Calculate x in the equation : `(velocity)^x = (pressure )^(3//2)xx(density)^(-3//2)`

Answer

To solve the equation \((\text{velocity})^x = (\text{pressure})^{\frac{3}{2}} \times (\text{density})^{-\frac{3}{2}}\), we will use dimensional analysis.

### Step 1: Write the dimensions of each physical quantity

1. **Velocity**:
- The dimension of velocity is given by:
\[ [\text{velocity}] = [L^1 T^{-1}] = M^0 L^1 T^{-1} \]

2. **Pressure**:
- Pressure is defined as force per unit area. The dimension of force is \(M L T^{-2}\), and area is \(L^2\). Therefore:
\[ [\text{pressure}] = \frac{[M L T^{-2}]}{[L^2]} = M^1 L^{-1} T^{-2} \]

3. **Density**:
- Density is mass per unit volume. The dimension of volume is \(L^3\). Thus:
\[ [\text{density}] = \frac{[M]}{[L^3]} = M^1 L^{-3} T^0 \]

### Step 2: Substitute the dimensions into the equation

Now, we can rewrite the equation using the dimensions we found:

\[ [\text{velocity}]^x = [\text{pressure}]^{\frac{3}{2}} \times [\text{density}]^{-\frac{3}{2}} \]

Substituting the dimensions:

\[ (M^0 L^1 T^{-1})^x = (M^1 L^{-1} T^{-2})^{\frac{3}{2}} \times (M^1 L^{-3} T^0)^{-\frac{3}{2}} \]

### Step 3: Simplify the right side

Calculating the right side:

1. For pressure:
\[ (M^1 L^{-1} T^{-2})^{\frac{3}{2}} = M^{\frac{3}{2}} L^{-\frac{3}{2}} T^{-3} \]

2. For density:
\[ (M^1 L^{-3} T^0)^{-\frac{3}{2}} = M^{-\frac{3}{2}} L^{\frac{9}{2}} T^0 \]

Now combine these results:

\[ M^{\frac{3}{2}} L^{-\frac{3}{2}} T^{-3} \times M^{-\frac{3}{2}} L^{\frac{9}{2}} T^0 = M^{\frac{3}{2} - \frac{3}{2}} L^{-\frac{3}{2} + \frac{9}{2}} T^{-3} = M^0 L^{3} T^{-3} \]

### Step 4: Set the dimensions equal

Now we have:

\[ M^0 L^x T^{-x} = M^0 L^3 T^{-3} \]

### Step 5: Compare the powers of dimensions

From the equation, we can compare the powers of \(L\) and \(T\):

1. For \(L\):
\[ x = 3 \]

2. For \(T\):
\[ -x = -3 \implies x = 3 \]

### Conclusion

Thus, the value of \(x\) is:

\[ \boxed{3} \]