A tunnel is dug inside the earth across one of its diameters. Radius of earth is R and its mass is M . A particle is projected inside the tunnel with velocity R 2 GM from one of its ends then maximum velocity attained by the particle in the subsequent motion is R nGM (assuming tunnel to be frictionless). Find n .

maximum velocity will be at its centre. From energy conservation 2 1 m ( R 2 GM ) + m ( R − GM ) = 2 1 mv 1 2 + m ( 2 − 3 R GM ) 2 1 mv 1 2 = 2 3 R GMm V 1 = R 3 GM

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A tunnel is dug inside the earth across one of its diameters. Radius of earth is R and its mass is M . A particle is projected inside the tunnel with velocity $\frac{2 GM}{R}$ from one of its ends then maximum velocity attained by the particle in the subsequent motion is $\frac{nGM}{R}$ (assuming tunnel to be frictionless). Find n .

Answer:

Solution:

maximum velocity will be at its centre. From energy conservation

$\frac{1}{2} m (\frac{2 GM}{R}) + m (\frac{- GM}{R}) = \frac{1}{2} mv_{1}^{2} + m (\frac{- 3}{2} \frac{GM}{R}) \frac{1}{2} mv_{1}^{2} = \frac{3}{2} \frac{GMm}{R} V_{1} = \frac{3 GM}{R}$

Stream:JEESubject:PhysicsTopic:GravitationSubtopic:Field & Force

⏱ 2mℹ️ Source: QB

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