JEE Main & Advanced Sample Paper JEE Main Sample Paper-39

  • question_answer
    DIRECTIONS: Read the following passage and answer the questions that follows:
    Two metallic plates A and B, each of area  \[5\times {{10}^{-4}}{{m}^{2}},\] are placed parallel to each other at a separation of 1 cm. Plate B carries a positive charge of \[33.7\times {{10}^{-12}}C.\]A mono-chromatic beam of light, with photons of energy 5 eV each, starts falling on plate A at \[t=0\]so that \[{{10}^{6}}\]photons fall on it per square meter per second. Assume that one photoelectron is emitted for every l06 incident photons. Also assume that all the emitted photoelectrons are collected by plate B and the work function of plate A remains constant at the value 2 eV.
    Magnitude of electric field between plates A and B at \[t=10\,\,\sec \]

    A)  1000 V/m                   

    B)  2000 V/m

    C)  1500 V/m                   

    D)  2500 V/m

    Correct Answer: B

    Solution :

     Charge on plate B at t = 10 sec \[{{Q}_{b}}=33.7\times {{10}^{-12}}-5\times {{10}^{7}}\times 1.6\times {{10}^{-19}}\] \[=25.7\times {{10}^{-12}}C\]also \[{{Q}_{a}}=8\times {{10}^{-12}}C\] \[E=\frac{{{\sigma }_{B}}}{2{{\varepsilon }_{0}}}-\frac{{{\sigma }_{A}}}{2{{\varepsilon }_{0}}}=\frac{1}{2A{{\varepsilon }_{0}}}({{Q}_{B}}-{{Q}_{A}})\] \[=\frac{17.7\times {{10}^{-12}}}{5\times {{10}^{-4}}\times 8.85\times {{10}^{-12}}}=2000\text{V/C}\]


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