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Q.

The length of a simple pendulum, measured by an instrument of least count 4 mm, is found to be 40 cm. Its time period is measured by taking 200 oscillations using a watch of resolution 2s. The time period of found to be 0.4 s. Find the fractional error in the determination of the acceleration due to gravity using this data.

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answer is 0.06.

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Detailed Solution

Δℓℓ=0.440ΔTT=2200×0.4Time Period, T=2πℓg⇒g=4π2ℓT2 ⇒ Δgg=(Δℓℓ+2.Δtt)⇒ Δgg=  (0.440+2×2200×0.4)=0.06
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