Question

A pendulum clock loses $12s$ a day if the temperature is $40{}^\circ C$ and gains $4s$ a day if the temperature is $20{}^\circ C$. The temperature at which the clock will show correct time, and the coefficient of linear expansion $(\alpha )$ of the metal of the pendulum shaft are respectively
$\begin{aligned} & A)25{}^\circ C;\alpha =1.85\times {{10}^{-5}}/{}^\circ C \\\ & B)60{}^\circ C;\alpha =1.85\times {{10}^{-4}}/{}^\circ C \\\ & C)30{}^\circ C;\alpha =1.85\times {{10}^{-3}}/{}^\circ C \\\ & D)55{}^\circ C;\alpha =1.85\times {{10}^{-2}}/{}^\circ C \\\ \end{aligned}$

Answer 1

The metal of a pendulum shaft undergoes thermal expansion, when kept at different temperatures. This linear expansion further causes change in time recorded by the pendulum. The time lost or gained by a pendulum clock is proportional to the coefficient of linear expansion of the shaft and change in temperature from its actual temperature, at which the pendulum records the correct time.
Formula used:
${{T}_{L/G}}(s)=43200\alpha \Delta \theta (s)$

Complete answer:
The metal of a pendulum shaft undergoes thermal expansion, when kept at different temperatures. This linear expansion further causes change in time recorded by the pendulum. The time lost or gained by a pendulum clock, when kept at a different temperature from the temperature, at which the pendulum records correct time, is given by
${{T}_{L/G}}(s)=43200\alpha \Delta \theta (s)$
where
${{T}_{L/G}}$ is the time lost or gained when the pendulum is kept at a different temperature
$\alpha$ is the coefficient of linear expansion of pendulum shaft
$\Delta \theta$ is the change in temperature from the temperature at which the pendulum records correct time
Let this be equation 1.
Coming to our question, we are provided with two cases. The first case involves a pendulum clock, which is said to lose $12s$a day, if the temperature is $40{}^\circ C$. The second case involves the same pendulum, which is said to gain $4s$ a day, if the temperature is $20{}^\circ C$. We are required to determine the temperature at which the clock will show correct time as well as the coefficient of linear expansion $(\alpha )$ of the metal of the pendulum shaft.
Using equation 1, time lost in the first case is given by
${{T}_{L}}(s)=43200\alpha \Delta \theta (s)\Rightarrow 12s=43200\alpha \times (40{}^\circ C-\theta {}^\circ C)s$
where,
${{T}_{L}}=12s$ is the time lost when the pendulum is kept at a temperature $40{}^\circ C$
$\theta {}^\circ C$ is the temperature at which the pendulum shows correct time
$\Delta \theta =40{}^\circ C-\theta {}^\circ C$ is the change in temperature, when the first case is considered
$\alpha$ is the coefficient of linear expansion of pendulum shaft
Let this be equation 2.
Similarly, using equation 1, time gained in the second case is given by
${{T}_{G}}(s)=43200\alpha \Delta \theta (s)\Rightarrow 4s=43200\alpha \times (\theta {}^\circ C-20{}^\circ C)s$
where
${{T}_{G}}=4s$is the time gained when the pendulum is kept at a temperature $20{}^\circ C$
$\theta {}^\circ C$ is the temperature at which the pendulum shows correct time
$\Delta \theta =\theta {}^\circ C-20{}^\circ C$ is the change in temperature when the second case is considered
$\alpha$ is the coefficient of linear expansion of pendulum shaft
Let this be equation 3.
Dividing equation 2 by equation 3 and solving for $\theta$, we have
$\dfrac{{{T}_{L}}(s)}{{{T}_{G}}(s)}=\dfrac{12}{4}=\dfrac{43200\alpha \times (40{}^\circ C-\theta {}^\circ C)}{43200\alpha \times (\theta {}^\circ C-20{}^\circ C)}\Rightarrow 3=\dfrac{40-\theta }{\theta -20}\Rightarrow 3\theta -60=40-\theta \Rightarrow \theta =25{}^\circ C$
Let this be equation 4.
Substituting equation 4 in equation 2, we have
$12=43200\alpha \times (40-\theta )\Rightarrow 12=43200\alpha \times (40-25)\Rightarrow 12=43200\alpha \times (15)\Rightarrow \alpha =1.85\times {{10}^{-5}}/{}^\circ C$
Let this be equation 5.
Therefore, from equation 4 and 5, we come to the conclusion that the temperature at which the clock will show correct time is $25{}^\circ C$ and the coefficient of linear expansion $(\alpha )$ of the metal of the pendulum shaft is $1.85\times {{10}^{-5}}/{}^\circ C$.

So, the correct answer is “Option A”.

Note:
When the pendulum clock is kept at a higher temperature than the temperature at which the pendulum shows correct time, the pendulum loses time. At the same time, when the pendulum clock is kept at a lower temperature than the temperature at which the pendulum shows correct time, the pendulum gains time. This explanation can be used to support equations 2 and 3 in the above solution. It is also to be noted that equation 1 is actually represented as:
$\begin{aligned} & {{T}_{L/G}}(day)=\dfrac{1}{2}\alpha \Delta \theta (day) \\\ & {{T}_{L/G}}(s)=\dfrac{1}{2}\alpha \Delta \theta \times 86400(s)=43200\alpha \Delta \theta (s) \\\ \end{aligned}$