JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Work, Power & Energy

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

3 0^{\circ}

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

4 5^{\circ}

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}

Answer: (A)

Solution

For a solid sphere of mass

m

, radius

r

, rolling without slipping down a length

L

of an incline at angle

θ

, energy conservation gives:

m g L sin θ = \frac{1}{2} m v^{2} + \frac{1}{2} I_{c m} ω^{2}

With

I_{c m} = \frac{2}{5} m r^{2}

and

ω = v / r

\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}

m g L sin θ = \frac{7}{10} m v^{2} \Rightarrow v^{2} = \frac{10 g L sin θ}{7}

, i.e.

v^{2} \propto sin θ

for fixed

L

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

v_{1}^{2} : v_{2}^{2} = 1 : 2

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Single correct option