JEE Main 2014 · 9 Apr · Shift 1 · Physics Q1

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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 :

1.7%

2.7%

4.4%

2.27%

Answer and worked solution

Official answer: B

$g=\dfrac{4\pi^2L}{T^2}$ , so
$\frac{\Delta g}g=\frac{\Delta L}L+2\frac{\Delta T}T=\frac{0.1}{20}+2\cdot\frac1{90}=0.005+0.022=0.027\approx2.7\%.$
(The 1 s least count applies to the 90 s total.)

Topics:Errors in measurement Simple pendulum experiment

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JEE Main 2014 · 9 Apr · Shift 1, Physics question 1