The period of oscillation of a simple pendulum is T = 2π√(l/g) where… | MHT CET 2026 (14 May, Shift 2)

Q20MHT CET 2026 · 14 MAY · SHIFT 2Medium

Errors in Measurement

The period of oscillation of a simple pendulum is $T = 2\pi\sqrt{\frac{l}{g}}$ where $l$ is about $100$ cm and has $1$ mm accuracy. The period is about $2$ s. The time of $100$ oscillations is measured by a stopwatch of least count $0.1$ s. The percentage error in ' $g$ ' is (gravitation acceleration)

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Answer: (c)

$g = \frac{4\pi^2 l}{T^2}$ , so $\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}$
$\frac{\Delta l}{l} = \frac{0.1\text{ cm}}{100\text{ cm}} = 0.1\%$
Time of 100 oscillations $= 200\text{ s}$ , so $\frac{\Delta T}{T} = \frac{0.1}{200} = 0.05\%$
$\frac{\Delta g}{g} = 0.1\% + 2(0.05\%) = 0.2\%$
So the answer is (c) $0.2\%$ .

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