Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Mathematics (Maths) Class 9

Created by: Full Circle

Class 9 : Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

The document Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

Note: Assume π = (22/7), unless stated otherwise.

Question 1. Find the volume of a sphere whose radius is (i) 7 cm (ii) 0.63 m
 Solution:
(i) Here, radius (r) = 7 cm

∴ Volume of the sphere = Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Thus, the required volume = 1437 (1/3)cm3

(ii) Here, radius (r) = 0.63 m

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

= 1.047816 m3 = 1.05 m3 (apπrox)

Thus, the required volume is 1.05 m3 (apπrox).


Question 2. Find the amount of water displaced by a solid spherical ball of diameter (i) 28 cm (ii) 0.21 m
 Solution: 
(i) Diameter of the ball = 28 cm
∴ Radius of the ball (r) =(28/2) cm = 14 cm

⇒ Volume of the spherical ball = 

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Thus, the amount of water displaced = 11498 (2/3) cm3.

(ii) Diameter of the solid spherical ball = 0.21 m

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev


Question 3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3?
 Solution
: Diameter of the metallic ball = 4.2 cm
⇒ Radius (r) = (42/2) cm = 2.1 cm

∴ Volume of the metallic ball = 

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

∵ Density = 8.9 g per cm3
∴ Mass of the ball = 8.9 x [Volume of the ball]

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev
= 345.39 g (apπrox.)
Thus, the mass of ball is 345.39 g (apπrox.).


Question 4. The diameter of the moon is apπroximately one-fourth of the diameter of the earth.
 What fraction of the volume of the earth is the volume of the moon?
 Solution:
Let diameter of the earth be 2r.

⇒ Radius of the earth =(2r/2)  = r
∵ Diameter of the moon = (1/4)(Diameter of the earth)
∴ Radius of the moon = (1/4)(Radius of the moon)
⇒ Radius of the moon = (1/4) (r) = (r/4)
∴ Volume of the earth = (4/3)πr 3

Volume of the moon = Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

∴ The required fraction is (1/64)
.

Question 5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
 Solution:
Diameter of the hemisphere = 10.5 cm
⇒ Radius of the hemisphere (r) = (10.5/2) cm = (105/20) cm
Volume of the hemispherical bowl = (2/3) πr 3

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

∴ Capacity of the hemispherical bowl = 303.1875 cm3

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev = 0.3031875 l = 0.303 l (apπrox.)                              [∵ 1000 cm3 = 1 l]
Thus, the capacity of the bowl = 0.303 l (apπrox.)


Question 6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
 Solution: 
Inner radius (r) = 1m
∵ Thickness = 1 cm = (1/100) m = 0.01 m
∴ Outer radius (R) = 1 m + 0.01 m = 1.01 m
Now, outer volume of the hemispherical bowl

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Inner volume of the hemispherical bowl

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

∴ Volume of the iron used = [Outer volume] – [Inner volume]

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

= 0.063487776 m3
= 0.06348 m3 (apπrox.)
Thus, the required volume of the iron = 0.06348 m3


Question 7. Find the volume of a sphere whose surface area is 154 cm2
 Solution: 
Let ‘r’ be the radius of the sphere.
∴ Its surface area = 4πr2
⇒ 4πr= 154

 

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Now, volume of the sphere = Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Thus, the required volume of the sphere = Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev  cm3


Question 8. A dome of a building is in the form of a hemisphere. From inside, it was white washed at the cost of Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev 4989.60. If the cost of white washing is Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev20 per square metre, find the (i) inside surface area of the dome, (ii) volume of the air inside the dome.
 Solution: 
Total cost of white washing = Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev 4989.60
Rate of white washing = Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev 20 per m

∴   Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Let ‘r’ be the radius of the hemispherical dome.
∴ Surface area = 2πr2
∴   2πr2 = 249.48

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

(i) To find the inside surface area of the dome:
∵ Radius of the hemisphere (r) = 6.3 m  
Surface area of a hemisphere = 2πr2 m2
∴ Surface area of the dome = 2 x (22/7) (6.3)2 m2

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Thus, the required surface area of the dome = 249.48 m

(ii) To find the volume of air in the dome:
Volume of a hemisphere = (2/3)πr 3

 ∴ Volume of the dome Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Thus, the required volume of air inside the dome is 523.9 m3 (apπrox.).


Question 9. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find the (i) radius r’ of the sphere, (ii) ratio of S and S'.
 Solution:
(i) To find r ':
∵ Radius of a small sphere = r

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Since, 

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

 

(ii) To find ratio of S and S ':
∵ Surface area of a small sphere = 4πr
∴ S= 4πrand S ' = 4p(3r)2

Now,  Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

 Thus, S : S' = 1 : 9


Question 10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule?
 Solution: 
Diameter of the spherical capsule = 3.5 mm
⇒ Radius of the spherical capsule (r) = (3.5/2) mm
∴ Volume of the spherical capsule  Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Thus, the required quantity of medicine = 22.46 mm3 (apπrox).

Complete Syllabus of Class 9

Dynamic Test

Content Category

Related Searches

ppt

,

Semester Notes

,

Viva Questions

,

Exam

,

Previous Year Questions with Solutions

,

Important questions

,

Sample Paper

,

shortcuts and tricks

,

Summary

,

past year papers

,

practice quizzes

,

Objective type Questions

,

video lectures

,

study material

,

pdf

,

Free

,

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

,

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

,

mock tests for examination

,

Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

,

Extra Questions

,

MCQs

;