An experiment is performed to obtain the value of acceleration due to… | JEE Main 2014 (9 Apr, Shift 1)

Q104JEE Main 2014 · 9 APR · SHIFT 1Difficult

Errors in Measurement

An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be:

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Answer: (b)

$g = \dfrac{4\pi^2 L}{T^2}$ , so $\dfrac{\Delta g}{g} = \dfrac{\Delta L}{L} + 2\dfrac{\Delta t}{t}$ .
$\dfrac{\Delta L}{L} = \dfrac{0.1}{20.0} = 0.5\%$
$2\dfrac{\Delta t}{t} = 2\times\dfrac{1}{90} = 2.22\%$
Total $\approx 2.7\%$ , answer (b).

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Errors in Measurement question from JEE Main 2014 (9 Apr, Shift 1)

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