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 = \(\frac{4}{3}\) πr^{3} = \(\frac{4}{3}\) x \(\frac{22}{7}\) x (7)^{3}

= \(\frac{4312}{3}\) = 1437 \(\frac{1}{3}\) cm^{3}

(ii) r = 0.63m

∴ Volume = \(\frac{4}{3}\) πr^{3} = \(\frac{4}{3}\) x \(\frac{22}{7}\) x (0.63)^{3}

= 1.05 m^{3} (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

= \(\frac{4}{3}\)πr^{3} = \(\frac{4}{3}\) x \(\frac{22}{7}\) x (14)^{3}

= \(\frac{34496}{3}\) cm^{3} = 11498 \(\frac{2}{3}\) cm^{3}

2. Diameter = 0.21 m

∴ Radius (r) = \(\frac{0.21}{2}\) m = 0.105 m

∴ Amount of water displaced = \(\frac{4}{3}\)πr^{3}

= \(\frac{4}{3}\) x \(\frac{22}{7}\) x (0.105)^{3}

= 0.004851 m^{3}

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 cm^{3}?

Solution:

1. Diameter = 4.2 m

Radius (r) = \(\frac{4.2}{2}\) cm = 2.1 cm

∴Volume = \(\frac{4}{3}\)πr^{3} = \(\frac{4}{3}\) x \(\frac{22}{7}\) x (2.1)^{3} = 38.808 cm^{3}

2. Density = 8.9 g per cm^{3}

∴ 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 (V_{1}) = \(\frac{4}{3}\)πr^{3}

Volume of the moon (V_{2})

= \(\frac{4}{3}\)π (\(\frac{r}{4}\))^{3} = \(\frac{1}{64}\)(\(\frac{4}{3}\) πr^{3}) = \(\frac{1}{64}\)V_{1}

= \(\frac{1}{64}\) (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 = \(\frac{10.5}{2}\)πr^{3}

= \(\frac{2}{3}\) x \(\frac{22}{7}\) x (5.25)^{3} cm^{3}

= 303 cm^{3} (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

= \(\frac{2}{3}\)πr^{3}(R^{3} – r^{3})

= \(\frac{2}{3}\) x \(\frac{22}{7}\) x [(1.01)^{3} – 1^{3}]

= 0.06348 m^{3} (approx.)

Question 7.

Find the volume of a sphere whose surface area is 154 cm^{2}.

Solution:

Let the radius of the sphere be r cm.

Surface area = 154 cm^{2}

4πr^{2} = 154

4 x \(\frac{22}{7}\) x r^{2} = 154

r^{2} = \(\frac{154 \times 7}{4 \times 22}\)

r^{2} = \(\frac{49}{4}\)

r = \(\sqrt {494}\)

r = \(\frac{7}{2}\) cm

∴ Volume of the sphere = \(\sqrt {43}\)πr^{3}

= \(\frac {4}{3}\) x \(\frac{22}{7}\) x (\(\frac{7}{2}\))^{3}

= \(\frac{539}{3}\) cm^{3}

= 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 m^{2}

2. Let the radius of the hemisphere be r m.

Inside surface area = 249.48 m^{2}

2πr^{3} = 249.48

2 x \(\frac{22}{7}\) x r^{2} = 249.48

r^{2} = \(\frac{249.48 \times 7}{2 \times 22}\)

r^{2} = 39.69

r = \(\sqrt {39.69}\)

r = 6.3m

∴ Volume of the air inside the dome = \(\frac {2}{3}\)πr^{3}

= \(\frac {2}{3}\) x \(\frac {22}{7}\) x (6.3)^{3}

= 523.9 m^{3} (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 = \(\frac {4}{3}\)πr^{3}

∴ Volume of 27 solid iron spheres

= 27 (\(\frac {4}{3}\)πr^{3}) = 36πr^{3}

∴ Volume of the new sphere = 36πr^{3}

Let the radius of the new sphere be r’.

Then,

Volume of the new sphere = \(\frac {4}{3}\)πr^{3}

According to the question,

\(\frac {4}{3}\)πr’^{3} = 36πr^{3}

r’^{3} = \(\frac{\left(36 \pi r^{3}\right) \times 3}{4 \pi}\)

r’^{3} = 27r^{3}

r’ = (27r^{3})^{1/3}

r’= (3 x 3 x 3 x r^{3})^{1/3}

r’ = 3r

Hence, the radius r’ of the new sphere is 3r.

2. The surface area of an iron sphere, S = 4πr^{2}

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 mm^{3}) 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

= \(\frac {4}{3}\)πr^{3} = \(\frac {4}{3}\) x \(\frac {22}{7}\) x (1.75)^{3 }mm^{3}

= 22.46 mm^{3} (approx.)

Hence, 22.46 mm^{3} (approx.) of medicine is needed to fill this capsule.