## Properties Of Bulk Matter

## Expansion Of Solid And Liquids Short Question And Answers

**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**

- γ = 3α
- γ > 3α
- γ <3α
- γ > 3α³

**Answer:**

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**

- Inversely proportional to α
- Inversely proportional to Y
- Directly proportional to ΔT/Y
- Independent of L

**Answer:**

Longitudinal thermal stress = YαΔT.

The option 4 is correct.

**Question 4. A solid rectangular sheet has two different coefficients of linear expansion α _{1} and α_{2} along its length and breadth respectively. The coefficient of surface expansion is (for α_{1} t<<1 1, α_{2 }t<<l)**

- \(\frac{\alpha_1+\alpha_2}{2}\)
- \(2\left(\alpha_1+\alpha_2\right)\)
- \(\frac{4 \alpha_1 \alpha_2}{\alpha_1+\alpha_2}\)
- \(\alpha_1+\alpha_2\)

**Answer:**

Let, initial length and breadth of rectangular sheet are a_{1} and b_{1} respectively.

If the final length and breadth are a_{2} and b_{2} 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

**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 10 ^{11} N • m^{-2} and coefficient of linear expansion is 1.1 x 10^{-5} K^{-1})**

- 2.2x 10
^{8}Pa - 2.2x 10
^{9}Pa - 2.2 x 10
^{7}Pa - 2.2 x 10
^{6}Pa

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

Now, pressure = stress = Yx strain

= 2x 10^{11} x 1.1 x 10^{-5 }x 100

= 2.2 x 10^{8} 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**

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

**Answer:**

Increase or decrease in time for change in temperature ΔT,

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

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

T_{0} = 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 T_{0} 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.

**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:**

- \(3 P K \alpha\)
- \(\frac{P}{3 \alpha K}\)
- \(\frac{P}{\alpha K}\)
- \(\frac{3 a}{P K}\)

**Answer:**

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 α _{1} and α_{2}. Lengths of brass and steel rods are l_{1} and l_{2} respectively. If (l_{2 }– l_{1}) is maintained same at all temperatures, which one of the following relations holds good?**

- \(\alpha_1 l_2^2=\alpha_2 l_1^2\)
- \(\alpha_1^2 l_2=\alpha_2^2 l_1\)
- \(\alpha_1 l_1=\alpha_2 l_2\)
- \(\alpha_1 l_2=\alpha_2 l_1\)

**Answer:**

Let, lengths of brass rod at temperatures t_{1 }and t_{2} are l_{2 }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 t_{1 }and t_{2 }are l_{2} 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.

**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 a_{1},b_{1} be the length and breadth, respectively, of the rectangular sheet at the initial temperature t_{1}.

Then, S_{1} = a_{1}b_{1} = initial area

Now, the temperature is raised to t_{2}, where t_{2 } – t_{1} =t

The new values of the length and breadth are, respectively, a_{2} and b_{2}; area S_{2} = a_{2}b_{2}.

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

S_{2} = S_{1}(1 +βt)

On comparison, we have β = 2α.