# RS Aggarwal Class 9 Solutions Chapter 13 - Volume And Surface Area Ex 13C(13.3)

## RS Aggarwal Class 9 Chapter 13 – Volume And Surface Area Ex 13C(13.3) Solutions Free PDF

The RS Aggarwal Class 9 Solutions are prepared with an aim to help students facilitate their preparation for the exam. We at BYJUâ€™S provide accurate solutions of the RS Aggarwal Class 10 Maths textbook and explains each and every topic in an easy manner and give detailed solutions to difficult questions. So, by referring to these solutions, you can improve your performance and score better marks.

The solutions are prepared by highly skilled subject experts are in accordance with the latest syllabus of the CBSE. Students of Class 9 are advised to practice from RS Aggarwal Class 9 Solutions Chapter 13 – Volume And Surface Area Ex 13C(13.3) so that they can solve all the questions that will be asked in the board exams.

Q.1: Find the volume and surface area of a sphere whose radius is (i) 3.5 cm (ii) 4.2 cm (iii) 5 m

Solution:

(i) Radius of sphere = 3.5 cm

∴ Volume of the sphere = 43πr3$\frac{4}{3} \pi r^{3}$ = 43×227×3.5×3.5×3.5$\frac{4}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 3.5$ = 179.67 cm3

∴ Surface area of the sphere = 4𝜋r2 = 4×227×3.5×3.5$4 \times \frac{22}{7} \times 3.5 \times 3.5$ = 154 cm2

(ii) Radius of the sphere = 4.2 cm

∴ Volume of the sphere = 43πr3$\frac{4}{3} \pi r^{3}$ = 43×227×4.2×4.2×4.2$\frac{4}{3} \times \frac{22}{7} \times 4.2 \times 4.2 \times 4.2$ = 310.464 cm3

∴ Surface area of the sphere = 4𝜋r2 = 4×227×4.2×4.2$4 \times \frac{22}{7} \times 4.2 \times 4.2$ = 221.76 cm2

(iii) Radius of the sphere = 5 m

∴ Volume of the sphere = 43πr3$\frac{4}{3} \pi r^{3}$

= 43×227×5×5×5$\frac{4}{3} \times \frac{22}{7} \times 5 \times 5 \times 5$ = 523.81 m3

∴ Surface area of the sphere = 4𝜋r2= 4×227×5×5$4 \times \frac{22}{7} \times 5 \times 5$ = 314.28 cm2

Q.2: The volume of a sphere is 38808 cm3. Find its radius and hence its surface area.

Solution:

Volume of the sphere = 43πr3$\frac{4}{3} \pi r^{3}$

38808=43×227×r3$\Rightarrow 38808 = \frac{4}{3} \times \frac{22}{7} \times r^{3}$ (because, volume = 38808 cm3)

r3=38808×3×788=9261$\Rightarrow r^{3} = \frac{38808 \times 3 \times 7}{88} = 9261$
r=21cm$\Rightarrow r = 21 cm$

∴ Surface area of the sphere = 4𝜋r2 = 4×227×21×21$4 \times \frac{22}{7} \times 21 \times 21$ = 5544 cm2

Q.3: Find the surface area of a sphere whose volume is 606.375 m3

Solution:

Volume of the sphere = 606.375 m3

Volume of the sphere = 43πr3$\frac{4}{3} \pi r^{3}$

606.375=43×227×r3$\Rightarrow 606.375 = \frac{4}{3} \times \frac{22}{7} \times r^{3}$ (because, volume = 606.375 cm3)

r3=606.375×3×74×22=144.703125$\Rightarrow r^{3} = \frac{606.375 \times 3 \times 7}{4 \times 22} = 144.703125$
r=5.25m$\Rightarrow r = 5.25 m$

∴ Surface area of the sphere = 4𝜋r2 = 4×227×5.25×5.25$4 \times \frac{22}{7} \times 5.25 \times 5.25$ = 346.5 m2

Q.4: The surface area of a sphere is 394.24 m2. Find its radius and volume.

Solution:

Let the radius of the sphere be ‘r’ m.

Then, its surface area = 4𝜋r2

∴ 4𝜋r2 = 394.24 [given]

4×227×r2=394.24$4 \times \frac{22}{7} \times r^{2} = 394.24$
r2=394.24×74×22=31.36$r^{2} = \frac{394.24 \times 7}{4 \times 22} = 31.36$
r=31.36=5.6m$r = \sqrt{31.36} = 5.6 m$

∴ Radius of the sphere = 5.6 m

Volume of the sphere = 43πr3$\frac{4}{3} \pi r^{3}$

43×227×5.6×5.6×5.6$\frac{4}{3} \times \frac{22}{7} \times 5.6 \times 5.6 \times 5.6$ = 735.91 m3

∴ Volume of the sphere = 735.91 m3

Q.5: The surface area of a sphere is (576𝜋) cm2. Find its volume.

Solution:

Surface area of sphere = 4𝜋r2

∴ 4𝜋r2 = 576𝜋 [surface area = 576𝜋 cm2]

r2=576π4π$\Rightarrow r^{2} = \frac{576\pi }{4\pi }$
r=144=12cm$\Rightarrow r = \sqrt{144} = 12 cm$

∴ Volume of the sphere = 43πr3$\frac{4}{3} \pi r^{3}$

= 43×227×12×12×12$\frac{4}{3} \times \frac{22}{7} \times 12 \times 12 \times 12$ = 2304𝜋 cm3

∴ Volume of the sphere = 2304𝜋 cm3

Q.6: The outer diameter of a spherical shell is 12 cm and its inner diameter is 8 cm. Find the volume of metal contained in the shell. Also, find its outer surface area.

Solution:

Outer diameter of the spherical shell = 12 cm

Inner diameter of spherical shell = 8 cm

Now, Volume of the outer shell = 43πr3$\frac{4}{3} \pi r^{3}$ = 43×227×6×6×6$\frac{4}{3} \times \frac{22}{7} \times 6 \times 6 \times 6$ = 905.15 cm3

And, Volume of the inner shell = 43πr3$\frac{4}{3} \pi r^{3}$ = 43×227×4×4×4$\frac{4}{3} \times \frac{22}{7} \times 4 \times 4 \times 4$ = 268.20 cm3

Volume of metal contained in the shell = (Volume of outer) – (Volume of inner) = (905.15) -(268.20) cm3 = 636.95 cm3

∴ Outer Surface area = 4𝜋r2 = 4×227×6×6$4 \times \frac{22}{7} \times 6 \times 6$ = 452.57 cm2

Q.7: How many lead shots, each of 3 mm in diameter can be made from a cuboid with dimensions (12 cm × 11 cm × 9 cm)?

Solution:

Here, diameter of the lead shot = 3 mm

Hence, radius = 32mm=0.32cm$\frac{3}{2} mm = \frac{0.3}{2} cm$

Now, number of lead shots = VolumeofthecuboidVolumeof1leadshot$\frac{Volume \, of \, the \, cuboid}{Volume \, of \, 1 \, lead \, shot}$

= 12×11×943×227×(0.32)3$\frac{12 \times 11 \times 9}{ \frac{4}{3} \times \frac{22}{7} \times (\frac{0.3}{2})^{3}}$ = 12×11×943×227×0.0278$\frac{12 \times 11 \times 9}{ \frac{4}{3} \times \frac{22}{7} \times \frac{0.027}{8}}$ = 12×11×9×3×7×84×22×0.027=84000$\frac{12 \times 11 \times 9 \times 3 \times 7 \times 8}{4 \times 22 \times 0.027} = 84000$

∴ Number of lead shots = 84000.

Solution:

But, number of lead balls = VolumeofthesphereVolumeof1leadball$\frac{Volume \, of \, the \, sphere}{Volume \, of \, 1 \, lead \, ball}$

= (43×π×R3)cm3(43×π×R3)cm3$\frac{ ( \frac{4}{3} \times \pi \times R^{3} ) cm^{3}}{( \frac{4}{3} \times \pi \times R^{3} ) cm^{3}}$

= 43×227×8343×227×13$\frac{ \frac{4}{3} \times \frac{22}{7} \times 8^{3}}{ \frac{4}{3} \times \frac{22}{7} \times 1^{3}}$

= 43×227×51243×227×1=512$\frac{ \frac{4}{3} \times \frac{22}{7} \times 512}{ \frac{4}{3} \times \frac{22}{7} \times 1} = 512$

∴ Number of lead balls = 512

Q.9: A solid sphere of radius 3 cm is melted and then recast into smaller spherical balls, each of diameter 0.6 cm. Find the number of small balls thus obtained.

Solution:

Here, radius of sphere = 3 cm

Diameter of spherical ball = 0.6 cm

Radius of spherical ball = 0.3 cm

Number of balls = VolumeofthesphereVolumeof1smallball$\frac{Volume \, of \, the \, sphere}{Volume \, of \, 1 \,small \, ball}$

= 43×227×3343×227×0.33$\frac{ \frac{4}{3} \times \frac{22}{7} \times 3^{3}}{ \frac{4}{3} \times \frac{22}{7} \times 0.3^{3}}$ = 43×227×2743×227×0.027=1000$\frac{ \frac{4}{3} \times \frac{22}{7} \times 27}{ \frac{4}{3} \times \frac{22}{7} \times 0.027} = 1000$

∴ Number of small balls obtained = 1000

Q.10: A metallic sphere of radius 10.5 cm is melted and then recast into smaller cones, each of radius 3.5 cm and height 3 cm. How many cones are obtained?

Solution:

Here, radius of sphere = 10.5 cm = 212$\frac{21}{2}$ cm

Radius of smaller cone = 3.5 cm = 72$\frac{7}{2}$ cm and height = 3 cm

Now, number of cones = VolumeofthesphereVolumeof1smallcone$\frac{ Volume \; of \; the \; sphere}{Volume \; of \; 1 \; small \; cone}$

= 43×227×(212)313×227×(72)2×3$\frac{ \frac{4}{3} \times \frac{22}{7} \times (\frac{21}{2})^{3}}{ \frac{1}{3} \times \frac{22}{7} \times (\frac{7}{2})^{2} \times 3}$ = 43×9261813×(72)2×3=96216494$\frac{ \frac{4}{3} \times \frac{9261}{8}}{ \frac{1}{3} \times (\frac{7}{2})^{2} \times 3} = \frac{ \frac{9621}{6}}{ \frac{49}{4}} Â$ = 92616×449=126$\frac{9261}{6} \times \frac{4}{49} = 126$

∴ Number of cones obtained = 126

Q.11: How many spheres 12 cm in diameter can be made from a metallic cylinder of diameter 8 cm and height 90 cm?

Solution:

Diameter of the sphere = 12 cm

Therefore, Volume of the sphere = 43πr3$\frac{4}{3} \pi r^{3}$

= 43×227×6×6×6$\frac{4}{3} \times \frac{22}{7} \times 6 \times 6 \times 6$ . . . . . . . . . . (i)

Diameter of cylinder = 8 cm

Hence, the radius = 4 cm

Height of the cylinder = 90 cm

Therefore, Volume of the outer shell = πr3$\pi r^{3}$

= 227×4×4×90$\frac{22}{7} \times 4 \times 4 \times 90$ . . . . . .(ii)

Number of spheres = VolumeofcylinderVolumeofsphere$\frac{ Volume \, of \, cylinder}{Volume \, of \, sphere}$

NUmber of spheres = 227×4×4×9043×227×6×6×6$\frac{ \frac{22}{7} \times 4 \times 4 \times 90}{ \frac{4}{3} \times \frac{22}{7} \times 6 \times 6 \times 6}$

Solving the above equation, we get, number of spheres = 5.

Q.12: The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 mm. Find the length of the wire.

Solution:

Here, diameter of a sphere = 6 cm

Diameter of wire = 2 mm

Hence, its radius = 1 mm = 0.1 cm

Let the required length of the wire be ‘h’ cm.

Then, π×(r)2×h=43×π×(R)3$\pi \times (r)^{2} \times h = \frac{4}{3} \times \pi \times (R)^{3}$

π×(0.1)@×h=43×π×(3)2$\Rightarrow \pi \times (0.1)^{@} \times h = \frac{4}{3} \times \pi \times (3)^{2}$
h=43×π×27i×(0.1)2$\Rightarrow h = \frac{ \frac{4}{3} \times \pi \times 27}{ \\i \times (0.1)^{2}}$

= 4×90.01cm=360.01$\frac{4 \times 9}{0.01} cm = \frac{36}{0.01}$ = 3600 cm = 36 m

Hence, the length of the wire = 36 m

Q.13: The diameter of a copper sphere is 18 am. It is melted and drawn into a long wire of uniform cross section. If the length of the wire is 108 m, find its diameter.

Solution:

Here, diameter of sphere = 18 cm

Length of the wire = 108 m = 10800 cm

Then,

43×π×(r)3=π×r2×10800$\frac{4}{3} \times \pi \times (r)^{3} = \pi \times r^{2} \times 10800$
r2=43×π×729π×10800$\Rightarrow r^{2} = \frac{ \frac{4}{3} \times \pi \times 729}{ \pi \times 10800}$
r2=4×24310800=97210800=9100$\Rightarrow r^{2} = \frac{4 \times 243}{10800} = \frac{972}{10800} = \frac{9}{100}$
r=9100=310=0.3$\Rightarrow r = \sqrt{9}{100} = \frac{3}{10} = 0.3$

So, the diameter = (2) (0.3) = 0.6 cm.

Q.14: A sphere of diameter 15.6 cm is melted and cast into the right circular cone of height 31.2 cm. Find the diameter of the base of the cone.

Solution:

Here, diameter of sphere = 15.6 cm

∴ Radius of sphere = 15.62$\frac{15.6}{2}$ cm = 7.8 cm

And, height of cone = 31.2 cm

Then, 43π×R3=13π×r2×h$\frac{4}{3} \pi \times R^{3} = \frac{1}{3} \pi \times r^{2} \times h$

43π×(7.8)3=13π×r2×31.2$\Rightarrow \frac{4}{3} \pi \times (7.8)^{3} = \frac{1}{3} \pi \times r^{2} \times 31.2$
r2=43π×(7.8)313π×31.2$\Rightarrow r^{2} = \frac{ \frac{4}{3} \pi \times (7.8)^{3}}{ \frac{1}{3} \pi \times 31.2}$
r2=4×474.55231.2=60.84=(7.8)2$\Rightarrow r^{2} = \frac{4 \times 474.552}{31.2} = 60.84 = (7.8)^{2}$
r=7.8cm$\Rightarrow r = 7.8 cm$

∴ Diameter of cone = (2) (7.8) cm = 15.6 cm.

Q.15: A spherical cannon ball 28 cm in diameter is melted and cast into a right circular cone mould, whose base is 35 cm in diameter. Find the height of the cone.

Solution:

Here, diameter of sphere = 28 cm

∴ radius of sphere = 282$\frac{28}{2}$ cm = 14 cm

Diameter of cone = 35 cm

∴ the radius of cone = 352$\frac{35}{2}$ cm = 17.5 cm

∴  43π×R3=13π×r2×h$\frac{4}{3} \pi \times R^{3} = \frac{1}{3} \pi \times r^{2} \times h$

h=43π×14313π×17.52$\Rightarrow h = \frac{ \frac{4}{3} \pi \times 14^{3}}{ \frac{1}{3} \pi \times 17.5^{2}}$
h=4×2744306.25cm=35.84$\Rightarrow h = \frac{4 \times 2744}{306.25} cm = 35.84$

∴ Height of the cone = 35.84 cm

Q.16: A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5 cm and 2 cm. Find the radius of the third ball.

Solution:

Let the radius of the third ball be ‘r’ cm

Then, 43×π×(3)3=43π(32)3+43×π×(2)3+43×π×(r)3$\frac{4}{3} \times \pi \times (3)^{3} = \frac{4}{3} \pi (\frac{3}{2})^{3} + \frac{4}{3} \times \pi \times (2)^{3} + \frac{4}{3} \times \pi \times (r)^{3}$

43×π×27=43π278+43×π×8+43×π×(r)3$\Rightarrow \frac{4}{3} \times \pi \times 27 = \frac{4}{3} \pi \frac{27}{8} + \frac{4}{3} \times \pi \times 8 + \frac{4}{3} \times \pi \times (r)^{3}$

27 = 278+8+r3$\frac{27}{8} + 8 + r^{3}$

i.e. r3=27278+8$r^{3} = { 27- { \frac{27}{8} + 8}}$

=2727+648$27 â€“ \frac{27 + 64}{8}$=27918$27 â€“ \frac{91}{8}$= 216918$\frac{216 â€“ 91}{8}$

i.e. r3=1258r3=(52)3$r^{3} = \frac{125}{8} \Rightarrow r^{3} = (\frac{5}{2})^{3}$

r = 52=2.5cm$\frac{5}{2} = 2.5 cm$

∴ Radius of the third ball = 2.5 cm.

Q.17: The radii of two spheres are in the ratio 1 : 2. Find the ratio of their surface areas.

Solution:

Let the radii of two spheres be ‘x’ and ‘2x’ and their respective surface areas be S1 and S2

Then, S1S2=4πx24π(2x)2$\frac{S_{1}}{S_{2}} = \frac{ 4 \pi x^{2}}{ 4 \pi (2x)^{2}}$

=x24x2=14$= \frac{x^{2}}{4x^{2}} = \frac{1}{4}$

∴ The ratio of their surface areas = 1: 4.

Q.18: The surface areas of two spheres are in the ratio 1 : 2. Find the ratio of their volumes.

Solution:

Let the radii of two spheres be ‘r’ and ‘R’

Then, 4πr24πR2=14$\frac{4 \pi r^{2}}{ 4 \pi R^{2}} = \frac{1}{4}$

(rR)2=(12)2rR=12$\Rightarrow ( \frac{r}{R})^{2} = (\frac{1}{2})^{2} \Rightarrow \frac{r}{R} = \frac{1}{2}$

Let V1 and V2 be the volumes of the respective spheres whose radii are r and R.

V1V2=43πr343πr3=(rR)3=(12)3=18$\frac{V_{1}}{V_{2}} = \frac{ \frac{4}{3} \pi r^{3}}{ \frac{4}{3} \pi r^{3}} = ( \frac{r}{R})^{3} = (\frac{1}{2})^{3} = \frac{1}{8}$

∴ The ratio of their volume = 1: 8.

Q.19: A cylindrical tube of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball?

Solution:

Let the radius of the ball be ‘r’ cm and ‘R’ be the radius of the cylindrical tub.

Then,

43π×r3=π×R2×h$\frac{4}{3} \pi \times r^{3} = \pi \times R^{2} \times h$
43π×(r)3=π×122×6.75$\Rightarrow \frac{4}{3} \pi \times (r)^{3} = \pi \times 12^{2} \times 6.75$
r3=π×144×6.7543π$\Rightarrow r^{3} = \frac{ \pi \times 144 \times 6.75}{ \frac{4}{3} \pi}$
r3=972×34=29164=729$\Rightarrow r^{3} = \frac{972 \times 3}{4} = \frac{2916}{4} = 729$

$\Rightarrow$ r = 9 cm

∴ The Radius of the ball = 9 cm.

Q.20: A cylindrical bucket with base radius 15 cm is filled with water up to a height of 20 cm. A heavy iron spherical ball of radius 9 cm is dropped into the bucket to submerge completely in the water. Find the increase in the level of water.

Solution:

Radius of the cylindrical bucket = 15 cm

Height of the cylindrical bucket = 20 cm

Volume of the water in the bucket = 𝜋 × 15 × 15 × 20 cm3

Radius of the spherical ball = 9 cm

Volume of the spherical ball = 43×π×9×9×9$\frac{4}{3} \times \pi \times 9 \times 9 \times 9$ cm3 . . . . . . . . (1)

Increase in the water level = h cm

Volume of the increased water level = 𝜋 × 15 × 15 × h cm3 . . . . . . . (2)

Equating (1) and (2), we have,

𝜋 × 15 × 15 × h = 43×π×9×9×9$\frac{4}{3} \times \pi \times 9 \times 9 \times 9$

h = 43×π×9×9×9π×15×15$\frac{\frac{4}{3} \times \pi \times 9 \times 9 \times 9}{ \pi \times 15 \times 15}$

h = 4.32 cm

Q.21: A hemispherical of the lead of radius 9 cm is cast into the right circular cone of height 72 cm. Find the radius of the base of the cone.

Solution:

Radius of hemisphere = 9 cm

Height of cone = 72 cm

Let the radius of the base of cone be ‘r’ cm

Then, 13×π×r2×h=23×π×R3$\frac{1}{3} \times \pi \times r^{2} \times h = \frac{2}{3} \times \pi \times R^{3}$

13×π×r2×72=23×π×93$\Rightarrow \frac{1}{3} \times \pi \times r^{2} \times 72 = \frac{2}{3} \times \pi \times 9^{3}$
r2=23×π×7213×π×r2×72=2×72972$\Rightarrow r^{2} = \frac{ \frac{2}{3} \times \pi \times 72}{ \frac{1}{3} \times \pi \times r^{2} \times 72} = \frac{2 \times 729}{72}$
r2=145872=20.25$\Rightarrow r^{2} = \frac{1458}{72} = 20.25$
r=4.5cm$\Rightarrow r = 4.5 cm$

∴ The radius of the base of the cone = 4.5 cm.

Q.22: A hemispherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into the cylindrical shaped small bottle of diameter 3 am and height 4 cm. How many bottles are required to empty the bowl?

Solution:

Here, internal radius of hemisphere bowl ® = 9 cm

Diameter of bottle = 3 cm

radius(r)=32cm$\Rightarrow radius (r) = \frac{3}{2} cm$

And , height of bottle = 4 cm

Number of bottles = VolumeofthebowlVolumeofeachbottle$\frac{ Volume \, of \, the \, bowl}{ Volume \, of \, each \, bottle}$

= 23×π×R3π×r2×h$\frac{\frac{2}{3}\times \pi \times R^{3}}{\pi \times r^{2}\times h}$

= 23×π×93π×(32)2×4$\frac{ \frac{2}{3} \times \pi \times 9^{3}}{ \pi \times (\frac{3}{2})^{2} \times 4}$

= 23×9×9×994×4$\frac{\frac{2}{3} \times 9 \times 9 \times 9}{\frac{9}{4} \times 4}$

= 2×3×819=54$\frac{2 \times 3 \times 81}{9} = 54$

∴ The number of bottles required = 54

Q.23: A hollow spherical shell is made of a metal of density 4.5 g/cm2$g/cm^{2}$. If its internal and external radii are 8 cm and 9 cm respectively, find the weight of the shell.

Solution:

Internal radius (r) = 8 cm

External radius (R) = 9 cm

Density of metal = 4.5 g per cm3

∴ weight of the shell = 43π×[(R)3(r)3]×density$\frac{4}{3} \pi \times [ (R)^{3} â€“ (r)^{3} ] \times density$

= 43×227×[729512]×4.51000$\frac{4}{3} \times \frac{22}{7} \times [ 729 â€“ 512] \times \frac{4.5}{1000}$ kg.

= 43×227×217×4.51000$\frac{4}{3} \times \frac{22}{7} \times 217 \times \frac{4.5}{1000}$ kg.

= 8593221000$\frac{85932}{21000}$ kg = 4.092 kg

∴ weight of the shell = 4.092 kg.