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JEE Main 2021 (24 Feb, Shift 2) · Errors in Measurement

Errors in Measurement question from JEE Main 2021 (24 Feb, Shift 2)

Q88JEE Main 2021 · 24 FEB · SHIFT 2Difficult

Errors in Measurement

The period of oscillation of a simple pendulum is $T = 2\pi\sqrt{\frac{L}{g}}$ . Measured value of 'L' is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of 'g' will be:

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Answer: (c)

$g = \dfrac{4\pi^2 L}{T^2}$ , so $\dfrac{\Delta g}{g} = \dfrac{\Delta L}{L} + 2\dfrac{\Delta T}{T}$ .
$\dfrac{\Delta L}{L} = \dfrac{0.001}{1.0} = 0.1\%$
$2\dfrac{\Delta T}{T} = 2\times\dfrac{0.01}{1.95} = 1.03\%$
Total $= 1.13\%$ , answer (c).

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