The ratio of weight of a man in a stationery lift and weight when the lift is moving downward with a uniform…
The ratio of weight of a man in a stationery lift and weight when the lift is moving downward with a uniform acceleration ' $a$ ' is $3: 2$. Then the value of ' $a$ ' is
$\frac{3}{2} \mathrm{~g}$
$\frac{\mathrm{g}}{3}$
$\frac{2}{3} \mathrm{~g}$
g
Solution
Weight of a man when the lift is stationary, $\mathrm{W}_1=\mathrm{mg}$.
Weight when the lift is going down, $\mathrm{W}_2=\mathrm{m}(\mathrm{g}-\mathrm{a})$
$\begin{array}{ll}
\therefore & \frac{W_1}{W_2}=\frac{m g}{m(g-a)}=\frac{3}{2} \\
\therefore & \frac{g}{g-a}=\frac{3}{2} \\
\therefore & 2 g=3 g-3 a \\
\therefore & g=3 a \\
\therefore & a=\frac{g}{3}
\end{array}$
....(Given)