Question

A pendulum clock loses $12{\text{s}}$ a day if the temperature is $40^\circ {\text{C}}$ and gains $4{\text{s}}$ a day if the temperature is $20^\circ {\text{C}}$. Find the temperature at which the clock shows the correct time and the coefficient of linear expansion $\alpha$ of the metal of the pendulum shaft.
A) $25^\circ {\text{C}}$ ; $\alpha = 1 \cdot 85 \times {10^{ - 5}}^\circ {{\text{C}}^{ - 1}}$
B) $60^\circ {\text{C}}$ ; $\alpha = 1 \cdot 85 \times {10^{ - 4}}^\circ {{\text{C}}^{ - 1}}$
C) $30^\circ {\text{C}}$ ; $\alpha = 1 \cdot 85 \times {10^{ - 3}}^\circ {{\text{C}}^{ - 1}}$
D) $55^\circ {\text{C}}$ ; $\alpha = 1 \cdot 85 \times {10^{ - 2}}^\circ {{\text{C}}^{ - 1}}$

Answer 1

As the temperature increases, the pendulum shaft will expand causing the pendulum to run slower and if the temperature decreases, the shaft will contract which causes the clock to run faster. This change in the length of the shaft due to the change in the temperature is referred to as linear expansion. Two temperatures are mentioned in the question. We have to find a temperature in between these two temperatures so that the time shown by the pendulum clock is the right time.

Formulas used:
-The linear expansion of a body is given by, $\dfrac{{\Delta l}}{l} = \alpha \left( {{\theta _1} - \theta } \right)$ where $\Delta l$ is the change in length, $l$ is the original length of the body, $\alpha$ is the coefficient of thermal expansion of the body and $\left( {{\theta _1} - \theta } \right)$ is the temperature change corresponding to two temperatures $\theta$ , ${\theta _1}$.
-The period of a pendulum is given by, $T = 2\pi \sqrt {\dfrac{l}{g}}$ where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.

Complete step by step answer.
Step 1: List the parameters given in the question.
At ${\theta _1} = 40^\circ {\text{C}}$, the time lost by the pendulum clock is $\Delta {T_1} = 12{\text{s}}$.
At ${\theta _2} = 20^\circ {\text{C}}$, the time gained by the pendulum clock is $\Delta {T_2} = 4{\text{s}}$.
Let $\theta$ be the temperature at which the clock shows the correct time.
Step 2: Express the relation for the period and linear expansion of the pendulum for the two temperatures.
Let $l$ be the original length of the pendulum shaft and $\Delta {l_1}$ be the change in length when the temperature decreases from ${\theta _1} = 40^\circ {\text{C}}$ to $\theta$ .
Let $\Delta {l_2}$ be the change in length when the temperature decreases from $\theta$ to ${\theta _2} = 20^\circ {\text{C}}$ .
Then the linear expansion corresponding to the change in temperature from ${\theta _1} = 40^\circ {\text{C}}$ to $\theta$ is expressed as $\dfrac{{\Delta {l_1}}}{l} = \alpha \left( {40 - \theta } \right)$ -------- (1)
Similarly, the linear expansion corresponding to the change in temperature from $\theta$ to ${\theta _2} = 20^\circ {\text{C}}$ is expressed as $\dfrac{{\Delta {l_2}}}{l} = \alpha \left( {\theta - 20} \right)$ -------- (2)
Now the change in the period of the pendulum corresponding to the change in temperature from ${\theta _1} = 40^\circ {\text{C}}$ to $\theta$ is expressed as $\dfrac{{\Delta {T_1}}}{T} = \dfrac{1}{2}\left( {\dfrac{{\Delta {l_1}}}{l}} \right)$ -------- (3) where $T$ is the time of the pendulum for one day.
Substituting equation (1) in (3) we get, $\dfrac{{\Delta {T_1}}}{T} = \dfrac{1}{2}\alpha \left( {40 - \theta } \right)$ -------- (4)
Similarly, the change in the period of the pendulum corresponding to the change in temperature from $\theta$ to ${\theta _2} = 20^\circ {\text{C}}$ is expressed as $\dfrac{{\Delta {T_2}}}{T} = \dfrac{1}{2}\left( {\dfrac{{\Delta {l_2}}}{l}} \right)$ -------- (5)
Substituting equation (2) in (5) we get, $\dfrac{{\Delta {T_2}}}{T} = \dfrac{1}{2}\alpha \left( {\theta - 20} \right)$ -------- (6)
Step 3: Using equations (4) and (6) obtain the required temperature $\theta$ .
Diving equation (4) by (6) we get, $\dfrac{{\Delta {T_1}}}{{\Delta {T_2}}} = \dfrac{{\dfrac{1}{2}\alpha \left( {40 - \theta } \right)}}{{\dfrac{1}{2}\alpha \left( {\theta - 20} \right)}}$
$\Rightarrow \dfrac{{\Delta {T_1}}}{{\Delta {T_2}}} = \dfrac{{\left( {40 - \theta } \right)}}{{\left( {\theta - 20} \right)}}$ --------- (7)
Substituting for $\Delta {T_1} = 12{\text{s}}$ and $\Delta {T_2} = 4{\text{s}}$ in equation (7) we get, $\dfrac{{12}}{4} = \dfrac{{40 - \theta }}{{\theta - 20}}$
$\Rightarrow 12\theta - \left( {12 \times 20} \right) = \left( {4 \times 40} \right) - 4\theta$
On simplifying we get, $\theta = \dfrac{{160 + 240}}{8} = 25^\circ {\text{C}}$
Thus the temperature at which the clock shows the right time is obtained as $\theta = 25^\circ {\text{C}}$ .
Now in the given options, we see that option A gives the required temperature as $25^\circ {\text{C}}$ .
So the correct option is A and the coefficient of linear expansion of the shaft is $\alpha = 1 \cdot 85 \times {10^{ - 5}}^\circ {{\text{C}}^{ - 1}}$

Note: We can also find the coefficient of linear expansion by calculation instead of picking the odd one out in the given options.
We know that for a day the period of the pendulum will be $T = \left( {24 \times 3600} \right){\text{s}}$.
Then substituting the values $T = \left( {24 \times 3600} \right){\text{s}}$ , $\Delta {T_1} = 12{\text{s}}$ and $\theta = 25^\circ {\text{C}}$ in equation (4) we get, $\dfrac{{12}}{{\left( {24 \times 3600} \right)}} = \dfrac{1}{2}\alpha \left( {40 - 25} \right) = \dfrac{{15\alpha }}{2}$
$\Rightarrow \alpha = \dfrac{{2 \times 12}}{{24 \times 3600 \times 15}} = 1 \cdot 85 \times {10^{ - 5}}^\circ {{\text{C}}^{ - 1}}$
Thus we calculate the coefficient of linear expansion as $\alpha = 1 \cdot 85 \times {10^{ - 5}}^\circ {{\text{C}}^{ - 1}}$ .

So the correct option is A.