Surface Areas and Volumes Test

Result

Surface Areas and Volumes Test - 43

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The addition of curved surface and circular surfaces are called

A
area

B
perimeter

C
lateral surface area

D
total surface area

Solution

The total surface area includes curved surface area and two circular surfaces.
Question 2
1 / -0

Find the surface area of a cube in square inches, whose each edge measures $$3$$ inches.

A
9

B
18

C
27

D
36

E
54

Solution

Surface area of cube$$=6a^2$$
Where '$$a$$' is the edge of the cube
Surface area $$=$$ $$6\times (3)^2$$
$$\Rightarrow 6\times 9=54$$ square inches
Question 3
1 / -0

What is the surface area of a cylinder that has a top and bottom if the height is $$8.5$$ units and the diameter is $$3$$ units?

A
$$18\pi$$ square units

B
$$30\pi$$ square units

C
$$32\pi$$ square units

D
$$36\pi$$ square units

E
$$54\pi$$ square units

Solution

Given: height $$= 8.5$$, diameter $$= 3$$
$$\therefore$$ radius $$=$$ $$\dfrac{D}{2}=\dfrac{3}{2}=1.5$$
Surface area $$=$$ $$2\pi rh +2\pi r^2$$
$$=2\pi\times1.5\times 8.5+2\pi\times 1.5^2$$
$$=2\pi(12.75+2.25)$$
$$=2\pi(15)$$
$$=30\pi$$ sq.units
Question 4
1 / -0

How much more water is needed to fill the tank to its brim?

A
2400 $$\displaystyle { cm }^{ 3 }$$

B
3200 $$\displaystyle { cm }^{ 3 }$$

C
3600 $$\displaystyle { cm }^{ 3 }$$

D
6000 $$\displaystyle { cm }^{ 3 }$$

Solution

Capacity of the tank $$=$$ Volume of tank

$$= 20\times 20\times 15 = 6000\ cm^3$$

Volume of water filled in the tank $$= 20\times 8\times 15$$

$$ = 2400\ cm^3$$

Volume yet to be filled in tank to its brim $$= 6000\ cm^3 - 2400\ cm^3$$

$$ = 3600 \ cm^3$$
Question 5
1 / -0

If the circumference of the inner edge of a hemispherical bowl is $$\dfrac {132}{7} cm$$, then what is its capacity?

A
$$12\pi cm^{3}$$

B
$$18\pi cm^{3}$$

C
$$24\pi cm^{3}$$

D
$$36\pi cm^{3}$$

Solution

$$2\pi r = \dfrac {132}{7}\Rightarrow r = \dfrac {132}{7}\times \dfrac {1}{2\times \dfrac {22}{7}}$$
$$r = 3$$
$$Capacity = \dfrac {2}{3}.\pi r^{3} = \dfrac {2}{3}.\pi 3^{3} = 18\pi$$
Question 6
1 / -0

A closed metallic cylindrical box is 1.25 m high and its base radius is 35 cm. If the sheet metal costs Rs. 80 per m$$^2$$, the cost of the material used in the box is

A
Rs. 281.60

B
Rs. 290

C
Rs. 340.50

D
Rs. 500

Solution

T.S.A. of cyl $$ = 2\pi rh+2\pi r^{2} = 2\pi \times \left ( \dfrac{35}{100} \right )\times 1.25+2\pi \left ( \dfrac{35}{100} \right )^{2} $$
$$ \Rightarrow TSA = 3.52\,m^{2} $$ & as cost of $$ 1\,m^{2} $$ is Rs 80
$$ \Rightarrow $$ Total cost $$ = 3.52\times 80 = Rs\,281.6 $$
Question 7
1 / -0

Find the total surface area of a hemisphere of radius $$10$$ cm.

A
$$942.86$$ cm$$^2$$

B
$$900$$ cm$$^2$$

C
$$300$$ cm$$^2$$

D
$$592.86$$ cm$$^2$$

Solution

T.S.A of hemisphere is $$ (2\pi r^{2}+\pi r^{2}) $$
$$ \Rightarrow T.S.A = 3\pi r^{2} = 3\pi (10)^{2} = \boxed{942.86} $$
Question 8
1 / -0

The curved surface area of a right circular cone of radius $$11.3$$ cm is $$355$$ cm$$^2$$. What is its slant height?

A
$$8$$ cm

B
$$9$$ cm

C
$$10$$ cm

D
$$11$$ cm

Solution

Given, radius $$r=11.3 \text{ cm}$$

and curved surface area $$=355 \text{ cm}^2$$.

We know, curved surface area of a cone $$ =\pi rl $$

$$11.3 \pi l= 355$$

$$l = \dfrac{355}{\pi \times 11.3}$$

$$ {l = 10\text{ cm}} $$

Therefore, option $$C$$ is correct.
Question 9
1 / -0

Each edge in a cube is $$5\ cm$$. What is the surface area in square cm?

A
$$25$$

B
$$30$$

C
$$100$$

D
$$125$$

E
$$150$$

Solution

Surface area of a cube is $$6a^2$$ where $$a$$ is an edge of the cube.

The surface area of the cube with edge $$5$$cm is given by:

$$A=6a^{ 2 }=6\times (5)^{ 2 }=6\times 25=150cm^{ 2 }$$
Question 10
1 / -0

The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

A
$$5:16$$

B
$$1:16$$

C
$$3:16$$

D
None of these

Solution

Let the diameter of earth be $$D$$ & of moon be $$d$$.

$$\therefore $$ According to the statement, $$ d=\dfrac { D }{ 4 } $$.

Then, $$ \dfrac{d}{2}=\dfrac { D }{ 2\times4 } $$.

$$\Rightarrow r=\dfrac { R }{ 4 }$$

$$ \Rightarrow \dfrac { r }{ R } =\dfrac { 1 }{ 4 } $$.

$$\therefore $$ The ratio of their surface areas is $$= \dfrac { 4\pi { r }^{ 2 } }{ 4\pi { R }^{ 2 } } ={ \left( \dfrac { r }{ R } \right) }^{ 2 }={ \left( \dfrac { 1 }{ 4 } \right) }^{ 2 }=\dfrac { 1 }{ 16 } =1:16$$.

Hence, option $$B$$ is correct.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Surface Areas and Volumes Test - 43

Result

Surface Areas and Volumes Test - 43

Score

-

Rank

-

SHARING IS CARING

Question 1

area

perimeter

lateral surface area

total surface area

Solution

Question 2

9

18

27

36

54

Solution

Question 3

$$18\pi$$ square units

$$30\pi$$ square units

$$32\pi$$ square units

$$36\pi$$ square units

$$54\pi$$ square units

Solution

Question 4

2400 $$\displaystyle { cm }^{ 3 }$$

3200 $$\displaystyle { cm }^{ 3 }$$

3600 $$\displaystyle { cm }^{ 3 }$$

6000 $$\displaystyle { cm }^{ 3 }$$

Solution

Question 5

$$12\pi cm^{3}$$

$$18\pi cm^{3}$$

$$24\pi cm^{3}$$

$$36\pi cm^{3}$$

Solution

Question 6

Rs. 281.60

Rs. 290

Rs. 340.50

Rs. 500

Solution

Question 7

$$942.86$$ cm$$^2$$

$$900$$ cm$$^2$$

$$300$$ cm$$^2$$

$$592.86$$ cm$$^2$$

Solution

Question 8

$$8$$ cm

$$9$$ cm

$$10$$ cm

$$11$$ cm

Solution

Question 9

$$25$$

$$30$$

$$100$$

$$125$$

$$150$$

Solution

Question 10

$$5:16$$

$$1:16$$

$$3:16$$

None of these

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.