JEE Main 2016 · 3 Apr · Shift 1 · Physics Q7

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PhysicsSingle correct+4 / -1Question 7 of 90

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

25°C;

\alpha=1.85\times10^{-5}

/°C

60°C;

\alpha=1.85\times10^{-4}

/°C

30°C;

\alpha=1.85\times10^{-3}

/°C

55°C;

\alpha=1.85\times10^{-2}

/°C

Answer and worked solution

Official answer: A

$T\propto\sqrt l$ , so the fractional loss per day is $\frac12\alpha\Delta\theta$ . With the correct time at θ:
$\frac{12}{86400}=\frac12\alpha(40-\theta),\qquad\frac4{86400}=\frac12\alpha(\theta-20).$
Dividing, $40-\theta=3(\theta-20)$ , so θ = 25°C. Then $\alpha=\frac{8}{5\times86400}\approx1.85\times10^{-5}$ /°C.

Topics:Temperature and thermal expansion Simple and compound pendulum

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JEE Main 2016 · 3 Apr · Shift 1, Physics question 7