A man is standing on a weighing machine kept on the floor of an elevator. The ratio of measured weight of the man when elevator is at rest and when it is moving with downward acceleration a is 3 : 2. The value of a is :

Options

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

Explanation

To solve the problem, we need to analyze the forces acting on the man standing on the weighing machine in the elevator.

### Step-by-Step Solution:

1. **Understanding Forces When the Elevator is at Rest:**
When the elevator is at rest, the only forces acting on the man are his weight (mg, acting downwards) and the normal force (N₀, acting upwards). Since the elevator is not accelerating, these forces are balanced:
\[ N₀ = mg \]

2. **Understanding Forces When the Elevator is Moving Downward with Acceleration a:**
When the elevator is moving downward with an acceleration \( a \), the forces acting on the man are still his weight (mg) and the normal force (N₁, acting upwards). In this case, the net force acting on the man can be described by Newton's second law:
\[ mg - N₁ = ma \]
Rearranging this gives:
\[ N₁ = mg - ma \]

3. **Setting Up the Ratio:**
According to the problem, the ratio of the measured weight of the man when the elevator is at rest (N₀) to when it is moving downward with acceleration \( a \) (N₁) is given as:
\[ \frac{N₀}{N₁} = \frac{3}{2} \]
Substituting the expressions for \( N₀ \) and \( N₁ \):
\[ \frac{mg}{mg - ma} = \frac{3}{2} \]

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

5. **Rearranging the Equation:**
Rearranging the equation to isolate terms involving \( a \):
\[ 3ma = 3mg - 2mg \]
Simplifying this gives:
\[ 3ma = mg \]

6. **Solving for a:**
Dividing both sides by \( 3m \) (assuming \( m \neq 0 \)):
\[ a = \frac{g}{3} \]

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

---