Surface Areas and Volumes | Exercise 12.2

Question 7

A solid consisting of a right circular cone of height $120\text{ cm}$ and radius $60\text{ cm}$ standing on a hemisphere of radius $60\text{ cm}$ is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is $60\text{ cm}$ and its height is $180\text{ cm}$ .

Solution

Understand the Question

A solid consisting of a cone mounted on a hemisphere is placed inside a cylinder completely full of water.
When the solid is immersed, it displaces an amount of water equal to its own volume.
The volume of water left in the cylinder is: $\text{Volume of water left} = \text{Volume of cylinder} - \text{Volume of solid}$
All three shapes share a common radius $r = 60\text{ cm}$ .

Step 1 · Identify Dimensions and Volume Expressions

Given dimensions:

Radius of cylinder, cone, and hemisphere: $r = 60\text{ cm}$
Height of cone: $h_{\text{cone}} = 120\text{ cm}$
Height of cylinder: $h_{\text{cyl}} = 180\text{ cm}$

Volume formulas:

$\text{Volume of cylinder} = \pi r^2 h_{\text{cyl}} = \pi \times 60^2 \times 180$
$\text{Volume of solid} = V_{\text{cone}} + V_{\text{hemisphere}} = \dfrac{1}{3}\pi r^2 h_{\text{cone}} + \dfrac{2}{3}\pi r^3$

V_{\text{solid}} = \dfrac{1}{3}\pi \times 60^2 \times 120 + \dfrac{2}{3}\pi \times 60^3

Step 2 · Calculate Volume of Water Left

Subtract the solid's volume from the cylinder's volume:

\begin{aligned} V_{\text{left}} &= V_{\text{cylinder}} - V_{\text{solid}} \\[0.6em] &= \pi \times 60^2 \times 180 - \left( \dfrac{1}{3}\pi \times 60^2 \times 120 + \dfrac{2}{3}\pi \times 60^3 \right) \end{aligned}

Factoring out $\pi \times 60^2$ :

V_{\text{left}} = \pi \times 60^2 \left[ 180 - \left( \dfrac{1}{3} \times 120 + \dfrac{2}{3} \times 60 \right) \right]

Since $\dfrac{1}{3} \times 120 = 40$ and $\dfrac{2}{3} \times 60 = 40$ :

\begin{aligned} V_{\text{left}} &= \pi \times 60^2 \left[ 180 - (40 + 40) \right] \\[0.6em] &= \pi \times 60^2 \times (180 - 80) \\[0.6em] &= \pi \times 3600 \times 100 \\[0.6em] &= 360000\pi\text{ cm}^3 \end{aligned}

Using $\pi = \dfrac{22}{7}$ :

\begin{aligned} V_{\text{left}} &= 360000 \times \dfrac{22}{7} \\[0.6em] &= \dfrac{7920000}{7} \\[0.6em] &\approx 1131428.57\text{ cm}^3 \end{aligned}

Step 3 · Convert Volume to Cubic Metres

Since $1\text{ m}^3 = 10^6\text{ cm}^3 = 1000000\text{ cm}^3$ :

\begin{aligned} V_{\text{left}} &= \dfrac{1131428.57}{1000000} \\[0.6em] &\approx 1.131\text{ m}^3 \end{aligned}

Answer

$1.131\text{ m}^3$

Common Mistakes

Unit Conversion Error: Dividing by $100$ or $10000$ instead of $10^6 = 1000000$ when converting $\text{cm}^3$ to $\text{m}^3$ .
Total Height Confusion: Note that the total height of the solid is $h_{\text{cone}} + r_{\text{hemi}} = 120\text{ cm} + 60\text{ cm} = 180\text{ cm}$ , which matches the height of the cylinder perfectly.
Hemisphere Volume Formula: Incorrectly using $\dfrac{4}{3}\pi r^3$ (full sphere) instead of $\dfrac{2}{3}\pi r^3$ for the hemisphere.

Question 7

More questions in Exercise 12.2