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 (g- Acceleration due to gravity on the earth)

Correct Option: g/3

To find the value of the downward acceleration of the lift, let us analyze the forces acting on the man in both scenarios.

Case 1: The lift is stationary
When the lift is stationary (at rest), the apparent weight of the man, $W_1$, is equal to his actual gravitational weight.
If $m$ is the mass of the man and $g$ is the acceleration due to gravity, then:

$W_1=mg$

Case 2: The lift is moving downward with a uniform acceleration 'a'
When the lift accelerates downwards with acceleration $a$, an observer inside the lift experiences a pseudo force acting upwards. The apparent weight of the man, $W_2$, decreases and is given by:

$W_2=m(g-a)$

Finding the Acceleration 'a'
We are given that the ratio of the weight in the stationary lift to the weight when moving downward is 3 : 2:

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

Substitute the expressions for $W_1$ and $W_2$ into the ratio:

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

Cancel the mass $m$ from both the numerator and the denominator:

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

Cross-multiply to solve for $a$:

$2g=3(g-a)$

$2g=3g-3a$

Rearrange the equation to isolate the term with $a$:

$3a=3g-2g$

$3a=g$

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

Thus, the value of the downward uniform acceleration $a$ is $\frac{g}{3}$.