NEET Physics Solutions For Class 11 Chapter 1 Thermal Expansion

NEET Physics Class 11 Chapter 1 Thermal Expansion

Most materials expand when their temperature is increased. Railway roads tracks, and bridges all have some means of compensating for thermal expansion.

• When a homogeneous object expands, the distance between any two points on the object increases. Figure 5 shows a block of metal with a hole in it.
• The expanded object is like a photographic enlargement. In the hole expands in the same proportion as the metal, it does not get smaller

• At the atomic level, thermal expansion may be understood by considering how the potential energy of the atoms varies with distance.
• The equilibrium position of an atom will be at the minimum of the potential energy well if the well is symmetric. At a given temperature each atom vibrates about its equilibrium position and its average remains at the minimum point.
• If the shape of the well is not symmetrical the average position of an atom will not be at the minimum point.
• When the temperature is raised the amplitude of the vibrations increases and the average position is located at a greater interatomic separation. This increased separation is manifested as expansion of the material as shown in figure 6.
• Almost all solids and liquids expand as their temperature increases. If allowed. Solids can change in length, area, or volume, while liquids change in their volume.

Question 1. A rectangular plate has a circular cavity as shown in the figure. If we increase its temperature then which dimension will increase in the following figure?

Answer:

Distance between any two points on an object increases with an increase in temperature. So, all dimensions a, b, c, and d will increase

Question 2. In the given figure, when temperature is increased then which of the following increases

1. R₁
2. R₂
3. R₂– R₁
4. None of these

Answer: All of the above

– – – – – represents expanded Boundary

——— represents the original Boundary

As the intermolecular distance between atoms increases on heating hence the inner and outer perimeter increases. Also if the atomic arrangement in radial direction is observed then we can say that it also increases hence all A, B, C are true.

NEET Physics Class 11 Chapter 1 Linear Expansion

When the rod is heated, its increase in length ΔL is proportional to its original length L₀ and change in temperature ΔT where ΔT is in ºC or K.

dL = αL₀dT ⇒ ΔL = α L₀ Δ T If α ΔT << 1

⇒ \(\alpha=\frac{\Delta L}{L_0 \Delta T}\) where α is called the coefficient of linear expansion whose unit is ºC−1 or K−1.

L = L₀(1 + α Δ T). Where L is the length after heating the rod.

Variation of a with temperature and distance

If α varies with distance, α = ax + b.

The total expansion = ∫(ax + b) ΔT dx.

If α varies with temperature, α = f (T). Then ΔL = ∫α L0dT

Note: Actually thermal expansion is always 3-D expansion. When the other two dimensions of an object are negligible with respect to one, then observations are significant only in one dimension and it is known as linear expansion.

Question 1. What is the percentage change in length of a 1m iron rod if its temperature changes by 100ºC? α for iron is 2 × 10–5/ºC.

Answer:

percentage change in length due to temperature change

⇒ \(\% \Delta \ell=\frac{\Delta \ell}{\ell}\)× 100 = αΔθ × 100 = 2 × 10–5 × 100 × 100 = 0.2%

Thermal Stress Of A Material: If a rod is free to expand then there will be no stress and strain.

• Stress and strain is produced only when an object is restricted to expand or contract according to change in temperature.
• When the temperature of the rod is decreased or increased under constrained condition , compressive or tensile stresses are developed in the rod. These stresses are known as thermal stresses.

Strain = \(\frac{\Delta \mathrm{L}}{\mathrm{L}_0}=\frac{\text { final length }- \text { original length }}{\text { original length }}=\alpha \Delta \mathrm{T}\)

Note: Original and final length should be at the same temperature.

Consider a rod of length l₀ which is fixed between two rigid supports separated at a distance l₀ now if the temperature of the rod is increased by Δθ then the strain produced

strain = \(
$$
=\frac{\text { length of the rod at new temperature- natural length of the rod at new temperature }}{\text { natural length of the rod at new temperature }}
$$\)

⇒ \(\frac{\ell_0-\ell_0(1+\alpha \Delta \theta)}{\ell_0(1+\alpha \Delta \theta)}=\frac{-\ell_0 \alpha \Delta \theta}{\ell_0(1+\alpha \Delta \theta)}\)

∵α is very small. So, Δθ << 1

strain = – α Δ θ (negative sign in the answer represents that the length of the rod is less than the natural length which means is compressed at the ends.)

Question 2. In the given figure a rod is free at one end and the other end is fixed. When we change the temperature of a rod by Δθ, the strain produced in the rod will be

1. αΔθ
2. \(\frac{1}{2}\)αΔθ
3. Zero
4. Information incomplete

Answer: 3. Zero

Here rod is free to expand from one side so by changing temperature no strain will be produced in the rod. Hence ans. is (3)

Question 3. An iron ring measuring 15.00 cm in diameter is to be shrunk on a pulley which is 15.05 cm in diameter. All measurements refer to the room temperature 20°C. To what minimum temperature should the ring be heated to make the job possible? Calculate the strain developed in the ring when it comes to room temperature. Coefficient of linear expansion of iron = 12 × 10^-6/°C.

Answer:

The ring should be heated to increase its diameter from 15.00 cm to 15.05 cm.

Using l2= l1(1 + α Δθ), ⇒ Δθ = \(\Delta \theta=\frac{\left(\ell_2-\ell_1\right)}{\ell_1 \alpha}\)

⇒ \(\frac{0.05 \mathrm{~cm}}{15.00 \mathrm{~cm} \times 12 \times 10^{-6} /{ }^{\circ} \mathrm{C}}=278^{\circ} \mathrm{C}\)

The temperature = 20°C + 278°C = 298°C.

The strain developed = \(\frac{\ell_2-\ell_1}{\ell_1}\) = 3.33 × 10–3

Question 4. A steel rod of length 1m rests on a smooth horizontal base. If it is heated from 0°C to 100°C, what is the longitudinal strain developed?

Answer: In the absence of external force no strain or stress will be created hear rod is free to move. 9

Question 5. A steel rod is clamped at its two ends and rests on a fixed horizontal base. The rod is in natural length at 20°C. Find the longitudinal strain developed in the rod if the temperature rises to 50°C. Coefficient of linear expansion of steel = 1.2 × 10^-5/°C.

Answer:  As we know strain

strain = \(\frac{\text { change in length }}{\text { original length }}=\frac{\Delta \ell}{\ell_0}\)

∴ Strain = αΔθ

= 1.2 × 10^-5 × (50 – 20) = 3.6 × 10^-4

here strain is compressive strain because the final length is smaller than the initial length.

Question 6. A steel wire of a cross-sectional area of 0.5 mm² is held between two fixed supports. If the wire is just taut at 20°C, determine the tension when the temperature falls to 0°C. The coefficient of linear expansion of steel is 1.2 × 10^-5/°C and Young’s modulus is 2.0 × 10¹¹N/m².

Answer:

here final length is more than the original length so that strain is tensile and tensile force is given by

F = AY α Δ t = 0.5 × 10^-6 × 2 × 10¹¹ × 1.2 × 10^-5 × 20 = 24 N

Variation Of Time Period Of Pendulum Clocks:

The time represented by the clock hands of a pendulum clock depends on the number of oscillations performed by the pendulum every time it reaches its extreme position the second hand of the clock advances by one second that means second hand moves by two seconds when one oscillation incomplete

Let T = \(2 \pi \sqrt{\frac{L_0}{g}}\) at temperature θ0 and T′ = \(2 \pi \sqrt{\frac{L}{g}}\) at temperature θ.

⇒ \(\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\sqrt{\frac{\mathrm{L}^{\prime}}{\mathrm{L}}}=\sqrt{\frac{\mathrm{L}[1+\alpha \Delta \theta]}{\mathrm{L}}}\) = 1+\(\frac{1}{2}\)2α Δ θ (Binomial approximation is used)

Therefore change (loss or gain) in time per unit time lapsed is

⇒ \(\frac{\mathrm{T}^{\prime}-\mathrm{T}}{\mathrm{T}}=\frac{1}{2} \alpha \Delta \theta\)

gain or loss in time in duration of ‘t’ in

Δt = \(\frac{1}{2}\)α Δθ t, if T is the correct time then 2

1. θ < θ0, T′ < T clock becomes fast and gains time
2. θ > θ0, T′ > T clock becomes slow and loses time

Question 7. A pendulum clock consists of an iron rod connected to a small, heavy bob. If it is designed to keep the correct time at 20°C, how fast or slow will it go in 24 hours at 40°C? Coefficient of linear expansion of iron = 1.2 × 10^-6 /°C.

Answer:

The time difference that occurred in 24 hours (86400 seconds) is given by

Linear Expansion of iron = 1.2 x 10^-6 /°C.

Δt = \(\frac{1}{2}\) 2α Δθ t

= \(\frac{1}{2}\)× 1.2 × 10^-6 × 20 × 86400 = 1.04 sec.

This is a loss of time as θ is greater than θ₀. As the temperature increases, the time period also increases. Thus, the clock goes slow.

Measurement of length by metallic scale:

Case 1

When an object is expanded only (figure 9a)

l₂= l₁{1 + α₀(θ₂– θ₁)

l₁= actual length of object at θ₁ºC = measure length of object at θ₁ºC.

l₂= actual length of object at θ₂ºC = measure length of object at θ₂ºC.

α₀= linear expansion coefficient of an object.

Case 2

When only the measuring instrument is expanded actual length of the object will not change but a measured value (MV) decreases.

MV = l₁{ 1 – α_s(θ₂– θ₁)}

α_s= linear expansion coefficient of measuring instrument.

at θ₁ C MV = 3

at θ₁ C MV = 2.2

Case 3

If both expanded simultaneously

MV = {1 + (α₀– α_s) (θ₂– θ₁)

1. If α₀> α_s, then the measured value is more the actual value at θ₁ºC
2. If α₀< α_s, then the measured value is less the actual value at θ₁ºC

at θ₁ºC MV = 3.4

θ₂ºC MV = 4.1

Measured value = calibrated value × {1 + α Δ θ}

where α = α₀– α_s

α₀= coefficient of linear expansion of object material, α_s = coefficient of linear expansion of scale material.

Δθ = θ − θ_C

θ = temperature at the time of measurement θ_C= temperature at the time of calibration. For scale, true measurement = scale reading [1 + α (θ − θ₀)]

If θ > θ₁ true measurement > scale reading

θ < θ₀ true measurement < scale reading

Question 8. A bar measured with a vernier caliper is found to be 180mm long. The temperature during the measurement is 10ºC. The measurement error will be if the scale of the vernier caliper has been graduated at a temperature of 20ºC : (α = 1.1 × 10-5°C^-1. Assume that the length of the bar does not change.)

1. 1.98 × 10^-1 mm
2. 1.98 × 10^-2 mm
3. 1.98 × 10^-3 mm
4. 1.98 × 10^-4 mm

Answer: True measurement = scale reading [1 + α (θ − θ₀)]

= 180 × {1+ (10 – 20) × (–1.1 × 10^-5) }

measurement error = true measurement – scale reading

= 180 × {1+ (10 – 20) × (–1.1 × 10^-5) } – 180

= 1.98 × 10^-2 mm