Exercise 11.2 Practice
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Overview
This page provides comprehensive Ch 11: Surface Areas and Volumes - Exercise 11.2 Practice. Free NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas And Volumes Exercise 11-2. Step-by-step explained answers for CBSE Board exams. Download PDF and practice now.
Surface Area of Spheres and Hemispheres
Q1: Surface Area of Sphere
Find the surface area of a sphere of radius:
(i) $14 \text{ cm}$
(ii) $21 \text{ cm}$
(i) $14 \text{ cm}$
(ii) $21 \text{ cm}$
Formula: Surface Area of Sphere $= 4\pi r^2$.
(i) Radius $r = 14 \text{ cm}$:
Area $= 4 \times \frac{22}{7} \times 14 \times 14$
$= 4 \times 22 \times 2 \times 14$
$= 88 \times 28 = 2464 \text{ cm}^2$.
Area $= 4 \times \frac{22}{7} \times 14 \times 14$
$= 4 \times 22 \times 2 \times 14$
$= 88 \times 28 = 2464 \text{ cm}^2$.
(ii) Radius $r = 21 \text{ cm}$:
Area $= 4 \times \frac{22}{7} \times 21 \times 21$
$= 4 \times 22 \times 3 \times 21$
$= 88 \times 63 = 5544 \text{ cm}^2$.
Area $= 4 \times \frac{22}{7} \times 21 \times 21$
$= 4 \times 22 \times 3 \times 21$
$= 88 \times 63 = 5544 \text{ cm}^2$.
(i) $2464 \text{ cm}^2$, (ii) $5544 \text{ cm}^2$
Q2: Surface Area from Diameter
Find the surface area of a sphere of diameter:
(i) $28 \text{ cm}$
(ii) $7 \text{ cm}$
(i) $28 \text{ cm}$
(ii) $7 \text{ cm}$
(i) Diameter $d = 28 \text{ cm} \Rightarrow r = 14 \text{ cm}$:
Area $= 4\pi r^2 = 4 \times \frac{22}{7} \times 14 \times 14$
$= 2464 \text{ cm}^2$.
Area $= 4\pi r^2 = 4 \times \frac{22}{7} \times 14 \times 14$
$= 2464 \text{ cm}^2$.
(ii) Diameter $d = 7 \text{ cm} \Rightarrow r = 3.5 \text{ cm}$:
Area $= 4 \times \frac{22}{7} \times 3.5 \times 3.5$
$= 4 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$
$= 22 \times 7 = 154 \text{ cm}^2$.
Area $= 4 \times \frac{22}{7} \times 3.5 \times 3.5$
$= 4 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$
$= 22 \times 7 = 154 \text{ cm}^2$.
(i) $2464 \text{ cm}^2$, (ii) $154 \text{ cm}^2$
Q3: Hemisphere TSA
Find the total surface area of a hemisphere of radius $20 \text{ cm}$. (Use $\pi = 3.14$)
Radius $r = 20 \text{ cm}$.
Total Surface Area (TSA) of Hemisphere $= 3\pi r^2$.
Total Surface Area (TSA) of Hemisphere $= 3\pi r^2$.
TSA $= 3 \times 3.14 \times (20)^2$
$= 3 \times 3.14 \times 400$
$= 1200 \times 3.14$
$= 3768 \text{ cm}^2$.
$= 3 \times 3.14 \times 400$
$= 1200 \times 3.14$
$= 3768 \text{ cm}^2$.
TSA $= 3768 \text{ cm}^2$
Q4: Balloon Expansion
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.
Let $r_1 = 7 \text{ cm}$ and $r_2 = 14 \text{ cm}$.
Surface Area $S_1 = 4\pi r_1^2$.
Surface Area $S_2 = 4\pi r_2^2$.
Surface Area $S_2 = 4\pi r_2^2$.
Ratio $\frac{S_1}{S_2} = \frac{4\pi r_1^2}{4\pi r_2^2} = \left(\frac{r_1}{r_2}\right)^2$
$= \left(\frac{7}{14}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
$= \left(\frac{7}{14}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
Ratio $= 1:4$
Q5: Tin-Plating Cost
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 $100 \text{ cm}^2$.
Inner diameter $= 10.5 \text{ cm} \Rightarrow r = 5.25 \text{ cm}$.
Inner Curved Surface Area $= 2\pi r^2$
$= 2 \times \frac{22}{7} \times 5.25 \times 5.25$
$= 44 \times 0.75 \times 5.25$
$= 33 \times 5.25 = 173.25 \text{ cm}^2$.
$= 2 \times \frac{22}{7} \times 5.25 \times 5.25$
$= 44 \times 0.75 \times 5.25$
$= 33 \times 5.25 = 173.25 \text{ cm}^2$.
Cost $= \frac{\text{Area}}{100} \times \text{Rate}$
$= \frac{173.25}{100} \times 16$
$= 1.7325 \times 16 = 27.72$.
$= \frac{173.25}{100} \times 16$
$= 1.7325 \times 16 = 27.72$.
Cost = ₹$27.72$
Q6: Find Radius
Find the radius of a sphere whose surface area is $616 \text{ cm}^2$.
Surface Area $= 4\pi r^2 = 616$.
$4 \times \frac{22}{7} \times r^2 = 616$.
$4 \times \frac{22}{7} \times r^2 = 616$.
$r^2 = \frac{616 \times 7}{4 \times 22}$
$r^2 = \frac{616 \times 7}{88} = 7 \times 7 = 49$.
$r^2 = \frac{616 \times 7}{88} = 7 \times 7 = 49$.
$r = \sqrt{49} = 7 \text{ cm}$.
Radius $= 7 \text{ cm}$
Q7: Moon vs Earth
The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of
their surface areas.
Moon
Earth
Let diameter of earth be $d$. Radius of earth $R = \frac{d}{2}$.
Diameter of moon $= \frac{1}{4}d$. Radius of moon $r = \frac{1}{2}(\frac{1}{4}d) = \frac{d}{8}$.
Diameter of moon $= \frac{1}{4}d$. Radius of moon $r = \frac{1}{2}(\frac{1}{4}d) = \frac{d}{8}$.
Ratio of Surface Areas $= \frac{4\pi r^2}{4\pi R^2} = \left(\frac{r}{R}\right)^2$.
$\frac{r}{R} = \frac{d/8}{d/2} = \frac{2}{8} = \frac{1}{4}$.
Ratio $= \left(\frac{1}{4}\right)^2 = \frac{1}{16}$.
Ratio $= \left(\frac{1}{4}\right)^2 = \frac{1}{16}$.
Ratio $= 1:16$
Q8: Cylinder Enclosing Sphere
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 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:
Since the cylinder just encloses the sphere, its radius is $r$ and height $h = 2r$ (diameter of sphere).
$S_2 = 2\pi r h = 2\pi r (2r) = 4\pi r^2$.
Since the cylinder just encloses the sphere, its radius is $r$ and height $h = 2r$ (diameter of sphere).
$S_2 = 2\pi r h = 2\pi r (2r) = 4\pi r^2$.
(iii) Ratio:
$\frac{S_1}{S_2} = \frac{4\pi r^2}{4\pi r^2} = 1$.
$\frac{S_1}{S_2} = \frac{4\pi r^2}{4\pi r^2} = 1$.
Ratio $= 1:1$
Q9: Dome Painting
A hemispherical dome of a building needs to be painted. If the circumference of the base of the dome is
$17.6 \text{ m}$, find the cost of painting it, given the cost of painting is ₹$5$ per $100 \text{
cm}^2$.
Circumference of base $= 2\pi r = 17.6 \text{ m}$.
$2 \times \frac{22}{7} \times r = 17.6$
$44r = 17.6 \times 7 = 123.2$
$r = \frac{123.2}{44} = 2.8 \text{ m}$.
$2 \times \frac{22}{7} \times r = 17.6$
$44r = 17.6 \times 7 = 123.2$
$r = \frac{123.2}{44} = 2.8 \text{ m}$.
CSA of Dome $= 2\pi r^2$
$= 2 \times \frac{22}{7} \times 2.8 \times 2.8$
$= 44 \times 0.4 \times 2.8$
$= 17.6 \times 2.8 = 49.28 \text{ m}^2$.
$= 2 \times \frac{22}{7} \times 2.8 \times 2.8$
$= 44 \times 0.4 \times 2.8$
$= 17.6 \times 2.8 = 49.28 \text{ m}^2$.
Cost rate $= $ ₹$5$ per $100 \text{ cm}^2$.
$1 \text{ m}^2 = 10000 \text{ cm}^2$.
Rate per $\text{m}^2 = \frac{5}{100} \times 10000 = $ ₹$500$.
$1 \text{ m}^2 = 10000 \text{ cm}^2$.
Rate per $\text{m}^2 = \frac{5}{100} \times 10000 = $ ₹$500$.
Total Cost $= 49.28 \times 500 = $ ₹$24640$.
Cost = ₹$24640$