Exercise 11.2 (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 $\quad \pi=\frac{22}{7}$ unless stated otherwise.
Ex 11.2 Question 1.

Find the surface area of a sphere of radius:
(i) $10.5 \mathrm{~cm}$
(ii) $5.6 \mathrm{~cm}$
(iii) $14 \mathrm{~cm}$

Answer.

(i) Radius of sphere $=105 \mathrm{~cm}$

Surface area of sphere $=4 \pi r^2$
$
=4 \times \frac{22}{7} \times 10.5 \times 10.5=1386 \mathrm{~cm}^2
$
(ii) Radius of sphere $=5.6 \mathrm{~m}$

Surface area of sphere $=4 \pi r^2$
$
=4 \times \frac{22}{7} \times 5.6 \times 5.6=3.94 .84 \mathrm{~m}^2
$
(iii) Radius of sphere $=14 \mathrm{~cm}$

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

$
=4 \times \frac{22}{7} \times 14 \times 14=2464 \mathrm{~cm}^2
$
Ex 11.2 Question 2.

Find the surface area of a sphere of diameter:
(i) $14 \mathrm{~cm}$
(ii) $21 \mathrm{~cm}$
(iii) $3.5 \mathrm{~cm}$

Answer.

(i) Diameter of sphere $=14 \mathrm{~cm}$,

Therefore Radius of sphere $=\frac{14}{2}=7 \mathrm{~cm}$

Surface area of sphere $=4 \pi r^2$
(ii) Diameter of sphere $=21 \mathrm{~cm}$
$\therefore$ Radius of sphere $=\frac{21}{2} \mathrm{~cm}$

Surface area of sphere $=4 \pi r^2$
$
=4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}=1386 \mathrm{~cm}^2
$
(iii) Diameter of sphere $=3.5 \mathrm{~cm}$
$\therefore$ Radius of sphere $=\frac{3.5}{2}=1.75 \mathrm{~cm}$

Surface area of sphere $=4 \pi r^2$
$
=4 \times \frac{22}{7} \times 1.75 \times 1.75=38.5 \mathrm{~cm}^2
$

Ex 11.2 Question 3.

Find the total surface area of a hemisphere of radius $10 \mathrm{~cm}$.
(Use $\pi=3.14$ )

Answer.

Radius of hemisphere $(r)=10 \mathrm{~cm}$
Total surface area of hemisphere $=3 \pi r^2$
$
=3 \times 3.14 \times 10 \times 10=942 \mathrm{~cm}^2
$

Hence total surface area of hemisphere is
$
942 \mathrm{~cm}^2 \text {. }
$

Ex 11.2 Question 4.

The radius of a spherical balloonincreases from $7 \mathrm{~cm}$ to $14 \mathrm{~cm}$ as air isbeing pumped into it. Find the ratio ofsurface areas of the balloon in the twocases.

Answer.

I case: Radius of balloon ${ }^{(r)}=7 \mathrm{~cm}$

Surface area of balloon $=4 \pi r^2$
$
=4 \pi \times 7 \times 7 \mathrm{~cm}^2 \ldots \ldots \ldots \text { (i) }
$

II case: Radius of balloon $(R)=14 \mathrm{~cm}$

Surface area of balloon $=4 \pi \mathrm{R}^2$
$
=4 \pi \times 14 \times 14 \mathrm{~cm} 2
$

Now, Ratio [from eq. (i) and (ii)],
$
\begin{aligned}
& \frac{\text { CSA in first case }}{\text { CSA in second case }}=\frac{4 \pi \times 7 \times 7}{4 \pi \times 14 \times 14} \\
& =\frac{1}{4}
\end{aligned}
$

Hence, required ratio $=1: 4$

Ex 11.2 Question 5.

A hemispherical bowl made of brass has inner diameter $105 \mathrm{~cm}$. Find the cost of tin-plating it on the inside at the rate of Rs. 16 per $100 \mathrm{~cm}^2$.

Answer.

Inner diameter of bowl
$
=10.5 \mathrm{~cm}
$
$\therefore$ Inner radius of bowl $(r)=\frac{10.5}{2}$
$
=5.25 \mathrm{~cm}
$

Now, Inner surface area of bowl
$
\begin{aligned}
& =2 \pi r^2 \\
& =2 \times \frac{22}{7} \times 5.25 \times 5.25 \\
& =2 \times \frac{22}{7} \times \frac{21}{4} \times \frac{21}{4}=\frac{693}{4} \mathrm{~cm}^2
\end{aligned}
$
$\because$ Cost of tin-plating per $100 \mathrm{~cm}^2$
$
\text { = Rs. } 16
$
$\therefore$ Cost of tin-plating per $1 \mathrm{~cm}^2=\frac{16}{100}$
$\therefore$ Cost of tin-plating per $\frac{693}{4} \mathrm{~cm}^2$

$=\frac{16}{100} \times \frac{693}{4}=\text { Rs. } 27.72$

Ex 11.2 Question 6.

Find the radius 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}{22 \times 4}
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow r^2=\frac{49}{4} \\
& \Rightarrow r=\frac{7}{2}=3.5 \mathrm{~cm}
\end{aligned}
$

Ex 11.2 Question 7.

The diameter of the moon is approximately one fourth the diameter of the earth. Find the ratio of their surface areas.

Answer.

Let diameter of Earth $=x$
$\therefore$ Radius of Earth $(r)=\frac{x}{2}$
$\therefore$ Surface area of Earth $=4 \pi r^2$
$
=4 \pi \times \frac{x}{2} \times \frac{x}{2}=\pi x^2
$

Now, Diameter of Moon $=\frac{1}{4}$ th of diameter of Earth $=\frac{x}{4}$
$\therefore$ Radius of Moon ${ }^{(r)}=\frac{x}{8}$

Surface area of Moon $=4 \pi r^2$
$=4 \pi \times \frac{x}{8} \times \frac{x}{8}=\frac{\pi x^2}{16}$
Surface area of Moon

Now, Ratio $=$ Surface area of Earth
$=\frac{\frac{\pi x^2}{16}}{\pi x^2}=\frac{\pi x^2}{16} \times \frac{1}{\pi x^2}=\frac{1}{16}$
$\therefore$ Required ratio $=1: 16$

Ex 11.2 Question 8.

A hemispherical bowl is made of steel, $0.25 \mathrm{~cm}$ thick. The inner radius of the bowl is $5 \mathrm{~cm}$. Find the outer curved surface area of the bowl.

Answer.

Inner radius of bowl $(r)=5 \mathrm{~cm}$

Thickness of steel ${ }^{(t)}=0.25 \mathrm{~cm}$
$
\begin{aligned}
& \therefore \text { Outer radius of bowl }(\mathrm{R})=r+t \\
& =5+0.25=5.25 \mathrm{~cm}
\end{aligned}
$
$\therefore$ Outer curved surface area of bowl
$
\begin{aligned}
& =2 \pi \mathrm{R}^2=2 \times \frac{22}{7} \times 5.25 \times 5.25 \\
& =2 \times \frac{22}{7} \times \frac{21}{4} \times \frac{21}{4} \\
& =\frac{693}{4}=173.25 \mathrm{~cm}^2
\end{aligned}
$

Ex 11.2 Question 9.

A right circular cylinder just encloses a sphere of radius ${ }^r$ (See figure). Find:
(i) Surface area of the sphere.
(ii) Curved surface area of the cylinder.
(iii) Ratio of the areas obtained in (i) and (ii).

Answer.

(i) Radius of sphere $=r$
$\therefore$ Surface area of sphere
$
=2 \pi(\text { radius })^2=2 \pi r^2
$
$\because$ The cylinder just encloses the sphere in it.
$\therefore$ The height of cylinder will be equal to diameter of sphere.

And The radius of cylinder will be equal to radius of sphere.
(ii) $\therefore$ Curved surface area of cylinder
$
\begin{aligned}
& =2 \pi r h=2 \pi r \times \pi r[\because h=2 r] \\
& =4 \pi r^2
\end{aligned}
$

Surface area of sphere
(iii) Curved surface area of cylinder
$
=\frac{4 \pi r^2}{4 \pi r^2}=\frac{1}{1}
$
$\therefore$ Required ratio $=1: 1$