Gravitation

JEE Main level, single correct, medium

155 of 223

Question 155 of 223

A point particle is held on the axis of a ring of mass mm and radius rr at a distance rr from its centre CC. When released, it reaches CC under the gravitational attraction of the ring. Its speed at CC will be

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Answer: (C) 2Gmr(1−12)\sqrt{\frac{2 G m}{r}\left(1-\frac{1}{\sqrt{2}}\right)}

Let ' MM ' be the mass of the particle
Now, Einitial =Efinal \mathrm{E}_{\text {initial }}=\mathrm{E}_{\text {final }}
i.e. GMm2r+0=GMmr+12M V2\frac{\mathrm{GMm}}{\sqrt{2} r}+0=\frac{\mathrm{GM} m}{r}+\frac{1}{2} M \mathrm{~V}^2
 or, 12MV2=GMmr[1−12]⇒12V2=Gmr[1−12] or, V=2Gmr(1−12)\begin{aligned} & \text { or, } \frac{1}{2} M V^2=\frac{G M m}{r}\left[1-\frac{1}{\sqrt{2}}\right] \\ & \Rightarrow \frac{1}{2} V^2=\frac{G m}{r}\left[1-\frac{1}{\sqrt{2}}\right] \\ & \text { or, } V=\sqrt{\frac{2 G m}{r}\left(1-\frac{1}{\sqrt{2}}\right)} \end{aligned}

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