KEAM 2021 Physics and Chemistry Solved Paper

Let the air filled capacitor and identical capacitor with dielectric medium having capacitance $C_{1}$ and $C_{2}$ is as shown in the figure belowGiven, dielectric constant, $K=5$ Since, the capacitors are connected in series, so charge on each capacitor remains same, while potential difference is distributed inversely as the ratio of capacitance, i.e. $V_{1}:V_{2}=\frac{1}{C_{1}}:\frac{1}{C_{2}}$ As we know, for a parallel plate capacitor, capacitance, $C=\frac{\varepsilon_{0} A}{d}$, where $A$ is the area of plates of capacitor and $d$ is the distance between the plates. $\therefore$ For air filled capacitor, $C_{1}=\frac{\varepsilon_{0} A}{d}$.....(i) Similarly, $C_{2}=\frac{K \varepsilon_{0} A}{d}$.....(ii) ( $\because$ the capacitors are identical) On dividing Eq. (i) by Eq. (ii), we get $\frac{C_{1}}{C_{2}}=\frac{\frac{\varepsilon_{0} A}{d}}{\frac{K \varepsilon_{0} A}{d}}$ $=\frac{\varepsilon_{0} A}{d} \times \frac{d}{K \varepsilon_{0} A}$ $=\frac{1}{K}$ or $C_{1}:C_{2}=1:K=1:5$ $\Rightarrow V_{1}:V_{2}=\frac{1}{C_{1}}:\frac{1}{C_{2}}=5:1$ OT $\frac{V_{1}}{V_{2}}=\frac{5}{1} \Rightarrow V_{1}=5V_{2}$...(iii) As, $V_{1}+V_{2}=12$ (given) $5V_{2}+V_{2}=12$ [from Eq. (iii)] or $6V_{2}=12$ $\Rightarrow V_{2}=2\,\text{V}$ and $V_{1}=5V_{2}=5\times 2=10\,\text{V}$