# Exercise E13.8 Surface Areas and Volumes NCERT Solutions Class 9

Go back to  'Surface Areas and Volumes'

## Chapter 13 Ex.13.8 Question 1

Find the volume of a sphere whose radius is

(i) $$7 \rm cm$$

(ii) $$0.63 \rm m$$

### Solution

Reasoning:

Volume of the sphere \begin{align} = \frac{4}{3} \end{align} $$\pi r^{3}$$

What is  known?

What is unknown?

Volume of the sphere.

Steps:

(i) Radius  $$=7 \rm cm$$

Volume of the sphere

\begin{align}&= {\frac{4}{3}} \\ &= \frac{4}{3} \times \frac{{22}}{7} \times {7^3}\\ &= 1437.33\,\,c{m^3} \end{align}

(ii) Radius $$=0.63 \rm m$$

Volume of the sphere

\begin{align}&=\frac{4}{3} \pi r^{3}\\&= \frac{4}{3} \times \frac{{22}}{7} \times {(0.63)^3}\\ &= 1.05\,\,{m^3}\end{align}

Volume ($$i$$) \begin{align} = 1437.33\,\,c{m^3} \end{align}

Volume ($$ii$$) \begin{align} = 1.05\,\,{m^3} \end{align}

## Chapter 13 Ex.13.8 Question 2

Find the amount of water displaced by a solid spherical ball of diameter.

(i) $$28\, \rm{cm}$$

(ii) $$0.21 \,\rm{m}$$

### Solution

Reasoning:

The amount of water displaced by a solid spherical ball is nothing but its volume.

Volume of the sphere \begin{align} = \frac{4}{3} \end{align} $$\pi r^{3}$$

What is known?

Diameter of the sphere.

What is unknown?

Amount of water displaced by the solid.

Steps:

1. Diameter

Radius \begin{align}\,r = \frac{{28}}{2} = 14\,\,cm \end{align}

Amount of water displaced by the solid sphere $$=$$ Volume of the sphere.

\begin{align}&= \frac{4}{3}\pi {r^3} \\&= \frac{4}{3} \times \frac{{22}}{7} \times {(14)^3}\\ &= 11498.66\,\,\,c{m^3} \end{align}

1. Diameter

Radius\begin{align}\,\,\,r = \frac{{0.21}}{2} = 0.105\,\,m \end{align}

Amount of water displaced by the solid sphere $$=$$ Volume of the sphere.

\begin{align}&= \frac{4}{3}\pi {r^3} \\&= \frac{4}{3} \times \frac{{22}}{7} \times {(0.105)^3}\\ &= .004851\,\,{m^3} \end{align}

1. Amount of water displaced\begin{align}\,\,\, = 11498.66\,\,c{m^3} \end{align}
2. Amount of water displaced \begin{align}\,\, = .004851\,\,{m^3} \end{align}

## Chapter 13 Ex.13.8 Question 3

The diameter of a metallic ball is $$4.2\;\rm{ cm}$$. What is the mass of the ball, if the density of the metal is $$8.9$$ g per\begin{align}\,c{m^3} \end{align}?

### Solution

Reasoning:

Volume of the sphere \begin{align} = \frac{4}{3} \end{align} $$\pi r^{3}$$ and mass is the product of density times volume.

What is the known?

Diameter of the sphere and density of the metal per \begin{align}{m^3} \end{align}.

What is the unknown?

Mass of the ball.

Steps:

Diameter

Radius \begin{align}\,r = \frac{{4.2}}{2} = 2.1\,\,cm \end{align}

Volume of the sphere

\begin{align}&= \frac{4}{3} \times \frac{{22}}{7} \times {(2.1)^3}\\ &= 38.808\,\,c{m^3}\\ &= \frac{4}{3}\pi {r^3} \end{align}

Density = $$8.9$$ g per \begin{align}c{m^3} \end{align}

Mass$$=$$volume$$×$$density.

\begin{align} &= 38.808\,\, \times \,8.9\\ &= 345.39\,g \end{align}

Mass of the ball \begin{align} = 345.39\,g \end{align}

## Chapter 13 Ex.13.8 Question 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?

### Solution

Reasoning:

Volume of the sphere \begin{align} = \frac{4}{3} \end{align} $$\pi r^{3}$$. So, the fraction of the volume of the earth is the volume of the moon is the ratio of the volume of the earth to the volume of the moon

What is known?

Ratio between the diameters of moon and earth.

What is unknown?

Fraction of the volume of the earth is the volume of the moon

Steps:

Diameter of the Earth $$=2r$$

Radius of the Earth \begin{align} = \frac{1}{2}(2r) = r \end{align}

Diameter of moon \begin{align} = \frac{1}{4}(2r) = \frac{1}{2}r \end{align}

Radius of the moon \begin{align} = \frac{1}{2}(\frac{r}{2}) = \frac{r}{4} \end{align}

Volume of the earth \begin{align}({v_1}) = \frac{4}{3} \end{align}

Volume of the moon \begin{align}({v_2}) = \frac{4}{3}\pi {(\frac{r}{4})^3} \end{align}

$= \frac{1}{{64}}\left( {\frac{4}{3}\pi {r^3}} \right)$

\begin{align} = \frac{1}{{64}} \end{align} Volume of earth

Hence the volume of the moon is \begin{align}\frac{1}{{64}} \end{align}th fraction of the volume of the earth.

## Chapter 13 Ex.13.8 Question 5

How many liters of milk can a hemispherical bowl of diameter $$10.5\; \rm{cm}$$ hold?

### Solution

Reasoning:

Volume of the hemisphere \begin{align}=\frac{2}{3} \pi r^{3}\end{align}

What is known?

Diameter of the bowl.

What is unknown?

Quantity of milk which the hemispherical bowl can hold.

Steps:

\begin{align}\text{Diameter} &=2 r=10.5 \mathrm{cm} \\ r &=\frac{10.5}{2}=5.25 \end{align}

Amount of water displaced

\begin{align}&= \frac{2}{3}\pi {r^3} \\ &= \frac{2}{3} \times \frac{{22}}{7} \times {(5.25)^3}\\ &= 303.19\,\,c{m^3}\\ &= 0.303\,l\,\,\,\,\,\,\,\,\,\,\,[\because 1000\,\,c{m^3} = 1\,\,l] \end{align}

Answer: $$0.303$$ litres of milk can be held in the bowl.

## Chapter 13 Ex.13.8 Question 6

A hemispherical tank is made up of an iron sheet $$1\rm{cm}$$thick. If the inner radius is $$1\rm{m}$$, then find the volume of the iron used to make the tank.

### Solution

Reasoning:

Volume of the hemisphere \begin{align} = \frac{2}{3} \end{align} $$\pi r^{3}$$ $$3$$ where r will be equal to the summation of thickness of sheet and inner radius of tank.

What is known?

Inner radius and thickness of the iron sheet.

What is unknown?

Volume of the iron used.

Steps:

Inner radius$$(𝑟) = 1 \,\rm{𝑚}$$
Thickness of the sheet $$= 1 \,\rm{𝑐𝑚} = .01 \,\rm{𝑚}$$

Outer radius (R) $$=$$ inner radius$$+$$thickness

\begin{align}={1 \mathrm{m}+0.01 \mathrm{m}=1.01 \mathrm{m}}\end{align}

Volume of the iron used to make the tank

\begin{align}&=\frac{2}{3} \pi\left(R^{3}-r^{3}\right)\\&={\frac{2}{3} \times \frac{22}{7} \times\left[1.01^{3}-1^{3}\right]} \\ &={.06348 \mathrm{m}^{3}}\end{align}

Volume of the iron used\begin{align} = .06348\,\,{m^3} \end{align}.

## Chapter 13 Ex.13.8 Question 7

Find the volume of a sphere whose surface area is \begin{align}154\,\,c{m^2} \end{align}.

### Solution

Reasoning:

Surface area of the sphere = $$4\pi r^2$$
Volume of the sphere=\begin{align}\frac{4}{3}\pi r^3\end{align}

What is known?

Surface area of the sphere.

What is unknown?

Volume of the sphere.

Steps:

Let the radius of the sphere be .

Surface area of the sphere

\begin{align}&={4 \pi r^{2}} \\ {\text { Surface area }}&={154 \mathrm{cm}^{2}} \\ r^{2} &=\frac{154}{4} \times \frac{7}{22} \\ r^{2} &=\frac{49}{4} \\ r &=\frac{7}{2} \mathrm{cm} \end{align}

Volume of the sphere

\begin{align}&={\frac{4}{3} \pi r^{3}} \\ &=\frac{4}{3} \times \frac{22}{7} \times\left(\frac{7}{2}\right)^{3} \\ &=\frac{539}{3} \mathrm{cm}^{3} \end{align}

Volume of the sphere \begin{align} = 179\frac{2}{3}\,\,\,c{m^3} \end{align}

## Chapter 13 Ex.13.8 Question 8

A dome of a building is in the form of a hemisphere. From inside, it was white washed at the cost of Rs $$4989.60$$. If the cost of white washing is Rs $$2$$ per square meter, find the

(i) inside surface area of the dome,

(ii) Volume of the air inside the dome.

### Solution

Reasoning:

Surface area of the sphere $$={4 \pi r^{2}}$$

Volume of the sphere $$={\frac{4}{3} \pi r^{3}}$$

What is known?

Total cost and per square meter.

What is unknown?

Inside surface area of the done.

Volume of the width the doom.

Steps:

Total cost for white wash $$\text { Rs } 4989.6$$

Cost per $$m^{2}{=2.00}$$

Let the radius of the hemisphere be r.

Inside surface area

\begin{align}&={2494.8 \mathrm{m}^{2}} \\ {2 \pi r^{2}}&={2494.8} \\ 2 \times \frac{22}{7} \times r^{2} &=2494.8 \\ r^{2} &=\frac{2494.8 \times 7}{2 \times 22} \\ r^{2} &=396.9 \\ r &=\sqrt{396.9} \\ r &=19.9 m \end{align}

Volume of the air inside the doom $$=$$  Volume of the hemisphere $$={\frac{2}{3} \pi r^{3}}$$

\begin{align}&=\frac{2}{3} \times \frac{22}{7} \times(19.9)^{3} \\ &=16511.73 \mathrm{m}^{3}\end{align}

Inner surface area of the doom $$=2494.8$$ \begin{align}{m^2} \end{align}

Volume of the air inside the doom $$=16511.73$$\begin{align}{m^3} \end{align}

## Chapter 13 Ex.13.8 Question 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

1. Radius $$r'$$ of the new sphere,
2. Ratio of $$S$$ and $$S’$$.

### Solution

Reasoning:

Volume of the sphere \begin{align} = \frac{4}{3} \end{align} $$\pi r^{3}$$

What is known?

Number of spheres melted.

What is unknown?

Ratio of $$S$$ and $$S’$$.

Steps:

Volume of a solid iron sphere $${=\frac{4}{3} \pi r^{3}}$$

Volume of $$27$$ solid spheres $${=27\left(\frac{4}{3} \pi r^{3}\right)}={36 \pi r^{3}}$$

Volume of new sphere $${=36 \pi r^{3}}$$

Let the radius of the new sphere$${{= r' }}$$

Volume of the new sphere $${=\frac{4}{3} \pi r^{3}}$$

According to the question,

\begin{align} \frac{4}{3} \pi r^{\prime 3}&= 36 \pi r^{3} \\ r^{3} &=\frac{\left(36 r^{3}\right)(3)}{4} \\ &=27 r^{3} \\ r^{\prime} &=\sqrt[3]{\left(27 r^{3}\right)} \\ r^{\prime} &=3 r \end{align}

Radius of the new sphere $$r=3r$$

\begin{align}{S}&={4 \pi r^{2}} \\ {s^{1}}&={4 \pi(3 r)^{2}} \\ {\frac{S}{s^{1}}}&={\frac{4 \pi r^{2}}{4 \pi(3 r)^{2}}} \\ {\frac{S}{s^{1}}}&={\frac{1}{3^{2}}}={\frac{1}{9}}\end{align}

So ratio of S:S’ is $$1:9$$

Answer: Radius r of new sphere = $$3r.$$

Ratio of S and S’$$= 1:9$$

## Chapter 13 Ex.13.8 Question 10

A capsule of medicine is in the shape of a sphere of diameter $$3.5 \,\rm{mm}$$.How much Medicine (in \begin{align}m{m^3} \end{align}) is needed to fill this capsule?.

### Solution

Reasoning:

Volume of the sphere \begin{align} = \frac{4}{3} \end{align} $$\pi r^{3}$$

What is known?

Diameter of the capsule.

What is unknown?

Medicine needed to fill the capsule in \begin{align}m{m^3} \end{align}.

Steps:

Volume of the sphere$$=\frac{4}{3} \pi r^{3}$$

Diameter$$=2 r=3.5 \rm{mm}$$

Radius$$=r=\frac{3.5}{2}=1.75 \rm{mm}$$

Capacity of capsule

\begin{align}&=\frac{3}{3} \pi r^{3} \\ &=\frac{4}{3} \times \frac{22}{7} \times(1.75)^{3} \\ &=22.46 \mathrm{mm}^{3} \end{align}

$$22.46\; \mathrm{mm}^{3}$$ of medicine is needed to fill this capsule.