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Physics · 2021

JEE Main · 18 March 2021, Shift 1 · Q9

The time period of a simple pendulum is given by T =2 π √(l/g). The measured value of the length of pendulum is 10 cm known to a 1 mm accuracy. The…

The time period of a simple pendulum is given by $\displaystyle \mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$. The measured value of the length of pendulum is $\displaystyle 10$ cm known to a $\displaystyle 1$ mm accuracy. The time for $\displaystyle 200$ oscillations of the pendulum is found to be $\displaystyle 100$ second using a clock of $\displaystyle 1$ s resolution. The percentage accuracy in the determination of ' $\displaystyle g$ ' using this pendulum is ' $\displaystyle x$ '. The value of ' $\displaystyle x$ ' to the nearest integer is,
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JEE Main 2021 Physics question, with the answer from NTA’s final answer key. Where our answers come from.