Q134 / 180 · COMEDK 2019 · Physics
Specific rotation of sugar solution is $0.01$ SI units. $200 \mathrm{kgm}^{-3}$ of impure sugar solution is taken in a polarimeter tube of length $0.25 \mathrm{~m}$ and an optical rotation of $0.4 \mathrm{rad}$ is observed.
The percentage of purity of sugar in the sample is
- Correct answerA$80 \%$
- B$89 \%$
- C$11 \%$
- D$20 \%$
Solution
Given, specific rotation of sugar solution, $\alpha=0.01$ units
Length of polarimeter tube, $l=0.25 \mathrm{~m}$
Optical rotation, $\theta=0.4 \mathrm{rad}$
Concentration of solution, $c=$ ?
As we know, specific rotation
$=\frac{\text { Optical rotation }}{\text { Length of tube } \times \text { Concentration of sol. }}$
$$
\begin{aligned}
\alpha &=\frac{\theta}{l \times c} \\
c &=\frac{\theta}{l \times \alpha} \\
&=\frac{0.4}{0.01 \times 0.25} \\
&=160 \mathrm{~kg} / \mathrm{m}^{3}
\end{aligned}
$$
Concentration of sugar solution $=200 \mathrm{~kg} / \mathrm{m}^{-3}$ (given)
Percentage impurity of sugar in the sample
$$
\begin{aligned}
&=\frac{160}{200} \times 100 \% \\
&=80 \%
\end{aligned}
$$
Hence, the correct answer is option: 'A'.