Question

A clock with a metallic pendulum at ${15^ \circ }C$ runs faster by a $5\,s$ each day and at ${30^\circ }C$, runs slow by a $10\,s$. Find the coefficient of linear expansion of the metal. (nearly in ${10^{ - 6}}/^\circ C$)
$
  \left( A \right)\,\,2 \\
  \left( B \right)\,\,4 \\
  \left( C \right)\,\,6 \\
  \left( D \right)\,\,8 \\
 $

Answer 1

Hint: In the question, we know that the time and temperature. By substituting the values in the equation of time period we get the value for the coefficient of linear expansion of the metal.

Formulae Used:
The expression for finding the time period is:
$T = \,2\pi \sqrt {\dfrac{l}{g}} $
Where
$l$ be the length of the pendulum, $g$ be the acceleration due to gravity and $t$ be the time period.

 Complete step-by-step solution:
We know that, $t = 5\,s\,and\,10\,s$
$T = \,2\pi \sqrt {\dfrac{l}{g}} $
Squaring on both sides:
\[{T^2} = \,{\left( {4\pi } \right)^2}\left( {\dfrac{l}{g}} \right)\]
Differentiating the above equation we get,
\[2TdT = \dfrac{{4{\pi ^2}}}{g}dl = \left( {\dfrac{{{T^2}}}{l}} \right)dl\]
$\dfrac{{dT}}{T} = \dfrac{1}{2}\dfrac{{dl}}{l} = \dfrac{1}{2}\dfrac{{l\alpha \Delta t}}{l}$
Where,
$\Delta t$ be the change in temperature.
Now we have to find the value of gain or loss per day, we get
Loss or gain per day $ = \,\dfrac{1}{2}\alpha \Delta t\, \times \,T\,$
At $15^\circ C$, we find the gain percentage we get,
Here, $t = 5\,s$
\[\;5 = \dfrac{\alpha }{2}\left( {t - 15} \right) \times 86400.......\left( 1 \right)\]
At ${30^\circ }C$ , we find the loss percentage we get,
Here, $t = 10\,s$
\[10 = \dfrac{\alpha }{2}\left( {t - 30} \right) \times 86400.......\left( 2 \right)\]
Dividing the equation $\dfrac{2}{1}$ we get,
$\dfrac{2}{1} = \,\dfrac{{10 = \dfrac{\alpha }{2}\left( {t - 30} \right) \times 86400}}{{\;5 = \dfrac{\alpha }{2}\left( {t - 15} \right) \times 86400}}$
Here $t$ is the value at which the clock shows correct time.
Simplify the above equation we get,
\[2 = \dfrac{{30 - t}}{{t - 15}}\]
Solving the above equation we get the value of $t$,
\[t = \,{20^\circ }C\,\]
Therefore, we get the $t$ value as ${20^\circ }C$
Substitute the value of $t$ in the equation $\left( 1 \right)$
\[5 = \dfrac{\alpha }{2}\left( {20 - 15} \right) \times 86400\]
$\alpha = 2.31\, \times \,{10^{ - 5}}/^\circ C$
Therefore, the coefficient of linear expansion of the metal is $2$.
Hence from the above option, option $\left( A \right)$ is correct.

Note:- In the question, we solved using the equation of time period. From that the formula we have to perform the squaring and differentiating operations on the equation. We have to find the value of temperature, by using the before the time and after the time so we get the value of coefficient of linear expansion. Using this formula, we can also calculate the value of length of the pendulum.