A solid object is made of a cone on a hemisphere. The cone's slant height is 10 cm and its curved surface area is 60π cm². The hemisphere's radius is the same as the cone's radius. This object is plac

Here's the solution for the geometry problem:

Given:

• Cylinder height ($H$) = $24 \text{ cm}$
• Cylinder radius ($R$) = $10 \text{ cm}$
• Initial water height ($h_w$) = $14.4 \text{ cm}$
• Solid object: a cone on a hemisphere.
  • Cone slant height ($l$) = $10 \text{ cm}$
  • Cone curved surface area ($\text{CSA}_c$) = $60\pi \text{ cm}^2$
  • Hemisphere radius ($r_h$) = Cone radius ($r_c$)
• Take $\pi = 3.14$

Part 1: Find the rise in the water level.

Step 1: Find the radius of the cone ($r_c$).
The curved surface area of a cone is given by $\text{CSA}_c = \pi r_c l$.

$$60\pi \text{ cm}^2 = \pi \times r_c \times 10 \text{ cm}$$ $$60 = 10 r_c$$ $$r_c = \frac{60}{10} \text{ cm}$$ $$r_c = 6 \text{ cm}$$

Step 2: Find the height of the cone ($h_c$).
Using the Pythagorean theorem for the cone: $l^2 = r_c^2 + h_c^2$.

$$(10 \text{ cm})^2 = (6 \text{ cm})^2 + h_c^2$$ $$100 \text{ cm}^2 = 36 \text{ cm}^2 + h_c^2$$ $$h_c^2 = 100 \text{ cm}^2 - 36 \text{ cm}^2$$ $$h_c^2 = 64 \text{ cm}^2$$ $$h_c = \sqrt{64 \text{ cm}^2}$$ $$h_c = 8 \text{ cm}$$

Step 3: Determine the radius of the hemisphere ($r_h$).
The hemisphere has the same radius as the cone.

$$r_h = r_c = 6 \text{ cm}$$

Step 4: Calculate the volume of the cone ($V_c$).
The volume of a cone is given by $V_c = \frac{1}{3}\pi r_c^2 h_c$.

$$V_c = \frac{1}{3} \times 3.14 \times (6 \text{ cm})^2 \times 8 \text{ cm}$$ $$V_c = \frac{1}{3} \times 3.14 \times 36 \text{ cm}^2 \times 8 \text{ cm}$$ $$V_c = 3.14 \times 12 \times 8 \text{ cm}^3$$ $$V_c = 3.14 \times 96 \text{ cm}^3$$ $$V_c = 301.44 \text{ cm}^3$$

Step 5: Calculate the volume of the hemisphere ($V_h$).
The volume of a hemisphere is given by $V_h = \frac{2}{3}\pi r_h^3$.

$$V_h = \frac{2}{3} \times 3.14 \times (6 \text{ cm})^3$$ $$V_h = \frac{2}{3} \times 3.14 \times 216 \text{ cm}^3$$ $$V_h = 2 \times 3.14 \times 72 \text{ cm}^3$$ $$V_h = 3.14 \times 144 \text{ cm}^3$$ $$V_h = 452.16 \text{ cm}^3$$

Step 6: Calculate the total volume of the solid object ($V_s$).

$$V_s = V_c + V_h$$ $$V_s = 301.44 \text{ cm}^3 + 452.16 \text{ cm}^3$$ $$V_s = 753.60 \text{ cm}^3$$

Step 7: Calculate the rise in water level ($\Delta h$).
The volume of water displaced is equal to the volume of the solid object. This displaced volume forms a cylinder with radius $R$ and height $\Delta h$.
Volume of displaced water = $\pi R^2 \Delta h$.

$$753.60 \text{ cm}^3 = 3.14 \times (10 \text{ cm})^2 \times \Delta h$$ $$753.60 \text{ cm}^3 = 3.14 \times 100 \text{ cm}^2 \times \Delta h$$ $$753.60 \text{ cm}^3 = 314 \text{ cm}^2 \times \Delta h$$ $$\Delta h = \frac{753.60 \text{ cm}^3}{314 \text{ cm}^2}$$ $$\Delta h = 2.4 \text{ cm}$$

The rise in the water level is $\boxed{\text{2.4 cm}}$.

Part 2: Find the amount of water still to be filled to the brim.

Step 1: Calculate the new water level ($h_{new}$).

$$h_{new} = h_w + \Delta h$$ $$h_{new} = 14.4 \text{ cm} + 2.4 \text{ cm}$$ $$h_{new} = 16.8 \text{ cm}$$

(Note: The total height of the solid is $h_c + r_h = 8 \text{ cm} + 6 \text{ cm} = 14 \text{ cm}$. Since $16.8 \text{ cm} > 14 \text{ cm}$, the solid is indeed completely immersed.)

Step 2: Calculate the remaining height to be filled.

$$h_{fill} = H - h_{new}$$ $$h_{fill} = 24 \text{ cm} - 16.8 \text{ cm}$$ $$h_{fill} = 7.2 \text{ cm}$$

Step 3: Calculate the volume of water still to be filled ($V_{fill}$).

$$V_{fill} = \pi R^2 h_{fill}$$ $$$$