RS Aggarwal Solutions Class 9 Ex 13C

RS Aggarwal Class 9 Ex 13C Chapter 13

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$\frac{ \frac{4}{3} \times \frac{9261}{8}}{ \frac{1}{3} \times (\frac{7}{2})^{2} \times 3} = \frac{ \frac{9621}{6}}{ \frac{49}{4}}$ = 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$27 – \frac{27 + 64}{8}$=27918$27 – \frac{91}{8}$= 216918$\frac{216 – 91}{8}$

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$\frac{4}{3} \pi \times [ (R)^{3} – (r)^{3} ] \times density$

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

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The third most abundant gas in the atmosphere is carbon dioxide.