A cylindrical overhead tank of radius 2 m and height 7 m is to be filled from an underground tank of size 5.5 m × 4 m × 6 m. How much portion of the underground tank is still filled with water after filling the overhead tank completely ?

The correct option is $\frac{1}{3}$.

To find the portion of the underground tank that remains filled with water after completely filling the cylindrical overhead tank, we need to calculate the volumes of both tanks and find the remaining water volume.

Step 1: Calculate the volume of the underground tank
The underground tank is a cuboid with dimensions length = 5.5 m, width = 4 m, and height = 6 m.
The formula for the volume of a cuboid is:
$V_{\text{underground}}=\text{length}\times \text{width}\times \text{height}$

Substituting the given values:
$V_{\text{underground}}=5.5\times 4\times 6=132{\text{m}}^3$

Step 2: Calculate the volume of the cylindrical overhead tank
The overhead tank is a cylinder with radius $r=2\text{m}$ and height $h=7\text{m}$.
The formula for the volume of a cylinder is:
$V_{\text{overhead}}=\pi r^{2}h$
Using $\pi \approx \frac{22}{7}$:
$V_{\text{overhead}}=\frac{22}{7}\times 2^2\times 7$

Simplifying the expression by cancelling the 7 in the numerator and denominator:
$V_{\text{overhead}}=22\times 4=88{\text{m}}^3$

Step 3: Calculate the volume of water left in the underground tank
The volume of water remaining in the underground tank after filling the overhead tank completely is the difference between the two volumes:
$V_{\text{remaining}}=V_{\text{underground}}-V_{\text{overhead}}$
$V_{\text{remaining}}=132-88=44{\text{m}}^3$

Step 4: Find the portion of the underground tank that is still filled with water
The fraction of the underground tank that remains filled is the ratio of the remaining volume to the total volume of the underground tank:
$\text{Portion remaining}=\frac{V_{\text{remaining}}}{V_{\text{underground}}}$
$\text{Portion remaining}=\frac{44}{132}$

Dividing both the numerator and the denominator by their greatest common divisor, 44:
$\text{Portion remaining}=\frac{1}{3}$

Thus, $\frac{1}{3}$ portion of the underground tank is still filled with water.