(D) The volume of water in the overhead tank equals the volume of water removed from the sump.
Volume of the overhead tank (cylinder) $= \pi r^2 h = 3.14 \times 0.6 \, m \times 0.6 \, m \times 0.95 \, m = 1.07388 \, m^3$.
Volume of the sump (cuboid) $= l \times b \times h = 1.57 \, m \times 1.44 \, m \times 0.95 \, m = 2.14776 \, m^3$.
Volume of water left in the sump $= 2.14776 \, m^3 - 1.07388 \, m^3 = 1.07388 \, m^3$.
Let the height of the water left in the sump be $H$. Since the base area of the sump remains constant,$l \times b \times H = 1.07388 \, m^3$.
$1.57 \, m \times 1.44 \, m \times H = 1.07388 \, m^3$.
$2.2608 \, m^2 \times H = 1.07388 \, m^3$.
$H = \frac{1.07388}{2.2608} \, m = 0.475 \, m = 47.5 \, cm$.
Ratio of capacities $= \frac{\text{Capacity of tank}}{\text{Capacity of sump}} = \frac{1.07388}{2.14776} = \frac{1}{2}$.
Thus,the capacity of the tank is half the capacity of the sump.