ML Aggarwal Mensuration Exe-17.3 Class 10 ICSE Maths Solutions

ML Aggarwal Mensuration Exe-17.3 Class 10 ICSE Maths Solutions. We Provide Step by Step Answer of  Exe-17.3 Questions for Mensuration as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10.

ML Aggarwal Mensuration Exe-17.3 Class 10 ICSE Maths Solutions

Board ICSE
Publications Avichal Publishig Company (APC)
Subject Maths
Class 10th
Chapter-17 Mensuration
Writer ML Aggarwal
Book Name Understanding
Topics Solution of Exe-17.3 Questions
Edition 2022-2023

Mensuration Exe-17.3

ML Aggarwal Class 10 ICSE Maths Solutions

Page 412

Question 1. Find the surface area of a sphere of radius 14 cm

Answer :

Given radius of the sphere, r = 14 cm

Surface area of the sphere = 4πr2

= 4 × (22/7) × 142

= 4 × 22 × 14 × 2

= 2464 cm3

Hence the surface area of the sphere is 2464 cm2.

Question 2. Find the surface area of a sphere of diameter  21 cm 

Answer :

Given diameter of the sphere, d = 21 cm

Radius, r = d/2 = 21/2 = 10.5

Surface area of the sphere = 4πr2

= 4 × (22/7) × 10.52

= 1386 cm2

Hence the surface area of the sphere is 1386 cm2.

Question 3. A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm3, find the mass of the shot-put.

Answer :

Radius of the metallic shot-put = 4.9 cm

Volume of the sphere, V = (4/3)πr3

V = (4/3) × (22/7) × 4.93

⇒ V = 493.005

⇒ V = 493 cm3 (approx)

Given Density = 7.8 g per cm3

Density = Mass/Volume

Mass = Density × Volume

= 7.8 × 493

3845.4 g

Hence, the mass of the shot put is 3845.4 g.

Question 4. Find the diameter of a sphere whose surface area is 154 cm2.

Answer :

Surface area of a sphere = 154 cm2

Surface area of the sphere = 4πr2

4 × (22/7) × r2 = 154

⇒ r2 = 154 × 7/(22 × 4) = 49/4

⇒ r = √49/2

Diameter = 2 × r = 2 × √49/2 = √49 = 7

Hence the diameter of the sphere is 7 cm.


Mensuration Exe-17.3

ML Aggarwal Class 10 ICSE Maths Solutions

Page 413

Question 5. Find:

(i) the curved surface area.
(ii) the total surface area of a hemisphere of radius 21 cm.

Answer :

Radius of a hemisphere = 21 cm

(i) Curved surface area = 2πr2

Curved surface area of the hemisphere = 2πr2

= 2 × (22/7) × 212

= 2 × 22 × 3 × 21

= 2772 cm2

Hence the curved surface area of the hemisphere is 2772 cm2.

(ii) Total surface area of the hemisphere = 3πr2

= 3 × (22/7) × 212

= 3 × 22 × 3 × 21

= 4158 cm2

Hence, the total surface area of the hemisphere is 4158 cm2.

Question 6. A hemispherical brass bowl has inner- diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2.

Answer :

The inner diameter of hemispherical bowl = 10.5 cm

Radius, r = d/2 = 10.5/2 = 5.25 cm

Curved surface area of the bowl = 2πr2

= 2 × (22/7) × 5.252

= 173.25 cm2

Rate of tin plating = Rs.16 per 100 cm2

So total cost = 173.25 × 16/100 = 27.72

Hence the cost of tin plating the bowl on the inside is Rs. 27.72.

Question 7. The radius of a spherical balloon increases from 7 cm to 14 cm as air is jumped into it. Find the ratio of the surface areas of the balloon in two cases.

Answer :

Original radius of balloon = 7 cm
Radius after filling the air in it = 14 cm
The surface area of balloon, the original position

Surface area of the sphere = 4πr2

Ratio of surface areas of the balloons = 4πr2/4πR2

= r2/R2

= 72/142

= 1/4

Hence the ratio of the surface areas of the spheres is 1:4.

Question 8. A sphere and a cube have the same surface. Show that the ratio of the volume of the sphere to that of the cube is √6 : √π

Answer :

Let the edge of a cube = a
Surface area = 6a2
and surface area of sphere = 6a2

Given sphere and cube has same surface area.

4πr2 = 6a2

⇒ πr2/a2 = 6/4

∴ r/a = √6/2√π

Volume of the sphere, V= (4/3)πr3

Volume of the cube, V= a3

∴ V1/V2 = 4πr3/3a3

⇒ V1/V2 = 4π√63/(3 × 2√π)3

⇒ V1/V2 = (4π × 6√6)/(3 × 8π × √π)

∴ V1/V2 = √6/√π

Hence proved.

Question 9.

(a) If the ratio of the radii of two sphere is 3 : 7, find :
(i) the ratio of their volumes.
(ii) the ratio of their surface areas.
(b) If the ratio of the volumes of the two sphere is 125 : 64, find the ratio of their surface areas.

Answer :

(a) Ratio in radii of two spheres = 3 : 7

Volume of sphere = (4/3)πr3

Ratio of the volumes = (4/3)πr13/(4/3)πr23

= r13/ r23

= 33/73

= 27/343

Hence, the ratio of their volumes is 27 : 343.

(ii) Surface area of a sphere = 4πr2

Ratio of surface areas of the spheres = 4πr12/4πr22

= r12/r22

= 32/72

= 9/49

Hence, the ratio of the surface areas is 9:49.

(b) Given ratio of volume of two spheres = 125/64

(4/3)πr13/(4/3)πr23 = 125/64

⇒ r13/ r23 = 125/64

Taking cube root on both sides

r1/r2 = 5/4

Ratio of surface areas of the spheres = 4πr12/4πr22

= r12/r22

= 52/42

= 25/16

Hence, the ratio of the surface areas is 25:16.

Question 10. Find the volume of a sphere whose surface area is 154 cm².

Answer :

Given that
Surface area of a sphere = 154 cm²

4πr2 = 154

⇒ 4 × (22/7) × r2 = 154

⇒ r2 = (154 × 7)/(4 × 22)

⇒ r2 = 49/4

⇒ r = 7/2

Volume of the sphere = (4/3)πr3

= (4/3) × (22/7) × (7/2)3

= 539/3

= 179.666

= 179.67 cm3 (approx)

Hence the volume of the sphere is 179.67 cm3

Question 11. If the volume of a sphere is 179.2/3. Find its radius and surface area.

Answer :

Given volume of the sphere is 179.2/3

(4/3)πr3 = 179.2/3 = 539/3

⇒ (4/3) × (22/7) × r= 539/3

⇒ r3 = (539 × 3 × 7)/(4 × 22 × 3)

⇒ r3 = 49 × 7/8

⇒ r3 = (7 × 7 × 7)/(2 × 2 × 2)

Taking cube root on both sides, we get

r = 7/2 = 3.5 cm

Surface area of the sphere = 4r2

(4/3) × (22/7) × (7/2)2

= 22 × 7

= 154 cm2

Hence, the radius and the surface area of the sphere is 3.5 cm and 154 cm2 respectively.

Question 12. A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?

Answer :

Radius of a hemispherical bowl (r) = 3.5 cm
= 7/2 cm

Volume of the hemisphere = (2/3)πr3

= (2/3) × (22/7) × (7/2)3

= 11 × 49/6

= 539/6

= 89. 5/6 cm3

Hence, the volume of the hemispherical bowl is 89.5/6 cm3.

(ML Aggarwal Mensuration Exe-17.3 Class 10)

Question 13. The surface area of a solid sphere is 1256 cm². It is cut into two hemispheres. Find the total surface area and the volume of a hemisphere. Take π = 3.14.

Answer :

Given surface area of the sphere = 1256 cm2

4πr2 = 1256

⇒ 4 × 3.14 × r2 = 1256

⇒ r2 = 1256 × /3.14 × 4

⇒ r2 = 100

⇒ r = 10 cm

Total surface area of the hemisphere = 3πr2

= 3 × 3.14 × 102

= 3 × 3.14 × 100

= 942 cm2

Hence,

the total surface area of the hemisphere is 942 cm2.

Volume of the hemisphere = (2/3)πr3

= (2/3) × 3.14 × 103

= (2/3) × 3.14 × 1000

= (2/3) × 3140

= 6280/3

= 2093.1/3 cm3

Hence, the volume of the hemisphere is 2093.1/3 cm3

-: End of ML Aggarwal Mensuration Exe-17.3 Class 10 ICSE Maths Solutions :–

Return to :- ML Aggarwal Solutions for ICSE Class-10

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