NCERT Solutions for Class 9 Maths Chapter 11 Exercise 11.4 - Surface Area and Volumes

Q1 (ii) Find the volume of a sphere whose radius is $\small 0.63\hspace{1mm}m$

Answer:

Given,

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

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

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

$\\ = 4\times22\times 0.03\times 0.63\times 0.63$

$\\ = 1.048\ m^3$

$\\ = 1.05\ m^3\ \ \ (approx.)$

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 = $r = \frac{28}{2}\ cm = 14\ cm$

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

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

$\\ = \frac{4}{3}\times22\times 2\times 14\times 14$

$\\ = \frac{34469}{3} \\ = 11489\frac{2}{3}\ cm^3$

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

Q2 (ii) Find the amount of water displaced by a solid spherical ball of diameter $\small 0.21\hspace{1mm}m$

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 = $\frac{4}{3}\times\frac{22}{7}\times \left(\frac{0.21}{2} \right )^3$

$\\ = 4\times22\times \frac{0.01\times 0.21\times 0.21}{8}$

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

$\\ =0.004851\ m^3$

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

Q3 The diameter of a metallic ball is $\small 4.2\hspace{1mm}cm$ . What is the mass of the ball, if the density of the metal is $\small 8.9\hspace{1mm}g$ per $\small cm^3$ ?

Answer:

Given,

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

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

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

$\\ = 4\times22\times 0.1\times 2.1\times 2.1$

$\\ =38.808\ cm^3$

Now, the density of the metal is $\small 8.9\hspace{1mm}g$ per $\small cm^3$ ,which means,

Mass of $\small 1\ cm^3$ of the metallic sphere = $\small 8.9\hspace{1mm}g$

Mass of $38.808\ cm^3$ of the metallic sphere = $\small (8.9\times38.808)\ g$

$\small \approx 345.39\ g$

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 diameter 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 = $\frac{Volume\ of\ the\ Earth}{Volume\ of\ the\ Moon}$

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

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 $\small 10.5\hspace{1mm}cm$ 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 = $\frac{2}{3}\times\frac{22}{7}\times \left (\frac{10.5}{2} \right )^3$

$= \frac{2}{3\times8}\times22\times1.5\times10.5\times10.5$

$= 303.1875\ cm^3$

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

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 = $r_1 = 1\ m$

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

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

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

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

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

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

$= \frac{44}{21}\times0.030301$

$= 0.06348\ m^3\ \ (approx)$

Q7 Find the volume of a sphere whose surface area is $\small 154\hspace{1mm}cm^2$ .

Answer:

Given,

The surface area of the sphere = $\small 154\hspace{1mm}cm^2$

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

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

$\\ \Rightarrow 4\times\frac{22}{7}\times r^2 = 14\times11 \\ \Rightarrow r^2 = \frac{7\times7}{4} \\ \Rightarrow r = \frac{7}{2} \\ \Rightarrow r = 3.5\ cm$

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

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

$= 179\frac{2}{3}\ cm^3$

Q8 (i) A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of $\small Rs\hspace{1mm}4989.60$ . If the cost of white-washing is Rs 20 per square metre, find the inside surface area of the dome

Answer:

Given,

$\small Rs\hspace{1mm}20$ is the cost of white-washing $1\ m^2$ of the inside area

$\small Rs\hspace{1mm}4989.60$ is the cost of white-washing $\frac{1}{20}\times4989.60\ m^2 = 249.48\ m^2$ of inside area

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

Q8 (ii) A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of $\small Rs\hspace{1mm}4989.60$ . 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 $r\ m$

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

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

$\\ \therefore 2\pi r^2 = 249.48 \\ \Rightarrow r^2 = \frac{249.48\times7}{2\times22} \\ \Rightarrow r = 6.3\ m$

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

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

$= 523.908\ m^3$

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

Answer:

Given,

The radius of a small sphere = $r$

The radius of the bigger sphere = $r'$

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

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

According to question,

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

$\\ \Rightarrow r'^3 = 27\times r^3 \\ \Rightarrow r' = 3\times r$

$\therefore r' = 3r$

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

Answer:

Given,

The radius of a small sphere = $r$

The surface area of a small sphere = $S$

The radius of the bigger sphere = $r'$

The surface area of the bigger sphere = $S'$

And, $r' = 3r$

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

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

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

Therefore, the required ratio is $1:9$

Q10 A capsule of medicine is in the shape of a sphere of diameter $\small 3.5\hspace{1mm}mm$ . How much medicine (in $\small mm^3$ ) 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 = $\frac{4}{3}\pi r^3$

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

$= \frac{4}{3}\times22\times\frac{0.5\times3.5\times3.5}{8}$

$= 22.458\ mm^3 \approx 22.46\ mm^3\ \ (approx)$

Therefore, $22.46\ mm^3\ \ (approx)$ of medicine is needed to fill the capsule.