Question

A pendulum clock loses 12sec a day if the temperature is ${40^\circ }C$ and gains $4\,s$ 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.

Answer 1

Hint: We can use the equation to calculate the time gain or lowest with respect to the change in temperature to solve this problem. By writing the equation for time loss at ${40^\circ }C$ the equation for time gain at ${20^\circ }C$ and on comparing both these two equations, we will get the value of temperature at which clock will show the correct time.
Using this value of temperature, we can find the value of coefficient of linear expansion.

Complete step by step answer:
It is given that a pendulum clock loses $12\,s$ a day If temperature is ${40^\circ }C$ .
Let this temperature be denoted as ${\theta _1}$ .
$ \Rightarrow {\theta _1} = {40^\circ }C$
The gain in time when the temperature reaches ${20^\circ }C$ is $4\,s$ .
Let this temperature be ${\theta _2}$ .
$ \Rightarrow {\theta _2} = {20^\circ }C$
We need to find the temperature at which the clock will show the correct time and the linear coefficient of expansion of the metal of the pendulum.
First let us calculate the temperature at which the clock will show the correct time.
Time loss or gain per day is calculated using the equation.
 $T = \dfrac{1}{2}\alpha \Delta \theta \times t$
Where $\alpha $ is the coefficient of linear expansion, $\Delta \theta $ is the change in temperature and t is the time.
$ \Rightarrow T = \dfrac{1}{2}\alpha \Delta \theta \times 86400s$
Since, $t = 24 \times 60 \times 60\,s = 86400\,s$
On substituting the values, the equation becomes
$12 = \dfrac{1}{2}\alpha \left( {40 - \theta } \right) \times 86400$ …………….(1)
Here $\theta $ is the temperature at which the clock will perform correctly.
If we substitute the values in case of time gain, we get
$4 = \dfrac{1}{2}\alpha \left( {\theta - 20} \right) \times 86400$ …………..(2)
Now, let us divide equation 1 by 2
$ \Rightarrow 3 = \dfrac{{40 - \theta }}{{\theta - 20}}$
$ \Rightarrow 3\theta - 60 = 40 - \theta $
$ \Rightarrow \theta = {25^0}C$
This is the temperature at which the clock shows the correct time.
In order to find the value of coefficient of linear expansion, let us substitute the values of $\theta $ in equation 1.
$ \Rightarrow 12 = \dfrac{1}{2} \times \alpha \times \left( {40 - 25} \right) \times 86400$
$ \Rightarrow \alpha = \dfrac{{24}}{{15 \times 86400}}$
$ \Rightarrow \alpha = 1 \cdot 85 \times {10^{ - 5}}{/^0}C$
This is the coefficient of linear expansion.

$\therefore $ The temperature at which the clock shows the correct time is $25^\circ$. The coefficient of linear expansion is $\alpha = 1 \cdot 85 \times {10^{ - 5}}{/^0}C$.

Note:
We know that metals expand on heating. So, at high temperature the time period of oscillation of the pendulum will be more since length is more. Which means time will run slowly and that is why there is time loss at high temperature. Whereas, when the temperature is lowered the length of the pendulum is decreased and it oscillates faster. Hence time will run faster and thus there will be gain in time.