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JEE Main 2020 (2 Sep, Shift 1) · Torque, Couple & Equilibrium of Rigid Bodies

Torque, Couple & Equilibrium of Rigid Bodies question from JEE Main 2020 (2 Sep, Shift 1)

Q29JEE Main 2020 · 2 SEP · SHIFT 1Difficult

Torque, Couple & Equilibrium of Rigid Bodies

A uniform cylinder of mass M and radius R is to be pulled over a step of height a (a < R) by applying a force F at its centre ‘O’ perpendicular to the plane through the axes of the cylinder on the edge of the step (see figure). The minimum value of F required is:

Figure

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Answer: (a)

  1. Just before lift-off the floor reaction is zero; take moments about the step edge P (the cylinder pivots about it).
  2. FF at O is perpendicular to OP, so its arm is RR (the largest possible), which gives the minimum FF.
  3. Weight arm = horizontal distance of O from P =R2−(R−a)2= \sqrt{R^2 - (R-a)^2}
  4. FR=MgR2−(R−a)2⇒F=Mg1−(R−aR)2F R = Mg\sqrt{R^2 - (R-a)^2} \Rightarrow F = Mg\sqrt{1 - \left(\frac{R-a}{R}\right)^2}
  5. So the answer is (a).
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