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?

The correct option is 63 pF.

Step 1: Understand the geometry of the container and capacitor setup
The container has dimensions:
- Base dimensions: $l=50\text{ cm}$ and $d=5\text{ cm}$
- Height: $H=50\text{ cm}$
- Total volume of the container: $V=50\text{ cm}\times 5\text{ cm}\times 50\text{ cm}=12500{\text{ cm}}^3$

The conducting walls form a parallel-plate capacitor with:
- Plate separation: $d=5\text{ cm}=0.05\text{ m}$
- Total length of the plates along the base: $l=50\text{ cm}=0.5\text{ m}$
- Total height of the plates: $H=50\text{ cm}=0.5\text{ m}$

Step 2: Calculate the height of the liquid after 10 seconds
The liquid is poured at a rate of $\frac{dV}{dt}$.
Volume of liquid added in $t=10\text{ s}$: $V_{\text{liquid}}=250\times 10=2500{\text{ cm}}^3$

Since the cross-sectional area of the base is: $A_{\text{base}}=50\text{ cm}\times 5\text{ cm}=250{\text{ cm}}^2$

The height of the liquid column $h$ inside the container after 10 seconds is: $h=\frac{V_{\text{liquid}}}{A_{\text{base}}}=\frac{2500}{250}=10\text{ cm}=0.1\text{ m}$

Step 3: Analyze the combination of capacitors
The container is now filled up to height $h=10\text{ cm}$ with liquid (dielectric constant $K=3$) and the remaining height $H-h=40\text{ cm}$ is air ($K=1$).
Because both regions share the same potential difference across the conducting plates, this acts as two capacitors connected in parallel: $C=C_1+C_2$

where:
- $C_1$ is the capacitance of the region filled with liquid of height $h=10\text{ cm}$:
$C_1=\frac{K{\epsilon }_0(l\times h)}{d}$
- $C_2$ is the capacitance of the region filled with air of height $H-h=40\text{ cm}$:
$C_2=\frac{{\epsilon }_0(l\times (H-h\left)\right)}{d}$

Step 4: Calculate the total capacitance
Adding $C_1$ and $C_2$: $C=\frac{{\epsilon }_{0}l}{d}[Kh+(H-h\left)\right]$

Substitute the known values:
- ${\epsilon }_0\approx 8.85\times {10}^{-12}\text{ F m}{}^{-1}$
- $l=0.5\text{ m}$
- $d=0.05\text{ m}$
- $Kh+(H-h)=3\times 0.1+0.4=0.3+0.4=0.7\text{ m}$

Now perform the calculation: $C=\frac{8.85\times {10}^{-12}\times 0.5}{0.05}\times 0.7$

Simplifying the fraction: $\frac{0.5}{0.05}=10$

Thus, we get: $C=8.85\times {10}^{-12}\times 10\times 0.7=61.95\times {10}^{-12}\text{ F}\approx 62\text{–}63\text{ pF}$

Rounding to the closest given option, we get 63 pF.