$(A)\;C_0 \bigg( \frac{V_0}{V} \bigg)^{1/n}\\ (B)\;C_0 \bigg[\bigg(\frac{V}{V_0}\bigg )^n +1\bigg] \\ (C)\; C_0 \bigg[\bigg(\frac{V_0}{v}\bigg)^{1/n}-1 \bigg] \\ (D)\;C_0 \bigg[\bigg(\frac{V}{V_0}\bigg)-1 \bigg]^n $

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Potential of C on 1st charge

$V_1 =\large\frac{C_0V_0}{(C+C_0)}$

On second charging ,

$V_2 =\large\frac{C_0V_1}{(C+C_0)}$

$\qquad= \bigg( \large\frac{C_0}{C+C_0}\bigg)^2$$ V_0$

Similarly after $n^{th}$ charging,

$V_n= \bigg( \large\frac{C_0}{C+C_0}\bigg)^n$$ V_0$

=>$ C= C_0 \bigg[\bigg(\frac{V_0}{v}\bigg)^{1/n}-1 \bigg]$

Hence C is the correct answer.

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