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 .............

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Correct Answer

(B) \(2\)

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\(v =\sqrt{ u ^{2}-2 gL }\)

\(\Delta v =\sqrt{ u ^{2}+ v ^{2}}\)

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

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

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

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