# Class 9 NCERT Solutions- Chapter 13 Surface Areas And Volumes – Exercise 13.8

**Question 1: Find the volume of a sphere whose radius is**

**(i) 7 cm (ii) 0.63 m (Assume π =22/7)**

**Solution:**

(i)Given: Radius of sphere, r = 7 cmBy using the formula, volume of sphere = (4/3) πr

^{3}= (4/3)×(22/7)×(7)

^{3}= 4312/3 cm

^{3}Therefore, the volume of the sphere is 4312/3 cm

^{3}

(ii)Given: Radius of sphere, r = 0.63 mBy using the formula, volume of sphere = (4/3) πr

^{3}= (4/3)×(22/7)×(0.63)

^{3}= 1.0478 m

^{3}Therefore, the volume of the sphere is 1.05 m

^{3}(approx).

**Question 2: Find the amount of water displaced by a solid spherical ball of diameter**

**(i) 28 cm (ii) 0.21 m (Assume π =22/7)**

**Solution:**

(i)Given: Diameter = 28 cmTherefore, the radius, r = 28/2 cm = 14cm

By using the formula, volume of the solid spherical ball = (4/3) πr

^{3}= (4/3)×(22/7)×(14)

^{3}= 34496/3 cm

^{3}Therefore, the vo

of the ball is 34496/3 cm^{3}

(ii)Given: Diameter = 0.21 mTherefore, the radius of the ball =0.21/2 m= 0.105 m

By using the formula, volume of the ball = (4/3 )πr

^{3}= (4/3)× (22/7)×(0.105)

^{3}= 0.004851 m

^{3}Therefore, the volume of the ball = 0.004851 m

^{3}

**Question 3: The diameter of a metallic ball is 4.2cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm**^{3}? (Assume π**=22/7)**

^{3}? (Assume

**Solution:**

Given: Diameter of a metallic ball = 4.2 cm

Therefore, the radius o

e metallic ball, r = 4.2/2 cm = 2.1 cmBy using the formula, volume of the metallic ball = 4/3 πr

^{3}= (4/3)×(22/7)×(2.1)

^{3}= 38.808 cm

^{3}Now by using the relationship between, density, mass, and volume,

Density = Mass/Volume

Mass = Density × volume

= (8.9×38.808) g

= 345.3912 g

Therefore, the mass of the ball is 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 us assume the diameter of the earth as “d”.

Therefore, the radius of the earth will be d/2

The diameter of the moon will be d/4

The radius of the moon will be d/8

Finding the volume of the moon.

By using the formula, volume of the moon = (4/3) πr

^{3 }= (4/3) π (d/8)

^{3 }= 4/3π(d

>^{3}/512Finding the volume of the earth :

By using the formula, volume of the earth = (4/3) πr

^{3}= (4/3) π (d/2)

^{3}= 4/3π(d

^{3}/8)The fraction of the volume of the earth to the volume of the moon

Volume of moon/volume of earth = 4/3π(d

^{3}/512)/4/3π(d^{3}/8)= 1/64

Therefore, the volume of moon is 1/64

^{th}of the volume of earth.

**Question 5: How many **liters** of milk can a hemispherical bowl of diameter 10.5cm hold? (Assume **π **= 22/7)**

**Solution:**

Given

ameter of hemispherical bowl = 10.5 cmTherefore, the radius of hemispherical bowl, r = 10.5/2 cm = 5.25 cm

By using the formula, volume of the hemispherical bowl = (2/3) πr

^{3}= (2/3)×(22/7)×(5.25)

^{3}= 303.1875 cm

^{3}Therefore, the volume of the hemispherical bowl is 303.1875 cm

^{3}Capacity of the bowl = (303.1875)/1000 L

= 0.303 liters(approx.)

Therefore, the hemispherical bowl can hold 0.303 liters of milk.

**Question 6: A hemispherical tank is made up of an iron sheet 1cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank. (Assume **π** = 22/7)ng>**

**Solution:**

Given: Inner Radius of the tank, (r) = 1m

Also, the thickness of hemispherical tank = 0.01m

Therefore, the outer Radius (R) = 1 + 0.01 = 1.01m

By using the formula, volume of the iron used in the tank = (2/3) π(R^{3}– r^{3})

= (2/3)×(22/7) × (1.01^{3}– 1^{3})

= 0.06348 m^{3}

Therefore, the volume of the iron used in the hemispherical tank is 0.06348 m^{3}.

**Question 7: Find the volume of a sphere whose surface area is 154 cm**^{2}. (Assume π** = 22/7)**

**Question 7: Find the volume of a sphere whose surface area is 154 cm**π

^{2}. (Assume**= 22/7)**

**Solution:**

Given:ace area of sphere =154 cm^{2}Let us assume

rto be the radius of a sphere.By using the formula, Surface area of sphere = 4πr

^{2}4πr

^{2}= 154 cm^{2}r

^{2}= (154×7)/(4 ×22)r = 7/2

Radius r = 7/2 cm

Now,

By using the formula, volume of the sphere = (4/3) πr

^{3}= (4/3)×(22/7)×(7/2)

^{3}= 539/3 cm

^{3}Therefore, the volume of a sphere is 539/3 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 Rs. 4989.60. If the cost of white-washing is Rs20 per square meter, find the**

**Question 8: A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs. 4989.60. If the cost of white-washing is Rs20 per square meter, find the**

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

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

**(Assume π = 22/7)**

**Solution:**

(i)Given: Cost of white-washing the dome from inside = Rs 4989.60

Also, Cost of white-washing 1m^{2}area = Rs 20

Therefore, CSA of the inner side of dome = 498.96/2 m^{2<}>= 249.48 m

^{2}

(ii)Let us assume the inner radius of the hemispherical dome be r.CSA of the inner side of dome = 249.48 m

^{2}By using the formula, CSA of a hemisphere = 2πr

^{2}2πr

^{2}= 249.482×(22/7)×r

^{2}= 249.4r

^{2}= (249.48×7)/(2×22)r

^{2}= 39.6r = 6.3

Hence, the radius r = 6.3 m

Volume of air inside the dome = Volume of hemispherical dome

By using the formula, volume of the hemisphere = 2/3 πr

^{3}= (2/3)×(22/7)×(6.3)

^{3}= 523.908 m

^{3}= 523.9(approx.)

Therefore, the volume of air inside the dome is 523.9 m

^{3}.

**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**

**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 new sphere,**

**(ii) ratio of S and S’.**

**Solution:**

By using the formula, volumehe solid sphere = (4/3)πr^{3}Therefore, the volume of twenty-seven solid sphere = 27×(4/3)πr

^{3}= 36 πr^{3}

(i)Given: new solid iron sphere radius = r’By using the formula, volume of this new sphere = (4/3)π(r’)

^{3}(4/3)π(r’)

^{3}= 36 πr^{3}(r’)

^{3}= 27r^{3}r’= 3r

Radius of new sphere = 3r

(ii)We know that,Surface area of iron sphere of radius r, S =4πr

^{2}Surface area of iron sph

of radius r’= 4π (r’)^{2}S/S’ = (4πr

^{2})/( 4π(r’)^{2})S/S’ = r

^{2}/(3r’)^{2}= r

^{2}/9r^{2}= 1/9

Therefore, the ratio of S and S’ is 1: 9.

**Question 10: A capsule of medicine is in the shape of a sphere of diameter 3.5mm. How much medicine (in mm**^{3}) is needed to fill this capsule? (Assume π** = 22/7)**

**Question 10: A capsule of medicine is in the shape of a sphere of diameter 3.5mm. How much medicine (in mm**π

^{3}) is needed to fill this capsule? (Assume**= 22/7)**

**Solution:**

Given: Diameter of capsule = 3.5 mm

Therefore, the radius of capsule, r = (3.5/2) mm = 1.75mm

By using the formula, volume of spherical capsule = 4/3 πr^{3}

= (4/3)×(22/7)×(1.75)^{3 }

= 22.458 mm^{3}

Therefore, the volume of the spherical capsule is 22.46 mm^{3}.