CET Karnataka Engineering CET - Karnataka Engineering 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\[60\,kg\,{{m}^{-3}}\]and specific   rotation 0.01 rad\[{{m}^{2}}\,k{{g}^{-1}}.\]The second tube contains laevo rotatory solution of concentration \[30\,kg\,{{m}^{-5}}\]and specific rotation 0\[0.02\,\]rad \[{{m}^{2}}k{{g}^{-1}}.\]The net rotation produced is:

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

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

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

    D)  \[{{20}^{o}}\]

    Correct Answer: A

    Solution :

    Specific rotation: S of a substance at a given temperature t tor 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 \[1\text{ }gm/cc\] \[\therefore \]  \[\] \[Q=\]rotation in degree \[l=\]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 \[d{{m}^{2}}k{{g}^{-1}}\] \[{{l}_{1}}=2.9\,d\] \[{{Q}_{1}}={{S}_{1}}{{C}_{1}}{{l}_{1}}=(7.74\times {{10}^{-4}})\]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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