A stone tide to a string of length L is whirled in a vertical circle with other end of string at cen

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5.Work, Energy, Power and Collision

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A stone tide to a string of length $L$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed $u$. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is $\sqrt{ x \left( u ^{2}- gL \right)}$. The value of $x$ is .............

A

$3$

B

$2$

C

$1$

D

$5$

(JEE MAIN-2022)

Solution

$v =\sqrt{ u ^{2}-2 gL }$

$\Delta v =\sqrt{ u ^{2}+ v ^{2}}$

$\Delta v =\sqrt{ u ^{2}+ v ^{2}-2 gL }$

$\Delta v =\sqrt{2 u ^{2}-2 gL }$

$\Delta v =\sqrt{2\left( u ^{2}- gL \right)} x =2$

Standard 11

Physics

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