Surface Areas and Volumes | Exercise 12.2

Question 5

A vessel is in the form of an inverted cone. Its height is $8\text{ cm}$ and the radius of its top, which is open, is $5\text{ cm}$ . It is filled with water up to the brim. When lead shots, each of which is a sphere of radius $0.5\text{ cm}$ are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Solution

Understand the Question

A vessel in the shape of an inverted cone is filled to the brim with water.
When spherical lead shots are immersed in the vessel, the volume of water that overflows is equal to the total volume of all the lead shots dropped (by Archimedes' principle).
Given that $\dfrac{1}{4}$ of the total water flows out: $\text{Volume of } n \text{ lead shots} = \dfrac{1}{4} \times \text{Volume of the cone}$
We find the total number of lead shots $n$ by dividing the volume of displaced water by the volume of a single spherical lead shot.

Step 1 · Find the Volume of the Conical Vessel

For the conical vessel, height $h = 8\text{ cm}$ and radius $r_1 = 5\text{ cm}$ . Diagram 1

\begin{aligned} V_{\text{cone}} &= \dfrac{1}{3} \pi r_1^2 h \\[0.6em] &= \dfrac{1}{3} \times \pi \times (5)^2 \times 8 \\[0.6em] &= \dfrac{1}{3} \times \pi \times 25 \times 8 \\[0.6em] &= \dfrac{200\pi}{3} \text{ cm}^3 \end{aligned}

Step 2 · Calculate the Volume of Displaced Water

Since one-fourth of the water flows out

\begin{aligned} V_{\text{out}} &= \dfrac{1}{4} \times V_{\text{cone}} \\[0.6em] &= \dfrac{1}{4} \times \dfrac{200\pi}{3} \\[0.6em] &= \dfrac{50\pi}{3} \text{ cm}^3 \end{aligned}

Step 3 · Find the Volume of One Lead Shot

Each lead shot is spherical with radius $r_2 = 0.5\text{ cm}$ .

\begin{aligned} V_{\text{shot}} &= \dfrac{4}{3} \pi r_2^3 \\[0.6em] &= \dfrac{4}{3} \times \pi \times (0.5)^3 \\[0.6em] &= \dfrac{4}{3} \times \pi \times 0.125 \\[0.6em] &= \dfrac{0.5\pi}{3} \text{ cm}^3 \end{aligned}

Step 4 · Calculate the Number of Lead Shots

Let $n$ be the number of lead shots dropped.

\begin{aligned} n \times V_{\text{shot}} &= V_{\text{out}} \\[0.6em] n \times \dfrac{0.5\pi}{3} &= \dfrac{50\pi}{3} \\[0.6em] n &= \dfrac{\dfrac{50\pi}{3}}{\dfrac{0.5\pi}{3}} \\[1.1em] &= \dfrac{50}{0.5} \\[0.6em] &= 100 \end{aligned}

Answer

100

Common Mistakes

Premature Numerical Evaluation of $\pi$ : Substituting $\pi \approx \dfrac{22}{7}$ or $3.14$ in intermediate steps leads to tedious computations. Leaving $\pi$ untouched lets it cancel out completely in the final division.
Volume Displaced vs. Total Volume: Equating the total volume of shots to the full cone volume instead of $\dfrac{1}{4}$ of the cone volume.
Cube of Decimals: Incorrectly computing $(0.5)^3$ as $0.125$ or $\left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$ .

Question 5

More questions in Exercise 12.2