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MHT CET202617 April 2026Evening ShiftPhysicsRotational MotionActual

A mass tied to a string is whirled in a horizontal circular path with a constant angular velocity and its angular momentum is 'L'. If the length of the string is now halved, keeping angular velocity same then the angular momentum will be

Options

  1. AL 4
  2. BL 2
  3. CL
  4. D2L

Correct answer

A. L 4

Step-by-step solution

The angular momentum of a particle of mass m moving in a circle of radius r with angular velocity is given by L = I = mr^2 . Given that the initial angular momentum is L = mr^2 . When the length of the string is halved, the new radius becomes r' = r 2 . The angular velocity is kept constant. The new angular momentum is L' = m(r')^2 = m ( r 2 )^2 = mr^2 4 . Substituting the initial angular momentum, we get L' = L 4 . Answer: L 4

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