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JEE Mains · Physics · STD 11 - 5. work,energy,power and collision

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

  1. A \(3\)
  2. B \(2\)
  3. C \(1\)
  4. D \(5\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(2\)

Step-by-step Solution

Detailed explanation

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