Physics · 2021
JEE Main · 24 February 2021, Shift 2 · Q1
The period of oscillation of a simple pendulum is T=2 π √(L/g). Measured value of ' L ' is 1.0 m from meter scale having a minimum division of 1 mm…
The period of oscillation of a simple pendulum is $\displaystyle T=2 \pi \sqrt{\frac{L}{g}}$. Measured value of ' $\displaystyle L$ ' is $\displaystyle 1.0$ m from meter scale having a minimum division of $\displaystyle 1$ mm and time of one complete oscillation is $\displaystyle 1.95$ s measured from stopwatch of $\displaystyle 0.01$ s resolution. The percentage error in the determination of ' $\displaystyle g$ ' will be :
Official answer
(3)
1.$\displaystyle 13$%
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JEE Main 2021 Physics question, with the answer from NTA’s final answer key. Where our answers come from.