JEE PYQ: Mathematics In Physics Question 12
Question 12 - 2020 (05 Sep Shift 1)
A physical quantity $z$ depends on four observables $a$, $b$, $c$ and $d$, as $z = \frac{a^2 b^{2/3}}{\sqrt{c}d^3}$. The percentages of error in the measurement of $a$, $b$, $c$ and $d$ are 2%, 1.5%, 4% and 2.5% respectively. The percentage of error in $z$ is:
(1) 12.25%
(2) 16.5%
(3) 13.5%
(4) 14.5%
Show Answer
Answer: (4)
Solution
Given: $Z = \frac{a^2 b^{2/3}}{\sqrt{c}d^3}$ Percentage error in $Z$, $\frac{\Delta Z}{Z} = \frac{2\Delta a}{a} + \frac{2}{3}\frac{\Delta b}{b} + \frac{1}{2}\frac{\Delta c}{c} + \frac{3\Delta d}{d}$ $= 2 \times 2 + \frac{2}{3} \times 1.5 + \frac{1}{2} \times 4 + 3 \times 2.5 = 14.5%$
BETA
