Gravitation

JEE Main level, single correct, medium

67 of 223

Question 67 of 223

TT is the time period of simple pendulum on the earth's surface. Its time period becomes xTxT when taken to a height RR (equal to earth's radius) above the earth's surface. Then, the value of xx will be:

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Answer: (B) 22

Time period of a simple pendulum at Earth's surface is given by
T=2πlgT=2\pi \sqrt{\dfrac{l}{g}}.
When distance from centre of Earth, i.e., at h=Rh=R, acceleration due to gravity is g′=g(1+hR)2=g(2)2=g4g'=\dfrac{g}{{\left(1+\dfrac{h}{R}\right)}^{2}}=\dfrac{g}{{\left(2\right)}^{2}}=\dfrac{g}{4}
Now, time period is xT=2πl(g4)xT=2\pi \sqrt{\dfrac{l}{\left(\dfrac{g}{4}\right)}}
⇒xT=2×2πlg\Rightarrow xT=2\times 2\pi \sqrt{\dfrac{l}{g}}
⇒xT=2T⇒x=2\Rightarrow xT=2T \Rightarrow x=2
Thus, x=2x=2.

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