A physical quantity P is described by the relation `P= a^(1/2)b^(2)c^(3)d^(-4)`. If the relative errors in the measurement of a,b,c and d respectively, are 2%, 1%, 3% asd 5%, then the relative error in P will be :

Options

A. `8%`
B. `12%`
C. `32%`
D. `25%`

Explanation

To find the relative error in the physical quantity \( P \) defined by the relation

\[ P = a^{1/2} b^{2} c^{3} d^{-4} \]

we will use the formula for the propagation of errors. The relative error in \( P \) can be expressed in terms of the relative errors in \( a \), \( b \), \( c \), and \( d \).

### Step 1: Write the formula for relative error in \( P \)

The relative error in \( P \) is given by:

\[ \frac{\Delta P}{P} = \frac{1}{2} \frac{\Delta a}{a} + 2 \frac{\Delta b}{b} + 3 \frac{\Delta c}{c} - 4 \frac{\Delta d}{d} \]

where \( \Delta a/a \), \( \Delta b/b \), \( \Delta c/c \), and \( \Delta d/d \) are the relative errors in \( a \), \( b \), \( c \), and \( d \) respectively.

### Step 2: Substitute the given relative errors

From the problem, we have the following relative errors:
- \( \frac{\Delta a}{a} = 2\% = 0.02 \)
- \( \frac{\Delta b}{b} = 1\% = 0.01 \)
- \( \frac{\Delta c}{c} = 3\% = 0.03 \)
- \( \frac{\Delta d}{d} = 5\% = 0.05 \)

Now, substituting these values into the formula:

\[ \frac{\Delta P}{P} = \frac{1}{2}(0.02) + 2(0.01) + 3(0.03) - 4(0.05) \]

### Step 3: Calculate each term

Calculating each term:

1. \( \frac{1}{2}(0.02) = 0.01 \)
2. \( 2(0.01) = 0.02 \)
3. \( 3(0.03) = 0.09 \)
4. \( -4(0.05) = -0.20 \)

### Step 4: Sum all the terms

Now, we sum these results:

\[ \frac{\Delta P}{P} = 0.01 + 0.02 + 0.09 - 0.20 \]

Calculating this gives:

\[ \frac{\Delta P}{P} = 0.01 + 0.02 + 0.09 - 0.20 = 0.12 - 0.20 = -0.08 \]

### Step 5: Convert to percentage

To convert this to a percentage, we multiply by 100:

\[ \frac{\Delta P}{P} \times 100 = -0.08 \times 100 = -8\% \]

### Final Result

The relative error in \( P \) is \( -8\% \).

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