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JEE Main 2020 (8 Jan, Shift 2) · Simple Pendulum

Simple Pendulum question from JEE Main 2020 (8 Jan, Shift 2)

Q70JEE Main 2020 · 8 JAN · SHIFT 2Medium

Simple Pendulum

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 cm and a stop watch with 1s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is:

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Answer: (a)

$g = \frac{4\pi^2 L}{T^2} \Rightarrow \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta t}{t}$ .
$\frac{\Delta L}{L} = \frac{0.1}{25.0} = 0.4\%$ (least count 1 mm).
$2\frac{\Delta t}{t} = 2 \times \frac{1}{50} = 4\%$ .
So the answer is (a) $4.40\%$ .

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