In a simple pendulum experiment for determination of acceleration due… | JEE Main 2019 (8 Apr, Shift 2)

Q96JEE Main 2019 · 8 APR · SHIFT 2Difficult

Errors in Measurement

In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to :-

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Answer: (d)

$g = \frac{4\pi^2 l}{T^2}$ with $T = t/20$ , so $\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta t}{t}$
$\frac{\Delta l}{l} = \frac{0.1}{55.0} = 0.18\%$
$2\frac{\Delta t}{t} = 2\left(\frac{1}{30}\right) = 6.67\%$
Total $\approx 6.85\% \approx 6.8\%$
So the answer is (d) $6.8\%$ .

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Errors in Measurement question from JEE Main 2019 (8 Apr, Shift 2)

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