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
A
g sinθ
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B
a sinθ
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C
(g−a)sin θ
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D
(g+a)sin θ
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Solution
The correct option is D(g+a)sin θ Lift is descending with retardation. Therefore acceleration a is upwards or pseudo force ma is upwards. Relative to lift force in downward diretion is m(g + a) in plane of mg. Therefore in smooth plane acceleration will be (g+a)sinθ not g sinθ.