in the figure. Locate the center of mass of the system.

Answer:

taking the x and y axes as shown.

Coordinates of body A = (0,0)

Coordinates of body B = (4,0)

Coordinates of body C = (0,3)

x – coordinate of c.m. = \(\frac{m_A x_A+m_B x_B+M_C r_C}{m_A+m_B+m_C}\)

⇒ \(\frac{0.5 \times 0+1.0 \times 4+1.5 \times 0}{0.5+1.0+1.5}\)

⇒ \(\frac{4}{3} \frac{\mathrm{cm}}{\mathrm{kg}}\)

= 1.33cm

similarly y – coordinates of c.m. = \(\frac{0.5 \times 0+1.0 \times 0+1.5 \times 3}{0.5+1.0+1.5}\)

⇒ \(\frac{4.5}{3}\)

= 1.5 cm

So, a center of mass is 1.33 cm right and 1.5 cm above particle A.

Question 3. A block A (mass = 4M) is placed on the top of a wedge B of base length l (mass = 20 M) as shown in the figure. When the system is released from rest. Find the distance moved by wedge B till block A reaches the lowest end of the wedge. Assume all surfaces are frictionless.

Answer:

The initial position of the center of mass

⇒ \(\frac{X_B M_B+X_A M_A}{M_B+M_B}\)

⇒ \(\frac{X_B \cdot 20 M+\ell .4 M}{24 M}\)

⇒ \(\frac{5 \mathrm{X}_{\mathrm{B}}+\ell}{6}\)

The final position of the center of mass

⇒ \(\frac{\left(X_B+x\right) 20 M+4 M x}{24 M}\)

⇒ \(\frac{5\left(X_B+x\right)+x}{6}\)

since there is no horizontal force on the system center of mass initially = center of mass finally.

5X_B+ l = 5X_B+ 5x + x

l = 6x

⇒ \(\frac{\ell}{6}\)

Question 4. An isolated particle of mass m is moving in a horizontal xy plane, along the x-axis. At a certain height above ground, it suddenly explodes into two fragments of masses m/4 and 3m/4. An instant later, the smaller fragment is at y = + 15 cm. Find the position of the heavier fragment at this instant.
Answer:

As the particle is moving along the x-axis, so, the y-coordinate of COM is zero.

⇒ \(Y_M M=Y_{\frac{M}{4}}\left(\frac{M}{4}\right)+Y_{\frac{M M}{4}}\left(\frac{3 M}{4}\right)\)

⇒ \(0 \times M=15\left(\frac{M}{4}\right)+Y_{\frac{3 M}{4}}\left(\frac{3 M}{4}\right)\)

⇒ \(\frac{Y_{3 M}}{4}=-5 \mathrm{~cm}\)

Question 5. A shell at rest at origin explodes into three fragments of masses 1 kg, 2 kg, and m kg. The fragments of masses 1 kg and 2 kg fly off with speeds of 12 m/s along the x-axis and 8 m/s along the axis respectively. If m kg flies off with speed 40 m/s then find the total mass of the shell.
Answer:

As initial velocity = 0, Initial momentum = (1 + 2 + m) × 0 = 0

Finally, let velocity of M = \(\vec{v}\)

We know \(|\vec{V}|\) = 40 m/s.

Initial momentum = final momentum

0 = \(1 \times 12 \hat{i}+2 \times 8 \hat{j}+m \vec{V}\)

⇒ \(\vec{V}=\frac{(12 \hat{i}+16 \hat{j})}{m}\)

⇒ \(|\vec{V}|=\sqrt{\frac{(12)^2+(16)^2}{m^2}}\)

⇒ \(\frac{1}{m} \sqrt{(12)^2+(16)^2}\) = 40 {given}

⇒ \(m=\sqrt{\frac{(12)^2+(16)^2}{40}}\)

= 0.5 kg Total mass = 1 + 2 + 0.5 = 3.5 kg

Question 6. A block moving horizontally on a smooth surface with a speed of 20 m/s bursts into two equal parts continuing in the same direction. If one of the parts moves at 30 m/s, with what speed does the second part move and what is the fractional change in the kinetic energy of the system?
Answer:

Applying momentum conservation:

⇒ \(m \times 20=\frac{m}{2} V+\frac{m}{2} \times 30\)

⇒ \(20=\frac{V}{2}+15\)

So, V = 10 m/s

initial kinetic energy = \(\frac{1}{2}\)m × (20)2 = 200 m

final kinetic energy = \(\frac{1}{2} \cdot \frac{m}{2} \cdot(10)^2+\frac{1}{2} \times \frac{m}{2}(30)^2\)

= 25 m + 225 m = 250 m

fractional change in kinetic energy = \(\frac{(\text { final K. E) }- \text { (initial K. E) }}{\text { initial K.E }}\)

⇒ \(\frac{250 m-200 m}{200 m}\)

⇒ \(\frac{1}{4}\)

Question 7. A block at rest explodes into three equal parts. Two parts start moving along the X and Y axes respectively with equal speeds of 10 m/s. Find the initial velocity of the third part.
Answer:

Let total mass = 3 m,

initial linear momentum = 3m × 0

Let velocity of third part = \(\overrightarrow{\mathrm{V}}\)

Using conservation of linear momentum:

⇒ \(m \times 10 \hat{i}+m \times 10 \hat{j}+m \vec{V}=0\)

So, \(\vec{V}=(-10 \hat{i}-10 \hat{j}) \mathrm{m} / \mathrm{sec} .\)

⇒ \(|\vec{V}|=\sqrt{(10)^2+(10)^2}=10 \sqrt{2}\) making angle 135o below x-axis

Question 8. Blocks A and B have masses of 40 kg and 60 kg respectively. They are placed on a smooth surface and the spring connected between them is stretched by 1.5m. If they are released from rest, determine the speeds of both blocks at the instant the spring becomes unstretched.

Answer:

Let, both blocks start moving with velocity V₁ and V₂ as shown in Figure

Since no horizontal force on the system so, applying momentum

conservation

0 = 40 V₁– 60 V₂

2V₁ = 3V₂ =……..(1)

Now applying energy conservation, Loss in potential energy = gain in kinetic energy

⇒ \(\frac{1}{2} k x^2=\frac{1}{2} m_1 V_1^2+\frac{1}{2} m_2 V_2^2\)

⇒ \(\frac{1}{2} \times 600 \times(1.5)^2=\frac{1}{2} \times 40 \times V_1{ }^2+\frac{1}{2} \times 60 \times V_2{ }^2\) …….

Solving the equation and we get, V1= 4.5 m/s, V2= 3 m/s.

Question 9. Find the mass of the rocket as a function of time, if it moves with a constant acceleration a, in the absence of external forces. The gas escaped with a constant velocity u relative to the rocket and its initial mass was m0₁.
Answer:

Using, Fnet = \(V_{\text {rel }}\left(\frac{-d m}{d t}\right)\)

⇒ \(F_{\text {net }}=-u \frac{d m}{d t}\) …….(1)

F_net = ma ……(2)

Solving equation and

⇒ \(\mathrm{ma}=-\mathrm{u} \frac{\mathrm{dm}}{\mathrm{dt}}\)

⇒ \(\int_{m_0}^m \frac{d m}{m}=\int_0^t \frac{-a d t}{u}\)

⇒ \(\ln \frac{m}{m_0}=\frac{-a t}{u}\)

⇒ \(\frac{m}{m_0}=e^{-a t / u}\)

⇒ \(\mathrm{m}=\mathrm{m}_0 \mathrm{e}^{-\frac{a t}{u}}\)

Question 10. A ball is approaching the ground with speed u. If the coefficient of restitution is e then find out:

1. The velocity just after a collision.
2. The impulse exerted by the normal is due to the ground on the ball.

Answer:

e = \(\frac{\text { velocity of separation }}{\text { velocity of approach }}\)

⇒ \(\frac{v}{u}\)

velocity after collision = V = eu ……..(1)

Impulse exerted by the normal due to ground on the ball = change in momentum of the ball. = {final momentum} – {initial momentum}

= {m v} – {– mu}

= mv + mu = m {u + eu} = mu {1 + e}

 

Elasticity And Viscosity Solids

The materials having a definite shape and volume are known as solids. All solids have the property of elasticity by virtue of which solids behave as incompressible substances and exhibit rigidity and mechanical strength. Solids are classified into two categories namely Crystalline solids and amorphous solids (or glassy solids).

Crystalline Solids: A solid in which atoms or molecules are arranged in a regular three-dimensional pattern is known as a crystalline solid shown in figure (1) For example quartz, mica, sugar, copper sulphate, sodium chloride, potassium iodide, cesium chloride, carbon, etc.

Amorphous Solids: A solid in which atoms or molecules are not arranged in a regular manner is known as an amorphous solid shown in figure (2) For example: talc powder, glass, rubber, plastics, etc.

Unit Cell And Crystal Lattice

Unit cell is the building block of a crystal. It is defined as the smallest pattern of atoms in a lattice, the repetition of which in three dimensions forms a crystal lattice.

Crystal lattice: It is defined as a regular arrangement of a large number of points in space, each point representing the position of an atom or a group of atoms in a crystal. The crystal lattice is shown in the figure.

Elasticity And Plasticity

The property of a material body by virtue of which it regains its original configuration (i.e. shape and size) when the external deforming force is removed is called elasticity.

• The property of the material body by virtue of which it does not regain its original configuration when the external force is removed is called plasticity.

Deforming Force: An external force applied to a body that changes its size or shape or both is called a deforming force.

Perfectly Elastic Body: A body is said to be perfectly elastic if it completely regains its original form when the deforming force is removed. Since no material can regain completely its original form the concept of a perfectly elastic body is only an ideal concept. A quartz fiber is the nearest approach to the perfectly elastic body.

Perfectly Plastic Body: A body is said to be perfectly plastic if it does not regain its original form even slightly when the deforming force is removed.

• Since every material partially regains its original form on the removal of the deforming force, the concept of a perfectly plastic body is only an ideal concept. Paraffin wax and wet clay are the nearest approaches to perfectly plastic bodies.

Cause of Elasticity: In a solid, atoms and molecules are arranged in such a way that each atom/molecule is acted upon by the forces due to the neighboring atom/molecules. These forces are known as intermolecular forces.

• When no deforming force is applied to the body, each atom/molecule of the solid (i.e. body) is in its equilibrium position and the intermolecular forces between the molecules of the solid are zero.
• On applying the deforming force on the body, the molecules either come closer or go far apart from each other. As a result of this, the atoms/molecules are atoms displaced from their equilibrium position.
• In other words, equilibrium intermolecular forces get disturbed or changed, and restoring forces are developed on the molecules. When the deforming force is removed, these restoring forces bring the molecules of the solid to their respective equilibrium positions and hence the solid (or the body) regains its original form.

Stress

When deforming force is applied to the body then the upto a certain limit equal restoring force in the opposite direction is developed inside the body. The restoring forces per unit area developed in the body is called stress.

stress = \(=\frac{\text { restoring force }}{\text { Area of the body }}=\frac{F}{A}\)

The unit of stress is N/m² or Nm^-2. There are three types of stress

1. Longitudinal or Normal stress: When an object is one dimensional or linear then force acting per unit area is called longitudinal stress. It is of two types :

1. Compressive stress
2. Tensile stress

Longitudinal or Normal stress Examples:

Consider a block of solid as shown in the figure. Let a force F be applied to the face which has area A. Resolve \(\vec{F}\) into two components:

F_n= F sin θ is called normal force and F_t= F cos θ called tangential force.

∴ Normal (tensile) stress = \(\frac{F_{\mathrm{n}}}{\mathrm{A}}=\frac{\mathrm{F} \sin \theta}{\mathrm{A}}\)

2. Tangential or shear stress: It is defined as the restoring force acting per unit area tangential to the surface of the body.

Tangential (shear) stress = \(\frac{F_t}{A}=\frac{F \cos \theta}{A}\)

The effect of stress is to produce distortion or a change in size, volume, and shape
(i.e. configuration of the body).

3. Bulk stress or All-around stress or Pressure: When force F is acting all along the surface normal (ΔA) to the area, then force acting per unit area is known as pressure. The effect of pressure is to produce volume change. The shape of the body may or may not change depending upon the homogeneity of the body.

Question1. Find out longitudinal stress and tangential stress on a fixed block

Answer:

Longitudinal or normal stress ⇒ \(\sigma_{\mathrm{n}}=\frac{100 \sin 30^{\circ}}{5 \times 2}=5 \mathrm{~N} / \mathrm{m}^2\)

Tangential stress ⇒ \(\sigma_{\mathrm{t}}=\frac{100 \cos 30^{\circ}}{5 \times 2}=5 \sqrt{3} \mathrm{~N} / \mathrm{m}^2\)

Question 2. Find out Bulk stress on the spherical object of radius\(\frac{10}{\pi}\) cm if the area and mass of the piston are 50 cm² and 50 kg respectively for a cylinder filled with gas. 

Answer:

⇒ \(p_{\text {gas }}=\frac{m g}{A}+p_a=\frac{50 \times 10}{50 \times 10^{-4}}+1 \times 10^5\)

⇒ \(2 \times 10^5 \mathrm{~N} / \mathrm{m}^2\)

Bulk stress = p_gas= 2 × 10⁵ N/m²

Strain

The ratio of the change in configuration (i.e. shape, length, or volume) to the original configuration of the body is called strain

i.e. Strain,\(\epsilon=\frac{\text { change in configuration }}{\text { original configuration }}\)

Types Of Strain: There are three types of strain

Longitudinal Strain: This type of strain is produced when the deforming force causes a change in the length of the body. It is defined as the ratio of the change in length to the original length of the body.

Consider A Wire Of Length L: When the wire is stretched by a force F, then let the change in length of the wire be ΔL as shown in the Figure.

∴ Longitudinal strain, \(\epsilon_{\ell}=\frac{\text { change in length }}{\text { original length }}\)

or Longitudinal strain = \(\frac{\Delta \mathrm{L}}{\mathrm{L}}\)

Volume Strain: This type of strain is produced when the deforming force produces a change in the volume of the body as shown in the Figure. It is defined as the ratio of the change in volume to the original volume of the body.

If the upper face is displaced through x keeping the lower face fixed at a distance of l then

If ΔV = change in volume

V = original volume

Δ tangentialdisplacement

⇒ \(\epsilon_{\mathrm{v}}=\text { volume strain }\)

⇒ \(\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{\text { tangential displacement }}{\text { transverse distance }}\)

Shear Strain: This type of strain is produced when the deforming force causes a change in the shape of the body. It is defined as the angle (θ) through which a face originally perpendicular to the fixed face is turned as shown in the Figure.

⇒ \(\tan \phi \text { or } \phi=\frac{\mathrm{x}}{\ell}\)

Hooke’s Law And Modulus Of Elasticity

According to this law, within the elastic limit, stress is proportional to the strain.

i.e. stress ∝ strain

or stress = constant × strain or \(\frac{\text { stress }}{\text { strain }}=\text { Modulus of Elasticity. }\)

This Constant Is Called The Modulus Of Elasticity.

Thus, the modulus of elasticity is defined as the ratio of the stress to the strain.

The modulus of elasticity depends on the nature of the material of the body and is independent of its dimensions (i.e. length, volume, etc.).

Unit: The Sl unit of modulus of elasticity is Nm^-2 or Pascal (Pa).

Types Of Modulus Of Elasticity

Corresponding to the three types of strain there are three types of modulus of elasticity.

1. Young’s modulus of elasticity (Y)
2. Bulk modulus of elasticity (B)
3. Modulus of rigidity (η).

Young’s modulus of elasticity

It is defined as the ratio of the normal stress to the longitudinal strain.

i.e. Young’s modulus (Y) = \(\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}\)

Normal stress = F/A,

Longitudinal strain = ΔL/L

Y = \(Y=\frac{F / A}{\Delta L / L}=\frac{F L}{A \Delta L}\)

Question 1. Find out the shift in points B, C, and D in the compound wire shown in Figure.

Answer:

⇒ \(\Delta L_B=\Delta L_{A B}=\frac{F L}{A Y}=\frac{M g L}{A Y}\)

⇒ \(\frac{10 \times 10 \times 0.1}{10^{-7} \times 2.5 \times 10^{10}}=4 \times 10^{-3} \mathrm{~m}=4 \mathrm{~mm}\)

⇒ \(\Delta \mathrm{L}_{\mathrm{c}}=\Delta \mathrm{L}_{\mathrm{B}}+\Delta \mathrm{L}_{\mathrm{BC}}=4 \times 10^{-3}+\frac{100 \times 0.2}{10^{-7} \times 4 \times 10^{10}}\)

⇒ \(4 \times 10^{-3}+5 \times 10^{-3}=9 \mathrm{~mm}\)

⇒ \(\Delta \mathrm{L}_{\mathrm{D}}=\Delta \mathrm{L}_{\mathrm{c}}+\Delta \mathrm{L}_{\mathrm{CD}}=9 \times 10^{-3}+\frac{100 \times 0.15}{10^{-7} \times 1 \times 10^{10}}\)

⇒ \(9 \times 10^{-3}+15 \times 10^{-3}=24 \mathrm{~mm}\)

Elongation Of Rod Under Its Self Weight

Let rod has a self-weight ‘W’, area of cross-section ‘A” and length ‘L’. Considering on element of length dx at a distance ‘x’ from the bottom. then \(\mathrm{T}=\frac{\mathrm{W}}{\mathrm{L}} \mathrm{x}\) is down word force on dx elongation in ‘dx’ element = \(\frac{\text { T.dx }}{\mathrm{AY}}\)

Total elongation s = \(\int_0^L \frac{T d x}{A Y}=\int_0^L \frac{W x d x}{L A Y}=\frac{W L}{2 A Y}\)

Note: One can do this directly by considering total weight at C.M. and using effective length L/2.

Question 1. Find out the elongation in the block shown in Figure. If mass, area of cross-section, and Young modulus of a block are m, A, and Y respectively.

Answer:

Acceleration, a = \(\frac{T d x}{A Y}\) then T = m′a

where ⇒ m′ = \(\frac{m}{L} x\)

T = \(\frac{m}{L} \times \frac{F}{m}=\frac{F x}{L}\)

Elongation in element ‘dx’ = \(\frac{T d x}{A Y}\) total elongation, δ = \(\int_0^L \frac{T d x}{A Y}=\int_0^L \frac{F x d x}{A L Y}=\frac{F L}{2 A Y}\)

Note: Try this problem, if friction is given between block and surface (µ = friction coefficient), and Case :

F < µmg

F > µmg

In both cases answer will be \(\frac{F L}{2 A Y}\)

Bulk Modulus:

It is defined as the ratio of the normal stress to the volume strain i.e. B = \(\frac{\text { Pressure }}{\text { Volume strain }}\)

The stress is the normal force applied per unit area and is equal to the pressure applied (p). p

⇒ \(B=\frac{p}{\frac{-\Delta V}{V}}=-\frac{p V}{\Delta V}\)

A negative sign shows that an increase in pressure (p) causes a decrease in volume (ΔV).

Compressibility: The reciprocal of bulk modulus of elasticity is called compressibility K. The Unit of compressibility in Sl is N^-1 m² or pascal^-1(Pa^-1).

The bulk modulus of solids is about fifty times that of liquids, and for gases it is 10–8 times of solids.

B_Solids > B_liquids > B_gases

Isothermal modulus of elasticity of gas B = P (pressure of gas)

Adiabatic modulus of elasticity of gas B = γ × P where γ = \(\) = ratio of specific beats.

Modulus of Rigidity:

It is defined as the ratio of the tangential stress to the shear strain. Let us consider a cube whose lower face is fixed and a tangential force F acts on the upper face whose area is A.

∴ Tangential stress = F/A.

Let the vertical sides of the cube shift through an angle θ, called shear strain

∴ The modulus of rigidity is given by

η = \(\frac{\text { Tangential stress }}{\text { Shear strain }}\)

or \(\eta=\frac{F / A}{\phi}=\frac{F}{A \phi}\)

Question 2. A rubber cube of side 5 cm has one side fixed while a tangential force equal to 1800 N is applied to the opposite face to find the shearing strain and the lateral displacement of the strained face. The modulus of rigidity for rubber is 2.4 × 10⁶ N/m².
Answer:

L = 5 × 10^-2 m ⇒ \(\frac{F}{A}=\eta \frac{x}{L}\)

Strain θ = \(\frac{F}{A \eta}\)

⇒ \(\frac{1800}{25 \times 10^{-4} \times 2.4 \times 10^6}\)

⇒ \(\frac{180}{25 \times 24}\)

⇒ \(\frac{3}{10}\)

=0.3 radian

⇒ \(\frac{x}{L}=0.3\)

x = 0.3 × 5 × 10^-2

= 1.5 × 10^-2 m

= 1.5 mm

Variation Of Strain With Stress

When a wire is stretched by a load, it is seen that for a small value of the load, the extension produced in the wire is proportional to the load. On removing the load, the wire returns to its original length.

• The wire regains its original dimensions only when the load applied is less or equal to a certain limit. This limit is called the elastic limit.
• Thus, the elastic limit is the maximum stress on whose removal, the bodies regain their original dimensions.
• In the shown figure, this type of behavior is represented by the OB portion of the graph. Till A the stress is proportional to strain and from A to B if deforming forces are removed then the wire comes to its original length but here stress is not proportional to strain.

OA → Limit of Proportionality

OB → Elastic limit

C → Yield Point

CD → Plastic behavior

D → Ultimate point

DE → Fracture

As we go beyond point B, then even for a very small increase in stress, the strain produced is very large. This type of behaviour is observed around point C and at this stage the wire begins to flow like a viscous fluid.

• The point C is called yield point. If the stress is further increased, then the wire breaks off at point D called the breaking point.
• The stress corresponding to this point is called breaking stress or tensile strength of the material of the wire. A material for which the plastic range CD is relatively high is called ductile material.
• These materials get permanently deformed before breaking. The materials for which the plastic range is relatively small are called brittle materials. These materials break as soon as the elastic limit is crossed.

Important Points

• Breaking stress = Breaking force/area of cross-section.
• Breaking stress is constant for a material
• Breaking force depends upon the area of the section of the wire of a given material.
• The working stress is always kept lower than that of breaking stress so the safety factor = breaking stress/working stress may have a large value.
• Breaking strain = elongation or compression/original dimension.
• Breaking strain is constant for the material.

Elastic After Effect

We know that some material bodies take some time to regain their original configuration when the deforming force is removed. The delay in regaining the original configuration by the bodies on the removal of deforming force is called elastic after effect.

• The elastic after-effect is negligibly small for quartz fiber and phosphor bronze. For this reason, the suspensions made from quartz and phosphor bronze are used in galvanometers and electrometers.
• For glass fibre elastic after effect is very large. It takes hours for glass fiber to return to its original state on removal of a deforming force.

Elastic Fatigue

The loss of strength of the material due to repeated strains on the material is called elastic fatigue. That is why bridges are declared unsafe after a long time of their use.

Analogy Of Rod As A Spring

⇒ \(\mathrm{Y}=\frac{\text { stress }}{\text { strain }}\)

⇒ \(\mathrm{Y}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \ell}\)

or \(\mathrm{F}=\frac{\mathrm{AY}}{\ell} \Delta \ell\)

⇒ \(\frac{\mathrm{AY}}{\ell}\) = constant, depending on the type of material and geometry of the rod. F = kΔl

where k = \(\frac{\mathrm{AY}}{\ell}\) = equivalent spring constant.

for the system of rods shown in the figure

1. The replaced spring system is shown in the figure
2. Two spring in series]. Figure
3. Represents an equivalent spring system.

Figure (4) represents another combination of rods and their replaced spring system.

Question 1. A mass ‘m’ is attached with rods as shown in the figure. This mass is slightly stretched and released whether the motion of the mass is S.H.M., if yes then find out the time period.

Answer:

⇒ \(k_{e q}=\frac{k_1 k_2}{k_1+k_2}\)

where \(\mathrm{k}_1=\frac{\mathrm{A}_1 \mathrm{Y}_1}{\ell_1}\)

⇒ \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_{\mathrm{eq}}}}\)

and \(\mathrm{k}_2=\frac{\mathrm{A}_2 \mathrm{Y}_2}{\ell_2}\)

Elastic Potential Energy Stored In A Stretched Wire Or In A Rod

Strain energy stored in equivalent spring

U = \(\frac{1}{2} k x^2\)

where x = \(\frac{\mathrm{F} \ell}{\mathrm{AY}}\)

k = \(\frac{\mathrm{AY}}{\ell}\)

U = \(\frac{1}{2} \frac{A Y}{\ell} \frac{F^2 \ell^2}{A^2 Y^2}\)

⇒ \(\frac{1}{2} \frac{F^2 \ell}{A Y}\)

equation can be re-arranged

⇒ \(\mathrm{U}=\frac{1}{2} \frac{\mathrm{F}^2}{\mathrm{~A}^2} \times \frac{\ell \mathrm{A}}{\mathrm{Y}}\) [lA = volume of rod, F/A = stress]

⇒ \(\mathrm{U}=\frac{1}{2} \frac{(\text { stress) })^2}{\mathrm{Y}} \times \text { volume }\) again, U = \(\frac{1}{2} \frac{\mathrm{F}}{\mathrm{A}} \times \frac{\mathrm{F}}{\mathrm{AY}} \times \mathrm{A} \ell\)

[Stain = \(\frac{\mathrm{F}}{\mathrm{AY}}\)]

⇒ \(\mathrm{U}=\frac{1}{2} \text { stress } \times \text { strain } \times \text { volume }\) again, U = \(\frac{1}{2} \frac{F^2}{A^2 Y^2} A \ell Y\)

⇒ \(U=\frac{1}{2} Y(\text { strain })^2 \times \text { volume }\)

strain energy density ⇒ \(\frac{\text { strain energy }}{\text { volume }}\)

⇒ \(\frac{1}{2} \frac{\text { (stress) }^2}{Y}\)

⇒ \(\frac{1}{2} Y(\text { strain })^2\)

⇒ \(\frac{1}{2} \text { stress } \times \text { strain }\)

Posson’s Ratio (σ)

Within the elastic limit, the ratio between the lateral strain and the linear strain is a constant. This constant is called Poisson’s ratio.

σ = \(\frac{\text { lateral strain }}{\text { longitudinal strain }}\)

Lateral Strain: Change in length of the body from its initial length perpendicular to deforming force lateral strain = \(-\frac{\Delta D}{D}=\frac{D-(D-\Delta D)}{D}=\beta\)

Longitudinal Strain: Change in length of the body from its initial length in the direction of deforming force

Longitudinal strain \(\varepsilon_{\ell}=\frac{\text { change in length }}{\text { original length }}=\frac{\Delta \mathrm{L}}{\mathrm{L}}=\alpha\)

⇒ \(\sigma=\frac{\beta}{\alpha}\)

theoretical limit of σ –1 < σ < 0.5

The experimental value of σ lies between 0.2 and 0.4

Four important Relations between Y, B, η and σ

1. \(\eta=\frac{Y}{2(1+\sigma)}\)
2. \(\frac{9}{Y}=\frac{3}{\eta}+\frac{1}{B}\)
3. \(\sigma=\frac{3 B-2 \eta}{6 B+2 \eta}\)
4. \(B=\frac{Y}{3(1-2 \sigma)}\)

Thermal Stress :

If the temperature of the rod is increased by ΔT, then the change in length

Δl = l α ΔT strain = \(\frac{\text { stress }}{\text { strain }}\)

But due to rigid support, there is no strain. Supports provide force on stresses to keep the length of the rod the same

Y = \(\frac{\text { stress }}{\text { strain }}\)

thermal stress = Y strain = Y α ΔT

⇒ \(\frac{F}{A}=Y \alpha \Delta T\)

F = AY α ΔT

Applications Of Elasticity

Some of the important applications of the elasticity of the materials are discussed below:

1. The material used in bridges loses its elastic strength with time, and bridges are declared unsafe after long use.
2. To estimate the maximum height of a mountain :

The pressure at the base of the mountain = hρg = stress. The elastic limit of a typical rock is 3 × 10⁸ N m^-2

The stress must be less than the elastic limits, otherwise the rock begins to flow.

⇒ \(h<\frac{3 \times 10^8}{\rho g}<\frac{3 \times 10^8}{3 \times 10^3 \times 10}<10^4 \mathrm{~m}\)

∵ \(\left(\rho=3 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3} ; g=10 \mathrm{~ms}^{-2}\right)\)

h = 10km

It may be noted that the height of Mount Everest is nearly 9 km.

Torsion Constant Of A Wire

C = \(\frac{\pi \eta r^4}{2 \ell}\) Where η is the modulus of rigidity r and l are the radius and length of the wire respectively.

1. Toque required for twisting by angle θ, τ = Cθ.
2. Work done in twisting by angle θ, W = \(\frac{1}{2} C \theta^2\)

Viscosity

When a solid body slides over another solid body, a frictional force begins to act between them. This force opposes the relative motion of the bodies.

• Similarly, when a layer of a liquid slides over another layer of the same liquid, a frictional force acts between them which opposes the relative motion between the layers.
• This force is called ‘internal frictional force’. This is due to intermolecular forces. Suppose a liquid is flowing in streamlined motion on a fixed horizontal surface AB.
• The layer of the liquid that is in contact with the surface is at rest, while the velocity of other layers increases with distance from the fixed surface. In the Figure, the lengths of the arrows represent the increasing velocity of the layers.
• Thus there is a relative motion between adjacent layers of the liquid. Let us consider three parallel layers a, b, and c. Their velocities are in increasing order. The layer a tends to retard the layer b, while b tends to retard c.
• Thus each layer tends to decrease the velocity of the layer above it. Similarly, each layer tends to increase the velocity of the layer below it.
• This means that in between any two layers of the liquid, internal tangential forces act which try to destroy the relative motion between the layers. These forces are called ‘viscous forces’.
• If the flow of the liquid is to be maintained, an external force must be applied to overcome the dragging viscous forces. In the absence of the external force, the viscous forces would soon bring the liquid to rest.
• The property of the liquid by virtue of which it opposes the relative motion between its adjacent layers is known as ‘viscosity’.

The property of viscosity is seen in the following examples :

A stirred liquid, when left, comes to rest on account of viscosity. Thicker liquids like honey, coaltar, glycerine, etc. have a larger viscosity than thinner ones like water.

• If we pour coaltar and water on a table, the coaltar will stop flowing soon while the water will flow upto quite a large distance.
• If we pour water and honey in separate funnels, water comes out readily from the hole in the funnel while honey takes enough time to do so.
• This is because honey is much more viscous than water. As honey tends to flow down under gravity, the relative motion between its layers is opposed strongly.
• We can walk fast in the air, but not in water. The reason is again viscosity which is very small for air but comparatively much larger for water.
• The cloud particles fall down very slowly because of the viscosity of air and hence appear floating in the sky.
• Viscosity comes into play only when there is a relative motion between the layers of the same material. This is why it does not act in solids.

Flow Of Liquid In A Tube Critical Velocity

When a liquid flows ‘in a tube, the viscous forces oppose the flow of the liquid, Hence a pressure difference is applied between the ends of the tube which maintains the flow of the liquid.

• If all particles of the liquid passing through a particular point in the tube move along the same path, the flow” of the liquid is called ‘stream-lined flow’.
• This occurs only when the velocity of flow of the liquid is below a certain limiting value called ‘critical velocity’. When the velocity of flow exceeds the critical velocity, the flow is no longer stream-lined but becomes turbulent.
• In this type of flow, the motion of the liquid becomes zig-zag, and eddy-currents are developed in it as shown in Figure.

• Reynold observed that the critical velocity for a liquid flowing in a tube is vc= Rη/ρa. where ρ is density η is the viscosity of the liquid, a is the radius of the tube and R is ‘Reynold’s number’ (whose value for a narrow tube and for water is about 1000).
• When the velocity of the flow of the liquid is less than the critical velocity, then the flow of the liquid is controlled by the viscosity, the density having no effect on it.
• But when the velocity of flow is larger than the critical velocity, then the flow is mainly governed by the density, and the effect of viscosity becomes less important.
• It is because of this reason that when a volcano erupts, then the lava coming out of it flows speedily in spite of being very thick (of large viscosity).

Velocity Gradient And Coefficient Of Viscosity

The property of a liquid by virtue of which an opposing force (internal friction) comes into play whenever there is a relative motion between the different layers of the liquid is called viscosity.

Consider a flow of a liquid over the horizontal solid surface as shown in Figure. Let us consider two layers AB and CD moving with velocities \(\vec{v}\)and \([\vec{v}+d \vec{v}/latex] a distance z and (z + dz) respectively from the fixed solid surface.

According to Newton, the viscous drag or backward force (F) between these layers happens to be.

1. Directly proportional to the area (A) of the layer and
2. Directly proportional to the velocity gradient [latex]\left(\frac{d v}{d x}\right)\) between the layers. Defined as a change in velocity per unit perpendicular separation between the layers. Thus

⇒ \(F \propto A \frac{d v}{d z} \text { or } F=-\eta A \frac{d v}{d z}\)……..(1)

η is called the Coefficient of viscosity. A negative sign shows that the direction of viscous drag (F) is just opposite to the direction of the motion of the liquid.

Similarities And Differences Between Viscosity And Solid Friction

Similarities: Viscosity and solid friction are similar as

1. Both oppose the relative motion. Whereas viscosity opposes the relative motion between two adjacent liquid layers, solid friction opposes the relative motion between two solid layers.
2. Both come into play, whenever there is relative motion between layers of liquid or solid surfaces as the case may be.
3. Both are due to molecular attractions.

Differences between them are given below

Some Applications Of Viscosity

Knowledge of the viscosity of various liquids and gases has been put to use in daily life. Some applications of its knowledge are discussed as under:

1. As the viscosity of liquids varies with temperature, the proper choice of lubricant is made depending on season.
2. Liquids of high viscosity are used in shock absorbers and buffers at railway stations.
3. The phenomenon of viscosity of air and liquid is used to dampen the motion of some instruments.
4. The knowledge of the coefficient of viscosity of organic liquids is used in determining the molecular weight and shape of the organic molecules.
5. It finds an important use in the circulation of blood through arteries and veins of the human body.

Units Of Coefficient Of Viscosity

From the above formula, we have \(\eta=\frac{F}{A\left(\Delta v_x / \Delta z\right)}\)

∴ dimensions of \(\eta=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^2\right]\left[\mathrm{LT}^{-1} / \mathrm{L}\right]}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^2 \mathrm{~T}^{-1}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]\)

Its unit is kg/(meter-second)

In the C.G.S. system, the unit of coefficient of viscosity is dyne s cm–2 and is called poise. In SI the unit of coefficient of viscosity is N sm–2 and is called decompose.

1 decapoise = 1 N sm^-2 = (10⁵ dyne) × s × (10² cm)^-2 = 10 dyne s cm^-2 = 10 poise

Question 1. A man is rowing a boat with a constant velocity ‘v₀’ in a river the contact area of the boat is ‘A’ and the coefficient of viscosity is η. The depth of the river is ‘D’. Find the force required to row the boat.
Answer:

F – F_T= m a_res

As the boat moves with constant velocity a_res = 0

F = F_T

But \(F_T=\eta A \frac{d v}{d z}\)

but \(\frac{d v}{d z}=\frac{v_0-0}{D}=\frac{v_0}{D}\)

then F = F_T = \(\frac{\eta A v_0}{D}\)

Question 2. A cubical block (of side 2m) of mass 20 kg slides on an inclined plane lubricated with the oil of viscosity η = 10^-1 poise with a constant velocity of 10 m/sec. (g = 10 m/sec²) find out the thickness of a layer of liquid.

Answer:

F = F’\(\eta A \frac{d v}{d z}=m g \sin \theta ; \frac{d v}{d z}=\frac{v}{h}\)

∴ 20 × 10 × sin 30° = η × 4 × \(\frac{10}{h}\)

⇒ \(\frac{40 \times 10^{-2}}{100}\) – [η = 10^-1 poise = 10^-2 N-sec-m^-2] = 4 × 10^-3 m = 4 mm

Effect Of Temperature On The Viscosity

The viscosity of liquids decreases with an increase in temperature and increases with a decrease in temperature. That is, \(\eta \propto \frac{1}{\sqrt{T}}\) T. On the other hand, the value of viscosity of gases increases with the increase in temperature and vice-versa. That is, η ∝ T.

Stoke’S Law

Stokes proved that the viscous drag (F) on a spherical body of radius r moving with velocity v in a fluid of viscosity η is given by F = 6 π η r v. This is called Stokes’ law.

Terminal Velocity

When a body is dropped in a viscous fluid, it is first accelerated. As its velocity increases, viscous force also increases so ultimately its acceleration becomes zero and it attains a constant velocity called terminal velocity.

Calculation Of Terminal Velocity

Let us consider a small ball, whose radius is r and density is ρ, is falling freely in a liquid (or gas), whose density is σ and coefficient of viscosity η. When it attains a terminal velocity v. It is subjected to two forces:

1. Effective weight force acting downward

⇒ \(V(\rho-\sigma) g=\frac{4}{3} \pi r^3(\rho-\sigma) g\)

2. A viscous force acting upward

= 6 π η rv.

Since the ball is moving with a constant velocity v i.e., there is no acceleration in it, the net force acting on it must be zero. That is

6πηrv = \(\frac{4}{3} p r^3(\rho-\sigma) g\)

or \(v=\frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}\)

Thus, the terminal velocity of the ball is directly proportional to the square of its radius Important point

The air bubble in water always goes up. This is because the density of air (ρ) is less than the density of water (σ).

So the terminal velocity for the air bubble is Negative, which implies that the air bubble will go up. Positive terminal velocity means the body will fall down.

Question 1. A spherical ball is moving with terminal velocity inside a liquid. Determine the relationship between the rate of heat loss with the radius of the ball.
Answer:

Rate of heat loss = power = F × v = 6 π η r v × v = 6 π η r v2 = 6p η r \(\left[\frac{2}{9} \frac{g r^2\left(\rho_0-\rho_{\ell}\right)}{\eta}\right]^2\)

Rate of heat loss α r⁵

Question 2. A drop of water of radius 0.0015 mm is falling in the air. If the coefficient of viscosity of air is 1.8 × 10^-5 kg /(m-s), what will be the terminal velocity of the drop? (density of water = 1.0 × 10³ kg/m² and g = 9.8 N/kg.) The density of air can be neglected.
Answer:

By Stoke’s law, the terminal velocity of a water drop of radius r is given by 2

⇒ \(v=\frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}\)

where ρ is the density of water, σ is the density of air, and η the coefficient of viscosity of air. Here σ is negligible and r = 0.0015 mm = 1.5 × 10^-3 mm = 1.5 × 10^-6 m. Substituting the values:

⇒ \(v=\frac{2}{9} \times \frac{\left(1.5 \times 10^{-6}\right)^2 \times\left(1.0 \times 10^3\right) \times 9.8}{1.8 \times 10^{-5}}\)

= 2.72×10^-4

Question 3. A metallic sphere of radius 1.0 × 10^-3 m and density 1.0 × 10⁴ kg/m³ enters a tank of water, after a free fall through a distance of h in the earth’s gravitational field. If its velocity remains unchanged after entering water, determine the value of h. Given : coefficient of viscosity of water = 1.0 × 10^-3 N-s/m², g = 10 m/s² and density of water = 1.0 × 10³ kg/m³.
Answer:

The velocity attained by the sphere in falling freely from a height h is

ν = \(\sqrt{2 \mathrm{gh}}\)….(1)

This is the terminal velocity of the sphere in water. Hence by Stoke’s law, we have 2

ν = \(\frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}\)…….(2)

where r is the radius of the sphere, ρ is the density of the material of the sphere σ (= 1.0 × 10³ kg/m³) is the density of water and η is the coefficient of viscosity of water.

∴ ν = \(\frac{2 \times\left(1.0 \times 10^{-3}\right)^2\left(1.0 \times 10^4-1.0 \times 10^3\right) \times 10}{9 \times 1.0 \times 10^{-3}}\)

= 20 m/s

from equation (1), we have h = \(\frac{v^2}{2 g}=\frac{20 \times 20}{2 \times 10}\)

= 20 m

Applications of Stokes’ Formula

In determining the Electronic Charge by Millikan’s Oil Drop Experiment: Stokes’ formula is used in Millikan’s method for determining the electronic charge. In this method, the formula is applied to find out the radii of small oil drops by measuring their terminal velocity in the air.

The velocity of Raindrops: Raindrops are formed by the condensation of water vapor on dust particles. When they fall under gravity, their motion is opposed by the viscous drag in the air.

• As the velocity of their fall increases, the viscous drag also increases and finally becomes equal to the effective force of gravity.
• The drops then attain a (constant) terminal velocity which is directly proportional to the square of the radius of the drops.
• In the beginning, the raindrops are very small in size and so they fall with such a small velocity that they appear floating in the sky as clouds. As they grow in size by further condensation, then they reach the earth with appreciable velocity,

Parachute: When a soldier with a parachute jumps from a flying airplane, he descends very slowly in the air.

• In the beginning, the soldier falls with gravity acceleration g, but soon the acceleration goes on decreasing rapidly when a parachute is fully opened.
• Therefore, in the beginning, the speed of the falling soldier increases somewhat rapidly but then very slowly.
• Due to the viscosity of air, the acceleration of the soldier becomes ultimately zero and the soldier then falls with a constant terminal speed. The figure shows the speed of the falling soldier with time.

 

NEET Physics Class 11 Chapter 8 Heat Transfer Introduction

Heat is energy in transit that flows due to temperature difference; from a body at a higher temperature to a body at a lower temperature. This transfer of heat from one body to the other takes place through three processes.

1. Conduction
2. Convection
3. Radiation

NEET Physics Class 11 Chapter 8 Conduction

The process of transmission of heat energy in which heat is transferred from one particle of the medium to the other, but each particle of the medium stays at its position is called conduction, for example, if you hold an iron rod with one of its end on a fire for some time, the handle will get heated.

• The heat is transferred from the fire to the handle by conduction along the length of the iron rod. The vibrational amplitude of atoms and electrons of the iron rod at the hot end takes on relatively higher values due to the higher temperature of their environment.
• These increased vibrational amplitudes are transferred along the rod, from atom to atom during collision between adjacent atoms. In this way, a region of rising temperature extends itself along the rod to your hand.

Consider a slab of face area A, Lateral thickness L, whose faces have temperatures THand TC(TH> TC).

Now consider two cross sections in the slab at positions A and B separated by a lateral distance of dx. Let the temperature of face A be T and that of face B be T + ΔT. Then experiments show that Q, the amount of heat crossing the area A of the slab at position x in time t is given by

⇒ \(\frac{Q}{t}=-K A \frac{d T}{d x}\)

Here K is a constant depending on the material of the slab and is named the thermal conductivity of the material, and the quantity \(\left(\frac{d T}{d x}\right)\) is called temperature gradient. The (–) sign in the equation heat flows from high to low temperature (ΔT is a –ve quantity)

NEET Physics Class 11 Chapter 8 Steady State

If the temperature of a cross-section at any position x in the above slab remains constant with time (remember, it does vary with position x), the slab is said to be in a steady state.

• Remember steady-state is distinct from thermal equilibrium for which temperature at any position (x) in the slab must be the same.
• For a conductor in a steady state, there is no absorption or emission of heat at any cross-section. (as the temperature at each point remains constant with time).
• The left and right faces are maintained at constant temperatures TH and TC respectively, and all other faces must be covered with adiabatic walls so that no heat escapes through them and the same amount of heat flows through each cross-section in a given Interval of time.

Hence Q₁= Q = Q₂. Consequently, the temperature gradient is constant throughout the slab.

Hence, \(\frac{d T}{d x}=\frac{\Delta T}{L}=\frac{T_f-T_i}{L}=\frac{T_C-T_H}{L}\)

and \(\frac{Q}{t}=-\mathrm{KA} \frac{\Delta T}{L} \Rightarrow \frac{Q}{t}\)

⇒ \(\mathrm{KAQ}\left(\frac{T_H-T_C}{L}\right)\)

Here Q is the amount of heat flowing through a cross-section of slab at any position in a time interval of t.

Question 1. One face of an aluminum cube of edge 2 meter is maintained at 100ºC and the other end is maintained at 0ºC. All other surfaces are covered by adiabatic walls. Find the amount of heat flowing through the cube in 5 seconds. (The thermal conductivity of aluminum is 209 W/m–ºC)
Answer:

Heat will flow from the end at 100ºC to the end at 0ºC.

Area of a cross-section perpendicular to the direction of heat flow,

A = 4m²

then \(\frac{Q}{t}=\mathrm{KA} \frac{\left(T_H-T_C\right)}{L}\)

Q = \(\frac{\left(209 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}\right)\left(4 \mathrm{~m}^2\right)\left(100^{\circ} \mathrm{C}-0^{\circ} \mathrm{C}\right)(5 \mathrm{sec})}{2 \mathrm{~m}}\)

= 209 KJ

NEET Physics Class 11 Chapter 8 Thermal Resistance To Conduction

If you are interested in insulating your house from cold weather or for that matter keeping the meal hot in your tiffin box, you are more interested in poor heat conductors, rather than good conductors. For this reason, the concept of thermal resistance R has been introduced.

For a slab of cross-section A, Lateral thickness L, and thermal conductivity K,

Resistance \(R=\frac{L}{K A}\)

In terms of R, the amount of heat flowing through a slab in steady-state (in time t)

⇒ \(\frac{Q}{t}=\frac{\left(T_H-T_L\right)}{R}\)

If we name as thermal current ir

then, \(i_T=\frac{T_H-T_L}{R}\)

This is mathematically equivalent to OHM’s law, with temperature donning the role of electric potential. Hence results derived from OHM’s law are also valid for thermal conduction.

Moreover, for a slab in the steady state, we have seen earlier that the thermal current it remains the same at each cross-section. This is analogous to Kirchoff’s current law in electricity, which can now be very conveniently applied to thermal conduction.

Question 2. Three identical rods of length 1m each, having cross-section area of 1cm2 each and made of Aluminium, copper, and steel respectively are maintained at temperatures of 12ºC, 4ºC, and 50ºC respectively at their separate ends. Find the temperature of their common junction.

[ KCu = 400 W/m-K , K_Al = 200 W/m-K , K_steel = 50 W/m-K ]

Answer:

⇒ \(\mathrm{R}_{\mathrm{Al}}=\frac{L}{K A}=\frac{1}{10^{-4} \times 209}=\frac{10^4}{209}\)

Similarly = \(R_{\text {steel }}=\frac{10^4}{46} \text { and } R_{\text {copper }}=\frac{10^4}{385}\)

Let the temperature of common junction = T then from Kirchoff’s Junction law.

i_Al + i_steel + iCu = 0

⇒ \(\frac{T-12}{R_{A I}}+\frac{T-51}{R_{\text {steel }}}+\frac{T-u}{R_{C u}}=0\)

⇒ (T – 12) 200 + (T – 50) 50 + (T – 4) 400 = 0

⇒ 4(T – 12) + (T – 50) + 8 (T – 4) = 0

⇒ 13T = 48 + 50 + 32 = 130

⇒ T = 10ºC

 

NEET Physics Class 11 Chapter 8 Growth Of Ice On Ponds

When atmospheric temperature falls below 0°C the water in the lake will start freezing. Let at any time t, the thickness of ice in the lake be y and atmospheric temperature is –θ°C. The temperature of water in contact with the lower surface of ice will be 0ºC.

the area of the lake = A

heat escaping through ice in time dt is

Now due to the escaping of this heat if the thickness of water in contact with the lower surface of ice freezes,

⇒ \(d Q_1=K A \frac{[0-(-\theta)]}{y} d t\)

dQ₂= mL = ρ(dy A)L [as m = ρV = ρA dy]

But as dQ₁= dQ₂, the rate of growth of ice will be

⇒ \(\frac{d y}{d t}=\frac{K \theta}{\rho L} \times \frac{1}{y}\)

and so the time taken by ice to grow a thickness y, \(t=\frac{\rho L}{K \theta} \int_0^y y \quad d y=\frac{1}{2} \frac{\rho L}{K \theta} \quad y^2\)

Time taken to double and triple the thickness will be in the ratio t₁: t₂: t₃:: 1² : 2²: 3², i.e., t₁: t₂: t₃:: 1 : 4: 9 and the time intervals to change thickness from 0 to y, from y to 2y and so on will be in the ratio Δt₁: Δt2: Δt₃: : (1² – 0² ) : (2² – 1² ) : (3² – 2² ), i.e., Δt₁: Δt₂: Δt₃:: 1 : 3: 5.

Can you now see how the following facts can be explained by thermal conduction?

1. In winter, iron chairs appear to be colder than the wooden chairs.
2. Ice is covered in gunny bags to prevent melting.
3. Woolen clothes are warmer.
4. We feel warmer in a fur coat.
5. Two thin blankets are warmer than a single blanket of double the thickness.
6. Birds often swell their feathers in winter.
7. A new quilt is warmer than an old one.
8. Kettles are provided with wooden handles.
9. Eskimos make double-walled ice houses.
10. Thermos flask is made double-walled.

NEET Physics Class 11 Chapter 8 Convection

When heat is transferred from one point to the other through the actual movement of heated particles, the process of heat transfer is called convection.

• In liquids and gases, some heat may be transported through conduction. But most of the transfer of heat in them occurs through the process of convection.
• Convection occurs through the aid of the earth’s gravity. Normally the portion of fluid at greater temperature is less dense, while that at lower temperature is denser. Hence hot fluids rise while colder fluids sink, accounting for convection. In the absence of gravity, convection would not be possible.
• Also, the anomalous behavior of water (its density increases with temperature in the range of 0-4ºC) gives rise to interesting consequences vis-a-vis the process of convection. One of these interesting consequences is the presence of aquatic life in temperate and polar waters. The other is the rain cycle.

Can you now see how the following facts can be explained by thermal convection?

1. Oceans freeze top-down and not bottom-up. (this fact is singularly responsible for the presence of aquatic life in temperate and polar waters.)
2. The temperature at the bottom of deep oceans is invariably 4ºC, whether it is winter or summer.
3. You cannot illuminate the interior of a lift in free fall or an artificial satellite of earth with a candle.
4. You can Illuminate your room with a candle.

NEET Physics Class 11 Chapter 8 Radiation

The process of the transfer of heat from one place to another place without heating the intervening medium is called radiation. The term radiation used here is another word for electromagnetic waves. These waves are formed due to the superposition of electric and magnetic fields perpendicular to each other and carry energy.

Properties of Radiation:

1. All objects emit radiation simply because their temperature is above absolute zero, and all objects absorb some of the radiation that falls on them from other objects.
2. Maxwell based on his electromagnetic theory proved that all radiations are electromagnetic waves and their sources are vibrations of charged particles in atoms and molecules.
3. More radiations are emitted at higher temperatures of a body and less at lower temperatures.
4. The wavelength corresponding to the maximum emission of radiation shifts from a longer wavelength to a shorter wavelength as the temperature increases. Due to this, the color of a body appears to be changing. Radiations from a body at NTP have predominantly infrared waves.
5. Thermal radiation travels with the speed of light and moves in a straight line.
6. Radiations are electromagnetic waves and can also travel through a vacuum.
7. Similar to light, thermal radiations can be reflected, refracted, diffracted, and polarized.
8. Radiation from a point source obeys the inverse square law (intensity α ).

NEET Physics Class 11 Chapter 8 Prevost Theory Of Heat Exchange

According to this theory, all bodies radiate thermal radiation at all temperatures. The amount of thermal radiation radiated per unit of time depends on the nature of the emitting surface, its area, and its temperature. The rate is faster at higher temperatures.

• Besides, a body also absorbs part of the thermal radiation emitted by the surrounding bodies when this radiation falls on it. If a body radiates more than what it absorbs, its temperature falls.
• If a body radiates less than what it absorbs, its temperature rises. And if the temperature of a body is equal to the temperature of its surroundings it radiates at the same rate as it absorbs.

NEET Physics Class 11 Chapter 8 Perfectly Black Body And Black Body Radiation (Fery’s Black Body)

A perfectly black body absorbs all the heat radiations of whatever wavelength, is incident on it. It neither reflects nor transmits any of the incident radiation and therefore appears black whatever the color of the incident radiation.

In actual practice, no natural object possesses strictly the properties of a perfectly black body.

• But the lamp-black and platinum black are a good approximation of the black body. They absorb about 99 % of the incident radiation. The most simple and commonly used black body was designed by Fery.
• It consists of an enclosure with a small opening which is painted black from inside. The opening acts as a perfect black body.
• Any radiation that falls on the opening goes inside and has very little chance of escaping the enclosure before getting absorbed through multiple reflections. The cone opposite to the opening ensures that no radiation is reflected directly.

NEET Physics Class 11 Chapter 8 Absorption, Reflection, And Emission Of Radiations

Q = Q_r+ Q_t+ Q_a

1 = \(\frac{Q_r}{Q}+\frac{Q_t}{Q}+\frac{Q_a}{Q}\)

where r = reflecting power , a = absorptive power

and t = transmission power.

1. r = 0, t = 0, a = 1, perfect black body
2. r = 1, t = 0, a = 0, perfect reflector
3. r = 0, t = 1, a = 0, perfect transmitter

Absorptive Power :

In particular, the absorptive power of a body can be defined as the fraction of incident radiation that is absorbed by the body.

a = \(\frac{\text { Energy absorbed }}{\text { Energy incident }}\)

As all the radiation incident on a black body is absorbed, a = 1 for a black body.

Emissive Power:

Energy radiated per unit time per unit area along the normal to the area is known as emissive power.

⇒ \(\frac{Q}{\Delta A \Delta t}\)

(Notice that, unlike absorptive power, emissive power is not a dimensionless quantity).

Spectral Emissive Power (Eλ) :

Emissive power per unit wavelength range at wavelength λ is known as spectral emissive power, Eλ. If E is the total emissive power and Eλ is spectral emissive power, they are related as follows,

⇒ \(\mathrm{E}=\int_0^{\infty} E_\lambda \mathrm{d} \lambda\)

and \(\frac{\mathrm{dE}}{\mathrm{d} \lambda}=\mathrm{E}_\lambda\)

Emissivity:

⇒ \(\mathrm{e}=\frac{\text { Emissive power of } \mathrm{a} \text { body at temperature } \mathrm{T}}{\text { Emissive power of } \mathrm{a} \text { black body at same temperature } \mathrm{T}}\)

⇒ \(\frac{E}{E_0}\)

NEET Physics Class 11 Chapter 8 Kirchoff’s Law

The ratio of the emissive power to the absorptive power for the radiation of a given wavelength is the same for all substances at the same temperature and is equal to the emissive power of a perfectly black body for the same wavelength and temperature.

⇒ \(\frac{E(\text { body })}{a(\text { body })}\) = E(black body)

Hence we can conclude that good emitters are also good absorbers.

Applications Of Kirchhoff’s Law

If a body emits strongly the radiation of a particular wavelength, it must also absorb the same radiation strongly.

1. Let a piece of china with some dark painting on it be first heated to nearly 1300 K and then examined in a dark room. It will be observed that the dark paintings appear much brighter than the white portion. This is because the paintings being better absorbers also emit much more light.
2. The silvered surface of a thermos flask does not absorb much heat from outside. This stops ice from melting quickly. Also, the silvered surface does not radiate much heat from the inside. This prevents hot liquids from becoming cold quickly.
3. A red glass appears red at room temperature. This is because it absorbs green light strongly. However, if it is heated in a furnace, it glows with green light. This is because it emits green light strongly at a higher temperature.
4. When white light is passed through sodium vapors and the spectrum of transmitted light is seen, we find two dark lines in the yellow region. These dark lines are due to absorption of radiation by sodium vapors which it emits when heated.

Fraunhofer lines are dark lines in the spectrum of the sun. When white light emitted from the central core of the sun (photosphere) passes through its atmosphere (chromosphere) radiations of those wavelengths will be absorbed by the gases present there which they usually emit (as a good emitter is a good absorber) resulting in dark lines in the spectrum of sun.

At the time of a solar eclipse, direct light rays emitted from the photosphere cannot reach the earth and only rays from the chromosphere can reach the earth’s surface. At that time we observe bright Fraunhofer lines.

NEET Physics Class 11 Chapter 8 Nature Of Thermal Radiations : (Wien’s Displacement Law)

From the energy distribution curve of black body radiation, the following conclusions can be drawn:

1. The higher the temperature of a body, the higher the area under the curve i.e. more amount of energy is emitted by the body at a higher temperature.
2. The energy emitted by the body at different temperatures is not uniform. For both long and short wavelengths, the energy emitted is very small.
3. For a given temperature, there is a particular wavelength (λ_m) for which the energy emitted (Eλ) is maximum.
4. With an increase in the temperature of the black body, the maxima of the curves shift towards shorter wavelengths.

From the study of the energy distribution of black body radiation discussed above, it was established experimentally that the wavelength (λ_m) corresponding to the maximum intensity of emission decreases inversely with an increase in the temperature of the black body. i.e.

λ_m ∝ or λ_m T = b

This is called Wien’s displacement law.

Here b = 0.282 cm-K, is the Wien’s constant.

Question 1. Solar radiation is found to have an intensity maximum near the wavelength range of 470 nm. Assuming the surface of the sun to be perfectly absorbing (a = 1), calculate the temperature of the solar surface.
Solution :

Since a =1, the sun can be assumed to be emitting as a black body from Wien’s law for a black body

λ_m. T = b

⇒ T = \(\frac{0.282(\mathrm{~cm}-\mathrm{K})}{\left(470 \times 10^{-7} \mathrm{~cm}\right)}\)

= ~ 6125 K.

 

Circular Motion

Fundamental parameter of circular motion

Radius Vector: The vector joining the center of the circle and the center of the particle performing circular motion is called the radius vector.

It has constant magnitude and variable direction

Angular Displacement (δθ or θ)

The angle described by the radius vector is called angular displacement.

Infinitesimal angular displacement is a vector quantity. However, finite angular displacement is a scalar quantity.

S.I Unit Radian

Dimension: M⁰L⁰T⁰

1 radian = \(\frac{360}{2 \pi}\)

No. Of revolution = \(\frac{\text { angular displacement }}{2 \pi}\)

In 1 revolution Δθ = 360º = 2π radian

In N revolution Δθ = 360º × N = 2πN radian

Clockwise rotation is taken as a negative

Anticlockwise rotation is taken as a positive

Question 1. If a particle completes one and a half revolutions along the circumference of a circle then its angular displacement is –

1. 0
2. π
3. 2π
4. 2π

Answer: = 3π

Angular Velocity (ω):

• The rate of change of angular displacement with time is called angular velocity. It is a vector quantity.
• The angle traced per unit time by the radius vector is called angular speed.

Instantaneous angular velocity = \(=\omega=\lim _{\delta t \rightarrow 0} \frac{\delta \theta}{\delta \mathrm{t}} \text { or } \omega=\frac{\mathrm{d} \theta}{\mathrm{dt}}\)

Average angular velocity

= \(\bar{\omega}=\frac{\theta_2-\theta_1}{\mathrm{t}_2-\mathrm{t}_1}=\frac{\Delta \theta}{\Delta \mathrm{t}}\)

S.I. Unit: rad/sec

Angular Velocity Dimension: M⁰L⁰T^-1

Angular Velocity Direction: Infinitesimal angular displacement, angular velocity, and angular acceleration are vector quantities whose direction is given by the right-hand rule.

Right-hand Rule: Imagine the axis of rotation to be held in the right hand with fingers curled around the axis and the thumb stretched along the axis. If the curled fingers denote the sense of rotation, then the thumb denotes the direction of the angular velocity (or angular acceleration of infinitesimal angular displacement.

Angular Acceleration (a):

The rate of change of angular velocity with time is called angular acceleration. Average angular acceleration

⇒ \(\bar{\alpha}=\frac{\omega_2-\omega_1}{t_2-t_1}=\frac{\Delta \omega}{\Delta t}\)

Instantaneous angular acceleration

⇒ \(\alpha=\frac{d \omega}{d t}=\frac{d^2 \theta}{d t^2}\)

It is a vector quantity, whose direction is along the change in direction of angular velocity.

S.I. Unit: radian/sec²

Dimension: M⁰L⁰T^-2

Relation Between Angular Velocity And Linear Velocity:

Suppose the particle moves along a circular path from point A to point B in infinitesimally small time δt. As, δt → 0, δθ → 0

∴ arc AB = chord AB i.e. displacement of the particle is along a straight line.

∴ Linear velocity, v = \(v=\lim _{\delta t\rightarrow 0} \frac{\delta s}{\delta t}\)

But, δs = r.δθ

∴ v = \(v=\lim _{\delta t \rightarrow 0} \frac{r \cdot \delta s}{\delta t}=r \lim _{\delta t \rightarrow 0} \frac{\delta \theta}{\delta t}\)

But, \(\lim _{\delta t \rightarrow 0} \frac{\delta \theta}{\delta \mathrm{t}}=\omega\) = angular velocity

V= r. ω [for circular motion only]

i.e. (linear velocity) = (Radian) × (angular velocity)

In vector notation, \(\vec{v}=\vec{\omega} \times \vec{r}\) [in general]

The linear velocity of a particle performing circular motion is the vector product of its angular velocity and radius vector.

Relation Between Angular Acceleration And Linear Acceleration

For perfect circular motion, we know

v = ω r

on differentiating with respect to time

we get \(\frac{d v}{d t}=r \frac{d \omega}{d t}\)

a = r α

In vector form = \(\overrightarrow{\mathrm{a}}=\vec{\alpha} \times \overrightarrow{\mathrm{r}}\) ×(linear acc.) = (angular acc) × (radius)

Types Of Circular Motion

Uniform Circular Motion: The motion of a particle along the circumference of a circle with a constant speed is called uniform circular motion. Uniform circular motion is an accelerated motion. In the case of uniform circular motion :

Speed remains constant. v = constant

and v = ω r

angular velocity ω = constant

Motion will be periodic with time period = \(T=\frac{2 \pi}{\omega}=\frac{2 \pi r}{v}\)

Frequency Of Uniform Circular Motion: The number of revolutions performed per unit of time by the particle performing uniform circular motion is called the frequency (n)

∴ n = \(n=\frac{1}{T}=\frac{v}{2 \pi r}=\frac{\omega}{2 \pi}\)

S.I. unit of frequency is Hz.

As ω = constant, from ω = ω₀+ αt

angular acceleration α = 0

As a_t = αr, tangential acc. a_t= 0

As a_t = 0, a = \(a=\left(a_r^2+a_t^2\right)^{1 / 2}\) yields a = ar, i.e. acceleration is not zero but along radius towards the center and has magnitude

⇒ \(\mathrm{a}=\mathrm{a}_{\mathrm{r}}=\left(\mathrm{v}^2 / \mathrm{r}\right)=\mathrm{r} \omega^2\)

Speed and magnitude of acceleration are constant. but their directions are always changing so velocity and acceleration are not constant.

The direction of \(\overrightarrow{\mathrm{v}}\) is always along the tangent while that of \(\overrightarrow{\mathrm{a}_{\mathrm{r}}}\) along the radius \(\vec{v} \perp \vec{a}_r\)

If the moving body comes to rest, i.e. \(\vec{v} \rightarrow 0\), and if radial acceleration vanishes, the body will fly off along the tangent. So a tangential velocity and a radial acceleration (hence force) is a must for uniform circular motion.

As \(\vec{F}=\frac{m v^2}{r}\) ≠ 0, the body is not in equilibrium and the linear momentum of the particle moving on the circle is not conserved. However, as the force is control, i.e.,

⇒ \(\vec{\tau}=0\), so angular momentum is conserved, i.e.,

⇒ \(\overrightarrow{\mathrm{p}}\) ≠ constant but

⇒ \(\overrightarrow{\mathrm{L}}\) = constant

The work done by a centripetal force is always zero as it is perpendicular to velocity and hence displacement. By work-energy theorem as work done = change in kinetic energy ΔK = 0

So K (kinetic energy) remains constant

For example., Planets revolving around the sun, the motion of an electron around the nucleus in an atom

In one-dimensional motion, acceleration is always parallel to velocity and changes only the magnitude of the velocity vector.

In uniform circular motion, acceleration is always perpendicular to velocity and changes only the direction of the velocity vector.

In the more general case, like projectile motion, acceleration is neither parallel nor perpendicular to the figure that summarizes these three cases.

If a particle moving with uniform speed v on a circle of radius r suffers angular displacement θ in time Δt then change in its velocity.

⇒ \(\Delta \vec{v}=\Delta \vec{v}_2-\Delta \vec{v}_1\)

⇒ \(\vec{v}_1=\vec{v}_1 \hat{i}\)

⇒ \(\vec{v}_2=\vec{v}_2 \cos \theta \hat{i}+\vec{v}_2 \sin \theta \hat{j}\)

⇒ \(\Delta \vec{v}=\left(\vec{v}_2 \cos \theta-\vec{v}_1\right) \hat{i}+\vec{v}_2 \sin ^2 \hat{j}\)

⇒ \(|\Delta \vec{v}|=\sqrt{\left(\vec{v}_2 \cos \theta-\vec{v}_1\right)^2+\vec{v}_2 \sin ^2}\)

⇒ \(|\Delta \vec{v}|=\sqrt{2 v^2-2 v^2 \cos \theta}=\sqrt{2 v^2(1-\cos \theta)}=\sqrt{2 v^2\left(2 \sin ^2 \frac{\theta}{2}\right)}\)

⇒ \(v_1=v_2=v\)

⇒ \(|\Delta \vec{v}|=2 v \sin \frac{\theta}{2}\)

Question 1. A particle is moving in a circle of radius r centered at O with constant speed v. What is the change in velocity in moving from A to B? Given ∠AOB = 40º.
Answer:

⇒ \(|\Delta \vec{v}|=2 v \sin 40^{\circ} / 2=2 \mathrm{v} \sin 20^{\circ}\)

Non-Uniform Circular Motion:

A circular motion in which both the direction and magnitude of the velocity change is called nonuniform circular motion.

• A merry-go-round is spinning up from rest to full speed, or a ball whirling around in a vertical circle. The acceleration is neither parallel nor perpendicular to the velocity.
• We can resolve the acceleration vector into two components:

Radial Acceleration: ar perpendicular to the velocity ⇒ changes only the directions of velocity Acts just like the acceleration in a uniform circular motion.

⇒ \(a_c=\text { or } \quad a_r=\frac{v^2}{r}\)

Centripetal force: \(F_c=\frac{m v^2}{r}=m \omega^2 r\)

Tangential acceleration: a_r parallel to the velocity (since it is tangent to the path)

⇒ changes in the magnitude of the velocity act just like one-dimensional acceleration

⇒ \(a_t=\frac{d v}{d t}\)

Tangential acceleration : \(a_t=\frac{d v}{d t}\) where \(v=\frac{d s}{d t}\) and s = length of arc

Tangential acceleration: F_t = ma_t

The net acceleration vector is obtained by vector addition of these two components.

⇒ \(a=\sqrt{a_r^2+a_t^2}\)

In non-uniform circular motion :

speed \(|\vec{v}|\) ≠ constant angular velocity ω ≠ constant

i.e. speed ≠ constant i.e. angular velocity ≠ constant

In any instant

⇒ v = magnitude of the velocity of a particle

⇒ r = radius of circular path

⇒ ω = angular velocity of a particle

then, at that instant v = r ω

The net force on the particle

⇒ \(\vec{F}=\vec{F}_c+\vec{F}_t \Rightarrow F=\sqrt{F_c^2+F_t^2}\)

If θ is the angle made by F = Fc,

then tan θ = \(=\frac{F_t}{F_c} \Rightarrow \theta=\tan^{-1}\left[\frac{F_{\mathrm{t}}}{F_c}\right]\)

[Note angle between Fcand Ftis 90º] Angle between F and Ftis (90º – θ)

Net acceleration: \(a=\sqrt{a_c^2+a_1^2}=\frac{F_{\mathrm{net}}}{m}\)

The angle made by ‘a’ with ac, tan θ = \(\frac{a_t}{a_c}=\frac{F_t}{F_c}\)

Special Note:

In both uniform and non-uniform circular motion Fc is perpendicular to velocity.

So work done by centripetal force will be zero in both cases.

In uniform circular motion Ft= 0, as = at= 0, so work done will be zero by tangential force.

But in non-uniform circular motion Ft≠ 0, the work done by tangential force is non-zero.

Rate of work done by net force in non-uniform circular motion = rate of work done by tangential force

⇒ \(P=\frac{d W}{d t}=\vec{F}_t \cdot \vec{v}=\vec{F}_t \cdot \frac{d \vec{x}}{d t}\)

In a circle tangent and radius are always normal to each other, so

⇒ \(\vec{a}_{\mathrm{t}} \perp \vec{a}_{\mathrm{r}}\)

Net acceleration in case of circular motion \(a=a_r^2=a_t^2\)

Here it must be noted that at governs the magnitude of \(\vec{v}\) while ar its direction of motion so that

If ar = 0 and at = 0 a → 0 ⇒ motion is uniform translatory

If ar = 0 and at ≠ 0 a → at ⇒ motion is accelerated translatory

If ar ≠ 0 and at = 0 a → ar ⇒ motion is uniform circular

If ar ≠ 0 and at ≠ 0 a → \(a \rightarrow \sqrt{a_r^2+a_1^2}\) ⇒ motion is non-uniform circular.

Question 2. A road makes a 90º bend with a radius of 190 m. A car enters the bend moving at 20 m/s. Finding this too fast, the driver decelerates at 0.92 m/s². Determine the acceleration of the car when its speed rounding the bend has dropped to 15 m/s.
Answer:

Since it is rounding a curve, the car has a radial acceleration associated with its changing direction, in addition to the tangential deceleration that changes its speed. We are given that at = 0.92 m/s²; since the car is slowing down, the tangential acceleration is directed opposite the velocity.

The radial acceleration is \(a_r=\frac{v^2}{r}=\frac{(15 \mathrm{~m} / \mathrm{s})^2}{190 \mathrm{~m}}=1.2 \mathrm{~m} / \mathrm{s}^{21}\)

Magnitude of net acceleration,

⇒ \(a=\sqrt{a_r^2+a_t^2}=\left[(1.2 \mathrm{~m} / \mathrm{s})^2+(0.92 \mathrm{~m} / \mathrm{s})^2\right]^{1 / 2}=1.5 \mathrm{~m} / \mathrm{s}^2\)

and points at an angle \(\theta=\tan ^{-1}\left(\frac{\mathrm{a}_{\mathrm{r}}}{\mathrm{a}_{\mathrm{t}}}\right)=\tan ^{-1}\left(\frac{1.2 \mathrm{~m} / \mathrm{s}^2}{0.92 \mathrm{~m} / \mathrm{s}^2}\right)=53^{\circ}\)

relative to the tangent line to the circle.

Question 3. A particle is constrained to move in a circular path of radius r = 6m. Its velocity varies with time according to the relation v = 2t (m/s). Determine its

1. Centripetal acceleration,
2. Tangential acceleration,
3. Instantaneous acceleration at
   1. t = 0 sec. and
   2. t = 3 sec.

Answer:

At = 0,

v = 0, Thus a_r = 0

but \(\frac{d v}{d t}=2\) thus at = 2 m/s2 and a = \(\sqrt{a_t^2+a_r^2}=2 \mathrm{~m} / \mathrm{s}^2\)

At t = 3 sec. v = 6 m/s so \(a_r=\frac{v^2}{r}=\frac{(6)^2}{6}=6 \mathrm{~m} / \mathrm{s}^2\)

and \(a_t=\frac{d v}{d t}=2 \mathrm{~m} / \mathrm{s}^2\) Therefore, \(a=a=\sqrt{2^2+6^2}=\sqrt{40} \mathrm{~m} / \mathrm{s}^2\)

Question 4. The kinetic energy of a particle moving along a circle of radius r depends on the distance covered as K = As² where A is a constant. Find the force acting on the particle as a function of s.
Answer:

According to the given Question

⇒ \(\frac{1}{2} m v^2=A s^2 \text { or } v=s \sqrt{\frac{2 A}{m}}\) ……….(1)

So \(a_{\mathrm{r}}=\frac{\mathrm{v}^2}{\mathrm{r}}=\frac{2 \mathrm{A} \mathrm{s}^2}{\mathrm{mr}}\) ………….(2)

Furthermore as at = \(a_t=\frac{d v}{d t}=\frac{d v}{d s} \cdot \frac{d s}{d t}=v \frac{d v}{d s}\) …………(3)

from eqn. (1), ⇒ ……….. (4)

Substitute values from eqn(1) and eqn(4) in eqn(3)

⇒ \(a_t=\left[s \sqrt{\frac{2 A}{m}}\right]\left[\sqrt{\frac{2 A}{m}}\right]=\frac{2 A s}{m}\)

so \(a=\sqrt{a_r^2+a_t^2}=\sqrt{\left[\frac{2 A s^2}{m r}\right]^2+\left[\frac{2 A s}{m}\right]^2}\)

i.e. \(\mathrm{a}=\frac{2 \mathrm{As}}{\mathrm{m}} \sqrt{1+[\mathrm{s} / \mathrm{r}]^2}\)

so \(\mathrm{F}=\mathrm{ma}=2 \mathrm{As} \sqrt{1+[\mathrm{s} / \mathrm{r}]^2}\)

Question 5. A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration varies with time t as ac= k²rt², where k is a constant. Determine the power delivered to a particle by the forces acting on it.
Answer:

If v is instantaneous velocity, centripetal acceleration ac = \(a_c=\frac{v^2}{r} \Rightarrow=\frac{v^2}{r} k^2 r^2 \Rightarrow v=k r t\)

In circular motion work done by centripetal force is always zero and work is done only by tangential force.

∵ Tangent acceleration \(a_t=\frac{d v}{d t}=\frac{d}{d t}(k r t)=k r\)

∴ Tangential force Ft= mat= mkr

Power P = \(F_t v=(m k r)(k r t)=m k^2 r^2 t\)

 

NEET Physics Class 11 Chapter 10 Mathematical Tools

Mathematics is the language of physics. It becomes easier to describe, understand, and apply the physical principles if one has a good knowledge of mathematics.

Mathematical Tools

To solve the problems of physics Newton made significant contributions to Mathematics by inventing differentiation and integration.

Appropriate Choice Of Tool Is Very Important

What are mathematical tools?

This chapter includes all the necessary mathematics knowledge that we require to possess to study physics efficiently Importance of mathematical tools. This is the most important chapter of physics as it will be repeatedly used in all the upcoming chapters.

Importance of Mathematical Tools in Physics: This chapter is the foundation of physics that needs to be studied. Here we will learn about the mathematics that will be involved in Physics.

NEET Physics Class 11 Chapter 10 Function

The function is a rule of relationship between two variables in which one is assumed to be dependent and the other independent variable, for example,

For example: The temperatures at which water boils depend on the elevation above sea level (the boiling point drops as you ascend). Here elevation above sea level is the independent and temperature is the dependent variable

For example:  The interest paid on a cash investment depends on the length of time the investment is held. Here time is the independent and interest is the dependent variable.

• In each of the above examples, the value of one variable quantity (dependent variable), which we might call y, depends on the value of another variable quantity (independent variable), which we might call x.
• Since the value of y is completely determined by the value of x, we say that y is a function of x and represent it mathematically as y = f(x).

Here f represents the function, x is the independent variable & y is the dependent variable.

• All possible values of independent variables (x) are called the domain of a function.
• All possible values of a dependent variable (y) are called a range of functions.
• Think of a function f as a kind of machine that produces an output value f(x) in its range whenever we feed it an input value x from its domain.

When we study circles, we usually call the area A and the radius r. Since area depends on radius, we say that A is a function of r, A = f(r). The equation A = πr² is a rule that tells how to calculate a unique (single) output value of A for each possible input value of the radius r.

A = f(r) = πr². (Here, the rule of relationship that describes the function may be square and multiplied by π).

If r = 1 A = π ; if r = 2A = 4π ; if r = 3 A = 9π

The set of all possible input values for the radius is called the domain of the function. The set of all output values of the area is the range of the function.

We usually denote functions in one of the two ways:

1. A formula such as y = x² uses a dependent variable, y, to denote the value of the function.
2. By giving a formula such as f(x) = x² that defines a function symbol f to name the function. Strictly speaking, we should call the function f and not f(x),

y = sin x. Here the function is sine, and x is the independent variable.

Question 1. The volume V of a ball (solid sphere) of radius r is given by the function V(r) = (4/3)π(r³). What is the volume of a ball with a radius of 3m?

Answer:

V= 4/3π(3)³ = 36π m³.

Function Of A Function:

Suppose we are given 2 functions, f (x) = x² and g (x) = x + 1

If we are required to find f(g(x))

i.e., value of f(x) at x = g(x)

f (g(x)) = (x + 1)² [put g (x) in place of x in f(x)]

g (f (x)) = x² + 1 [put f (x) in place of x in g (x)]

Question 2. Suppose that the function F is defined for all real numbers r by the formula F(r) = 2(r – 1) + 3. Evaluate F at the input values 0, 2, x + 2, and F(2).

Answer:

Given formula F(r) = 2(r – 1) + 3.

In each case, we substitute the given input value for r into the formula for F :

F(0) = 2(0 – 1) + 3 = – 2 + 3 = 1 ;

f(2) = 2(2 – 1) + 3 = 2(1) + 3 = 2 + 3 = 5

F(x + 2) = 2(x +2 – 1) + 3 = 2x + 5 ;

F(F(2)) = F(5) = 2(5 – 1) + 3 = 2(4) + 3 = 8 + 3 = 11.

Question 3. A function ƒ(x) is defined as ƒ(x) = x² + 3, Find ƒ(0), ƒ(1), ƒ(x²), ƒ(x+1) and ƒ(ƒ(1)).

Answer:

f(x) = x² + 3

We have to find f(0), f(1), f(x²), f(x + 2) and f(f(1))

ƒ(0) = 0² + 3 = 3 ;

ƒ(1)= 1² + 3 = 4 ;

ƒ(x²) = (x²)²+3 = x⁴+3

ƒ(x+1) = (x + 1)² + 3 = x² + 2x +1 + 3 = x² + 2x + 4;

ƒ(ƒ(1)) = ƒ(4)= 4²+3 = 16 + 3 = 19

Question 4. If function F is defined for all real numbers x by the formula F(x) = x². Evaluate F at the input values 0, 2, x + 2, and F

Answer:

F(x) = x²

We have to find f(0), f(2), f(x + 2) and f(f(2)).

F(0) = 0 ;

F(2)= 2² = 4 ;

F(x+2) = (x+2)² = x² + 8x + 4;

F(F(2)) = F(4) = 4² = 16

NEET Physics Class 11 Chapter 10 Mathematical Tools – Geometry

Formulae For Determination Of Area:

1. Area of a square = (side)²
2. Area of rectangle = length × breadth
3. Area of a triangle = \(\frac{1}{2}\) (base × height)
4. Area of trapezoid = \(\frac{1}{2}\)(distance between parallel side) × (sum of parallel side)
5. Area enclosed by a circle = π r²(r = radius)
6. Surface area of a sphere = 4πr²(r = radius)
7. Area of a parallelogram = base × height
8. Area of curved surface of cylinder = 2πr l(r = radius and l = length)
9. Area of ellipse = π ab (a and b are semi-major and semi-minor axes respectively)
10. Surface area of a cube = 6(side)²
11. Total surface area of cone = \(\pi r^2+\pi r \ell\) where \(\pi r \ell=\pi r \sqrt{r^2+h^2}\)lateral area

Formulae For Determination Of Volume:

1. Volume of rectangular slab = length × breadth × height = abt
2. Volume of a cube = (side)³
3. Volume of a sphere = \(\frac{4}{3} \pi r^3 \ell\) (r = radius)
4. Volume of a cylinder = \(\pi r^2 \lambda \ell\) (r = radius and l is length)
5. Volume of a cone = \(\frac{1}{3} \pi r^2 h\) (r = radius and h is height) 3

Note:

⇒ \(\pi=\frac{22}{7}\)= 3.14, π² = 9.8776 ≈10

and \(\frac{1}{\pi}\)

= 0.3182 ≈0.3

NEET Physics Class 11 Chapter 10 Mathematical Tools – Differentiation

Finite Difference

The finite difference between two values of a physical quantity is represented by Δ notation. For example:

The difference in two values of y is written as Δy as given in the table below.

Infinitely Small Difference

The infinitely small difference means a very small difference. And this difference is represented by ‘d’ notation instead of ‘Δ’.

For example, an infinitely small difference in the values of y is written as ‘dy’

if y₂= 100 and y₁= 99.99999999……..

then dy = 0.000000……………….00001

NEET Physics Class 11 Chapter 10 Definition Of Differentiation

Another name for differentiation is derivative. Suppose y is a function of x or y = f(x) Differentiation of y concerning x is denoted by the symbol f’(x)

where F(x) = \(\frac{d y}{d x}\)

dx is a very small change in x and dy corresponds very small change in y.

Notation: There are many ways to denote the derivative of a function y = f(x). Besides f ’(x), the most common notations are these:

NEET Physics Class 11 Chapter 10 Mathematical Tools – Slope Of A Line

It is the tan of the angle made by a line with the positive direction of the x-axis, measured in an anticlockwise direction.

Slope = tan θ ( In 1^st quadrant tan θ is +ve and 2nd quadrant tan θ is –ve )

In Figure – 1 slope is positive In Figure – 2 slope is negative θ < 90° (1st quadrant) θ > 90° (2nd quadrant)

NEET Physics Class 11 Chapter 10 Mathematical Tools – Average Rates Of Change

Given an arbitrary function y = f(x) we calculate the average rate of change of y concerning x over the interval (x, x + Δx) by dividing the change in the value of y, i.e. Δy = f(x + Δx) – f(x), by the length of interval Δx over which the change occurred.

The average rate of change of y concerning x over the interval

[x, x + Δx] = \(\frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}\)

Geometrically, \(\frac{\Delta y}{\Delta x}=\frac{Q R}{P R}\)= tan θ = Slope of the line PQ

therefore we can say that the average rate of change of y concerning x is equal to the slope of the line joining P and Q.

In triangle QPR tanθ = \(\frac{\Delta y}{\Delta x}\)

The Derivative Of A Function

We know that, average rate of change of y w.r.t. x is \(\frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}\)

If the limit of this ratio exists as Δx → 0, then it is called the derivative of the given function f(x) and is denoted as

\(f^{\prime}(x)=\frac{d y}{d x}=\Delta x \rightarrow 0 \frac{\lim _x(x+\Delta x)-f(x)}{\Delta x}\)

 

NEET Physics Class 11 Chapter 10 Mathematical Tools – Geometrical Meaning Of Differentiation

The geometrical meaning of differentiation is very useful in the analysis of graphs in physics. To understand the geometrical meaning of derivatives we should know the secant and tangent to a curve

Secant And Tangent To A Curve

Secant: A secant to a curve is a straight line, which intersects the curve at any two points.

Tangent: A tangent is a straight line, which touches the curve at a particular point. Tangent is a limiting case of secant which intersects the curve at two overlapping points.

1. In the figure-1 shown, if the value of Δx is gradually reduced then the point Q will move nearer to the point P.
2. If the process is continuously repeated (Figure – 2) value of Δx will be infinitely small and the secant PQ to the given curve will become a tangent at point P.

Therefore \(_{\Delta x \rightarrow 0}\left(\frac{\Delta y}{\Delta x}\right)=\frac{d y}{d x}=\tan \theta\)= tan θ

we can say that differentiation of y with respect to x, i.e. \(\left(\frac{d y}{d x}\right)\) is equal to slope of the tangent at point P (x, y) or tanθ = \(\frac{d y}{d x}\)

(the average rate of change of y from x to x + Δx is identical to the slope of the second PQ.)

NEET Physics Class 11 Chapter 10 Integration

In mathematics, for each mathematical operation, there has been defined an inverse operation. For example- The inverse operation of addition is subtraction, the inverse operation of multiplication is division and the inverse operation of a square is a square root. Similarly, there is an inverse operation for differentiation which is known as integration

Antiderivatives Or Indefinite Integrals Definitions:

A function F(x) is an antiderivative of a function f(x) if F´(x) = f(x) for all x in the domain of f. The set of all antiderivatives of f is the indefinite integral of f concerning x, denoted by

The symbol ∫is an integral sign. The function f is the integrand of the integral and x is the variable of integration.

For example f(x) = x³ then f′(x) = 3x²

So the integral of 3x² is x³

Similarly if f(x) = x³ + 4 then f′(x) = 3x²

So the integral of 3x² is x³ + 4

there for general integral of 3x² is x³ + c where c is a constant

One antiderivative F of a function f, the other antiderivatives of f differ from F by a constant. We indicate this in integral notation in the following way:

∫f(x)dx F(x) C. = + ∫………….(1)

The constant C is the constant of integration or arbitrary constant, Equation (1) is read, “The indefinite integral of f concerning x is F(x) + C.” When we find F(x)+ C, we say that we have integrated f and evaluated the integral.

Example. Evaluate ∫2x dx.

Answer:

The formula x² + C generates all the antiderivatives of the function 2x. The function x² + 1, x² – π, and \(x^2+\sqrt{2}\) are all antiderivatives of the function 2x, as you can check by differentiation.

Many of the indefinite integrals needed in scientific work are found by reversing derivative formulas.

NEET Physics Class 11 Chapter 10 Integral Formulas

Indefinite Integral

⇒ \(\int x^n d x=\frac{x^{n+1}}{n+1}+C, n \neq-1, n \text { rational }\)

⇒ \(\int d x=\int 1 d x=x+C \text { (special case) }\)

⇒ \(\int \sin (A x+B) d x=\frac{-\cos (A x+B)}{A}+C\)

⇒ \(\int \cos k x d x=\frac{\sin k x}{k}+C\)

Reversed Derivation Formula

⇒ \(\frac{d}{d x}\left(\frac{x^{n+1}}{n+1}\right)=x^n\)

⇒ \(\frac{d}{d x}(x)=1\)

⇒ \(\frac{d}{d x}\left(-\frac{\cos k x}{k}\right)=\sin k x\)

⇒ \(\frac{d}{d x}\left(\frac{\sin k x}{k}\right)=\cos k x\)

Question 1. Examples based on the above formulas:

1. \(\int x^5 d x=\frac{x^6}{6}+C\) Formula 1 with n = 5
2. \(\int \frac{1}{\sqrt{x}} d x=\int x^{-1 / 2} d x=2 x^{1 / 2}+C=2 \sqrt{x}+C\) Formula 1 with n = –1/2
3. \(\int \sin 2 x d x=\frac{-\cos 2 x}{2}+C\) Formula 2 with k = 2
4. \(\int \cos \frac{x}{2} d x=\int \cos \frac{1}{2} x d x=\frac{\sin (1 / 2) x}{1 / 2}+C=2 \sin \frac{x}{2}+C\) Formula 3 with k = 1/2

Question 2. Right: ∫x cosx dx = x sin x + cos x + C

Reason: The derivative of the right-hand side is the integrand:

Check: \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x sin x + cos x + C) = x cos x + sin x – sin x + 0 = x cos x

Wrong: ∫x cosx dx = x sin x + C

Reason: The derivative of the right-hand side is not the integrand:

Check: \(\frac{\mathrm{d}}{\mathrm{dx}}\)(x sin x + C) = x cos x + sin x + 0 x cos x.

NEET Physics Class 11 Chapter 1 Calorimetry And Thermal Expansion Heat

The energy that is being transferred between two bodies or between adjacent parts of a body as a result of temperature difference is called heat. Thus, heat is a form of energy.

• It is energy in transit whenever temperature differences exist. Once it is transferred, it becomes the internal energy of the receiving body.
• It should be clearly understood that the word “heat” is meaningful only as long as the energy is being transferred.
• Thus, expressions like “heat in a body” or “heat of a body” are meaningless. Heat transfer from a body at high temperature to low temperature.

When we say that a body is heated it means that its molecules begin to move with greater kinetic energy. So, it is the energy of molecular motions.

S.Ι. unit of heat energy is joule (J). Another practical unit of heat energy is calorie (cal).

1 calorie = 4.18 joules.

1 calorie: The amount of heat needed to increase the temperature of 1 gm of water from 14.5 to 15.5 ºC at one atmospheric pressure is 1 calorie.

Mechanical Equivalent of Heat

In the early days, the heat was not recognized as a form of energy. The heat was supposed to be something needed to raise the temperature of a body or to change its phase. The calorie was defined as the unit of heat.

• A number of experiments were performed to show that the temperature may also be increased by doing mechanical work on the system.
• These experiments established that heat is equivalent to mechanical energy and measured how much mechanical energy is equivalent to a calorie. If mechanical work W produces the same temperature change as heat H, we write,

W = JH

where J is called the mechanical equivalent of heat. J is expressed in joule/calorie. The value of J gives how many joules of mechanical work is needed to raise the temperature of 1 g of water by 1°C. It is a conversion factor and not a physical quantity.

Question 1. What is the change in potential energy (in calories) of a 10 kg mass when it falls through 10 m?

Solution: Change in potential energy

Mass = 10 kg

Gravitation = 10

Height = 10 m

ΔU = mgh = 10 × 10 × 10 = 1000 J

⇒ \(\frac{1000}{4.186} \text { cal }\)

NEET Physics Class 11 Chapter 1 Specific Heat

The specific heat of a substance is equal to the heat gained or released by that substance to raise or fall its temperature by 1ºC for a unit mass of a substance.

When a body is heated, it gains heat. On the other hand, heat is lost when the body is cooled. The gain or loss of heat is directly proportional to:

1. The mass of the body ΔQ ∝ m
2. Rise or fall of temperature of the body ΔQ ∝ Δ T

ΔQ ∝ m Δ T or ΔQ = m s Δ T

or dQ = m s d T or Q = m ∫s d T.

where s is a constant and is known as the specific heat of the body s = \(\frac{Q}{m \Delta T}\) joule/kg-kelvin and C.G.S. unit is cal./gm °C.

Qm TΔ. S.Ι. unit of s is

Specific heat of water : S = 4200 J/kgºC = 1000 cal/kgºC = 1 Kcal/kgºC = 1 cal/gmºC

Specific heat of steam = half of specific heat of water = specific heat of ice

Question 1. Calculate the heat required to increase the temperate of 1 kg water by 20ºC

Solution :

Heat required = ΔQ = msΔθ

∵ S = 1 cal/gmºC = 1 Kcal/kgºC

= 1 × 20 = 20 Kcal.

Heat capacity or Thermal capacity: The heat capacity of a body is defined as the amount of heat required to raise the temperature of that body by 1°. If ‘m’ is the mass and ‘s’ the specific heat of the body, then

Heat capacity = m s.

Units of heat capacity in the CGS system is, cal ºC^-1; the SI unit is, JK^-1

Important Points:

1. We know, s = \(\frac{Q}{m \Delta T}\), if the substance undergoes the change of state which occurs at constant temperature (Isothermal ΔT = 0), then s = Q/0 = ∞. Thus the specific heat of a substance when it melts or boils at constant temperature is infinite.
2. If the temperature of the substance changes without the transfer of heat adiabatic (Q = 0) then s = \(\frac{Q}{m \Delta T}\) = 0. Thus when liquid in the thermos flask is shaken, its temperature increases without the transfer of heat, and hence the specific heat of liquid in the thermos flask is zero.
3. To raise the temperature of saturated water vapors, heat (Q) is withdrawn. Hence, the specific heat of saturated water vapors is negative. (This is for your information only and not in the course)
4. The slight variation of specific heat of water with temperature is shown in the graph at 1-atmosphere pressure. Its variation is less than 1% over the interval from 0 to 100ºC.

Relation between Specific heat and Water equivalent: It is the amount of water that requires the same amount of heat for the same temperature rise as that of the object

ms ΔT = m_W S_W ΔT ⇒ mW = \(\frac{\mathrm{ms}}{\mathrm{s}_{\mathrm{W}}}\)

In calorie s_W = 1

∴ m_W = ms

m_W is also represented by W

So, W = ms.

Phase change: Heat required for the change of phase or state,

Q = mL, L = latent heat.

Latent heat (L): The heat supplied to a substance that changes its state at constant temperature is called latent heat of the body.

Latent heat of Fusion (L_f ): The heat supplied to a substance which changes it from solid to liquid state at its melting point and 1 atm. pressure is called latent heat of fusion. The latent heat of the fusion of ice is 80 kcal/kg

Latent heat of vaporization (L_v): The heat supplied to a substance that changes it from liquid to vapor state at its boiling point and 1 atm. pressure is called latent heat of vaporization. The latent heat of vaporization of water is 540 kcal kg^-1.

If in question latent heat of ice or steam is not mentioned and to solve the problem, it is required to assume them, we should consider the following values.

Latent heat of ice: L = 80 cal/gm = 80 Kcal/kg = 4200 × 80 J/kg

Latent heat of steam: L = 540 cal/gm = 540 Kcal/kg = 4200 × 540 J/kg

The given behavior of a solid substance when it is continuously heated is shown. Various parts show.

OA − solid state, AB − solid + liquid state (Phase change)

BC − liquid state, CD − liquid + vapor state (Phase change)

DE − vapor state

When there is no change of state for example., OA, BC, DE

Slope, \(\frac{\Delta T}{\Delta \mathrm{Q}}=\frac{1}{\mathrm{~ms}} \quad \Rightarrow \quad \frac{\Delta \mathrm{T}}{\Delta \mathrm{Q}} \propto \frac{1}{\mathrm{~s}}\)

The mass (m) of a substance is constant. So, the slope of T – Q graph is inversely proportional to specific heat. In the given diagram.

(slope) OA > (slope) DE

then (s)_OA < (s)_DE

when there is a change of state for example., AB and CD

ΔQ = mL

If (length of AB) > (length of CD)

then (latent heat of AB) > (latent heat of CD)

Question 2. Find the amount of heat released when 1 kg steam at 200ºC is converted into –20ºC ice.

Answer:

Heat required ΔQ = heat release to convert steam at 200 ºC into 100ºC steam + heat release to convert 100ºC steam into 100ºC water + heat release to convert 100º water into 0ºC water + heat release to convert 0 ºC water into 0ºC ice + 0ºC ice converted to – 20ºC ice.

⇒ \(\Delta Q=1 \times \frac{1}{2} \times 100+540 \times 1+1 \times 1 \times 100+1 \times 80+1 \times \frac{1}{2}=780(\mathrm{Kcal})\)

NEET Physics Class 11 Chapter 1 Superficial Or Areal Expansion

When a solid is heated and its area increases, then the thermal expansion is called superficial or areal expansion. Consider a solid plate of area A0. When it is heated, the change in the area of the plate is directly proportional to the original area A₀ and the change in temperature ΔT.

dA = βA₀dT or ΔA = β A₀ Δ T; β is called real expansion

β = \(\frac{\Delta \mathrm{A}}{\mathrm{A}_0 \Delta \mathrm{T}}\) Unit of β is ºC^-1 or K^-1.

A = A₀(1 + β Δ T)

Where A is an area of the plate after heating.

it follows that β = 2α.

Question 1. A plane lamina has an area of 2m² at 10ºC then what is its area at 110ºC Its superficial expansion is 2 × 10^-5/C

Answer:

Area of plane lamina = 2m²

Superficial expansion = 2 × 10^-5/C

A = A₀ ( 1 + β Δ θ ) = 2 {1 + 2 × 10⁵ × (110 – 10)}

= 2 × {1 + 2 × 10^-3} m²

NEET Physics Class 11 Chapter 1 Volume Or Cubical Expansion

When a solid is heated and its volume increases, then the expansion is called volume expansion or cubical expansion. Let us consider a solid or liquid whose original volume is V₀. When it is heated to a new volume, then the change ΔV

dV = γV₀dT or ΔV = γ V₀ Δ T

⇒ \(\gamma=\frac{\Delta V}{V_0 \Delta T}\) Unit of γ is ºC^-1 or K^-1.

V = V₀(1 + γ Δ T)

where V is the volume of the body after heating.

It can be shown easily that γ = 3α for isotropic solids.

Question 1. The volume of the glass vessel is 1000 cc at 20°C. What volume of mercury should be poured into it at this temperature so that the volume of the remaining space does not change with temperature? The coefficient of cubical expansion of mercury and glass is 1.8 × 10^-4/°C and 9.0 × 10^-6/°C respectively.

Answer:

Let the volume of the glass vessel at 20ºC be Vgand volume of mercury at 20ºC be Vm so the volume of remaining space is = V_g– V_m

It is given constant so that

V_g– V_m= V_g’ – V’_m

where V₀‘ and V_m‘ are final volumes.

V_g– V_m= V_g{1 + γ_g Δθ} – V_m{1 + γ_Hg Δθ} ⇒ V_gγ_g= V_mγ_Hg

⇒ V_m = \(V_m=\frac{100 \times 9 \times 10^{-6}}{1.8 \times 10^{-4}} V_m=50 \mathrm{cc}\)

Relation Between α, β And γ

1. For isotropic solids: α : β : γ = 1 : 2 : 3 or \(\frac{\alpha}{1}=\frac{\beta}{2}=\frac{\gamma}{3}\)
2. For non-isotropic solid β = α₁+ α₂ and γ = α₁+ α₂+ α₃. Here α₁, α_2, and α₃ are coefficients of linear expansion in the X, Y, and Z direction.

Question 1. If the percentage change in length is 1% with the change in temperature of a cuboid object (l × 2l × 3l) then what is the percentage change in its area and volume?

Answer:

Percentage change in length with change in temperature = %l

⇒ \(\frac{\Delta \ell}{\ell} \times 100=\alpha \Delta \theta \times 100=1\)

change in area ⇒ % A = \(\frac{\Delta \mathrm{A}}{\mathrm{A}} \times 100=\beta \Delta \theta \times 100\)

⇒ 2 (α Δ θ × 100) % A = 2%

Change in volume

% V = \(\) × 100 = V Δ θ × 100 = 3 (α Δ θ × 100) % V = 3 %

NEET Physics Class 11 Chapter 1 Variation Of Density With Temperature

As we know mass = volume × density.

The mass of a substance does not change with a change in temperature so with an increase in temperature, volume increases so density decreases and vice-versa.

d = \(\frac{d_0}{(1+\gamma \Delta T)}\)

For solids values of γ are generally small so we can write d = d₀(1 − γ ΔT) (using binomial expansion).

Note:

1. γ for liquids is of the order of 10−3.
2. Anomalous expansion of water:

For water density increases from 0 ºC to 4 ºC so γ is negative and for 4 ºC to higher temperature γ is positive. At 4 ºC density is maximum. This anomalous behavior of water is due to the presence of three types of molecules i.e. H₂O, (H₂O)₂and (H₂O)₃ having different volumes/masses at different temperatures.

This anomalous behavior of water causes ice to form first at the surface of a lake in cold weather. As winter approaches, the water temperature decreases initially at the surface. The water there sinks because of its increased density.

Consequently, the surface reaches 00C first and the lake becomes covered with ice. Aquatic life is able to survive the cold winter as the lake bottom remains unfrozen at a temperature of about 4°C.

Question 1. The densities of wood and benzene at 0°C are 880 kg/m³ and 900 kg/m³ respectively. The coefficients of volume expansion are 1.2 × 10^-3/°C for wood and 1.5 × 10^-3/°C for benzene. At what temperature will a piece of wood just sink in benzene?

Answer:

At just sink gravitation force = up thrust force

⇒ mg = F_B

⇒ Vp₁g = Vp₂g

⇒ p₁= p₂

⇒ \(\frac{880}{1+1.2 \times 10^{-3} \theta}=\frac{900}{1+1.5 \times 10^{-3} \theta}\)

⇒ θ = 83º C

NEET Physics Class 11 Chapter 1 Apparent Expansion Of A Liquid In A Container

Initially container was full. When temperature increases by ΔT,

volume of liquid V_L= V₀(1 + γ_L Δ T)

volume of container V_C= V₀(1 + γ_C Δ T)

So overflow volume of liquid relative to the container

ΔV = V_L− V_C ΔV = V₀(γ_L− γ_C) ΔT

So, the coefficient of apparent expansion of liquid w.r.t. Container

γ_apparent = γ_L− γ_C.

In case of expansion of liquid + container system:

if γ_L> γ_C ⎯→ level of liquid rise

if γ_L< γ_C ⎯→ level of liquid fall

Increase in height of liquid level in a tube when the bulb was initially completely filled

h = \(h=\frac{\text { volume of liquid }}{\text { area of tube }}=\frac{V_0\left(1+\gamma_L \Delta T\right)}{A_0\left(1+2 \alpha_S \Delta T\right)}=h_0\left\{1+\left(\gamma_L-2 \alpha_S\right) \Delta T\right\}\)

h = h₀{ 1 + ( γ_L– 2α_S) ΔT}

where h₀= original height of liquid in a container

α_S= linear coefficient of expansion of container.

Question 1. A glass vessel of volume 100 cm³ is filled with mercury and is heated from 25°C to 75°C. What volume of mercury will overflow? The coefficient of linear expansion of glass = 1.8 × 10^-6/°C and the coefficient of volume expansion of mercury is 1.8 × 10^-4/°C.

Answer:

ΔV = V₀(γ_L– γ_C) ΔT = 100 × {1.8 × 10^-4 – 3 × 1.8 × 10^-6} × 50

ΔV = 0.87 cm³

Variation Of Force Of Buoyancy With Temperature

If the body is submerged completely inside the liquid

For solid, Buoyancy force F_B= V₀ dL g

V₀= Volume of the solid inside the liquid,

d_L= density of liquid

Volume of body after increase its temperature V = V₀[1 + γ_S Δθ],

Density of body after increase its temperature d′_L = \(\frac{d_{\mathrm{L}}}{\left[1+\gamma_{\mathrm{L}} \Delta \theta\right]}\)

Buoyancy force of body after increasing its temperature, \(F_B^{\prime}=V d_L^{\prime} g \quad \Rightarrow \frac{F_B^{\prime}}{F_B}=\frac{\left[1+\gamma_S \Delta \theta\right]}{\left[1+\gamma_L \Delta \theta\right]}\)

If γ_S< γ_L then F′_B< F_B

(Buoyant force decreases) or apparent weight of the body in liquid gets increased

[W − F′_B> W − F_B].

Question 1. A body is floating on the liquid if we increase temperature then what changes occur in buoyancy force? (Assume the body is always in floating condition)

Answer:

The body is in equilibrium

So, mg = B (Boyant Force)

The gravitational force does not change with a change in temperature. So buoyancy force remains constant.

By increasing temperature density of the liquid decreases so the volume of the body inside the liquid increases to keep the Buoyance force constant and equal to the gravitational force)

Question 1. In the previous question discuss the case when the body move downward, upwards and remains at same position when we increases temperature.

Answer:

Let f = fraction of volume of body submerged in liquid

⇒ \(\mathrm{f}=\frac{\text { volume of body submerged in liquid }}{\text { total volume of body }}\)

⇒ \(f_1=\frac{v_1}{v_0} \text { at } \theta_1{ }^{\circ} \mathrm{C}\)

⇒ \(f_2=\frac{v_2}{v_0\left(1+3 \alpha_S \Delta \theta\right)}\) at θ2ºC

for equilibrium mg = B = v1d1g = v2d2g. d

so v₂ = \(\frac{v_1 d_1}{d_2}\)

⇒ \(d_2=\frac{d_1}{1+\gamma_L \Delta \theta}=v_1\left(1+\gamma_L \Delta \theta\right)\)

⇒ \(f_2=\frac{v_1\left(1+\gamma_L \Delta \theta\right)}{v_0\left(1+3 \alpha_s \Delta \theta\right)}\)

where Δθ = θ₂– θ₁

Case 1: Body move downward if f₂> f₁

means γ_L> 3α_S

Case 2: Body moves upwards if f₂< f₁

means γ_L< 3α_S

Case 3: The body remains in the same position

if f₂= f₁

means γ_L= 3α_S

Bimetallic Strip

Two strips of different metals are welded together to form a bimetallic strip, when heated uniformly it bends in the form of an arc, and the metal with a greater coefficient of linear expansion lies on the convex side. The radius of the arc is thus formed by bimetal:

⇒ \(\ell_0\left(1+\alpha_1 \Delta \theta\right)=\left(\mathrm{R}-\frac{\mathrm{d}}{2}\right) \theta\)

⇒ \(\ell_0\left(1+\alpha_2 \Delta \theta\right)=\left(R+\frac{d}{2}\right) \theta\)

⇒ \(\frac{1+\alpha_2 \Delta \theta}{1+\alpha_1 \Delta \theta}=\frac{R+\frac{d}{2}}{R-\frac{d}{2}}\)

⇒ \(R=\frac{d}{\left(\alpha_2-\alpha_1\right) \Delta \theta}\)

⇒ \(\Delta \theta=\text { change in temperature }=\theta_2-\theta_1\)

• A bimetallic strip, consisting of a strip of brass and a strip of steel welded together as shown in figure (3). At temperature T₀ in figure and figure  (2). The strip bends as shown at temperatures above the reference temperature.
• Below the reference temperature, the strip bends the other way. Many thermostats operate on this principle, making and breaking an electrical constant as the temperature rises and falls as shown in Figure (2).

Applications Of Thermal Expansion

1. A small gap is left between two iron rails of the railway.
2. Iron rings are slipped on the wooden wheels by heating the iron rings
3. The stopper of a glass bottle jammed in its neck can be taken out by heating the neck.
4. The pendulum of a clock is made of invar [an alloy of zinc and copper].

Temperature

Temperature may be defined as the degree of hotness or coldness of a body. Heat energy flows from a body at a higher temperature to that at a lower temperature until their temperatures become equal. At this stage, the bodies are said to be in thermal equilibrium.

Measurement of Temperature: The branch of thermodynamics which deals with the measurement of temperature is called thermometry.

• A thermometer is a device used to measure the temperature of a body.
• The substances like liquids and gases that are used in the thermometer are called thermometric substances. Suppose a physical quantity X varies linearly with temperature then

⇒ \(X_t=X_0(1+\alpha \mathrm{t}) \Rightarrow \quad t=\frac{X_t-X_0}{X_0 \alpha}\)

α is constant and X0 is the value of X at the reference temperature set at 0º. In the case of absolute scale

⇒ \(\frac{X_T}{X_0}=\frac{T}{T_0} \quad \Rightarrow \quad T=T_0 \frac{X_T}{X_0}=273.16 \frac{X_T}{X_0} K\)

Different Scales of Temperature: A thermometer can be graduated into the following scales.

1. The Centigrade or Celsius scale (ºC)
2. The Fahrenheit scale (ºF)
3. The Reaumer scale (ºR)
4. Kelvin scale of temperature (K)

Comparison Between Different Temperature Scales:

The formula for the conversion between different temperature scales is:

⇒ \(\frac{\mathrm{K}-273}{100}=\frac{\mathrm{C}}{100}=\frac{\mathrm{F}-32}{180}=\frac{\mathrm{R}}{80}\)

A general formula for the conversion of temperature from one scale to another:

⇒ \(=\frac{\text { Temp. on other scale }\left(\mathrm{S}_1\right) \text {-Lower fixed point }\left(\mathrm{S}_1\right)}{\text { Upper fixed point }\left(\mathrm{S}_2\right) \text {-Lower fixed point }\left(\mathrm{S}_1\right)}\)

⇒ \(=\frac{\text { Temp. on other scale }\left(\mathrm{S}_2\right) \text {-Lower fixed point }\left(\mathrm{S}_2\right)}{\text { Upper fixed point }\left(\mathrm{S}_2\right) \text {-Lower fixed point }\left(\mathrm{S}_2\right)}\)

Thermometers: Thermometers are devices that are used to measure temperatures. All thermometers are based on the principle that some physical property of a system changes as the system temperature changes.

Required properties of good thermometric substance.

1. Non-sticky (absence of adhesive force)
2. Low melting point (in comparison with room temperature)
3. High boiling temperature
4. The coefficient of volumetric expansion should be high (to increase accuracy in measurement).
5. Heat capacity should be low.
6. Conductivity should be high Mercury (Hg) suitably exhibits the above properties.

Types Of Thermometers

The Constant-volume Gas Thermometer: The standard thermometer, against which all other thermometers are calibrated, is based on the pressure of a gas in a fixed volume.

The figure shows such a constant volume gas thermometer; it consists of a gas-filled bulb connected by a tube to a mercury manometer.

T = \((273.16 \mathrm{~K})\left(\lim _{{gas} \rightarrow 0} \frac{p}{p_3}\right)\)

P = Pressure at the temperature being measured,

P₃ = pressure when the bulb is in a triple point cell.

Question 1. The readings of a thermometer at 0ºC and 100ºC are 50 cm and 75 cm of mercury column respectively. Find the temperature at which its reading is 80 cm of mercury column.

Answer:

By using formula

⇒ \(\frac{80-50}{75-50}=\frac{T-0}{100-0}\)

⇒ T = 120ºC

Question 2. A bullet of mass 10 gm in moving with speed 400m/s. Find its kinetic energy in calories.

Answer:

⇒ \(\Delta k=\frac{1}{2} \times \frac{10}{1000} \times 400 \times 400=800\)

⇒ \(\frac{800}{4.2}=\mathrm{Cal}\) = 191.11 Cal.

Question 3. Calculate the amount of heat required to convert 1 kg steam from 100ºC to 200ºC steam.

Answer:

Heat req. = 1 × \(\frac{1}{2}\)× 100 = 50 kcal

Question 4. Calculate the heat required to raise the temperature of 1 g of water through 1ºC.

Answer:

Heat req. = 1 × 10^-3 × 1 × 1 = 1 × 10^-3 kcal

Question 5. 420 J of energy supplied to 10 g of water will raise its temperature by

Answer:

⇒ \(\frac{420 \times 10^{-3}}{4.20}=10 \times 10^{-3} \times 1 \times \Delta \mathrm{t}=10^{\circ} \mathrm{C}\)

Question 6. The ratio of the densities of the two bodies is 3: 4 and the ratio of specific heats is 4 : 3 . Find the ratio of their thermal capacities for unit volume.

Answer:

⇒ \(\frac{\rho_1}{\rho_2}=\frac{3}{4}, \frac{s_1}{s_2}=\frac{4}{3}\)

⇒ \(\theta=\frac{\mathrm{m} \times \mathrm{s}}{\mathrm{m} / \mathrm{\rho}}\)

⇒ \(\frac{\theta_1}{\theta_2}=\frac{s_1}{s_2} \times \frac{\rho_1}{\rho_2}=1: 1\)

Question 7. Heat is released by 1 kg steam at 150ºC if it converts into 1 kg water at 50ºC.

Answer:

H = 1 ×\(\frac{1}{2}\)× 50 + 1 × 540 + 1 × 1 × 50

= 540+75 = 615

Heat release = 615 Kcal.

Question 8. 200 gm water is filled in calorimetry of negligible heat capacity. It is heated till its temperature is increased by 20ºC. Find the heat supplied to the water.

Answer:

H = 200 × 10^-3 × 1 × 20 = 4 Kcal.

Heat supplied = 4000 cal = 4 Kcal

Question 9. A bullet of mass 5 gm is moving with a speed of 400 m/s. strike a target and energy. Then calculate the rise of the temperature of the bullet. Assuming all the loss in kinetic energy is converted into heat energy of the bullet if its specific heat is. 500J/kgºC.

Answer:

Kinetic energy = 12× 5 × 10^-3 × 400 × 400 = 5 × 10^-3 × 500 × ΔT

ΔT = 160º C

The rise in temperature is 160 ºC

Question 10. 1 kg ice at –10ºC is mixed with 1 kg water at 100ºC. Then find the equilibrium temperature and mixture content.

Answer:

Heat gained by 1 kg ice at – 10ºC to convert into 0ºC ice = 1 × \(\frac{1}{2}\)x10 = 5Kcal = 5000 cal

In the thermal equilibrium

5 + 1 × 80 + 1 × T = 1 × (100 – T)

85 = 100 – 2T ⇒ 2T = 15

⇒ \(\theta=\frac{15}{2}\) = 7.5 ºC, water, entire ice melts.

Question 11. 1 kg ice at –10º is mixed with 1kg water at 50ºC. Then find the equilibrium temperature and mixture content.

Answer:

The heat required by ice at –10ºC to convert it into 0ºC water

1 × \(\frac{1}{2}\)× 10 + 1 × 50 = 55 Kcal

The heat released by 1 kg of water to reduce its temperature from

50ºC to 0ºC = 1 × 1 × 50 = 50 kcal

Heat required > Heat released so, ice will not completely melt. Let m g ice melt then for equilibrium.

1 × \(\frac{1}{2}\) × 10 + 80 m = 50 m ⇒ 80 m = 45 ⇒ m = \(\frac{45}{80}\)

⇒ \(\text { Content of mixture }\left\{\begin{array}{ll}
\text { water } & \left(1+\frac{45}{80}\right) \mathrm{kg} \\
\text { ice } & \left(1-\frac{45}{80}\right) \mathrm{kg}
\end{array}\right\} \text { and temperature is } 0^{\circ} \mathrm{C}\)

Question 12. A small ring having a small gap is shown in the figure on heating what will happen to the size of the gap?

Answer:

The gap will also increase. The reason is the same as in the above example.

Question 13. A concrete slab has a length of 10 m on a winter night when the temperature is 0°C. Find the length of the slab on a summer day when the temperature is 35°C. The coefficient of linear expansion of concrete is 1.0 × 10^-5/°C.

Answer:

l1= 10(1 + 1 × 10^-5× 35) = 10.0035 m

Question 14. A steel rod is clamped at its two ends and rests on a fixed horizontal base. The rod is unstrained at 20°C. Find the longitudinal strain developed in the rod if the temperature rises to 50°C. Coefficient of linear expansion of steel = 1.2 × 10^-5/°C.
Δ αΔt

Answer:

⇒ \(\frac{\Delta \ell}{\ell}=\frac{\ell_0 \alpha \Delta t}{\ell_0}\)= – 3.6 × 10^-4

Question 15. If the rod is initially compressed by Δl length then what is the strain on the rod when the temperature

1. Is increased by Δθ
2. Is decreased by Δθ.

Answer:

1. Strain = \(\frac{\Delta \ell}{\ell}+\alpha \Delta \theta\)
2. Strain = \(\left|\frac{\Delta \ell}{\ell}-\alpha \Delta \theta\right|\)

Question 16. A pendulum clock with having copper rod keeps the correct time at 20°C. It gains 15 seconds per day if cooled to 0°C. Calculate the coefficient of linear expansion of copper.

Answer:

⇒ \(\frac{15}{24 \times 60 \times 60}=\frac{1}{2} \alpha \times 20\)

⇒ \(\alpha=\frac{1}{16 \times 3600}=1.7 \times 10^{-5} /{ }^{\circ} \mathrm{C}\)

Question 17. A meter scale made of steel is calibrated at 20°C to give the correct reading. Find the distance between the 50 cm mark and the 51 cm mark if the scale is used at 10°C. The coefficient of linear expansion of steel is 1.1 × 10^-5/°C.

Answer:

lt = 1 (1 – 1.1 × 10^-5 × 10) = 0.99989 cm

Question 18. A uniform solid brass sphere rotates with an angular speed ω₀ about a diameter. If its temperature is now increased by 100ºC. What will be its new angular speed? (Given α_B = 2.0 × 10^-5 perºC)

1. \(\frac{\omega_0}{1-0.002}\)
2. \(\frac{\omega_0}{1+0.002}\)
3. \(\frac{\omega_0}{1+0.004}\)
4. \(\frac{\omega_0}{1-0.004}\)

Answer:

⇒ \(\mathrm{I}_0 \omega_0=\mathrm{I}_{\mathrm{t}} \omega \mathrm{t}\)

⇒ \(M r_0{ }^2 \omega_0={Mr}_0{ }^2(1+2 \alpha \Delta \mathrm{T}) \omega_{\mathrm{t}}\)

⇒ \(\omega_t=\frac{\omega_0}{1+0.004}\)

Question 19. The volume occupied by a thin-wall brass vessel and the volume of a solid brass sphere is the same and equal to 1,000 cm³ at 0ºC. How much will the volume of the vessel and that of the sphere change upon heating to 20ºC? The coefficient of linear expansion of brass is α = 1.9 × 10^-5.

Answer:

V = V₀(1 + 3α ΔT) = 1.14 cm³ ⇒ 1.14 cm³ for both

Question 20. A thin copper wire of length L increases in length by 1%, when heated from temperature T₁ to T₂. What is the percentage change in area when a thin copper plate having dimensions 2L × L is heated from T₁ to T₂?

1. 1%
2. 3%
3. 4%
4. 2%

Answer: \(\ell_{\mathrm{f}}=\mathrm{L}(1+\alpha \Delta \mathrm{t})=\frac{\mathrm{L}_{\mathrm{f}}}{\mathrm{L}} \times 100=(1+\alpha \Delta t) \times 100=1 \%\)

⇒ \(A=2 L \times L(1+2 \alpha \Delta t)=\frac{A_f}{2 L \times L} \times 100=(1+2 \alpha \Delta t) \times 100=2 \%\)

Question 21. The density of water at 0°C is 0.998 g/cm³ and at 4°C is 1.000 g/cm³. Calculate the average coefficient of volume expansion of water in the temperature range of 0 to 4°C.

Answer: \(d_t=\frac{d_0}{1+\gamma \Delta t}\)

⇒ \(1=\frac{0.998}{1+\gamma \times 4}\)

= – 5 × 10^-4/°C

Question 22. A glass vessel measures exactly 10 cm × 10 cm × 10 cm at 0°C. it is filled completely with mercury at this temperature. When the temperature is raised to 10°C, 1.6 cm³ of mercury overflows. Calculate the coefficient of volume expansion of mercury. Coefficient of linear expansion of glass = 6.5 × 10^-6/°C

Answer:

ΔV = V_Hg – V_V

1.6 = 10³(γ_L× 10 – 10³ × 3 × 6.5 × 10^-6 × 10

γ_L = (1.6 + 0.195) × 10^-4 = 1.795 × 10^-4/°C

Question 23. A metal ball immersed in alcohol weighs W₁ at 00C and W₂ at 500C. The coefficient of cubical expansion of the metal is less than alcohol. Assuming that the density of the metal is large compared to that of the alcohol, find which of W₁ and W₂ is greater.

Answer:

⇒ \(\gamma_{\mathrm{M}}<\gamma_{\ell} \text { so, } \frac{F_{\mathrm{B}}^{\prime}}{F_{\mathrm{B}}}=\frac{\left[1+\gamma_{\mathrm{S}} \Delta \theta\right]}{\left[1+\gamma_{\ell} \Delta \theta\right]} F_{\mathrm{B}}^{\prime}<F_{\mathrm{B}}\)

So Apprent weight increased so, W₂> W₁

Question 24. In the figure strip brass or steel have a higher coefficient of linear expansion.

Answer:

Brass Strip; a strip of higher α is on the convex side.

Question 25. The upper and lower fixed points of a faulty thermometer are 5ºC and 105º C. If the thermometer reads 25ºC, what is the actual temperature?

Answer:

⇒ \(\frac{25-5}{100}=\frac{C-0}{100}\)

C = 20ºC

Question 26. At what temperature is the Fahrenheit scale reading equal to twice Celsius?

Answer:

⇒ \(\frac{F-32}{180}=\frac{C-0}{100}=\)

⇒ \(\frac{C-0}{100}\)

1 × -160 = 9x

x = 160º

Question 27. The temperature of a patient is 40º C. Find the temperature on a Fahrenheit scale.

Answer:

⇒ \(\frac{F-32}{180}\)

⇒ \(\frac{40-0}{100}\)

F = 104ºC