Q63 / 75 · JEE Main 2020 (08 Jan Shift 2) · Physics

A simple pendulum is being used to determine the value of gravitational acceleration $g$ at a certain place. The length of the pendulum is $25.0 c m$ and a stopwatch with $1 s$ resolution measures the time taken for $40$ oscillations to be $50 s$ . The accuracy in $g$ is:

Solution

Time period of the simple pendulum, $T = 2 π \sqrt{\frac{l}{g}} \Rightarrow g = \frac{4 π^{2} l}{T^{2}}$

$\frac{∆ g}{g} = \frac{2 ∆ T}{T} + \frac{∆ L}{L} \Rightarrow \frac{∆ g}{g} = 2 (\frac{1}{50}) + \frac{0.1}{25.0}$

$\Rightarrow \frac{∆ g}{g} = 4.4 %$

Hence, the correct answer is option: 'C'.