A physical quantity $$P$$ is described by the relation $$P = a^{\frac{1}{2}} b^2 c^3 d^{-4}$$. 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:

We are given the physical quantity

$$P = a^{\frac{1}{2}}\, b^{2}\, c^{3}\, d^{-4}.$$

First, recall the rule for propagation of relative errors in a product of powers. If a quantity $$Q$$ is expressed as

$$Q = x^{p}\, y^{q}\, z^{r}\ldots,$$

then the relative error in $$Q$$ is obtained from

$$\frac{\Delta Q}{Q} = |p|\frac{\Delta x}{x} + |q|\frac{\Delta y}{y} + |r|\frac{\Delta z}{z} + \ldots.$$

Here $$\Delta x/x,\; \Delta y/y,\; \Delta z/z$$ are the relative errors (usually quoted as percentages) in the measurements of $$x,\; y,\; z,$$ and the absolute values of the exponents are taken because an error is always a positive quantity.

Now we match our expression for $$P$$ with the rule:

Comparing $$P = a^{\frac{1}{2}}\, b^{2}\, c^{3}\, d^{-4}$$ with $$Q = x^{p}\, y^{q}\, z^{r}\ldots,$$ we identify the exponents:

$$p = \frac{1}{2}, \qquad q = 2, \qquad r = 3, \qquad s = -4.$$

The given relative errors are

$$\frac{\Delta a}{a} = 2\% ,\qquad \frac{\Delta b}{b} = 1\% ,\qquad \frac{\Delta c}{c} = 3\% ,\qquad \frac{\Delta d}{d} = 5\% .$$

Substituting the absolute values of the exponents and the corresponding relative errors into the formula, we obtain

$$\frac{\Delta P}{P} = \Bigl|\frac{1}{2}\Bigr|\frac{\Delta a}{a} + |2|\frac{\Delta b}{b} + |3|\frac{\Delta c}{c} + |{-4}|\frac{\Delta d}{d}.$$

Writing every term explicitly,

$$\frac{\Delta P}{P} = \frac{1}{2}\times 2\% \;+\; 2\times 1\% \;+\; 3\times 3\% \;+\; 4\times 5\%.$$

Now we perform each multiplication step by step:

$$\frac{1}{2}\times 2\% = 1\%,$$

$$2\times 1\% = 2\%,$$

$$3\times 3\% = 9\%,$$

$$4\times 5\% = 20\%.$$

Add all these individual contributions to find the total relative error:

$$\frac{\Delta P}{P} = 1\% + 2\% + 9\% + 20\%.$$

The additions go as follows:

$$1\% + 2\% = 3\%,$$

$$3\% + 9\% = 12\%,$$

$$12\% + 20\% = 32\%.$$

So the relative error in $$P$$ is

$$\frac{\Delta P}{P} = 32\%.$$

Hence, the correct answer is Option D.