Step-by-step Solution:

1. Modeling the Capacitor System The problem describes a parallel plate capacitor partially filled with a dielectric liquid whose level is changing. This setup can be modeled as two capacitors connected in series.

   • One capacitor ($C_1$) is formed by the part of the volume filled with the liquid dielectric.
   • The other capacitor ($C_2$) is formed by the part of the volume filled with air (or vacuum, with dielectric constant $K_{air} = 1$).

   The plates are parallel with area A=1 and separation d. Let x be the thickness of the liquid layer at any given time t. The thickness of the air layer will then be d-x.

2. Capacitance of the Components

   • The capacitance of the liquid-filled part ($C_1$) with dielectric constant K=2 and thickness x is: $C_1 = {{K{\varepsilon _0}A} \over x} = {{2{\varepsilon _0}(1)} \over x} = {{2{\varepsilon _0}} \over x}$
   • The capacitance of the air-filled part ($C_2$) with dielectric constant $K_{air}=1$ and thickness d-x is: $C_2 = {{K_{air}{\varepsilon _0}A} \over {d - x}} = {{1 \cdot {\varepsilon _0}(1)} \over {d - x}} = {{{\varepsilon _0}} \over {d - x}}$

3. Equivalent Capacitance Since these two components are in series, the reciprocal of the equivalent capacitance $C_{eq}$ is the sum of the reciprocals of the individual capacitances: ${1 \over {{C_{eq}}}} = {1 \over {{C_1}}} + {1 \over {{C_2}}}$ Substituting the expressions for $C_1$ and $C_2$: ${1 \over {{C_{eq}}}} = {x \over {2{\varepsilon _0}}} + {{d - x} \over {{\varepsilon _0}}}$ To add these fractions, we find a common denominator, $2ε_0$: ${1 \over {{C_{eq}}}} = {x \over {2{\varepsilon _0}}} + {{2(d - x)} \over {2{\varepsilon _0}}} = {{x + 2d - 2x} \over {2{\varepsilon _0}}} = {{2d - x} \over {2{\varepsilon _0}}}$ Inverting this gives the equivalent capacitance as a function of x: ${C_{eq}}(x) = {{2{\varepsilon _0}} \over {2d - x}}$

4. Liquid Level as a Function of Time The initial level (thickness) of the liquid is given as x(0) = d/3. The liquid level decreases at a constant speed V. Therefore, the thickness x at time t is given by: $x(t) = x(0) - Vt = {d \over 3} - Vt$

5. Equivalent Capacitance as a Function of Time Substitute the expression for x(t) into the equation for $C_{eq}(x)$: ${C_{eq}}(t) = {{2{\varepsilon _0}} \over {2d - \left( {{d \over 3} - Vt} \right)}}$ ${C_{eq}}(t) = {{2{\varepsilon _0}} \over {2d - {d \over 3} + Vt}}$ Simplify the denominator: ${C_{eq}}(t) = {{2{\varepsilon _0}} \over {{{6d - d} \over 3} + Vt}} = {{2{\varepsilon _0}} \over {{{5d} \over 3} + Vt}}$ To remove the fraction in the denominator, multiply the numerator and denominator by 3: ${C_{eq}}(t) = {{3 \times 2{\varepsilon _0}} \over {3 \times \left( {{{5d} \over 3} + Vt} \right)}} = {{6{\varepsilon _0}} \over {5d + 3Vt}}$

6. Time Constant as a Function of Time The time constant τ of an RC circuit is given by $τ = R C_{eq}$. Substituting the expression for $C_{eq}(t)$: $\tau (t) = R \cdot {{6{\varepsilon _0}} \over {5d + 3Vt}} = {{6{\varepsilon _0}R} \over {5d + 3Vt}}$

7. Conclusion The derived expression for the time constant is $τ(t) = {{6{\varepsilon _0}R} / {5d + 3Vt}}$. This matches option A.

Evaluating the Options:

• A: ${{6{\varepsilon _0}R} \over {5d + 3Vt}}$: This matches our derived result.
• B: ${{(15d + 9Vt){\varepsilon _0}R} \over {2{d^2} - 3dVt - 9{V^2}{t^2}}}$: Incorrect functional form.
• C: ${{6{\varepsilon _0}R} \over {5d - 3Vt}}$: This would correspond to an increasing liquid level, which contradicts the problem statement.
• D: ${{(15d - 9Vt){\varepsilon _0}R} \over {2{d^2} + 3dVt - 9{V^2}{t^2}}}$: Incorrect functional form.