Expansion Of Solid And Liquids Short Answer Questions

Properties Of Bulk Matter

Expansion Of Solid And Liquids Short Question And Answers

Short Answer Questions on Thermal Expansion

Question 1. The liquid with a co-efficient of volume expansion (γ) is filled in a container of a material having the coefficient of linear expansion a. If the liquid overflows on heating, then

1. γ = 3α
2. γ > 3α
3. γ <3α
4. γ > 3α³

Answer:

Given

The liquid with a co-efficient of volume expansion (γ) is filled in a container of a material having the coefficient of linear expansion a.

Apparent coefficient of volume expansion, γ’= γ-3α

Now, γ’ is positive when the liquid overflows on heating.

So the condition is γ – 3α > 0 or, ϒ > 3α

The option 2 is correct.

Question 2. When a copper sphere is heated, which physical quantity of the sphere will show maximum percentage change?
Answer:

When a copper sphere is heated, its volume will show maximum change in percentage.

Question 3. A metal rod is fixed rigidly at two ends so as to prevent its thermal expansion. If L, a and Y respectively denote the length of the rod, coefficient of linear thermal expansion and Young’s modulus of its material, then for an increase in temperature of the rod by ΔT, the longitudinal stress developed in the rod is

1. Inversely proportional to α
2. Inversely proportional to Y
3. Directly proportional to ΔT/Y
4. Independent of L

Answer:

Given

A metal rod is fixed rigidly at two ends so as to prevent its thermal expansion. If L, a and Y respectively denote the length of the rod, coefficient of linear thermal expansion and Young’s modulus of its material, then for an increase in temperature of the rod by ΔT,

Longitudinal thermal stress = YαΔT.

The option 4 is correct.

Key Concepts in Solid and Liquid Expansion

Question 4. A solid rectangular sheet has two different coefficients of linear expansion α₁ and α₂ along its length and breadth respectively. The coefficient of surface expansion is (for α₁ t<<1 1, α₂ t<<l)

1. \(\frac{\alpha_1+\alpha_2}{2}\)
2. \(2\left(\alpha_1+\alpha_2\right)\)
3. \(\frac{4 \alpha_1 \alpha_2}{\alpha_1+\alpha_2}\)
4. \(\alpha_1+\alpha_2\)

Answer:

Given

A solid rectangular sheet has two different coefficients of linear expansion α₁ and α₂ along its length and breadth respectively.

Let, initial length and breadth of rectangular sheet are a₁ and b₁ respectively.

If the final length and breadth are a₂ and b₂ respectively in increase in temperature t° C, then

⇒ \(a_2=a_1\left(1+a_1 t\right) \text { and } b_2=b_1\left(1+\alpha_2 t\right)\)

∴ \(a_2 b_2=a_1 b_1\left(1+\alpha_1 t\right)\left(1+\alpha_2 t\right)\)

or, \(a_1 b_1(1+\beta t)=a_1 b_1\left\{1+\left(\alpha_1+\alpha_2\right) t+\alpha_1 \alpha_2 t^2\right\}\)

or, \(1+\beta t=1+\left(\alpha_1+\alpha_2\right) t\)

[neglecting the term \(\alpha_1 \alpha_2 t^2\)]

∴ \(\beta=\alpha_1+\alpha_2\)

The option 4 is correct

Real-Life Applications of Thermal Expansion

Question 5. The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100 °C is (For steel, Young’s modulus is 2 x 10¹¹ N • m^-2 and coefficient of linear expansion is 1.1 x 10^-5 K^-1)

1. 2.2x 10⁸ Pa
2. 2.2x 10⁹ Pa
3. 2.2 x 10⁷ Pa
4. 2.2 x 10⁶ Pa

Answer: As length is constant, strain =ΔL/L = αΔQ

Now, pressure = stress = Yx strain

= 2x 10¹¹ x 1.1 x 10^-5 x 100

= 2.2 x 10⁸ Pa

The option 1 is correct.

Question 6. A pendulum clock loses 12 s a day if the temperature is 40°C and gains 4 s a day if the temperature is 20°C. The temperature at which the clock will show correct time, and the coefficient of linear expansion (α) of the metal of the pendulum shaft are respectively

1. 25°C and 1.85 x 10^-5 l °C
2. 60°C and 1.85 x 10^-4 l °C
3. 30°C and 1.85 x 10^-3 l°C
4. 55°C and 1.85 x 10^-2 l °C

Answer:

Given

A pendulum clock loses 12 s a day if the temperature is 40°C and gains 4 s a day if the temperature is 20°C. The temperature at which the clock will show correct time,

Increase or decrease in time for change in temperature ΔT,

⇒ \(\Delta t =\frac{1}{2} \alpha \Delta T t\)

∴ 12 = \(\frac{1}{2} \alpha\left(40-T_0\right) \times 1\)

T₀ = temperature at which clock gives accurate breading, t = 1 day]

and 4 = \(\frac{1}{2} \alpha\left(T_0-20\right) \times 1\)….(2)

(1)+(2) we get,

3 = \(\frac{40-T_0}{T_0-20} \quad \text { or, } 3 T_0-60=40-T_0\)

or, \(T_0=\frac{100}{4}=25^{\circ} \mathrm{C}\)

Putting the value of T₀ in equation (1) we get,

⇒ \(15 \alpha=\frac{24}{24 \times 60 \times 60}\)

or, \(\alpha=\frac{1}{15 \times 60 \times 60}=1.85 \times 10^{-5} /{ }^{\circ} \mathrm{C}\)

The option 1 is correct.

Coefficient of Expansion Short Answer Questions

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Question 7. An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and α is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by:

1. \(3 P K \alpha\)
2. \(\frac{P}{3 \alpha K}\)
3. \(\frac{P}{\alpha K}\)
4. \(\frac{3 a}{P K}\)

Answer:

Given

An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and α is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating

Bulk modulus \(K=\frac{P}{\left(\frac{\Delta V}{V}\right)}\)

∴ \(\frac{\Delta V}{V}=\frac{P}{K}[\Delta V= change in volume ]\)

If increase in temperature Δt brings the cube to its original size, then

∴ \(\Delta V=V \cdot \gamma \Delta t \quad \text { or, } \Delta t=\frac{\Delta V}{V} \cdot \frac{1}{\gamma}=\frac{\Delta V}{V} \cdot \frac{1}{3 \alpha}=\frac{P}{3 K \alpha}\)

The option 2 is correct.

Question 8. 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 same at all temperatures, which one of the following relations holds good?

1. \(\alpha_1 l_2^2=\alpha_2 l_1^2\)
2. \(\alpha_1^2 l_2=\alpha_2^2 l_1\)
3. \(\alpha_1 l_1=\alpha_2 l_2\)
4. \(\alpha_1 l_2=\alpha_2 l_1\)

Answer:

Given

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 same at all temperatures,

Let, lengths of brass rod at temperatures t₁ and t₂ are l₂ and l’₂ respectively.

∴ \(l_1^{\prime}=l_1\left\{1+\alpha_1\left(t_2-t_1\right)\right\}\)

Again, let lengths of steel rod at temperatures t₁ and t₂ are l₂ and l’₂ respectively.

∴ \(l_2^{\prime}=l_2\left\{1+\alpha_2\left(t_2-t_1\right)\right\}\)

According to the question, \(l_2-l_1=l_2^{\prime}-l_1^{\prime}\)

or, \(l_2-l_1=l_2\left\{1+\alpha_2\left(t_2-t_1\right)\right\}-l_1\left\{1+\alpha_1\left(t_2-t_1\right)\right\}\)

or, \(l_2-l_1=l_2+l_2 a_2\left(t_2-t_1\right)-l_1-l_1 a_1\left(t_2-t_1\right)\)

or, \(l_2-l_1=\left(l_2-l_1\right)+\left(l_2-t_1\right)\left(l_2 a_2-l_1 a_1\right)\)

or, \(\left(t_2-t_1\right)\left(l_2 a_2-l_1 a_1\right)=0\)

as \(\left(t_2-t_1\right) \neq 0\)

∴ \(\left(l_2 \alpha_2-l_1 \alpha_1\right)=0 \quad or, l_1 \alpha_1=l_2 a_2\)

The option 3 is correct.

Thermal Expansion Problems with Solutions

Question 9. Show that the coefficient of superficial expansion of a rectangular sheet of the solid is twice its coefficient of linear expansion.
Answer:

Let a₁,b₁ be the length and breadth, respectively, of the rectangular sheet at the initial temperature t₁.

Then, S₁ = a₁b₁ = initial area

Now, the temperature is raised to t₂, where t₂  – t₁ =t

The new values of the length and breadth are, respectively, a₂ and b₂; area S₂ = a₂b₂.

If α be the coefficient of linear expansion, then

⇒ \(a_2=a_1(1+\alpha t) \text { and } b_2=b_1(1+\alpha t)\)

∴ \(S_2=a_2 b_2=a_1 b_1(1+\alpha t)^2=S_1(1+2 \alpha t)\)

(neglecting α²t² as α<<1)

On the other hand, if β be the coefficient of superficial expansion, then

S₂ = S₁(1 +βt)

On comparison, we have β = 2α.