Q147 / 150 · MHT CET 2026 (11 April Shift 2) · Physics
A quantity P is related as $X^{-2}Y^{-3/2}Z^{2/5}$ where X,Y,Z are independent parameters which have fractional errors of $0.1, 0.2$ and $0.5$ respectively in measurement. The maximum fractional error in P is
- Correct answerA$0.7$
- B$0.1$
- C$0.8$
- D$0.6$
Solution
Given $P = X^{-2}Y^{-3/2}Z^{2/5}$
The maximum fractional error in $P$ is given by
$\dfrac{\Delta P}{P} = |-2|\dfrac{\Delta X}{X} + \left|-\dfrac{3}{2}\right|\dfrac{\Delta Y}{Y} + \left|\dfrac{2}{5}\right|\dfrac{\Delta Z}{Z}$
Substituting the given fractional errors $\dfrac{\Delta X}{X} = 0.1$, $\dfrac{\Delta Y}{Y} = 0.2$, and $\dfrac{\Delta Z}{Z} = 0.5$:
$\dfrac{\Delta P}{P} = 2(0.1) + \dfrac{3}{2}(0.2) + \dfrac{2}{5}(0.5)$
$\dfrac{\Delta P}{P} = 0.2 + 0.3 + 0.2 = 0.7$
Answer: $0.7$
The maximum fractional error in $P$ is given by
$\dfrac{\Delta P}{P} = |-2|\dfrac{\Delta X}{X} + \left|-\dfrac{3}{2}\right|\dfrac{\Delta Y}{Y} + \left|\dfrac{2}{5}\right|\dfrac{\Delta Z}{Z}$
Substituting the given fractional errors $\dfrac{\Delta X}{X} = 0.1$, $\dfrac{\Delta Y}{Y} = 0.2$, and $\dfrac{\Delta Z}{Z} = 0.5$:
$\dfrac{\Delta P}{P} = 2(0.1) + \dfrac{3}{2}(0.2) + \dfrac{2}{5}(0.5)$
$\dfrac{\Delta P}{P} = 0.2 + 0.3 + 0.2 = 0.7$
Answer: $0.7$
Hence, the correct answer is option: 'A'.