Q121 / 160 · AP EAMCET 2022 (07 Jul Shift 2) · Physics
A physical quantity $P$ is related to four observables $a, b, c$ and $d$ as $P=\frac{\sqrt{a b} \cdot d^\alpha}{\sqrt{c}}(\alpha$ is constant) The percentage errors in $a, b, c$ and $d$ are $0.5 \%$ in each. If the percentage error in $P$ is $2 \%$, then $\alpha$ is
- Correct answerA$\frac{5}{2}$
- B$\frac{2}{5}$
- C$\frac{3}{4}$
- D$\frac{3}{2}$
Solution
As, $P=\frac{a^{1 / 2} \cdot b^{1 / 2} \cdot d^\alpha}{c^{1 / 2}}$
So $\%$ error in $P=\frac{\Delta P}{P} \times 100$
$\begin{aligned}
=\frac{1}{2} & \left(\frac{\Delta a}{a}\right) \times 100+\frac{1}{2}\left(\frac{\Delta b}{b} \times 100\right) \\
& +\alpha\left(\frac{\Delta d}{d} \times 100\right)+\frac{1}{2}\left(\frac{\Delta c}{c} \times 100\right)
\end{aligned}$
$\Rightarrow \%$ error in $P=\frac{1}{2} \times \%$ error in $a+\frac{1}{2} \times \%$ error in
$b+\frac{1}{2} \times \%$ error in $c+\alpha \times \%$ error in $d$
Now given,
\% Error in $P=2 \%$
$\%$ Error in $a=\%$ error in $b$
$=\%$ Error in $c=\%$ error in $d=0.5 \%$
Substituting these values in Eq. (i), we get
$\begin{aligned}
2 & =\frac{1}{2} \times 0.5+\frac{1}{2} \times 0.5+\frac{1}{2} \times 0.5+\alpha \times 0.5 \\
\Rightarrow 2 & =\frac{1}{2}\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\alpha\right) \Rightarrow 4=\frac{3}{2}+\alpha
\end{aligned}$
or $\alpha=4-\frac{3}{2}=\frac{5}{2}$
$\therefore \alpha$ is $5 / 2$.
Hence, the correct answer is option: 'A'.