A simple pendulum is given the minimum velocity at the lowest point so that it can just complete the vertical circle. Column-1 Column-2AVelocity at the lowest pointPglBVelocity at highest pointQ2glCVelocity when string becomes horizontalR3glDVelocity when the string makes 60° angle with verticalS4gl T5gl

v2=u2−2gh =(5gl)2−2gh v=5gl−2ghLowest point h=0,v=5glHighest point h=2l,v=glHorizontal h=l,v=3glAt, 60°,h=l2,v=4gl

A simple pendulum is given the minimum velocity at the lowest point so that it can just complete the vertical circle. Column-1 Column-2AVelocity at the lowest pointPglBVelocity at highest pointQ2glCVelocity when string becomes horizontalR3glDVelocity when the string makes 60° angle with verticalS4gl T5gl

v2=u2−2gh =(5gl)2−2gh v=5gl−2ghLowest point h=0,v=5glHighest point h=2l,v=glHorizontal h=l,v=3glAt, 60°,h=l2,v=4gl

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A simple pendulum is given the minimum velocity at the lowest point so that it can just complete the vertical circle.
Column-1 Column-2
A Velocity at the lowest point P $\sqrt{g l}$
B Velocity at highest point Q $\sqrt{2 g l}$
C Velocity when string becomes horizontal R $\sqrt{3 g l}$
D Velocity when the string makes $60 °$ angle with vertical S $\sqrt{4 g l}$
T $\sqrt{5 g l}$

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$A - P, B - Q, C - R, D - S$

$A - Q, B - P, C - Q, D - R$

$A - T, B - R, C - P, D - Q$

$A - T, B - P, C - R, D - S$

answer is D.

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Detailed Solution

$v^{2} = u^{2} - 2 g h = {(\sqrt{5 g l})}^{2} - 2 g h v = \sqrt{5 g l - 2 g h}$

Lowest point $h = 0, v = \sqrt{5 g l}$
Highest point $h = 2 l, v = \sqrt{g l}$
Horizontal $h = l, v = \sqrt{3 g l}$
At, $60 °, h = \frac{l}{2}, v = \sqrt{4 g l}$

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