Questions & Answers

Question

Answers

A. $0$

B. ${90}^{0}$

C. ${60}^{0}$

D. ${30}^{0}$

Answer
Verified

It is given in the question that the length, l of both the polarimeters is 0.29m, the concentration of the solution, ${C}_{D}$ and the specific rotation, ${S}_{D}$ in the dextro rotatory is $60\,kg{m}^{-3}$ and $0.01\,rad{m}^{2}{kg}^{-1}$ respectively and the concentration of the solution, ${C}_{L}$ and the specific rotation, ${S}_{L}$ in the laevo rotatory is $30\,kg{m}^{-3}$ and $0.02\,rad{m}^{2}{kg}^{-1}$ respectively.

We know that the rotation produced, $\theta$ is given as the product of the length of the polarimeter, l, concentration of the solution, C and the specific rotation, S.

$\theta = lCS$ ----(1)

Now, for net rotation, let us assume the sign convention for the rotations. We consider all the rotation in the dextro rotatory to be positive and all the rotations in the laevo rotatory to be negative.

Therefore, net rotation, $ { \theta }_{ r }\quad =\quad { \theta }_{ D }\quad -\quad { \theta }_{ L }$

where, ${\theta}_{D}$ is the rotation produced in dextro rotatory and ${\theta}_{L}$ is the rotation produced in laevo rotatory.

From equation (1), we can write that,

${ \theta }_{ r }\quad =\quad l{ C }_{ D }{ S }_{ D }\quad -\quad l{ C }_{ L }{ S }_{ L } $

or, ${ \theta }_{ r }\quad =\quad l({ C }_{ D }{ S }_{ D }\quad -\quad { C }_{ L }{ S }_{ L })$

Now, substituting the values of l, ${C}_{D}, {S}_{D}, {C}_{L}, and {S}_{L}$ in the above equation, we get

${ \theta }_{ r }\quad =\quad 0.29[(60\quad \times \quad 0.01)\quad -\quad (30\quad \times \quad 0.02)]$

$\implies { \theta }_{ r }\quad =\quad { 0 }$

Therefore, the net rotation is ${0}$.