EXERCISE 11.2
$\text { Assume } \pi=\frac{22}{7} \text {, unless stated otherwise. }$
1. Find the surface area of a sphere of radius:
(i) $10.5 \mathrm{~cm}$
(ii) $5.6 \mathrm{~cm}$
(iii) $14 \mathrm{~cm}$
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Solution
(i) Radius ( $r$ ) of sphere $=10.5 cm$
Surface area of sphere $=4 n r^{2}$
$=[4 \times \frac{22}{7} \times(10.5)^{2}] cm^{2}$
$=(4 \times \frac{22}{7} \times 10.5 \times 10.5) cm^{2}$
$=(88 \times 1.5 \times 10.5) cm^{2}$
$=1386 cm^{2}$
Therefore, the surface area of a sphere having radius $10.5 cm$ is $1386 cm^{2}$.
(ii) Radius( $r$ ) of sphere $=5.6 cm$
Surface area of sphere $=4 \pi r^{2}$$=[4 \times \frac{22}{7} \times(5.6)^{2}] cm^{2}$
$=(88 \times 0.8 \times 5.6) cm^{2}$
$=394.24 cm^{2}$
Therefore, the surface area of a sphere having radius $5.6 cm$ is $394.24 cm^{2}$.
(iii) Radius ( $r$ ) of sphere $=14 cm$
Surface area of sphere $=4 \pi r^{2}$
$=[4 \times \frac{22}{7} \times(14)^{2}] cm^{2}$
$=(4 \times 44 \times 14) cm^{2}$
$=2464 cm^{2}$
Therefore, the surface area of a sphere having radius $14 cm$ is $2464 cm^{2}$.
2. Find the surface area of a sphere of diameter:
(i) $14 \mathrm{~cm}$
(ii) $21 \mathrm{~cm}$
(iii) $3.5 \mathrm{~m}$
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Solution
(i) $Radius(r)$ of sphere $=\frac{\text{ Diameter }}{2}=(\frac{14}{2}) cm=7 cm$
Surface area of sphere $=4 n r^{2}$
$=(4 \times \frac{22}{7} \times(7)^{2}) cm^{2}$$=(88 \times 7) cm^{2}$$=616 cm^{2}$
Therefore, the surface area of a sphere having diameter $14 cm$ is $616 cm^{2}$.
(ii) Radius ( $r$ ) of sphere $=\frac{21}{2}=10.5 cm$
Surface area of sphere $=4 n r^{2}$
$=[4 \times \frac{22}{7} \times(10.5)^{2}] cm^{2}$
$=1386 cm^{2}$
Therefore, the surface area of a sphere having diameter $21 cm$ is $1386 cm^{2}$.
(iii) Radius ( $r$ ) of sphere $=\frac{\frac{3.5}{2}=1.75}{} m$
Surface area of sphere $=4 n r^{2}$
$=[4 \times \frac{22}{7} \times(1.75)^{2}] m^{2}$
$=38.5 m^{2}$
Therefore, the surface area of the sphere having diameter $3.5 m$ is $38.5 m^{2}$.
3. Find the total surface area of a hemisphere of radius $10 \mathrm{~cm}$. (Use $\pi=3.14$ )
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Solution
Radius ( $r$ ) of hemisphere $=10 cm$
Total surface area of hemisphere $=$ CSA of hemisphere + Area of circular end of hemisphere
$\begin{aligned}& =2 \pi r^{2}+\pi r^{2} \\& =3 \pi r^{2} \\& =[3 \times 3.14 \times(10)^{2}] cm^{2} \\& =942 cm^{2}\end{aligned}$
Therefore, the total surface area of such a hemisphere is $942 cm^{2}$.
4. The radius of a spherical balloon increases from $7 \mathrm{~cm}$ to $14 \mathrm{~cm}$ as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
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Solution
Radius ( $r_1$ ) of spherical balloon $=7 cm$
Radius $(r_2.$ ) of spherical balloon, when air is pumped into it $=14 cm$
$\begin{aligned}\text{ Required ratio } & =\frac{\text{ Ini }}{\text{ Surface area af }} \\& =\frac{4 \pi r_1^{2}}{4 \pi r_2^{2}}=(\frac{r_1}{r_2})^{2} \\& =(\frac{7}{14})^{2}=\frac{1}{4}\end{aligned}$
Therefore, the ratio between the surface areas in these two cases is $1: 4$.
5. A hemispherical bowl made of brass has inner diameter $10.5 \mathrm{~cm}$. Find the cost of tin-plating it on the inside at the rate of ₹ 16 per $100 \mathrm{~cm}^{2}$.
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Solution
$=(\frac{10.5}{2}) cm=5.25 cm$
Inner radius ( $r$ ) of hemispherical bowl
Surface area of hemispherical bowl $=2 \pi r^{2}$
$=[2 \times \frac{22}{7} \times(5.25)^{2}] cm^{2}$
$=173.25 cm^{2}$
Cost of tin-plating $100 cm^{2}$ area $=$ Rs 16
Cost of tin-plating $173.25 cm^{2}$ area $=Rs(\frac{16 \times 173.25}{100})=Rs 27.72$
Therefore, the cost of tin-plating the inner side of the hemispherical bowl is Rs 27.72 .
6. Find the radius of a sphere whose surface area is $154 \mathrm{~cm}^{2}$.
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Solution
Let the radius of the sphere be $r$.
Surface area of sphere $=154$
$\begin{aligned}& \therefore 4 \pi r^{2}=154 cm^{2} \\& r^{2}=(\frac{154 \times 7}{4 \times 22}) cm^{2}=(\frac{7 \times 7}{2 \times 2}) cm^{2} \\& r=(\frac{7}{2}) cm=3.5 cm\end{aligned}$
Therefore, the radius of the sphere whose surface area is $154 cm^{2}$ is $3.5 cm$.
7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
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Solution
Let the diameter of earth be $d$. Therefore, the diameter of moon will be $\frac{d}{4}$.
$\begin{aligned}& \text{ Radius of earth }=\frac{\frac{d}{2}}{\frac{1}{2} \times \frac{d}{4}=\frac{d}{8}} \\& \text{ Radius of moon }=\frac{4 \pi(\frac{d}{8})^{2}}{4 \pi(\frac{d}{2})^{2}} \\& \text{ Surface area of moon }=\frac{4 \pi(\frac{d}{8})^{2}}{4 \pi(\frac{d}{2})^{2}} \\& \text{ Surface area of earth }=\frac{4}{\text{ Required ratio }} \\& =\frac{4}{64}=\frac{1}{16}\end{aligned}$
Therefore, the ratio between their surface areas will be $1: 16$.
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.
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Solution
Inner radius of hemispherical bowl $=5 cm$
Thickness of the bowl $=0.25 cm$
$\therefore$ Outer radius $(r)$ of hemispherical bowl $=(5+0.25) cm$
$=5.25 cm$
Outer CSA of hemispherical bowl $=2 \pi r^{2}$
$=2 \times \frac{22}{7} \times(5.25 cm)^{2}=173.25 cm^{2}$
Therefore, the outer curved surface area of the bowl is $173.25 cm^{2}$.
9. A right circular cylinder just encloses a sphere of radius $r$ (see Fig. 11.10). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).
Fig. 11.10
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Solution
(i) Surface area of sphere $=4 \pi r^{2}$
(ii) Height of cylinder $=r+r=2 r$
Radius of cylinder $=r$
CSA of cylinder $=2 nrh$
$=2 \pi r(2 r)$
$=4 \Gamma r^{2}$
(iii)
Required ratio $=\frac{\text{ Surface area of sphere }}{\text{ CSA of cylinder }}$
$=\frac{4 \pi r^{2}}{4 \pi r^{2}}$
$=\frac{1}{1}$
Therefore, the ratio between these two surface areas is $1: 1$.
