The time period of a simple pendulum is given by T = 2π√(l/g). The… | JEE Main 2021 (18 Mar, Shift 1)

Q86JEE Main 2021 · 18 MAR · SHIFT 1Difficult

Errors in Measurement

The time period of a simple pendulum is given by $T = 2\pi\sqrt{\frac{l}{g}}$ . The measured value of the length of pendulum is 10 cm known to a 1mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to be nearest integer is :-

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Answer: (b)

$g = \dfrac{4\pi^2 l}{T^2}$ , so $\dfrac{\Delta g}{g} = \dfrac{\Delta l}{l} + 2\dfrac{\Delta T}{T}$ .
$\dfrac{\Delta l}{l} = \dfrac{0.1\text{ cm}}{10\text{ cm}} = 1\%$
$\dfrac{\Delta T}{T} = \dfrac{\Delta t}{t} = \dfrac{1}{100} = 1\%$ , so $2\dfrac{\Delta T}{T} = 2\%$
$x = 1\% + 2\% = 3\%$ , answer (b).

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