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Mathematics in Physics

JEE Main level, single correct, medium

95 of 121

Question 95 of 121

The period of oscillation of a simple pendulum is $T=2\pi \sqrt{\dfrac{L}{g}}.$ Measured value of $L$ is $1.0m$ from meter scale having a minimum division of $1mm$ and time of one complete oscillation is $1.95s$ measured from stopwatch of $0.01s$ resolution. The percentage error in the determination of $'g'$ will be:

Show answer and solution

Answer: (D) $1.13%$

T=2\pi \sqrt{\dfrac{\ell }{g}}

g=\dfrac{4{\pi }^{2}\ell }{{T}^{2}}

\dfrac{\Delta g}{g}=\dfrac{\Delta \ell }{\ell }+\dfrac{2\Delta T}{T}

\dfrac{\Delta g}{g}=\dfrac{1\times {10}^{-3}}{1}+2\times \dfrac{0.01}{1.95}

\dfrac{\Delta g}{g}=0.0113

1.13%

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JEE Main 2025 Apr 2 Shift 1 question paper

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