•  It's given that each side of Q is two times that of P.

•  Let's take the side of cube P as ‘a’. Then Volume of cube P is  x = a3

•  Then the side of cube Q as ‘2a’. Then Volume of cube Q is

  y = (2a)3 = 8a3

•  Also, we know mass of Q is two times that of P

Mass of P= m

Mass of Q, n = 2m

•  According to question: u = m/x = m/ (a3)

•  v = n/y = 2m/ (8a3) = m/ (4a3) = {m / (a3)} / 4 = u/4

•  v = u/4

Option A: 1/3_
To solve this problem, we need to calculate the volume of both the overhead tank and the underground tank, and then determine how much of the underground tank remains filled after transferring water to the overhead tank.
     Step 1: Calculate the volume of the cylindrical overhead tank.
     The formula for the volume of a cylinder is given by:
     \[
     V = \pi r^2 h
     \]
     where \( r \) is the radius and \( h \) is the height.
     For the overhead tank:
         ○ Radius \( r = 2 \) m
         ○ Height \( h = 7 \) m
     \[
     V_{\text{overhead}} = \pi \times (2)^2 \times 7 = 28\pi \, \text{m}^3
     \]
     Step 2: Calculate the volume of the underground tank.
     The underground tank is a rectangular prism, so its volume is calculated as:
     \[
     V = \text{length} \times \text{width} \times \text{height}
     \]
     For the underground tank:
         ○ Length = 5.5 m
         ○ Width = 4 m
         ○ Height = 6 m
     \[
     V_{\text{underground}} = 5.5 \times 4 \times 6 = 132 \, \text{m}^3
     \]
     Step 3: Determine the remaining volume in the underground tank.
     After filling the overhead tank, the volume of water removed from the underground tank is \( 28\pi \, \text{m}^3 \).
     The remaining volume in the underground tank is:
     \[
     V_{\text{remaining}} = V_{\text{underground}} - V_{\text{overhead}} = 132 - 28\pi
     \]
     Step 4: Calculate the portion of the underground tank still filled with water.
     To find the portion of the underground tank still filled, we divide the remaining volume by the total volume of the underground tank:
     \[
     \text{Portion filled} = \frac{V_{\text{remaining}}}{V_{\text{underground}}} = \frac{132 - 28\pi}{132}
     \]
     Approximating \(\pi \approx 3.14\):
     \[
     28\pi \approx 28 \times 3.14 = 87.92
     \]
     \[
     V_{\text{remaining}} \approx 132 - 87.92 = 44.08
     \]
     \[
     \text{Portion filled} \approx \frac{44.08}{132} \approx \frac{1}{3}
     \]
     Therefore, the correct answer is Option A: 1/3.