A physical quantity y is given by `y=(P^2Q^(3//2))/(R^4S^(1//2))`
The percentage error in P,Q ,R and S are 1%,2%, 4% and 2% respectively. Find the percentage error in y.

Answer

To find the percentage error in the physical quantity \( y \) given by the equation:

\[ y = \frac{P^2 Q^{3/2}}{R^4 S^{1/2}} \]

we will use the formula for the propagation of errors. The percentage error in \( y \) can be calculated using the following steps:

### Step 1: Identify the formula for percentage error
The percentage error in a quantity that is a function of several variables can be calculated using the formula:

\[ \frac{\Delta y}{y} = \left| n_1 \frac{\Delta P}{P} \right| + \left| n_2 \frac{\Delta Q}{Q} \right| + \left| n_3 \frac{\Delta R}{R} \right| + \left| n_4 \frac{\Delta S}{S} \right| \]

where \( n_1, n_2, n_3, n_4 \) are the powers of \( P, Q, R, S \) in the equation for \( y \).

### Step 2: Write down the powers of each variable
From the equation \( y = \frac{P^2 Q^{3/2}}{R^4 S^{1/2}} \), we can identify the powers:
- For \( P \), \( n_1 = 2 \)
- For \( Q \), \( n_2 = \frac{3}{2} \)
- For \( R \), \( n_3 = -4 \) (since it is in the denominator)
- For \( S \), \( n_4 = -\frac{1}{2} \) (since it is also in the denominator)

### Step 3: Write down the given percentage errors
The percentage errors for each variable are given as follows:
- \( \frac{\Delta P}{P} \times 100 = 1\% \) ⇒ \( \frac{\Delta P}{P} = 0.01 \)
- \( \frac{\Delta Q}{Q} \times 100 = 2\% \) ⇒ \( \frac{\Delta Q}{Q} = 0.02 \)
- \( \frac{\Delta R}{R} \times 100 = 4\% \) ⇒ \( \frac{\Delta R}{R} = 0.04 \)
- \( \frac{\Delta S}{S} \times 100 = 2\% \) ⇒ \( \frac{\Delta S}{S} = 0.02 \)

### Step 4: Substitute the values into the error propagation formula
Now we can substitute these values into the error propagation formula:

\[ \frac{\Delta y}{y} = 2 \cdot 0.01 + \frac{3}{2} \cdot 0.02 + 4 \cdot 0.04 + \frac{1}{2} \cdot 0.02 \]

### Step 5: Calculate each term
Calculating each term:
- \( 2 \cdot 0.01 = 0.02 \)
- \( \frac{3}{2} \cdot 0.02 = 0.03 \)
- \( 4 \cdot 0.04 = 0.16 \)
- \( \frac{1}{2} \cdot 0.02 = 0.01 \)

### Step 6: Sum the contributions
Now, we sum all these contributions:

\[ \frac{\Delta y}{y} = 0.02 + 0.03 + 0.16 + 0.01 = 0.22 \]

### Step 7: Convert to percentage
To convert this to a percentage, we multiply by 100:

\[ \Delta y \text{ (percentage)} = 0.22 \times 100 = 22\% \]

### Final Result
Thus, the percentage error in \( y \) is:

\[ \boxed{22\%} \]

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