The percentage errors in quantities P, Q, R and S are 0.5%, 1%, 3% and 1.5% respectively in the measurement of a physical quantity $$A = \frac{P^3 Q^2}{\sqrt{RS}}$$. The maximum percentage error in the value of A will be:

We have the physical quantity

$$A=\dfrac{P^{3}\,Q^{2}}{\sqrt{R\,S}}$$

In order to find the maximum (or worst-case) percentage error in a quantity that is a product or quotient of measured variables raised to various powers, we first recall the standard result from error analysis:

If $$X = P^{\alpha}\,Q^{\beta}\,R^{\gamma}\ldots$$, then the maximum percentage error in $$X$$ is obtained by adding (in absolute value) the individual percentage errors each multiplied by the absolute value of its respective power. Symbolically,

$$\frac{\Delta X}{X}\times100\% = |\alpha|\,\frac{\Delta P}{P}\times100\% \;+\; |\beta|\,\frac{\Delta Q}{Q}\times100\% \;+\; |\gamma|\,\frac{\Delta R}{R}\times100\% \;+\;\cdots$$

Applying this rule to the given expression, we first rewrite $$A$$ in the form of a pure product with explicit powers:

$$A = P^{3}\,Q^{2}\,(R\,S)^{-1/2} = P^{3}\,Q^{2}\,R^{-1/2}\,S^{-1/2}$$

Thus the powers (in magnitude) of the individual quantities are

$$\begin{aligned} P &: 3,\\[2pt] Q &: 2,\\[2pt] R &: \frac12,\\[2pt] S &: \frac12. \end{aligned}$$

The stated percentage errors in the measurements are

$$\begin{aligned} \text{Error in }P &= 0.5\%,\\[2pt] \text{Error in }Q &= 1\%,\\[2pt] \text{Error in }R &= 3\%,\\[2pt] \text{Error in }S &= 1.5\%. \end{aligned}$$

We now multiply each percentage error by the corresponding power (always using the absolute value of that power) and add the results:

$$\begin{aligned} \text{Contribution from }P &: 3\times0.5\% = 1.5\%,\\[4pt] \text{Contribution from }Q &: 2\times1\% = 2.0\%,\\[4pt] \text{Contribution from }R &: \tfrac12\times3\% = 1.5\%,\\[4pt] \text{Contribution from }S &: \tfrac12\times1.5\% = 0.75\%. \end{aligned}$$

Adding these individual contributions gives the calculated maximum percentage error:

$$1.5\% + 2.0\% + 1.5\% + 0.75\% = 5.75\%.$$

Since a quoted “maximum” error must never underestimate the true possible deviation, it is customary in laboratory work to round this result up to the next convenient half-percent so that the specified limit is certainly not exceeded. Rounding $$5.75\%$$ upward therefore gives

$$5.75\%\;\longrightarrow\;6.5\%.$$

Hence, the correct answer is Option A.