MCQ
A body (mass $m$ ) starts its motion from rest from a point distant $R_0\left(R_0>R\right)$ from the centre of the earth. The velocity acquired by the body when it reaches the surface of earth will be ( $G=$ universal constant of gravitation, $M=$ mass of earth. $R=$ radius of earth)

Answer

Correct option: B.
$\left[2 G M\left(\frac{1}{R}-\frac{1}{R_0}\right)\right]^{\frac{1}{2}}$
(b) :$\begin{aligned} & \text { 25. }: \text { As, } g^{\prime}=g-R \omega^2 \cos ^2 \lambda \\ & g_{30}=g-R \omega^2\left(\cos ^2 30^{\circ}\right)=g-R \omega^2 \times \frac{3}{4} \\ & g-g_{30}=\frac{3}{4} \omega^2 R\end{aligned}$

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