Assume π = ^{22}/_{7} , unless stated otherwise.

**Find the volume of a sphere whose radius is**

**(i) 7 cm**

**(ii) 0.63 m**

Solution:

(i) Volume of sphere = ^{4}/_{3} πr^{3} = ^{22}/_{7} x ^{4}/_{3} x 7^{3} = 1437.33 m^{3}

(ii) Volume of sphere = ^{4}/_{3} πr^{3} = ^{22}/_{7} x ^{4}/_{3} x (0.63)^{3} = 1.05m^{3}

**Find the amount of water displaced by a solid spherical ball of diameter**

**(i) 28 cm**

**(ii) 0.21 m**

Solution:

(i)d = 28 cm

r = ^{d}/_{2} = ^{28}/_{2} = 14 cm

Volume of sphere = ^{4}/_{3} πr^{3} = ^{22}/_{7} x ^{4}/_{3} x 14^{3} = 11498.67 m^{3}

(ii) d = 0.21 m

r = ^{d}/_{2} = ^{0.21}/_{2} = 0.105 m

Volume of sphere = ^{4}/_{3} πr^{3} = ^{22}/_{7} x ^{4}/_{3} x (0.105)^{3} = 0.004851 m^{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:

d = 4.2 cm

r = ^{d}/_{2} = ^{4.2}/_{2} = 2.1 cm

Volume of sphere = ^{4}/_{3} πr^{3} = ^{22}/_{7} x ^{4}/_{3} x (2.1)^{3} = 38.808 cm^{3}

Density of metal for 1 cm^{3} = 8.9 g

Density of metal for 38.808 cm^{3} = 38.808 x 8.9 = 345.39g

**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 diameter of earth = d and radius of earth = ^{d}/_{2}

Volume of earth = ^{4}/_{3} πr^{3} = ^{4}/_{3} x ^{22}/_{7} x (^{d}/_{2})^{3}

Given, the diameter of the moon is approximately one-fourth of the diameter of the earth.

i.e., diameter of moon = ^{1}/_{4} x d = ^{d}/_{4}

Then radius of moon = ^{d}/_{4} x ^{1}/_{2} = ^{d}/_{8}

Volume of moon = ^{4}/_{3} πr^{3} = ^{4}/_{3} x ^{22}/_{7} x (^{d}/_{8})^{3}

Fraction of the volume of the earth is the volume of the moon = [^{4}/_{3} x ^{22}/_{7} x (^{d}/_{2})^{3}]/[^{ 4}/_{3} x ^{22}/_{7} x (^{d}/_{8})^{3}]

= (^{1}/_{2})^{3}/(^{1}/_{8})^{3}

= (^{1}/_{8})/(^{1}/_{512})

= ^{512}/_{8} = 64

⇒Volume of the moon = ^{1}/_{64} x volume of earth

Hence, volume of the moon is ^{1}/_{64} of volume of earth.

**How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?**

Solution:

d = 10.5 cm

r = ^{d}/_{2} = ^{10.5}/_{2} = 5.25 cm

Volume of the hemisphere bowl = ^{2}/_{3} πr^{3}

= ^{2}/_{3} x ^{22}/_{7} x (5.25)^{3}

= 303.1875 cm^{3}

Therefore, the hemispherical bowl can hold ^{303.1875}/_{1000} litres = 0.303 litres of milk.

**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 of the tank, r = 1m

Thickness of the iron sheet = 1 m = 0.01 m

⸫External radius of the tank , R = (1+ 0.01) = 1.01 m

Volume of the iron used to make the tank = ^{2}/_{3} π(R^{3} – r^{3})

= ^{2}/_{3} x ^{22}/_{7} x (1.01^{3} – 1^{3})

= 0.0635m^{3}

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

Solution:

Surface area if the given sphere = 154cm^{2}

We know, Surface area of the given sphere = 4πr^{2} = 154

4x^{22}/_{7}x r^{2} = 154

r^{2} = ^{154×7}/_{4×22} = ^{49}/_{4}

r = ^{7}/_{2}

Volume of sphere = ^{4}/_{3} πr^{3} = ^{4}/_{3} x ^{22}/_{7} x (^{7}/_{2})^{3} = 179.67cm^{3}

**A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs 498.96. If the cost of white-washing is Rs 2.00 per square metre, find the**

**(i) inside surface area of the dome,**

**(ii) volume of the air inside the dome.**

Solution:

(i) inside surface area of the dome = ^{total cost}/_{cost of white washing per m}^{2} = ^{498.96}/_{2} = 249.48 m^{2}

(ii) Let the radius of the dome be ‘r’ m

Then 2πr^{2} = 249.48

2 x ^{22}/_{7} x r^{2} = 249.48

r^{2} = ^{249.48×7}/_{2×22} = ^{3969}/_{100}

r = ^{63}/_{10} cm

volume of the air inside the dome = ^{2}/_{3}πr^{3} = ^{2}/_{3} x ^{22}/_{7} x (6.3)^{3} = 523.908 cm^{3}

- T
**wenty 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) Volume of a sphere of radius, r = ^{4}/_{3} πr^{3}

Volume of 27 such spheres = 27 x ^{4}/_{3} πr^{3} = 36 πr^{3}

Volume of the sphere with radius r’ = ^{4}/_{3} πr^{3}

⇒ 36 πr^{3 } = ^{4}/_{3} πr^{3}

⇒ 27r^{3} = r’^{3}

⇒ r’ = ∛(27r^{3})

⇒ r’ = 3r

**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:

d = 3.5mm

r = ^{3.5}/_{2} = 1.75 mm

Volume of sphere = ^{4}/_{3} πr^{3}

= ^{4}/_{3} x ^{22}/_{7} x (1.75)^{3}

= 22.46 mm^{3}

Therefore, 22.46mm^{3} medicine is needed to fill this capsule.

## 1 thought on “Surface Area and Volumes – Exercise 13.8 – Class IX”

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