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BHU PMT BHU PMT (Mains) Solved Paper-2007

  • question_answer
    Separation between two parallel plates facing each other is 2 cm and surface area\[{{l}^{2}}=100\] \[c{{m}^{2}}\]. If\[{{10}^{6}}\]electrons of velocity\[10{{\,}^{8}}m/s\]projected into the gap between plates of potential difference 400 V, the deflection of an electron is

    A)  17.6 mm                             

    B)  1.76 mm

    C)  0.176 mm          

    D)  zero

    Correct Answer: B

    Solution :

                     The reflection of electron is given by \[y=0+\frac{1}{2}a{{t}^{2}}\] \[=\frac{1}{2}\left( \frac{eE}{m} \right){{\left( \frac{l}{v} \right)}^{2}}\] \[=\frac{eE{{l}^{2}}}{2m{{v}^{2}}}=\frac{eV{{l}^{2}}}{2md{{v}^{2}}}\] \[=\frac{1.6\times {{10}^{-19}}\times 400\times {{10}^{-2}}}{2\times 9.1\times {{10}^{-31}}\times 2\times {{10}^{-2}}\times {{({{10}^{8}})}^{2}}}\] \[=0.176\times {{10}^{-2}}m\] \[=0.176\text{ }cm=1.76\text{ }mm\]


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