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JCECE Engineering JCECE Engineering Solved Paper-2014

  • question_answer
    Cathode rays of velocity \[{{10}^{6}}m{{s}^{-1}}\] describe an approximate circular path of radius \[1\,\,m\] in an electric field\[300\,\,Vc{{m}^{-1}}\]. If the velocity of the cathode rays are doubled. The value of electric field so that the rays describe the same circular path, will be

    A) \[2400\,\,V\,\,c{{m}^{-1}}\]                       

    B) \[600\,\,V\,\,c{{m}^{-1}}\]

    C) \[1200\,\,V\,\,c{{m}^{-1}}\]                       

    D)  \[12000\,\,V\,\,c{{m}^{-1}}\]

    Correct Answer: C

    Solution :

    Cathode rays are composed of electrons, when they move in electric field a force                 \[F=eE\]                                                               ... (i) Its acts on them and provides the necessary centripetal force the particles                 \[F=\frac{m{{v}^{2}}}{r}\]                                            ? (ii) From Eqs. (i) and (ii) we get                 \[eE=\frac{m{{v}^{2}}}{r}\]                 \[r=\frac{m{{v}^{2}}}{eE}=\frac{m{{({{10}^{6}})}^{2}}}{e(300)}\]                                ... (iii) when, velocity is doubled same circular path is followed. Hence, radius is same                 \[r=\frac{m{{(2\times {{10}^{6}})}^{2}}}{eE}\]                                    ... (iv) Equating Eqs. (iii) and (iv) we get                 \[\frac{m\times {{({{10}^{6}})}^{2}}}{e(300)}=\frac{m\times {{(2\times {{10}^{6}})}^{2}}}{eE}\]                 \[E=300\times 4=1200\,\,V/m\]


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