A large vessel is filled with water having dielectric constant $K = 81$ . Two square plates each of length $10 c m$ and separated by $1 c m$ are held parallel to each other with their lower ends just touching the surface of water. The plates are charged by applying a voltage $V$ and then disconnected from the voltage source. The liquid rises by $1 c m$ between the plates. The capacitor has capacitance $C_{0}$ and $C$ without and with water respectively. Then

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$C = 80 p F$

$C = 0.689 n F$

$V = 11100 V$

$C_{0} = 8.85 p F$

answer is A, C, D.

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Detailed Solution

Capacitance without water is

$C_{0} = \frac{ε_{0} A}{d} = \frac{ε_{0} \times 0.01}{0.01} C_{0} = 8.85 \times 10^{- 12} F$

When water rises by $1 c m$

$C = \frac{ε_{0} \times 81 \times 0.001}{0.01} + \frac{ε_{0} (0.01 - 0.001)}{0.01} C = 8.1 ε_{0} + 0.9 ε_{0} = 9 ε_{0} C = 9 (8.85 \times 10^{- 12} F) = 80 p F$

Let $m$ be the mass of water that rises to a height of $1 cm$ , then

$m = (0.001 \times 0.01) \times 1000 = 0.01 k g$

By Law of Conservation of Energy, we have

$\frac{q^{2}}{2 C_{0}} = \frac{q^{2}}{2 C} + \frac{m g h}{2} \frac{q^{2}}{C_{0}} = \frac{q^{2}}{C} + m g h$

Now , $q = C_{0} V$

$\frac{V^{2} C_{0}^{2}}{C_{0}} = \frac{V^{2} C_{0}^{2}}{C} + m g h V^{2} C_{0} (1 - \frac{C_{0}}{C}) = m g h V^{2} ε_{0} (1 - \frac{ε_{0}}{9 ε_{0}}) = 0.01 \times 9.8 \times 0.01 V \approx 11100 V$