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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A hole is drilled from the surface of earth to its center. A particle is dropped from rest at the surface of earth . The speed of the particle when it reaches the center of the earth in terms of escape velocity of the surface of earth $v_e$ is

\[(a)\;\frac{v_e}{2} \quad (b)\;v_e \quad (c)\; \sqrt 2 v_e \quad (d)\;\frac{v_e}{\sqrt 2}\]
Can you answer this question?
 
 

1 Answer

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Kinetic energy at center= Potential energy at surface-potential energy at center
$\large\frac{1}{2}$$mv^2=-\large\frac{GMm}{R}-\bigg(\frac{-3}{2}\frac{GMm}{R}\bigg)$
When $M$-Mass of earth
$\qquad R$-Radius of earth
$\qquad G$- Gravitational constant
$\large\frac{1}{2}$$mv^2=\large\frac{1}{2}\frac{GMm}{R}$
$v^2=\sqrt {\large\frac{GM}{R}}$$=\sqrt {gR}=\large\frac{v_e}{\sqrt 2}$
Hence d is the correct answer.

 

answered Aug 5, 2013 by meena.p
edited Feb 10, 2014 by meena.p
 

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