Question

A simple pendulum made a bob of mass $m$ and a metallic wire of negligible mass has a time period $2\;s$ at $T=0$. If the temperature of the wire is increased and the corresponding change in its time period is plotted against its temperature, the resulting graph is a line of slope $S$. If the coefficient of linear expression of metal is $\alpha$ then the value of $S$ is
\[\begin{align}
  & A.2\alpha \\
 & B.\frac{1}{\alpha } \\
 & C.\alpha \\
 & D.\frac{\alpha }{2} \\
\end{align}\]

Answer 1

Hint: We know that the time period of an oscillation is dependent on the length of the wire. Thus we can see that the time period gets affected, by the time taken by the oscillation changes, due to the coefficient of linear expression of metal.
Formula used:
Time period of a simple pendulum $T=2\pi \sqrt{\dfrac{L}{g}}$ and $L\prime=L(1+\alpha(T-T_{0}))$

Complete answer:
We know that the time period of a second’s pendulum is $t=$2 seconds, the time period $t=2\pi
\sqrt{\dfrac{L}{g}}$,where $L$ is length of the simple pendulum and $g$ is gravity.
We know that expansion means a change in volume or length, more specifically increase in volume or length. This is majorly due to increase in pressure. Thus, we can say that linear expansion is the change in volume due to change in length majorly due some temperature, at constant pressure.
Thus, the coefficient of linear expansion $\alpha_{L}$, is the rate at which length $d\;L$ changes with respect to temperature $d\;T$. Mathematically, it is given as $\alpha_{L}=\dfrac{dL}{dT}$
Clearly, the length of the pendulum due to the change in temperature, from $T_{0}$ to $T$ is given as $L\prime=L(1+\alpha(T-T_{0}))$
Then the new time period is given as $t\prime=2\pi
\sqrt{\dfrac{L\prime}{g}}$
$\implies t\prime=2\pi\sqrt{\dfrac{L(1+\alpha(T-T_{0}))}{g}}$
$\implies t\prime =t(1+\alpha(T-T_{0}))^{\dfrac{1}{2}}$
$\implies t\prime =2(1+\alpha(T-T_{0}))^{\dfrac{1}{2}}$
Let us assume that there is a small change in length,
$\implies t\prime \approx 2(1+\alpha(T-T_{0}))$
Differentiating with respect to time $T$
$\implies \dfrac{dt}{dT}=\alpha$

So, the correct answer is “Option C”.

Note:
This is an easy sum, provided one knows the basic formulas used. Also, it is important to assume that the change in length is small to find the new time period. If it is taken with the power, when differentiating $t\prime$ with respect to time $T$, we will get a big equation when solved, and will get the same answer. However, not that the process is tedious and time consuming.