Question

A steel pendulum clock manufactured at 32$^\circ$C and working at 47$^\circ$C is nearly (Coefficient of linear expansion of steel = $12 \times 10^{-6}$ /$^\circ$C)

• A. 7.8 s slow per day
• B. 7.8 s fast per day
• C. 15.6 s slow per day
• D. 15.6 s fast per day

Hint:

For a pendulum clock, the time lost or gained per day due to a temperature change $\Delta\theta$ can be calculated directly using the formula $\Delta t = \frac{1}{2}\alpha (\Delta\theta) \times 86400$. If the temperature increases, the clock runs slow (loses time). If the temperature decreases, it runs fast (gains time).

Answer 1

The Correct Option is A

Step 1: State the formula for the period of a simple pendulum.
The period $T$ of a simple pendulum is given by $T = 2\pi\sqrt{\frac{L}{g}}$, where $L$ is its length.

Step 2: Find the fractional change in the period due to thermal expansion.
From the formula, $T \propto \sqrt{L}$. The fractional change in the period is given by: \[ \frac{\Delta T}{T} = \frac{1}{2}\frac{\Delta L}{L}. \] The fractional change in length due to a temperature change $\Delta\theta$ is given by $\frac{\Delta L}{L} = \alpha \Delta\theta$, where $\alpha$ is the coefficient of linear expansion. Combining these, we get: \[ \frac{\Delta T}{T} = \frac{1}{2}\alpha \Delta\theta. \]

Step 3: Calculate the fractional change in the period.
Given values: $\alpha = 12 \times 10^{-6}$ /$^\circ$C. The change in temperature is $\Delta\theta = 47^\circ\text{C} - 32^\circ\text{C} = 15^\circ\text{C}$. \[ \frac{\Delta T}{T} = \frac{1}{2} (12 \times 10^{-6}) (15) = 6 \times 15 \times 10^{-6} = 90 \times 10^{-6}. \]

Step 4: Calculate the total time lost or gained in one day.
The total time change in a day is the fractional change multiplied by the total number of seconds in a day. Number of seconds in a day = $24 \text{ hours} \times 60 \text{ min/hr} \times 60 \text{ s/min} = 86400 \text{ s}$. \[ \Delta t_{day} = \left(\frac{\Delta T}{T}\right) \times 86400 = (90 \times 10^{-6}) \times 86400. \] \[ \Delta t_{day} = 9 \times 10^{-5} \times 8.64 \times 10^4 = 9 \times 8.64 \times 10^{-1} = 77.76 \times 10^{-1} = 7.776 \text{ s}. \] This is approximately 7.8 s. Since the temperature increased, the length $L$ increased, and the period $T$ also increased. A longer period means the clock runs slower, so it loses time. The clock is 7.8 s slow per day.