JEE Main 2022 · Physics · Fluid Mechanics, Elasticity and Surface Tension · medium

The pendulum of a clock has a brass rod of linear expansion coefficient 2×10^-5 ^circC^-1 and keeps correct time at 20^circC. The room temperature…

Numerical answer

The pendulum of a clock has a brass rod of linear expansion coefficient

2\times10^{-5}\ {}^\circ\mathrm{C}^{-1}

and keeps correct time at

20^\circ\mathrm{C}

. The room temperature rises to

50^\circ\mathrm{C}

. The clock loses time. The time lost per day, in seconds (to the nearest integer), is

Show answer and solution

Answer: 26

The period

T=2\pi\sqrt{L/g}

\dfrac{\Delta T}{T}=\dfrac12\dfrac{\Delta L}{L}=\dfrac12\alpha\Delta\theta=\dfrac12(2\times10^{-5})(30)=3\times10^{-4}

. The period is longer, so the clock runs slow. In one day of

86400\ \mathrm{s}

it loses

86400\times3\times10^{-4}=25.92\ \mathrm{s}\approx26\ \mathrm{s}

. The factor

\tfrac12

is essential; omitting it gives

52\ \mathrm{s}

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