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 30° with the horizontal. The speed of the particle at the bottom of the plane is v₁. If the angle of inclination is increased to 45° while keeping L constant. Then the new speed of the sphere at the bottom of the plane is v₂. The ratio v₁²: v₂² is

Solution

Physics sir · 4 steps

Same length L, so only the height changes: v² follows sin θ directly.

1. Height of the incline. The sphere travels the full length L along the slope, so it drops a height h = L sin θ.
2. Energy with rolling. Rolling without slipping means no energy is lost to friction. The lost potential energy becomes translational plus rotational kinetic energy. For a solid sphere I = 2/5mr² and ω = v/r.

   mgL sin θ = 1/2mv² + 1/2(2/5mr²)v²/r² = 7/10mv²

3. Take the ratio. This gives v² = 10/7gL sin θ. The factor 10/7gL is the same on both inclines, so it cancels: v₁²/v₂² = (sin 30°)/(sin 45°) = (1/2)/(1/√2) = 1/√2.
4. So. v₁² : v₂² = 1 : √2, which is option (c).

The idea to keep. Rolling only changes the constant in front, here 7/10. When you compare two inclines of the same length, v² ∝ h = L sin θ, and that constant cancels out.

Answer from the NTA final answer key. Solution by PrepWiser.