The document Ex 13.8 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.

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**Note:** Assume π = (22/7), unless stated otherwise.

**Question 1. Find the volume of a sphere whose radius is (i) 7 cm (ii) 0.63 m Solution:** (i) Here, radius (r) = 7 cm

∴ Volume of the sphere =

Thus, the required volume = 1437 (1/3)cm^{3}

(ii) Here, radius (r) = 0.63 m

= 1.047816 m^{3} = 1.05 m^{3} (apπrox)

Thus, the required volume is 1.05 m^{3} (apπrox).

**Question 2. Find the amount of water displaced by a solid spherical ball of diameter (i) 28 cm (ii) 0.21 m Solution: **(i) Diameter of the ball = 28 cm

∴ Radius of the ball (r) =(28/2) cm = 14 cm

⇒ Volume of the spherical ball =

Thus, the amount of water displaced = 11498 (2/3) cm^{3}.

(ii) Diameter of the solid spherical ball = 0.21 m

**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**: Diameter of the metallic ball = 4.2 cm

⇒ Radius (r) = (42/2) cm = 2.1 cm

∴ Volume of the metallic ball =

∵ Density = 8.9 g per cm^{3}

∴ Mass of the ball = 8.9 x [Volume of the ball]

= 345.39 g (apπrox.)

Thus, the mass of ball is 345.39 g (apπrox.).

**Question 4. The diameter of the moon is apπroximately 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 the earth be 2r.

⇒ Radius of the earth =(2r/2) = r

∵ Diameter of the moon = (1/4)(Diameter of the earth)

∴ Radius of the moon = (1/4)(Radius of the moon)

⇒ Radius of the moon = (1/4) (r) = (r/4)

∴ Volume of the earth = (4/3)πr ^{3}

Volume of the moon =

∴ The required fraction is (1/64)

.

**Question 5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold? Solution:** Diameter of the hemisphere = 10.5 cm

⇒ Radius of the hemisphere (r) = (10.5/2) cm = (105/20) cm

Volume of the hemispherical bowl = (2/3) πr

∴ Capacity of the hemispherical bowl = 303.1875 cm^{3}

= 0.3031875 l = 0.303 l (apπrox.) [∵ 1000 cm^{3} = 1 l]

Thus, the capacity of the bowl = 0.303 l (apπrox.)

**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) = 1m

∵ Thickness = 1 cm = (1/100) m = 0.01 m

∴ Outer radius (R) = 1 m + 0.01 m = 1.01 m

Now, outer volume of the hemispherical bowl

Inner volume of the hemispherical bowl

∴ Volume of the iron used = [Outer volume] – [Inner volume]

= 0.063487776 m^{3}

= 0.06348 m^{3} (apπrox.)

Thus, the required volume of the iron = 0.06348 m^{3}

**Question 7. Find the volume of a sphere whose surface area is 154 cm ^{2} Solution: **Let ‘r’ be the radius of the sphere.

∴ Its surface area = 4πr

⇒ 4πr

Now, volume of the sphere =

Thus, the required volume of the sphere = 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 **** 4989.60. If the cost of white washing is ****20 per square metre, find the (i) inside surface area of the dome, (ii) volume of the air inside the dome. Solution: **Total cost of white washing = 4989.60

Rate of white washing = 20 per m

∴

Let ‘r’ be the radius of the hemispherical dome.

∴ Surface area = 2πr^{2}

∴ 2πr^{2} = 249.48

(i) To find the inside surface area of the dome:

∵ Radius of the hemisphere (r) = 6.3 m

Surface area of a hemisphere = 2πr^{2} m^{2}

∴ Surface area of the dome = 2 x (22/7) (6.3)^{2} m^{2}

Thus, the required surface area of the dome = 249.48 m^{2 }

(ii) To find the volume of air in the dome:

Volume of a hemisphere = (2/3)πr ^{3}

∴ Volume of the dome

Thus, the required volume of air inside the dome is 523.9 m^{3} (apπrox.).

**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 sphere, (ii) ratio of S and S'. Solution:** (i) To find r ':

∵ Radius of a small sphere = r

Since,

(ii) To find ratio of S and S ':

∵ Surface area of a small sphere = 4πr^{2 }

∴ S= 4πr^{2 }and S ' = 4p(3r)^{2}

Now,

Thus, S : S' = 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 spherical capsule = 3.5 mm

⇒ Radius of the spherical capsule (r) = (3.5/2) mm

∴ Volume of the spherical capsule

Thus, the required quantity of medicine = 22.46 mm^{3} (apπrox).

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