A physical quantity X is related to four measurable quantities a, b, c and d as follows $X=a^2{b}^3{c}^{5/2}{d}^{-2}$
The percentage error in the measurement of a, b, c and d are $1\%,2\%,2\%$ and $4\%$ respectively. What is the percentage error in quantity X?

Step 1: Given that:

The physical quantity X is given as;

$X=a^2{b}^3{c}^{\frac{5}{2}}{d}^{-2}$

Percentage error in a that is $\frac{\Delta a}{a}\times 100\%$ = $1\%$

Percentage error in b that is $\frac{\Delta b}{b}\times 100\%$ = $2\%$

Percentage error in c that is $\frac{\Delta c}{c}\times 100\%$ = $2\%$

Percentage error in d that is $\frac{\Delta d}{d}\times 100\%$ = $4\%$

Step 2: Calculation of percentage error in X:
It is given that

$X=a^2{b}^3{c}^{\frac{5}{2}}{d}^{-2}$ , then

Taking log on both sides, we get

$\log X=\log \left(a^2{b}^3{c}^{\frac{5}{2}}{d}^{-2}\right)$

$\log X=\log a^2+\log b^3+\log c^{\frac{5}{2}}+\log d^{-2}$

$\log X=2\log a+3\log b+\frac{5}{2}\log c+(-2)\log d$

$\log X=2\log a+3\log b+\frac{5}{2}\log c-2\log d$

Differentiating both sides, we get;

$\frac{\Delta X}{X}=2\frac{\Delta a}{a}+3\frac{\Delta b}{b}+\frac{5}{2}\frac{\Delta c}{c}-2\frac{\Delta d}{d}$

${\left(\frac{\Delta X}{X}\right)}_{Max}=2\frac{\Delta a}{a}+3\frac{\Delta b}{b}+\frac{5}{2}\frac{\Delta c}{c}+2\frac{\Delta d}{d}$

For percentage error we get

${\left(\frac{\Delta X}{X}\right)}_{Max}\times 100\%=2\frac{\Delta a}{a}\times 100\%+3\frac{\Delta b}{b}\times 100\%+\frac{5}{2}\frac{\Delta c}{c}\times 100\%+2\frac{\Delta d}{d}\times 100\%$

${\left(\frac{\Delta X}{X}\right)}_{Max}\times 100\%=2\times 1\%+3\times 2\%+\frac{5}{2}\times 2\%+2\times 4\%$

${\left(\frac{\Delta X}{X}\right)}_{Max}\times 100\%=2\%+6\%+5\%+8\%$

${\left(\frac{\Delta X}{X}\right)}_{Max}\times 100\%=21\%$

Thus,

The percentage error in the value of X is 21%.

Hence option A) 21% is the correct option.