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A point particle is held on the axis of a ring of mass $m$ and radius $r$ at a distance $r$ from its centre C. When released, it reaches $\mathrm{C}$ under the gravitational attraction of the ring. Its speed at $C$ will be
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Verified Answer
The correct answer is:
$\sqrt{\frac{2 \mathrm{G} m}{r}\left(1-\frac{1}{\sqrt{2}}\right)}$
Let ' $\mathrm{M}$ ' be the mass of the particle
Now, $E_{\text {initial }}=E_{\text {final }}$
i.e. $\frac{\mathrm{GMm}}{\sqrt{2} r}+0=\frac{\mathrm{GM} m}{r}+\frac{1}{2} \mathrm{MV}^{2}$
or, $\frac{1}{2} \mathrm{MV}^{2}=\frac{\mathrm{GM} m}{r}\left[1-\frac{1}{\sqrt{2}}\right]$
$\Rightarrow \frac{1}{2} \mathrm{~V}^{2}=\frac{\mathrm{G} m}{r}\left[1-\frac{1}{\sqrt{2}}\right]$ or,
$\mathrm{V}=\sqrt{\frac{2 \mathrm{Gm}}{r}\left(1-\frac{1}{\sqrt{2}}\right)}$
Now, $E_{\text {initial }}=E_{\text {final }}$
i.e. $\frac{\mathrm{GMm}}{\sqrt{2} r}+0=\frac{\mathrm{GM} m}{r}+\frac{1}{2} \mathrm{MV}^{2}$
or, $\frac{1}{2} \mathrm{MV}^{2}=\frac{\mathrm{GM} m}{r}\left[1-\frac{1}{\sqrt{2}}\right]$
$\Rightarrow \frac{1}{2} \mathrm{~V}^{2}=\frac{\mathrm{G} m}{r}\left[1-\frac{1}{\sqrt{2}}\right]$ or,
$\mathrm{V}=\sqrt{\frac{2 \mathrm{Gm}}{r}\left(1-\frac{1}{\sqrt{2}}\right)}$
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