__Exercise 13.8 Page No: 236__

__Exercise 13.8 Page No: 236__

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

**(i) 7 cm (ii) 0.63 m**

**(Assume π =22/7)**

**Solution:**

(i) Radius of sphere, r = 7 cm

Using, Volume of sphere = (4/3) πr^{3}

= (4/3)×(22/7)×7^{3}

= 4312/3

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

(ii) Radius of sphere, r = 0.63 m

Using, volume of sphere = (4/3) πr^{3}

= (4/3)×(22/7)×0.63^{3}

= 1.0478

Hence, volume of the sphere is 1.05 m^{3 }(approx).

**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) Diameter = 28 cm

Radius, r = 28/2 cm = 14cm

Volume of the solid spherical ball = (4/3) πr^{3}

Volume of the ball = (4/3)×(22/7)×14^{3 }= 34496/3

Hence, volume of the ball is 34496/3 cm^{3}

(ii) Diameter = 0.21 m

Radius of the ball =0.21/2 m= 0.105 m

Volume of the ball = (4/3 )πr^{3}

Volume of the ball = (4/3)× (22/7)×0.105^{3} m^{3}

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

**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)**

**Solution:**

Given,

Diameter of a metallic ball = 4.2 cm

Radius(r) of the metallic ball, r = 4.2/2 cm = 2.1 cm

Volume formula = 4/3 πr^{3}

Volume of the metallic ball = (4/3)×(22/7)×2.1 cm^{3}

Volume of the metallic ball = 38.808 cm^{3}

Now, using relationship between, density, mass and volume,

Density = Mass/Volume

Mass = Density × volume

= (8.9×38.808) g

= 345.3912 g

Mass of the ball is 345.39 g (approx).

**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 diameter of earth be “d”. Therefore, the radius of earth will be will be d/2

Diameter of moon will be d/4 and the radius of moon will be d/8

Find the volume of the moon :

Volume of the moon = (4/3) πr^{3 }= (4/3) π (d/8)^{3} = 4/3π(d^{3}/512)

Find the volume of the earth :

Volume of the earth = (4/3) πr^{3}= (4/3) π (d/2)^{3} = 4/3π(d^{3}/8)

Fraction of the volume of the earth is the volume of the moon

Answer: Volume of moon is of the 1/64 volume of earth.

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

**Solution:**

Diameter of hemispherical bowl = 10.5 cm

Radius of hemispherical bowl, r = 10.5/2 cm = 5.25 cm

Formula for volume of the hemispherical bowl = (2/3) πr^{3}

Volume of the hemispherical bowl = (2/3)×(22/7)×5.25^{3} = 303.1875

Volume of the hemispherical bowl is 303.1875 cm^{3}

Capacity of the bowl = (303.1875)/1000 L = 0.303 litres(approx.)

Therefore, hemispherical bowl can hold 0.303 litres of milk.

**6. A hemi spherical 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)**

**Solution:**

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

Outer Radius (R ) = 1.01m

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

Put values,

Volume of the iron used in the hemispherical tank = (2/3)×(22/7)×(1.01^{3}– 1^{3}) = 0.06348

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

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

**Solution:**

Let r be the radius of a sphere.

Surface area of sphere = 4πr^{2}

4πr^{2 }= 154 cm^{2} (given)

r^{2} = (154×7)/(4 ×22)

r = 7/2

Radius is 7/2 cm

Now,

Volume of the sphere = (4/3) πr^{3}

**8. A dome of a building is in the form of a hemi sphere. From inside, it was white-washed at the cost of Rs. 4989.60. If the cost of white-washing isRs20 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) Cost of white-washing the dome from inside = Rs 4989.60

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

CSA of the inner side of dome = 498.96/2 m^{2 }= 249.48 m^{2}

(ii) Let the inner radius of the hemispherical dome be r.

CSA of inner side of dome = 249.48 m^{2} (from (i))

Formula to find CSA of a hemi sphere = 2πr^{2}

2πr^{2} = 249.48

2×(22/7)×r^{2 }= 249.48

r^{2 }= (249.48×7)/(2×22)

r^{2 }= 39.69

r = 6.3

So, radius is 6.3 m

Volume of air inside the dome = Volume of hemispherical dome

Using formula, volume of the hemisphere = 2/3 πr^{3}

= (2/3)×(22/7)×6.3×6.3×6.3

= 523.908

= 523.9(approx.)

Answer: Volume of air inside the dome is 523.9 m^{3}.

**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 Sand S’.**

**Solution:**

Volume of the solid sphere = (4/3)πr^{3}

Volume of twenty seven solid sphere = 27×(4/3)πr^{3} = 36 π r^{3}

(i) New solid iron sphere radius = r’

Volume of this new sphere = (4/3)π(r’)^{3}

(4/3)π(r’)^{3 }= 36 π r^{3}

(r’)^{3 }= 27r^{3}

r’= 3r

Radius of new sphere will be 3r (thrice the radius of original sphere)

(ii) Surface area of iron sphere of radius r, S =4πr^{2}

Surface area of iron sphere of radius r’= 4π (r’)^{2}

Now

S/S’ = (4πr^{2})/( 4π (r’)^{2})

S/S’ = r^{2}/(3r’)^{2} = 1/9

The ratio of S and S’ is 1: 9.

**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)**

**Solution:**

Diameter of capsule = 3.5 mm

Radius of capsule, say r = diameter/ 2 = (3.5/2) mm = 1.75mm

Volume of spherical capsule = 4/3 πr^{3}

Volume of spherical capsule = (4/3)×(22/7)×(1.75)^{3} = 22.458

Answer: The volume of the spherical capsule is 22.46 mm^{3}.