n beads are resting on a smooth horizontal wire which is circular at the end with radius R as shown. The masses of the beads are m, $\frac{m}{2}, \frac{m}{4}, ..... \frac{m}{2^{n - 1}},$ respectively. Find the minimum velocity that should be imparted to the first bead of mass m such that the n^th bead will fall in the tank shown in the figure. Assuming that elastic collisions.

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${(\frac{3}{4})}^{n - 1} \sqrt{5 g R}$

${(\frac{4}{3})}^{n - 1} \sqrt{5 g R}$

${(\frac{3}{4})}^{n - 1} 2 \sqrt{g R}$

${(\frac{4}{3})}^{n - 1} 2 \sqrt{g R}$

answer is C.

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Detailed Solution

Let required minimum velocity of m is x
Velocity of second bead after collision will be
$v_{2} = \frac{m_{2} - m_{1}}{m_{1} + m_{2}} u_{2} + \frac{2 m_{1}}{m_{1} + m_{2}} u_{1} = 0 + \frac{2 m}{m + \frac{m}{2}} x$
$v_{2} = \frac{4}{3} x$
Similarly, you get velocity of third bead after collision as $v_{1} = {(\frac{4}{3})}^{2} x$
So, velocity of nth bead after collision will be
$v_{n} = {(\frac{4}{3})}^{n - 1} x$
But ${(v_{n})}_{\min} = \sqrt{4 g R}$
$\sqrt{4 g R} = {(\frac{4}{3})}^{n - 1} x \Rightarrow x = {(\frac{3}{4})}^{n - 1} \sqrt{4 g R}$
Note: At the maximum height in the limiting case, the nth bead can momentarily stationary.
$∴ \frac{1}{2} m_{n} {(v_{n})}_{\min}^{2} = m_{n} g \times 2 R$
${(v_{n})}_{\min} = 2 \sqrt{g R}$