A lift is moving down with acceleration a. A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively
A
g,g
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B
g−a,g−a
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C
g−a,g
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D
a,g
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Solution
The correct option is Cg−a,g Man in lift is in non-inertial frame. By applying psuedoforce on the ball,we get net acceleration =g−a w.r.t. lift,
As earth is assumed to be inertial frame, no pseudo force acts. So FBD is