Experimental Physics

JEE Main level, single correct, medium

61 of 70

Question 61 of 70


In an experiment to determine the gravitational acceleration gg of a place with the help of a simple pendulum, the measured time period squared is plotted against the string length of the pendulum in the figure. What is the value of gg at the place?

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Answer: (B) 9.87ms−29.87m{s}^{-2}

T=2πLgT=2\pi \sqrt{\dfrac{L}{g}}
⇒g=4π2LT2\Rightarrow g=\dfrac{4{\pi }^{2}L}{{T}^{2}}
T2L=4π2g=tanθ=\dfrac{{T}^{2}}{L}=\dfrac{4{\pi }^{2}}{g}=tan\theta =slope of graph
⇒g=4π2tanθ\Rightarrow g=\dfrac{4{\pi }^{2}}{tan\theta }
∵tanθ=8−22−0.5∵tan\theta =\dfrac{8-2}{\begin{matrix}2-0.5\end{matrix}}=61.5=4=\dfrac{6}{1.5}=4
⇒g=4π24=(3.14)2\Rightarrow g=\dfrac{4{\pi }^{2}}{4}={(3.14)}^{2}=π2=(3.14)2=9.87ms−2={\pi }^{2}={\left(3.14\right)}^{2}=9.87m{s}^{-2}

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