Physics · 2014
JEE Main · 19 April 2014, Shift 1 · Q30
In an experiment for determining the gravitational acceleration g of a place with the help of a simple pendulum, the measured time period square is…
In an experiment for determining the gravitational acceleration $\displaystyle g$ of a place with the help of a simple pendulum, the measured time period square is plotted against the string length of the pendulum in the figure.What is the value of g at the place ?Official answer
(2)
$\displaystyle 9.87 \mathrm{~m} / \mathrm{s}^2$
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JEE Main 2014 Physics question, with the answer from NTA’s final answer key. Where our answers come from.