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JEE Main 202523 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Morning Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2025 · 23 January · Morning Shift · Physics · Rotational Motion

A solid sphere of mass ' m ' and radius ' r ' is allowed to roll...

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 isA solid sphere of mass ' m ' and radius ' r ' is allowed to roll without slipping from the highest point of an inclined

Options

  1. A1:21: \sqrt{2}
  2. B1:31: \sqrt{3}
  3. C1:31: 3
  4. D1:21: 2

Correct answer

A. 1:21: \sqrt{2}

Step-by-step solution

A solid sphere of mass ' m ' and radius ' r ' is allowed to roll without slipping from the highest point of an inclined
using WET Wg=kf−kiMg Lsin⁡θ=kf−ki\begin{aligned} & \mathrm{W}_{\mathrm{g}}=\mathrm{k}_{\mathrm{f}}-\mathrm{k}_{\mathrm{i}} \\ & \mathrm{Mg} \mathrm{~L} \sin \theta=\mathrm{k}_{\mathrm{f}}-\mathrm{k}_{\mathrm{i}} \end{aligned} K.E. in pure rolling 12mVcm2+12\frac{1}{2} \mathrm{mV}_{\mathrm{cm}}^2+\frac{1}{2}\mathrm{I}cmω2_{\mathrm{cm}} \omega^2=12mV2+12×25mR2 V2R2710mV2\begin{aligned} & =\frac{1}{2} \mathrm{mV}^2+\frac{1}{2} \times \frac{2}{5} \mathrm{mR}^2 \frac{\mathrm{~V}^2}{\mathrm{R}^2} \\ & \frac{7}{10} \mathrm{mV}^2 \end{aligned}mgLsin⁡θ=710mVf2−0\mathrm{mgL} \sin \theta=\frac{7}{10} \mathrm{mV}_{\mathrm{f}}^2-0Vf2∝sin⁡θ( V1 V2)2=sin⁡θ1sin⁡θ2=sin⁡30∘sin⁡45∘=12\begin{aligned} & \mathrm{V}_{\mathrm{f}}^2 \propto \sin \theta \\ & \left(\frac{\mathrm{~V}_1}{\mathrm{~V}_2}\right)^2=\frac{\sin \theta_1}{\sin \theta_2}=\frac{\sin 30^{\circ}}{\sin 45^{\circ}}=\frac{1}{\sqrt{2}} \end{aligned}$$
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