The measured value of the length of a simple pendulum is $20 \mathrm{~cm}$ with $2 \mathrm{~mm}$ accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is $\mathrm{N} \%$. The value of $\mathrm{N}$ is:

[JEE Main 2024, 31 Jan (Shift 2)]

Question

The measured value of the length of a simple pendulum is $20 \mathrm{~cm}$ with $2 \mathrm{~mm}$ accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is $\mathrm{N} \%$. The value of $\mathrm{N}$ is:

[JEE Main 2024, 31 Jan (Shift 2)]

YESPYQ · Accepted Answer

Correct answer: c) 6. \begin{equation} \begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{~g}}} \ & \mathrm{~g}=\frac{4 \pi^2 \ell}{\mathrm{~T}^2} \ & \text { Percentage error, } \frac{\Delta \mathrm{g}}{\mathrm{~g}}=\frac{\Delta \ell}{\ell}+\frac{2 \Delta \mathrm{~T}}{\mathrm{~T}} \ & =\frac{0.2}{20}+2\left(\frac{1}{40}\right)=\frac{1.2}{20} \end{aligned} \end{equation} percentage change = $\frac{1.2}{20}\times 100=6%$

The measured value of the length of a simple pendulum is \(20 \mathrm{~cm}\) with \(2 \mathrm{~mm}\) accuracy. The time …

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