Question

A capacitor $C_{1}$ of capacitance $C$ is charged to a potential difference $V_{0}$. The terminals of the charged capacitor are then connected to an uncharged capacitor $C_{2}$ of capacitance $C / 2$.

Column I		Column II
A	Final energy of capacitor $C_{1}$	1	$-(1 / 6) C V_{0}^{2}$
B	Final energy of capacitor $C_{2}$	2	$(1 / 6) C V_{0}^{2}$
C	Final energy of the system	3	$(1/3) C V_{0}^{2}$
D	Change in energy on joining the capacitors	4	$(2/9) C V_{0}^{2}$
		5	$(1 / 9) C V_{0}^{2}$

• A. A-(1), B-(2), C-(3), D-(5)
• B. A-(4), B-(5), C-(3), D-(1)
• C. A-(5), B-(4), C-(3), D-(1)
• D. A-(1), B-(2), C-(3), D-(4)

Answer 1

Answer: (B)

$A \rightarrow 4 ; B \rightarrow 5 ; C \rightarrow 3 ; D \rightarrow 1$
As, $\frac{Q_{1}}{C_{1}}=\frac{Q_{2}}{C_{2}}, Q_{2}=\left(\frac{C_{2}}{C_{1}}\right) Q_{1}=\frac{(C / 2)}{C} Q_{1}=\frac{Q_{1}}{2}$
Further, as
$Q=Q_{1}+Q_{2}=Q_{1}+\frac{Q_{1}}{2}$
$Q_{1}=\frac{2}{3} Q \text { and } Q_{2}=\frac{1}{3} Q$
$U_{1} =\frac{Q_{1}^{2}}{2 C_{1}}=\frac{\left(\frac{2}{3} Q\right)^{2}}{2 C}=\frac{2}{9}\left(\frac{Q^{2}}{C}\right)$
$=\frac{2}{9}\left(\frac{C^{2} V_{0}^{2}}{C}\right)=\frac{2}{9} C V_{0}^{2}$ $\left(\text { as } Q=C V_{0}\right)( A \rightarrow 4)$
$( B \rightarrow 5) U_{2}=\frac{Q_{2}^{2}}{2 C_{2}}=\frac{\left(\frac{1}{3} Q\right)^{2}}{2(C / 2)}$
$=\frac{1}{9}\left(\frac{Q^{2}}{C}\right)=\frac{1}{9} C V_{0}^{2}$
$( C \rightarrow 3) U_{\text {final }}=U_{1}+U_{2}=\frac{1}{3} C V_{0}^{2}$
$( D \rightarrow 1) U_{\text {initial }}=\frac{Q^{2}}{2 C}=\frac{\left(C V_{0}\right)^{2}}{2 C}=\frac{1}{2} C V_{0}^{2}$
Change in energy, $\Delta U=-\frac{1}{6} C V_{0}^{2}$