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Q.

A block is placed on the top of a smooth inclined plane of inclination θ kept on the floor of a lift. When the lift is descending with a retardation a, the block is released. The acceleration of the block relative to the incline is

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a

g sinθ

b

a sinθ

c

(g-a) sinθ

d

(g+a) sinθ

answer is C.

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Detailed Solution

Since the lift is accelerating downward , vertically upward pseudo force ma will be acting on the block and hence resultant force acting on the block along the incline relative to the lift is mg-asin θ .Hence, the acceleration of the block relative to the incline is (g-a) sinθ.
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