A parallel plate capacitor $C$ with plates of unit area and separation $d$ is filled with a liquid of…
A parallel plate capacitor $C$ with plates of unit area and separation $d$ is filled with a liquid of dielectric constant $K=2$. The level of liquid is $\frac{d}{3}$ initially. Suppose the liquid level decreases at a constant speed $v$, the time constant as a function of time $t$ is
$\frac{6 \varepsilon_0 R}{5 d+3 v t}$
$\frac{(15 d+9 v t) \varepsilon_0 R}{2 d^2-3 d v t-9 v^2 t^2}$
$\frac{6 \varepsilon_0 R}{5 d-3 v t}$
$\frac{(15 d-9 v t) \varepsilon_0 R}{2 d^2+3 d v t-9 v^2 t^2}$
Solution
After time $t$, thickness of liquid will remain $\left(\frac{d}{3}-v t\right)$.
Now, time constant as function of time
$
\begin{aligned}
\tau_c & =C R \\
& \left.=\frac{\varepsilon_0(1) \cdot R}{\left(d-\frac{d}{3}+v t\right)+\frac{d / 3-v t}{2}} \quad \quad \text { Applying } C=\frac{\varepsilon_0 A}{d-t+\frac{t}{k}}\right) \\
& =\frac{6 \varepsilon_0 R}{5 d+3 v t}
\end{aligned}
$
$\therefore$ correct option is (a)
