Question

A physical quantity $P$ is described by the relation $p = {a^{\dfrac{1}{2}}}{b^2}{c^3}{d^{ - 4}}$. If the relative errors in the measurement of $a,b,c$and $d$respectively $2,1,3$and $5\%$, then the relative error in $P$ will be:
A. $8\%$
B. $25\%$
C. $12\%$
D. $32\%$

Answer 1

Hint Calculate the relative error by multiplying the powers of each of the variables with their percentage errors and adding their individual values.

Complete step by step answer
Percent error is calculated by dividing the difference between measured and known value with the known value, multiplied by $100\%$
Relative error is connected with the notion of correct significant digits or correct significant figures. The significant digits in a number are the first nonzero digit and all succeeding digits.
It gives us an idea about the size of the error in a particular physical quantity. The relative error does not have a unit because it is a ratio of the same physical quantity. Since it is a proportion, so we can express it as a percentage by multiplying the relative error by $100\%$
On the other hand absolute error is just the difference between the measured value and the true value unlike the relative error which is a ratio. So the absolute error has a unit.
The relative error in $P$ is given as
$\dfrac{{\Delta P}}{P} = \dfrac{1}{2}\dfrac{{\Delta a}}{a} + 2\dfrac{{\Delta b}}{b} + 3\dfrac{{\Delta c}}{c} + 4\dfrac{{\Delta d}}{d} \\\ \Rightarrow \dfrac{{\Delta P}}{P} = \left( {\dfrac{1}{2} \times 2} \right) + \left( {2 \times 1} \right) + \left( {3 \times 3} \right) + \left( {4 \times 5} \right) \\\ \Rightarrow \dfrac{{\Delta P}}{P} = (1 + 2 + 9 + 20) \\\ \Rightarrow \dfrac{{\Delta P}}{P} = 32\% \\\$
Therefore the relative error in $P$ is $32\%$

So, the correct answer is D.

Note Relative error has two features- Its value becomes undefined when the true value of the variable is zero.
-Relative error is only applicable when measured on a ratio scale.