Water enters continuously at a constant rate of 12π cm³/s. The total volume of water pumped into the cone to reach a depth of 12 cm is 220π cm³. Dividing this by the filling rate gives the total elapsed time: t = (220π)/(12π) = (55)/(3) seconds, which is 18.33 seconds.

Water enters continuously at a constant rate of 12π cm³/s. The total volume of water pumped into the cone to reach a depth of 12 cm is 220π cm³. Dividing this by the filling rate gives the total elapsed time: t = (220π)/(12π) = (55)/(3) seconds, which is 18.33 seconds.

An empty inverted right circular conical vessel of base radius 12 cm and vertical height 18 cm is being filled with water at a constant rate of $12\pi \text{ cm}^3/\text{s}$ .At the exact instant when the water level reaches a vertical depth of 6 cm, a small spherical lead shot of radius 3 cm is dropped into the cone, sinking completely to the bottom.Find the total time elapsed from the start of filling until the water level rises to a vertical depth of 12 cm.Determine how much the water level rises instantly when the lead shot is dropped.

Answer: Total time elapsed is $\frac{55}{3}\text{ s}$ (or $18\frac{1}{3}\text{ s}$ ), and the water level rises instantly by $(\sqrt[3]{459} - 6)\text{ cm} \approx 1.71\text{ cm}$ .

Step-by-step solution

Step 1: Establish the relationship between radius and height

For an inverted right circular cone with top radius $R = 12\text{ cm}$ and height $H = 18\text{ cm}$ , the ratio of the water surface radius $r$ to the water depth $h$ is constant by similar triangles. Therefore, the radius at any vertical height $h$ is given by $r = \frac{2}{3}h$ .

Step 2: Express the conical volume as a function of depth

Substituting $r = \frac{2}{3}h$ into the formula for the volume of a cone gives the total capacity of the liquid cone as a function of its vertical height $h$ . This gives $V(h) = \frac{4\pi}{27}h^3$ .

Step 3: Calculate the volume of the lead shot and the instantaneous height rise

When $h = 6\text{ cm}$ , the volume of water is $V(6) = \frac{4\pi}{27}(216) = 32\pi\text{ cm}^3$ . The volume of the lead shot with radius $3\text{ cm}$ is $\frac{4}{3}\pi (3)^3 = 36\pi\text{ cm}^3$ . When dropped, the sphere is completely submerged, so the new combined volume is $68\pi\text{ cm}^3$ . Solving $\frac{4\pi}{27}(h')^3 = 68\pi$ yields $(h')^3 = 459$ , so $h' = \sqrt[3]{459}\text{ cm}$ . The instant rise is $\sqrt[3]{459} - 6\text{ cm} \approx 1.71\text{ cm}$ .

Step 4: Calculate total volume of water needed to reach 12 cm depth

The total volume inside the cone up to a vertical height of $12\text{ cm}$ is $256\pi\text{ cm}^3$ . Since the lead shot remains fully submerged at the bottom and occupies $36\pi\text{ cm}^3$ , the volume of water required is $V_{\text{water}} = 256\pi - 36\pi = 220\pi\text{ cm}^3$ .

Step 5: Find the total elapsed time

Water enters continuously at a constant rate of $12\pi\text{ cm}^3/\text{s}$ . The total volume of water pumped into the cone to reach a depth of $12\text{ cm}$ is $220\pi\text{ cm}^3$ . Dividing this by the filling rate gives the total elapsed time: $t = \frac{220\pi}{12\pi} = \frac{55}{3}\text{ seconds}$ , which is $18.33\text{ seconds}$ .

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