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

Gujarat Board GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

Question 1.
Find the volume of a sphere whose radius is ∝
(i) 7 cm
(ii) 0.63 m
Solution:
(i) r = 7cm
∴ Volume = \(\frac{4}{3}\) πr³ = \(\frac{4}{3}\) x \(\frac{22}{7}\) x (7)³
= \(\frac{4312}{3}\) = 1437 \(\frac{1}{3}\) cm³

(ii) r = 0.63m
∴ Volume = \(\frac{4}{3}\) πr³ = \(\frac{4}{3}\) x \(\frac{22}{7}\) x (0.63)³
= 1.05 m³ (approx.)

Question 2.
Find the amount of water displaced by a solid
spherical ball of diameter
1. 28 cm
2. 0.21m
Solution:
1. Diameter = 28 cm
∴ Radius (r) = \(\frac{28}{2}\)cm = 14 cm
∴ Amount of water displaced
= \(\frac{4}{3}\)πr³ = \(\frac{4}{3}\) x \(\frac{22}{7}\) x (14)³
= \(\frac{34496}{3}\) cm³ = 11498 \(\frac{2}{3}\) cm³

2. Diameter = 0.21 m
∴ Radius (r) = \(\frac{0.21}{2}\) m = 0.105 m
∴ Amount of water displaced = \(\frac{4}{3}\)πr³
= \(\frac{4}{3}\) x \(\frac{22}{7}\) x (0.105)³
= 0.004851 m³

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:
1. Diameter = 4.2 m
Radius (r) = \(\frac{4.2}{2}\) cm = 2.1 cm
∴Volume = \(\frac{4}{3}\)πr³ = \(\frac{4}{3}\) x \(\frac{22}{7}\) x (2.1)³ = 38.808 cm³

2. Density = 8.9 g per cm³
∴ Mass of the ball = Volume x Density
= 38.808 x 8.9
= 345.39 g (approx.)

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 radius of the earth be r.
Then, diameter of the earth = 2r
∴ Diameter of the moon = \(\frac{1}{4}\)(2r) = \(\frac{r}{2}\)
∴ Radius of the moon = \(\frac{r}{2}\)(\(\frac{r}{2}\)) = \(\frac{r}{4}\)
Volume of the earth (V₁) = \(\frac{4}{3}\)πr³
Volume of the moon (V₂)
= \(\frac{4}{3}\)π (\(\frac{r}{4}\))³ = \(\frac{1}{64}\)(\(\frac{4}{3}\) πr³) = \(\frac{1}{64}\)V₁
= \(\frac{1}{64}\) (Volume of the earth)
Hence, the volume of the moon is \(\frac{1}{64}\) th fraction of the volume of the earth.

Question 5.
How many liters of milk can a hemispherical bowl of diameter 10.5 cm hold?
Solution:
Diameter = 10.5 cm
∴ Radius (r) = \(\frac{10.5}{2}\) = 5.25 cm
∴ Amount of milk = \(\frac{10.5}{2}\)πr³
= \(\frac{2}{3}\) x \(\frac{22}{7}\) x (5.25)³ cm³
= 303 cm³ (approx.)
= 0.303 L (approx.)

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 of iron sheet = 1 cm = 0.01 m
∴ Outer radius (R) = Inner radius (r) + Thickness of iron sheet
= 1 m + 0.01 m = 1.01 m
∴ Volume of the iron used to make the tank
= \(\frac{2}{3}\)πr³(R³ – r³)
= \(\frac{2}{3}\) x \(\frac{22}{7}\) x [(1.01)³ – 1³]
= 0.06348 m³ (approx.)

Question 7.
Find the volume of a sphere whose surface area is 154 cm².
Solution:
Let the radius of the sphere be r cm.
Surface area = 154 cm²
4πr² = 154
4 x \(\frac{22}{7}\) x r² = 154
r² = \(\frac{154 \times 7}{4 \times 22}\)
r² = \(\frac{49}{4}\)
r = \(\sqrt {494}\)
r = \(\frac{7}{2}\) cm

∴ Volume of the sphere = \(\sqrt {43}\)πr³
= \(\frac {4}{3}\) x \(\frac{22}{7}\) x (\(\frac{7}{2}\))³
= \(\frac{539}{3}\) cm³
= 179 \(\frac{2}{3}\) cm^3.

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 whitewashing is 2.00 per square meter, find the
1. inside surface area of the dome,
2. the volume of the air inside the dome.
Solution:
1. Inside surface area of the dome
= \(\frac{498.96}{2.00}\) = 249.48 m²

2. Let the radius of the hemisphere be r m.
Inside surface area = 249.48 m²
2πr³ = 249.48
2 x \(\frac{22}{7}\) x r² = 249.48
r² = \(\frac{249.48 \times 7}{2 \times 22}\)
r² = 39.69
r = \(\sqrt {39.69}\)
r = 6.3m

∴ Volume of the air inside the dome = \(\frac {2}{3}\)πr³
= \(\frac {2}{3}\) x \(\frac {22}{7}\) x (6.3)³
= 523.9 m³ (approx.)

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
1. radius r’ of the new sphere,
2. the ratio of S and S’.
Solution:
1. Volume of a solid iron sphere = \(\frac {4}{3}\)πr³
∴ Volume of 27 solid iron spheres
= 27 (\(\frac {4}{3}\)πr³) = 36πr³
∴ Volume of the new sphere = 36πr³
Let the radius of the new sphere be r’.

Then,
Volume of the new sphere = \(\frac {4}{3}\)πr³
According to the question,
\(\frac {4}{3}\)πr’³ = 36πr³
r’³ = \(\frac{\left(36 \pi r^{3}\right) \times 3}{4 \pi}\)
r’³ = 27r³
r’ = (27r³)^1/3
r’= (3 x 3 x 3 x r³)^1/3
r’ = 3r
Hence, the radius r’ of the new sphere is 3r.

2. The surface area of an iron sphere, S = 4πr²
And the surface area of new sphere,
S’ = 4π(3r)²
∴ \(\frac {S}{S’}\) = \(\frac{4 \pi r^{2}}{4 \pi(3 r)^{2}}\) = \(\frac {1}{9}\) = 1:9
Hence, the ratio of S and S’ is 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 capsule = 3.5 mm
∴ Radius of the capsule (r) = \(\frac {3.5}{2}\)mm = 1.75 mm
∴ Capacity of the capsule
= \(\frac {4}{3}\)πr³ = \(\frac {4}{3}\) x \(\frac {22}{7}\) x (1.75)³ mm³
= 22.46 mm³ (approx.)
Hence, 22.46 mm³ (approx.) of medicine is needed to fill this capsule.