Question

A clock pendulum made of invar has a period of \(0.5\,s\) at \(20^\circ C\). If the clock is used in a place where temperature averages to \(30^\circ C\), how much time does the clock lose in each oscillation? (For invar, \(\alpha = 9 \times 10^{-7}/^\circ C\), \(g =\) constant)

• A. \(2.25 \times 10^{-6}\,s\)
• B. \(2.5 \times 10^{-6}\,s\)
• C. \(5 \times 10^{-7}\,s\)
• D. \(1.125 \times 10^{-6}\,s\)

Hint:

For pendulum, \(\frac{\Delta T}{T} = \frac{1}{2}\alpha \Delta \theta\). Increase in temperature increases period, so clock loses time.

Answer 1

The Correct Option is A

Step 1: Use relation of time period with length.
For pendulum:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
So:
\[ T \propto \sqrt{L} \]
Step 2: Small change approximation.
\[ \frac{\Delta T}{T} = \frac{1}{2}\frac{\Delta L}{L} \]
Step 3: Thermal expansion of rod.
\[ \frac{\Delta L}{L} = \alpha \Delta \theta \]
Here:
\[ \Delta \theta = 30 - 20 = 10^\circ C \]
So:
\[ \frac{\Delta L}{L} = 9 \times 10^{-7} \times 10 = 9 \times 10^{-6} \]
Step 4: Find change in time period.
\[ \frac{\Delta T}{T} = \frac{1}{2}(9\times 10^{-6}) = 4.5\times 10^{-6} \]
Given \(T = 0.5s\):
\[ \Delta T = 0.5 \times 4.5\times 10^{-6} = 2.25 \times 10^{-6}s \]
Final Answer: \[ \boxed{2.25 \times 10^{-6}\,s} \]