Question

The weight of man in lift moving in upward direction with an acceleration 'a' is \(660\) N. When the lift moves in the downward direction with the same acceleration, his weight is found to be \(380\) N. The real weight of the man when the lift is at rest is

• A. \(400\) N
• B. \(460\) N
• C. \(520\) N
• D. \(640\) N

Hint:

Upward acceleration gives m(g+a), downward gives m(g-a); add the two.

Answer 1

The Correct Option is C

Step 1: Average argument:
The two apparent weights are $mg + ma$ and $mg - ma$. They sit symmetrically about $mg$.

Step 2: So mg is the midpoint:
$mg = \frac{660 + 380}{2} = 520$ N.

Step 3: Meaning:
The extra $140$ N is $ma$ in either direction.

Final Answer:
The weight at rest is 520 N, option (C). \[ \boxed{520\text{ N}} \]