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The period of oscillation of a simple pendulum is T=2π√L/g. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. The accuracy in the determination of g is

Asked in JEE Main 2015 · Error in g from a pendulum experiment

Answer: (4) 3%

Step-by-step solution

Given: L=20.0 cm with Δ L=1 mm; 100 oscillations in 90 s with a 1 s watch.

Length: (Δ L)/L=(0.1)/(20.0)=0.5%.

Time: 2(Δ T)/T=2×1/(90)=2.22%.

Total: 0.5%+2.22%=2.72%, which to the nearest option is 3%.

Why the other options are wrong

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