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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2014

  • question_answer
    Let \[\text{15}0\text{ }\mu \text{A}\] and \[\text{5 mA}\]be unit vectors at an angle \[\text{10 mA}\] \[\text{ }\!\!\beta\!\!\text{ }\] from each other. Then, \[\left( \frac{1}{V(volume)} \right)\], if

    A) \[\frac{3}{4}\text{m}/\text{s}\]               

    B) \[\frac{1}{3}\text{m}/\text{s}\]

    C) \[\frac{3}{2}\text{m}/\text{s}\]               

    D) \[\frac{2}{3}\text{m}/\text{s}\]

    Correct Answer: D

    Solution :

                    Given,\[=2.12\times {{10}^{8}}m/s\] \[X=r\tan \theta \]          \[V=\frac{dX}{dt}\] \[=r{{\cos }^{2}}\theta \left( \frac{d\theta }{dt} \right)\]               \[\sqrt{1+1+2\cos 2\alpha }<1\] \[\frac{d\theta }{dt}=\omega \]                \[\sqrt{2(1+\cos 2\alpha )}<1\] \[v=\omega r{{\sec }^{2}}\theta \]          \[\phi ={{45}^{o}}\,\,so,\,\,\theta ={{45}^{o}}\] \[\therefore \]  \[v=0.1\times 3\times {{\sec }^{2}}{{45}^{o}}\] \[=0.1\times 3\times 2=0.6m/s\]              \[{{\left( \text{CD} \right)}^{\text{2}}}={{\left( \text{SC} \right)}^{\text{2}}}-{{\left( \text{SD} \right)}^{\text{2}}}\] \[{{X}^{2}}={{(L)}^{2}}-{{\left( \frac{L}{2} \right)}^{2}}\] \[{{X}^{2}}={{\frac{3L}{4}}^{2}}\]


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