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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Gravitation

A smooth tunnel is dug along the radius of earth that ends at center. A ball is released from the surface of earth along tunnel. What is the velocity when it strike the center?

\[(a)\;v=\sqrt {\frac{GR}{2M}} \quad (b)\;v=\sqrt {\frac{GM}{2R}} \quad (c)\;v=\sqrt {\frac{2GM}{R}} \quad (d)\;v=\sqrt {\frac{GM}{R}} \]

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1 Answer

Gain in KE= change in potential energy
$\large\frac{1}{2} $$mv^2 =m [V_A-V_B]$
$\qquad= m\bigg[\large\frac{-GM}{2R} -\bigg(\frac{-GM}{R}\bigg)\bigg]$
$\qquad= \large\frac{GMm}{2R}$
$v= \sqrt {\large\frac{GM}{R}}$
Hence d is the correct answer.
answered Aug 29, 2013 by meena.p
edited Jul 3, 2014 by lmohan717
 

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