Q110 / 150 · MHT CET 2020 (16 Oct Shift 2) · Physics

A body slides down a smooth inclined plane having angle ' $\theta$ ' and reaches the bottom with velocity 'v'. If a body is a sphere then its linear velocity at the bottom of the plane is

  • A
    $\sqrt{\frac{2}{7}} \mathrm{v}$
  • B
    $\sqrt{\frac{3}{7}} \mathrm{v}$
  • Correct answer
    C
    $\sqrt{\frac{5}{7}} \mathrm{v}$
  • D
    $\sqrt{\frac{9}{7}} \mathrm{v}$

Solution

The linear velocity of the body, $v=\sqrt{2 g h}$ The velocity of the sphere about its centre, $v_{C M}=\sqrt{\frac{2 g h}{1+\frac{K^{2}}{R^{2}}}}=\frac{v}{\sqrt{1+\frac{K^{2}}{R^{2}}}} \ldots(\text { i })$ For uniform solid sphere, $\frac{K^{2}}{R^{2}}=\frac{2}{5}$ Substituting value in Eq. (i), we get $v_{C M}=\frac{v}{\sqrt{1+\left(\frac{2}{5}\right)}}=\sqrt{\frac{5}{7} v}$

Hence, the correct answer is option: 'C'.

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