Question

If the formula for the physical quantity is $W = \dfrac{{{a^3}{b^3}}}{{{c^{\dfrac{1}{3}}}{d^{\dfrac{1}{2}}}}}$ and if percentage errors in the measurement of $a,b,c$ and $d$ are $1\% ,3\% ,3\%$ and $4\%$ respectively, Find the percentage error in $W$.

Answer 1

Percent error is defined as the difference between the actual value to the estimated value in comparison to an actual value and it is expressed as a percentage. In other words, we can say that the percent error is the relative error multiplied by 100. The error can be because of the measuring errors (tools or a human error) or because of the approximations or by the assumption used in calculating (rounding errors, for example). Regardless,
Steps to Calculate the Percent Error
Subtract an accepted value from the experimental value. Divide that answer by the accepted value. Multiply the answer we got by 100 and add the % symbol to express the answer as a percentage.

Complete step by step solution:
In the question they have given the equation that is
$W = \dfrac{{{a^4}{b^3}}}{{{c^{\dfrac{1}{3}}}{d^{\dfrac{1}{2}}}}}$
Now we have to calculate the percentage error therefore the above equation becomes
$\dfrac{{\Delta W}}{W} = \dfrac{{4\dfrac{{\Delta a}}{a}\% \times 3\dfrac{{\Delta b}}{b}\% }}{{\dfrac{1}{3}\dfrac{{\Delta c}}{c}\% \times \dfrac{1}{2}\dfrac{{\Delta d}}{d}\% }}$
Now we have to convert the above equation into percentage so we have to multiply $100$
$\dfrac{{\Delta W}}{W} \times 100 = \dfrac{{4\dfrac{{\Delta a}}{a}\% \times 3\dfrac{{\Delta b}}{b}\% }}{{\dfrac{1}{3}\dfrac{{\Delta c}}{c}\% \times \dfrac{1}{2}\dfrac{{\Delta d}}{d}\% }} \times 100$
In the question they have also given the percentage errors in the measurement of $a,b,c$ and $d$ are $1\% ,3\% ,3\%$ and $4\%$ respectively,
After applying it to the above equation
$\dfrac{{\Delta W}}{W}\% = \dfrac{{4 \times 1 \times 3 \times 3}}{{\dfrac{1}{3} \times 3 \times \dfrac{1}{2} \times 4}} \times 100$
After further simplification of the above equation we get
$\dfrac{{\Delta W}}{W}\% = 0.18 \times 100$
After multiplying $0.18$ with $100$ we get
$\dfrac{{\Delta W}}{W}\% = 18\%$

Hence, the percentage error in $W$ is $18\%$

Note: A physical quantity is defined as a property of a material or system that can be quantified by a measurement. A physical quantity can be expressed as a combination of the numerical value and a unit. For example, the physical quantity length can be quantified as $Xm$, where $X$ is the numerical value and $m$ is the unit.