A pendulum clock of time period $2\text{sec}$ gives the correct time at ${30}^{\circ }\text{C}$. The pendulum is made of iron. How many seconds will it lose or gain per day when the temperature falls to $0^{\circ }\text{C}$ ? ${\alpha }_{Fe}=1.2\times {10}^{-5}{(}^{\circ }\text{C}{)}^{-1}$

The correct option is A $15.5\text{sec}$
Time period of a pendulum is given by,
$T=2\pi \sqrt{\frac{L}{g}}$
Here, as the length changes, time period of the pendulum also changes, given by,
$\frac{\Delta T}{T}=\frac{1}{2}\left(\frac{\Delta l}{l}\right).....\left(i\right)$

Now, the length of the simple pendulum changes due to change in temperature, according to the relation,
$\frac{\Delta l}{l}=\alpha \left(\Delta \theta \right).....\left(ii\right)$

From $(i$) and $\left(ii\right)$, we can write,
$\frac{\Delta T}{T}=\frac{1}{2}\alpha \left(\Delta \theta \right)$
As the temperature is decreasing, the pendulum will move faster, and the clock will gain time.

Time gained per unit time period $=\frac{\Delta T}{T}$
$\therefore$ Time gained in 1 day $=\frac{\Delta T}{T}\times 24\times 3600$
$=\frac{1}{2}\alpha \left(\Delta \theta \right)\times 24\times 3600$
$=\frac{1}{2}\times 1.2\times {10}^{-5}\times 30\times 24\times 3600$
$=15.55\text{sec}$

Hence, $\left(A\right)$ is the correct answer.

Why this Question :- Time period of a pendulum depends upon the length as, $T\propto \sqrt{L}$ As the temp increases, the length also increases; as a result time period of the pendulum increases, So the clock loses time. Similarly, if the temp decreases, time period of the pendulum decreases so that it moves faster and the clock gains time.