# VOLUME OF 3D SHAPES QUESTIONS WITH ANSWERS

Volume of 3D Shapes Questions with Answers :

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

## Volume Formulas for 3D Shapes

 Volume  =  Πr2h Volume  =  Πh(R2 - r2) Volume  =  (1/3) Πr2h Volume  =  (4/3) Πr3Volume of hollow sphere  =  (4/3) Π(R3 - r3) Volume  =  (2/3) Πr3 Volume = (2/3)Π(R3 - r3) Volume = (1/3)Πh(R2 + r2 + Rr)

## Volume of 3D Shapes Questions with Answers

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)πr2h

=  (1/3)π 6(8)

=  96π cm3

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

r = 8 cm and h = 6 cm

=  (1/3)πr2h

=  π 8(6)

=  128π cm3

Difference  =  128π - 96π

=  32π cm3

=  32(22/7)

=  100.5 cm3

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 cm3 and 5040 cm3. Find the ratio of heights.

Solution :

Let "h1" and "h2" be the heights of 1st and 2nd cone.

Volume of 1st cone  =  3600 cm3

(1/3)πr2h1  =  3600 cm3

Volume of 2nd cone  =  5040 cm3

(1/3)πr2h2  =  5040 cm3

h1 : h2  =  3600 : 5040

h1 / h2  =  3600 / 5040

h1 : h2  =  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 1st sphere (r1) =  4x

Radius of 2nd sphere  (r2)  =  7x

Volume of sphere  =  (4/3) πr3

(4/3) πr13 (4/3) πr13

(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 : 4 .

Solution :

Total surface area of sphere = 4πr12

Total surface area of hemisphere = 3πr22

4πr12= 3πr22

r12= (3/4)r22

r1  = (√3/2)r2

Volume of sphere  =  (4/3) πr13

Volume of hemisphere  =  (2/3) πr13

(4/3) πr13 : (2/3) πr23

2((√3/2))3 : r23

2(3√3/8)  r23r23

3√3 : 4

Hence it is proved.

Question 9 :

The outer and the inner surface areas of a spherical copper shell are 576π cm2 and 324π cm2 respectively. Find the volume of the material required to make the shell.

Solution :

Outer curved surface area of sphere  =  4πR2

Inner curved surface area of sphere  =  4πr2

4πR=  576π

R2  =  576/4  =  144

R  =  12

4πr2  =  324π

r2  =  324/4  =  81

r  =  9

Volume of sphere  =  (4/3) π (R3r3)

=  (4/3) π (123 - 93)

=  (4/3) π (1728 - 729)

=  (4/3) π (999)

=  1332(22/7)

=  4186.28 cm3

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+ r2 + Rr)

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

= (1/3)Π (16)(20+ 82 + 20(8)) - Πr2

=  (22/7) [(1/3)(16)(20+ 82 + 20(8)) - 64]

=  (22/7) [(1/3)(16)(624) - 64]

=  (22/7) [3264]

=  10258.28 cm3

1000 cm3  =  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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