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# Light passes successively through two polarimeter tubes each of length 0.29 m. The first tube contains a dextrorotatory solution of concentration, $60\, kg{m}^{-3}$ and specific rotation $0.01\, rad{m}^{2}{kg}^{-1}$. The second tube contains a laevorotatory solution of concentration, $30\, kg{m}^{-3}$ and specific rotation $0.02\, rad{m}^{2}{kg}^{-1}$. The net rotation produced is:A. $0$B. ${90}^{0}$C. ${60}^{0}$D. ${30}^{0}$

Hint: The dextrorotation refers to the rotation of the plane-polarized light to the right side and laevorotatory refers to the rotation of the plane-polarized light to the left side.

Formula used: The formula that is used here is $\theta = lCS$.

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 ${0}$.

Hence, the correct answer is option (A).

Note: It is important to note the units of the specific rotation. In this case, the answer was 0, so it doesn't make a difference, but if some other values were given and the answer doesn't come out to be 0, then the required conversions from one unit to another needs to be done in order to obtain the correct answer.