A smooth narrow tube in form of an arc AB of a circle of centre ‘O’ and radius ‘R’ is fixed so that A is vertically above ‘O’ and OB is horizontal. Particles P of mass ‘m’ and Q of mass ‘2m’ with a light inextensible string of lengthπr2 connecting them are placed in side the tube with P at A and Q at B and released from rest. Assuming the string remains taut during motion, find the speed of particles when P reaches B

mgR+2mg(πR2) =12(3m)v2 gR+gπR=32v23v2=2gR+2gπR 3v2=2gR(1+π)⇒v=2gR3(1+π)

A smooth narrow tube in form of an arc AB of a circle of centre ‘O’ and radius ‘R’ is fixed so that A is vertically above ‘O’ and OB is horizontal. Particles P of mass ‘m’ and Q of mass ‘2m’ with a light inextensible string of lengthπr2 connecting them are placed in side the tube with P at A and Q at B and released from rest. Assuming the string remains taut during motion, find the speed of particles when P reaches B

mgR+2mg(πR2) =12(3m)v2 gR+gπR=32v23v2=2gR+2gπR 3v2=2gR(1+π)⇒v=2gR3(1+π)

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A smooth narrow tube in form of an arc AB of a circle of centre ‘O’ and radius ‘R’ is fixed so that A is vertically above ‘O’ and OB is horizontal. Particles P of mass ‘m’ and Q of mass ‘2m’ with a light inextensible string of length $\frac{π r}{2}$ connecting them are placed in side the tube with P at A and Q at B and released from rest. Assuming the string remains taut during motion, find the speed of particles when P reaches B

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\sqrt{\frac{2}{3} (1 + π) gr}

\sqrt{\frac{3}{2} (π + 2) gr}

\sqrt{\frac{2}{3} (π - 1) gr}

\sqrt{\frac{2}{3} (π - 2) gr}

answer is A.

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Detailed Solution

$mgR + 2 mg (\frac{π R}{2})$

$= \frac{1}{2} (3 m) v^{2}$

$\begin{matrix} g R + g π R = \frac{3}{2} v^{2} \\ 3 v^{2} = 2 g R + 2 g π R \end{matrix}$

$3 v^{2} = 2 g R (1 + π) \Rightarrow v = \sqrt{\frac{2 g R}{3} (1 + π)}$

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