Physics - NLM Question with Solution | TestHub

Weight of the man in a stationary lift is $W_1=mg$.

Apparent weight of the man in a lift moving downward with uniform acceleration $a$ is $W_2=m(g-a)$.

Given that the ratio of the weights is $3:2$, we have:

$\frac{W_1}{W_2}=\frac{3}{2}$

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

$2g=3(g-a)$

$2g=3g-3a$

$3a=g$

$a=\frac{g}{3}$

Answer: $\frac{\mathrm{g}}{3}$