JEE Main 2014 · 19 Apr · Shift 1 · Physics Q30

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PhysicsSingle correct+4 / -1Question 30 of 90

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 ?

A straight line of T² (in s²) against L (in m) through the points (0.5, 2.0), (1.0, 4.0), (1.5, 6.0) and (2.0, 8.0).

9.81 m/s

^2

9.87 m/s

^2

9.91 m/s

^2

10.0 m/s

^2

Answer and worked solution

Official answer: B

$T^2=\dfrac{4\pi^2}gL$ . The slope is $\dfrac{8.0}{2.0}=4$ s²/m, so
$g=\frac{4\pi^2}4=\pi^2\approx9.87\text{ m/s}^2.$

Topics:Simple pendulum experiment

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JEE Main 2014 · 19 Apr · Shift 1, Physics question 30