1. Write the given relation

$P=\frac{a^2 b^3}{c\sqrt d}=\frac{a^2 b^3}{c d^{1/2}}$

2. Use the rule for percentage error

For a quantity of the form

$Q = x^{m} y^{n} z^{p}$

the maximum percentage error is

$\frac{\Delta Q}{Q}\times 100 = |m|\frac{\Delta x}{x}\times 100 + |n|\frac{\Delta y}{y}\times 100 + |p|\frac{\Delta z}{z}\times 100$

So for

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

the percentage error is

$\%\text{ error in }P = 2(\%\text{ error in }a) + 3(\%\text{ error in }b) + 1(\%\text{ error in }c) + \frac{1}{2}(\%\text{ error in }d)$

3. Substitute the given values

• Error in $a = 1\%$
• Error in $b = 2\%$
• Error in $c = 3\%$
• Error in $d = 4\%$

Therefore,

$\%\text{ error in }P = 2(1) + 3(2) + 1(3) + \frac{1}{2}(4)$

$= 2 + 6 + 3 + 2$

$= 13\%$

4. Match with the options

$13\%$ corresponds to Option B.

5. Comparison with stored correct answer

Stored correct answer = B

Our derived answer = B

So they agree.