A pendulum clock gives correct time at $20°$C at a place where $g = 10\ \text{m/s}^2$. The pendulum consists of a light steel rod connected to a heavy ball. If it is taken to a different place where $g = 10.01\ \text{m/s}^2$, at what temperature will the pendulum give correct time? ($\alpha$ of steel $= 10^{-5}\ °\text{C}^{-1}$)
Answer: (D) $120°$C
Concept. $T = 2\pi\sqrt{\dfrac{L}{g}}$ stays correct if $\dfrac{L}{g}$ is unchanged, so $L$ must grow by the same fraction as $g$.
Step 1. Fractional change in $g$. $\dfrac{\Delta g}{g} = \dfrac{0.01}{10} = 0.001$.
Step 2. Temperature rise needed.
$$\alpha\Delta T = 0.001 \;\Rightarrow\; \Delta T = \frac{0.001}{10^{-5}} = 100°\text{C}$$
Step 3. Required temperature. $20 + 100 = 120°$C.
Answer. $120°$C (Option D).
Solution by Sreeraj P, M.Sc Physics