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The period of oscillation of a simple pendulum is $T = 2 π \sqrt{L / g}$ . Measured value of $L$ is $20.0 c m$ known to $1 m m$ accuracy and time for 100 oscillations of the pendulum is found to be 90s using a wrist watch of 1s resolution. What is the relative error in the determination of ‘g’?

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0.027

0.27

2.7

0.0027

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detailed solution

Correct option is A

g=4π2lT2⇒Δgg=Δll+2(ΔTT) =0.120+2(190)=0.005+0.022=0.027

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The period of oscillation of a simple pendulum isT=2πL/g . Measured value of L is 20.0cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s using a wrist watch of 1s resolution. What is the relative error in the determination of ‘g’?

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Ready to Test Your Skills?

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