A pendulum clock loses 12 s a day if the temperature is 40°C and… | JEE Main 2016

Q68JEE Main 2016Difficult

Thermometry & Thermal Expansion

A pendulum clock loses $12\text{ s}$ a day if the temperature is $40^{\circ}\text{C}$ and gains $4\text{ s}$ a day if the temperature is $20^{\circ}\text{C}.$ 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:

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Answer: (c)

Time lost or gained per day: $\Delta t = \dfrac{1}{2}\alpha(\theta - \theta_0) \times 86400$ .
$12 = \tfrac{1}{2}\alpha(40-\theta_0)(86400)$ and $4 = \tfrac{1}{2}\alpha(\theta_0-20)(86400)$
Dividing: $3(\theta_0 - 20) = 40 - \theta_0 \Rightarrow \theta_0 = 25^\circ\text{C}$
Adding: $16 = \tfrac{1}{2}\alpha(20)(86400) \Rightarrow \alpha \approx 1.85\times10^{-5}/^\circ\text{C}$
So the answer is (c).

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