Q81 / 90 · JEE Main 2022 (25 Jun Shift 2) · Physics

For $z = a^{2} x^{3} y^{\frac{1}{2}}$ , where ' $a$ ' is a constant. If percentage error in measurement of ' $x$ ' and ' $y$ ' are $4 %$ and $12 %$ , respectively, then the percentage error for ' $z$ ' will be _____ $%$ .

Correct answer: 18

Solution

Given: $z = a^{2} x^{3} y^{\frac{1}{2}}$

Relative error is given by

$\frac{∆ z}{z} = [3 \times \frac{∆ x}{x} + \frac{1}{2} \times \frac{∆ y}{y}]$

Now, maximum percentage error in $z$ is

$\frac{∆ z}{z} \times 100 = [3 (4) + \frac{1}{2} (2)] \times 100$

$= 18 %$ .

Hence, the correct answer is 18.