Question

A clock with an iron pendulum keeps the correct time at \(14^{\circ }\text{C}\). If the room temperature is \(20^{\circ }\text{C}\), the error in seconds per day will be nearly (coefficient of linear expansion of iron is \(1.2\times 10^{-5}/{}^{\circ }\text{C}\) )

• A. \(2.62\text{ s}\)
• B. \(3.11\text{ s}\)
• C. \(4.34\text{ s}\)
• D. \(6.25\text{ s}\)

Hint:

A longer pendulum has a longer period, so the clock loses time; use the small change relation for the period.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For a pendulum \(T=2\pi\sqrt{L/g}\), so \(\dfrac{\Delta T}{T}=\dfrac{1}{2}\dfrac{\Delta L}{L}=\dfrac{1}{2}\alpha\Delta\theta\).

Step 2: Direction of error
At the higher temperature the rod gets longer, the period increases, and the clock runs slow.

Step 3: Calculation
Time lost per day \(=\dfrac{1}{2}\alpha\Delta\theta\times86400\)
\(=\dfrac{1}{2}(1.2\times10^{-5})(6)(86400)\)

Step 4: Result
\(=3.6\times10^{-5}\times86400\approx3.11\text{ s}\)

Final Answer:
The clock is in error by about \(3.11\text{ s}\) per day. \[ \boxed{\text{(B) }3.11\text{ s}} \]