Chapter 11: Surface Areas and Volumes

(i) Radius of the sphere, r = 7 cm

Using, Volume of the sphere = (4/3) πr³

= (4/3)×(22/7)×7³

= 4312/3

Hence, volume of the sphere is 4312/3 cm³

(ii) Radius of the sphere, r = 0.63 m

Using, volume of sphere = (4/3) πr³

= (4/3)×(22/7)×0.63³

= 1.0478

Hence, volume of the sphere is 1.05 m³ (approx).

(i) Diameter = 28 cm

Radius, r = 28/2 cm = 14cm

Volume of the solid spherical ball = (4/3) πr³

Volume of the ball = (4/3)×(22/7)×14³ = 34496/3

Hence, volume of the ball is 34496/3 cm³

(ii) Diameter = 0.21 m

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

Volume of the ball = (4/3 )πr³

Volume of the ball = (4/3)× (22/7)×0.105³ m³

Hence, volume of the ball = 0.004851 m³

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 πr³

Volume of the metallic ball = (4/3)×(22/7)×2.1 cm³

Volume of the metallic ball = 38.808 cm³

Now, using the 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).

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

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

Find the volume of the moon.

Volume of the moon = (4/3) πr³ = (4/3) π (d/8)³ = 4/3π(d³/512)

Find the volume of the earth

Volume of the earth = (4/3) πr³= (4/3) π (d/2)³ = 4/3π(d³/8)

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

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

Diameter of the hemispherical bowl = 10.5 cm

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

Formula for volume of the hemispherical bowl = (2/3) πr³

Volume of the hemispherical bowl = (2/3)×(22/7)×5.25³ = 303.1875

Volume of the hemispherical bowl is 303.1875 cm³

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

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

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

Outer Radius (R ) = 1.01m

Volume of the iron used in the tank = (2/3) π(R³– r³)

Put values,

Volume of the iron used in the hemispherical tank = (2/3)×(22/7)×(1.01³– 1³) = 0.06348

So, volume of the iron used in the hemispherical tank is 0.06348 m³.

Let r be the radius of a sphere.

Surface area of the sphere = 4πr²

4πr² = 154 cm² (given)

r² = (154×7)/(4 ×22)

r = 7/2

The radius is 7/2 cm.

Now,

Volume of the sphere = (4/3) πr³

(i) Cost of whitewashing the dome from inside = Rs 4989.60

Cost of whitewashing 1m² area = Rs 20

CSA of the inner side of dome = 498.96/2 m²  = 249.48 m²

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

CSA of the inner side of dome = 249.48 m² (from (i))

Formula to find CSA of a hemisphere = 2πr²

2πr² = 249.48

2×(22/7)×r² = 249.48

r² = (249.48×7)/(2×22)

r² = 39.69

r = 6.3

So, the radius is 6.3 m.

Volume of air inside the dome = Volume of hemispherical dome

Using the formula, the volume of the hemisphere = 2/3 πr³

= (2/3)×(22/7)×6.3×6.3×6.3

= 523.908

= 523.9(approx.)

Answer: The volume of air inside the dome is 523.9 m³.

Volume of the solid sphere = (4/3)πr³

Volume of twenty seven solid sphere = 27×(4/3)πr³ = 36 π r³

(i) New solid iron sphere radius = r’

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

(4/3)π(r’)³ = 36 π r³

(r’)³ = 27r³

r’= 3r

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

(ii) Surface area of the iron sphere of radius r, S =4πr²

Surface area of the iron sphere of radius r’= 4π (r’)²

Now

S/S’ = (4πr²)/( 4π (r’)²)

S/S’ = r²/(3r’)² = 1/9

The ratio of S and S’ is 1: 9.

Diameter of the capsule = 3.5 mm

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

Volume of the spherical capsule = 4/3 πr³

Volume of the spherical capsule = (4/3)×(22/7)×(1.75)³ = 22.458

Answer: The volume of the spherical capsule is 22.46 mm³.