A pendulum clock loses 12 s a… - JEE Main 2016 Physics PYQ

JEE Main 2016 — Physics Thermodynamics

2016mcqmedium

A pendulum clock loses 12 s a day if the temperature is ${40}^{o}$ C and gains 4s a day if the temperature is ${20}^{o}$ C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion $(\alpha )$ of the metal of the pendulum shaft are respectively:

Official previous-year question

Held on 3 Apr 2016 · Verified 6 Jul 2026.

Options

A
${30}^{o}C;\alpha =1.85\times {10}^{-3}$ / ${}^{o}C$
B
${55}^{o}C;\alpha =1.85\times {10}^{-2}$ / ${}^{o}C$
C
${25}^{o}C;\alpha =1.85\times {10}^{-5}$ / ${}^{o}C$
D
${60}^{o}C;\alpha =1.85\times {10}^{-4}$ / ${}^{o}C$

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Solution

$\frac{\Delta T}{T}=\frac{1}{2}\frac{\Delta l}{l}=\frac{1}{2}\alpha \Delta \theta$

$\frac{12}{4}=\frac{40-\theta }{\theta -20}$

$3\theta -60=40-\theta$

$4\theta =100$

$\theta ={25}^{o}C$

$Also, \Delta T=\frac{1}{2}\alpha \Delta \theta \times T$

$4=\frac{1}{2}\alpha 5\times 86400$

$\frac{8}{5\times 86400}=\alpha$

$\alpha =1.85\times {10}^{-5}$ ${/}^{o}C$

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Options

${30}^{o}C;\alpha =1.85\times {10}^{-3}$ / ${}^{o}C$

${55}^{o}C;\alpha =1.85\times {10}^{-2}$ / ${}^{o}C$

${25}^{o}C;\alpha =1.85\times {10}^{-5}$ / ${}^{o}C$

${60}^{o}C;\alpha =1.85\times {10}^{-4}$ / ${}^{o}C$

Solution

$\frac{\Delta T}{T}=\frac{1}{2}\frac{\Delta l}{l}=\frac{1}{2}\alpha \Delta \theta$

$\frac{12}{4}=\frac{40-\theta }{\theta -20}$

$3\theta -60=40-\theta$

$4\theta =100$

$\theta ={25}^{o}C$

$Also, \Delta T=\frac{1}{2}\alpha \Delta \theta \times T$

$4=\frac{1}{2}\alpha 5\times 86400$

$\frac{8}{5\times 86400}=\alpha$

$\alpha =1.85\times {10}^{-5}$ ${/}^{o}C$

Options

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Solution

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Did you get this right?

Solution

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JEE Main Thermodynamics in other years

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Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

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