A block 'A' takes 2 s to slide down a frictionless incline of 30° and length 'l', kept inside a lift going up with uniform velocity 'v'. If the incline is changed to 45°, the time taken by the block, to slide down the incline, will be approximately :

Solution

Physics sir · 3 steps

A lift moving at steady speed changes nothing, so only sin θ decides the sliding time.

1. Lift does nothing. The lift moves with uniform velocity, so it has no acceleration. Inside it, gravity acts exactly as on the ground, with g_(eff) = g.
2. Time on incline. On a frictionless incline the acceleration along the slope is g sin θ. Starting from rest, l = 1/2g sin θ t², so t = √(2l/(g sin θ)). The length l is the same both times, which means t ∝ 1/√(sin θ).

   t₂/t₁ = √((sin 30°)/(sin 45°)) = √(0.5/0.707) = √0.707 ≈ 0.841

3. So. t₂ = 2 s × 0.841 ≈ 1.68 s. The steeper incline gives a larger acceleration, so the block reaches the bottom sooner. The answer is (c).

The idea to keep. A lift moving at constant velocity is an inertial frame, so treat it like the ground. For the same incline length, the sliding time goes as 1/√(sin θ).

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