Gravitation

JEE Main level, single correct, medium

53 of 223

Question 53 of 223

If RR is the radius of the earth and the acceleration due to gravity on the surface of earth is g=π2ms−2g={\pi }^{2}m{s}^{-2}, then the length of the second's pendulum at a height h=2Rh=2R from the surface of earth will be:

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Answer: (B) 19m\dfrac{1}{9}m

Acceleration due to gravity at the given height, g′=GM(3R)2=19GM(R)2=19gg'=\dfrac{GM}{{\left(3R\right)}^{2}}=\dfrac{1}{9}\dfrac{GM}{{\left(R\right)}^{2}}=\dfrac{1}{9}g, where gg is acceleration due to gravity at the surface of the Earth.
Now, time period is given by, T=2πlg′T=2\pi \sqrt{\dfrac{l}{g'}}
Also the time period of second pendulum is 22 sec.
Therefore, T=2sec⁡T=2\sec
Then, 2=2πlg×92=2\pi \sqrt{\dfrac{l}{g}\times 9}
⇒l=19m\Rightarrow l=\dfrac{1}{9}m

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