Question

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Answer 1

$\huge{\text{\underline{Answer:-}}}$

Let U be the speed of first ball, then for a strike between first and second ball.

$Mu = Mv_1+ \frac{Mv_2}{2} - - - > (1) \\ \frac{1}{ 2 }{Mu}^{2} \frac{1}{2} {Mv}^{2}_1 + \frac{1}{2} \frac{M}{2} {v}^{2}_2 \: - - - > (2)$

Solving (1) and (2)

$v_2 = \frac{4}{3} u$

Similarly for second and third ball

$\frac{M}{2} \times \frac{4}{3}u = \frac{M}{2}v_2 + \frac{M}{ {2}^{2} } {v_3}^{2}$

$Solving = (a_r)bar \: (u)$

$In \: this \: way, v_n = ({ \frac{v}{3}) }^{n - 1} u$

$For\:nth\:ball\:to\:loop\:v_n = \sqrt{5gR}$

$i.e., \: ({\frac{4}{3} })^{n - 1} u = \sqrt{5gR} \\ or \: u = ({\frac{3}{4} })^{n - 1} \sqrt{5gR}$

∴ The correct option is (a)

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Answer 2

Answer:

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my all answers are correct