Short Discussion Questions

1. Why do clocks slow in summer and fast in winter?

Answer: The time period of a pendulum clock is given by: $T = 2 \pi \sqrt{ \dfrac{l}{g} }$. In summer, due to thermal expansion, the length of the pendulum increases and hence, the time period increases. Due to increased time period, clocks slow in summer. While in winter, the length of pendulum contracts due to decrease in temperature and the clocks become fast due to decreased time period.

2. Why does water level fall initially in a vessel when it is heated?

Answer: Water is not heated directly. It is heated by placing it in a metal container. During initial heating, the water does not get the heat and all the heat is absorbed by the container itself. Due to this, the container expands but the liquid doesn't. Due to this, there is extra volume in the container and the water level in a vessel falls initially when it is heated.

3. What is the effect of temperature on density of liquid?

Answer: We have,
$\rho_2 = \dfrac{ \rho _1}{1 + \gamma \triangle \theta}$

$or, \rho_1 = \rho_2 (1 + \gamma \triangle \theta}$

This implies that the density of liquid before heating was more than the density of liquid after heating. Therefore, increase in temperature causes decrease in the density of liquid.

4. What is the relation between linear, superficial and cubical expansivity?

Answer: The relation between linear, superficial and cubical expansivity is mentioned below:

$ \alpha : \beta : \gamma = 1 : 2:3$

5. What do you mean by differential expansion? Under what condition it's value is zero?

Answer: Differential expansion is the observed change in lengths of two materials of different inital lengths after heating them through the same range of temperature.

When $l_1 \propto \dfrac{1}{\alpha}$, then $l_1 \alpha_1 = l_2 \alpha_2$ the value of differential expansion is zero.

6. Water pipes burst in winter, why?

Answer: In winter, the temperature falls causing contractions in the dimensions of various substances. Accordingly, in winter, water pipes contract but water flowing inside those pipes expand due to anomalous expansion of water (between 4 to 0 degree celsius). So, there is excessive volume of water flowing in limited volume of pipe. Thus, water pipes burst in winter.

7. Liquid has only cubical expansion, why?

Answer: Liquid doesn't have its own shape and takes the shape of the container it is placed in. So, it does not have linear of superficial expansion. However, it does have a fixed volume so, it has only cubical expansion.

8. Many automobile engines have cast iron cylinder and aluminum pistons. What kinds of problems could occur if the engine gets too hot?

Answer:  Due to overheating of the engines, there becomes signficant expansion in the dimensions of the iron cylinder and alumnium pistons which could damage the engine and even stop the moving automobile instantly.

9. During winter, water pipes burst, why? Would the mercury thermometer break if the temperature went below the freezing point of mercury? Why or why not?

Answer: Water pipes burst in winter due to increase of volume of water due to its anomalous expansion (from 4 degree to 0 degree Celsius) and decrease in volume of water pipe due to decrease in temperature.

No, the mercury thermometer would not break because mercury does not show any anomalous behavior like water so, it would just keep on contracting as the temperature is conntinuously reduced.

10. Two bodies made of same material have the same external dimensions and appearance, but one is solid and the other is hollow. When their temperature is increased, is the overall expansion same or different?

Answer: The expansion of any solid depends upon its initial dimension, coefficient of cubical expansivity and change in temperature range. Since, all of the above-mentioned physical conditions are the same for given two bodies so, they will show uniform increase in their dimensions for the same increase in tempearture.

11. The pendulum shaft of a pendulum clock is made of brass. What is the fractional change in lenght of the shaft when it cools form 19.2 degree C to 7 degree C? (Coefficient of thermal expansion of brass = 2 * 10^(-5) /K)

Solution:
Let the initial length of the pendulum shaft be L1.

Let the final length of the pendulum shaft be L2. 

Given,
$\alpha = 2 * 10^{-5} K^{-1}$
$\theta _1$ = 19.2 degree C
$\theta_2$ = 7 degree C
$\triangle \theta = 7 - 19.2$ = -12.2 Celsius degree

Now,
Fractional increase in length of the pendulum shaft $\triangle L$ is given by

$= \dfrac{|\triangle L|}{L1} * 100$

$= \dfrac{|L1 \alpha * \triangle \theta|}{L1} *100$

$= |2 * 10^{-5} * (-12.2)| * 100$

$= 

Long Discussion Questions

3. Define the term coefficients of real and apparent expansion of liquid and prove that $\gamma_r = \gamma_a  + \gamma_g$.

Solution:
Coefficient of real expansion of liquid ($\gamma_r$): It is the ratio of change in real volume of a liquid to the product of the original volume and difference in temperature.

Coefficient of apparent expansion of liquid ($\gamma_r$): It is the ratio of change in apparent volume of a liquid to the product of the original volume and difference in temperature.

To prove: $\gamma_r = \gamma_a + \gamma_g$

Let us assume that a vessel of volume V has liquid filled to its maximum height. On heating, the apparent expansion in volume is liquid is observed to be $\triangle V_a$ and the expansion of vessel itself was $V_g$.

Let the real expansion of liquid be $V_r$

From expansion of liquid, we know,

Real expansion of liquid = Apparent expansion of liquid + Expansion of vessel

$or, \triangle V_r = \triangle V_a  + \triangle V_g$

[From formula]
$or, V \gamma_r \triangle \theta = V \gamma_a \triangle \theta +V \gamma_g \triangle \theta$

$\therefore \gamma_r = \gamma_a + \gamma_g$