A container has a base of 50 cm × 5 cm and height 50 cm, as shown in the figure. It has two parallel electrically conducting walls each of area 50 cm × 50 cm. The remaining walls of the container are thin and non-conducting. The container is being filled with a liquid of dielectric constant 3 at a uniform rate of 250 cm3 s −1 . What is the value of the capacitance of the container after 10 seconds? [Given: Permittivity of free space 𝜖0 = 9 × 10−12 C 2N −1m−2 , the effects of the non-conducting walls on the capacitance are negligible] (A) 27 pF (B) 63 pF (C) 81 pF (D) 135 p

 To determine the capacitance of the container after 10 seconds, we need to follow these steps:

1. Calculate the Volume of Liquid Added:
   The liquid is being filled at a rate of $250{\text{cm}}^3\mathrm{/}\text{s}$. After 10 seconds, the volume $V$ of the liquid added is:

   $$V=\text{rate}\times \text{time}=250{\text{cm}}^3\mathrm{/}\text{s}\times 10\text{s}=2500{\text{cm}}^3$$

2. Determine the Height of the Liquid:
   The base area $A$ of the container is $50\text{cm}\times 5\text{cm}=250{\text{cm}}^2$. The height $h$ of the liquid is:

   $$h=\frac{V}{A}=\frac{2500{\text{cm}}^3}{250{\text{cm}}^2}=10\text{cm}$$

3. Calculate the Capacitance:
   The capacitance $C$ of a parallel plate capacitor with a dielectric is given by:

   $$C=\frac{\kappa {ϵ}_{0}A}{d}$$

   where:

   • $\kappa$ is the dielectric constant of the liquid ($\kappa =3$),

   • ${ϵ}_0$ is the permittivity of free space (${ϵ}_0=9\times 10^{-12}{\text{C}}^2\mathrm{/}\text{N}\cdot {\text{m}}^2$),

   • $A$ is the area of the conducting plates ($A=50\text{cm}\times 50\text{cm}=2500{\text{cm}}^2=0.25{\text{m}}^2$),

   • $d$ is the distance between the plates ($d=5\text{cm}=0.05\text{m}$).

   Plugging in the values:

   $$C=\frac{3\times 9\times 10^{-12}\times 0.25}{0.05}=\frac{6.75\times 10^{-12}}{0.05}=135\times 10^{-12}\text{F}=135\text{pF}$$

Therefore, the capacitance of the container after 10 seconds is:

$$\boxed{{\displaystyle 135\text{pF}}}$$