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.


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