NEET Sample Paper NEET Sample Test Paper-14

  • question_answer Two identical capacitors 1 and 2 are connected in series to a battery as shown in the figure. Capacitor 2 contains a dielectric slab of dielectric constant K as shown. \[{{Q}_{1}}\]and \[{{Q}_{2}}\]are the charges stored in-the capacitors. Now the dielectric slab is removed and the corresponding charges are \[Q_{1}^{'}\]and\[Q_{2}^{'}.\]Then:

    A)  \[\frac{Q_{1}^{'}}{{{Q}_{1}}}=\frac{K+1}{K}\]                    

    B)  \[\frac{Q_{2}^{'}}{{{Q}_{2}}}=\frac{K+1}{2}\]

    C)  \[\frac{Q_{2}^{'}}{{{Q}_{2}}}=\frac{K+1}{2K}\]                  

    D)  \[\frac{Q_{1}^{'}}{{{Q}_{1}}}=\frac{K}{2}\]

    Correct Answer: C

    Solution :

    Let C be the capacity of each condenser without slab. \[\therefore \] When slab from capacitor 2 is removed, net capacity \[=C/2\therefore \,Q_{2}^{'}=Q_{2}^{'}=\frac{{{C}_{\varepsilon }}}{2}\] Before the slab is removed\[{{C}_{1}}=C\] and \[{{C}_{2}}=KC\]             \[{{C}_{Net}}=\frac{KC}{K+1}\]             \[{{Q}_{1}}={{Q}_{2}}=\frac{K{{C}_{\varepsilon }}}{K+1}\]             \[\therefore \]      \[\frac{Q_{2}^{'}}{{{Q}_{2}}}=\frac{{{C}_{\varepsilon }}(K+1)}{2(K{{C}_{\varepsilon }})}=\frac{K+1}{2K}.\] Hence, the correction option is [c].


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