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Chapter 11: Surface Areas and Volumes

    Surface Area of a Right Circular Cone

    Surface Area of a Sphere

    Volume of a Right Circular Cone

    Volume of a Sphere

    Exercise 11.1

    Exercise 11.2

    Exercise 11.3

    Exercise 11.4

    Summary

    Enhanced Curriculum Support

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Chapter 11: Surface Areas and Volumes > Exercise 11.4

Exercise 11.4

Assume π = 227, unless stated otherwise.

Find the volume of a sphere whose radius is:

(i) 7 cm (ii) 0.63 m

Instructions

We know: Volume of the sphere =
(i) r = 7 cm: Volume = 43×227×73 = cm3 (Round off to two decimal places)
(i) r = 0.63 m: Volume = 43×227×0.633 = m3 (Round off to two decimal places)
We have found all the answers.

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

(i) 28 cm (ii) 0.21 m

Instructions

Amount of water displaced by a solid spherical ball is equal to the of the sphere.
Volume of the solid spherical ball =
(i) Diameter = 28 cm i.e. r = cm
Volume of the ball = 43×227×143 = cm3 (Round off to two decimal places)
(ii) Diameter = 0.21 m i.e. r = m (Round off to three decimal places)
Volume of the ball = 43×227×0.1053 = m3 (Round off to five decimal places)
We have found all the answers.

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 cm3 ?

Instructions

Radius of the metallic ball = cm with volume =
Volume of the metallic ball = 43×227×2.13 = cm3 (Round off to two decimal places)
We know: Density =
Thus, Mass = Density × volume = = (8.9 × 38.81) g = g (Round off to one decimal place)
So, mass of the ball is 345.4 g.

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?

Instructions

If the diameter of the earth is “d” then, the radius of the earth will be .
If diameter of the moon will be d4 then radius of the moon will be .
Volume of the moon = = × π × =
Volume of the earth = 43πr3 = 43πd23 =
Fraction of the volume of the earth is the volume of the moon = 43πd351243πd38 = =

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

Instructions

We have: Radius of the hemispherical bowl = cm with volume of the hemispherical bowl =
Volume of the hemispherical bowl = 23×227×5.253 = cm3 (Round off to one decimal place)
So, Capacity of the bowl = / L = litres (Round off to three decimal places)
Therefore, the hemispherical bowl can hold 0.303 litres of milk.

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.

Instructions

Finding volume of metal used

  • Volume of hemisphere : where r is the radius.
  • But since, we need to find the volume of the metal used, the volume becomes where a and b are the outer and inner radii, respectively.
  • The outer radius a = m
  • Substituting values in the formula
  • We get the volume of metal = m3(Round off to four decimal places)
  • We have found the desired answer.

7. Find the volume of a sphere whose surface area is 154 cm2.

Instructions

Surface area of the sphere = = 154
r2 = 154×74×22 i.e. r = cm
Volume of the sphere = = × × = cm3 (Round off to two decimal places)
We have found the answer.

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 Rs. 20 per square metre, find the:

(i) inside surface area of the dome, (ii) volume of the air inside the dome

Instructions

(i) Cost of whitewashing the dome from inside = Rs while cost of whitewashing 1 m2 area = Rs
Thus, CSA of the inner side of dome = m2 (Round off to two decimal places)
(ii) Assume the inner radius of the hemispherical dome to be r.
We know that: CSA of the inner side of dome = 249.48 m2 and the formula to find CSA of a hemisphere =
So, 2πr2 = 249.48 i.e. 2×227×r2 = 249.48 which gives us: r = m (Round off to one decimal place)
Volume of air inside the dome = Volume of hemispherical dome and the volume of the hemisphere =
Substituting: Volume = = 23×227×6.33 = m3 (Round off to one decimal place)
The volume of air inside the dome is 523.9 m3.

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′

Instructions

Finding the above answers

  • Volume of sphere : where r is the radius.
  • Thus, volume of twenty seven solid sphere =
  • Let the new sphere radius = r1
  • By solving for the new radius
  • We get r1 =
  • Now, finding the required ratio: SS1
  • The ratio SS1 =
  • We have found the desired answer.

10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule?

Instructions

Radius of the capsule (r) = mm
We know: Volume of the spherical capsule = = × × = mm3 (Round off to two decimal places)
The volume of the spherical capsule is 22.46 mm3.