1. The problem describes an inverted cone filled with fluid, with density $\rho = 1000$ kg/m³, gravitational acceleration $g = 9.8$ m/s², cone height $H = 2.5$ m, and top radius $R = 0.7$ m. 2. To analyze the fluid pressure or force at a depth $y$ (measured from the vertex at $y=0$ upwards), we need to relate the radius of the cone at depth $y$ to $y$ itself. 3. The radius varies linearly from 0 at the vertex ($y=0$) to 0.7 m at the top ($y=2.5$ m), so the radius $r(y)$ at depth $y$ is given by $$r(y) = \frac{R}{H} y = \frac{0.7}{2.5} y = 0.28 y.$$ 4. The cross-sectional area $A(y)$ of the fluid surface at height $y$ is the area of a circle with radius $r(y)$: $$A(y) = \pi (r(y))^2 = \pi (0.28 y)^2 = 0.0784 \pi y^2.$$ 5. The fluid pressure at depth $y$ from the top surface is $p(y) = \rho g (H - y)$ since pressure increases with depth. 6. If we need to find force on a horizontal surface at height $y$, it is $F(y) = p(y) A(y) = \rho g (H - y) A(y)$. 7. Substitute known values to get $$F(y) = 1000 \times 9.8 \times (2.5 - y) \times 0.0784 \pi y^2 = 768.3 \pi y^2 (2.5 - y).$$ This formula allows calculation of forces or pressure-related quantities at any height within the cone. Final formula for force as a function of height $y$ is: $$F(y) = 768.3 \pi y^2 (2.5 - y).$$ This provides a complete relationship based on the given cone and fluid parameters.