# 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 $$\frac{2}{3}$$ 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 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. 