Q 11-10-012NEETJEE MainEAMCET 2007 (Engineering)Medium
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
$T = 2\pi\sqrt{\dfrac{L}{g}}$ stays the same if $\dfrac{L}{g}$ stays the same. $g$ has increased by a factor $1.001$, so $L$ must increase by the same factor:
$$\alpha\Delta T = 0.001 \;\Rightarrow\; \Delta T = \frac{0.001}{10^{-5}} = 100°\text{C}$$
Required temperature $= 20 + 100 = 120°$C.
Solution by Sreeraj P, M.Sc Physics