A physical quantity `rho` is related to four variables `alpha, beta, gamma` and `eta` as
`rho=((alpha^3)(beta^2))/(etasqrt( gamma))`
The percentage errors of measurements in `alpha, beta, gamma` and `eta` are `1%, 3%, 4%` and `2%` respectively. Find the percentage error in `rho`.

Explanation

To find the percentage error in the physical quantity \( \rho \) given its relationship with the variables \( \alpha, \beta, \gamma, \) and \( \eta \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship**: The physical quantity \( \rho \) is given by the formula:
\[ \rho = \frac{\alpha^3 \beta^2}{\eta \sqrt{\gamma}} \]

2. **Identify the percentage errors**: The percentage errors in the measurements are:
- \( \alpha \): \( 1\% \)
- \( \beta \): \( 3\% \)
- \( \gamma \): \( 4\% \)
- \( \eta \): \( 2\% \)

3. **Use the formula for percentage error**: The percentage error in \( \rho \) can be calculated using the formula for propagation of errors in multiplication and division:
\[ \frac{\Delta \rho}{\rho} \times 100 = n_1 \frac{\Delta \alpha}{\alpha} + n_2 \frac{\Delta \beta}{\beta} - n_3 \frac{\Delta \eta}{\eta} - \frac{1}{2} n_4 \frac{\Delta \gamma}{\gamma} \]
where \( n_1, n_2, n_3, n_4 \) are the powers of \( \alpha, \beta, \eta, \gamma \) in the equation for \( \rho \).

4. **Substitute the powers**: From the equation of \( \rho \):
- \( n_1 = 3 \) (for \( \alpha^3 \))
- \( n_2 = 2 \) (for \( \beta^2 \))
- \( n_3 = 1 \) (for \( \eta \))
- \( n_4 = \frac{1}{2} \) (for \( \sqrt{\gamma} \))

5. **Calculate the contribution of each variable**:
- Contribution from \( \alpha \):
\[ 3 \times 1\% = 3\% \]
- Contribution from \( \beta \):
\[ 2 \times 3\% = 6\% \]
- Contribution from \( \eta \):
\[ -1 \times 2\% = -2\% \]
- Contribution from \( \gamma \):
\[ -\frac{1}{2} \times 4\% = -2\% \]

6. **Sum the contributions**:
\[ \text{Total percentage error in } \rho = 3\% + 6\% - 2\% - 2\% = 5\% \]

### Final Result:
The percentage error in \( \rho \) is \( 5\% \).