NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.8

Question 1.
Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 m
Solution:
(i) We have radius of sphere = 7 cm
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Q1
∴ Volume of sphere = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7} \times 7 \times 7 \times 7\)
= 1437 \(\frac {1}{3}\) cm3

(ii) We have
Radius of sphere = 0.63 m
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Q1.1
∴ Volume of sphere = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7} \times 0.63 \times 0.63 \times 0.63\)
= 1.0478 m3
= 1.05 m3 (approx)

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

Question 2.
Find the amount of water displaced by a solid spherical ball of diameter.
(i) 28 cm
(ii) 0.21 m
Solution:
(i) We have given that
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Q2
Diameter of spherical ball = 28 cm
∴ Radius of spherical ball = \(\frac {28}{2}\) = 14 cm
Volume of spherical ball = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14\)
= 11498\(\frac {2}{3}\) cm3
Therefore, amount of water displaced by solid sphere is 11498\(\frac {2}{3}\) cm3.

(ii) Diameter of spherical ball = 0.21 m
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Q2.1
∴ Radius of spherical a ball = \(\frac {0.21}{2}\) = 0.105 m
So, volume of spherical ball = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7}\) × 0.105 × 0.105 × 0.105
= 0.004851 m3
Therefore, the amount of water displaced by the solid sphere is 0.004851 m3

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

Question 3.
The diameter of a metallic ball is 4.2. What is the mass of the ball, if the density of the metal is 8.9 g per cm3?
Solution:
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Q3
We have given that diameter of metallic ball = 4.2 cm
Radius of metallic ball = \(\frac {4.2}{2}\) = 2.1 cm
volume of metallic ball = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7}\) × 2.1 × 2.1 × 2.1
= 38.808 cm3
The density of the metal is 8.9 g per cm3.
Mass of the metallic ball = 38.808 × 8.9 = 345.39 g (approx)

Question 4.
The diameter of the moon is approximately one-fourth the diameter of the earth. What fraction of the volume of the earth is the volume of the moon.
Solution:
Let the diameter of earth = x
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Q4
∴ Radius of earth = \(\frac{x}{2}\)
Volume of earth = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \pi \times \frac{x}{2} \times \frac{x}{2} \times \frac{x}{2}\)
= \(\frac{\pi x^{3}}{6}\)
Again, the diameter of the moon is one-fourth of the diameter of the earth.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Q4.1
Diameter of moon = \(\frac{x}{4}\)
Radius of moon = \(\frac{x}{8}\)
∴ Volume of moon = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \pi \times \frac{x}{8} \times \frac{x}{8} \times \frac{x}{8}\)
= \(\frac{x^{3} \pi}{384}\)
Therefore, fraction of the volume of the moon and the volume of earth is = \(\frac{\frac{\pi x^{3}}{384}}{\frac{\pi x^{3}}{6}}=\frac{6}{384}=\frac{1}{64}\)

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

Question 5.
How many liters of milk can a hemispherical bowl of diameter 10.5 cm hold.
Solution:
We have a diameter of hemispherical bowl = 10.5 cm
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Q5
∴ Radius of hemispherical = \(\frac{10.5}{2}\) = 5.25 cm
∴ Volume of hemisphere = \(\frac{2}{3} \pi r^{3}\)
= \(\frac{2}{3} \times \frac{22}{7}\) × 5.25 × 5.25 × 5.25
= 303.187 cm3
We know that 1000 cm3 = 1 litre
∴ 303.18 cm3 = 0.303 litre (approx)
Therefore, a hemispherical bowl of diameter 10.2 cm holds 0.303 liters of milk.

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:
We have given that
The inner radius of hemispherical tank = 1 m
and thickness of iron sheet = 1 cm = 0.01 m
The outer radius of hemispherical tank = 1.01 m
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Q6
So, volume of iron used to make the tank = Volume of out spherical tank – Volume of inner spherical tank
= \(\left.\frac{2}{3} \pi(1.01)^{3}\right)-\frac{2}{3} \pi(1)^{3}\)
= \(\frac{2}{3}\) × π × 1.030301 – \(\frac{2}{3}\) × π × 1
= \(\frac{2}{3}\) × π × (1.030301 – 1)
= \(\frac{2}{3} \times \frac{22}{7} \times 0.030301\)
= 0.06348 m3 (approx)
Therefore, the volume of the iron used to make the tank is 0.06348 m3 (approx)

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

Question 7.
Find the volume of a sphere whose surface area is 154 cm2.
Solution:
We have given that
The surface area of a sphere is 154 cm2
but, the surface area of sphere = 4πr2
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Q7
Volume of sphere = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7}\) × 3.5 × 3.5 × 3.5
= 179\(\frac{2}{3}\) cm3

Question 8.
A dome of a building is in the form of a hemisphere. From inside it was while washed at the cost of Rs. 498.96. If the cost of whitewashing is Rs. 2.00 per square meter find the:
(i) Inside the surface area of the dome.
(ii) Volume of the air inside the dome.
Solution:
(i) We have given that Rs. 2 is the rate of whitewashing of 1 m2 area
Rs. 1 is the rate of white-washing \(\frac{1}{2}\) m2 area
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Q8
Rs. 498.96 is the rate of white-washing = \(\frac{1}{2}\) × 498.96 = 249.48 m2 area.
Therefore, according to question, inside surface area of dome = 249.48 m2

(ii) We have,
surface area of hemispherical dome = 249.48 m2
but surface area of hemispherical dome = 2πr2
⇒ 2πr2 = 249.48
⇒ r2 = \(\frac{249.48 \times 7}{2 \times 22}\)
⇒ r2 = 39.69
⇒ r = √39.69 = 6.3 m
Therefore,
Volume of hemispherical dome = \(\frac{2}{3} \pi r^{3}\)
= \(\frac{2}{3} \times \frac{22}{7}\) × 6.3 × 6.3 × 6.3
= 523.908 m3
Therefore, the volume of the air inside the dome is 523.9 m3 (approx).

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

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:
We have given that
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Q9
The radius of sphere = r
Surface area(s) = 4πr2
and volume of sphere = \(\frac{4}{3} \pi r^{3}\)
Volume of 27 such speres = \(\frac{4}{3} \times \pi r^{3} \times 27\) = 36πr3
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Q9.1
According to question Volume of 27 small sphere = volume of big sphere
or, 36πr3 = \(\frac{4}{3} \pi\left(r^{\prime}\right)^{3}\) (∵ radius = r’)
⇒ (r’)3 = \(\frac{36 \pi r^{3} \times 3}{4 \times \pi}\)
(∴ volume of big sphere = volume of 27 small sphere)
⇒ (r’)3 = 27r3
⇒ r’ = 3r
Therefore, radius r’ of new sphere is equal to 3r.

(ii) Surface area of new-sphere (s’) =4π(r’)2
and surface area of small sphere = 4πr2
∴ Ratio between them = \(\left(\frac{4 \pi r^{2}}{4 \pi\left(r^{\prime}\right)^{2}}\right)\)
= \(\left(\frac{4 \pi r^{2}}{4 \pi \times 3 r \times 3 r}\right)\) (∴ r’ = 3r)
= \(\frac{1}{9}\)
= 1 : 9

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

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:
We have given
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Q10
Diameter of spherical capsule = 3.5 mm
∴ Radius of spherical capsule = \(\frac{3.5}{2}\) = 1.75 mm
∴ Volume of spherical capsule = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7}\) × 1.75 × 1.75 × 1.75
= 22.4583 mm3 (approx)
Therefore, medicine is needed to fill the capsule is 22.4583 mm3 (approx)

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