$A$ physical quantity $X$ is related to four measurable quantities $a, b, c$ and $d$ as follows: $X = a^2b^3c^{\frac{5}{2}}d^{-2}$. The percentage errors in the measurement of $a, b, c$ and $d$ are $1\%$,$2\%$,$3\%$ and $4\%$ respectively. What is the percentage error in quantity $X$? If the value of $X$ calculated on the basis of the above relation is $2.763$,to what value should you round off the result?

(D) Given the physical quantity $X = a^2 b^3 c^{\frac{5}{2}} d^{-2}$.
The maximum percentage error in $X$ is given by the formula:
$\frac{\Delta X}{X} \times 100 = \left[ 2 \left( \frac{\Delta a}{a} \times 100 \right) + 3 \left( \frac{\Delta b}{b} \times 100 \right) + \frac{5}{2} \left( \frac{\Delta c}{c} \times 100 \right) + 2 \left( \frac{\Delta d}{d} \times 100 \right) \right]$
Substituting the given percentage errors:
$= [2(1\%) + 3(2\%) + 2.5(3\%) + 2(4\%)]$
$= [2 + 6 + 7.5 + 8] \% = 23.5 \%$
Thus,the percentage error in $X$ is $23.5 \%$.
To round off the value $2.763$,we look at the relative error,which is $23.5 \% = 0.235$. Since the error is in the first decimal place,we round off the result to two significant figures. Therefore,$2.763$ rounded to two significant figures is $2.8$.