Question

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 $2\%, 1\%, 3\%$ and $5\%$, then the relative error in $P$ will be :

• A. 8%
• B. 12%
• C. 32%
• D. 25%

Answer 1

The Correct Option is C

The given problem involves calculating the relative error in a derived physical quantity \(P\). The formula for \(P\) is given by:

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

To find the relative error in \(P\), we need to apply the principle of error propagation for multiplication and powers, which states that if a quantity \(Q\) is given by:

Q = A^m B^n C^p \ldots

Then, the relative error in \(Q\) can be estimated as:

\frac{\Delta Q}{Q} = |m|\frac{\Delta A}{A} + |n|\frac{\Delta B}{B} + |p|\frac{\Delta C}{C} + \ldots

Applying this rule to our expression for \(P\), we get:

\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}

We are given the relative errors:

• \(\frac{\Delta a}{a} = 2\% = 0.02\)
• \(\frac{\Delta b}{b} = 1\% = 0.01\)
• \(\frac{\Delta c}{c} = 3\% = 0.03\)
• \(\frac{\Delta d}{d} = 5\% = 0.05\)

Substituting these values into the error formula for \(P\):

\frac{\Delta P}{P} = \frac{1}{2} \times 0.02 + 2 \times 0.01 + 3 \times 0.03 + 4 \times 0.05

Calculate each term:

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

Summing these contributions:

0.01 + 0.02 + 0.09 + 0.20 = 0.32

Thus, the relative error in \(P\) is \(0.32\), or expressed as a percentage:

32\%

Therefore, the correct answer is 32%.