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JEE Main 2014 — Experimental Physics Question with Solution

JEE Main 2014 (19 Apr Online)

Question


In an experiment to determine the gravitational acceleration g 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 g at the place?

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Show full solutionCorrect option: B
Correct answer
B9.87 m s-2

Step-by-step explanation

T=2πLg

⇒   g=4π2LT2

T2L=4π2g=tanθ=slope of graph

⇒   g=4π2tanθ

∵  tanθ=8−22−0.5=61.5=4

⇒   g=4π24=(3.14)2=π2=3.142=9.87 m s-2

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About this question

This is a previous-year question from JEE Main 2014, covering the Experimental Physics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.