A smooth and bottomless tunnel is dug through the centre of earth. A particle is released from the surface of earth into the tunnel. Time to reach centre of tunnel is (approximately) equal to (where $R =$ Radius of earth).

56.4 m i n u t e

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84.35 m i n u t e

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42.3 m i n u t e

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28.2 m i n u t e

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Solution

The correct option is C $42.3 m i n u t e$
Force on particle at any $x$ is $\frac{G M^{'} m}{x^{2}}$
But $\frac{M^{'}}{M} = \frac{x^{3}}{R^{3}}$
$\frac{G M x^{3} m}{R^{3} x^{2}} = (\frac{G M m}{R^{3}}) x$
Acceleration of particle, $a = \frac{G M}{R^{3}} x = \frac{g}{R} x$
It is an SHM motion, $T = 2 π \sqrt{\frac{R}{g}}$
Time to reach center of earth $= π \sqrt{\frac{R}{g}}$
$= 3.14 \times \sqrt{\frac{6400 \times 10^{3}}{9.81}} = 42.283$ minutes

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A smooth and bottomless tunnel is dug through the centre of earth. A particle is released from the surface of earth into the tunnel. Time to reach centre of tunnel is (approximately) equal to (where R= Radius of earth).

The correct option is C 42.3 minuteForce on particle at any x is GM′mx2But M′M=x3R3GMx3mR3x2=(GMmR3)xAcceleration of particle, a=GMR3x=gRxIt is an SHM motion, T=2π√RgTime to reach center of earth =π√Rg=3.14×√6400×1039.81=42.283 minutes

A smooth and bottomless tunnel is dug through the centre of earth. A particle is released from the surface of earth into the tunnel. Time to reach centre of tunnel is (approximately) equal to (where $R =$ Radius of earth).