99 Percentile Qs Bank for JEE MainPhysicsLaws of Motion
A triangular prism of mass M with a block of mass m placed on it, is released from rest on a smooth inclined plane of inclination θ . The block does not slip on the prism. Then
Options
- AThe acceleration of the prism is g   cos   θ .
- BThe acceleration of the prism is g   tan   θ .
- CThe minimum coefficient of friction between the block and the prism is μ m i n = c o t   θ .
- DThe minimum coefficient of friction between the block and the prism is μ m i n = tan   θ .
Correct answer
D. The minimum coefficient of friction between the block and the prism is μ m i n = tan   θ .
Step-by-step solution
As the small block does not slip over the prism, We can consider the mass of small block into prism. Where, N is normal reaction and here ( m   +   M ) g   sin   θ is the only force in the line of inclination acting downwards F n e t = m a ⇒ m + M g sin θ = m + M a ⇒ a = g sin θ ∴   g   sin   θ will be its acceleration ∴ option 1 and 2 are wrong Now FBD of block with respect to prism Where, m g   sin   θ is the pseudo for