Q132 / 150 · MHT CET 2026 (20 April Shift 2) · Physics
A physical quantity $x$ is related as $x = \sqrt{a}\ b^2 c^3 d^{-4}$. Relative errors in the quantities $a$, $b$, $c$ and $d$ are 2%, 1%, 3% and 4% respectively. Relative error in $x$ will be
- A30%
- Correct answerB28%
- C24%
- D18%
Solution
Given $x = a^{1/2} b^2 c^3 d^{-4}$
The maximum relative percentage error in $x$ is given by
$\dfrac{\Delta x}{x} \times 100 = \dfrac{1}{2} \left(\dfrac{\Delta a}{a} \times 100\right) + 2 \left(\dfrac{\Delta b}{b} \times 100\right) + 3 \left(\dfrac{\Delta c}{c} \times 100\right) + 4 \left(\dfrac{\Delta d}{d} \times 100\right)$
Substituting the given percentage errors:
$\dfrac{\Delta x}{x} \times 100 = \dfrac{1}{2}(2\%) + 2(1\%) + 3(3\%) + 4(4\%)$
$\dfrac{\Delta x}{x} \times 100 = 1\% + 2\% + 9\% + 16\% = 28\%$
Answer: 28%
The maximum relative percentage error in $x$ is given by
$\dfrac{\Delta x}{x} \times 100 = \dfrac{1}{2} \left(\dfrac{\Delta a}{a} \times 100\right) + 2 \left(\dfrac{\Delta b}{b} \times 100\right) + 3 \left(\dfrac{\Delta c}{c} \times 100\right) + 4 \left(\dfrac{\Delta d}{d} \times 100\right)$
Substituting the given percentage errors:
$\dfrac{\Delta x}{x} \times 100 = \dfrac{1}{2}(2\%) + 2(1\%) + 3(3\%) + 4(4\%)$
$\dfrac{\Delta x}{x} \times 100 = 1\% + 2\% + 9\% + 16\% = 28\%$
Answer: 28%
Hence, the correct answer is option: 'B'.