Manipal Engineering Manipal Engineering Solved Paper-2014

  • question_answer
    Two substances of densities \[{{x}^{2}}-{{y}^{2}}={{a}^{2}}\]and \[{{y}^{2}}=4ax.\]are mixed in equal volume and the relative density of mixture is 4. When they are mixed in equal masses, the relative density of the mixture is 3. The values of \[{{y}^{2}}=(x-a){{x}^{3}}\] and \[{{y}^{2}}(x-a)={{x}^{3}}\]are

    A) \[{{x}^{2}}(x-a)={{x}^{3}}\]and \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{\sin x}}-1}{x}\]               

    B) \[(p\wedge q)\vee (q\Leftrightarrow p)\]and \[\left| \begin{matrix}    {{e}^{a}} & {{e}^{2a}} & {{e}^{3a}}-1  \\    {{e}^{b}} & {{e}^{2b}} & {{e}^{3a}}-1  \\    {{e}^{c}} & {{e}^{2c}} & {{e}^{3c}}-1  \\ \end{matrix} \right|\]

    C)  \[fog=|\sin x|\]and \[gof={{\sin }^{2}}\sqrt{x},\]            

    D) None of these

    Correct Answer: A

    Solution :

    When substances are mixed in equal volume then density \[VT\left( \frac{1}{r}-\frac{1}{R} \right)\] \[VT\left( \frac{1}{{{r}^{2}}}-\frac{1}{{{R}^{2}}} \right)\]\[{{\rho }_{1}}\]                                ...(i) When substances are mixed in equal masses then density \[{{\rho }_{2}}\] \[{{\rho }_{1}}\]\[{{\rho }_{2}}\]                                               ...(ii) On solving (i) and (ii) we get\[{{\rho }_{1}}=6\] and \[{{\rho }_{2}}=2\]


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