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JEE Main201911 Jan 2019Evening ShiftPhysicsRotational MotionActual

a string is wound around a hollow cylinder of mass 5 ~kg and radius 0.5 ~m . If the string is now pulled with a horizontal force of 40 ~N , and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string)

Options

  1. A20 rad / s ²
  2. B16 rad / s ²
  3. C12 rad / s ²
  4. D10 rad / s ²

Correct answer

B. 16 rad / s ²

Step-by-step solution

From newton's second law 40+ f = m ( R ) ...(i) Taking torque about 0 we get 40 R - f R = I 40 R - f R = mR ² 40- f = mR ...(ii) Solving equation (i) and (ii) = 40 mR =16 rad / s ²

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