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BITSAT 2021
Exam
BITSAT
Year
2021
Subject
Physics
Chapter
Gravitation

BITSAT 2021 · Physics · Gravitation

A point particle is held on the axis of a ring of mass m and radius...

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

Options

  1. A2Gmr(2−1)\sqrt{\frac{2 \mathrm{G} m}{r}(\sqrt{2}-1)}
  2. BGmr\sqrt{\frac{\mathrm{G} m}{r}}
  3. C\sqrt{\frac{2 \mathrm{G} m}{r}\left(1-\frac{1}{\sqrt{2}}\right)}
  4. D2Gmr\sqrt{\frac{2 \mathrm{G} m}{r}}

Correct answer

C. \sqrt{\frac{2 \mathrm{G} m}{r}\left(1-\frac{1}{\sqrt{2}}\right)}

Step-by-step solution

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