Surface Areas & Volumes
NCERT Solutions • Class 9 Maths • Chapter 11Exercise 11.4
1. Find the volume of a sphere whose radius is: (i) $7 \text{ cm}$ (ii) $0.63 \text{ m}$
(i) $r = 7 \text{ cm}$. Formula: $V = \frac{4}{3} \pi r^3$
$V = \frac{4}{3} \times \frac{22}{7} \times 7 \times 7 \times 7$
$= \frac{4 \times 22 \times 49}{3} = \frac{4312}{3} \approx$ $1437.33 \text{ cm}^3$.
$V = \frac{4}{3} \times \frac{22}{7} \times 7 \times 7 \times 7$
$= \frac{4 \times 22 \times 49}{3} = \frac{4312}{3} \approx$ $1437.33 \text{ cm}^3$.
(ii) $r = 0.63 \text{ m}$.
$V = \frac{4}{3} \times \frac{22}{7} \times (0.63)^3$
$= \frac{4}{3} \times 22 \times 0.09 \times 0.63 \times 0.63 \approx$ $1.05 \text{ m}^3$.
$V = \frac{4}{3} \times \frac{22}{7} \times (0.63)^3$
$= \frac{4}{3} \times 22 \times 0.09 \times 0.63 \times 0.63 \approx$ $1.05 \text{ m}^3$.
2. Find the amount of water displaced by a solid spherical ball of diameter: (i) $28 \text{ cm}$ (ii) $0.21 \text{ m}$
(i) $d = 28 \text{ cm} \Rightarrow r = 14 \text{ cm}$.
Water displaced = Volume of sphere = $\frac{4}{3} \pi r^3$
$V = \frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14 = \frac{34496}{3}$
$\approx$ $11498.66 \text{ cm}^3$.
Water displaced = Volume of sphere = $\frac{4}{3} \pi r^3$
$V = \frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14 = \frac{34496}{3}$
$\approx$ $11498.66 \text{ cm}^3$.
(ii) $d = 0.21 \text{ m} \Rightarrow r = 0.105 \text{ m}$.
$V = \frac{4}{3} \times \frac{22}{7} \times (0.105)^3$
$\approx$ $0.004851 \text{ m}^3$.
$V = \frac{4}{3} \times \frac{22}{7} \times (0.105)^3$
$\approx$ $0.004851 \text{ m}^3$.
3. The diameter of a metallic ball is $4.2 \text{ cm}$. What is the mass of the ball, if the density of the metal is $8.9 \text{ g per cm}^3$?
Step 1: Calculate Volume
$d = 4.2 \text{ cm} \Rightarrow r = 2.1 \text{ cm}$.
$V = \frac{4}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1$
$= 4 \times 22 \times 0.1 \times 2.1 \times 2.1 = 38.808 \text{ cm}^3$.
$d = 4.2 \text{ cm} \Rightarrow r = 2.1 \text{ cm}$.
$V = \frac{4}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1$
$= 4 \times 22 \times 0.1 \times 2.1 \times 2.1 = 38.808 \text{ cm}^3$.
Step 2: Calculate Mass
Mass = Density $\times$ Volume
$= 8.9 \times 38.808$
$\approx$ $345.39 \text{ g}$.
Mass = Density $\times$ Volume
$= 8.9 \times 38.808$
$\approx$ $345.39 \text{ g}$.
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4. The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
Let diameter of earth be $d$. Radius $r_e = \frac{d}{2}$.
Diameter of moon $= \frac{d}{4}$. Radius $r_m = \frac{d}{8}$.
$\text{Ratio} = \frac{V_{moon}}{V_{earth}} = \frac{\frac{4}{3}\pi r_m^3}{\frac{4}{3}\pi r_e^3} = \frac{r_m^3}{r_e^3}$
$= \frac{(d/8)^3}{(d/2)^3} = \frac{d^3/512}{d^3/8} = \frac{8}{512} = \frac{1}{64}$.
The volume of the moon is $\frac{1}{64}$ of the volume of the earth.
Diameter of moon $= \frac{d}{4}$. Radius $r_m = \frac{d}{8}$.
$\text{Ratio} = \frac{V_{moon}}{V_{earth}} = \frac{\frac{4}{3}\pi r_m^3}{\frac{4}{3}\pi r_e^3} = \frac{r_m^3}{r_e^3}$
$= \frac{(d/8)^3}{(d/2)^3} = \frac{d^3/512}{d^3/8} = \frac{8}{512} = \frac{1}{64}$.
The volume of the moon is $\frac{1}{64}$ of the volume of the earth.
5. How many litres of milk can a hemispherical bowl of diameter $10.5 \text{ cm}$ hold?
$r = \frac{10.5}{2} = 5.25 \text{ cm}$.
Volume of Hemisphere $= \frac{2}{3} \pi r^3$
$= \frac{2}{3} \times \frac{22}{7} \times 5.25 \times 5.25 \times 5.25$
$\approx 303.1875 \text{ cm}^3$.
Capacity in Litres $= \frac{303.1875}{1000} \approx$ $0.303 \text{ L}$.
Volume of Hemisphere $= \frac{2}{3} \pi r^3$
$= \frac{2}{3} \times \frac{22}{7} \times 5.25 \times 5.25 \times 5.25$
$\approx 303.1875 \text{ cm}^3$.
Capacity in Litres $= \frac{303.1875}{1000} \approx$ $0.303 \text{ L}$.
6. A hemispherical tank is made up of an iron sheet $1 \text{ cm}$ thick. If the inner radius is $1 \text{ m}$, then find the volume of the iron used to make the tank.
Inner radius $r = 1 \text{ m}$.
Thickness $= 1 \text{ cm} = 0.01 \text{ m}$.
Outer radius $R = 1 + 0.01 = 1.01 \text{ m}$.
Volume of iron $= \frac{2}{3} \pi (R^3 – r^3)$
$= \frac{2}{3} \times \frac{22}{7} \times (1.01^3 – 1^3)$
$= \frac{44}{21} \times (1.030301 – 1) = \frac{44}{21} \times 0.030301$
$\approx$ $0.06348 \text{ m}^3$.
Thickness $= 1 \text{ cm} = 0.01 \text{ m}$.
Outer radius $R = 1 + 0.01 = 1.01 \text{ m}$.
Volume of iron $= \frac{2}{3} \pi (R^3 – r^3)$
$= \frac{2}{3} \times \frac{22}{7} \times (1.01^3 – 1^3)$
$= \frac{44}{21} \times (1.030301 – 1) = \frac{44}{21} \times 0.030301$
$\approx$ $0.06348 \text{ m}^3$.
7. Find the volume of a sphere whose surface area is $154 \text{ cm}^2$.
Step 1: Find Radius
$4 \pi r^2 = 154 \Rightarrow 4 \times \frac{22}{7} \times r^2 = 154$
$r^2 = \frac{154 \times 7}{88} = \frac{49}{4} \Rightarrow r = \frac{7}{2} = 3.5 \text{ cm}$.
Step 2: Find Volume
$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 3.5$
$= \frac{4}{3} \times 22 \times 0.5 \times 12.25 \approx$ $179.67 \text{ cm}^3$.
$4 \pi r^2 = 154 \Rightarrow 4 \times \frac{22}{7} \times r^2 = 154$
$r^2 = \frac{154 \times 7}{88} = \frac{49}{4} \Rightarrow r = \frac{7}{2} = 3.5 \text{ cm}$.
Step 2: Find Volume
$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 3.5$
$= \frac{4}{3} \times 22 \times 0.5 \times 12.25 \approx$ $179.67 \text{ cm}^3$.
8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹ 4989.60. If the cost of white-washing is ₹ 20 per square metre, find the (i) inside surface area of the dome, (ii) volume of the air inside the dome.
(i) Inside Surface Area:
Area $= \frac{\text{Total Cost}}{\text{Rate}} = \frac{4989.60}{20} =$ $249.48 \text{ m}^2$.
Area $= \frac{\text{Total Cost}}{\text{Rate}} = \frac{4989.60}{20} =$ $249.48 \text{ m}^2$.
(ii) Volume of Air:
First find radius: $2 \pi r^2 = 249.48$
$2 \times \frac{22}{7} \times r^2 = 249.48 \Rightarrow r^2 = \frac{249.48 \times 7}{44} = 39.69$
$r = \sqrt{39.69} = 6.3 \text{ m}$.
Volume $= \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (6.3)^3$
$\approx$ $523.9 \text{ m}^3$.
First find radius: $2 \pi r^2 = 249.48$
$2 \times \frac{22}{7} \times r^2 = 249.48 \Rightarrow r^2 = \frac{249.48 \times 7}{44} = 39.69$
$r = \sqrt{39.69} = 6.3 \text{ m}$.
Volume $= \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (6.3)^3$
$\approx$ $523.9 \text{ m}^3$.
9. Twenty seven solid iron spheres, each of radius $r$ and surface area $S$ are melted to form a sphere with surface area $S’$. Find the (i) radius $r’$ of the new sphere, (ii) ratio of $S$ and $S’$.
(i) Volume of 27 spheres = Volume of new sphere.
$27 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (r’)^3$
$27 r^3 = (r’)^3$
$r’ = \sqrt[3]{27 r^3} =$ $3r$.
$27 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (r’)^3$
$27 r^3 = (r’)^3$
$r’ = \sqrt[3]{27 r^3} =$ $3r$.
(ii) Ratio $S : S’$
$\frac{S}{S’} = \frac{4 \pi r^2}{4 \pi (r’)^2} = \frac{r^2}{(3r)^2}$
$= \frac{r^2}{9r^2} = \frac{1}{9}$.
Ratio is $1:9$.
$\frac{S}{S’} = \frac{4 \pi r^2}{4 \pi (r’)^2} = \frac{r^2}{(3r)^2}$
$= \frac{r^2}{9r^2} = \frac{1}{9}$.
Ratio is $1:9$.
10. A capsule of medicine is in the shape of a sphere of diameter $3.5 \text{ mm}$. How much medicine (in $\text{mm}^3$) is needed to fill this capsule?
$r = \frac{3.5}{2} = 1.75 \text{ mm}$.
Volume $= \frac{4}{3} \pi r^3$
$= \frac{4}{3} \times \frac{22}{7} \times 1.75 \times 1.75 \times 1.75$
$\approx$ $22.46 \text{ mm}^3$.
Volume $= \frac{4}{3} \pi r^3$
$= \frac{4}{3} \times \frac{22}{7} \times 1.75 \times 1.75 \times 1.75$
$\approx$ $22.46 \text{ mm}^3$.