There is a clock which gives correct time at $20^o$C is subjected to $40^o$C. The coefficient of linear expansion of the pendulum is $12\times 10^{-6}$ per $^oC$, how much is gain or loss in time?

Given

$\alpha=12 \times 10^{-6} \stackrel{\circ}{C}^{-1}$

correct time at $T=20^{\circ} \mathrm{C}$

Let initial length of pendulum $=l_{0}$

length of pendulun at $T=40^{\circ} \mathrm{C}=l^{\prime}$

we know $l^{\prime}=l_{0}(1+\alpha \Delta \theta) \quad \Delta \theta$ is change in

Time period $\Rightarrow T_{\text {new }}=2 \pi \sqrt{\frac{l^{\prime}}{g}}$ temperature

substitute $l^{\prime}$ in expression of Thew

$T_{\text {new }}=2 \pi \sqrt{\frac{l_{0}(1+\alpha \Delta \theta)}{g}}$

$\quad T_{0}=2 \pi \sqrt{\frac{l_{0}}{g}}$

$T_{\text {new }}=2 \pi \sqrt{\frac{l_{0}}{g}} \cdot(\sqrt{1+\alpha \Delta \theta})=T_{0} \sqrt{1+\alpha \Delta \theta}$

using binomial approximation $\sqrt{1+\alpha \Delta \theta}=1+\frac{1}{2} \alpha \Delta \theta$

$\begin{array}{l} T_{\text {new }}=T_{0}\left(1+\frac{1}{2} \alpha \Delta \theta\right) \\ \Rightarrow \frac{T_{\text {new }}-T_{0}}{T_{0}}=\frac{1}{2} \alpha \Delta \theta \Rightarrow \frac{\Delta T}{T_{0}}=\frac{1}{2} \alpha \Delta \theta \end{array}$

change intime for onesecond $\frac{\Delta T}{T_{0}}=\frac{1}{2} \alpha \Delta \theta$

Change in time for $t$ second $=\frac{1}{2} \alpha \Delta \theta \times t$

$\begin{array}{l} t \text { in second }=\text { seconds in } 24 \mathrm{hr} . \\ =24 \times 60 \times 60=(24 \times 3600) \mathrm{sec} \\ \Delta \theta=40^{\circ} \mathrm{C}-20^{\circ} \mathrm{C}=20^{\circ} \end{array}$

so $\Delta t$ for one day $=\frac{1}{2} \times 12 \times 10^{-6} \times 20 \times 24 \times 3600$

$=10.37 \mathrm{~s}$

Time period increases by 10.3 second so there is loss in time of $10.37 \mathrm{sec}$ ond

hence option ' $A$ ' is correct.