Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

The amount of water displaced by a solid spherical ball of diameter $$4.2$$ cm, when it is completely immersed in water, is

A
$$38.808$$ cm$$^3 $$

B
$$14.808$$ cm$$^3 $$

C
$$58.808$$ cm$$^3 $$

D
$$45.808$$ cm$$^3 $$

Solution

The amount of water displaced by a solid spherical ball when it is completely immersed in water is equal to its volume.
Volume of a sphere of radius $$r$$ is $$\dfrac { 4 }{ 3 } \pi { r }^{ 3 } $$
As the diameter of the ball is $$4.2$$ cm, its radius $$r = 2.1$$ cm
Hence, volume of water displaced $$ = \dfrac { 4 }{ 3 } \pi { r }^{ 3 } = \dfrac { 4 }{ 3 } \times \dfrac { 22 }{ 7 } \times 2.1\times 2.1\times 2.1 = 38.808$$ cm$$^{ 3 } $$
Question 2
1 / -0

If two solid hemispheres of same base radius r are joined together along their bases, then curved surface area of this new solid is

A
$$4\pi r^{2}$$

B
$$6\pi r^{2}$$

C
$$3\pi r^{2}$$

D
$$8\pi r^{2}$$

Solution

We know that,
Curved surface area of a hemisphere $$=2\pi r^2$$
So, when two solids hemispheres of same base radius are joined together along their bases, then
Curved surface area of newly formed solid sphere$$=2\times2\pi r^2=4\pi r^2$$
Hence, A is the correct option.
Question 3
1 / -0

In a cylinder, radius is doubled and height is halved, curved surface area will

A
halved

B
doubled

C
same

D
four times

Solution

Curved surface area of cylinder is $$2 \pi r h$$.
Given, $$R=2r,H=\dfrac{h}{2}$$
Therefore, curved surface area of the cylinder$$=2\pi RH$$
$$=2\pi(2r)\left (\dfrac{h}{2}\right)$$
$$=2\pi rh$$
Question 4
1 / -0

The number of dimensions, a solid has :

A
1

B
2

C
3

D
0

Solution

A solid has length, width and height. Hence, it is 3-Dimensional.
Question 5
1 / -0

Area of the base of a solid hemisphere is 36$$\pi\ cm^2$$. Then its volume is :

A
$$288\pi \ cm^{3}$$

B
$$108\pi \ cm^{3}$$

C
$$144\pi\ cm^{3}$$

D
$$72\pi \ cm^{3}$$

Solution

Let the radius of the hemisphere be $$'r'$$ cm.
Area of base $$=36\pi { cm }^{ 2 }$$
$$\Rightarrow \pi { r }^{ 2 }=36\pi $$
$$\Rightarrow { r }^{ 2 }=36$$
$$\Rightarrow r=6cm$$
$$\therefore $$ Volume of hemisphere $$=\dfrac { 2 }{ 3 } \pi { r }^{ 3 }$$
$$=\dfrac { 2 }{ 3 } \times \pi \times 6\times 6\times 6$$
$$=144\pi { cm }^{ 3 }$$
Question 6
1 / -0

Area of base of a solid hemisphere is $$36 \pi\ sq cm.$$ Then its volume is :

A
$$288 \pi\ cm^{3}$$

B
$$108 \pi\ cm^{3}$$

C
$$144 \pi\ cm^{3}$$

D
$$72 \pi\ cm^{3}$$

Solution

Area of base of solid hemisphere $$=36\pi $$
$$\pi { r }^{ 2 }=36\pi $$
$$r=6\ cm$$
Volume of hemisphere $$=\dfrac { 2 }{ 3 } \pi { r }^{ 3 }=\dfrac { 2 }{ 3 } \times 6\times 6\times 6\times \pi =144\pi\ { cm }^{ 3 }$$
Question 7
1 / -0

The volume of a sphere of radius $$r$$ is:

A
$$\dfrac{4}{3}\pi r^{3}$$

B
$$2\pi r^{2}$$

C
$$\dfrac{2}{3}\pi r^{3}$$

D
$$4\pi r^{2}$$

Solution

$$\textbf{Step-1:Write the relevant formula of volume of sphere using given radius.}$$
$$\text{We have given,}$$
$$\text{Radius = r unit.}$$
$$\therefore$$ $$\text{Volume of a sphere}$$ $$=\cfrac { 4 }{ 3 } \pi { r }^{ 3 }$$ $$\text{Cubic unit.}$$
$$\textbf{Hence, option - A}$$
Question 8
1 / -0

Given dimensions of a cuboid as $$ l = 3\ cm, b = 2\ cm$$ and $$h = 1\ cm.$$ Find the volume.

A
$$12\ cm^3$$

B
$$6\ cm^3$$

C
$$4\ cm^3$$

D
$$3\ cm^3$$

Solution

Given: $$l=3\ cm$$
$$ b=2\ cm$$
$$h=1\ cm $$

Volume of cuboid $$=lbh$$

$$=3\times 2\times 1$$

$$=6 \ { cm }^{ 3 }$$
Question 9
1 / -0

The radius of a solid hemisphere is $$2r$$, then its total surface area will be :

A
$$12\pi r^2$$

B
$$8\pi r^2$$

C
$$6\pi r^2$$

D
$$3\pi r^2$$

Solution

Radius of solid hemisphere $$=2r$$
Total surface area of hemisphere $$=2\pi { r }^{ 2 }+\pi { r }^{ 2 }$$
  $$=3\pi { r }^{ 2 }$$
  $$=3\pi \times { \left( 2r \right) }^{ 2 }$$
  $$=12\pi { r }^{ 2 }$$
Question 10
1 / -0

If the surface area of a sphere is $$144 \pi \ cm^{2}$$, then its radius is:

A
$$6 cm$$

B
$$8 cm$$

C
$$12 cm$$

D
$$10 cm$$

Solution

Let radius of sphere be $$r$$ cm.
We know, surface area of sphere $$=4\pi { r }^{ 2 }$$
$$\implies$$ $$144\pi =4\pi { r }^{ 2 }$$
$$\implies$$ $$ { { r }^{ 2 } }=\cfrac { 144\pi }{ 4\pi } =36$$
$$\implies$$ $$ r=\sqrt { 36 } =6 cm$$ .
Therefore, option $$A$$ is correct.

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Re-Attempt Weekly Quiz Competition

Surface Areas and Volumes Test - 23

Result

Surface Areas and Volumes Test - 23

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SHARING IS CARING

Question 1

$$38.808$$ cm$$^3 $$

$$14.808$$ cm$$^3 $$

$$58.808$$ cm$$^3 $$

$$45.808$$ cm$$^3 $$

Solution

Question 2

$$4\pi r^{2}$$

$$6\pi r^{2}$$

$$3\pi r^{2}$$

$$8\pi r^{2}$$

Solution

Question 3

halved

doubled

same

four times

Solution

Question 4

1

2

3

0

Solution

Question 5

$$288\pi \ cm^{3}$$

$$108\pi \ cm^{3}$$

$$144\pi\ cm^{3}$$

$$72\pi \ cm^{3}$$

Solution

Question 6

$$288 \pi\ cm^{3}$$

$$108 \pi\ cm^{3}$$

$$144 \pi\ cm^{3}$$

$$72 \pi\ cm^{3}$$

Solution

Question 7

$$\dfrac{4}{3}\pi r^{3}$$

$$2\pi r^{2}$$

$$\dfrac{2}{3}\pi r^{3}$$

$$4\pi r^{2}$$

Solution

Question 8

$$12\ cm^3$$

$$6\ cm^3$$

$$4\ cm^3$$

$$3\ cm^3$$

Solution

Question 9

$$12\pi r^2$$

$$8\pi r^2$$

$$6\pi r^2$$

$$3\pi r^2$$

Solution

Question 10

$$6 cm$$

$$8 cm$$

$$12 cm$$

$$10 cm$$

Solution

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