The ratio of the weight of the man in stationary lift and in a lift accelerating upwards with uniform acceleration is 2 : 3 . The acceleration of lift is

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The weight of the man in a stationary lift is $W_{1} = m g$ .

The apparent weight of the man in a lift accelerating upwards with uniform acceleration $a$ is $W_{2} = m (g + a)$ .

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

$\frac{W _{1}}{W _{2}} = \frac{2}{3}$

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

$\frac{g}{g + a} = \frac{2}{3}$

$3 g = 2 (g + a)$

$3 g = 2 g + 2 a$

$2 a = g$

$a = \frac{g}{2}$

Answer: $\frac{g}{2}$

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