NEET Physics Class 11

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.

 

NEET Physics Class 11 Chapter 8 Slabs In Parallel And Series

Slabs In Series (in steady state)

Consider a composite slab consisting of two materials having different thicknesses L₁ and L₂, different cross-sectional areas A₁ and A_2, and different thermal conductivities K₁ and K₂. The temperature at the outer surface at the ends is maintained at T_H and T_C, and all lateral surfaces are covered by an adiabatic coating.

Let the temperature at the junction be T, since a steady state has been achieved thermal current through each slab will be equal. Then thermal current through the first slab.

⇒ \(\mathrm{i}=\frac{Q}{t}=\frac{T_H-T}{R_1}\)

or \(\mathrm{T}_{\mathrm{H}}-\mathrm{T}=\mathrm{iR} \mathrm{R}_1\)……(1)

and that through the second slab,

⇒ \(\mathrm{i}=\frac{Q}{t}\)

⇒ \(\frac{T-T_C}{R_2}\)

or \(\mathrm{T}-\mathrm{T}_{\mathrm{c}}=\mathrm{i} \mathrm{R}_2\)…..(2)

adding eqn. (1) and eqn (2)

T_H– T_L= (R₁+ R₂) i

or \(\mathrm{i}=\frac{T_H-T_C}{R_1+R_2}\)

Thus these two slabs are equivalent to a single slab of thermal resistance R1+ R2. If more than two slabs are joined in series and are allowed to attain a steady state, then equivalent thermal resistance is given by

R = R₁+ R₂+ R₃+ ……. (3)

Question 1. The figure shows the cross-section of the outer wall of a house built in a hill resort to keep the house insulated from the freezing temperature outside. The wall consists of teak wood of thickness L₁ and brick of thickness (L₂= 5L₁), sandwiching two layers of an unknown material with identical thermal conductivities and thickness. The thermal conductivity of teak wood is K1and that of brick is (K₂= 5K). Heat conduction through the wall has reached a steady state with the temperature of three surfaces being known. (T₁= 25ºC, T₂= 20ºC and T₅= –20ºC). Find the interface temperature T₄ and T₃.

Answer: Let the interface area be A. Then thermal resistance of wood,

⇒ \(\mathrm{R}_1=\frac{L_1}{K_1 A}\)

and that of brick wall \(\mathrm{R}_2=\frac{L_2}{K_2 A}=\frac{5 L_1}{5 K_1 A}=\mathrm{R}_1\)

Let the thermal resistance of each sandwiched layer = R. Then the above wall can be visualized as a circuit

thermal current through each wall is the same.

Hence \(\frac{25-20}{R_1}=\frac{20-T_3}{R}\)

⇒ \(\frac{T_3-T_4}{R}=\frac{T_4+20}{R_1}\)

⇒ 25 – 20 = T4+ 20

⇒ T₄= –15ºC

also, 20 – T₃= T₃– T₄

⇒ \(\mathrm{T}_3=\frac{20+T_4}{2}\)

= 2.5ºC

Question 2. In example 3, K₁= 0.125 W/m–ºC, K₂= 5K₁= 0.625 W/m–ºC, and the thermal conductivity of the unknown material is K = 0.25 W/mºC. L₁= 4cm, L₂= 5L₁= 20cm and L = 10cm. If the house consists of a single room with a total wall area of 100 m₂, then find the power of the electric heater being used in the room.
Answer:

R₁ = R₂ = \(\frac{\left(4 \times 10^{-2} \mathrm{~m}\right)}{\left(0.125 \mathrm{w} / \mathrm{m}-{ }^{\circ} \mathrm{C}\right)\left(100 \mathrm{~m}^2\right)}\)

= 32 × 10^-4ºC/w

R = \(\frac{\left(10 \times 10^{-2} \mathrm{~m}\right)}{\left(0.25 \mathrm{~W} / \mathrm{m}-{ }^{\circ} \mathrm{C}\left(100 \mathrm{~m}^2\right)\right.}\)

= 40 × 10^-4ºC/w

The equivalent thermal resistance of the entire wall = R₁+ R₂+ 2R = 144×10^-4ºC/W

∴ Net heat current, i.e. amount of heat flowing out of the house per second

⇒ \(\frac{T_H-T_C}{R}=\frac{25^{\circ} \mathrm{C}-\left(-20^{\circ} \mathrm{C}\right)}{144 \times 10^{-40} \mathrm{C} / \mathrm{w}}=\frac{45 \times 10^4}{144}\)watt = 3.12 Kwatt

Hence the heater must supply 3.12 kW to compensate for the outflow of heat.

Slabs In Parallel :

Consider two slabs held between the same heat reservoirs, their thermal conductivities K₁ and K_2, and cross-sectional areas A₁ and A₂ Heat reservoir

then \(\mathrm{R}_1=\frac{L}{K_1 A_1}\), \(\mathrm{R}_2=\frac{L}{K_2 A_2}\)

thermal current through slab 1 and that through slab 2 Net heat current from the hot to cold reservoir

⇒ \(i_1=\frac{T_H-T_C}{R_1}\)

Comparing with i = \(i_1+i_2=\left(T_H-T_C\right)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\),

⇒ \(\mathrm{i}=\frac{T_H-T_C}{R_{e q}}\) we get, \(\)

⇒ \(\frac{1}{R_{c q}}=\frac{1}{R_1}+\frac{1}{R_2}\)

If more than two rods are joined in parallel, the equivalent thermal resistance is given by

⇒ \(\frac{1}{R_{e q}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)

Question 3. Figure shows a copper rod joined to a steel rod. The rods have equal length and equal cross sectional area. The free end of the copper rod is kept at 0ºC and that of the steel rod is kept at 100ºC. Find the temperature of the junction of the rod. Conductivity of copper = 390 W/mºC Conductivity of steel = 46 W/m ºC 0ºC Copper Steel 100ºC
Answer:

Heat current in first rod (copper) = \(\frac{390 \times A(A-\theta)}{\ell}\)

Here θ is the temperature of the junction and A and l are the area and length of the copper rod.

Heat current in second rod (steel) = \(\frac{46 \times A(\theta-100)}{\ell}\)

In series combination. heat current remains the same. So,

⇒ \(\frac{390 \times A(0-\theta)}{\ell}\)

⇒ \(\frac{46 \times A(\theta-100)}{\ell}\)

⇒ -390 θ = 46θ-4600

⇒ 436 θ = 4600 θ = 10.6ºC

Question 4. An aluminum rod and a copper rod of equal length 1m and cross-sectional area 1cm₂ are welded together as shown in the figure. One end is kept at a temperature of 20ºC and the other at 60ºC. Calculate the amount of heat taken out per second from the hot end. The thermal conductivity of aluminum is 200 W/mºC and of copper is 390 W/mºC.

⇒ \(20^{\circ} \mathrm{C} \begin{array}{|l|}
\hline \text { Aluminium } \\
\hline \text { Copper } \\
\hline\end{array} 60^{\circ} \mathrm{C}\)

Answer: Heat current through the \(\frac{200 \times\left(1 \times 10^{-4}\right)}{1}\) = (60-20)

Heat current through the copper rod = \(\frac{390 \times\left(1 \times 10^{-4}\right)}{1}\) . (60-20)

Total heat = 200 × 10–4 × 40 + 390 × 10–4× 40 = 590 × 40 × 10–4= 2.36 Joule

Question 5. The three rods shown in the figure have identical geometrical dimensions. Heat flows from the hot end at the rate of 40W in an arrangement

1. Find the rate of heat flow when the rods are joined in arrangement
2. The thermal conductivity of aluminum and copper are 200 W/mºC and 400 W/mºC respectively.

Answer:

In the arrangement

The three rods are joined in series. The rate of flow of heat,

⇒ \(\frac{d \theta}{d t}=\frac{K A\left(\theta_1-\theta_2\right)}{\ell}=\frac{\theta_1-\theta_2}{R}\)

But, R = R₁+ R₂+ R₃[In series]

∴ 40 = \(\frac{100-0}{R_1+R_2+R_1}\)

40 = \(\frac{100}{\frac{\ell}{K_1 A}+\frac{\ell}{K_2 A}+\frac{\ell}{K A}}\)

40 = \(\frac{100}{\frac{\ell}{A}\left[\frac{2}{K_1}+\frac{1}{K_2}\right]}\)

⇒ \(\frac{\ell}{A}\left[\frac{2}{200}+\frac{1}{400}\right]=\frac{100}{40}\)

⇒ \(\frac{\ell}{A}\)

= 200 per m

In Figure two rods are all in parallel and the resultant of both is in series with the first rod

∴ \(\frac{d Q}{d t}=\frac{\theta_1-\theta_2}{R}\)

But R = \(\mathrm{R}_1+\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}}\)

⇒ \(\frac{d Q}{d t}=\frac{100-0}{\frac{\ell}{K_1 A}+\frac{1}{\frac{K_1 A}{\ell}+\frac{K_2 A}{\ell}}}\)

⇒ \(\frac{100-0}{\frac{\ell}{A}\left[\frac{1}{K_1}+\frac{1}{K_1+K_2}\right]}\)

⇒ \(\frac{600 \times 100}{200 \times 4}\)

= 75W

Question 6. A metal rod of length 20cm and diameter 2 cm is covered with a nonconducting substance. One of its ends is maintained at 100ºC while the other end is put at 0ºC. It is found that 25 g of ice melts in 5 min. Calculate the coefficient of thermal conductivity of the metal. Latent heat of ice = 80 cal gram^-1
Answer:

Here, the length of the rod,

Δx = 20 cm = 20 × 10^-2m

Diameter = 2cm,

Radius = r = 1 cm = 10^-2m

Area of cross-section

a = πr²= π(10^-2)²= π × 10^-4 sq. m

ΔT = 100 – 0 = 100ºC

Mass of ice melted, m = 25g

As L = 80 cal/ g

∴ Heat conducted, ΔQ = mL = 25 × 80 = 2000 cal = 2000 × 4.2 J

Δt = 5 min = 300 s

From = \(\frac{\Delta Q}{\Delta t}=\mathrm{KA} \frac{\Delta T}{\Delta x}\)

K = \(\frac{2000 \times 4.2 \times 20 \times 10^{-2}}{300 \times 10^{-4} \pi \times 100}\)

= 1.78Js^-1m^-1ºC^-1

Question 7. Two thin concentric shells made from copper with radius r₁ and r₂(r₂> r₁) have a material of thermal conductivity K filled between them. The inner and outer spheres are maintained at temperatures T_H and T_C respectively by keeping a heater of power P at the center of the two spheres. Find the value of P.
Answer:

Heat flowing per second through each cross-section of the sphere = P = i. Thermal resistance of the spherical shell of radius x and thickness dx,

dR = \(\frac{d x}{K .4 \pi x^2}\)

⇒ \(\mathrm{R}=\int_n^2 \frac{d x}{4 \pi x^2 \cdot K}=\frac{1}{4 \pi K}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\)

thermal current i = P = \(\frac{T_H-T_C}{R}=\frac{4 \pi K\left(T_H-T_C\right) r_1 r_2}{\left(r_2-r_1\right)}\)

NEET Physics Class 11 Chapter 8 Stefan-Boltzmann’S Law

According to this law, the amount of radiation emitted per unit of time from area A of a black body at absolute temperature T is directly proportional to the fourth power of the temperature.

u = σ A T⁴

where σ is Stefan’s constant = 5.67 x 10^-8 W/m² k⁴

A body that is not black absorbs and hence emits less radiation than that given by the equation.

For such a body, u e A T⁴

where e = emissivity (which is equal to absorptive power) which lies between 0 to 1

With the surroundings of temperature T0, net energy radiated by an area A per unit of time.

⇒ \(\Delta u=u-u_0=e \quad \sigma A\left(T^4-T_0^4\right)\)

Question 2. A black body at 2000K emits radiation with λ_m= 1250 nm. If the radiation coming from the star SIRUS λ_mis 71 nm, then the temperature of this star is …….
Answer:

Using Wien’s displacement law

⇒ \(\frac{T_2}{T_1}=\frac{\left(\lambda_m\right)_1}{\left(\lambda_m\right)_2}\)

∴ \(\mathrm{T}_2=\frac{2000 \times 1250 \times 10^{-3}}{71 \times 10^{-9}}\)

T₂ = 35.211 K

Question 3. At 1600 K maximum radiation is emitted at a wavelength of 2µm. Then the corresponding wavelength at 2000 K will be –
Answer:

Using T₁= T₂

∴ \(\frac{\lambda_{m_2} T_1}{T_2}\)

∴ \(\frac{2 \times 10^{-6} \times 1600}{2000}\)

= 1.6 µm

Question 4. If the temperature of a body is increased by 50%, then the increase in the amount of radiation emitted by it will be
Answer: Percentage increase in the amount of radiation emitted

∴ \(\frac{E_2-E_1}{E_1} \times 100\)

⇒ \(\frac{\left(1.5 T_1\right)^4-T_1^4}{T_1^4} \times 100\)

⇒ \(\frac{E_2-E_1}{E_1} \times 100\)

⇒ \(\left[(1.5)^4-1\right] \times 10\)

⇒ \(\frac{E_2-E_1}{E_1} \times 100\)

= 400%

Question 5. A blackened platinum wire of length 5cm and perimeter 0.02 cm is maintained at a temperature of 300K. Then at what rate the wire is losing its energy? (Take σ = 57 × 10^-8units)
Solution :

The rate of radiation heat loss is

⇒ \(\frac{d Q}{d t}=\mathrm{eA} \sigma \mathrm{T}^4 \text { (watts) }\)

for blackened surface e = 1

and A = (2πr)l = P_erimeter × length

∴ A = 0.02 × 5 × 10^-4

Thus

⇒ \(\frac{d Q}{d t}\)= 0.02 × 5 × 10^-4× 5.7 × 10^-8× (300)⁴

⇒ \(\frac{d Q}{d t}\)= 46.2W

Question 6. A hot black body emits energy at the rate of 16 J m^-2 s^-1 and its most intense radiation corresponds to 20,000 Å. When the temperature of this body is further increased and its most intense radiation corresponds to 10,000 Å, then find the value of energy radiated in Jm^-2 s^-1.
Answer:

Wein’s displacement law is :

λ_m.T = b i.e. \(\mathrm{T} \propto \frac{1}{\lambda_m}\)

Here, λ_m becomes half.

∴ Temperature doubles.

Also E = σT⁴

⇒ \(\frac{e_1}{e_2}=\left(\frac{T_1}{T_2}\right)^4\)

⇒ \(\mathrm{E}_2=. \mathrm{E}_1 \frac{e_1}{e_2}=\left(\frac{T_1}{T_2}\right)^4 \mathrm{e}_1=(2)^4 .16\)

= 16.16 = 256 J m^-2 s^-1

 

NEET Physics Class 11 Chapter 8 Newton’s Law Of Cooling

For a small temperature difference between a body and its surroundings, the rate of cooling of the body is directly proportional to the temperature difference and the surface area exposed.

⇒ \(\frac{d Q}{d t} \propto\left(\theta-\theta_0\right)\), where θ and θ0are temperature corresponding to object and surroundings.

From above expression \(\frac{d \theta}{d t}=-k\left(\theta-\theta_0\right)\)

This expression represents Newton’s law of cooling. It can be derived directly from Stefan’s law, which gives,

⇒ \(\mathrm{k}=\frac{4 \mathrm{e} \sigma \theta_0^3}{\mathrm{mc}} \mathrm{A}\)

Now \(\frac{d \theta}{d t}=-k\left[\begin{array}{ll}
\theta & \left.-\theta_0\right]
\end{array}\right.\)

⇒ \(\int_{\theta_i}^{\theta_f} \frac{d \theta}{\left(\theta-\theta_0\right)}=\int_0^t-k d t\)

where \(\theta_{\mathrm{i}}=\ln \frac{\left(\theta_f-\theta_0\right)}{\left(\theta_i-\theta_0\right)}\) initial temperature of object and

θf = final temperature of object.

⇒ –kt ⇒ (θf− θ0) = (θi– θ0) e–kt

⇒ θf= θ0+ (θi– θ0) e –kt

Limitations of Newton’s Law of Cooling:

1. The difference in temperature between the body and surroundings must be small
2. The loss of heat from the body should be by radiation only.
3. The temperature of the surroundings must remain constant during the cooling of the body.

Approximate Method For Applying Newton’s Law Of Cooling

Sometimes when we need only approximate values from Newton’s law, we can assume a constant rate of cooling, which is equal to the rate of cooling corresponding to the average temperature of the body during the interval.

⇒ \(\left\langle\frac{d \theta}{d t}\right\rangle-\mathrm{k}\left(<\theta>-\theta_0\right)\)

If θ_i and θ_f are the initial and final temperatures of the body then,

⇒ \(<\theta>=\frac{\theta_i+\theta_f}{2}\)

Remember equation is only an approximation and an equation must be used for exact values.

Comparison Of Specific Heat Of Two Liquids Using Newton’s Law Of Cooling:

If an equal volume of two liquids of densities and specific heats ρ₁, s_1, and ρ₂, s₂ respectively are filled in calorimeters having the same surface area and finish, cool from the same initial temperature θ₁ to the same final temperature θ₂ with the same temperature of surroundings, i.e., θ₀, in time intervals t₁and t₂ respectively and water equivalent of calorimeter is w. According to Newton’s law of cooling

⇒ \(\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\text {liq }}=\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\text {water }}\)

⇒ \(\frac{\left(w+m_1 s_1\right)\left(\theta_1-\theta_2\right)}{t_1}\)

⇒ \(\frac{\left(w+m_2 s_2\right)\left(\theta_1-\theta_2\right)}{t_2}\)

or \(\frac{w+m_1 s_1}{t_1}=\frac{w+m_2 s_2}{t_2}\)

If the water equivalent of calorimeter w is negligible then

⇒ \(\frac{m_1 s_1}{t_1}=\frac{m_2 s_2}{t_2}\)

So, \(\frac{m_1 s_1}{m_2 s_2}=\frac{t_1}{t_2} \quad \text { or } \frac{\rho_1 s_1}{\rho_2 s_2}=\frac{t_1}{t_2}\)(v₁= v₂, volume are equal) with the help of this eqn. we can find the specific heat of the liquid.

Question 1. A body at a temperature of 40ºC is kept in a surrounding constant temperature of 20ºC. It is observed that its temperature falls to 35ºC in 10 minutes. Find how much more time will it take for the body to attain a temperature of 30ºC.
Answer:

Δθ_f = Δθ_ie^-kt

for the interval in which temperature falls from 40 to 35º C.

(35 – 20) = (40 – 20) e^-k.10

⇒ \(\mathrm{e}^{-10 \mathrm{k}}=\frac{3}{4}\)

⇒ \(K=\frac{\ln \frac{4}{3}}{10}\)

for the next interval (30 – 20) = (35 – 20)e^-kt

⇒ \(\mathrm{e}^{-10 \mathrm{k}}=\frac{2}{3}\)

⇒ \(\mathrm{kt}=\mathrm{ln} \frac{3}{2}\)

⇒ \(\frac{\left(\ln \frac{4}{3}\right) t}{10}=\ell n\)

⇒ \(\mathrm{t}=10 \frac{\left(\ln \frac{3}{2}\right)}{\left(\ln \frac{4}{3}\right)}\) minute = 14.096 min

Alternative : (by approximate method) for the interval in which temperature falls from 40 to 35ºC 40 35

<θ> = \(\frac{40+35}{2}\) = 37.5ºC

⇒ \(\left\langle\frac{d \theta}{d t}\right\rangle=-\mathrm{k}\left(<\theta>-\theta_0\right)\)

⇒ \(\frac{\left(35^{\circ} \mathrm{C}-40^{\circ} \mathrm{C}\right)}{10(\mathrm{~min})}\)

= –K(37.5ºC – 20ºC)

K = \(\frac{1}{35}\)(min^-1)

for the interval in which temperature falls from 35ºC to 30ºC

<θ> = \(\frac{\left(30^{\circ} \mathrm{C}-35^{\circ} \mathrm{C}\right)}{t}\) = (32.5º) C

= – K(32.5ºC – 20ºC)

⇒ required time, t = \(\)

⇒ \(\frac{5}{12.5} \times 35 \mathrm{~min}\)

= 14 min

Question 2. Two liquids of the same volume are cooled under the same conditions from 65ºC to 50ºC. The time taken is 200sec and 480sec. If the ratio of their specific heats is 2 : 3 then find the ratio of their densities. (neglect the water equivalent of a calorimeter)
Answer:

From Newton’s law of cooling

⇒ \(\left(\frac{m_1 s_1+w_1}{t_1}\right)\left(\theta_1-\theta_2\right)\)

⇒ \(\left(\frac{m_2 s_2+w_2}{t_2}\right)\left(\theta_1-\theta_2\right)\)

here w₁= w₂= 0

⇒ \(\frac{m_1 s_1}{t_1}=\frac{m_2 s_2}{t_2}\)

⇒ \(\frac{V d_1 s_1}{t_1}=\frac{V d_2 s_2}{t_2}\)

⇒ \(\frac{d_1}{d_2}=\frac{t_1 s_2}{t_2 s_1}\)

⇒ \(\frac{200}{480} \times \frac{3}{2}\)

⇒ \(\frac{5}{8}\)

Question 3. A calorimeter of water equivalent to 5 × 10^-3 kg contains 25 × 10^-3 kg of water. It takes 3 minutes to cool from 28°C to 21°C. When the same calorimeter is filled with 30 × 10^-3 kg of turpentine oil then it takes 2 minutes to cool from 28°C to 21°C. Find out the specific heat of turpentine oil.
Answer:

⇒ \(\mathrm{R}_{\text {water }}=\mathrm{R}_{\text {turpentine }} \frac{\left(m_1+w\right)}{t_1}=\frac{\left(m_2 s_2+w\right)}{t_2}\)

or \(\frac{\left(25 \times 10^{-3}+5 \times 10^{-3}\right)}{3}=\frac{30 \times 10^{-3} s_2+5 \times 10^{-3}}{2}\)

10 = \(\frac{30 s_2+5}{2}, 20\)

⇒ \(30 s_2+5\)

∴ specific heat of turpentine s₂= 1/2 = 0.5 kcal/kg/°C

Question 4. A man, the surface area of whose skin is 2m², is sitting in a room where the air temperature is 20°C. If his skin temperature is 28°C, find the rate at which his body loses heat. The emissivity of his skin = 0.97.
Answer:

Absolute room temperature (T₀) = 20 + 273 = 293 K

Absolute skin temperature (T) = 28 + 273 = 301 K

Rate of heat loss = σ e A (T⁴ – T₀⁴)

= 5.67 × 10^-8 × 0.97 × 2 × {(301)⁴ – (293)⁴} = 92.2 W

Question 5. Compare the rate of loss of heat from a metal sphere of the temperature 827°C, with the rate of loss of heat from the same sphere at 427 °C, if the temperature of surroundings is 27°C.
Answer:

Given : T₁= 827 °C = 1100 K, T₂= 427 °C = 700 K and T₀= 27 °C = 300 K

According to Steafan’s law of radiation, \(\frac{d Q}{d t}\)

⇒ \(\sigma \mathrm{Ae}\left(\mathrm{T}^4-\mathrm{T}_0{ }^4\right)\)

⇒ \(\frac{\left(\frac{d Q}{d t}\right)_1}{\left(\frac{d Q}{d t}\right)_2}=\frac{\left(T_1^4-T_0^4\right)}{\left(T_2^4-T_0^4\right)}\)

⇒ \(\frac{\left[(1100)^4-(300)^4\right]}{\left[(700)^4-(300)^4\right]}\) = 6.276

⇒ \(\left(\frac{d Q}{d t}\right)_1:\left(\frac{d Q}{d t}\right)_2\) = 6.276:1

Question 6. Two spheres of the same material have radii of 6 cm and 9 cm respectively. They are heated to the same temperature and allowed to cool in the same enclosure. Compare their initial rates of emission of heat and initial rates of fall of temperature.
Answer:

Given : r₁= 6 cm r₂= 9 cm,

∴ \(\frac{r_1}{r_2}=\frac{2}{3}\)

According to Stefan’s law of radiation, the rate of emission of heat by an ordinary body,

⇒ \(\mathrm{R}_{\mathrm{h}}=\left(\frac{d Q}{d t}\right)=\sigma \mathrm{AeT}^4\)

or \(R_h \propto r^2\) (A=4σr²)

⇒ \(\frac{R_{h 1}}{R_{h 2}}=\left(\frac{r_1}{r_2}\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)

⇒ \(\frac{R_{F T 1}}{R_{F T 2}}=\frac{r_2}{r_1}=\frac{3}{2}\)

[Rate of fall of temp. \(\mathrm{R}_{\mathrm{FT}}, \frac{d \theta}{d t}, \frac{\sigma A e\left(T^4-T_0^4\right)}{m s J}\) = or \(R_{F T} \propto \frac{A}{m} \text { or } R_{F T} \propto \frac{1}{r}\)]

∴ Initial rates of emission of heat are in the ratio 4: 9 and initial rates of fall of temperature are in the ratio 3: 2.

Question 7. The filament of an evacuated light bulb has a length of 10 cm, a diameter of 0.2 mm, and an emissivity of 0.2, calculate the power it radiates at 2000 K. (σ = 5.67 × 10^-8 W/m² K⁴)
Answer:

l = 10 cm = 0.1 m, d = 0.2 mm, r = 0.1 mm = 1 × 10^-4 m, e = 0.2, T = 2000 K, σ = 5.67 × 10^-8 W/m² K⁴

According to Stefan’s law of radiation, the rate of emission of heat for an ordinary body (filament), E = σAeT⁴ = σ(2 π r l) eT⁴

= 5.67 × 10^-8 × 2 × 3.14 × 1 × 10^-4 × 0.1 × 0.2 × (2000)⁴

= 11.4 W

∴ Power radiated by the filament = 11.4 W [A = 2πrl]

Question 8. The energy radiated from a black body at a temperature of 727°C is E. By what factor does the radiated energy increase if the temperature is raised to 2227°C?
Solution :

⇒ \(\frac{E_2}{E_1}=\left[\frac{T_2}{T_1}\right]^4\)

⇒ \(\left[\frac{2227+273}{727+273}\right]^4=\left[\frac{2500}{1000}\right]^4\)

= 39

Question 9. An ice box made of 1.5 cm thick styrofoam has dimensions of 60 cm × 30cm. It contains ice at 0ºC and is kept in a room at 40ºC. Find the rate at which the ice is melting. Latent heat of fusion of ice = 3.36 × 10⁵J/kg. and thermal conductivity of stryrofoam = 0.4 W/m-ºC.
Answer:

The total surface area of the walls

= 2(60 cm × 60 cm + 60 cm × 30 cm + 60 cm × 30 cm)

= 1.44 m²

The thickness of the wall = 1.5 cm = 0.015m

The rate of heat flow into the box is

⇒ \(\frac{\Delta Q}{\Delta t}=\frac{K A\left(\theta_1-\theta_2\right)}{x}\)

⇒ \(\frac{\left(0.04 \mathrm{~W} / \mathrm{m}-{ }^{\circ} \mathrm{C}\right)\left(1.44 \mathrm{~m}^2\right)\left(40^{\circ} \mathrm{C}\right)}{0.015 \mathrm{~m}}\)

= 154 W.

The rate at which the ice melts is = 0.46 g/s

Question 10. A black body emits 10 watts per cm² at 327ºC. The sun radiates 10⁵ watts per cm². Then what is the temperature of the sun?
Answer:

∴ \(\frac{E_{\text {sun }}}{E_{\text {body }}}=\left(\frac{E_{\text {sun }}}{E_{\text {body }}}\right)^4\)

∴ \(\frac{T_{\text {sun }}}{T_{\text {body }}}=\left(\frac{10^5}{10}\right)^{1 / 4}\)

= 6000 K

∴ T_sun = 6000 K

Question 11. A bulb made of a tungsten filament of a surface is 0.5 cm² is heated to a temperature of 3000k when operated at 220V. The emissivity of the filament is e = 0.35 and take σ = 5.7 × 10^-8 mks units. Then the wattage of the bulb is …..(calculate)
Answer:

The emissive power watt/m² is

E = eσT⁴

Therefore the power of the bulb is

P = E x area (Watts)

∴ P = eσT⁴A

∴ P = 0.35 × 0.5 × 10^-4× 5, 7 × 10^-8× (3000)⁴

⇒ P = 80.8 W

Question 12. In the above example, if the temperature of the filament falls to 2000k due to a drop in mains voltage, then what will be the wattage of the bulb?
Answer:

Now the power of the bulb will be such that

⇒ \(\frac{P_2}{P_1}=\left(\frac{T_2}{T_1}\right)^4\)

Thus \(P_2=P_1 \times\left(\frac{2}{3}\right)^4\)

∴ \(P_2=80.8 \times \frac{16}{81}\)

Thus P₂= P1x

∴ P₂= 80.8 x

⇒ P₂= 15.96

Question 13. A liquid takes 30 seconds to cool from 95ºC to 90ºC and 70 seconds to cool from 55 to 50ºC. Find the room temperature and the time it will take to cool from 50ºC to 45ºC
Answer:

From the first date

⇒ \(\frac{95-90}{30}=\mathrm{k}\left(\frac{95 \times 90}{2}-\theta_0\right)=\mathrm{k}\)….(1)

From the second data \(\frac{55-50}{70}=\mathrm{k}\left(\frac{55-50}{0}-\theta_0\right)\) = k….(2)

Dividing (1) and (2) we get

⇒ \(\frac{7}{3}=\frac{92.5-\theta_0}{52.5-\theta_0}\)

⇒ \(\theta_0=22.5^{\circ}\) ……….(3)

Let the time taken in cooling from 50ºC to 45ºC is t, then

⇒ \(\frac{50-45}{t}\)

⇒ \(\mathrm{k}\left(\frac{50-45}{2}-\theta_0\right)\)….(4)

Using θ₀= 22.5ºC, and dividing (1) by (2) we get

⇒ \(\frac{t}{30}=\left(\frac{92.5-22.5}{47.5-22.5}\right)\)

t = 84 sec

Question 14. A blackened metal disc is held normal to the sun’s rays, Both of its surfaces are exposed to the atmosphere if the distance of the earth from the sun is 216 times the radius of the sun and the temperature of the sun is 6000K, the temperature of the disc in the steady state will be
Answer:

In the steady state, the heat received from the sun will be equal to the heat radiated out. Heat received from the sun will be on one side only and it will radiate from both sides.

∴ \(\mathrm{A} \sigma\left(\frac{R s}{d}\right)^2 \mathrm{~T}^4\)=2AσT⁴

⇒ \(\frac{R s}{d}=\frac{1}{216}\)

∴ T′ = \(\frac{T}{(216)^{1 / 2} 2^{1 / 4}}\)

⇒ \(\frac{6000}{14.7 \times 1.189}\)

= 345K

∴ T′ = 70ºC

Question 15. Behaving like a black body sun emits maximum radiation at wavelength 0.48µm. The mean radius of the sun is 6.96 × 10⁸m. Stefan’s constant is 5.67 × 10^-8wm^-2k^-4and wien’s constant is 0.293 cm-k. The loss of mass per second by the emission of radiation from the sun is
Answer: Using Wien’s law

T = T = \(\frac{b}{\lambda_m}=\frac{0.293 \times 10^{-2}}{0.48 \times 10^{-6}}\)

= 6104 K

Energy given out by sun per second

= AσT⁴=4π (6.96 × 10⁸)²× 5.67 × 10^-8(6104)⁴

⇒ 49.285 × 10²⁵J

Loss of mass per second

m = \(\frac{E}{c^2}=\frac{49.285 \times 10^{25}}{9 \times 10^{16}}\)

m = 5.4×10⁹kg/s

Question 16. 50g of water and an equal volume of alcohol (relative density 0.8) are placed one after the other in the same calorimeter. They are found to cool from 60ºC to 50ºC in 2 minutes and 1 minute respectively if the water equivalent of the calorimeter is 2g then what is the specific heat of the alcohol?
Answer:

Given t_w= 2min, telco = 1 min

m_w= 50g, malco = 50 × 0.8 = 40g

S_w= 1 in cgs units, w = 2g

Therefore,

⇒ \(\mathrm{S}_{\mathrm{alco}}=\frac{1}{m_{\text {alco }}}\left\{\frac{t_{\text {alco }}}{t_w}\left[m_w+W\right]-W\right\}\)

⇒ \(\mathrm{S}_{\mathrm{alco}}=\frac{1}{40}\left\{\frac{1}{2}[50+2]-2\right\}\)

⇒ \(\mathrm{S}_{\mathrm{alco}}=\frac{24}{50}\)

S_alco = 0.6 cgs units = 0.6 cal/gºC

 

NEET Physics Class 11 Chapter 8 Heat Transfer Multiple Choice Question And Answers

Question 1. A wall has two layers A and B, each made of different material. Both the layers have the same thickness. The thermal conductivity for A is twice that of B. Under steady state, the temperature difference across the whole wall is 36°C. Then the temperature difference across layer A is

1. 6°C
2. 12°C
3. 18°C
4. 24°C

Answer: 2. 12°C

Question 2. A heat flux of 4000 J/s is to be passed through a copper rod of length 10 cm and an area of cross-section 100 sq. cm. The thermal conductivity of copper is 400 W/mC. The two ends of this rod must be kept at a temperature difference of–

1. 1ºC
2. 10ºC
3. 100ºC
4. 1000ºC

Answer: 3. 100ºC

Question 3. If two conducting slabs of thickness d₁ and d₂, and thermal conductivity K₁ and K₂ are placed together face to face as shown in the figure the steady state temperatures of outer surfaces are θ₁ and θ₂. The temperature of the common surface is–

1. \(\frac{K_1 \theta_1 d_1+K_2 \theta_2 d_2}{K_1 d_1+K_2 d_2}\)
2. \(\frac{K_1 \theta_1+K_2 \theta_2}{K_1+K_2}\)
3. \(\frac{K_1 \theta_1+K_2 \theta_2}{\theta_1+\theta_2}\)
4. \(\frac{K_1 \theta_1 d_2+K_2 \theta_2 d_1}{K_1 d_2+K_2 d_1}\)

Answer: 4. \(\frac{K_1 \theta_1 d_2+K_2 \theta_2 d_1}{K_1 d_2+K_2 d_1}\)

Question 4. Which of the following qualities suit a cooking utensil?

1. High specific heat and low thermal conductivity
2. High specific heat and high thermal conductivity
3. Low specific heat and low thermal conductivity
4. Low specific heat and high thermal conductivity

Answer: 4. Low specific heat and high thermal conductivity

Question 5. The lengths and radii of two rods made of the same material are in the ratios 1: 2 and 2 : 3 respectively; If the temperature difference between the ends of the two rods is the same, then in the steady state, the amount of heat flowing per second through them will be in the ratio:

1. 1 : 3
2. 4 : 3
3. 8: 9
4. 3: 2

Answer: 3. 8: 9

Question 6. Two metal cubes with 3 cm-edges of copper and aluminum are arranged as shown in the figure (K_CU =385 W/m-K, K_AL = 209 W/m-K).

1. The total thermal current from one reservoir to the other is :

1. 1.42 × 10³ W
2. 2.53 × 10³ W
3. 1.53 × 10⁴ W
4. 2.53 × 10⁴ W

Answer: 1. 1.42 × 10³ W

2. The ratio of the thermal current carried by the copper cube to that carried by the aluminum cube is: –

1. 1.79
2. 1.69
3. 1.54
4. 1.84

Answer: 4. 1.84

Question 7. Two rods having thermal conductivities in the ratio of 5 : 3 and having equal length and equal cross section are joined face-to-face (series combination). If the temperature of the free end of the first rod is 100ºC and the free end of the second rod is 20ºC, the temperature of the junction is–

1. 50ºC
2. 70ºC
3. 85ºC
4. 90ºC

Answer: 2. 70ºC

Question 8. One end of a metal rod of length 1.0m and area of cross-section 100 cm² is maintained at 100ºC. If the other end of the rod is maintained at 0ºC, the quantity of heat transmitted through the rod per minute will be (coefficient of thermal conductivity of the material of rod = 100W/Kg/K)

1. 3 × 103 J
2. 6 × 103 J
3. 9 × 103 J
4. 12 × 103 J

Answer: 2. 6 × 103 J

Question 9. The coefficients of thermal conductivity of a metal depends on

1. Temperature difference between the two sides
2. Thickness of the metal plate
3. Area of the plate
4. None of the above

Answer: 4. None of the above

Question 10. Two identical square rods of metal are welded end to end as shown in the figure

1. Assume that 10 cal of heat flows through the rods in 2 min. Now the rods are welded as shown in the figure.
2. The time it would take for 10 cal to flow through the rods now, is

1. 0.75 min
2. 0.5 min
3. 1.5 min
4. 1 min

Answer: 2. 0.5 min

Question 11. The area of cross-section of two rods of equal lengths are A₁ and A₂ and thermal conductivities are K₁ and K₂. Specific heats are S₁ and S₂. Condition for equal heat flow is–

1. \(\mathrm{K}_1=\mathrm{K}_2\)
2. \(\mathrm{K}_1 \mathrm{~S}_1=\mathrm{K}_2 \mathrm{~S}_2\)
3. \(\frac{K_1}{A_1 S_1}=\frac{K_2}{A_2 S_2}\)
4. \(\mathrm{K}_1 \mathrm{~A}_1=\mathrm{K}_2 \mathrm{~A}_2\)

Answer: 4. \(\mathrm{K}_1 \mathrm{~A}_1=\mathrm{K}_2 \mathrm{~A}_2\)

Question 12. If two metallic plates of equal thickness, equal cross-section area, and thermal conductivities K₁ and K₂ are put together face to face (series combination) and a common plate is constructed, then the equivalent thermal conductivity of this plate will be

1. \(\frac{K_1 \quad K_2}{K_1+K_2}\)
2. \(\frac{2 K_1 K_2}{K_1+K_2}\)
3. \(\frac{\left(K_1^2+K_2^2\right)^{3 / 2}}{K_1 K_2}\)
4. \(\frac{\left(K_1^2+K_2^2\right)^{3 / 2}}{2 K_1 K_2}\)

Answer: 2. \(\frac{2 K_1 K_2}{K_1+K_2}\)

Question 13. Consider a compound slab consisting of two different materials having equal thicknesses, equal cross-section area, and thermal conductivities k and 2k respectively. If they are connected in parallel combination, the equivalent thermal conductivity of the slab is–

1. \(\sqrt{2}\)
2. 3k
3. \(\frac{4}{3} \mathrm{k}\)
4. \(\frac{2}{3} \mathrm{k}\)

Answer: 3. \(\frac{4}{3} \mathrm{k}\)

Question 14. The two ends of a rod of length L and a uniform cross-sectional area A kept at two temperatures T₁ and T₂(T₁>T₂ ). The rate of heat transfer,\(\frac{d Q}{d t}\) through the rod in a steady state is given by

1. \(\frac{d Q}{d t}=\frac{K L\left(T_1-T_2\right)}{A}\)
2. \(\frac{d Q}{d t}=\frac{K\left(T_1-T_2\right)}{L A}\)
3. \(\frac{d Q}{d t}=K L A\left(T_1-T_2\right)\)
4. \(\frac{d Q}{d t}=\frac{K A\left(T_1-T_2\right)}{L}\)

Answer: 4. \(\frac{d Q}{d t}=\frac{K A\left(T_1-T_2\right)}{L}\)

Question 15. A square is made of four rods of the same material one of the diagonals of a square is at a temperature difference of 100°C, then the temperature difference of the second diagonal :

1. 0°C
2. \(\frac{100}{\ell}\)
3. \(\frac{100}{2 \ell}\)
4. 100°C

Answer: 1. 0°C

Question 16. Three rods made of the same material and having the same cross-section are joined as shown in Fig. Each rod is of the same length. The left and right ends are kept at 0°C and 90°C respectively. The temperature of the junction of the three rods will be:

1. 45°C
2. 60°C
3. 30°C
4. 20°C

Answer: 2. 60°C

Question 17. Two containers, one containing ice at 0°C and the other containing boiling water at 100°C are connected by two identical rods. When rods are in parallel the rate of heat transfer is Q₁ and when rods are in series, the rate of heat transfer is Q₂. Then Q₂ /Q₁ will be:

1. 2: 1
2. 1: 2
3. 4: 1
4. 1: 4

Answer: 4. 1: 4

Question 18. 2 litre water at 27°C is heated by a 1 kW heater in an open container. On average heat is lost to surroundings at the rate of 160 J/s. The time required for the temperature to reach 77°C is

1. 8 min 20 sec
2. 10 min
3. 7 min
4. 14 min

Answer: 1. 8 min 20 sec

Question 19. If the temperature difference on the two sides of a wall increases from 100°C to 200°C, its thermal conductivity

1. Remains unchanged
2. Is doubled
3. Is halved
4. Becomes four times

Answer: 1. Remains unchanged

Question 20. A cylindrical rod having temperature T₁ and T₂ at its ends. The rate of flow of heat is Q₁cal/sec. If all the linear dimensions are doubled keeping the temperature constant then the rate of flow of heat Q₂ will be–

1. 4Q₁
2. 2Q₁
3. \(\frac{Q_1}{4}\)
4. \(\frac{Q_1}{2}\)

Answer: 2. 2Q₁

Question 21. One end of a thermally insulated rod is kept at a temperature of T₁ and the other at T₂. The rod is composed of two sections of lengths L₁ and L₂ and thermal conductivities k₁ and k₂ respectively. The temperature at the interface of the sections is

1. \(\frac{\left(\begin{array}{ll}
   K_2 & L_2 T_1+K_1 L_1 T_2
   \end{array}\right)}{\left(K_1 L_1+K_2 L_2\right)}\)
2. \(\frac{\left(\begin{array}{ll}
   K_2 L_1 T_1+K_1 L_2 T_2
   \end{array}\right)}{\left(K_2 L_1+K_1 L_2\right)}\)
3. \(\frac{\left(\begin{array}{ll}
   K_1 & L_2 T_1+K_2 L_1 T_2
   \end{array}\right)}{\left(K_1 L_2+K_2 L_1\right)}\)
4. \(\frac{\left(\begin{array}{ll}
   K_1 L_1 T_1+K_2 L_2 T_2
   \end{array}\right)}{\left(K_1 L_1+K_2 L_2\right)}\)

Answer: 3. \(\frac{\left(\begin{array}{ll}
K_1 & L_2 T_1+K_2 L_1 T_2
\end{array}\right)}{\left(K_1 L_2+K_2 L_1\right)}\)

Question 22. Three rods A, B, and C of the same length and same cross-section area are joined as shown in the figure. Their thermal conductivities are in the ratio 1: 2: 1.5. If the open ends of A and C are at 200°C and 18°C respectively, the temperature at the junction of A and B in equilibrium is-

1. 156°C
2. 116°C
3. 74°C
4. 148°C

Answer: 2. 116°C

Question 23. In the above question, the temperature at the junction of B and C will be

1. 124°C
2. 124°K
3. 74°C
4. 74°K

Answer: 3. 74°C

Question 24. The ends of the two rods of different materials with their thermal conductivities, radii of cross-section, and lengths in the ratio 1: 2 are maintained at the same temperature difference. If the rate of flow of heat in the larger rod is 4 cal/sec., that in the shorter rod will be (in cal/sec)

1. 1
2. 2
3. 8
4. 16

Answer: 2. 2

Question 25. The coefficients of thermal conductivity of copper, mercury, and glass are respectively K_c, K_m, and Kg such that K_c> K_m> K_g. If the same quantity of heat is to flow per second per unit area of each and corresponding temperature gradients are X_c, X_m, and X_g.

1. X_c= X_m= K_g
2. X_c> X_m> X_g
3. X_c< X_m< X_g
4. X_m< X_c< X_g

Answer: 3. X_c< X_m< X_g

Question 26. A compound slab is composed of two parallel layers of different materials, with thicknesses of 3 cm and 2 cm. The temperatures of the outer faces of the compound slab are maintained at 100°C and 0°C. If conductivities are 0.036 cal/cm-sec-°C and 0.016 cal/cm-sec-°C then the temperature of the junction is-

1. 40°C
2. 60°C
3. 100°C
4. 50°C

Answer: 2. 60°C

Question 27. The intensity of heat radiation by a point source measured by a thermopile placed at a distance d is Ι, If the distance of the thermopile is doubled then the intensity of radiation will be

1. Ι
2. 2Ι
3. \(\frac{\mathrm{I}}{4}\)
4. \(\frac{\mathrm{I}}{2}\)

Answer: 3. \(\frac{\mathrm{I}}{4}\)

Question 28. Two rods of copper and brass of the same length and area of cross-section are joined as shown. One end is kept at 100°C and the other at 0°C. The temperature at the mid-point will be

1. More if A is at 100°C and B at 0°C
2. More if A is at 0°C and B at 100°C
3. Will be the same in both the above cases, but not 50°C
4. 50°C in both the above cases

Answer: 1. More if A is at 100°C and B at 0°C

Question 29. Two identical square rods of metal are welded end to end as shown in Fig.

1. 20 cal. of heat flows through it in 4 min. If the rods are welded as shown in Fig.
2. The same amount of heat will flow through the rods in

1. 1 min.
2. 2 min.
3. 3 min.
4. 16 min.

Answer: 1. 1 min.

Question 30. A wall consists of alternating blocks with length ‘d’ and coefficient of thermal conductivity k₁ and k₂. The cross sectional area of the blocks is the same. The equivalent coefficient of thermal conductivity of the wall between left and right is:-

1. \(\mathrm{K}_1+\mathrm{K}_2\)
2. \(\frac{\left(K_1+K_2\right)}{2}\)
3. \(\frac{K_1 K_2}{K_1+K_2}\)
4. \(\frac{2 K_1 K_2}{K_1+K_2}\)

Answer: 2. \(\frac{\left(K_1+K_2\right)}{2}\)

Question 31. Five rods of the same dimensions are arranged as shown in the fig. They have thermal conductivities, k₁, k₂, k₅, k_4, and k₃ when points A and B are maintained at different temperatures. No heat flows through the central rod if-

1. \(k_1 k_4=k_2 k_3\)
2. \(\mathrm{k}_1=\mathrm{k}_4 \text { and } \mathrm{k}_2=\mathrm{k}_3\)
3. \(\frac{k_1}{k_4}=\frac{k_2}{k_3}\)
4. \(k_1 k_2=k_3 k_4\)

Answer: 4. \(k_1 k_2=k_3 k_4\)

Question 32. Three metal rods made of copper, aluminum, and brass, each 20 cm long and 4 cm in diameter, are placed end to end with aluminum between the other two. The free ends of copper and brass are maintained at 100 and 0°C respectively. Assume that the thermal conductivity of copper is twice that of aluminum and four times that of brass. The equilibrium temperatures of the copper-aluminium and aluminium-brass junctions are respectively.

1. 68 °C and 75 °C
2. 75 °C and 68 °C
3. 57 °C and 86 °C
4. 86 °C and 57 °C

Answer: 3. 57 °C and 86 °C

Question 33. The coefficient of thermal conductivity of copper is nine times that of steel. In the composite cylindrical bar shown in the figure, what will be the temperature at the junction of copper and steel?

1. 75ºC
2. 67ºC
3. 33ºC
4. 25ºC

Answer: 1. 75ºC

Question 34. The heat conduction coefficient of copper is 9 times the heat conduction coefficient of steel. The junction temperature of the combined cylindrical rod shown in the figure will be.

1. 75°C
2. 67°C
3. 33°C
4. 25°C

Answer: 1. 75°C

Question 35. Water is usually heated by

1. Conduction
2. Convection
3. Radiation
4. All the above processes

Answer: 2. Convection

Question 36. In natural convection a heated portion of a liquid moves because-

1. Its molecular motion becomes aligned
2. Of molecular collisions within it
3. Its density is less than that of the surrounding fluid
4. Of currents of the surrounding fluid

Answer: 3. Its density is less than that of the surrounding fluid

Question 37. It is hotter at the same distance over the top of the fire than it is on the side of it mainly because

1. Heat is radiated upwards
2. Air conducts heat upwards
3. Convection takes more heat upwards
4. Conduction, convection, and radiation all contribute significantly to transferring heat upwards

Answer: 3. Convection takes more heat upwards

Question 38. Ventilators are provided at the top of the room

1. To bring oxygen for breathing
2. So that sunlight may enter the room
3. To maintain convection currents to keep the air fresh in the room
4. To provide an outlet for carbon dioxide

Answer: 3. To maintain convection currents to keep the air fresh in the room

Question 39. The mode of transmission of heat in which heat is carried by moving particles is:

1. Wave motion
2. Convection
3. Conduction
4. Radiation

Answer: 2. Convection

Question 40. The temperature of a piece of metal is increased from 27°C to 327°C. The rate of emission of heat by radiation by a metal will become-

1. Double
2. Four times
3. Eight times
4. Sixteen times

Answer: 4. Sixteen times

Question 41. Radiation emitted by a surface is directly proportional to-

1. The third power of its temperature
2. The fourth power of its temperature
3. Twice the power of its temperature
4. None of above

Answer: 2. Fourth power of its temperature

Question 42. If the temperature of the surface of the sun becomes half then the energy emitted by it to the earth per second will reduce to –

1. 1/2
2. 1/4
3. 1/16
4. 1/64

Answer: 3. 1/16

Question 43. If the distance between point sources and the screen is doubled then the intensity of light becomes-

1. Four times
2. Doubled
3. Half
4. One fourth

Answer: 4. One fourth

Question 44. At T = 200K a black body emits maximum energy at a wavelength of 14 μm. Then at T = 1000K, the body will emit maximum energy at a wavelength of-

1. 70 mm
2. 70 μm
3. 2.8 μm
4. 2.8 mm

Answer: 3. 2.8 μm

Question 45. If the temperature of a black body is raised by 50%, then the energy emitted per second will be increased by an order of-

1. 50%
2. 100%
3. 200%
4. 400%

Answer: 4. 400%

Question 46. What represents the color of the star-

1. Density
2. Distance
3. Energy
4. Temperature

Answer: 4. Temperature

Question 47. The black body spectrum is-

1. Continuous spectrum with black lines
2. Continuous spectrum with black bands
3. Continuous spectrum
4. None of the above

Answer: 3. Continuous spectrum

Question 48. There is a black spot on the body. If the body is heated and carried in a dark room then it glows more. This can be explained based on-

1. Newton’s law of cooling
2. Vien’s law
3. Kirchoff’s law
4. Stefan’s

Answer: 3. Kirchoff’s law

Question 49. A heated body emits radiation which has maximum intensity at frequency νm. If the temperature of the body is doubled:

1. The maximum intensity radiation will be at frequency 2 νm
2. The maximum intensity radiation will be at frequency νm.
3. The total emitted energy will increase by a factor of 2.
4. None of these

Answer: 1. The maximum intensity radiation will be at frequency 2 νm

Question 50. If λ_m denotes the wavelength at which the radiative emission from a black body at a temperature T K is maximum, then-

1. \(\lambda_{\mathrm{m}} \propto \mathrm{T}^4\)
2. \(\lambda_{\mathrm{m}}\) is independent of T
3. \(\lambda_{\mathrm{m}} \propto \mathrm{T}\)
4. \(\lambda_m \propto \mathrm{T}^{-1}\)

Answer: 4. \(\lambda_m \propto \mathrm{T}^{-1}\)

Question 51. A black body at 1227°C emits radiations with maximum intensity at a wavelength of 5000 Å. If the temperature of the body is increased by 1000°C, the maximum intensity will be at

1. 4000 Å
2. 5000Å
3. 6000 Å
4. 3000Å

Answer: 4. 3000Å

Question 52. A black body is at 727°C. It emits energy at a rate which is proportional to

1. (277)²
2. (1000)⁴
3. (1000)²
4. (727)⁴

Answer: 2. (1000)⁴

Question 53. If the temperature of the body increases by 10%, then the increase in radiated energy of the body is :

1. 10%
2. 40%
3. 46%
4. 1000%

Answer: 3. 46%

Question 54. Infrared radiations are detected by

1. Spectrometer
2. Pyrometer
3. Nanometer
4. Photometer

Answer: 2. Pyrometer

Question 55. The plots of intensity vs. wavelength for three black bodies at temperatures T₁, T_2, and T₃ respectively are as shown. Their temperatures are such that-

1. T₁> T₂> T₃
2. T₁> T₃> T₂
3. T2> T₃> T₁
4. T₃> T₂> T₁

Answer: 2. T₁> T₃> T₂

Question 56. In which of the following phenomenon heat convection does not take place

1. Land and sea breeze
2. Boiling of water
3. Heating of glass surface due to filament of the bulb
4. The air around the furnace

Answer: 3. Heating of glass surface due to a filament of the bulb

Question 57. The energy radiated by a black body is directly proportional to :

1. T²
2. T^-2
3. T⁴
4. T

Answer: 3. T⁴

Question 58. When a substance is gradually heated, its initial color is

1. Red
2. Green
3. Yellow
4. White

Answer: 1. Red

Question 59. If the temperature becomes double, the emitted radiation will be :

1. 16 times
2. 8 times
3. times
4. 32 times

Answer: 1. 16 times

Question 60. If at temperature T₁= 1000 K, the wavelength is 1.4 × 10^-6 m, then at what temperature the wavelength will be 2.8 × 10^-6m?

1. 2000 K
2. 500 K
3. 250 K
4. None of these

Answer: 2. 500 K

Question 61. A black body is heated from 27°C to 927°C the ratio of radiations emitted will be :

1. 1: 256
2. 1: 64
3. 1: 16
4. 1: 4

Answer: 1. 1: 256

Question 62. Water is used to cool the radiators of engines in cars because :

1. Of its low boiling point
2. Of its high specific heat
3. Of its low-density
4. Of its easy availability

Answer: 2. Of its high specific heat

Question 63. The color of the star indicates its :

1. Temperature
2. Distance
3. Velocity
4. Size

Answer: 1. Temperature

Question 64. The means of energy transfer in a vacuum are:

1. Irradiation
2. Convection
3. Radiation
4. Conduction

Answer: 3. Radiation

Question 65. The temperature of the black body increases from T to 2T. The factor by which the rate of emission will increase is

1. 4
2. 2
3. 16
4. 8

Answer: 3. 16

Question 66. Let there be four articles having colors blue, red, black, and white. When they are heated together and allowed to cool, which article will cool at the earliest?

1. Blue
2. Red
3. Black
4. White

Answer: 2. Red

Question 67. A piece of red glass when heated in dark to red hot states will appear to be :

1. White
2. Red
3. Green
4. Invisible

Answer: 3. Green

Question 68. What is the mode of heat transfer by which a hot cup of coffee loses most of its heat?

1. Condition
2. Convection
3. Evaporation
4. Radiation

Answer: 1. Condiction

Question 69. Which one of the following processes depends on gravity :

1. Conduction
2. Convection
3. Radiation
4. None of the above

Answer: 2. Convection

Question 70. For a black body at a temperature of 727°C, its radiating power is 60 watts and the temperature of surrounding is 227°C. If the temperature of the black body is changed to 1227°C then its radiating power will be

1. 304 W
2. 320 W
3. 240 W
4. 120 W

Answer: 2. 320 W

Question 71. Wien’s displacement law expresses a relation between-

1. Wavelength corresponds to maximum energy and temperature.
2. Radiation energy and wavelength
3. Temperature and wavelength
4. Color of light and temperature

Answer: 1. Wavelength corresponds to maximum energy and temperature.

Question 72. The unit of Stefan’s constant is-

1. Watt-m²-K⁴
2. Watt-m²/K⁴
3. Watt/m²-K
4. Watt/m²K⁴

Answer: 4. Watt/m²K⁴

Question 73. Which of the following radiations has the least wavelength?

1. γ-rays
2. β-rays
3. α-rays
4. X-rays

Answer: 1. γ-rays

Question 74. If the temperature of the sun were to increase from T to 2T and its radius from R to 2R, then the ratio of the radiant energy received on Earth to what it was previously would be

1. 4
2. 16
3. 32
4. 64

Answer: 4. 64

Question 75. Assuming the sun to be a spherical body of radius R at a temperature of T K, evaluate the total radiant power, incident on Earth, at a distance r from the Sun. (earth radius = r₀)

1. \(\frac{R^2 \sigma T^4}{r^2}\)
2. \(\frac{4 \pi r_0^2 \quad R^2 \sigma T^4}{r^2}\)
3. \(\frac{\pi r_0^2 \quad R^2 \sigma T^4}{r^2}\)
4. \(\frac{r_0^2 \quad R^2 \sigma T^4}{4 \pi r^2}\)

Answer: 3. \(\frac{\pi r_0^2 \quad R^2 \sigma T^4}{r^2}\)

Question 76. The energy emitted per second by a black body at 1227ºC is E. If the temperature of the black body is increased to 2727ºC, the energy emitted per second in terms of E is –

1. 16 E
2. E
3. 4E
4. 2E

Answer: 1. 16 E

Question 77. Temp. of a black body is 3000 k. When the black body cools, then change in wavelength Δ λ = 9 microns corresponding to maximum energy density. Now temp. of a black body is-

1. 300 K
2. 2700 K
3. 270 K
4. 1800 K

Answer: 1. 300 K

Question 78. If the radius of the sun is RS, the radius of the orbit of the earth about the sun is Reand σ is Stefan’s constant, then the amount of radiation falling per second on a unit area of the earth’s surface is-

1. \(\left(\frac{R_s}{R_e}\right)^2 \sigma \mathrm{T}^4\)
2. \(\left(\frac{R_e}{R_s}\right)^2 \sigma \mathrm{T}^4\)
3. \(\frac{\sigma}{T^4}\left(\frac{R_s}{R_e}\right)^2\)
4. \(\left(\frac{R_e}{R_s}\right)^2 \frac{T^4}{\sigma}\)

Answer: 1. \(\left(\frac{R_s}{R_e}\right)^2 \sigma \mathrm{T}^4\)

Question 79. Which of the following surfaces will absorb maximum radiant energy-

1. Black
2. Rough
3. Smooth white
4. Rough black

Answer: 4. Rough black

Question 80. After heating two pieces of iron, they are taken to a dark room. One of them appears red and another appears blue, then-

1. The temperature of the red piece will be higher.
2. The temperature of the blue piece will be higher.
3. The temperature of both pieces will be the same.
4. Nothing can be said about their temp.

Answer: 2. The temperature of the blue piece will be higher.

Question 81. If the temperature of a lamp is about 600K, then the wavelength at which maximum emission takes place will be (Wien’s constant b = 3 × 10^-3 m-K)

1. 500 A°
2. 5000 A°
3. 50000 A°
4. 500000 A°

Answer: 3. 50000 A°

Question 82. The rate of cooling of a sphere of thermal capacity 1000 cal/K is 400 J/s, and its rate of fall of temperature is-

1. 0.095 K/min
2. 0.62 K/min
3. 2.8 K/min
4. 5.7 K/min

Answer: 4. 5.7 K/min

Question 83. If maximum spectral emissivity at temperature T₁ K is at wavelength λ₁, then the wavelength of maximum emissivity at temperature T₂ K will be

1. \(\frac{\lambda_1 T_2}{2}\)
2. \(\lambda_1\left(\frac{T_1}{T_2}\right)^4\)
3. \(\lambda_1\left(\frac{T_1}{T_2}\right)^5\)
4. \(\frac{\lambda_1 \mathrm{~T}_1}{\mathrm{~T}_2}\)

Answer: 4. \(\frac{\lambda_1 \mathrm{~T}_1}{\mathrm{~T}_2}\)

Question 84. The spectral emissive power of a black body at a temperature of 6000K is maximum at λ_m= 5000 A°. If the temperature is increased by 10%, then the decrease in λ_m will be

1. 2.5%
2. 5.0%
3. 7.5%
4. 10%

Answer: 4. 10%

Question 85. The rate of emission of energy by a unit area of a body is 10 watts and that of the sun is 106 watts. The emissive power of the body is 0.1. If the temperature of the sun is 6000K, then the temperature of the body will be

1. 6000K
2. 600K
3. 60010K
4. (600 10)K

Answer: 2. 600K

Question 86. The ratio of masses of two copper spheres of identical surfaces is 8: 1. If their temperatures are 2000K and 1000K respectively then the ratio of energies radiated per second by the two is-

1. 128: 1
2. 64: 1
3. 16: 1
4. 4: 1

Answer: 2. 64: 1

Question 87. A solid body is heated upto very high temperatures. As we go on heating, its brightness increases and it appears white at the end. The sequence of the color observed as the temperature of the body increases will be

1. Yellow, green, red, white.
2. Green, yellow, red, white.
3. Red, green, yellow, white.
4. Red, yellow, green, white.

Answer: 4. Red, yellow, green, white.

Question 88. The effective area of a black body is 0.1 m² and its temperature is 100 K. The amount of radiation emitted by it per minute is –

1. 1.34 cal
2. 8.1 cal
3. 5.63 cal
4. 1.34 J

Answer: 2. 8.1 cal

Question 89. What is the energy of emitted radiation from the Sun when the temperature is doubled-

1. 2
2. 4
3. 8
4. 16
5. Answer: 4. 16

Question 90. Newton’s law of cooling is a special case of

1. Wien’s displacement law
2. Kirchoff’s law
3. Stefan’s law
4. Planck’s law

Answer: 3. Stefan’s law

Question 91. A hot liquid is kept in a big room. Its temperature is plotted as a function of time. Which of the following curves may represent the plot?

1. a
2. c
3. d
4. b

Answer: 1. a

Question 92. A body takes 4 minutes to cool from 100°C to 70°C. To cool from 70°C to 40°C it will take-(room temperature is 15°C)

1. 7 minutes
2. 6 minutes
3. 5 minutes
4. 4 minutes

Answer: 1. 7 minutes

Question 93. A cup of tea cools from 80°C to 60°C in one minute. The ambient temperature is 30°C. In cooling from 60°C to 50°C it will take-

1. 30 seconds
2. 60 seconds
3. 96 seconds
4. 48 seconds

Answer: 4. 48 seconds

Question 94. A hot liquid cools from 70°C to 60°C in 5 minutes. The time needed by the same liquid to cool from 60°C to 50°C will be

1. Less than 5 minutes
2. More than 5 minutes
3. Equal to 5 minutes
4. Less or more than 5 minutes depends on the density of the liquid

Answer: 2. More than 5 minutes

Question 95. Which of the following is a true statement?

1. A good absorber is a bad conductor
2. Each body emits and absorbs radiation at each temperature
3. In a black body energy of emitted radiation is equal for all wavelength
4. Planck’s law gives the relation between the maximum wavelength of black body radiation and its temperature.

Answer: 2. Each body emits and absorbs radiation at each temperature

Question 96. A body takes 10 minutes to cool down from 62°C to 50°C. If the temperature of the surroundings is 26°C then in the next 10 minutes temperature of the body will be :

1. 38°C
2. 40°C
3. 42°C
4. 44°C

Answer: 3. 42°C

Question 97. A body cools from 60°C to 50°C in 10 minutes. If the room temperature is 25°C and assuming Newton’s law of cooling to hold good, the temperature of the body at the end of the next 10 minutes will be :

1. 45°C
2. 41.67°C
3. 40°C
4. 38.5°C

Answer: 2. 41.67°C

Question 98. Two spheres of radii in the ratio 1: 2 and densities in the ratio 2: 1 and of the same specific heat, are heated to the same temperature and left in the same surrounding. Their rate of cooling will be in the ratio :

1. 2: 1
2. 1: 1
3. 1: 2
4. 1: 4

Answer: 2. 1: 1

Question 99. The formation of ice is started in a lake with water at 0°C. When the atmospheric temperature is –10°C. If the time taken for 1 cm of ice to be formed is 7 hours, the time taken for the thickness of ice to increase from 1cm to 2 cm is :

1. Less than 7 hours
2. 7 hours
3. More than 14 hours
4. More than 7 hours but less than 14 hours

Answer: 3. More than 14 hours

Question 100. Two circular discs A and B with equal radii are blackened. They are heated to the same temperature and are cooled under identical conditions. What inference do you draw from their cooling curves?

1. A and B have the same specific heats
2. The specific heat of A is less
3. The specific heat of B is less
4. Nothing can be said

Answer: 2. Specific heat of A is less

Question 101. According to Newton’s law of cooling, the rate of cooling of a body is proportional to (Δθ)n, where Δθ is the difference between the temperature of the body and the surroundings, and n is equal to

1. 2
2. 3
3. 4
4. 1

Answer: 4. 1

Question 102. A liquid cools down from 70° C to 60°C in 5 min. The time taken to cool it from 60°C to 50°C will be

1. 5 min
2. Lesser than 5 min
3. Greater than 5 min
4. Lesser or greater than 5 minutes depending upon the density of the liquid

Answer: 3. Greater than 5 min

Question 103. The heat capacities of three liquids A, B, and C of the same volumes are in the ratio 3: 2: 1. They are allowed to cool in the same surroundings and same conditions for the same temperature difference. Which of these will cool first?

1. A
2. B
3. C
4. All will cool at the same time

Answer: 3. C

Question 104. The temperature of a room is 30°C. A body kept in it takes 4 minutes to cool from 61°C to 59°C. The time taken by the body to cool from 51°C to 49°C will be

1. 4 min.
2. 5 min.
3. 6 min.
4. 8 min.

Answer: 3. 6 min.

Question 105. Two cylindrical conductors A and B of the same metallic material have their diameters in the ratio 1: 2 and lengths in the ratio 2: 1. If the temperature difference between their ends is the same, the ratio of heat conducted respectively by A and B per second is,

1. 1: 2
2. 1: 4
3. 1: 16
4. 1: 8

Answer: 4. 1: 8

Question 106. According to Kirchoff’s law-

1. a_λe_λ= E_λ
2. E_λa_λ= e_λ
3. a_λ= e_λE_λ
4. E_λ, a_λ, e_λ= const.

Answer: 2. E_λa_λ= e_λ

Question 107. A spherical solid black body of radius ‘r’ radiates power ‘H’ and its rate of cooling is ‘C’. If the density is constant then which of the following is/are true?

1. H ∝ r and c ∝ r²
2. H ∝ r² and c ∝
3. H ∝ r and c ∝ 2
4. H ∝ r² and c ∝ r²

Answer: 2. H ∝ r² and c ∝

Question 108. Which of the following is nearest to Blackbody-

1. An enclosure with a small hole
2. Carbon black
3. Ebonite
4. None of these

Answer: 1. An enclosure with a small hole

Question 109. Which of the following processes is reversible?

1. Transfer of heat by radiation
2. Electrical heating of nichrome wire
3. Transfer of heat by conduction
4. Isothermal compression

Answer: 4. Isothermal compression

Question 110. Assuming the sun to have a spherical outer surface of radius r, radiating like a black body at temperature t°C, the power received by a unit surface, (normal to the incident rays) at a distance R from the center of the sun is (considering solar constant to be uniform)

1. \(\frac{4 \pi r^2 t^4}{R^2}\)
2. \(\frac{r^2 \sigma(t+273)^4}{4 \pi R^2}\)
3. \(\frac{16 \pi^2 r^2 \sigma t^4}{R^2}\)
4. \(\frac{r^2 \sigma(t+273)^2}{R^2}\)

Answer: 4. \(\frac{r^2 \sigma(t+273)^2}{R^2}\)

Question 111. Which of the following is more close to a black body?

1. Blackboard paint
2. Green leaves
3. Black holes
4. Red roses

Answer: 1. Blackboard paint

Question 112. A black body is at a temperature of 2800 K. The energy of radiation emitted by this object with a wavelength between 499 nm and 500 nm is U₁, between 999 nm and 1000 nm is U_2, and between 1499 nm and 1500 nm is U₃. The Wien constant b = 2.88 × 106 nm K. Then

1. U₁= 0
2. U₃= 0
3. U₁> U₂
4. U₂>U₁

Answer: 4. U₂>U₁

Question 113. The temperature of bodies X and Y vary with time as shown in the figure. If the emissivity of bodies X and Y are e_X and e_Y and absorptive powers are A_X and A_Y, (assume other conditions are identical for both) then:

1. e_Y> e_X, A_Y> A_X
2. e_Y< e_X, A_Y< A_X
3. e_Y> e_X, A_Y< A_X
4. e_Y< e_X, A_Y> A_X

Answer: 1. e_Y> e_X, A_Y> A_X

Question 114. Three discs of the same material A, B, and C of radii 2 cm, 4 cm, and 6 cm respectively are coated with carbon black. Their wavelengths corresponding to maximum spectral radiancy are 300, 400, and 500 nm respectively then maximum power will be emitted by

1. A
2. B
3. C
4. Same for all

Answer: 2. B

Question 115. Three graphs marked 1, 2, and 3 represent the variation of maximum emissive power and wavelength of radiation of the sun, a welding arc, and a tungsten filament. Which of the following combinations is correct

1. 1- tungsten filament, 2 → welding arc, 3 → sun
2. 2- tungsten filament, 3 → welding arc, 1 → sun
3. 3- tungsten filament, 1 → welding arc, 2 → sun
4. 2- tungsten filament, 1 → welding arc, 3 → sun

Answer: 1. 1- tungsten filament, 2 → welding arc, 3 → sun

Question 116. Two rectangular blocks, having identical dimensions, can be arranged either in configuration Ι or in configuration ΙΙ as shown in the figure, One of the blocks has a thermal conductivity of k, and the other 2k. The temperature difference between the ends along the x-axis is the same in both configurations. It takes 9s to transport a certain amount of heat from the hot end to the cold end in configuration 1. The time to transport the same amount of heat in the configuration 2 is:

1. 2.0 s
2. 3.0 s
3. 4.5 s
4. 6.0 s

Answer: 1. 2.0 s

Question 117. Parallel rays of light of intensity Ι = 912 Wm^-2 are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Biltzmann constant σ = 5.7 × 10^-8 Wm^-2 K^-4 and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close to:

1. 330 K
2. 660 K
3. 990 K
4. 1550 K

Answer: 1. 330 K

Question 118. Two spherical stars A and B emit blackbody radiation. The radius of A is 400 times that of B and A emits 10⁴ times the power emitted from B. The ratio \(\left(\frac{\lambda_A}{\lambda_B}\right)\) to their wavelengths λ_A and λ_B at which the peaks occur in their respective radiation curves is:

1. 1
2. 2
3. 3
4. 4

Answer: 2. 2

Question 119. The earth radiates in the infrared region of the spectrum. The spectrum is correctly given by :

1. Rayleigh-Jeans law
2. Planck’s law of radiation
3. Stefan’s law of radiation
4. Wien’s law

Answer: 1. Rayleigh-Jeans law

Question 120. Two spheres of different materials one having a radius double of other and a wall thickness 1/4 of the other are filled with ice. If the time required to completely melt the ice is 25 min. for a larger radius sphere and 16 min. for a smaller radius sphere, then the ratio of the thermal conduction coefficient for the material of larger radius to that of the thermal conduction coefficient for the material of smaller radius sphere will be.

1. 4: 5
2. 5: 4
3. 8: 25
4. 1: 25

Answer: 3. 8: 25

Question 121. A black body is at room temperature. It is placed in a furnace, and it is observed that

1. In the beginning, it is seen most black, and later on, it is seen as the brightest.
2. It is always seen as black.
3. It can’t be resolved at any times
4. In the beginning, it is seen mostly black, and later on it can’t be resolved.

Answer: 1. In the beginning it is seen as the most black and later on it is seen brightest.

Question 122. If a liquid takes 30 sec. in cooling of 95°C to 90°C and 70 sec. in cooling of 55°C to 50°C then temp. of the room is-

1. 16.5 °C
2. 22.5 °C
3. 28.5 °C
4. 32.5 °C

Answer: 2. 22.5 °C

Question 123. A body takes 2 minutes in cooling from 365K to 361K. If the room temperature is 293K, then the time taken to cool from 344K to 342K will be

1. 1 min.
2. 1.2 min.
3. 1.4 min.
4. 1.8 min.

Answer: 3. 1.4 min.

Question 124. The reflection and absorption coefficients of a given surface at 0°C for a fixed wavelength are 0.5 (each). At the same temperature and wavelength, the transmission (coefficient) of the surface will be

1. 0.5
2. 1.0
3. Zero
4. In between zero and one

Answer: 3. Zero

Question 125. The earth receives radiation from the sun at the rate of 1400 watts/m². The distance from the center of the sun to the surface of the earth is 1.5 × 1011 m and the radius of the sun is 7.0 × 108 m. Treating the sun as a black body the temperature of the sun will be

1. 6000K
2. 5500K
3. 5800K
4. 6200K

Answer: 3. 5800K

Question 126. The rate of cooling of a heated solid sphere is R cal/min. If it is divided into two hemispheres the rate of cooling at the same temperature will become-

1. 1.25R cal/min.
2. 1.5R cal/min.
3. 1.75R cal/min.
4. 2.5R cal/min.

Answer: 2. 1.5R cal/min.

Question 127. Equal volumes of a liquid of relative density 1.02 and water are allowed to cool from 80°C to 60°C in the same surroundings. The times taken are 8 mts and 15 mts respectively. The specific heat of the liquid in cal/gm-°C is-

1. 0.52
2. 0.81
3. 1.02
4. 1.23

Answer: 1. 0.52

Question 128. Two identical calorimeters of negligible heat capacities are filled with two liquids A & B whose densities are in the ratio 4 : 3. The ratio of times taken in cooling from 80°C to 75°C is 5: 6. The ratio of their specific heats is-

1. 1: 2
2. 5: 6
3. 4 : 3
4. 5: 8

Answer: 4. 5: 8

Question 129. Blackened metal foil receives heat from a heated sphere placed at a distance r from it. It is found that foil receives power P. If the temperature and the distance of the sphere are doubled, then the power received by the foil will be

1. P
2. 2P
3. 8P
4. 4P

Answer: 4. 4P

Question 130. The temperature of the two outer surfaces of a composite slab, consisting of two materials K and 2K, and thickness x and 4x, respectively, are T₂ and T₁(T₂> T₁). The rate of heat transfer through the slab,\(\left(\frac{A\left(T_2-T_1\right) K}{x}\right) f\) with f equal to–

1. 1
2. 1/2
3. 2/3
4. 1/3

Answer: 4. 1/3

Question 131. If the radius of a star is R and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is Q?

1. Q /4πR²σ
2. (Q /4πR²σ)–1/2
3. (4πR²Q/σ)1/4
4. (Q/ 4πR²σ)1/4

Answer: 4. (Q/ 4πR²σ)1/4

Question 132. A slab of stone of area 0.36 m² and thickness 0.1 m is exposed on the lower surface to steam at 100°C. A block of ice at 0°C rests on the upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of slab is :(Given latent heat of fusion of ice = 3.36 × 105 J kg^-1) :

1. 1.24 J/m/s/°C
2. 1.29 J/m/s/°C
3. 2.05 J/m/s/°C
4. 1.02 J/m/s/°C

Answer: 1. 1.24 J/m/s/°C

Question 133. A piece of iron is heated in a flame. It first becomes dull red then becomes reddish yellow and finally turns to white hot. The correct explanation for the above observation is possible by using :

1. Wien’s Displacement Law
2. Kirchoff’s Law
3. Newton’s Law of Cooling
4. Stefan’s Law

Answer: 1. Wien’s displacement Law

Question 134. A certain quantity of water cools from 700C to 600C in the first 5 minutes and to 540C in the next 5 minutes. The temperature of the surroundings is;

1. 450C
2. 200C
3. 420C
4. 100C

Answer: 1. 450C

Question 135. On observing light from three different stars P, Q and R, it was found that the intensity of the violet colour is maximum in the spectrum of P, the intensity of the green colour is maximum in the spectrum of R and the intensity of the red colour is maximum in the spectrum in the spectrum of Q. If T_P, T_Q, and T_R are the respective absolute temperature of P, Q, and R, then it can be concluded from the above observations that :

1. T_P> T_R> T_Q
2. T_P< T_R< T_Q
3. T_P< T_Q<T_R
4. T_P >T_QT_R

Answer: 1. T_P> T_R> T_Q

Question 136. The two ends of a metal rod are maintained at temperatures 100ºC and 110ºC. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200ºC and 210ºC, the rate of heat flow will be :

1. 16.8 J/s
2. 8.0 J/s
3. 4.0 J/s
4. 44.0 J/s

Answer: 3. 4.0 J/s

Question 137. The coefficient of linear expansion of brass and steel rods are α₁ and α₂. Lengths of brass and steel rods are l₁ and l₂ respectively. If (l₂– l₁) is maintained the same at all temperatures, which one of the following relations holds good?

1. α₁l₁= α₂l₂
2. α₁l₂= α₂l₁
3. α₁l_2² = α₂l_1²
4. α_1²l₂= α_2² l₁

Answer: 1. α₁l₁= α₂l₂

Question 138. A refrigerator works between 4°C and 30°C. it is required to remove 600 calories of heat every second to keep the temperature of the refrigerated space constant. The power required is : (Take 1 cal = 4.2 Joules)

1. 2365 W
2. 2.365 W
3. 23.65 W
4. 236.5 W

Answer: 4. 236.5 W

Question 139. A piece of ice falls from a height h so that it melts completely. Only one-quarter of the heat produced is absorbed by the ice and all energy of ice gets converted into heat during its fall. The value of h is : [Latent heat of ice is 3.4 × 10⁵ J/Kg and g = 10 N/kg]

1. 68 km
2. 34 km
3. 544 km
4. 136 km

Answer: 4. 136 km

Question 140. A block body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is U₁ at wavelength 500 nm is U₂ and that at 1000 nm is U₃. Wien’s constant, b = 2.88 × 10⁶ nmK. Which of the following is correct?

1. U₂> U₁
2. U₁= 0
3. U₃= 0
4. U₁> U₂

Answer: 1. U₂> U₁

Question 141. Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100ºC, while the other one is at 0ºC. If the two bodies are brought into contact, then assuming no heat loss, the final common temperature is

1. 0º C
2. 50º C
3. More than 50º C
4. Less than 50º C but greater than 0º C

Answer: 3. More than 50º C

Question 142. A body cools from a temperature of 3T to 2T in 10 minutes. The room temperature is T. Assume that Newton’s law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be

1. T
2. \(\frac{7}{4} \mathrm{~T}\)
3. \(\frac{3}{2} \mathrm{~T}\)
4. \(\frac{4}{3} \mathrm{~T}\)

Answer: 3. \(\frac{3}{2} \mathrm{~T}\)

Question 143. Two rods A and B of different materials are welded together as shown in the figure. Their thermal conductivities are K₁ and K₂. The thermal conductivity of the composite rod will be

1. \(\frac{\mathrm{K}_1+\mathrm{K}_2}{2}\)
2. \(\frac{3\left(\mathrm{~K}_1+\mathrm{K}_2\right)}{2}\)
3. \(\mathrm{K}_1+\mathrm{K}_2\)
4. \(2\left(\mathrm{~K}_1+\mathrm{K}_2\right)\)

Answer: 1. \(\frac{\mathrm{K}_1+\mathrm{K}_2}{2}\)

Question 144. A spherical black body with a radius of 12 cm radiates 450-watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watts would be :

1. 225
2. 450
3. 1000
4. 1800

Answer: 4. 1800

Question 145. The power was radiated by a black body in P and it radiated maximum energy at wavelength, λ₀. If the temperature of the black body is now changed so that it radiates maximum energy at wavelength \(\frac{3}{4} \lambda_0\) the power radiated by it becomes nP. The value of n is :

1. \(\frac{3}{4}\)
2. \(\frac{81}{256}\)
3. \(\frac{256}{81}\)
4. \(\frac{4}{3}\)

Answer: 3. \(\frac{256}{81}\)

Question 146. A copper rod of 88 cm and an aluminum rod of unknown length have their increase in length independent of an increase in temperature. The length of aluminium rod is (α_Cu = 1.7 × 10^-5 K^-1 and α_Al = 2.2 × 10^-5 K^-1)

1. 68 cm
2. 6.8 cm
3. 113.9 cm
4. 88 cm

Answer: 1. 68 cm

Question 147. The unit of thermal conductivity is :

1. W m^-1 K^-1
2. J m K^-1
3. J m^-1 K^-1
4. W m K^-1

Answer: 1. W m^-1 K^-1

Question 148. An object kept in a large room having an air temperature of 25ºC takes 12 minutes to cool from 80ºC to 70ºC. The time taken to cool the same object from 70º to 60ºC would be nearly

1. 10 min
2. 12 min
3. 20 min
4. 15 min

Answer: 4. 15 min

Question 149. A deep rectangular pond of surface area A, containing water (density = ρ, specific heat capacity = s), is located in a region where the outside air temperature is at a steady value of –26ºC. The thickness of the ice layer in this pond, at a certain instant, is x. Taking the thermal conductivity of ice as K, and its specific latent heat of fusion as L, the rate of increase of the thickness of the ice layer, at this instant, would be given by

1. 26K/ρx(L-4s)
2. 26K/(ρx2L)
3. 26K/(ρxL)
4. 26K/ρx(L+4s)

Answer: 3. 26K/(ρxL)

Question 150. Three stars A, B, and C have surface temperatures T_A, T_B, and T_C respectively. Star A appears bluish, star B appears reddish, and star C is yellowish. Hence

1. T_A> T_B> T_C
2. T_B> T_C> T_A
3. T_C> T_B> T_A
4. T_A> T_C> T_B

Answer: 4. T_A> T_C> T_B

Question 151. An ideal gas equation can be written as \(\mathrm{P}=\frac{\rho R T}{M_0}\) where ρ and M₀ are respectively,

1. Mass density is, the mass of the gas
2. Number density, molar mass
3. Mass density, molar mass
4. Number density, the mass of the gas

Answer: 3. Mass density, molar mass

Question 152. A cylinder contains hydrogen gas at a pressure of 245 K Pa and a temperature of 270C density is (R=8.3 J mol^-1 K^-1)

1. 0.02 kg/m³
2. 0.5 kg/m³
3. 0.2 kg/m³
4. 0.1 / kg m

Answer: 3. 0.2 kg/m³

Question 153. A cup of coffee cools from 90°C to 80°C in two minutes, when the room temperature is 20°C. The time taken by a similar cup of coffee to cool from 80°C to 60°C at room temperature same at 20°C is

1. \(\frac{13}{5} t\)
2. \(\frac{10}{13} t\)
3. \(\frac{5}{13} t\)
4. \(\frac{13}{10} t\)

Answer: 1. \(\frac{13}{5} t\)

Question 154. A flask containing air at 27ºC is corked up at atmospheric pressure. The cork can be forced out by a pressure of 2.5 atmospheres. To what temperature the flask should be heated to do that?

1. 150 K
2. 300 K
3. 600 K
4. 750 K

Answer: 4. 750 K

Question 155. 1 kcal of heat flowing through a rod of iron per second. When the rod is cut down to 4 pieces then what will be the heat flowing through each piece per second having the same differential temperature (temperature gradient)?

1. (1/2) kcal
2. (1/4) kcal
3. 1 kcal
4. (1/15) kcal

Answer: 3. 1 kcal

Question 156. Black holes in orbit around a normal star are detected from the earth due to the frictional heating of infalling gas into the black hole, which can reach temperatures greater than 106K. Assuming that the infalling gas can be modelled as a blackbody radiator then the wavelength of maximum power lies

1. In the visible region
2. In the X-ray region
3. In the microwave region
4. In the gamma-ray region of the electromagnetic spectrum.

Answer: 2. In the X-ray region

Question 157. Two conductors having the same width and length, thickness d₁ and d₂ thermal conductivity K₁ and K₂ are placed one above the other. Find the equivalent thermal conductivity.

1. \(\frac{\left(d_1+d_2\right)\left(K_1 d_2+K_2 d_1\right)}{2\left(K_1+K_2\right)}\)
2. \(\frac{\left(d_1-d_2\right)\left(K_1 d_2+K_2 d_1\right)}{2\left(K_1+K_2\right)}\)
3. \(\frac{K_1 d_1+K_2 d_2}{d_1+d_2}\)
4. \(\frac{K_1+K_2}{d_1+d_2}\)

Answer: 3. \(\frac{K_1 d_1+K_2 d_2}{d_1+d_2}\)

Question 158. A long metallic bar carries heat from one of its ends to the other end under a steady-state. The variation of temperature θ along the length x of the bar from its hot end is best described by which of the following figures

Answer: 1

Question 159. If a piece of metal is heated to temperature θ and then allowed to cool in a room which is at temperature θ₀, the graph between the temperature T of the metal and time t will be closest to:

Answer: 3

Question 160. Three rods of Copper, brass, and steel are welded together to form a Y-shaped structure. Area of cross section of each rod = 4 cm². The end of the copper rod is maintained at 100°C whereas ends of brass and steel are kept at 0°C. Lengths of the copper, brass, and steel rods are 46, 13, and 12 cm respectively. The rods are thermally insulated from surroundings except at the ends. Thermal conductivities of copper, brass, and steel are 0.92, 0.26, and 0.12 CGS units respectively. The rate of heat flow through copper rod is:

1. 1.2 cal/s
2. 2.4 cal/s
3. 4.8 cal/s
4. 6.0 cal/s

Answer: 4. 6.0 cal/s

Question 161. An ideal gas undergoes a quasi-static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PVⁿ = constant, then n is given by (Here C_p and C_v are molar specific heat at constant pressure and constant volume, respectively ) :

1. \(n=\frac{C-C_p}{C-C_V}\)
2. \(n=\frac{C_p-C}{C-C_V}\)
3. \(n=\frac{C-C}{C-C p}\)
4. \(n=\frac{C_p}{C_V}\)

Answer: 1. \(n=\frac{C-C_p}{C-C_V}\)

Question 162. Temperature difference of 120°C is maintained between two ends of a uniform rod AB of length 2L. Another bent rod PQ, of the same cross-section as AB and length \(\frac{3 L}{2}\), is connected across AB (see figure). In a steady state, the temperature difference between P and Q will be close to :

1. 75°C
2. 45°C
3. 60°C
4. 35°C

Answer: 2. 45°C

Question 163. A heat source at T = 10³ K is connected to another heat reservoir at T = 10² K by a copper slab that is 1m thick. Given that the thermal conductivity of copper is 0.1 WK^-1 m^-1, the energy flux through it in the steady state is :

1. 200 Wm^-2
2. 90 Wm^-2
3. 65 Wm^-2
4. 120 Wm^-2

Answer: 2. 90 Wm^-2

Question 164. A thermometer graduated according to a linear scale reads a value of x₀ when in contact with ice. What is the temperature of an object in °C, if this thermometer in contact with the object reads x₀/2?

1. 35
2. 60
3. 25
4. 40

Answer: 3. 25

Question 165. A cylinder of radius R is surrounded by a cylindrical of inner radius R and outer radius 2R. The thermal conductivity of the material of the inner cylinder is K₁ and that of the outer cylinder is K₂. Assuming no loss of heat, the effective thermal conductivity of the system for the heart flowing along the length of the cylinder is:

1. \(\mathrm{K}_1+\mathrm{K}_2\)
2. \(\frac{2 K_1+3 K_2}{2}\)
3. \(\frac{\mathrm{K}_1+\mathrm{K}_2}{2}\)
4. \(\frac{\mathrm{K}_1+3 \mathrm{~K}_2}{4}\)

Answer: 4. \(\frac{\mathrm{K}_1+3 \mathrm{~K}_2}{4}\)

Question 166. Two rods A and B of identical dimensions are at a temperature of 30°C. If A is heated upto 180°C and B upto T°C, then the new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4 : 3, then the value of T is :

1. 270°C
2. 200°C
3. 230°C
4. 250°C

Answer: 3. 230°C

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 = \(F_t v=(m k r)(k r t)=m k^2 r^2 t\)

 

Circular Motion Multiple Choice Question And Answers

Question 1. Two racing cars of masses m₁ and m₂ are moving in circles of radii r₁ and r₂ respectively; their speeds are such that they each make a complete circle in the same time t. The ratio of the angular speed of the first to the second car is :

1. m₁: m₂
2. r₁: r₂
3. 1: 1
4. m₁r₁: m₂r₂

Answer: 1. m₁: m₂

Question 2. A wheel is at rest. Its angular velocity increases uniformly and becomes 80 radians per second after 5 seconds. The total angular displacement is :

1. 800 rad
2. 400 rad
3. 200 rad
4. 100 rad

Answer: 3. 200 rad

Question 3. When a particle moves in a circle with a uniform speed

1. Its velocity and acceleration are both constant
2. Its velocity is constant but the acceleration changes
3. Its acceleration is constant but the velocity changes
4. Its velocity and acceleration both change

Answer: 4. Its velocity and acceleration both change

Question 4. The relation between an angular velocity, the position vector and the linear velocity of a particle moving in a circular path is.

1. \(\vec{\omega} \times \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{v}}\)
2. \(\vec{\omega} \cdot \vec{r}=\vec{v}\)
3. \(\overrightarrow{\mathbf{r}} \times \vec{\omega}=\overrightarrow{\mathrm{v}}\)
4. \(\vec{\omega} \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{v}}\)

Answer: 1. \(\vec{\omega} \times \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{v}}\)

Question 5. A wheel is of diameter 1m. If it makes 30 revolutions/sec., then the linear speed of a point on its circumference will be.

1. 30 π m/s
2. π m/s
3. 60π m/s
4. π/2 m/s

Answer: 1. 30π m/s

Question 6. In a uniform circular motion

1. Both the angular velocity and the angular momentum vary
2. The angular velocity varies but the angular momentum remains constant.
3. Both the angular velocity and the angular momentum stay constant
4. The angular momentum varies but the angular velocity remains constant.

Answer: 3. Both the angular velocity and the angular momentum stay constant

Question 7. The angular speed of a flywheel making 120 revolutions/minute is.

1. 2π rad/s
2. 4π2 rad/s
3. π rad/s
4. 4π rad/s

Answer: 4. 4π rad/s

Question 8. The angular velocity of the second needle in a watch is-

1. \(\frac{\pi}{30}\)
2. 2π
3. π
4. \(\frac{60}{\pi}\)

Answer: 1. \(\frac{\pi}{30}\)

Question 9. The average acceleration vector for a particle having a uniform circular motion is- 

1. A constant vector of magnitude \(\frac{v^2}{r}\)
2. A vector of magnitude \(\frac{v^2}{r}\) directed normal to the plane of the given uniform circular motion.
3. Equal to the instantaneous acceleration vector at the start of the motion.
4. A null vector.

Answer: 4. A null vector.

Question 10. The angular velocity of the minute hand of a clock is:

1. \(\frac{\pi}{30} \mathrm{rad} / \mathrm{s} \)
2. π rad/s
3. 2π rad/s
4. \(\frac{\pi}{1800} \mathrm{rad} / \mathrm{s}\)

Answer: 4. \(\frac{\pi}{1800} \mathrm{rad} / \mathrm{s}\)

Question 11. The second hand of a watch has a length of 6 cm. The speed of the endpoint and magnitude of the difference of velocities at two perpendicular positions will be :

1. 2π and 0 mm/s
2. \(2 \sqrt{2}\) π and 4.44 mm/s
3. \(2 \sqrt{2}\) π and 2π mm/s
4. 2π and \(2 \sqrt{2}\) π mm/s

Answer: 4. 2π and \(2 \sqrt{2}\) π mm/s

Question 12. An aeroplane revolves in a circle above the surface of the earth at a fixed height with a speed of 100 km/hr. The change in velocity after completing 1/2 revolution will be.

1. 200 km/hr
2. 150 km/hr
3. 300 km/hr
4. 400 km/hr

Answer: 1. 200 km/hr

Question 13. A particle moving on a circular path travels the first one-third part of the circumference in 2 sec and the next one-third part in 1 sec. The average angular velocity of the particle is (in rad/sec) –

1. \(\frac{2 \pi}{3}\)
2. \(\frac{\pi}{3}\)
3. \(\frac{4 \pi}{9}\)
4. \(\frac{5 \pi}{3}\)

Answer: 3. \(\frac{4 \pi}{9}\)

Question 14. A grind-stone starts revolving from rest, if its angular acceleration is 4.0 rad/sec² (uniform) then after 4 sec. What are its angular displacement and angular velocity respectively –

1. 32 rad, 16 rad/sec
2. 16 rad, 32 rad/sec
3. 64 rad, 32 rad/sec
4. 32 rad, 64 rad/sec

Answer: 1. 32 rad, 16 rad/sec

Question 15. Angular displacement of any particle is given θ = ω₀t +\(\frac{1}{2}\)αt² where ω₀ and α are constant if ω₀ = 1 rad/sec, α = 1.5 rad/sec² then in t = 2 sec. angular velocity will be (in rad/sec)

1. 1
2. 5
3. 3
4. 4

Answer: 4. 4

Question 16. A particle of mass M is revolving along a circle of radius R and another particle of mass m is revolving in a circle of radius r. If the periods of both particles are the same, then the ratio of their angular velocities is:

1. 1
2. \(\frac{R}{r}\)
3. \(\frac{r}{R}\)
4. \(\sqrt{\frac{R}{r}}\)

Answer: 1. 1

Question 17. In a uniform circular motion

1. Velocity and acceleration remain constant
2. Kinetic energy remains constant
3. Speed and acceleration changes
4. Only velocity changes and acceleration remains constant

Answer: 2. Kinetic energy remains constant

Question 18. Which of the following statements is false for a particle moving in a circle with a constant angular speed?

1. The velocity vector is tangent to the circle
2. The acceleration vector is tangent to the circle
3. The acceleration vector points to the center of the circle
4. The velocity and acceleration vectors are perpendicular to each other

Answer: 2. The acceleration vector is tangent to the circle

Question 19. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane, it follows that

1. Its velocity is constant
2. Its acceleration is constant
3. Its kinetic energy is constant
4. It moves in a straight line

Answer: 3. Its kinetic energy is constant

Question 20. A wheel is subjected to uniform angular acceleration about its axis. Initially, its angular velocity is zero. In the first 2 seconds, it rotates through an angle θ1. In the next 2 sec, it rotates through an additional angle θ \(\frac{\theta_2}{\theta_1}\) is

1. 1
2. 2
3. 3
4. 5

Answer: 3. 3

Question 21. If the equation for the displacement of a particle moving on a circular path is given by (θ)= 2t³ + 0.5, where θ is in radians and t in seconds, then the angular velocity of the particle after 2 sec from its start is

1. 8 rad/sec
2. 12 rad/sec
3. 24 rad/sec
4. 36 rad/sec

Answer: 3. 24 rad/sec

Question 22. For a particle in a non-uniform accelerated circular motion

1. Velocity is radial and acceleration is transverse only
2. Velocity is transverse and acceleration is radial only
3. Velocity is radial and acceleration has both radial and transverse components
4. Velocity is transverse and acceleration has both radial and transverse components

Answer: 4. Velocity is transverse and acceleration has both radial and transverse components

Question 23. Two particles P and Q are located at distances rP and rQ respectively from the axis of a rotating disc such that rP > rQ :

1. Both P and Q have the same acceleration
2. Both P and Q do not have any acceleration
3. P has greater acceleration than Q
4. Q has greater acceleration than P

Answer: 3. P has greater acceleration than Q

Question 24. Let ar and at represent radial and tangential acceleration. The motion of a particle may be circular if :

1. a_r = 0, a_t = 0
2. a_r = 0, a_t ≠ 0
3. a_r ≠ 0, a_t = 0
4. None of these

Answer: 3. a_r ≠ 0, a_t = 0

Question 25. A stone tied to one end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If a stone makes 25 revolutions in 14 seconds, the magnitude of the acceleration of the stone is :

1. 850 cm/s²
2. 996 cm/s²
3. 720 cm/s²
4. 650 cm/s²

Answer: 2. 996 cm/s²

Question 26. A body is moving in a circular path with acceleration a. If its velocity gets doubled, find the ratio of acceleration after and before the change :

1. 1: 4
2. 4: 1
3. 2: 1
4. 2: 1

Answer: 2. 4: 1

Question 27. A spaceman in training is rotated in a seat at the end of a horizontal arm of length 5m. If he can withstand acceleration upto 9 g then what is the maximum number of revolutions per second permissible? (Take g = 10 m/s²)

1. 13.5 rev/s
2. 1.35 rev/s
3. 0.675 rev/s
4. 6.75 rev/s

Answer: 3. 0.675 rev/s

Question 28. A particle of mass m is moving in a uniform circular motion. The momentum of the particle is

1. Constant over the entire path
2. Constantly changes and direction of change is along the tangent
3. Constantly changes and direction of change are along the radial direction
4. Constantly change and direction of change are along a direction which is the instantaneous vector sum of the radial and tangential direction

Answer: 3. Constantly changes and direction of change is along the radial direction

Question 29. A particle is going in a uniform helical and spiral path separately as shown in the figure with constant speed.

1. The velocity of the particle is constant in both cases
2. The acceleration of the particle is constant in both cases
3. The magnitude of acceleration is constant in (1) and decreasing in (2)
4. The magnitude of acceleration is decreasing continuously in both cases

Answer: 3. The magnitude of acceleration is constant in (1) and decreasing in (2)

Question 30. A car is traveling with linear velocity v on a circular road of radius r. If the speed is increasing at the rate of ‘a’ meter/sec², then the resultant acceleration will be –

1. \(\sqrt{\left[\frac{v^2}{r^2}-a^2\right]}\)
2. \(\sqrt{\left.\frac{v^4}{r^2}+a^2\right]}\)
3. \(\sqrt{\left[\frac{v^4}{r^2}-a^2\right]}\)
4. \(\sqrt{\left.\frac{v^2}{r^2}+a^2\right]}\)

Answer: 2. \(\sqrt{\left.\frac{v^4}{r^2}+a^2\right]}\)

Question 31. If the mass, speed & radius of rotation of a body moving on a circular path are increased by 50% then to keep the body moving in a circular path increase in force required will be –

1. 225%
2. 125%
3. 150%
4. 100%

Answer: 2. 125%

Question 32. A motorcycle driver doubles its velocity when he is having a turn. The force exerted outwardly will be.

1. Double
2. Half
3. 4 times
4. 1/4 times

Answer: 3. 4 times

Question 33. For a particle in circular motion, the centripetal acceleration is

1. Less than its tangential acceleration
2. Equal to its tangential acceleration
3. More than its tangential acceleration
4. May be more or less than its tangential acceleration

Answer: 4. May be more or less than its tangential acceleration

Question 34. If the radii of circular paths of two particles of the same masses are in the ratio of 1: 2, then in order to have the same centripetal force, their speeds should be in the ratio of:

1. 1: 4
2. 4: 1
3. 1 : \(\sqrt{2}\)
4. \(\sqrt{2}\): 1

Answer: 3. 1 : \(\sqrt{2}\)

Question 35. On a horizontal smooth surface, a mass of 2 kg is whirled in a horizontal circle by means of a string at an initial angular speed of 5 revolutions per minute. Keeping the radius constant the tension in the string is doubled. The new angular speed is near:

1. 14 rpm
2. 10 rpm
3. 2.25 rpm
4. 7 rpm

Answer: 4. 7 rpm

Question 36. If ar and at represent radial and tangential accelerations, the motion of a particle will be uniformly circular if

1. a_r = 0 and a_t = 0
2. a_r = 0 but a_t ≠ 0
3. a_r ≠ 0 but a_t = 0
4. a_r ≠ 0 and a_t ≠ 0

Answer: 3. a_r ≠ 0 but a_t = 0

Question 37. A string breaks if its tension exceeds 10 newtons. A stone of mass 250 gm tied to this string of length 10 cm is rotated in a horizontal circle. The maximum angular velocity of rotation can be.

1. 20 rad/s
2. 40 rad/s
3. 100 rad/s
4. 200 rad/s

Answer: 1. 20 rad/s

Question 38. A particle moving along a circular path due to a centripetal force having constant magnitude is an example of motion with :

1. Constant speed and velocity
2. Variable speed and velocity
3. Variable speed and constant
4. Velocity constant speed and variable velocity.

Answer: 4. Velocity constant speed and variable velocity.

Question 39. A stone of mass 0.5 kg tied with a string of length 1 meter is moving in a circular path with a speed of 4 m/sec. The tension acting on the string in Newton is –

1. 2
2. 8
3. 0.2
4. 0.8

Answer: 2. 8

Question 40. The formula for centripetal acceleration in a circular motion is.

1. \(\vec{\alpha} \times \overrightarrow{\mathbf{r}}\)
2. \(\vec{\omega} \times \overrightarrow{\mathrm{V}}\)
3. \(\vec{\alpha} \times \overrightarrow{\mathrm{V}}\)
4. \(\vec{\omega} \times \overrightarrow{\mathbf{r}}\)

Answer: 2. \(\vec{\omega} \times \overrightarrow{\mathrm{V}}\)

Question 41. A stone is moved around a horizontal circle with a 20 cm long string tied to it. If centripetal acceleration is 9.8 m/sec², then its angular velocity will be

1. 7 rad/s
2. 22/7 rad/s
3. 49 rad/s
4. 14 rad/s

Answer: 1. 7 rad/s

Question 42. A particle of mass m is executing a uniform motion along a circular path of radius r. If the magnitude of its linear momentum is p, the radial force acting on the particle will be.

1. pmr
2. rm/p
3. mp²/r
4. p²/mr

Answer: 4. p²/mr

Question 43. A particle moves in a circular orbit under the action of a central attractive force inversely proportional to the distance ‘r’. The speed of the particle is.

1. Proportional to r²
2. Independent of r
3. Proportional to r
4. Proportional to 1/r

Answer: 2. Independent of r

Question 44. A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to –k/r². The total kinetic energy of the particle is-

1. –k/r
2. k/r
3. k/2r
4. –k/2r

Answer: 3. k/2r

Question 45. A 500 kg car takes around a turn of radius of 50 m with a speed of 36 km/hr. The centripetal force acting on the car will be :

1. 1200 N
2. 1000 N
3. 750 N
4. 250 N

Answer: 2. 1000 N

Question 46. If the radii of circular paths of two particles of the same masses are in the ratio of 1: 2, then in order to have the same centripetal force, their speeds should be in the ratio of :

1. 1: 4
2. 4: 1
3. 1 : \(\sqrt{2}\)
4. \(\sqrt{2}\) : 1

Answer: 3. 1 : \(\sqrt{2}\)

Question 47. A particle is moving in a horizontal circle with constant speed. It has constant

1. Velocity
2. Acceleration
3. Kinetic energy
4. Displacement

Answer: 3. Kinetic energy

Question 48. A particle P will be equilibrium inside a hemispherical bowl of radius 0.5 m at a height 0.2 m from the bottom when the bowl is rotated at an angular speed (g = 10 m/sec²)-

1. \(10 / \sqrt{3} \mathrm{rad} / \mathrm{sec}\)
2. \(10 \sqrt{3} \mathrm{rad} / \mathrm{sec}\)
3. 10 rad/sec
4. \(\sqrt{20} \mathrm{rad} / \mathrm{sec}\)

Answer: 1. \(10 / \sqrt{3} \mathrm{rad} / \mathrm{sec}\)

Question 49. Three identical particles are joined together by a thread. All the three particles are moving on a smooth horizontal plane about point O. If the speed of the outermost particle is v₀, then the ratio of tensions in the three sections of the string is : (Assume that the string remains straight)

1. 3: 5: 7
2. 3: 4: 5
3. 7 : 11: 6
4. 3: 5: 6

Answer: 4. 3: 5: 6

Question 50. A heavy and big sphere is hanging with a string of length l, this sphere moves in a horizontal circular path making an angle θ with vertical then its time period is –

1. \(T=2 \pi \sqrt{\frac{\ell}{g}}\)
2. \(\mathrm{T}=2 \pi \sqrt{\frac{\ell \sin \theta}{\mathrm{g}}}\)
3. \(\mathrm{T}=2 \pi \sqrt{\frac{\cos \theta}{g}}\)
4. \(\mathrm{T}=2 \pi \sqrt{\frac{\ell}{g \cos \theta}}\)

Answer: 3. \(\mathrm{T}=2 \pi \sqrt{\frac{\cos \theta}{g}}\)

Question 51. A gramophone recorder rotates at an angular velocity of ω a coin is kept at a distance r from its center. If μ is static friction constant then the coil will rotate with gramophone if –

1. r > μ g > ω²
2. r = μ g/ω² only
3. r < μ g/ω²
4. r ≤ μ g/ω²

Answer: 4. r ≤ μ g/ω²

Question 52. A train A runs from east to west and another train B of the same mass runs from west to east at the same speed along the equator. A presses the track with a force F₁ and B presses the track with a force F₂.

1. F₁ > F₂
2. F₁ < F₂
3. F₁= F₂
4. The information is insufficient to find the relation between F₁ and F₂.

Answer: 1. F₁ > F₂

Question 53. A cyclist is moving on a circular track of radius 80 m with a velocity of 72 km/hr. He has to lean from the vertical approximately through an angle –

1. tan^-1(1/4)
2. tan^-1(1)
3. tan^-1(1/2)
4. tan^-1(2)

Answer: 3. tan^-1(1/2)

Question 54. A car of mass m is taking a circular turn of radius ‘r’ on a fictional level road with a speed v. In order that the car does not skid –

1. \(\frac{\mathrm{mv}^2}{\mathrm{r}} \geq \mu \mathrm{mg}\)
2. \(\frac{m v^2}{r} \leq \mu \mathrm{mg}\)
3. \(\frac{m v^2}{r}=\mu \mathrm{mg}\)
4. \(\frac{v}{r}=\mu \mathrm{mg}\)

Answer: 2. \(\frac{m v^2}{r} \leq \mu \mathrm{mg}\)

Question 55. A car travels at constant speed on a circular road on level ground. In the figure shown, Fair is the force of air resistance on the car. Which of the other forces best represents the horizontal force of the road on the car’s tires?

1. F_A
2. F_B
3. F_C
4. F_D

Answer: 2. F_B

Question 56. The driver of a car traveling at full speed suddenly sees a wall a distance r directly in front of him. To avoid a collision,

1. He should apply brakes sharply
2. He should turn the car sharply
3. He should apply brakes and then sharply turn
4. None of these

Answer: 1. He should apply brakes sharply

Question 57. A mass is supported on a frictionless horizontal surface. It is attached to a string and rotates about a fixed center at an angular velocity ω₀. If the length of the string and angular velocity are doubled, the tension in the string which was initially T₀0 is now –

1. T₀
2. T₀/2
3. 4T₀
4. 8T₀

Answer: 4. 8T₀

Question 58. Two masses M and m are attached to a vertical axis by weightless threads of combined length l. They are set in rotational motion in a horizontal plane about this axis with constant angular velocity ω. If the tensions in the threads are the same during motion, the distance of M from the axis is.

1. \(\frac{\mathrm{M} \ell}{\mathrm{M}+\mathrm{m}}\)
2. \(\frac{\mathrm{m} \ell}{\mathrm{M}+\mathrm{m}}\)
3. \(\frac{M+m}{M} \ell\)
4. \(\frac{\mathrm{M}+\mathrm{m}}{\mathrm{m}} \ell\)

Answer: 2. \(\frac{\mathrm{m} \ell}{\mathrm{M}+\mathrm{m}}\)

Question 59. A stone tied to the end of a string 1 m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44s, what is the magnitude and direction of acceleration of the stone?

1. \(\frac{\pi^2}{4} \mathrm{~ms}^{-2}\) and direction along the radius towards the centre 4
2. π²ms^-2 and direction along the radius away from the center
3. π²ms^-2 and direction along the radius towards the center
4. π²ms^-2 and direction along the tangent to the circle

Answer: 3. π²ms^-2 and direction along the radius towards the center

Question 60. The maximum velocity (in ms^-1) with which a car driver can traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is :

1. 60
2. 30
3. 15
4. 25

Answer: 2. 30

Question 61. A cylindrical vessel partially filled with water is rotated about its vertical central axis. It’s surface will

1. Rise equally
2. Rise from the sides
3. Rise from the middle
4. Lowered equally

Answer: 2. Rise from the sides

Question 62. A long horizontal rod has a bead that can slide along its length and is initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with a constant angular acceleration, α. If the coefficient of friction between the rod and the bead is μ, and gravity is neglected, then the time after which the bead starts slipping is-

1. \(\sqrt{\frac{\mu}{\alpha}}\)
2. \(\frac{\mu}{\sqrt{\alpha}}\)
3. \(\frac{1}{\sqrt{\mu \alpha}}\)
4. Infinitesimal

Answer: 1. \(\sqrt{\frac{\mu}{\alpha}}\)

Question 63. A ball of mass (m) 0.5 kg is attached to the end of a string having length (L) 0.5 m. The ball is rotated on a horizontal circular path about a vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of the angular velocity of the ball (in radian/s) is:

1. 9
2. 18
3. 27
4. 36

Answer: 4. 36

Question 64. A particle of mass m is moving with constant velocity \(\overrightarrow{\mathrm{V}}\) on a smooth horizontal surface. A constant force starts acting on a particle perpendicular to velocity v. The Radius of curvature after force F starts acting is:

1. \(\frac{m v^2}{F}\)
2. \(\frac{m v^2}{F \cos \theta}\)
3. \(\frac{m v^2}{F \sin \theta}\)
4. None of these

Answer: 1. \(\frac{m v^2}{F}\)

Question 65. A stone is projected with speed u and the angle of projection is θ. Find the radius of curvature at t = 0.

1. \(\frac{u^2 \cos ^2 \theta}{g}\)
2. \(\frac{u^2}{g \sin \theta}\)
3. \(\frac{u^2}{g \cos \theta}\)
4. \(\frac{u^2 \sin ^2 \theta}{g}\)

Answer: 3. \(\frac{u^2}{g \cos \theta}\)

Question 66. The velocity and acceleration vectors of a particle undergoing circular motion are \(\overrightarrow{\mathrm{v}}=2 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}\) and \(\vec{a}=2 \hat{i}+4 \hat{j} \mathrm{~m} / \mathrm{s}^2\) respectively at an instant of time. The radius of the circle is

1. 1m
2. 2m
3. 3m
4. 4m

Answer: 1. 1m

Question 67. A particle is projected horizontally from the top of a tower with a velocity v0. If v is its velocity at any instant, then the radius of curvature of the path of the particle at that instant is directly proportional to:

1. v³
2. v²
3. v
4. 1/v

Answer: 1. 1/v

Question 68. The tension in the string revolving in a vertical circle with a mass m at the end when it is at the lowest position.

1. \(\frac{m v^2}{r}\)
2. \(\frac{m v^2}{r}-m g\)
3. \(\frac{m v^2}{r}+m g\)
4. mg

Answer: 3. \(\frac{m v^2}{r}+m g\)

Question 69. A motorcycle is going on an overbridge of radius R. The driver maintains a constant speed. As the motorcycle is ascending on the overbridge, the normal force on it :

1. Increase
2. Decreases
3. Remains constant
4. First increases then decreases.

Answer: 1. Increase

Question 70. In a circus, a stuntman rides a motorbike in a circular track of radius R in the vertical plane. The minimum speed at the highest point of the track will be :

1. \(\sqrt{2 \mathrm{gR}}\)
2. 2gR
3. \(\sqrt{3 \mathrm{gR}}\)
4. \(\sqrt{g R}\)

Answer: 4. \(\sqrt{g R}\)

Question 71. A particle is moving in a vertical circle. The tensions in the string when passing through two positions at angles 30° and 60° from vertical (lowest positions) are T₁ and T₂ respectively. Then

1. T₁ = T₂
2. T₂ > T₁
3. T₁ > T₂
4. Tension in the string always remains the same

Answer: 3. T₁ > T₂

Question 72. A car moves at a constant speed on a road. The normal force by the road on the car is N_A and N_B when it is at points A and B respectively.

1. N_A = N_B
2. N_A > N_B
3. N_A < N_B
4. Insufficient

Answer: 2. N_A > N_B

Question 73. A heavy mass is attached to a thin wire and is whirled in a vertical circle. The wire is most likely to break.

1. When the mass is at the height point of the circle
2. When the mass is at the lowest point of the circle
3. When the wire is horizontal
4. At an angle of cos^-1(1/3) from the upward vertical

Answer: 2. When the mass is at the lowest point of the circle

Question 74. A hollow sphere has a radius of 6.4 m. The minimum velocity required by a motorcyclist at the bottom to complete the circle will be.

1. 17.7 m/s
2. 10.2 m/s
3. 12.4 m/s
4. 16.0 m/s

Answer: 1. 17.7 m/s

Question 75. A body of mass 100 g is rotating in a circular path of radius r with constant speed. The work done in one complete revolution is.

1. 100 rJ
2. (r/100) J
3. (100/r) J
4. Zero “

Answer: 4. Zero “

Question 76. A weightless thread can bear tension upto 3.7 kg wt. A stone of mass 500 gms is tied to it and revolved in a circular path of radius 4 m in a vertical plane. If g = 10 ms^-2, then the maximum angular velocity of the stone will be.

1. 4 radians/sec
2. 16 radians/sec
3. 21radians/sec
4. 2 radians/sec

Answer: 1. 4 radians/sec

Question 77. A small disc is on the top of a hemisphere of radius R. What is the smallest horizontal velocity v that should be given to the disc for it to leave the hemisphere and not slide down it? [There is no friction]

1. \(v=\sqrt{2 g R}\)
2. \(v=\sqrt{g R}\)
3. \(v=\frac{g}{R}\)
4. \(v=\sqrt{g^2 R}\)

Answer: 2. \(v=\sqrt{g R}\)

Question 78. The maximum velocity at the lowest point, so that the string just slacks at the highest point in a vertical circle of radius l.

1. \(\sqrt{g \ell}\)
2. \(\sqrt{3 \mathrm{~g} \ell}\)
3. \(\sqrt{5 \mathrm{~g} \ell}\)
4. \(\sqrt{7 g \ell}\)

Answer: 3. \(\sqrt{5 \mathrm{~g} \ell}\)

Question 79. A simple pendulum oscillates in a vertical plane. When it passes through the mean position, the tension in the string is 3 times the weight of the pendulum bob. What is the maximum displacement of the pendulum of the string with respect to the vertical?

1. 30º
2. 45º
3. 60º
4. 90º

Answer: 4. 90º

Question 80. A coin placed on a rotating turntable just slips if it is placed at a distance of 4 cm from the center. If the angular velocity of the turntable is doubled, it will just slip at a distance of

1. 1 cm
2. 2 cm
3. 4 cm
4. 8 cm

Answer: 1. 1 cm

Question 81. A cane filled with water is revolved in a vertical circle of radius 4 meter and the water just does not fall down. The time period of the revolution will be

1. 1 sec
2. 10 sec
3. 8 sec
4. 4 sec

Answer: 4. 4 sec

Question 82. A weightless rod of length 2l carries two equal masses ‘m’, one tied at the lower end of A and the other at the middle of the rod at B. The rod can rotate in a vertical plane about a fixed horizontal axis passing through C. The rod is released from rest in a horizontal position. The speed of mass B at the instant rod, becomes vertical is :

1. \(\sqrt{\frac{3 \mathrm{~g} \ell}{5}}\)
2. \(\sqrt{\frac{4 g \ell}{5}}\)
3. \(\sqrt{\frac{6 \mathrm{~g} \ell}{5}}\)
4. \(\sqrt{\frac{7 g \ell}{5}}\)

Answer: 3. \(\sqrt{\frac{6 \mathrm{~g} \ell}{5}}\)

Question 83. A body is suspended from a smooth horizontal nail by a string of length 0.25 meters. What minimum horizontal velocity should be given to it in the lowest position so that it may move in a complete vertical circle with the nail at the center?

1. 3.5 ms^-1
2. 4.9 ms^-1
3. 7\(\sqrt{2}\) ms^-1
4. \(\sqrt{9.8}\) ms^-1

Answer: 1. 3.5 ms^-1

Question 84. A block of mass m slides down along the surface of the bowl from the rim to the bottom as shown in Fig. The velocity of the block at the bottom will be –

1. \(\sqrt{\pi \mathrm{Rg}}\)
2. \(2 \sqrt{\pi \mathrm{Rg}}\)
3. \(\sqrt{2 \mathrm{Rg}}\)
4. \(\sqrt{g R}\)

Answer: 3. \(\sqrt{2 \mathrm{Rg}}\)

Question 85. A mass m is revolving in a vertical circle at the end of a string of length 20 cm. By how many times does the tension of the string at the lowest point exceed the tension at the topmost point –

1. 2 mg
2. 4 mg
3. 6 mg
4. 8 mg

Answer: 3. 6 mg

Question 86. A block follows the path as shown in the figure from height h. If the radius of the circular path is r, then the relation holds well to complete full circle.

1. h < 5r/2
2. h > 5r/2
3. h = 5r/2
4. h ≥ 5r/2

Answer: 4. h ≥ 5r/2

Question 87. A particle is kept at rest at the top of a sphere of diameter 42 m. When disturbed slightly, it slides down. At what height ‘h’ from the bottom, the particle will leave the sphere?

1. 14 m
2. 28 m
3. 35 m
4. 7 m

Answer: 3. 35 m

Question 88. A stone of 1 kg tied up with a 10/3 meter long string rotated in a vertical circle. If the ratio of maximum and minimum tension in the string is 4 then the speed of the stone at the highest point of the circular path will be – (g = 10 m/s²)

1. 20 m/s
2. \(10 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
3. \(5 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
4. 10 m/s

Answer: 4. 10 m/s

Question 89. A child is swinging a swing, Minimum and maximum heights of the swing from the earth’s surface are 0.75 m and 2 m respectively. The maximum velocity of this swing is :

1. 5 m/s
2. 10 m/s
3. 15 m/s
4. 20 m/s

Answer: 1. 5 m/s

Question 90. A stone is tied to a string of length l and is whirled in a vertical circle with the other end of the string as the center. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of the change in velocity as it reaches a position where the string is horizontal (g being acceleration due to gravity) is :

1. \(\sqrt{2\left(\mathrm{u}^2-\mathrm{g} \ell\right)}\)
2. \(\sqrt{\left(u^2-g \ell\right)}\)
3. \(u-\sqrt{\left(u^2-2 g \ell\right)}\)
4. \(\sqrt{2 g \ell}\)

Answer: 1. \(\sqrt{2\left(\mathrm{u}^2-\mathrm{g} \ell\right)}\)

Question 91. In a circus, a stuntman rides a motorbike in a circular track of radius R in the vertical plane. The minimum speed at the highest point of the track will be :

1. \(\sqrt{2 g R}\)
2. 2gR
3. \(\sqrt{3 \mathrm{gR}}\)
4. \(\sqrt{g R}\)

Answer: 4. \(\sqrt{g R}\)

Question 92. A particle of mass m begins to slide down a fixed smooth sphere from the top. What is its tangential acceleration when it breaks off the sphere?

1. \(\frac{2 \mathrm{~g}}{3}\)
2. \(\frac{\sqrt{5} g}{3}\)
3. g
4. \(\frac{g}{3}\)

Answer: 2. \(\frac{\sqrt{5} g}{3}\)

Question 93. A body of mass 1 kg is moving in a vertical circular path of radius 1m. The difference between the kinetic energies at its highest and lowest position is

1. 20J
2. 10J
3. \(4 \sqrt{5} \mathrm{~J}\)
4. \(10(\sqrt{5}-1) \mathrm{J}\)

Answer: 1. 20J

Question 94. A small block is shot into each of the four tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in –

Answer: 1.

Question 95. A simple pendulum is oscillating without damping. When the displacements of the bob is less than maximum, its acceleration vector \(\overrightarrow{\mathrm{a}}\) is correctly shown in

Answer: 3.

Question 96. A car moving on a horizontal road may be thrown out of the road in taking a turn:

1. By the gravitational force
2. Due to a lack of sufficient centripetal force
3. Due to friction between the road and the tire
4. Due to the reaction of the earth

Answer: 2. Due to a lack of sufficient centripetal force

Question 97. The magnitude of the centripetal force acting on a body of mass m executing uniform motion in a circle of radius r with speed υ is-

1. mυr
2. \(\frac{m v^2}{r}\)
3. \(\frac{v}{r^2 m}\)
4. \(\frac{v}{\mathrm{rm}}\)

Answer: 2. \(\frac{m v^2}{r}\)

Question 98. The radius of the curved road on the national highway is R. The Width of the road is b. The outer edge of the road is raised by h with respect to the inner edge so that a car with velocity υ can pass safely over it. The value of h is-

1. \(\frac{v^2 b}{R g}\)
2. \(\frac{v}{R g b}\)
3. \(\frac{v^2 R}{g}\)
4. \(\frac{u^2 b}{R}\)

Answer: 1. \(\frac{v^2 b}{R g}\)

Question 99. If the apparent weight of the bodies at the equator is to be zero, then the earth should rotate with angular velocity

1. \(\sqrt{\frac{\mathrm{g}}{\mathrm{R}}} \mathrm{rad} / \mathrm{sec}\)
2. \(\sqrt{\frac{2 g}{R}} \mathrm{rad} / \mathrm{sec}\)
3. \(\sqrt{\frac{\mathrm{g}}{2 \mathrm{R}}} \mathrm{rad} / \mathrm{sec}\)
4. \(\sqrt{\frac{3 \mathrm{~g}}{2 \mathrm{R}}} \mathrm{rad} / \mathrm{sec}\)

Answer: 1. \(\sqrt{\frac{\mathrm{g}}{\mathrm{R}}} \mathrm{rad} / \mathrm{sec}\)

Question 100. The road is 10 m wide. Its radius of curvature is 50 m. The outer edge is above the lower edge by a distance of 1.5 m. This road is most suited for the velocity

1. 2.5 m/sec
2. 4.5 m/sec
3. 6.5 m/sec
4. 8.5 m/sec

Answer: 4. 6.5 m/sec

Question 101. The radius of the curved road on the national highway is R. The Width of the road is b. The outer edge of the road is raised by h with respect to the inner edge so that a car with velocity v can pass safely over it. The value of h is

1. \(\frac{v^2 b}{R g}\)
2. \(\frac{\mathrm{v}}{\mathrm{Rgb}}\)
3. \(\frac{v^2 R}{g}\)
4. \(\frac{v^2 b}{R}\)

Answer: 1. \(\frac{v^2 b}{R g}\)

Question 102. A circular road of radius 1000 m has a banking angle of 45º. The maximum safe speed of a car having a mass of 2000 kg will be if the coefficient of friction between tire and road is 0.5

1. 172 m/s
2. 124 m/s
3. 99 m/s
4. 86 m/s

Answer: 1. 172 m/s

Question 103. A cane filled with water is revolved in a vertical circle of radius 4 meter and the water just does not fall down. The time period of the revolution will be

1. 1 sec
2. 10 sec
3. 8 sec
4. 4 sec

Answer: 4. 4 sec

Question 104. A motorcyclist moving with a velocity of 72 km/hour on a flat road takes a turn on the road at a point where the radius of curvature of the road is 20 meters. The acceleration due to gravity is 10 m/sec². In order to avoid skidding, he must not bend with respect to the vertical plane by an angle greater than-

1. θ = tan^-1 6
2. θ = tan^-1 2
3. θ = tan^-1 25.92
4. θ = tan^-1 4

Answer: 2. θ = tan^-1 2

Question 105. The kinetic energy k of a particle moving along a circle of radius R depends on the distance covered s as k = as2 where a is a constant. The force acting on the particle is

1. \(2 a \frac{s^2}{R}\)
2. \({2as}\left(1+\frac{s^2}{R^2}\right)^{1 / 2}\)
3. 2as
4. \(2 \mathrm{a} \frac{\mathrm{R}^2}{\mathrm{~s}}\)

Answer: 2. \({2as}\left(1+\frac{s^2}{R^2}\right)^{1 / 2}\)

Question 106. A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration ac is varying with time t as ac = k²rt² where k is a constant. The power delivered to the particle by the force acting on it is-

1. 2πmk²r²
2. mk²r²t
3. \(\frac{\left(m k^4 r^2 t^5\right)}{3}\)
4. Zero

Answer: 2. mk²r²t

Question 107. A small block slides with velocity 0.5 gron the horizontal frictionless surface as shown in the Figure. The block leaves the surface at point C. The angle θ in the Figure is :

1. cos^-1(4/9)
2. cos^-1(3/4)
3. cos^-1(1/2)
4. None of the above

Answer: 4. None of the above

Question 108. A particle moves along a circle of radius \(\left(\frac{20}{\pi}\right)\) with constant tangential acceleration. If the speed of the particle is 80 m/s at the end of the second revolution after motion has begun, the tangential acceleration is:

1. 160 π m/s²
2. 40 π m/s²
3. 40 m/s²
4. 640 π m/s²

Answer: 3. 40 m/s²

Question 109. Centrifugal force is an inertial force when considered by –

1. An observer at the center of circular motion
2. An outside observer
3. An observer who is moving with the particle that is experiencing the force
4. None of the above

Answer: 3. An observer who is moving with the particle which is experiencing the force

Question 110. A rod of length L is pivoted at one end and is rotated with a uniform angular velocity in a horizontal plane. Let T₁ and T₂ be the tensions at the points L/4 and 3L/4 away from the pivoted ends.

1. T₁ > T₂
2. T₂ > T₁
3. T₁ = T₂
4. The relation between T₁ and T₂ depends on whether the rod rotates clockwise or anticlockwise

Answer: 1. T₁ > T₂

Question 111. When a ceiling fan is switched off its angular velocity reduces to 50% while it makes 36 rotations. How many more rotations will it make before coming to rest (Assume uniform angular retardation)

1. 18
2. 12
3. 36
4. 48

Answer: 2. 12

Question 112. A particle is moving in the vertical plane. It is attached at one end of a string of length l whose other end is fixed. The velocity at the lowest point is u. The tension in the string \(\overrightarrow{\mathrm{T}}\) is and acceleration of the particle \(\overrightarrow{\mathrm{a}}\) is at any position. Then, \(\overrightarrow{\mathrm{T}}.\overrightarrow{\mathrm{a}}\) is zero at the highest point:

1. Only if \(u \leq \sqrt{2 \mathrm{~g} \ell}\)
2. If \(\sqrt{5 \mathrm{~g} \ell}\)
3. Only if \(\mathrm{u}=\sqrt{2 \mathrm{~g} \ell}\)
4. Only if \(u>\sqrt{2 g \ell}\)

Answer: 2. If \(\sqrt{5 \mathrm{~g} \ell}\)

Question 113. In the above question, \(\overrightarrow{\mathrm{T}} \cdot \overrightarrow{\mathrm{a}}\) T.a is non-negative at the lowest point for:

1. \(\mathrm{u} \leq \sqrt{2 \mathrm{~g} \ell}\)
2. \(\mathrm{u}=\sqrt{2 \mathrm{~g} \ell}\)
3. \(\mathrm{u}<\sqrt{2 \mathrm{~g} \ell}\)
4. Any value of u

Answer: 4. Any value of u

Question 114. In the above question, \(\overrightarrow{\mathrm{T}} . \vec{u}\) is zero for:

1. \(\mathrm{u} \leq \sqrt{2 \mathrm{~g} \ell}\)
2. \(\mathrm{u}=\sqrt{2 \mathrm{~g} \ell}\)
3. \(\mathrm{u} \geq \sqrt{2 \mathrm{~g} \ell}\)
4. Any value of u

Answer: 4. Any value of u

Question 115. A bob of mass M is suspended by a massless string of length L. The horizontal velocity V at position A is just sufficient to make it reach the point B. The angle θ at which the speed of the bob is half of that at A satisfies

1. \(\theta=\frac{\pi}{4}\)
2. \(\frac{\pi}{4}<\theta<\frac{\pi}{2}\)
3. \(\frac{\pi}{2}<\theta<\frac{3 \pi}{4}\)
4. \(\frac{3 \pi}{4}<\theta<\pi\)

Answer: 4. \(\frac{3 \pi}{4}<\theta<\pi\)

Question 116. If a particle of mass m is moving in a horizontal circle of radius r with a centripetal force \(\left(-\frac{\mathrm{K}}{\mathrm{r}^2}\right)\), the total energy is-

1. \(-\frac{\mathrm{K}}{2 \mathrm{r}}\)
2. \(-\frac{\mathrm{K}}{\mathrm{r}}\)
3. \(-\frac{2 \mathrm{~K}}{\mathrm{r}}\)
4. \(-\frac{4 K}{r}\)

Answer: 1. \(-\frac{\mathrm{K}}{2 \mathrm{r}}\)

Question 117. A particle moves in a circle of radius 5 cm with constant speed and time period 0.2 πs. The acceleration of the particle is :

1. 15 m/s²
2. 25 m/s²
3. 36 m/s²
4. 5 m/s²

Answer: 4. 5 m/s²

Question 118. A car of mass 1000 kg negotiates a banked curve of radius 90 m on a frictionless road. If the banking angle is 45º, the speed of the car is :

1. 20 ms^-1
2. 30 ms^-1
3. 5 ms^-1
4. 10 ms^-1

Answer: 2. 30 ms^-1

Question 119. A car of mass m is moving on a level circular track of radius R. If μs represents the static friction between the road and tires of the car, the maximum speed of the car in circular motion is given by :

1. \(\sqrt{\mu_{\mathrm{s}} \mathrm{mRg}}\)
2. \(\sqrt{\mathrm{Rg} / \mu_{\mathrm{s}}}\)
3. \(\sqrt{\mathrm{mRg} / \mu_{\mathrm{s}}}\)
4. \(\sqrt{\mu_{\mathrm{s}} \mathrm{Rg}}\)

Answer: 4. \(\sqrt{\mu_{\mathrm{s}} \mathrm{Rg}}\)

Question 120. Two stones of masses m and 2 m are whirled in horizontal circles the heavier one in radius \(\{r}{2}\) and the lighter one in radius r. The tangential speed of lighter stones is n times that of the value of heavier stones when they experience the same centripetal forces. The value of n is :

1. 3
2. 4
3. 1
4. 2

Answer: 4. 2

Question 121. The position vector of a particle \(\overrightarrow{\mathrm{R}}\) as a function of time is given by:

⇒ \(\vec{R}=4 \sin (2 \pi t) \hat{i}+4 \cos (2 \pi t)\)

Where R is in meters, t is seconds, and \(\hat{i} \text { and } \hat{j}\) denote unit vectors along x-and y-directions, respectively. Which one of the following statements is wrong for the motion of a particle?

1. Magnitude of acceleration vector is \(\frac{v^2}{R}\), where v is the velocity of particle
2. The magnitude of the velocity of the particle is 8 meters/second
3. path of the particle is a circle of radius 4 meters.
4. Acceleration vector is along – \(\vec{R}\)

Answer: 2. Magnitude of the velocity of the particle is 8 meters/second

Question 122. What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?

1. \(\sqrt{5 \mathrm{gR}}\)
2. \(\sqrt{g R}\)
3. \(\sqrt{2 g R}\)
4. \(\sqrt{3 \mathrm{gR}}\)

Answer: 1. \(\sqrt{5 \mathrm{gR}}\)

Question 123. A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 × 10^-4 J by the end of the second revolution after the beginning of the motion?

1. 0.2 m/s²
2. 0.1 m/s²
3. 0.15 m/s²
4. 0.18 m/s²

Answer: 2. 0.1 m/s²

Question 124. A car is negotiating a curved road of radius R. The road is banked at an angle θ. The coefficient of friction between the tires of the car and the road is μs. The maximum safe velocity on this road is:

1. \(\sqrt{\frac{\mathrm{g}}{\mathrm{R}^2} \frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}}+\tan \theta}}\)
2. \(\sqrt{g R^2 \frac{\mu_s+\tan \theta}{1-\mu_s+\tan \theta}}\)
3. \(\sqrt{g R \frac{\mu_s+\tan \theta}{1-\mu_s+\tan \theta}}\)
4. \(\sqrt{\frac{g}{R} \frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}}+\tan \theta}}\)

Answer: 3. \(\sqrt{g R \frac{\mu_s+\tan \theta}{1-\mu_s+\tan \theta}}\)

Question 125. In the given figure, a = 15 m/s² represents the total acceleration of a particle moving in the clockwise direction in a circle of radius R = 2.5 m at a given instant of time. The speed of the particle is

1. 6.2 m/s
2. 4.5 m/s
3. 5.0 m/s
4. 5.7 m/s

Answer: 4. 5.7 m/s

Question 126. One end of a string of length l is connected to a particle of mass ‘m’ and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in a circle with speed ‘v’ the net force on the particle (directed towards the center) will be (T represents the tension in the string)

1. T
2. \(\mathrm{T}+\frac{\mathrm{m} \mathrm{v}^2}{\ell}\)
3. \(\mathrm{T}-\frac{\mathrm{m} \mathrm{v}^2}{\ell}\)
4. zero

Answer: 1. T

Question 127. A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) Just completes a vertical circle of diameter AB = D. The height h is equal to

1. \(\frac{3}{2} D\)
2. \(\frac{5}{4} D\)
3. \(\frac{7}{5} D\)
4. D

Answer: 2. \(\frac{5}{4} D\)

Question 128. A mass m is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when:

1. Inclined at an angle of 60º from vertical
2. The mass is at the highest point
3. The wire is horizontal
4. The mass is at the lowest point

Answer: 4. The mass is at the lowest point

Question 129. A block of mass 10 kg in contact against the inner wall of a hollow cylindrical drum of radius 1m. The coefficient of friction between the block and the inner wall of the cylinder is 0.1. The minimum angular velocity needed for the cylinder to keep the block stationary when the cylinder is vertical and rotating about its axis will be (g = 10 m/s²)

1. 10 π rad/s
2. \(\sqrt{10} \mathrm{rad} / \mathrm{s}\)
3. \(\frac{10}{2 \pi} \mathrm{rad} / \mathrm{s}\)
4. 10 π rad/s

Answer: 4. 10 π rad/s

Question 130. A particle starting from rest, moves in a circle of radius ‘r’. It attains a velocity of V0 m/s in the nth round. Its angular acceleration will be:

1. \(\frac{V_0}{n} \mathrm{rad} / \mathrm{s}^2\)
2. \(\frac{V_0^2}{2 \pi \mathrm{nr}^2} \mathrm{rad} / \mathrm{s}^2\)
3. \(\frac{V_0^2}{4 \pi \mathrm{r}^2} \mathrm{rad} / \mathrm{s}^2\)
4. \(\frac{\mathrm{V}_0^2}{4 \pi \mathrm{nr}} \mathrm{rad} / \mathrm{s}^2\)

Answer: 3. \(\frac{V_0^2}{4 \pi \mathrm{r}^2} \mathrm{rad} / \mathrm{s}^2\)

Question 131. A po1nt P moves in a counter-clockwise direction on a circular path as shown in the figure. The movement of ‘P’ is such that it sweeps out a length s = t3 + 5, where s is in meters and t is in seconds. The radius of the path is 20 m. The acceleration of ‘P’ when t = 2 s is nearly.

1. 13 m/s²
2. 12 m/s²
3. 7.2 m/s²
4. 14 m/s²

Answer: 4. 14 m/s²

Question 132. For a particle in uniform circular motion, the acceleration \(\overrightarrow{\mathrm{a}}\) at a point P (R, θ) on the circle of radius R is (Here θ is measured from the x-axis)

1. \(-\frac{v^2}{R} \cos \theta \hat{i}+\frac{v^2}{R} \sin \theta \hat{j}\)
2. \(-\frac{v^2}{R} \sin \theta \hat{i}+\frac{v^2}{R} \cos \theta \hat{j}\)
3. \(-\frac{v^2}{R} \cos \theta \hat{i}-\frac{v^2}{R} \sin \theta \hat{j}\)
4. \(\frac{v^2}{R} \hat{i}+\frac{v^2}{R} \hat{j}\)

Answer: 3. \(-\frac{v^2}{R} \sin \theta \hat{i}+\frac{v^2}{R} \cos \theta \hat{j}\)

Question 133. Two cars of masses m₁ and m2 are moving in circles of radii r₁ and r₂, respectively. Their speeds are such that they make complete circles at the same time t. The ratio of their centripetal acceleration is:

1. m₁ r₁ : m₂r₂
2. m₁ : m₂
3. r₁ : r₂
4. 1: 1

Answer: 4. 1:1

Question 134. A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the n^th power of R. If the period of rotation of the particle is T, then:

1. \(\mathrm{T} \alpha \mathrm{R}^{(\mathrm{n}+1) / 2}\)
2. \(\mathrm{T} \alpha \mathrm{R}^{\mathrm{n} / 2}\)
3. \(\mathrm{T} \alpha \mathrm{R}^{3 / 2}\) For any n
4. \(T \alpha R^{\frac{n}{2}+1}\)

Answer: 1. \(\mathrm{T} \alpha \mathrm{R}^{(\mathrm{n}+1) / 2}\)

Question 135. A particle is moving along a circular path with a constant speed of 10 ms^-1. What is the magnitude of the change in the velocity of the particle, when it moves through an angle of 60° around the center of the circle?

1. Zero
2. 10 m/s
3. \(10 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
4. \(10 \sqrt{3} \mathrm{~m} / \mathrm{s}\)

Answer: 2. 10 m/s

Question 136. Two particles A and, B are moving on two concentric circles of radii R₁ and R₂ with equal angular speed ω. At t = 0, their positions and direction of motion are shown in the figure.

The relative velocity \(\vec{v}_A-\vec{v}_B \text { at } t=\frac{\pi}{2 \omega}\) is

1. \(\omega\left(R_2-R_1\right) \hat{i}\)
2. \(\omega\left(R_1-R_2\right) \hat{i}\)
3. \(-\omega\left(R_1+R_2\right) \hat{i}\)
4. \(\left(R_1+R_2\right) \hat{i}\)

Answer: 1. \(\omega\left(R_2-R_1\right) \hat{i}\)

 

A Centrifuge: A centrifuge works on the principle of centrifugal force.

• The centrifuge consists of two steel tubes suspended from the ends of a horizontal bar which can be rotated at high speed in a horizontal plane by an electric motor.
• The tubes are filled with the liquid and the bar is set into rotation.
• Due to rotational motion, the tubes get tied and finally become horizontal.
• Due to heavy mass, the heavier particles experience more centrifugal force than that of the liquid particles. Therefore, is then stopped so that the tubes become vertical.

Question 1. Two balls of equal masses are attached to a string at a distance of 1 m and 2 m from one end as shown in Fig. The string with masses is then moved in a horizontal circle with constant speed. Find the ratio of the tension T₁ and T₂.
Answer:

Let the balls of the two circles be r₁ and r₂. The linear speeds of the two masses are v₁= ωr₁, v₂= ωr₂

where ω is the angular speed of the circular motion. The tension in the strings is such that

⇒ \(T_2=\frac{m v_2^2}{r_2}=m \omega^2 r_2\)

\(T_1-T_2=\frac{m v_1^2}{r_1}=m \omega^2 r_1\)

∴ \(T_1=m \omega^2 r_1+T_2=m \omega^2\left(r_1+r_2\right)\)

∴ \(\frac{T_1}{T_2}=\frac{r_1+r_2}{r_2}=\frac{1+2}{2}=\frac{3}{2}\)

Conical Pendulum:

(This is the best Question of uniform circular motion)

A conical pendulum consists of a body attached to a string, such that it can revolve in a horizontal circle with uniform speed. The string traces out a cone in the space.

The force acting on the bob is

1. Tension T
2. Weight mg

The horizontal component T sinθ of the tension T provides the centripetal force and the vertical component T cos θ balances the weight to bob

∴ T sinθ = and T cosθ = mg

From these equations ………(1)

and tan θ = ………(2)

If h = height of conical pendulum tanθ = = ………(3)

From (2) and (3)

The time period of the revolution

Hints To Solve Numerical Problems (UCM)

1. First, show all forces acting on a particle
2. Resolve these forces along radius and tangent.
3. The resultant force along the radial direction provides the necessary centripetal force.
4. Resultant force along tangent = Ma_r= 0 (a_r= tangential acceleration)

Question 1. A vertical rod is rotating about its axis with a uniform angular speed ω. A simple pendulum of length l is attached to its upper end what is its inclination with the rod?
Answer:

Let the radius of the circle in which the bob is rotating be, the tension in the string is T, the weight of the bob mg, and the inclination of the string θ. Then T cos θ balances the weight mg and T sin θ provides the centripetal force necessary for circular motion.

That is –

T cos θ = mg and T sin θ = mω² x

but x = l sin θ

∴ T = mω² l

and \(\cos \theta=\frac{\mathrm{mg}}{\mathrm{T}}=\frac{\mathrm{mg}}{\mathrm{m} \omega^2 \ell}\) or \(\theta=\cos ^{-1}\left(\frac{g}{\omega^2 \ell}\right)\)

Question 2. A circular loop has a small bead that can slide on it without friction. The radius of the loop is r. Keeping the loop vertically it is rotated about a vertical diameter at a constant angular speed ω. What is the value of angle θ, when the bead is in dynamic equilibrium?
Answer:

Centripetal force is provided by the horizontal component of the normal reaction N. The vertical component balances the weight. Thus

N sin θ = mω²x and N cos θ = mg

Also x = r sin θ ⇒ N = mω²r

cos θ = \(\frac{g}{\omega^2 r}\) or \(\theta=\cos ^{-1}\left(\frac{g}{r \omega^2}\right)\)

Question 3. A particle of mass m slides down from the vertex of the semihemisphere, without any initial velocity. At what height from the horizontal will the particle leave the sphere?
Answer:

Let the particle leave the sphere at height h, \(\frac{\mathrm{mv}^2}{\mathrm{R}}=\mathrm{mg} \cos \theta-\mathrm{N}\)

When the particle leaves the sphere N = 0,

⇒ \(\frac{m v^2}{R}=m g \cos \theta \Rightarrow v^2=g R \cos \theta\)

According to law of conservation of energy ( K . E.+ P. E.) at A =( K . E.+ P. E.) at B

⇒ \(o+m g R=\frac{1}{2} m v^2+m g h \Rightarrow v^2=2 g(R-h)\)

From 1 and 2 h = \(\frac{2 R}{3}\) Also cosθ = 2/3

Question 4. A particle describes a horizontal circle of radius r in a funnel-type vessel of the frictionless surface with half one angle θ. If the mass of the particle is m, then in dynamical equilibrium the speed of the particle must be –
Answer:

The normal reaction N and weight mg are the only forces acting on the particle (inertial frame view), the N is making an angle \(\left(\frac{\pi}{2}-\theta\right)\) with the vertical.

The vertical component of N balances the weight mg and the horizontal component provides the centripetal force required for circular motion.

Thus

⇒ \(N \cos \left(\frac{\pi}{2}-\theta\right)=m g\)

⇒ \(N \sin \left(\frac{\pi}{2}-\theta\right)=\frac{m^2}{r}\)

or N sin θ = \(\frac{m v^2}{r} \mathrm{mg}\)

N cos θ = tan θ = \(\frac{\mathrm{rg}}{\mathrm{v}^2}\) on dividing we get ,

so \(v=\sqrt{\frac{\mathrm{rg}}{\tan \theta}}\)

Question 5. Prove that a motor car moving over a

1. Convex bridge is lighter than the same car resting on the same bridge.
2. The concave bridge is heavier than the same car resting on the same bridge.

Answer:

Apparent weight of car = N (normal reaction)

1. Convex bridge

The motion of the motor car over a convex bridge is the motion along the segment of a circle. The centripetal force is provided by the difference in weight mg of the car and the normal reaction N of the bridge.

∴ \(\mathrm{mg}-\mathrm{N}=\frac{\mathrm{mv}^2}{\mathrm{r}}\)

or N = mg – \(\frac{m v^2}{r}\)

Clearly N < mg, i.e., the apparent weight of the moving car is less than the weight of the stationary car.

2. Concave bridge N – mg = \(\frac{m v^2}{r}\)

Apparent weight N = mg + \(\frac{m v^2}{r}\)

Motion In Vertical Circle:

Motion of a body suspended by string: This is the best Question of non-uniform circular motion.

Suppose a particle of mass m is attached to an inexcusable light string of length r. The particle is moving in a vertical circle of radius r, about a fixed point O.

At lost point A velocity of particle = u (in a horizontal direction)

After covering ∠θ velocity of particle = v (at point B)

Resolve weight (mg) into two components

1. mg cos θ (along radial direction)
2. mg sin θ (tangential direction)

Then force T – mg cos θ provides the necessary centripetal force

T – mg cosθ = \(\frac{m v^2}{\mathrm{r}}\) ……(1)

Δ OCB cos θ = \(\frac{r-h}{r}\) ……(2)

or h = r (1 – cosθ)

By conservation of energy at points A and B

⇒ \(\frac{1}{2} m u^2=\frac{1}{2} m v^2+m g h \text { or } u^2=v^2+2 g h\)

or v² = u² + 2gh ……..(3)

Substitute value of cos θ and v² in equn. (1)

⇒ \(T-m g\left[\frac{r-h}{r}\right]=\frac{m}{r}\left(u^2-2 g h\right) \text { or } T=\frac{m}{r}\left[u^2-2 g h+g r-g h\right) \text { or } T=\frac{m}{r}\left[u^2+g r-3 g h\right]\) …………(4)

1. If velocity becomes zero at height h₁

⇒ \(O=u^2-2 g h,\) or \(\mathrm{h}_1\)

=\(\frac{\mathrm{u}^2}{2 \mathrm{~g}}\) ………..

2. If tension becomes zero at height h₂

⇒ \(\mathrm{O}=\frac{\mathrm{m}}{\mathrm{r}}\left[\mathrm{u}^2+\mathrm{gr}-3 \mathrm{g} \mathrm{h}_2\right]\)

or \(\mathrm{u}^2+\mathrm{gr}-3 \mathrm{gh}_2=0\)

or \(h_2=\frac{u^2+g r}{3 g}\) …….. (4)

3. Case of oscillation

It v = 0, T ≠ 0 then h₁< h₂

⇒ \(\frac{u^2}{2 g}<\frac{u^2+g r}{3 g}\)

⇒ \(3 u^2<2 u^2+2 g r\)

⇒ \(\mathrm{u}^2<2 \mathrm{gr}\)

⇒ \(u<\sqrt{2 g r}\)

4. Case Of Leaving The Circle

If v = 0, T = 0 then h₁ < h₂

⇒ \(\frac{u^2}{2 g}>\frac{u^2+g r}{3 g}\)

⇒ \(\sqrt{5 \mathrm{gr}}>\mathrm{u}>\sqrt{2 \mathrm{gr}}\)

Case of leaving the circle

⇒ \(3 \mathrm{u}^2>2 \mathrm{u}^2+2 \mathrm{gr}\)

⇒ \(\mathrm{u}^2>2 \mathrm{gr}\)

⇒ \(u>\sqrt{2 g r}\)

⇒ \(\sqrt{5 \mathrm{gr}}>\mathrm{u}>\sqrt{2 \mathrm{gr}}\)

5. Case Of Complete The Circle

Case of completing the circle or looping the loop

⇒ \(u \geq \sqrt{5 \mathrm{gr}}\)

T > 0

v ≠ 0

Special Note

The same conditions apply if a particle moves inside a smooth spherical shell of radius R. The only difference is that the tension is replaced by the normal reaction N.

This is shown in the figure given below \(v=\sqrt{g R} \quad N=0\)

1. Condition of looping the loop is u ≥ \(\sqrt{5 g R}\)

2. Condition of leaving the circle \(\sqrt{2 \mathrm{gR}}<\mathrm{u}<\sqrt{5 \mathrm{gR}}\)

3. Condition of oscillation is 0 < u ≥ \(\sqrt{2 g R}\)

Question 1. A ball is released from height h. Find the condition for the particle to complete the circular path.
Answer:

According to law of conservation of energy (K.E. + P.E) at A = (K.E. + P.E) at B

⇒ \(0+m g h=\frac{1}{2} m v^2+0\)

⇒ \(v=\sqrt{2 g h}\)

But velocity at the lowest point of the circle,

⇒ \(v \geq \sqrt{5 g R} \Rightarrow \sqrt{2 g h} \geq \sqrt{5 g R} \Rightarrow h \geq \frac{5 R}{2}\)

Question 2. A body weighing 0.4 kg is whirled in a vertical circle making 2 revolutions per second. If the radius of the circle is 1.2 m, find the tension in the string, when the body is

1. At the top of the circle
2. At the bottom of the circle. Given : g = 9.8 ms^-2 and π = 1.2 m

Answer:

Mass m = 0.4 kg time period = \(\frac{1}{2}\) second and radius, r = 1.2 m

Angular velocity,\(\omega=\frac{2 \pi}{1 / 2}=4 \pi \mathrm{rad} \mathrm{s}^{-1}=12.56 \mathrm{rad} \mathrm{s}^{-1}\)

At the top of the circle, \(\mathrm{T}=\frac{\mathrm{m} \mathrm{v}^2}{\mathrm{r}}-\mathrm{mg}=\mathrm{mr} \omega^2-\mathrm{mg}=\mathrm{m}\left(\mathrm{r} \omega^2-\mathrm{g}\right)\)

= 0.4 (1.2 × 12.56 × 12.56 – 9.8) N = 71.8 N

At the lowest point, T = m(rω² + g) = 79.64 m

Question 3. In a circus a motorcyclist moves in a vertical loop inside a ‘death well’ (a hollow spherical chamber with holes, so that the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when the is at the uppermost point, with no support from below. What is the minimum speed required to perform a vertical loop if the radius of the chamber is 25 m?
Answer:

When the motorcyclist is at the highest point of the death well, the normal reaction R on the motorcyclist by the ceiling of the chamber acts downwards. His weight mg also acts downwards. These two forces are balanced by the outward centrifugal force acting on him.

∴ R + mg = \(\frac{m v^2}{r}\) …….(1) r = radius of the circle

Here v is the speed of the motorcyclist and m is the mass of the motorcyclist (including the mass of the motorcycle). Because of the balancing of the forces, the motorcyclist does not fall down.

The minimum speed required to perform a vertical loop is given by equation (i), when R = 0

∴ mg = \(\frac{m v_{\text {min }}^2}{r} \text { or } v_{\min }^2=\mathrm{gr}\)

or \(v_{\min }=\sqrt{\mathrm{gr}}=\sqrt{9.8 \times 25} \mathrm{~ms}^{-1}=15.65 \mathrm{~ms}^{-1}\)

So, the minimum speed at the top required to perform a vertical loop is 15.65 ms–1.

Question 4. A 4kg ball is swung in a vertical circle at the end of a cord 1 m long. What is the maximum speed at which it can swing if the cord can sustain a maximum tension of 163.6 N?
Answer:

Maximum tension = \(T=\frac{m v^2}{r}\) +mg(at lowest point)

∴ \(\frac{m v^2}{r}=T-m g\)

or \(\frac{4 v^2}{1}\) = 163.6 – 4 × 9.8

Solving we get v = 6 m/sec

Question 5. A small body of mass m = 0.1 kg swings in a vertical circle at the end of a chord of length 1 m. Its speed is 2 m/s when the chord makes an angle θ = 30º with the vertical. Find the tension in the chord.
Answer:

The equation of motion is

⇒ \(T-m g \cos \theta=\frac{m v^2}{r}\) or \(T=m g \cos \theta+\frac{m v^2}{r}\)

Substituting the given values, we get

T = 0.1 × 9.8 × cos 30 + \(\frac{0.1 \times(2)^2}{1}=0.98 \times\left(\frac{\sqrt{3}}{2}\right)+0.4\)

= 0.85 + 0.4 = 1.25 N

Special Notes Important Point:

If a particle of mass m is connected to a light rod and whirled in a vertical circle of radius R, then to complete the circle, the minimum velocity of the particle at the bottommost points is not. Because in this case, the velocity of the particle at the topmost point can be zero also. Using the conservation of mechanical energy between points A and B (1) we get

⇒ \(\frac{1}{2} m\left(u^2-v^2\right)=m g h\)

or \(\frac{1}{2} m u^2=m g(2 R)\)

∴ u = \(2 \sqrt{g R}\)

Therefore, the minimum value of u in this case is \(2 \sqrt{g R}\)

The same is the case when a particle is compelled to move inside a smooth vertical tube

Particle Application Of Circular Motion

A Cyclist Making A Turn: Let a cyclist moving on a circular path of radius r bend away from the vertical by an angle θ. If R is the reaction of the ground, then R may be resolved into two components horizontal and vertical.

The vertical component R cos θ balances the weight mg of the cyclist and the horizontal component R sin θ provides the necessary centripetal force for circular motion.

R sin θ = \(\frac{m v^2}{r}\) ……. (1)

and R cos θ = mg ……. (2)

Dividing by (2), we get

⇒ \(\tan \theta \frac{\mathrm{v}^2}{\mathrm{rg}}\) ……. (3)

For less bedding of cyclists, his speed v should be smaller, and the radius r of a circular path should be greater. If μ is the coefficient of friction, then for no skidding of cycle (or overturning of cyclist)

⇒ \(\mu \geq \tan \theta\) …… (4)

⇒ \(\mu \geq \frac{v^2}{r g}\)

An Aeroplane Making A Turn

In order to make a circular turn, a plane must roll at some angle θ in such a manner that the horizontal component of the lift force L provides the necessary centripetal force for circular motion. The vertical component of the lift force balances the weight of the plane.

L sin θ = \(\frac{m v^2}{r}\)

and L cos θ = mg

or the angle θ should be such that tan θ = \(\frac{v^2}{rg}\)

Death Well And Rotor: Example of uniform circular motion In ‘Death Well’ a person drives a bicycle on the vertical surface of a large wooden well.

• In ‘Death Well’ walls are at rest while the person revolves.
• In a rotor, at a certain angular speed of the rotor, a person hangs resting against the wall without any floor.
• In the rotor, a person is at rest and the walls rotate.
• In both these cases friction balances the weight of a person while reaction provides the centripetal force necessary for circular motion i.e.

Force of fiction F_s= mg and Normal reaction \(F_N=\frac{m v^2}{r}\)

so \(\frac{F_N}{F_{\mathrm{s}}}=\frac{v^2}{r g}\) i.e., v = \(\sqrt{\frac{\mathrm{rgF}_{\mathrm{N}}}{\mathrm{F}_{\mathrm{s}}}}\)

Now for v to be minimum FS must be maximum, i.e., \(v_{\min }=\sqrt{\frac{\mathrm{gg}}{\mu}}\) [as F_SMax = μFN]

Question 6. A 62 kg woman is a passenger in a “rotor ride” at an amusement park. A drum of radius 5.0 m is spun with an angular velocity of 25 rpm. The woman is pressed against the wall of the rotating drum. Calculate the normal force of the drum of the woman (the centripetal force that prevents her from leaving her circular path).  While the drum rotates, the floor is lowered. A vertical static friction force supports the woman’s weight. What must the coefficient of friction be to support her weight? (ω = 25 rev/min, r = 5m)

Answer:

Normal force exerted by the drum on the woman towards the center

⇒ \(F_N=m a_c=m \omega^2 r=62 \mathrm{~kg} \times\left(25 \frac{\mathrm{rev}}{\mathrm{min}} \times \frac{2 \pi \mathrm{rad}}{1 \mathrm{rev}} \times \frac{1 \mathrm{~min}}{60 \mathrm{~s}}\right)^2 \times 5 \mathrm{~m}=2100 \mathrm{~N}\)

μ = FN = F mg N ………

dividing eqn. (2) be eq. (1)

⇒ \(\mu=\frac{g}{\omega^2 r}=\left(\frac{60}{2 \pi \times 25}\right)^2 \times \frac{10}{5}=0.292\)

Question 7. A 1.1 kg block slides on a horizontal frictionless surface in a circular path at the end of a 0.50 m long string.

1. Calculate the block’s speed if the tension in the string in 86 N.
2. By what percent does the tension change if the block speed decreases by 10 percent?

Answer:

1. Force diagram for the block. The upward normal force balances the block’s weight.

The tension force of the string on the block provides the centripetal force that keeps the 50 blocks moving in a circle. Newton’s second law for forces along the radial direction is ∑ F (in a radial direction) \(T=\frac{m v^2}{r}\)

or \(v=\sqrt{\frac{T r}{m}}=\sqrt{\frac{(80 \mathrm{~N})(0.50 \mathrm{~m})}{1.2 \mathrm{~kg}}}=5.0 \mathrm{~m} / \mathrm{s}\)

2. A 10 percent reduction in the speed results in a speed v’ = 5.4 m/s. The new tension is

⇒ \(T^{\prime}=\frac{m v^{\prime 2}}{\mathrm{r}}-\frac{(1.2 \mathrm{~kg})(5.4 \mathrm{~m} / \mathrm{s})^2}{0.50 \mathrm{~m}}=70 \mathrm{~N}\)

Thus, \(\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\frac{70 \mathrm{~N}}{86 \mathrm{~N}}=0.81\)

The percentage reduction in the tension is about 19%.

The same result is obtained using a proportionality method.

⇒ \(\frac{T^{\prime}}{T}=\frac{\left(\mathrm{mv}^{\prime 2} / \mathrm{r}\right)}{\left(\mathrm{mv}^2 / \mathrm{r}\right)}=\left(\frac{\mathrm{v}^{\prime}}{\mathrm{v}}\right)^2=\left(\frac{0.90}{\mathrm{v}}\right)^2=0.81\)

Looping The Loop: This is the best Question of nonuniform circular motion in a vertical plane.

For looping the pilot of the plane puts off the engine at the lowest point and traverses a vertical loop. (with variable velocity).

Question 8. An airplane moves at 64 m/s in a vertical loop of radius 120 m. Calculate the force of the plane’s seat on a 72 kg pilot while passing through the bottom part of the loop.

Answer:

Two forces act on the pilot his downward weight force w and the upward force of the aeroplane’s seat F_seat. Because the pilot moves in a circular path, these forces along the radial direction must, according to Newton’s second law (∑ F = ma), equal the pilot’s mass times his centripetal acceleration, where

⇒ \(a_c=v^2 /r\) we find F (in radial direction) = \(F_{\text {seat }}-w=\frac{m v^2}{r}\)

Remember that force pointing towards the center of the circle (Fseat) is positive and those pointing away from the center (w) are negative.

Substituting ω = mg and rearranging, we find that the force of the airplane seat on the pilot is

⇒ \(F_{\text {seat }}=m\left(\frac{\mathrm{v}^2}{\mathrm{r}}+\mathrm{g}\right)=72 \mathrm{~kg}\left[\frac{64(\mathrm{~m} / \mathrm{s})^2}{120 \mathrm{~m}}+9.8 \mathrm{~m} / \mathrm{s}^2\right]\)

= 72 kg (34.1 m/s² + 9.8m/s²) = 3160.8 N

The pilot in this Question feels very heavy. To keep him in the circular path, the seat must push the pilot upwards with a force of 3160 N, 4.5 times his normal weight. He experiences an acceleration of 4.5 g, that is, 4.5 times the acceleration of gravity.

A Car Taking A Turn On A Level Road: When a car takes a turn on a level road, the portion of the turn can be approximated by an arc of a circle of radius r.

If the car makes the turn at a constant speed v, then there must be some centripetal force acting on the car. This force is generated by the friction between the tire and the road. (A car tends to slip radially outward, so frictional force acts inwards)

μ_s is the coefficient of static friction

N = mg is the normal reaction of the surface

The maximum safe velocity v is –

⇒ \(\frac{m v^2}{r}-\mu_s N=\mu_s m g\)

or \(\mu_{\mathrm{s}}=\frac{v^2}{\mathrm{rg}}\)

or \(v=\sqrt{\mu_s r g}\)

It is independent of the mass of the car. The safe velocity is the same for all vehicles of larger and smaller mass.

Question 9. A car is traveling at 30 km/h in a circle of radius 60 m. What is the minimum value of μs for the car to make the turn without skidding?
Answer:

The minimum μSshould be that

⇒ \(\mu_{\mathrm{s}} \mathrm{mg}=\frac{\mathrm{mv}^2}{r}\) or \(\)

Here, \(v=30 \frac{\mathrm{km}}{\mathrm{h}}=\frac{30 \times 1000}{3600}=\frac{25}{3} \mathrm{~m} / \mathrm{s}\)

⇒ \(\mu_s=\frac{25}{3} \times \frac{25}{3} \times \frac{1}{60 \times 10}=0.115\)

For all values of μ_S greater than or equal to the above value, the car can make the turn without skidding. If the speed of the car is high so that the minimum μ_s is greater than the standard values (rubber tire on dry concrete μ_s= 1 and on wet concrete μ_s= 0.7), then the car will skid.

Banking Of Road: If a cyclist takes a turn, he can bend from his vertical position. This is not possible in the case of cars, trucks, or trains.

The tilting of the vehicle is achieved by raising the outer edge of the circular track, slightly above the inner edge. This is known as the banking of the curved track. The angle of inclination with the horizontal is called the angle of banking. If the driver moves with a slow velocity friction does not play any role in negotiating the turn. The various forces acting on the vehicle are :

1. Weight of the vehicle (mg) in the downward direction.
2. Normal reaction (N) perpendicular to the inclined surface of the road.

Resolve N in two components.

N cosθ, vertically upwards which balances the weight of the vehicle.

∴ N cosθ = mg …….(1)

N sin θ, in the horizontal direction which provides the necessary centripetal force.

∴ N sin θ = \(\frac{m v^2}{r}\) …….(2)

on dividing eqn. (2) by eqn. (1)

⇒ \(\frac{N \sin \theta}{N \cos \theta}=\frac{\frac{m v^2}{r}}{m g}\)

or \(\tan \theta=\frac{\mathrm{v}^2}{\mathrm{rg}}\)

∴ \(\theta=\tan ^{-1}\left(\frac{\mathrm{v}^2}{\mathrm{rg}}\right)\)

Where m is the mass of the vehicle, r is the radius of curvature of the road, v is the speed of the vehicle and θ is the banking angle (sinθ =h/b).

Factors that decide the value of the angle of banking are as follows:

Thus, there is no need for the mass of the vehicle to express the value of the angle of banking i.e. angle of banking ⇒ is not dependent on the mass of the vehicle.

∵ v² = gr tanθ

∴ v = \(\sqrt{g r \tan \theta}\) (maximum safe speed)

This gives the maximum safe speed of the vehicle. In actual practice, some frictional forces are always present. So, the maximum safe velocity is always much greater than that given by the above equation. While constructing the curved track, the value of θ is calculated for fixed values of V_Max and r. This explains why along the curved roads, the speed limit at which the curve is to be negotiated is clearly incited on sign boards.

The outer side of the road is raised by h = b × θ.

When θ i small, then tan θ ≈ sin θ = \(\frac{\mathrm{h}}{\mathrm{b}}\)

Also tan θ = \(\frac{v^2}{\mathrm{rg}}\)

⇒ \(\frac{v^2}{\mathrm{rg}}=\frac{\mathrm{h}}{\mathrm{b}} \text { or } \mathrm{h}=\frac{\mathrm{v}^2}{\mathrm{rg}} \times \mathrm{b}\)

Question 10. At what should a highway be banked for cars traveling at a speed of 100 km/h if the radius of the road is 400 m and no frictional forces are involved?
Answer:

The banking should be done at an angle θ such that

tan θ = \(\frac{v^2}{r g}=\frac{\frac{250}{9} \times \frac{250}{9}}{400 \times 10}\)

or \(\tan \theta=\frac{625}{81 \times 40}=0.19\)

or θ = tan–1 0.19 ≈ 0.19 radian

≈ 0.19 × 57.3º ≈ 11º

Question 11. The radius of curvature of a railway line at a place where the train is moving with a speed of 36 kmh^-1 is 1000 m, the distance between the two rails being 1.5 meters. Calculate the elevation of the outer rail above the inner rail so that there may be no side pressure on the rails.
Answer:

Velocity, \(v=36 \mathrm{~km} \mathrm{~h}^{-1}=\frac{36 \times 1000}{3600} \mathrm{~ms}^{-1}=10 \mathrm{~ms}^{-1}\)

radius, r = 1000 m ; tanθ = \(\frac{v^2}{\mathrm{rg}}=1000 \times 9.8=\frac{1}{9.8}\)

Let h be the height through which the outer rail is raised. Let l be the distance between the two rails.

Then, tan θ = \(\frac{h}{l}[latex]

[∵ θ is very small] or h = l tan θ

h = 1.5 × [latex]\frac{1}{98}[/frac] = 0.0153 m [∵ l = 1.5 m]

Question 12. An aircraft executes a horizontal loop at a speed of 720 km h–1 with its wing banked at 15º. Calculate the radius of the loop.
Answer:

Speed , v = [latex]720 \mathrm{~km} \mathrm{~h}^{-1}=\frac{720 \times 1000}{3600} \mathrm{~ms}^{-1}=200 \mathrm{~ms}^{-1}\)

and \(\tan \theta=\tan 15^{\circ}=0.2679\)

⇒ \(\tan \theta=\frac{\mathrm{v}^2}{\mathrm{rg}}\)

or \(r=\frac{v^2}{g \tan \theta}=\frac{200 \times 200}{9.8 \times 0.2679} \mathrm{~m}=1523.7 \mathrm{~m}=15.24 \mathrm{~km}\)

Question 13. A train rounds an unbanked circular bend of radius 30 m at a speed of 54 km h^-1. The mass of the train is 10⁶ kg. What provides the centripetal force required for this proposal? The engine or the rails? The outer or inner rails? Which rail will wear out faster, the outer or the inner rail? What is the angle of banking required to prevent wearing out of the rails?
Answer:

⇒ \(r=30 \mathrm{~m}, \mathrm{v}=54 \mathrm{~km} \mathrm{~h}^{-1}=\frac{54 \times 5}{18} \mathrm{~ms}^{-1}=15 \mathrm{~ms}^{-1} \mathrm{~m}=10^5 \mathrm{~kg}, \quad \theta=?\)

The centripetal force is provided by the lateral thrust by the outer rail on the flanges of the wheel of the train. The train causes an equal and opposite thrust on the outer rail (Newton’s third law of motion).

Thus, the outer rails wear out faster.

tan θ = \(\tan \theta=\frac{v^2}{\mathrm{rg}}=\frac{15 \times 15}{60 \times 9.8}=0.7653\)

∴ θ = tan–1(0.7653) = 37.43º 60 9.8

Special Points About Circular Motion

Centripetal force does not increase the kinetic energy of the particle moving in a circular path, hence the work done by the force is zero.

• Centrifuges are the apparatuses used to separate small and big particles from a liquid.
• The physical quantities that remain constant for a particle moving in the circular path are speed, kinetic energy, and angular momentum.
• If a body is moving on a curved road with a speed greater than the speed limit, the reaction at the inner wheel disappears and it will leave the ground first.
• On unbanked curved roads, the minimum radius of curvature of the curve for safe driving is r = v²/μg, where v is the speed of the vehicle and μ is the coefficient of friction.
• The skidding of a vehicle will occur if v²/r > μg i.e., skidding will take place if the speed is large, the curve is sharp and μ is small.
• If r is the radius of curvature of the speed breaker, then the maximum speed with which the vehicle can run on it without leaving contact with the ground is \(v=\sqrt{(g r)}\)
• While taking a turn on the level road sometimes vehicles overturn due to centrifugal force.

Points To Be Remember

Uniform Motion In A Circle –

Angular velocity = \(\omega=\frac{\mathrm{d} \theta}{\mathrm{dt}}=2 \pi \mathrm{n}=\frac{2 \pi}{\mathrm{T}}\)

Linear velocity \(v=\vec{\omega} \times \vec{r}\)

v = \(v=\omega r \text { when } \vec{\omega} \text { and } \vec{r}\) are perpendicular to each other.

Centripetal acceleration = \(a=\frac{v^2}{r}=\omega^2 r=\omega v=4 \pi^2 n^2 r\)

Equations Of Motion –

For constant angular acceleration –

1. \(\omega=\omega_0+\alpha \mathrm{t}\)
2. \(\theta=\omega_0 t+\frac{1}{2} \alpha t^2\)
3. \(\omega^2=\omega_0^2+2 \alpha \theta\)

Motion Of A Car On A Plane Circular Road –

For motion without skidding

⇒ \(\frac{M v_{\max }^2}{r}=\mu M_g, v_{\max } \sqrt{\mu \mathrm{g}}\)

Motion On A Banked Road – Angle of banking = θ

tan θ = \(\frac{h}{b}\)

Maximum safe speed at the bend \(\mathrm{v}_{\max }=\left[\frac{\mathrm{rg}(\mu+\tan \theta)}{1-(\mu \tan \theta)}\right]^{1 / 2}\)

If friction is negligible = \(\mathrm{v}_{\max }=\sqrt{\mathrm{rg} \tan \theta}=\sqrt{\frac{\mathrm{rhg}}{\mathrm{b}}} \text { and } \quad \tan \theta=\frac{\mathrm{v}^2 \max }{\mathrm{rg}}\)

Motion Of Cyclist On A Curve –

In equilibrium angle with vertical is θ then tan θ = \(\frac{v^2}{\mathrm{rg}}\)

Maximum safe speed = \(v_{\max }=\sqrt{\mu r g}\)

Motion In A Vertical Circle (particle tied to a string) –

At the top position – Tension \(T_A=m\left(\frac{v_A^2}{r}-g\right)\)

For T_A = 0, critical speed = \(\sqrt{\mathrm{gr}}\)

At the bottom – Tension \(T_B=m\left(\frac{v_B^2}{r}+g\right)\)

For completing the circular motion minimum speed at the bottom \(v_B=\sqrt{5 \mathrm{gr}}\)

Tension T_B = 6mg

Conical Pendulum (Motion in a horizontal circle)

Tension is string = \(\frac{\mathrm{mg} \ell}{\left(\ell^2-r^2\right)^{1 / 2}}\)

Angular velocity = \(\sqrt{\frac{\mathrm{g}}{\ell \cos \theta}}\)

Periodic time = \(2 \pi \sqrt{\frac{\ell \cos \theta}{g}}\)

⇒ \(2 \pi \sqrt{\frac{r}{g \tan \theta}}\)

Centripetal Force

Centripetal Force: In uniform circular motion the force acting on the particle along the radius and towards the center keeps the body moving along the circular path. This force is called centripetal force.

Centripetal Force Explanation:

1. Centripetal force is necessary for uniform circular motion.
2. It is along the radius and towards the center.
3. Centripetal force = [mass] × [centripetal acceleration] = \(\frac{m v^2}{r}=m r \omega^2\)

4. Centripetal force is due to known interaction. Therefore it is a real force. If an object tied to a string is revolved uniformly in a horizontal circle, the centripetal force is due to the tension imparted to the string by the hand.

When a satellite revolves in a circular orbit around the Earth, the centripetal force is due to the gravitational force of attraction between the satellite and the Earth.

In an atom, an electron revolves in a circular orbit around the nucleus. The centripetal force is due to the electrostatic force of attraction between the positively charged nucleus and the negatively charged.

Question 1. A stone of mass 1kg is whirled in a circular path of radius 1 m. Find out the tension in the string if the linear velocity is 10 m/s.
Answer: Tension

⇒ \(\frac{m v^2}{R}=\frac{1 \times(10)^2}{1}=100 \mathrm{~N}\)

Question 2. A satellite of mass 107 kg is revolving around the earth with a time period of 30 days at a height of 1600 km. Find out the force of attraction on satellite by Earth.
Answer:

Force = \(\mathrm{m} \omega^2 R \text { and } \frac{2 \pi}{\mathrm{T}}=\frac{2 \times 3.4}{30 \times 86400}=\frac{6.28}{2.59 \times 10^6}\)

Force = \(m \omega^2 r=\left(\frac{6.28}{2.59 \times 10^6}\right)^2 \times 10^7 \times(6400+1600) \times 10^3=2.34 \times 10^5 \mathrm{~N}\)

Centrifugal Force

The pseudo force experienced by a particle performing uniform circular motion due to an accelerated frame of reference which is along the radius and directed way from the center is called centrifugal force.

Centrifugal Force Explanation:

1. Centrifugal force is a pseudo force as it is experienced due to accelerated frame of reference. The interaction of origin and away from the centre.
2. It is along the radius and away from the centre.
3. The centrifugal force in having the same magnitude as that of centripetal force. But, its direction is opposite to that of centripetal force . It is not due to reaction of centripetal force because without action, reaction not possible, but centrifugal force can exists without centripetal force.
4. Magnitude of the centrifugal force is \(\frac{\mathrm{mv}^2}{\mathrm{r}} \text { or } \mathrm{mr} \omega^2\)

Note: Pseudo force acts in a noninertial frame i.e. accelerated frame of reference in which Neutron’s laws of motion do not hold good.

• When a car moving along a horizontal curve takes a turn, the person in the car experiences a push in the outward direction.
• The coin placed slightly away from the center of a rotating gramophone disc slips towards the edge of the disc.
• A cyclist moving fast along a curved road has to lean inwards to keep his balance.

Difference Between Centripetal Force And Centrifugal Force:

Applications Of Centrifugal Force: The centrifugal pump used to lift the waterworks on the principle of centrifugal force.

A cream separator used in the diary work works on the principle of centrifugal force. Centrifuge used for the separation of suspended particles from the liquid, works on the principle of centrifugal force. Centrifugal drier.

 

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