9th Maths 13.8
Chapter 13
Surface Areas and Volumes
NCERT Class 9th solution of Exercise 13.1
NCERT Class 9th solution of Exercise 13.2
NCERT Class 9th solution of Exercise 13.3
NCERT Class 9th solution of Exercise 13.4
NCERT Class 9th solution of Exercise 13.5
NCERT Class 9th solution of Exercise 13.6
NCERT Class 9th solution of Exercise 13.7
NCERT Class 9th solution of Exercise 14.1
Exercise 13.8
Assume π=227, unless stated otherwise.
Q1. Find the volume of a sphere whose radius, is
i) 7cm ii) 0.63m
Sol. :
i)
Volume of a sphere =43πr3
V=43×227×(7)3
V=43×227×7×7×7
V=4×22×493
V=43123
V=1437.33cm3 (approx)
Answer:
V=1437.33cm3.
ii)
Volume of a sphere =43πr3
V=43×227×(0.63)3
V=43×22×0.63×0.63×0.09
V=1.05m3
Answer:
V=1.05m3.
Q2. Find the amount of water displaced by a solid spherical ball of diameter
i) 28cm ii) 0.21m
Sol. :
i)
Given:
Radius r=282=14cm,
Solve:
Amount of water displaced = Volume of the spherical ball
=43πr3
=43×227×(14)3
=43×22×14×14×2
=344963
=11498.67cm3 (approx)
Answer:
Amount of water displaced 11498.67cm3
ii)
Given:
Radius r=0.212cm
Solve:
Amount of water displaced = Volume of the spherical ball
=43πr3
=43×227×(0.212)3
=43×227×0.212×0.212×0.212
=11×0.03×0.07×0.21
=0.004851m3
Answer:
Amount of water displaced 0.004851m3.
Q3. The diameter of metallic ball is 4.2cm. What is the mass of the ball, if the density of the metal is 8.9g per cm3?
Sol. :
Given:
Radius of metallic ball r=4.22=2.1cm
Density of metal is 8.9g per cm3
To find:
Mass of the metal ball.
Solve:
Volume of the metal ball =43πr3
=43×227×(2.1)3
=38.808cm3
Mass of the metal ball
= Volume of the metal ball×Density of metal ball
=38.808×8.9
=345.3912gm
Answer:
Mass of the metal ball 345.3912gm.
Q4. The diameter of moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
Sol. :
Given:
Radius of the earth re=D2
Radius of the moon rm=D2×4=D8
To Find:
Fraction of the Volume of the earth is the volume of the moon.
Solve:
=VmoonVearth
=43π(rm)343π(re)3
=(rm)3(re)3
=(D2D8)3
=2×2×28×8×8
=164
Answer:
Fraction of the Volume of the earth is the Volume of the moon 164.
Q5. How many litres of milk can a hemispherical bowl of diameter 10.5cm hold?
Sol. :
Given:
Radius of a bowl r=10.52cm
To find:
The volume of milk in litre.
Solve:
Volume of bowl =23πr3
=23×227×(10.52)3
=303.1875cm3
=303.2cm3
=0.3032litres [1cm3=11000litres]
Answer:
The volume of milk is 0.3032litres.
Q6. A hemispherical tank is made up of an iron sheet 1cm thick. If the inner radius is 1m, then find the volume of the iron used to make the tank.
Sol. :
Given:
The thickness of the iron sheet =1cm=0.01m
Inner Radius of the iron tank r2=1m
Outer Radius of the iron tank r1=1+0.01=1.01m
To find:
The volume of the iron used to make the tank.
Solve:
Volume of hemisphere =23π(r31-r32)
=23×227[(1.01)3-(1)3]
=4421(1.03-1)
=4421×0.03
=0.447
=0.06286m3 (approx)
Answer:
The volume of the iron used to make the tank 0.06286m3 (approx)`
Q7. Find the volume of a sphere whose surface area is 154cm2.
Sol. :
Given:
The surface area of the sphere is 154m3
To Find:
The volume of the sphere.
Solve:
The surface area of the sphere =4πr2
154
r^2 = (154times7)/(4times22)
r^2 = (7times7)/4
r^2 =(7/2)^2
r = 7/2
The volume of the sphere =4/3pir^3
=4/3times22/7times(7/2)^3
=(11times49)/3
=539/3
=179.67 cm^3 (approx)
Answer:
The volume of the sphere is 179.67 cm^3 (approx).
Q8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹498.96. If the cost of white-washing is ₹2.00 per square metre., find the
i) inside surface area of the dome, ii) volume of the air inside the dome.
Sol. :
Given:
The total cost of white-wash ₹ 498.96
The rate of the white-wash ₹ 2.00.
To Find:
i) inside surface area of the dome
ii) volume of the air inside the dome.
Solve:
i) inside surface area of the dome = (total cost)/rate
=498.96/2=249.48 m^2
Answer:
The inside surface area of the dome 249.48 m^2.
ii) The inside surface area of the dome =2pir^2
249.48 = 2times22/7times r^2
r^2 =(249.48times7)/44
r^2 =39.69
r^2 =(6.3)^2
r = 6.3
The volume of the air inside the dome=2/3pir^3
=2/3times22/7times(6.3)^3
=44times6.3times6.3times0.3
=523.9 m^3 (approx)
Answer:
The volume of the air inside the dome 523.9 m^3.
Q9. 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 new sphere, ii) ratio of S and S'.
Sol. :
Given:
Twenty-seven spheres each Radius r and
Surface area S,
new Surface area S'
To Find:
i) radius r' of the new sphere,
ii) ratio of S and S'
Solve:
i) The volume of new sphere = 27timesVolume of sphere
4/3pi(r')^3=27times4/3pi(r)^3
(r')^3=(3)^3times(r)^3
(r')^3=(3r)^3
r'=3r
Answer:
r'=3r
ii) Ratio =S:S'=4pir^2:4pir'^2
S: S'=r^2:r'^2
S: S'=r^2:(3r)^2 = r^2:9r^2=1:9
Answer:
The ratio S:S=1:9`.
Q10. A capsule of medicine is in the shape of a sphere of diameter 3.5mm. How much medicine (in mm) is needed to fill this capsule.
Sol. :
Given:
The radius of capsule 3.5/2mm
To Find:
Volume of capsule
Solve:
The volume of the sphere capsule =4/3pir^3
=4/3times22/7times(3.5/2)^3
=11/3times3.5times3.5times3.5
=22.46mm^3 (approx)
Answer:
The volume of the sphere capsule 22.46mm^3.
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