The lift is moving down with an 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

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  1. g, g
  2. g - a, g - a
  3. g - a, g
  4. a, g

Answer (Detailed Solution Below)

Option 3 : g - a, g
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Concept:

Apparent weight of a person inside a lift

(i) When the lift moves upward with acceleration a. Then the net upward force on the person is-

R - mg = ma

∴ Apparent weight, R = mg + ma = m (g + a)

So, when a lift accelerates upwards, the apparent weight of the person inside it increases.

(ii) When the lift moves downwards with acceleration a. Then the net downward force on the person is-

Mg - R = ma

∴ Apparent weight, R = mg - ma = m (g - a)

So, when a lift accelerates downwards, the apparent weight of the person inside it decreases.

(iii) When the lift is at rest or moving with uniform velocity v downward/upward. The acceleration a = 0, then the net force on the person is-

R - mg = m x 0 = 0

R = mg

∴ Apparent weight = Actual weight

(iv) When the lift falls freely. If the supporting cable of the lift breaks, the lift falls freely under gravity. Then a = g. The net downward force on the person is -

R = m(g - g) = 0.

Thus, the apparent weight of the man becomes zero. This is because both the man and the lift are moving downwards with same acceleration ‘g’ and so there are no forces of action and reaction between the person and the lift. Hence a person develops a feeling of weightlessness when he falls freely under gravity.

F1 J.S Madhu 02.06.20 D4

Explanation

∵ Lift is accelerating downwards, 

Acceleration of lift w.r.t. ground  alg = a

And, acceleration of ball w.r.t. ground  abg = g
∴ Acceleration of ball w.r.t. lift   abl = abg − alg
 abl = g−a
So, The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are, respectively
(g - a), g 
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