Question 66

The time period is given by $T = \frac{t}{n}$, where $t$ is the total time and $n$ is the number of oscillations. The error in time period is

$\Delta T = \frac{\Delta t}{n}$.

Given least count $\Delta t = 1 \text{ s}$ and $n = 20$, we get

$\Delta T = \frac{1}{20} = 0.05 \text{ s}$.

Thus, option A is correct. The formula for $g$ is $g = 4\pi^2 \frac{L}{T^2}$.

The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}$.

Since the length $L$ is given as exactly $1 \text{ m}$, we will take $\Delta L = 0$.

The time period $T = \frac{40}{20} = 2 \text{ s}$.

Therefore, $\frac{\Delta g}{g} = 2 \times \frac{0.05}{2} = 0.05$.

In percentage, this is $0.05 \times 100 = 5\%$. Thus, option C is correct.