MHT CET · Physics · Thermal Properties of Matter

A clock with iron pendulum keeps correct time at \(15^{\circ} \mathrm{C}\). If the room temperature is \(20^{\circ} \mathrm{C}\), the error in second per day will be nearly (coefficient of linear expansion of iron is \(\alpha=1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}\) )

A \(3.1 \mathrm{~s}\)
B \(1.3 \mathrm{~s}\)
C \(6.2 \mathrm{~s}\)
D \(2.6 \mathrm{~s}\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(2.6 \mathrm{~s}\)

Step-by-step Solution

Detailed explanation

Time period of the pendulum is, \(T=2 \pi \sqrt{\frac{l}{g}}\)
\(\therefore \Delta T=\frac{2 \pi}{\sqrt{g}} \times \frac{\Delta l}{2 \sqrt{l}}\)
We know, \(\frac{\Delta l}{l}=\alpha \Delta \theta\) the change in length due to thermal expansion.
\(\therefore \frac{\Delta T}{T}=\frac{1}{2} \frac{\Delta l}{l}=\frac{1}{2} \alpha \Delta \theta\)
Given, \(\quad \alpha=1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}, \quad g=9.8 \mathrm{~m} / \mathrm{s}, \quad T=86,400 \mathrm{~s} \quad\) and \(\Delta T=\alpha \Delta \theta T \approx 2.6 \mathrm{sec}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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