A block can slide on a smooth inclined plane of inclination θ, kept on the floor of a lift. When the lift is descending with a retardation a, the acceleration of the block relative to the incline is:
A
(g+a)sinθ
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B
(g−a)
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C
gsinθ
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D
(g−a)sinθ
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Solution
The correct option is C(g+a)sinθ Let the acceleration of the block kept on an incline is b and its mass is m.
Since the lift is retarding downwards with a magnitude a, a pseudo force will be applied on the block in downward direction as shown in the fbd. (ma+mg)sinθ=mb b=(a+g)sinθ