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. (a)`( 3)/( 2) g `
B. (b)`( g)/( 3)`
C. (c)g
D. (d)`( 2)/( 3) f `

Answer

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

### Step-by-Step Solution:

1. **Understanding the Forces in the Stationary Lift:**
- When the lift is stationary, the only force acting on the man is his weight, which is given by:
\[ W = mg \]
- Here, \( m \) is the mass of the man and \( g \) is the acceleration due to gravity.

2. **Understanding the Forces in the Moving Lift:**
- When the lift is moving downward with uniform acceleration \( a \), the effective weight of the man (the normal force \( N \)) can be expressed as:
\[ N = mg - ma \]
- This is because the downward acceleration reduces the normal force experienced by the man.

3. **Setting Up the Ratio:**
- According to the problem, the ratio of the weight of the man in the stationary lift to the normal force when the lift is moving downward is given as:
\[ \frac{W}{N} = \frac{3}{2} \]
- Substituting the expressions for \( W \) and \( N \):
\[ \frac{mg}{mg - ma} = \frac{3}{2} \]

4. **Cross-Multiplying to Solve for \( a \):**
- Cross-multiplying gives us:
\[ 2mg = 3(mg - ma) \]
- Expanding the right side:
\[ 2mg = 3mg - 3ma \]

5. **Rearranging the Equation:**
- Rearranging the equation to isolate terms involving \( a \):
\[ 2mg - 3mg = -3ma \]
\[ -mg = -3ma \]
- Dividing both sides by \( -m \) (assuming \( m \neq 0 \)):
\[ g = 3a \]

6. **Finding the Value of \( a \):**
- Solving for \( a \):
\[ a = \frac{g}{3} \]

### Final Answer:
The value of \( a \) is \( \frac{g}{3} \).