NCERT Solutions for Exercise 13.8 Class 9 Maths Chapter 13 - Surface Area and Volumes

Access Surface Area and Volumes Class 9 Maths Chapter 13 Exercise: 13.8

Q1 (i) Find the volume of a sphere whose radius is 7 cm

Answer:

Given,

The radius of the sphere = $r=7cm$

We know, Volume of a sphere = $\frac{4}{3}\pi r^3$

The required volume of the sphere = $\frac{4}{3}\times \frac{22}{7}\times (7{)}^3$

$=\frac{4}{3}\times 22\times 7\times 7$

$=\frac{4312}{3}$

$=1437\frac{1}{3}c{m}^3$

Q1 (ii) Find the volume of a sphere whose radius is $0.63m$

Answer:

Given,

The radius of the sphere = $r=0.63m$

We know, Volume of a sphere = $\frac{4}{3}\pi r^3$

The required volume of the sphere = $\frac{4}{3}\times \frac{22}{7}\times (0.63{)}^3$

$=4\times 22\times 0.03\times 0.63\times 0.63$

$=1.048m^3$

$=1.05m^3(approx.)$

Q2 (i) Find the amount of water displaced by a solid spherical ball of diameter 28 cm

Answer:

The solid spherical ball will displace water equal to its volume.

Given,

The radius of the sphere = $r=\frac{28}{2}cm=14cm$

We know, Volume of a sphere = $\frac{4}{3}\pi r^3$

$\therefore$ The required volume of the sphere = $\frac{4}{3}\times \frac{22}{7}\times (14{)}^3$

$=\frac{4}{3}\times 22\times 2\times 14\times 14$

$$\begin{gathered} =\frac{34469}{3} \\ =11489\frac{2}{3}c{m}^3 \end{gathered}$$

Therefore, the amount of water displaced will be $11489\frac{2}{3}c{m}^3$

Q2 (ii) Find the amount of water displaced by a solid spherical ball of diameter $0.21m$

Answer:

The solid spherical ball will displace water equal to its volume.

Given,

The radius of the sphere = $r=\frac{0.21}{2}m$

We know, Volume of a sphere = $\frac{4}{3}\pi r^3$

$\therefore$ The required volume of the sphere = $\frac{4}{3}\times \frac{22}{7}\times {\left(\frac{0.21}{2}\right)}^3$

$=4\times 22\times \frac{0.01\times 0.21\times 0.21}{8}$

$=11\times 0.01\times 0.21\times 0.21$

$=0.004851m^3$

Therefore, amount of water displaced will be $0.004851m^3$

Q3 The diameter of a metallic ball is $4.2cm$ . What is the mass of the ball, if the density of the metal is $8.9g$ per $c{m}^3$ ?

Answer:

Given,

The radius of the metallic sphere = $r=\frac{4.2}{2}cm=2.1cm$

We know, Volume of a sphere = $\frac{4}{3}\pi r^3$

$\therefore$ The required volume of the sphere = $\frac{4}{3}\times \frac{22}{7}\times {2.1}^3$

$=4\times 22\times 0.1\times 2.1\times 2.1$

$=38.808c{m}^3$

Now, the density of the metal is $8.9g$ per $c{m}^3$ ,which means,

Mass of $1c{m}^3$ of the metallic sphere = $8.9g$

Mass of $38.808c{m}^3$ of the metallic sphere = $(8.9\times 38.808)g$

$\approx 345.39g$

Q4 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?

Answer:

Given,

Let $d_e$ be the diameters of Earth

$\therefore$ The diameter of the Moon = $d_m=\frac{1}{4}{d}_e$

We know, Volume of a sphere =

$\frac{4}{3}\pi r^3=\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3=\frac{1}{6}\pi d^3$

$\therefore$ The ratio of the volumes = $\frac{VolumeoftheEarth}{VolumeoftheMoon}$

$$\begin{gathered} =\frac{\frac{1}{6}\pi d_e^3}{\frac{1}{6}\pi d_m^3} \\ =\frac{d_e^3}{(\frac{d_e}{4}{)}^3} \\ =64:1 \end{gathered}$$

Therefore, the required ratio of the volume of the moon to the volume of the earth is $1:64$

Q5 How many litres of milk can a hemispherical bowl of diameter $10.5cm$ hold?

Answer:

The radius of the hemispherical bowl = $r=\frac{10.5}{2}cm$

We know, Volume of a hemisphere = $\frac{2}{3}\pi r^3$

The volume of the given hemispherical bowl = $\frac{2}{3}\times \frac{22}{7}\times {\left(\frac{10.5}{2}\right)}^3$

$=\frac{2}{3\times 8}\times 22\times 1.5\times 10.5\times 10.5$

$=303.1875c{m}^3$

The capacity of the hemispherical bowl = $=\frac{303.1875}{1000}\approx 0.303litres(approx.)$

Q6 A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Answer:

Given,

Inner radius of the hemispherical tank = $r_1=1m$

Thickness of the tank = $1cm=0.01m$

$\therefore$ Outer radius = Internal radius + thickness = $r_2=(1+0.01)m=1.01m$

We know, Volume of a hemisphere = $\frac{2}{3}\pi r^3$

$\therefore$ Volume of the iron used = Outer volume - Inner volume

$=\frac{2}{3}\pi r_2^3-\frac{2}{3}\pi r_1^3$

$=\frac{2}{3}\times \frac{22}{7}\times ({1.01}^3-1^3)$

$=\frac{44}{21}\times 0.030301$

$=0.06348m^3(approx)$

Q7 Find the volume of a sphere whose surface area is $154c{m}^2$ .

Answer:

Given,

The surface area of the sphere = $154c{m}^2$

We know, Surface area of a sphere = $4\pi r^2$

$\therefore 4\pi r^2=154$

$$\begin{gathered} \Rightarrow 4\times \frac{22}{7}\times r^2=14\times 11 \\ \Rightarrow r^2=\frac{7\times 7}{4} \\ \Rightarrow r=\frac{7}{2} \\ \Rightarrow r=3.5cm \end{gathered}$$

$\therefore$ The volume of the sphere = $\frac{4}{3}\pi r^3$

$=\frac{4}{3}\times \frac{22}{7}\times (3.5{)}^3$

$=179\frac{2}{3}c{m}^3$

Q8 (i) A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of $Rs4989.60$ . If the cost of white-washing is Rs 20 per square metre, find the inside surface area of the dome

Answer:

Given,

$Rs20$ is the cost of white-washing $1m^2$ of the inside area

$Rs4989.60$ is the cost of white-washing $\frac{1}{20}\times 4989.60m^2=249.48m^2$ of inside area

(i) Therefore, the surface area of the inside of the dome is $249.48m^2$

Q8 (ii) A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of $Rs4989.60$ . If the cost of white-washing is Rs 20 per square metre, find the volume of the air inside the dome.

Answer:

Let the radius of the hemisphere be $rm$

Inside the surface area of the dome = $249.48m^2$

We know, Surface area of a hemisphere = $2\pi r^2$

$$\begin{gathered} \therefore 2\pi r^2=249.48 \\ \Rightarrow r^2=\frac{249.48\times 7}{2\times 22} \\ \Rightarrow r=6.3m \end{gathered}$$

$\therefore$ The volume of the hemisphere = $\frac{2}{3}\pi r^3$

$=\frac{2}{3}\times \frac{22}{7}\times (6.3{)}^3$

$=523.908m^3$

Q9 (i) Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area $S^{\prime }$ . Find the radius $r^{\prime }$ of the new sphere

Answer:

Given,

The radius of a small sphere = $r$

The radius of the bigger sphere = $r^{\prime }$

$\therefore$ The volume of each small sphere= $\frac{4}{3}\pi r^3$

And, Volume of the big sphere of radius $r^{\prime }$ = $\frac{4}{3}\pi r^{\prime 3}$

According to question,

$27\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi r^{\prime 3}$

$$\begin{gathered} \Rightarrow r^{\prime 3}=27\times r^3 \\ \Rightarrow r^{\prime }=3\times r \end{gathered}$$

$\therefore r^{\prime }=3r$

Q9 (ii) Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area $S^{\prime }$ . Find the ratio of S and $S^{\prime }$ .

Answer:

Given,

The radius of a small sphere = $r$

The surface area of a small sphere = $S$

The radius of the bigger sphere = $r^{\prime }$

The surface area of the bigger sphere = $S^{\prime }$

And, $r^{\prime }=3r$

We know, the surface area of a sphere = $4\pi r^2$

$\therefore$ The ratio of their surface areas = $\frac{4\pi r^{\prime 2}}{4\pi r^2}$

$$\begin{gathered} =\frac{(3r{)}^2}{r^2} \\ =9 \end{gathered}$$

Therefore, the required ratio is $1:9$

Q10 A capsule of medicine is in the shape of a sphere of diameter $3.5mm$ . How much medicine (in $m{m}^3$ ) is needed to fill this capsule?

Answer:

Given,

The radius of the spherical capsule = $r=\frac{3.5}{2}$

$\therefore$ The volume of the capsule = $\frac{4}{3}\pi r^3$

$=\frac{4}{3}\times \frac{22}{7}\times (\frac{3.5}{2}{)}^3$

$=\frac{4}{3}\times 22\times \frac{0.5\times 3.5\times 3.5}{8}$

$=22.458m{m}^3\approx 22.46m{m}^3(approx)$

Therefore, $22.46m{m}^3(approx)$ of medicine is needed to fill the capsule.