The ratio of the weight of a man in a stationary lift & when it is moving downward with uniform acceleration 'a' is 3 : 2. The value of 'a' is (g = acceleration due to gravity)

(3/2) g

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(2/3) g

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g/3

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Solution

The correct option is D g/3
$\frac{m g}{m g - m a} = \frac{3}{2}$
or, $2 g = 3 g - 3 a$
or, $g = 3 a$
$a = \frac{g}{3}$
hence,option D is correct.

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The ratio of the weight of a man in a stationary lift & when it is moving downward with uniform acceleration 'a' is 3 : 2. The value of 'a' is (g = acceleration due to gravity)

The correct option is D g/3mgmg−ma=32or, 2g=3g−3aor, g=3a a=g3hence,option D is correct.

The correct option is D g/3
$\frac{m g}{m g - m a} = \frac{3}{2}$
or, $2 g = 3 g - 3 a$
or, $g = 3 a$
$a = \frac{g}{3}$
hence,option D is correct.