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 of the earth)

Options

A. `(3)/(2)g`
B. `(g)/(3)`
C. `(2)/(3)g`
D. `g`

Answer

To solve the problem, we need to analyze the forces acting on the man in both scenarios: when the lift is stationary and when it is moving downward with uniform acceleration \( a \).

### Step 1: Identify the forces when the lift is stationary
When the lift is stationary, the only force acting on the man is his weight, which is given by:
\[ W_{\text{stationary}} = mg \]
where \( m \) is the mass of the man and \( g \) is the acceleration due to gravity.

### Step 2: Identify the forces when the lift is moving downward with acceleration \( a \)
When the lift is moving downward with uniform acceleration \( a \), the effective weight of the man (the normal force \( N \)) can be calculated by considering the downward acceleration. The forces acting on the man are:
- Weight acting downward: \( mg \)
- Pseudo force acting upward due to the downward acceleration of the lift: \( ma \)

Thus, the normal force (which is the apparent weight of the man in the lift) is given by:
\[ N = mg - ma \]

### Step 3: Set up the ratio of weights
According to the problem, the ratio of the weight of the man in the stationary lift to the weight of the man in the moving lift is given as:
\[ \frac{mg}{mg - ma} = \frac{3}{2} \]

### Step 4: Cross-multiply to solve for \( a \)
Cross-multiplying gives us:
\[ 2mg = 3(mg - ma) \]
Expanding the right side:
\[ 2mg = 3mg - 3ma \]

### Step 5: Rearranging the equation
Rearranging the equation to isolate terms involving \( a \):
\[ 2mg = 3mg - 3ma \implies 3ma = 3mg - 2mg \implies 3ma = mg \]

### Step 6: Solve for \( a \)
Now, divide both sides by \( 3m \):
\[ a = \frac{g}{3} \]

### Conclusion
Thus, the value of \( a \) is:
\[ a = \frac{g}{3} \]