**Volume of 3D Shapes Questions with Answers :**

Here we are going to see, some practice questions on volume of 3D shapes.

Volume = Πr |
Volume = Πh(R |

Volume = (1/3) Πr |
Volume = (4/3) Πr Volume of hollow sphere = (4/3) Π(R |

Volume = (2/3) Πr |
Volume = (2/3)Π(R |

Volume = (1/3)Πh(R |

To find questions from 1 to 3, please visit the page "VOLUME OF 3D SHAPES EXAMPLES".

**Question 5 :**

A right angled triangle whose sides are 6 cm, 8 cm and 10 cm is revolved about the sides containing the right angle in two ways. Find the difference in volumes of the two solids so formed.

**Solution :**

**The longest side of a triangle is 10 cm**

**If the triangle is revolved about 6 cm**

r = 6 cm and h = 8 cm

Volume of solid, when it is revolved about 6 cm,

= (1/3)**πr ^{2}h**

** = (1/3)****π**** 6**^{2 }(8)

** = 96****π**** cm ^{3}**

Volume of solid, when it is revolved about 8 cm,

r = 8 cm and h = 6 cm

= (1/3)**πr ^{2}h**

** = ****π**** 8**^{2 }(6)

** = 128****π cm ^{3}**

**Difference = 128****π**** - 96****π**

** = 32****π ****cm ^{3}**

** = 32(22/7)**

**= 100.5 ****cm ^{3}**

**Now let us see the solution of the next problem on "**Volume of 3D Shapes Questions with Answers".

**Question 6 :**

The volumes of two cones of same base radius are 3600 cm^{3} and 5040 cm^{3}. Find the ratio of heights.

**Solution :**

Let "h_{1}" and "h_{2}" be the heights of 1^{st} and 2^{nd} cone.

Volume of 1^{st} cone = 3600 cm^{3}

(1/3)**πr ^{2}h_{1}** = 3600 cm

Volume of 2^{nd} cone = 5040 cm^{3}

(1/3)**πr ^{2}h_{2}** = 5040 cm

h_{1} : h_{2} = 3600 : 5040

h_{1} / h_{2} = 3600 / 5040

h_{1} : h_{2} = 5 : 7

**Now let us see the solution of the next problem on "**Volume of 3D Shapes Questions with Answers".

**Question 7 :**

If the ratio of radii of two spheres is 4 : 7, find the ratio of their volumes.

**Solution :**

Radius of 1^{st} sphere (r_{1}) = 4x

Radius of 2^{nd} sphere (r_{2}) = 7x

Volume of sphere = (4/3) πr^{3}

(4/3) πr_{1}^{3 }: (4/3) πr_{1}^{3}

(4x)^{3} : (7x)^{3}

64 : 343

**Question 8:**

A solid sphere and a solid hemisphere have equal total surface area. Prove that the ratio of their volume is 3 √3 : 4 .

**Solution :**

Total surface area of sphere = 4πr_{1}^{2}

Total surface area of hemisphere = 3πr_{2}^{2}

4πr_{1}^{2}= 3πr_{2}^{2}

r_{1}^{2}= (3/4)r_{2}^{2}

r_{1} = (√3/2)r_{2}

Volume of sphere = (4/3) πr_{1}^{3}

Volume of hemisphere = (2/3) πr_{1}^{3}

(4/3) πr_{1}^{3 : }(2/3) πr_{2}^{3}

2((√3/2))^{3 : }r_{2}^{3}

2(3√3/8)^{ }^{ }r_{2}^{3}: r_{2}^{3}

3√3 : 4

Hence it is proved.

**Question 9 :**

The outer and the inner surface areas of a spherical copper shell are 576π cm^{2} and 324π cm^{2} respectively. Find the volume of the material required to make the shell.

**Solution :**

Outer curved surface area of sphere = 4πR^{2}

Inner curved surface area of sphere = 4πr^{2}

4πR^{2 }= 576π

R^{2} = 576/4 = 144

R = 12

4πr^{2 }= 324π

r^{2} = 324/4 = 81

r = 9

Volume of sphere = (4/3) π (R^{3} - r^{3})

= (4/3) π (12^{3} - 9^{3})

= (4/3) π (1728 - 729)

= (4/3) π (999)

= 1332(22/7)

= 4186.28 cm^{3}

**Question 10 :**

A container open at the top is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends are 8 cm and 20 cm respectively. Find the cost of milk which can completely fill a container at the rate of ₹40 per litre.

**Solution :**

Volume of milk in the frustum container

= (1/3)Πh(R^{2 }+ r^{2} + Rr)

h = 16 cm, r = 8 cm, R = 20 cm

= (1/3)Π (16)(20^{2 }+ 8^{2} + 20(8)) - Πr^{2}

= (22/7) [(1/3)(16)(20^{2 }+ 8^{2} + 20(8)) - 64]

= (22/7) [(1/3)(16)(624) - 64]

= (22/7) [3264]

= 10258.28 cm^{3}

1000 cm^{3} = 1 liter

= 10.25828 liter

Cost per liter = ₹40

= 40(10.258)

= ₹410.32

After having gone through the stuff given above, we hope that the students would have understood, "Volume of 3D Shapes Questions with Answers".

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