Verified Solution • 2025–26 Session moderate

QJEEM22Ph1517: JEE Main 2022 Physics - Class 11 JEEM22Ph1517 Solution: A solid spherical ball is rolling on a frictionless horizontal…

A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is

Multiple Choice Options

  • A 25{2 \over 5}
  • B 27{2 \over 7}
  • C 15{1 \over 5}
  • D 710{7 \over 10}

Step-by-Step Verified Solution

KER=12lw2K{E_R} = {1 \over 2}l{w^2}

=12×25×ω2×(mR2)= {1 \over 2} \times {2 \over 5} \times {\omega ^2} \times (m{R^2})

KEtotal=12×75×mR2×ω2K{E_{total}} = {1 \over 2} \times {7 \over 5} \times m{R^2} \times {\omega ^2}

∴\therefore KERKEtotal=27{{K{E_R}} \over {K{E_{total}}}} = {2 \over 7}

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