Explanation

(d) Case I : $l_{1} = 20cm,\ \theta_{1} = 38{^\circ}$
Connection $= C_{1}$
$\therefore$ Specific rotation, $\alpha_{1} = \frac{\theta_{1}}{l_{1}C_{1}} = \frac{(38{^\circ})}{20 \times C_{1}}$ Liquid $A$
Case II : $l_{2} = 30cm,\ \theta_{2} = - 24{^\circ}$, concentration $= C_{2}$
Specific rotation $\alpha_{2} = \frac{( - 24{^\circ})}{30 \times C_{2}}$. Liquid $B$
The mixture has 1 part $A$ and 2 part $B$
$\therefore C_{1}^{'}:C_{2}^{'} = 1:2$
$\therefore\theta = \left\lbrack \alpha_{1}C_{1}^{'} + \alpha_{2}C_{2}^{'} \right\rbrack l$
$= \left\{ \frac{38}{20 \times C_{1}} \times \frac{C_{1}}{3} + \frac{( - 24{^\circ})}{30 \times C_{2}} \times \frac{2C_{2}}{3} \right\} \times 30 = 19{^\circ} - 16{^\circ} = + 3{^\circ}$
The new angle of rotation is +3 in the right hand direction