JEE PYQ: Units & Measurements - Question ID 2bbc49f6c62e (JEE Main 2021)

The question involves a simple pendulum, whose time period $T$ is given by the fundamental relation: $T = 2\pi \sqrt{\frac{L}{g}}$ where

• $L$ = length of the pendulum,
• $g$ = acceleration due to gravity.

A pendulum clock keeps time by counting the number of oscillations. If the length $L$ changes slightly, the time period $T$ changes, and consequently the clock either gains or loses time over a day.

We are asked to find the error in time per day when $L$ increases by 0.1 %. This means we need to relate the fractional change in $L$ to the fractional change in $T$, then scale that fractional change to the total number of seconds in a day.

Step 1: Relate fractional changes in $L$ and $T$

From $T = 2\pi \sqrt{\frac{L}{g}}$, take the natural logarithm of both sides: $\ln T = \ln(2\pi) + \frac{1}{2}\ln L - \frac{1}{2}\ln g$ Differentiate both sides with respect to $L$ (treating $g$ as constant): $\frac{dT}{T} = \frac{1}{2}\frac{dL}{L}$ This shows that the fractional change in period $\frac{\Delta T}{T}$ is half the fractional change in length $\frac{\Delta L}{L}$.

Step 2: Plug in the given fractional change in $L$

Given $\frac{\Delta L}{L} = +0.1\% = +0.001$, we get: $\frac{\Delta T}{T} = \frac{1}{2} \times 0.001 = 0.0005$ This means the period increases by 0.05 %.

Step 3: Relate period change to time error over one day

A pendulum clock measures time by counting $N$ oscillations. In one day (86400 seconds), the number of oscillations is: $N = \frac{86400}{T}$ If the period changes to $T + \Delta T$, the clock will count the same number $N$ but the actual time elapsed will be: $N \times (T + \Delta T) = 86400 + N \Delta T$ The error in time is therefore: $\Delta t = N \Delta T = \frac{86400}{T} \times \Delta T = 86400 \times \frac{\Delta T}{T}$ Substitute $\frac{\Delta T}{T} = 0.0005$: $\Delta t = 86400 \times 0.0005 = 43.2 \text{ seconds}$ Since the period increased, the clock runs slower and loses 43.2 seconds per day.

1. Sign confusion: Students often forget whether the clock gains or loses time. A longer pendulum means a longer period, so the clock runs slower and loses time.
2. Fractional change misapplication: Some mistakenly use $\frac{\Delta T}{T} = \frac{\Delta L}{L}$ instead of half. Remember the square-root dependence halves the fractional change.
3. Unit conversion: Ensure the percentage change (0.1 %) is converted to decimal form (0.001) before multiplying.

Final Answer: The error in time per day is 43.2 s, which corresponds to option C.