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Question

At the lowest position the velocity of the bucket is $\sqrt{2 g L}$ $K E \quad$ of the bucket is MgL. Then it scoops some water so, its weight becomes

Vedclass · Accepted Answer

${\left( {\frac{M}{{M + m}}} \right)^2}L$

Vedclass · Answer

At the lowest position the velocity of the bucket is $\sqrt{2 g L}$ $K E \quad$ of the bucket is MgL. Then it scoops some water so, its weight becomes $M+m .$ Velocity of the bucket just after it scoops some water is

$M \sqrt{2 g L}=(M+m) v$

$v=\frac{M \sqrt{2 a L}}{M+m}$

maximum height reached is $\frac{v^{2}}{2 g}=\left(\frac{M}{M+m}\right)^{2} L$

$A$ small bucket of mass $M\, kg$ is attached to $a$ long inextensible cord of length $L\, m$ . The bucket is released from rest when the cord is in a horizontal position. At its lowest position, the bucket scoops up $m\, kg$ of water and swings up to a height $h$. The height $h$ in meters is

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