JEE MAIN - Physics (2024 - 31st January Evening Shift - No. 13)
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:
6
5
4
8
Explanation
$$\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \\ & \mathrm{g}=\frac{4 \pi^2 \ell}{\mathrm{T}^2} \\ & \frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{\Delta \ell}{\ell}+\frac{2 \Delta \mathrm{T}}{\mathrm{T}} \\ & =\frac{0.2}{20}+2\left(\frac{1}{40}\right) \\ & =\frac{1.2}{20} \end{aligned}$$
Percentage change $$=\frac{1.2}{20} \times 100=6 \%$$
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