In adiabatic process, the pressure is increased by `2//3%`. If `gamma=3//2` then the volume decreases by nearly –

Options

A. `(4)/(9)%`
B. `(2)/(3)%`
C. `1%`
D. `(9)/(4)%`

Explanation

To solve the problem, we need to determine the decrease in volume when the pressure is increased by \( \frac{2}{3} \% \) in an adiabatic process, given that \( \gamma = \frac{3}{2} \).

### Step-by-Step Solution:

1. **Understanding the Adiabatic Process**:
In an adiabatic process, the relationship between pressure (P), volume (V), and the adiabatic index (\( \gamma \)) is given by:
\[ P V^\gamma = \text{constant} \]

2. **Differentiating the Equation**:
To find the relationship between changes in pressure and volume, we differentiate the equation:
\[ d(P V^\gamma) = 0 \]
Using the product rule, we get:
\[ P \cdot d(V^\gamma) + V^\gamma \cdot dP = 0 \]
This can be rewritten as:
\[ P \cdot \gamma V^{\gamma - 1} dV + V^\gamma dP = 0 \]

3. **Rearranging the Equation**:
Rearranging gives us:
\[ \gamma P V^{\gamma - 1} dV = -V^\gamma dP \]
Dividing both sides by \( P V^\gamma \):
\[ \frac{dP}{P} = -\gamma \frac{dV}{V} \]

4. **Expressing Changes in Terms of Percentages**:
If we express the changes in terms of percentages:
\[ \frac{\Delta P}{P} \cdot 100 = -\gamma \frac{\Delta V}{V} \cdot 100 \]

5. **Substituting Given Values**:
We know from the problem that the pressure increases by \( \frac{2}{3} \% \):
\[ \frac{\Delta P}{P} \cdot 100 = \frac{2}{3} \]
Therefore:
\[ \frac{2}{3} = -\gamma \frac{\Delta V}{V} \cdot 100 \]

6. **Substituting \( \gamma \)**:
Substitute \( \gamma = \frac{3}{2} \):
\[ \frac{2}{3} = -\frac{3}{2} \frac{\Delta V}{V} \cdot 100 \]

7. **Solving for \( \Delta V/V \)**:
Rearranging gives:
\[ \frac{\Delta V}{V} = -\frac{2}{3} \cdot \frac{2}{3} = -\frac{4}{9} \]

8. **Interpreting the Result**:
The negative sign indicates a decrease in volume. Thus, the volume decreases by \( \frac{4}{9} \% \).

### Final Answer:
The volume decreases by nearly \( \frac{4}{9} \% \).