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JEE Advanced 2023 Paper 1 · Physics · Rotational Motion · medium

A thin hollow spherical shell of mass m and radius R (moment of inertia 2/3mR^2 about a diameter) rolls without slipping down a fixed rough incline…

One or more options correct

A thin hollow spherical shell of mass mm and radius RR (moment of inertia 23mR2\tfrac23mR^2 about a diameter) rolls without slipping down a fixed rough incline of angle θ\theta, where sin⁡θ=0.6\sin\theta=0.6 and cos⁡θ=0.8\cos\theta=0.8. Take gg as the acceleration due to gravity. Which of the following statements is/are correct?
  1. A. The acceleration of the centre of the shell is 9g25\dfrac{9g}{25}.
  2. B. The least coefficient of static friction for pure rolling is 0.300.30.
  3. C. The friction force on the shell equals 0.36 mg0.36\,mg.
  4. D. If released from the same height on a perfectly smooth incline of the same angle, the speed at the bottom would exceed the rolling speed by a factor 5/3\sqrt{5/3}.
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Answer: A, B, D

For rolling without slipping, mgsin⁡θ−f=mamg\sin\theta-f=ma and fR=Iα=23mR2 (a/R)fR=I\alpha=\tfrac23mR^2\,(a/R), so f=23maf=\tfrac23ma. Then a=gsin⁡θ1+2/3=35gsin⁡θ=0.36g=9g25a=\dfrac{g\sin\theta}{1+2/3}=\dfrac35g\sin\theta=0.36g=\dfrac{9g}{25}, so (A) is correct. The friction force is f=23m(0.36g)=0.24 mgf=\tfrac23m(0.36g)=0.24\,mg, not 0.36 mg0.36\,mg, so (C) is wrong (0.36mg0.36mg is mama, not ff). The normal force is N=mgcos⁡θ=0.8mgN=mg\cos\theta=0.8mg, so the minimum coefficient needed is μmin=f/N=0.24/0.8=0.30\mu_{min}=f/N=0.24/0.8=0.30, so (B) is correct. Energy: rolling gives v2=2gh1+2/3=65ghv^2=\dfrac{2gh}{1+2/3}=\dfrac65gh, while a frictionless slide gives v2=2ghv^2=2gh. The ratio of speeds (smooth to rolling) is 2/(6/5)=5/3\sqrt{2/(6/5)}=\sqrt{5/3}, so (D) is correct. Correct options: A, B and D.
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Original question written in the JEE Advanced 2023 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake