IIT JEE Advanced 2006 Paper 1

From the figure shown here, we have
cos θ =

4

5

; sin θ =

3

5

That is, θ = 37°. From the free-body diagram of the block shown here, we get
N = mg
N′ = masin37°
macosθ – μ(N + N′) = mb
where b is the acceleration of the block with respect to the disc.

Therefore,
ma(

4

5

)−

2

5

[mg+(ma×

3

5

)] = mb
20 - 10 = b
That is, b = 10 m/s2.