A body of mass 50 kg is lifted to a height of 20 m from the ground in the two different ways as shown in the figures. The ratio of work done against the gravity in both the respective cases, will be :

Solution

Physics sir · 4 steps

Gravity only cares how high you went, not the path you took, so both cases need the same work.

1. Straight up. In case 1 you pull against the full weight mg through a height of 20 m, so W₁ = mgh.
2. Along the ramp. In case 2 you only work against the component of gravity along the incline, mg sin 30° = mg/2. But the ramp is longer: L = h/(sin 30°) = 40 m.

   W₂ = (mg sin 30°) × h/(sin 30°) = mgh

3. Put in numbers. With g = 9.8 m s⁻², both are 50 × 9.8 × 20 = 9800 J. Half the force acts over twice the distance, which gives the same product.
4. So. W₁ : W₂ = 1 : 1, which is option (c).

The idea to keep. Gravity is a conservative force: the work done against it is mgh, fixed by the change in height alone. A ramp lowers the force you need but raises the distance by the same factor.

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