Q94 / 100 · BITSAT 2024 (Memory Based Paper 3) · Physics
You measure two quantities as $A=1.0 \mathrm{~m} \pm 0.2 \mathrm{~m}$, $B=2.0 \mathrm{~m} \pm 0.2 \mathrm{~m}$. We should report correct value for $\sqrt{A B}$ as
- A$1.4 \mathrm{~m} \pm 0.4 \mathrm{~m}$
- B$1.41 \mathrm{~m} \pm 0.15 \mathrm{~m}$
- C$1.4 \mathrm{~m} \pm 0.3 \mathrm{~m}$
- Correct answerD$1.4 \mathrm{~m} \pm 0.2 \mathrm{~m}$
Solution
Given, $A=1.0 \mathrm{~m} \pm 0.2 \mathrm{~m}, B=2.0 \mathrm{~m} \pm 0.2 \mathrm{~m}$
Let $Y=\sqrt{A B}=\sqrt{(1.0)(20)}=1.414 \mathrm{~m}$
Rounding off to two significant digits
$\begin{aligned}
Y & =1.4 \mathrm{~m} \\
\frac{\Delta Y}{Y} & =\frac{1}{2}\left[\frac{\Delta A}{A}+\frac{\Delta B}{B}\right] \\
& =\frac{1}{2}\left[\frac{0.2}{1.0}+\frac{0.2}{2.0}\right] \\
& =\frac{0.6}{2 \times 20} \\
\Rightarrow \quad \Delta Y & =\frac{0.6 Y}{2 \times 20}=\frac{0.6 \times 1.4}{2 \times 20}=0.212
\end{aligned}$
Rounding off to one significant digit,
$\Delta Y=0.2 \mathrm{~m}$
Hence, the correct answer is option: 'D'.