MHTCET 13 Apr 2026 Shift 1 Paper

Concept:When a dielectric slab of thickness $t$ is inserted between parallel plates, the effective capacitance depends on the reduced air gap and the dielectric constant $K$.Explanation:Original capacitance of the parallel plate capacitor is given by:$$C_0 = \frac{\varepsilon_0 A}{d}$$With a dielectric slab of thickness $t$, the new capacitance becomes:$$C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$$According to the question,$$C = \frac{7}{6} C_0$$Substituting the expressions:$$\frac{\varepsilon_0 A}{d - t + \frac{t}{K}} = \frac{7}{6} \cdot \frac{\varepsilon_0 A}{d}$$Cancelling $\varepsilon_0 A$ from both sides:$$\frac{1}{d - t + \frac{t}{K}} = \frac{7}{6d}$$Taking reciprocals:$$d - t + \frac{t}{K} = \frac{6d}{7}$$Given $t = \frac{2d}{3}$, we get:$$d - t = d - \frac{2d}{3} = \frac{d}{3}$$Substitute into the equation:$$\frac{d}{3} + \frac{2d}{3K} = \frac{6d}{7}$$Cancelling $d$ from every term:$$\frac{1}{3} + \frac{2}{3K} = \frac{6}{7}$$Multiplying throughout by 3:$$1 + \frac{2}{K} = \frac{18}{7}$$$$\frac{2}{K} = \frac{18}{7} - 1 = \frac{11}{7}$$Therefore,$$K = \frac{14}{11}$$Answer:The dielectric constant of the slab is $\frac{14}{11}$.Correct option: B.