CET Karnataka Medical CET - Karnataka Medical Solved Paper-2002

  • question_answer
    Light passes successively through two polarimeter tubes each of length 0.29 m. The first tube contains dextro rotatory solution of concentration \[60kg{{m}^{-3}}\]and specific rotation \[0.01\text{ }rad\text{ }{{m}^{2}}k{{g}^{-1}}.\] The second tube contains laevorotatory solution of concentration \[30\text{ }kg\text{ }{{m}^{-5}}\] and specific rotation \[0.02\text{ }rad\text{ }{{m}^{2}}k{{g}^{-1}}.\] The net rotation produced is:

    A) \[0{}^\circ \]

    B) \[15{}^\circ \]

    C) \[10{}^\circ \]

    D) \[20{}^\circ \]

    Correct Answer: A

    Solution :

     Specific rotation : S of a substance at a given temperature t for a given wavelength of light is defined as the rotation (in degree) produced by 1 decimeter length of substance in solution when its concentration in 1 gm/cc \[\therefore \] \[\] Q = rotation in degree I = length of solution in decimeter C = concentration of solution in gm/cc. \[{{C}_{1}}=60\times {{10}^{-2}}gm/cc\] \[{{S}_{1}}=0.01\times {{10}^{-1}}rad\,d{{m}^{2}}k{{g}^{-1}}\] \[{{l}_{1}}=2.9d\] \[{{Q}_{1}}={{S}_{1}}{{C}_{1}}{{l}_{1}}=(7.74\times {{10}^{-4}})\]degree \[{{Q}_{2}}={{S}_{2}}{{C}_{2}}{{l}_{2}}\] \[=0.02\times {{10}^{-1}}\times 3\times {{10}^{-2}}\times 2.9\] \[=1.74\times {{10}^{-4}}\]degree Net rotation \[\theta ={{\theta }_{1}}+{{\theta }_{2}}\] \[=3.84\times {{10}^{-4}}\] \[\therefore \] \[={{0.000384}^{o}}\] \[\therefore \] \[\theta ={{0}^{o}}\]


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