Surface Areas & Volumes
NCERT Solutions • Class 9 Maths • Chapter 11Exercise 11.2
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1. Find the surface area of a sphere of radius: (i) $10.5 \text{ cm}$ (ii) $5.6 \text{ cm}$ (iii) $14 \text{ cm}$
(i) $r = 10.5 \text{ cm}$. SA $= 4\pi r^2 = 4 \times \frac{22}{7} \times 10.5 \times 10.5$
$= 88 \times 1.5 \times 10.5 =$ $1386 \text{ cm}^2$.
$= 88 \times 1.5 \times 10.5 =$ $1386 \text{ cm}^2$.
(ii) $r = 5.6 \text{ cm}$. SA $= 4 \times \frac{22}{7} \times 5.6 \times 5.6$
$= 88 \times 0.8 \times 5.6 =$ $394.24 \text{ cm}^2$.
$= 88 \times 0.8 \times 5.6 =$ $394.24 \text{ cm}^2$.
(iii) $r = 14 \text{ cm}$. SA $= 4 \times \frac{22}{7} \times 14 \times 14$
$= 88 \times 2 \times 14 =$ $2464 \text{ cm}^2$.
$= 88 \times 2 \times 14 =$ $2464 \text{ cm}^2$.
2. Find the surface area of a sphere of diameter: (i) $14 \text{ cm}$ (ii) $21 \text{ cm}$ (iii) $3.5 \text{ m}$
(i) $d = 14 \text{ cm} \Rightarrow r = 7 \text{ cm}$.
SA $= 4 \times \frac{22}{7} \times 7 \times 7 = 88 \times 7 =$ $616 \text{ cm}^2$.
SA $= 4 \times \frac{22}{7} \times 7 \times 7 = 88 \times 7 =$ $616 \text{ cm}^2$.
(ii) $d = 21 \text{ cm} \Rightarrow r = 10.5 \text{ cm}$.
SA $= 4 \times \frac{22}{7} \times 10.5 \times 10.5 =$ $1386 \text{ cm}^2$.
SA $= 4 \times \frac{22}{7} \times 10.5 \times 10.5 =$ $1386 \text{ cm}^2$.
(iii) $d = 3.5 \text{ m} \Rightarrow r = 1.75 \text{ m}$.
SA $= 4 \times \frac{22}{7} \times 1.75 \times 1.75 = 88 \times 0.25 \times 1.75 =$ $38.5 \text{ m}^2$.
SA $= 4 \times \frac{22}{7} \times 1.75 \times 1.75 = 88 \times 0.25 \times 1.75 =$ $38.5 \text{ m}^2$.
3. Find the total surface area of a hemisphere of radius $10 \text{ cm}$. (Use $\pi = 3.14$)
Given: $r = 10 \text{ cm}$.
Formula: TSA of Hemisphere $= 3\pi r^2$
$= 3 \times 3.14 \times 10 \times 10$
$= 3 \times 314 =$ $942 \text{ cm}^2$.
Formula: TSA of Hemisphere $= 3\pi r^2$
$= 3 \times 3.14 \times 10 \times 10$
$= 3 \times 314 =$ $942 \text{ cm}^2$.
4. The radius of a spherical balloon increases from $7 \text{ cm}$ to $14 \text{ cm}$ as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Ratio Calculation:
$r_1 = 7 \text{ cm}, \quad r_2 = 14 \text{ cm}$.
Ratio $= \frac{4\pi r_1^2}{4\pi r_2^2} = \frac{r_1^2}{r_2^2} = (\frac{7}{14})^2$
$= (\frac{1}{2})^2 = \frac{1}{4}$.
Ratio is $1:4$.
$r_1 = 7 \text{ cm}, \quad r_2 = 14 \text{ cm}$.
Ratio $= \frac{4\pi r_1^2}{4\pi r_2^2} = \frac{r_1^2}{r_2^2} = (\frac{7}{14})^2$
$= (\frac{1}{2})^2 = \frac{1}{4}$.
Ratio is $1:4$.
5. A hemispherical bowl made of brass has inner diameter $10.5 \text{ cm}$. Find the cost of tin-plating it on the inside at the rate of ₹ 16 per $100 \text{ cm}^2$.
Step 1: Calculate Inner CSA
$r = \frac{10.5}{2} = 5.25 \text{ cm}$.
CSA $= 2\pi r^2 = 2 \times \frac{22}{7} \times 5.25 \times 5.25$
$= 44 \times 0.75 \times 5.25 = 173.25 \text{ cm}^2$.
$r = \frac{10.5}{2} = 5.25 \text{ cm}$.
CSA $= 2\pi r^2 = 2 \times \frac{22}{7} \times 5.25 \times 5.25$
$= 44 \times 0.75 \times 5.25 = 173.25 \text{ cm}^2$.
Step 2: Calculate Cost
Rate = ₹ 16 per $100 \text{ cm}^2$.
Cost $= \frac{173.25}{100} \times 16 = 1.7325 \times 16 =$ ₹ $27.72$.
Rate = ₹ 16 per $100 \text{ cm}^2$.
Cost $= \frac{173.25}{100} \times 16 = 1.7325 \times 16 =$ ₹ $27.72$.
6. Find the radius of a sphere whose surface area is $154 \text{ cm}^2$.
Calculation:
$4\pi r^2 = 154$
$4 \times \frac{22}{7} \times r^2 = 154$
$r^2 = \frac{154 \times 7}{88} = \frac{1078}{88} = 12.25$
$r = \sqrt{12.25} =$ $3.5 \text{ cm}$.
$4\pi r^2 = 154$
$4 \times \frac{22}{7} \times r^2 = 154$
$r^2 = \frac{154 \times 7}{88} = \frac{1078}{88} = 12.25$
$r = \sqrt{12.25} =$ $3.5 \text{ cm}$.
7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Let diameter of Earth $= d$. Radius of Earth $= R = \frac{d}{2}$.
Diameter of Moon $= \frac{d}{4}$. Radius of Moon $= r = \frac{d}{8}$.
Ratio of SA $= \frac{4\pi r^2}{4\pi R^2} = \frac{r^2}{R^2} = \frac{(d/8)^2}{(d/2)^2}$
$= \frac{d^2/64}{d^2/4} = \frac{4}{64} = \frac{1}{16}$.
Ratio is $1:16$.
Diameter of Moon $= \frac{d}{4}$. Radius of Moon $= r = \frac{d}{8}$.
Ratio of SA $= \frac{4\pi r^2}{4\pi R^2} = \frac{r^2}{R^2} = \frac{(d/8)^2}{(d/2)^2}$
$= \frac{d^2/64}{d^2/4} = \frac{4}{64} = \frac{1}{16}$.
Ratio is $1:16$.
8. A hemispherical bowl is made of steel, $0.25 \text{ cm}$ thick. The inner radius of the bowl is $5 \text{ cm}$. Find the outer curved surface area of the bowl.
Inner radius $r = 5 \text{ cm}$. Thickness $t = 0.25 \text{ cm}$.
Outer radius $R = r + t = 5 + 0.25 = 5.25 \text{ cm}$.
Outer CSA $= 2\pi R^2 = 2 \times \frac{22}{7} \times 5.25 \times 5.25$
$= 44 \times 0.75 \times 5.25 =$ $173.25 \text{ cm}^2$.
Outer radius $R = r + t = 5 + 0.25 = 5.25 \text{ cm}$.
Outer CSA $= 2\pi R^2 = 2 \times \frac{22}{7} \times 5.25 \times 5.25$
$= 44 \times 0.75 \times 5.25 =$ $173.25 \text{ cm}^2$.
9. A right circular cylinder just encloses a sphere of radius $r$. Find: (i) surface area of the sphere, (ii) curved surface area of the cylinder, (iii) ratio of the areas obtained in (i) and (ii).
(i) Surface Area of Sphere:
$S_1 = 4\pi r^2$.
$S_1 = 4\pi r^2$.
(ii) CSA of Cylinder:
The cylinder just encloses the sphere, so:
Radius of cylinder $= r$.
Height of cylinder $h = 2r$ (diameter of sphere).
$S_2 = 2\pi r h = 2\pi r(2r) = 4\pi r^2$.
The cylinder just encloses the sphere, so:
Radius of cylinder $= r$.
Height of cylinder $h = 2r$ (diameter of sphere).
$S_2 = 2\pi r h = 2\pi r(2r) = 4\pi r^2$.
(iii) Ratio:
Ratio $= \frac{S_1}{S_2} = \frac{4\pi r^2}{4\pi r^2} =$ $1:1$.
Ratio $= \frac{S_1}{S_2} = \frac{4\pi r^2}{4\pi r^2} =$ $1:1$.