NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes Ex 11.4
Ex
11.4 Class 9 Maths Question 1.
Find the volume of a sphere whose radius is:
(i) 7 cm
(ii) 0.63 cm
Solution:
(i) Here, radius (r) = 7 cm
Thus, the required volume = 143713 cm3
(ii) Here, radius (r) = 0.63 m
Thus, the required volume is 1.05 m3 (approx.)
Ex 11.4 Class 9 Maths 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
(ii)
Diameter of the ball = 0.21 m
⇒ Radius (r) = 0.21/2 m = 21/200 m
Thus, the amount of water displaced = 0.004851 m3.
Ex 11.4 Class 9 Maths 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 of the metallic ball (r) = 4.2/2 cm = 2.1 cm
Density of the metal = 8.9 g per cm3
∴ Mass of the ball = 8.9 × [Volume of the ball]
Thus, the mass of ball is 345.39 g (approx.)
Ex 11.4 Class 9 Maths Question 4.
The diameter of the 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?
Solution:
Let the diameter of the earth be 2r.
⇒ Radius of the earth = 2r/2 = r
Since, diameter of the moon = ¼ (Diameter of
the earth)
⇒ Radius of the moon = ¼ (Radius of the earth)
Radius of the moon = ¼ (r) = r/4
∴ Volume of the earth = 4/3 Ï€r3 and
Ex 11.4 Class 9 Maths Question 5.
How many litres of milk can a hemispherical bowl of
diameter 10.5 cm hold?
Solution:
Diameter of the hemispherical bowl = 10.5 cm
⇒ Radius of the hemispherical bowl (r) = 10.5/2 cm = 105/20 cm
Thus, the capacity of the bowl = 0.303 litres (approx.)
Ex 11.4 Class 9 Maths 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) = 1 m
∵ Thickness = 1 cm = 1/100 m
= 0 .01 m
Thus, the required volume of the iron used = 0.06348
m3 (approx.)
Ex 11.4 Class 9 Maths Question 7.
Find the volume of a sphere whose surface area is
154 cm2.
Solution:
Let ‘r’ be the radius of the sphere.
∴ Surface area of the sphere = 4Ï€r2
4Ï€r2 = 154 [Given]
Ex 11.4 Class 9 Maths Question 8.
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.
Solution:
(i) Total cost of white-washing = ₹ 498.96
Cost of 1 m² of white-washing = ₹ 2
∴ Area = Total/Cost/Cost of 1 m2 area = 498.96/2 = 249.48 m2
Thus, the required surface area of the dome is 249.48 m2.
(ii) Let ‘r’ be the radius of the hemispherical
dome.
∴ Inside surface area of the dome = 2Ï€r2
Now, volume of air inside the dome = Volume of a hemisphere
Ex 11.4 Class 9 Maths 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 new sphere,
(ii) ratio of S and S’.
Solution:
(i) The radius of the small sphere is r.
∴ Its volume = 4/3 Ï€r3
Volume of 27 small spheres 27 × [4/3 Ï€r3]
Let the radius of the new sphere be r’.
∴ Volume of the new sphere = 4/3 Ï€(r’)3
Hence, the radius of the new sphere is 3r.
(ii) Surface area of a sphere = 4Ï€r2
S = 4Ï€r2 and S’ = 4Ï€ (3r)2 [∵ r’ = 3r]
Thus, S : S’ = 1 : 9
Ex 11.4 Class 9 Maths 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
= 22.45833 mm3
Thus, the required quantity of medicine = 22.46
mm3 (approx.)
NCERT Solutions for Maths Class 10
NCERT Solutions for Maths Class 11