Gravitation

JEE Main level, single correct, medium

75 of 223

Question 75 of 223

The length of a seconds pendulum at a height h=2Rh=2R from earth surface will be:
(Given: R=R= Radius of earth and acceleration due to gravity at the surface of earth g=π2ms−2g={\pi }^{2}m{s}^{-2})

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

The acceleration due to gravity at height 2R2R from the surface of the Earth will be,
g′=gR2(R+h)2⇒g′=gR2(R+2R)2⇒g′=g9{g}^{'}=\dfrac{g{R}^{2}}{{\left(R+h\right)}^{2}} \Rightarrow {g}^{'}=\dfrac{g{R}^{2}}{{\left(R+2R\right)}^{2}} \Rightarrow {g}^{'}=\dfrac{g}{9}
Length of the second's pendulum at the surface of the Earth is 1m1m.
Since, time period of a pendulum is given by,
T=2πlg⇒T′T=l′l×gg′⇒1=l′l×gg′⇒l′=l9=19mT=2\pi \sqrt{\dfrac{l}{g}} \Rightarrow \dfrac{{T}^{'}}{T}=\sqrt{\dfrac{{l}^{'}}{l}\times \dfrac{g}{{g}^{'}}} \Rightarrow 1=\sqrt{\dfrac{{l}^{'}}{l}\times \dfrac{g}{{g}^{'}}} \Rightarrow {l}^{'}=\dfrac{l}{9}=\dfrac{1}{9}m

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