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JEE Main 2020 Physics question

PhysicsOscillationsJEE Main 2020

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 1 s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is :

Answer: (D)

Solution

g = 4π²L/T², so Δg/g = ΔL/L + 2ΔT/T = 0.1/25 + 2(1/50) = 0.044, i.e. 4.40 %,

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Question ID VK-011920