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Find the distance between the centre of gravity and centre of mass of a two-particle system attached to the ends of a light rod. Each particle has the same mass. Length of the rod is $R$ , where $R$ is the radius of the earth.

$R$
$\frac{R}{2}$
zero
$\frac{R}{4}$

A

\frac{R}{4}

B

\frac{R}{2}

C

R

D

zero

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Solution

Verified by Toppr

The centre of gravity of the Earth lie at its centre (Point p in picture)
The centre of mass of the rod is $\frac{m R + m .0}{2 m} = \frac{R}{2}$
The distance between CM of the rod and CG of the Earth is= $\frac{R}{2}$ [ $P S = \frac{R}{2}$ in the picture]

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