JEE Main & Advanced Sample Paper JEE Main - Mock Test - 34

  • question_answer
    The circuit contains two ideal cells connected as shown in the figure. Initially key is connected to position 1 and then pushed to position 2. Then match the List-1 with appropriate values in List-2.
    List-1 List-2
    P. The final charge on capacitor (in \[\mu C\]) 1. 2
    Q. The work done by 4 volt cell (in \[\mu J\]) 2. 4
    R. The gain in potential energy of capacitor (in \[\mu J\]) 3. 6
    S. The net heat produced in circuit (in (\[\mu J\]) 4. 8

    A)              \[P-4,Q-2,R-1,S-3\]                      

    B) \[P-4,Q-2,\text{ }R-3,\text{ }S-1\]

    C) \[P-2,\text{ }Q-4,\text{ }R-3,S-1\] 

    D) \[~P-2,Q-4,R-1,S-3\]

    Correct Answer: C

    Solution :

    [c] Initial charge on capacitor \[=C\times V=2\times 1=2\mu C\] Final charge on capacitor                                     \[=C\times V=4\times 1=4\mu C\] Net charge crossing cell of emf \[4V=4-2=2\mu C\]. Work done by cell of emf                         \[4V=q.V=2\times 4=8\mu J\] Gain in PE of capacitor \[=\frac{1}{2}C(V_{F}^{2}-V_{1}^{2})\]                         \[=\frac{1}{2}\,(1\mu F)\times 1[{{4}^{2}}-{{2}^{2}}]=6\mu J\] Net heat produced \[=8-6=2\mu J\]                   


You need to login to perform this action.
You will be redirected in 3 sec spinner