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JEE Main 2025 (23 Jan, Shift 1) · Rolling Motion

Rolling Motion question from JEE Main 2025 (23 Jan, Shift 1)

Q17JEE Main 2025 · 23 JAN · SHIFT 1Medium

Rolling Motion

A solid sphere of mass ' mm ' and radius ' rr ' is allowed to roll without slipping from the highest point of an inclined plane of length ' LL ' and makes an angle 30∘30^\circ with the horizontal. The speed of the particle at the bottom of the plane is v1v_1. If the angle of inclination is increased to 45∘45^\circ while keeping LL constant. Then the new speed of the sphere at the bottom of the plane is v2v_2. The ratio v12:v22v_1^{2} : v_2^{2} is

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Answer: (c)

  1. Rolling sphere: v2=107ghv^2 = \frac{10}{7}gh with h=Lsin⁡θh = L\sin\theta, so v2∝sin⁡θv^2 \propto \sin\theta.
  2. v12v22=sin⁡30∘sin⁡45∘=1/21/2=12\dfrac{v_1^2}{v_2^2} = \dfrac{\sin 30^\circ}{\sin 45^\circ} = \dfrac{1/2}{1/\sqrt{2}} = \dfrac{1}{\sqrt{2}}
  3. So the answer is (c) 1:21 : \sqrt{2}.
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