Two inclined planes are placed as shown in figure. A block is projected from the Point A of inclined plane AB along its surface with a velocity just sufficient to carry it to the top Point B at a height 10 m. After reaching the Point B the block slides down on inclined plane BC. Time it takes to reach to the point C from point A is t(√2+1) s. The value of t is ___________.
(use g=10 m / s² )

Solution

Physics sir · 4 steps

Going up to rest at B takes the same time as sliding down from B, so treat both legs as slides from rest.

1. Reverse the climb. On a smooth incline the block slows at g sin θ going up and speeds up at g sin θ coming down. It just stops at B, so the time from A to B equals the time to slide from rest at B down to A.
2. Time down an incline. The length of the incline is L = h/sin θ. Starting from rest with acceleration g sin θ, use L = 1/2(g sin θ)t² and solve for t.

   t = 1/(sin θ)√(2h/g), √(2h/g) = √((2 × 10)/10) = √2 s

3. Each leg. For AB, θ = 45°: t_(AB) = √2 × √2 = 2 s. For BC, θ = 30°: t_(BC) = 2 × √2 = 2√2 s.
4. So. Total time = 2 + 2√2 = 2(√2+1) s. Comparing with t(√2+1) gives t = 2.

The idea to keep. On a smooth incline, motion up to rest is the mirror image of sliding down from rest. The time to slide a height h is 1/(sin θ)√(2h/g), so a steeper incline is quicker.

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