In an experiment for determining the gravitational acceleration g of… | JEE Main 2014 (19 Apr, Shift 1)

Q78JEE Main 2014 · 19 APR · SHIFT 1Medium

Simple Pendulum

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 plotted against the string length of the pendulum in the figure.

what is the value of g at the place?

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Answer: (b)

$T^2 = \frac{4\pi^2}{g}L$ , so slope of the $T^2$ vs $L$ graph $= \frac{4\pi^2}{g}$ .
From the graph: slope $= \frac{8.0 - 4.0}{2.0 - 1.0} = 4\text{ s}^2/\text{m}$ .
$g = \frac{4\pi^2}{4} = \pi^2 \approx 9.87\text{ m/s}^2$ .
So the answer is (b) $9.87\text{ m/s}^2$ .

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