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 = 43×227×3.5×3.5×3.5 = 179.67 cm3

∴ Surface area of the sphere = 4𝜋r2 = 4×227×3.5×3.5 = 154 cm2

 

(ii) Radius of the sphere = 4.2 cm

∴ Volume of the sphere = 43πr3 = 43×227×4.2×4.2×4.2 = 310.464 cm3

∴ Surface area of the sphere = 4𝜋r2 = 4×227×4.2×4.2 = 221.76 cm2

 

(iii) Radius of the sphere = 5 m

∴ Volume of the sphere = 43πr3

= 43×227×5×5×5 = 523.81 m3

∴ Surface area of the sphere = 4𝜋r2= 4×227×5×5 = 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

38808=43×227×r3 (because, volume = 38808 cm3)

r3=38808×3×788=9261
r=21cm

∴ Surface area of the sphere = 4𝜋r2 = 4×227×21×21 = 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

606.375=43×227×r3 (because, volume = 606.375 cm3)

r3=606.375×3×74×22=144.703125
r=5.25m

∴ Surface area of the sphere = 4𝜋r2 = 4×227×5.25×5.25 = 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
r2=394.24×74×22=31.36
r=31.36=5.6m

∴ Radius of the sphere = 5.6 m

Volume of the sphere = 43πr3

43×227×5.6×5.6×5.6 = 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
r=144=12cm

∴ Volume of the sphere = 43πr3

= 43×227×12×12×12 = 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

Hence, radius = 6 cm

Inner diameter of spherical shell = 8 cm

Hence, radius = 4 cm

Now, Volume of the outer shell = 43πr3 = 43×227×6×6×6 = 905.15 cm3

And, Volume of the inner shell = 43πr3 = 43×227×4×4×4 = 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 = 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

Now, number of lead shots = VolumeofthecuboidVolumeof1leadshot

= 12×11×943×227×(0.32)3 = 12×11×943×227×0.0278 = 12×11×9×3×7×84×22×0.027=84000

∴ Number of lead shots = 84000.

 

Q.8: How many lead balls, each of radius 1 cm, can be made from a sphere of radius 8 cm?

Solution:

Here, radius of 1 lead ball = 1 cm and radius of sphere = 8 cm

But, number of lead balls = VolumeofthesphereVolumeof1leadball

= (43×π×R3)cm3(43×π×R3)cm3

= 43×227×8343×227×13

= 43×227×51243×227×1=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

= 43×227×3343×227×0.33 = 43×227×2743×227×0.027=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 cm

Radius of smaller cone = 3.5 cm = 72 cm and height = 3 cm

Now, number of cones = VolumeofthesphereVolumeof1smallcone

= 43×227×(212)313×227×(72)2×3 = 43×9261813×(72)2×3=96216494 = 92616×449=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

Hence, radius = 6 cm

Therefore, Volume of the sphere = 43πr3

= 43×227×6×6×6 . . . . . . . . . . (i)

Diameter of cylinder = 8 cm

Hence, the radius = 4 cm

Height of the cylinder = 90 cm

Therefore, Volume of the outer shell = πr3

= 227×4×4×90 . . . . . .(ii)

Number of spheres = VolumeofcylinderVolumeofsphere

NUmber of spheres = 227×4×4×9043×227×6×6×6

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

Hence, radius = 3 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

π×(0.1)@×h=43×π×(3)2
h=43×π×27i×(0.1)2

= 4×90.01cm=360.01 = 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

Hence, radius = 9 cm

Length of the wire = 108 m = 10800 cm

Then,

43×π×(r)3=π×r2×10800
r2=43×π×729π×10800
r2=4×24310800=97210800=9100
r=9100=310=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 cm = 7.8 cm

And, height of cone = 31.2 cm

Then, 43π×R3=13π×r2×h

43π×(7.8)3=13π×r2×31.2
r2=43π×(7.8)313π×31.2
r2=4×474.55231.2=60.84=(7.8)2
r=7.8cm

∴ 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 cm = 14 cm

Diameter of cone = 35 cm

∴ the radius of cone = 352 cm = 17.5 cm

∴  43π×R3=13π×r2×h

h=43π×14313π×17.52
h=4×2744306.25cm=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

43×π×27=43π278+43×π×8+43×π×(r)3

27 = 278+8+r3

i.e. r3=27278+8

=2727+648=27918= 216918

i.e. r3=1258r3=(52)3

r = 52=2.5cm

∴ 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

=x24x2=14

∴ 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

(rR)2=(12)2rR=12

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

V1V2=43πr343πr3=(rR)3=(12)3=18

∴ 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
43π×(r)3=π×122×6.75
r3=π×144×6.7543π
r3=972×34=29164=729

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 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

h = 43×π×9×9×9π×15×15

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

13×π×r2×72=23×π×93
r2=23×π×7213×π×r2×72=2×72972
r2=145872=20.25
r=4.5cm

∴ 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

And , height of bottle = 4 cm

Number of bottles = VolumeofthebowlVolumeofeachbottle

= 23×π×R3π×r2×h

= 23×π×93π×(32)2×4

= 23×9×9×994×4

= 2×3×819=54

∴ The number of bottles required = 54

 

Q.23: A hollow spherical shell is made of a metal of density 4.5 g/cm2. 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

= 43×227×[729512]×4.51000 kg.

= 43×227×217×4.51000 kg.

= 8593221000 kg = 4.092 kg

∴ weight of the shell = 4.092 kg.


Practise This Question

What are the factors for which crop improvement is desirable?

i. Higher yield
ii. Improved quality
iii. Good looks
iv. Disease resistance
v. More height
vi. Wider adaptability