Q92 / 120 · VITEEE 2009 · Physics
A clock pendulum made of invar has a period of $0.5 \mathrm{~s}$, at $20^{\circ} \mathrm{C}$. If the clock is used in a climate where the temperature averages to $30^{\circ} \mathrm{C}$, how much time does the clock lose in each oscillation? (For invar, $\alpha=9 \times 10^{-7} /{ }^{\circ} \mathrm{C}, \mathrm{g}=$ constan $\left.\mathrm{t}\right)$
- Correct answerA$2.25 \times 10^{-6} \mathrm{~s}$
- B$2.5 \times 10^{-7} \mathrm{~s}$
- C$5 \times 10^{-7} \mathrm{~s}$
- D$1.125 \times 10^{-6} \mathrm{~s}$
Solution
Time period of oscillation,
$T=2 \pi \sqrt{\frac{l}{g}} \Rightarrow \frac{d T}{T}=\frac{1}{2} \frac{d l}{l}$
$\begin{aligned} \text { As, } \frac{d l}{l} &=\alpha d t \\ \Rightarrow \frac{d T}{T} &=\frac{1}{2} \alpha d t=\frac{1}{2} \times 9 \times 10^{-7} \times(30-20) \\ &=4.5 \times 10^{-6} \end{aligned}$
$\begin{aligned} \therefore \text { Loss in time } &=4.5 \times 10^{-6} \times 0.5 \\ &=2.25 \times 10^{-6} \mathrm{~s} \end{aligned}$
Hence, the correct answer is option: 'A'.