Exercise 11.4 (Revised) - Chapter 13 - Surface Areas & Volumes - Ncert Solutions class 9 - Maths

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Chapter 11: Surface Areas & Volumes - NCERT Solutions for Class 9 Maths

Assume $\pi=\frac{22}{7}$ unless stated otherwise.
Ex 11.4 Question 1.

Find the volume of a sphere whose radius is (i) $7 \mathrm{~cm}$ and (ii) $0.63 \mathrm{~cm}$.

Answer.

(i) Radius of sphere ${ }^{(r)}=7 \mathrm{~cm}$

Volume of sphere $=\frac{4}{3} \pi r^3$
$=\frac{4}{3} \times \frac{22}{7} \times 7 \times 7 \times 7=\frac{4312}{3}$
$=1437 \frac{1}{3} \mathrm{~cm}^3$
(ii) Radius of sphere ${ }^{(r)}=0.63 \mathrm{~m}$

Volume of sphere $=\frac{4}{3} \pi r^3$
$=\frac{4}{3} \times \frac{22}{7} \times 0.63 \times 0.63 \times 0.63$
$=\frac{4}{3} \times \frac{22}{7} \times \frac{63}{100} \times \frac{63}{100} \times \frac{63}{100}$
$=1.047816 \mathrm{~m}^3=1.05 \mathrm{~m}^3$ (approx.)

Ex 11.4 Question 2.

Find the amount of water displaced by a solid spherical ball of diameter:
(i) $28 \mathrm{~cm}$
(ii) $0.21 \mathrm{~m}$

Answer.

(i) Diameter of spherical ball
$
=28 \mathrm{~cm}
$
$\therefore$ Radius of spherical ball $(r)=\frac{28}{2}$
$
=14 \mathrm{~cm}
$

According to question, Volume of water replaced $=$ Volume of spherical ball
$
\begin{aligned}
& =\frac{4}{3} \pi r^3 \\
& =\frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14=\frac{34496}{3} \\
& =11498 \frac{2}{3} \mathrm{~cm}^3
\end{aligned}
$
(ii) Diameter of spherical ball $=0.21 \mathrm{~m}$
$\therefore$ Radius of spherical ball $(r)$

$
=\frac{0.21}{2} \mathrm{~m}
$

According to question,

Volume of water replaced $=$ Volume of spherical ball
$
\begin{aligned}
& =\frac{4}{3} \pi r^3 \\
& =\frac{4}{3} \times \frac{22}{7} \times \frac{0.21}{2} \times \frac{0.21}{2} \times \frac{0.21}{2} \\
& =\frac{4}{3} \times \frac{22}{7} \times \frac{21}{200} \times \frac{21}{200} \times \frac{21}{200} \\
& =11 \times \frac{441}{100 \times 100 \times 100} \\
& =0.004851 \mathrm{~m}^3
\end{aligned}
$

Ex 11.4 Question 3.

The diameter of a metallic ball is $4.2 \mathrm{~cm}$. What is the mass of the ball, if the metal weighs
$8.9 \mathrm{~g}$ per $\mathrm{cm}^3$ ?

Answer.

Diameter of metallic ball $=4.2 \mathrm{~cm}$
$\therefore$ Radius of metallic ball $(r)=\frac{4.2}{2}$
$=2.1 \mathrm{~cm}$

Volume of metallic ball $=\frac{4}{3} \pi r^3$
$=\frac{4}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1$
$=\frac{4}{3} \times \frac{22}{7} \times \frac{21}{10} \times \frac{21}{10} \times \frac{21}{10}$
$=38.808 \mathrm{~cm}^3$

Density of metal $=8.9 \mathrm{~g}$ per $\mathrm{cm}^3$
$\because$ Mass of $1 \mathrm{~cm}^3=8.9 \mathrm{~g}$
$\therefore$ Mass of $38.808 \mathrm{~cm}^3=8.9 \times 38.808$
$=345.3912 \mathrm{~g}=345.39 \mathrm{~g}$ (approx).

Ex 11.4 Question 4.

The diameter of the moon is approximately one-fourth the diameter of the earth. What fraction is the volume of the moon of the volume of the earth?

Answer.

Let diameter of earth be $x$
$\therefore$ Radius of earth $(r)=\frac{x}{2}$

Now, Volume of earth $=\frac{4}{3} \pi r^3$
$[\because$ Earth is considered to be a sphere]
$=\frac{4}{3} \times \pi \times \frac{x}{2} \times \frac{x}{2} \times \frac{x}{2}=\frac{1}{8} \times \frac{4}{3} \pi x^3$

According to question,

Diameter of moon $=\frac{1}{4} \times$ Diameter of earth $=\frac{1}{4} \times x=\frac{x}{4}$
$\therefore$ Radius of moon $(R)=\frac{x}{8}$

Now, Volume of Moon $=\frac{4}{3} \pi \mathrm{R}^3$
[ $\because$ Moon is considered to be a sphere]
$
=\frac{4}{3} \times \pi \times \frac{x}{8} \times \frac{x}{8} \times \frac{x}{8}=\frac{1}{512} \times \frac{4}{3} \pi x^3
$

$
=\frac{1}{64} \times\left[\frac{1}{8} \times \frac{4}{3} \pi x^3\right]=\frac{1}{64} \times \text { Volume of Earth }
$
[From eq. (i)]
$\therefore$ Volume of moon is $\frac{1}{64}$ th the volume of earth.

Ex 11.4 Question 5.

How many litres of milk can a hemispherical bowl of diameter $\mathbf{1 0 . 5}$ hold?

Answer.

Diameter of hemispherical bowl
$
=10.5 \mathrm{~cm}
$
$\therefore$ Radius of hemispherical bowl $(r)$
$
=\frac{10.5}{2}=5.25 \mathrm{~cm}
$

Volume of milk in hemispherical bowl
$
\begin{aligned}
& =\frac{2}{3} \pi r^3 \\
& =\frac{2}{3} \times \frac{22}{7} \times 5.25 \times 5.25 \times 5.25 \\
& =\frac{2}{3} \times \frac{22}{7} \times \frac{525}{100} \times \frac{525}{100} \times \frac{525}{100} \\
& =11 \times \frac{21}{4} \times \frac{21}{4}=303.187 \mathrm{~cm}^3 \\
& =\frac{303.187}{1000} \text { liters }\left[\because 1000 \mathrm{~cm}^3=1 l\right] \\
& =0.303187 \text { liters } \\
& =0.303 \text { liters (approx.) } \\
&
\end{aligned}
$

Ex 11.4 Question 6.

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

Answer.

Inner radius of hemispherical tank ${ }^{(r)}=1 \mathrm{~m}=100 \mathrm{~cm}$

Thickness of sheet $=1 \mathrm{~cm}$
$\therefore$ Outer radius of hemispherical tank $(R)=100+1=101 \mathrm{~cm}$

Volume of iron of hemisphere
$
=\frac{2}{3} \pi\left[R^3-r^3\right]
$
$
\begin{aligned}
& =\frac{2}{3} \times \frac{22}{7} \times\left[(101)^3-(100)^3\right] \\
& =\frac{44}{21}[1030301-1000000] \\
& =63487.81 \mathrm{~cm}^3 \\
& =0.06348 \mathrm{~m}^3
\end{aligned}
$

Ex 11.4 Question 7.

Find the volume of a sphere whose surface area is $154 \mathrm{~cm}^2$.

Answer.

Surface area of sphere $=154 \mathrm{~cm}^2$
$
\begin{aligned}
& \Rightarrow 4 \pi r^2=154 \\
& \Rightarrow 4 \times \frac{22}{7} \times r^2=154 \\
& \Rightarrow r^2=\frac{154 \times 7}{4 \times 22}=\frac{49}{4} \\
& \Rightarrow r=\frac{7}{2} \mathrm{~cm}
\end{aligned}
$

Now, Volume of sphere
$
\begin{aligned}
& =\frac{4}{3} \pi r^3 \\
& =\frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} \\
& =\frac{1}{3} \times 11 \times 49=\frac{539}{3}=179 \frac{2}{3} \mathrm{~cm}^3 \\
&
\end{aligned}
$

Ex 11.4 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. 498.96. If the cost of white-washing is at the rate of Rs. 2.00 per square meter, find:
(i) the inner surface area of the dome.
(ii) the volume of the air inside the dome.

Answer.

Cost of white washing from inside $=$ Rs. 498.96

Rate of white washing $=$ Rs. 2
$\therefore$ Area white washed
$
=\frac{498.96}{2}=249.48 \mathrm{~cm}^2
$

Inside surface area of the dome
$=249.48 \mathrm{~cm}^2$
$\Rightarrow 2 \pi r^2 249.48$
$\Rightarrow r^2=\frac{249.48 \times 7}{2 \times 22}$
$=5.67 \times 7$

$
\Rightarrow r=6.3
$

So, Volume of the dome
$
\begin{aligned}
& =\frac{2}{3} \pi r^3=\frac{2}{3} \times \frac{22}{7} \times 6.3 \times 6.3 \times 6.3 \\
& =523.9 \mathrm{~cm}^3
\end{aligned}
$

Ex 11.4 Question 9.

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

Answer.

(i) Let radius of sphere be $r$ and radius of new sphere be R.
$27 \times$ Volume of sphere $=$ Volume of new sphere
$\Rightarrow 27 \times \frac{4}{3} \pi r^3=\frac{4}{3} \pi \mathrm{R}^3$
$\Rightarrow \sqrt[3]{3^3 \times r^3}=\mathrm{R}$
$\Rightarrow 3 r=R$
$\frac{\text { Surface area of sphere }(\mathrm{S})}{\text { Surface area of sphere }\left(\mathrm{S}^{\prime}\right)}=\frac{4 \pi r^2}{4 \pi \mathrm{R}^2}$
$
=\frac{r^2}{\mathrm{R}^2}=\frac{r^2}{(3 r)^2}=\frac{r^2}{9 r^2}=\frac{1}{9}
$

Ex 11.4 Question 10.

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

Answer.

Diameter of spherical capsule $=3.5 \mathrm{~mm}$

$\therefore$ Radius of spherical capsule $(r)$
$
=\frac{3.5}{2}=\frac{35}{20}=\frac{7}{4} \mathrm{~mm}
$

Medicine needed to fill the capsule
$
\begin{aligned}
& =\text { Volume of sphere }=\frac{4}{3} \pi r^3 \\
& =\frac{4}{3} \times \frac{22}{7} \times \frac{7}{4} \times \frac{7}{4} \times \frac{7}{4} \\
& =\frac{11 \times 7 \times 7}{3 \times 2 \times 4}=\frac{539}{34} \mathrm{~mm}^3 \\
& =22.46 \mathrm{~mm}^3 \text { (Approx.) }
\end{aligned}
$