JEE Main 2015 · 4 Apr · Shift 1 · Physics Q2

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The period of oscillation of a simple pendulum is $T=2\pi\sqrt{\dfrac Lg}$ . 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 1s resolution. The accuracy in the determination of g is :

Answer and worked solution

Official answer: B

$g=\dfrac{4\pi^2L}{T^2}$ , so
$\frac{\Delta g}g=\frac{\Delta L}L+2\frac{\Delta T}T=\frac{0.1}{20}+2\times\frac1{90}=0.5\%+2.2\%\approx3\%.$

Topics:Errors in measurement

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JEE Main 2015 · 4 Apr · Shift 1, Physics question 2