Question

A pendulum clock loses 10.8 seconds a day when the temperature is \( 38^\circ C \) and gains 10.8 seconds a day when the temperature is \( 18^\circ C \). The coefficient of linear expansion of the metal of the pendulum clock is:

• (1) 7 × 10^-5 °C^-1
• (2) 1.25 × 10^-5 °C^-1
• (3) 5 × 10^-5 °C^-1
• (4) 2.5 × 10^-5 °C^-1

Hint:

The time lost or gained by a pendulum clock due to temperature changes is proportional to the coefficient of linear expansion \( \alpha \). Use the relation: \[ \frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta T \] to find the coefficient.

Answer 1

The Correct Option is D

Step 1: Relation Between Time Period and Length of a Pendulum The time period of a pendulum is given by: \[ T = 2\pi \sqrt{\frac{L}{g}} \] where: - \( L \) is the length of the pendulum, - \( g \) is the acceleration due to gravity. Since the pendulum length increases with temperature, we consider the fractional change in length: \[ \frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta T \] where: - \( \alpha \) is the coefficient of linear expansion, - \( \Delta T \) is the temperature change. \vspace{0.5cm} 

Step 2: Calculate \( \alpha \) Given that the pendulum loses 10.8 seconds per day at \( 38^\circ C \) and gains 10.8 seconds per day at \( 18^\circ C \), the total time change per day is: \[ \Delta T = 38^\circ C - 18^\circ C = 20^\circ C \] Since time lost or gained per day is given by: \[ \frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta T \] Using: \[ \frac{10.8}{86400} = \frac{1}{2} \alpha \times 20 \] Solving for \( \alpha \): \[ \alpha = \frac{2 \times 10.8}{86400 \times 20} \]