Question

The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration $a$ is $3:2$. The value of $a$ is:

• A. $\frac{g}{3}$
• B. $\frac{g}{4}$
• C. $\frac{g}{5}$
• D. $\frac{g}{6}$

Answer 1

Answer: (A)

$\frac{g}{3}$

Let mass of man be $m$.

Thus its weight in stationary lift $W_{\text{stationary}}=mg$

Acceleration of the lift moving downwards is $a$.

Weight of the man in downward moving lift $W_{\text{moving}}=m(g-a)$

Given : $\frac{W_{\text{stationary}}}{W_{\text{moving}}}=\frac{3}{2}$

$\therefore$ $\frac{mg}{m(g-a)}=\frac{3}{2}$

Or $\frac{g}{g-a}=\frac{3}{2}$

$⟹a=\frac{g}{3}$