JEE Main 2025 Jan 23 Shift 1, Physics Q14: Rolling Motion

JEE Main2025Jan 23, Shift 1Physics

A solid sphere of mass ' $m$ ' and radius ' $r$ ' is allowed to roll without slipping from the highest point of an inclined plane of length ' $L$ ' and makes an angle 30º with the horizontal. The speed of the particle at the bottom of the plane is $v_{1}$ . If the angle of inclination is increased to 45º while keeping $L$ constant, then the new speed of the sphere at the bottom of the plane is $v_{2}$ . The ratio of $v_{1}^{2} : v_{2}^{2}$ is

A
$1 : 2$
B
$1 : 3$
C
$1 : 2$
D
$1 : 3$

Solution

For a rolling solid sphere, the kinetic energy is $K E = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} = \frac{1}{2} m v^{2} + \frac{1}{2} \cdot \frac{2}{5} m r^{2} \cdot \frac{v ^{2}}{r ^{2}} = \frac{7}{10} m v^{2}$ .

Energy conservation from rest at the top: $m g L sin θ = \frac{7}{10} m v^{2}$ , so $v^{2} \propto sin θ$ .

$\frac{v _{1}^{2}}{v _{2}^{2}} = \frac{sin 3 0 ^{\circ}}{sin 4 5 ^{\circ}} = \frac{1/2}{1/ 2} = \frac{1}{2}$ .

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JEE Main 2025 Jan 23 Shift 1, Physics Q14: Rolling Motion

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