Question

A physical quantity $z$ depends on four observables $a,b,c$ and $d$, as $z=\left[\frac{a^2{b}^{{\displaystyle \frac{2}{3}}}}{\sqrt{c}{d}^3}\right]$. The percentages of error in the measurement of $a,b,c$ and $d$ are $2\%,1.5\%,4\%,$ and $2.5\%$ respectively. The percentage of error in $z$ is

• A. $16.5\%$
• B. $12.25\%$
• C. $13.5\%$
• D. $14.5\%$

Answer 1

The correct option is D

$14.5\%$

Step 1. Given Data:

$z=\left[\frac{a^2{b}^{{\displaystyle \frac{2}{3}}}}{\sqrt{c}{d}^3}\right]$

and percentages of error in the measurement of $a,b,c$ and $d$ are $2\%,1.5\%,4\%,$ and $2.5\%$ respectively.

Step 2. Find the percentage error:

The above relation can also be written as:

$z=a^2{b}^{\frac{2}{3}}{c}^{-\frac{1}{2}}{d}^{-3}$

The percentage error is given by the formula:

$100\times \left(\frac{dz}{z}\right)=[2(\frac{da}{a})+(\frac{2}{3})(\frac{db}{b})+(\frac{1}{2})(\frac{dc}{c})+3\frac{d\left(d\right)}{d}]\times 100$

$=\left[\left(2\times 2\right)+\left(\frac{2}{3}\times 1.5\right)+\left(\frac{1}{2}\times 4\right)+\left(3\times 2.5\right)\right]$

$=\left[4+1+2+7.5\right]\%$

$=14.5\%$

Hence option (D) is correct.