Question

A physical quantity $x$ is calculated from the relation $x=a^3{b}^2/\sqrt{cd}$. Calculate percentage error in $x$, if $a,b,c$ and $d$ are measured respectively with an error of $1\%$, $3\%$, $4\%$ and $2\%$.

• A. $12\%$.
• B. $22\%$.
• C. $2\%$.
• D. $0.2\%$.

Answer 1

The correct option is A $12\%$.

$\text{Step 1 : Relative error expression}$

$x=\frac{a^3{b}^2}{\sqrt{cd}}$ $....\left(1\right)$

$\text{Percentage relative error can be determined by}$

$\frac{△x}{x}\times 100\%$

Now taking log both sides of equation $\left(1\right)$

$\log x=\log \left(a^3{b}^2\right)-\log (cd{)}^{1/2}$

$\Rightarrow \log x=3\log a+2\log b-\frac{1}{2}[\log c+\log d]$

Now differentiating both sides, we get Relative Error in $x$ as

$\frac{△x}{x}=\underset{–}{+}[3\frac{△a}{a}+2\frac{△b}{b}+\frac{1}{2}\frac{△c}{c}+\frac{1}{2}\frac{△d}{d}]$ $....\left(2\right)$

$\text{Note:}$ One should remember that all the errors are added even if the formula for physical quantity has negative powers of some quantities in it.

$\text{Step 2 : Percentage error calculation}$

Multiplying by 100 on both sides of equation $\left(2\right)$, We get percentage error in $x$ as:

$\frac{△x}{x}\times 100\%=\underset{–}{+}[3\frac{△a}{a}\times 100\%+2\frac{△b}{b}\times 100\%+\frac{1}{2}\frac{△c}{c}\times 100\%+\frac{1}{2}\frac{△d}{d}\times 100\%]$

$\Rightarrow \frac{△x}{x}\times 100\%=3\times 1+2\times 3+\frac{1}{2}\left(4\right)+\frac{1}{2}\left(2\right)$

$=(3+6+2+1)\%$

$=12\%$

Hence option $A$ is correct.