Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

Find the total surface area of the cuboids of length, breadth and height
$$12 \mathrm { cm } , 10 \mathrm { cm } , 5 \mathrm { cm }$$ respectively.

A
$$460 \ cm^2$$

B
$$490 \ cm^2$$

C
$$360 \ cm^2$$

D
$$230 \ cm^2$$

Solution
Question 2
1 / -0

If the diameter of the base of a cylinder is $$14$$ cm and its height is $$12$$ cm then the volume of the cylinder in $$m^3$$ is

A
$$10.1848$$ $$m^3$$

B
$$0.001858$$ $$m^3$$

C
$$0.001848$$ $$m^3$$

D
$$18.48$$ $$m^3$$

Solution

Diameter $$= 14$$ cm
Height $$= 12$$ cm

radius $$(r) = \dfrac {14}{ 2}= 7$$ cm

Volume of the cylinder = $$\pi r^{2}h$$

$$=\dfrac{22}{7}\times 7\times 7\times 12$$

$$=1848\ cm^{2}$$

$$=\dfrac {1848}{100} m^{3}$$

$$= 18.48\ m^{3}$$
Question 3
1 / -0

The largest sphere is carved out of a cube of edge 14 cm. The volume of this sphere is:

A
1370 $$\text{cm}^3$$

B
1800 $$\text{cm}^3$$

C
1437 $$\text{cm}^3$$

D
1734 $$\text{cm}^3$$

Solution

The largest sphere that can be carved out of a cube will have diameter equal to length of the side of the cube.
Diameter $$ = 14\, \text{cm}$$
$$ \Rightarrow$$ Radius, $$r = 7\ \text{cm} $$
Volume of the sphere, $$V=\dfrac{4}{3} \pi r^3$$
$$ \therefore V = \dfrac{4}{3}\pi r^{3} $$
$$= \dfrac{4}{3}\times \dfrac{22}{7}\times 7\times 7 \times 7\ \text{cm}^3$$

$$ = \dfrac{88}{3}\times 49\, \text{cm}^{3} $$
$$ = 1437.33\, \text{cm}^{3} $$
Question 4
1 / -0

The volume of a sphere whose diameter is $$14 ~cm$$ is

A
$$5226.66 \ cm^3$$

B
$$4281.33 \ cm^3$$

C
$$1437.33 \ cm^3$$

D
$$2382.66 \ cm^3$$

Solution
Question 5
1 / -0

Mark the correct alternative of the following.
The number of surfaces of a hollow cylindrical object is?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

The hollow cylinder has a total of 4 surfaces, inner and outer curved surfaces and the ring lies at the base of the cylinder and the top of the cylinder
Question 6
1 / -0

Mark the correct alternative of the following.
A cylinder with radius $$r$$ and height $$h$$ is closed on the top and bottom. Which of the following expressions represents the total surface area of this cylinder?

A
$$2\pi r(r+h)$$

B
$$\pi r(r+2h)$$

C
$$\pi r(2r+h)$$

D
$$2\pi r^2+h$$

Solution

Let $$S$$ be the total surface area of the closed cylinder with radius $$r$$ and height $$h$$, then,
Total surface area of this cylinder $$(S)= 2πrh + πr^2+ πr^2$$
$$\Rightarrow$$ $$S=2\pi r^2+2\pi r h$$
$$\Rightarrow$$ $$S=2\pi r(r+h)$$
Question 7
1 / -0

A right circular cylindrical tunnel of diameter $$2$$m and length $$40$$m is to be constructed from a sheet of iron. The area of the iron sheet required in $$m^2$$, is?

A
$$40\pi$$

B
$$80\pi$$

C
$$160\pi$$

D
$$200\pi$$

Solution

Let $$r$$ be the radius of the cylinder and $$h$$be its height.
It is given that,
$$d=2\,m$$
Then, $$r=\dfrac{2m}{2}=1\,m$$
$$h=40\,m$$
$$\therefore$$ Total surface area of sheet required is:
$$\Rightarrow$$ $$S=2\pi rh$$
$$=2\pi\times 1\times 40$$
$$=80\pi \,m^2$$
Question 8
1 / -0

Mark the correct alternative of the following.
The height of sand in a cylindrical-shaped can drop by $$3$$ inches when $$1$$ cubic foot of sand is pound out. What is the diameter of the cylinder (in inches)?

A
$$\dfrac{24}{\sqrt{\pi}}$$

B
$$\dfrac{48}{\sqrt{\pi}}$$

C
$$\dfrac{32}{\sqrt{\pi}}$$

D
$$\dfrac{48}{\pi}$$

Solution

Let $$r$$ be the radius of the cylinder.
It is given that, the height drops $$3$$ inches, when $$1$$ cubic foot of sand is poured out and $$1$$ foot $$=12\ \text{in}$$
So,
Volume of reduced cone = Volume of 1 cubic foot sand
$$\pi r^2 h=12\times 12\times 12\times 12$$
$$\pi r^2 \times 3=12\times 12\times 12$$
$$r^2=\dfrac{12\times 12\times 4}{\pi}$$

$$r=\dfrac{12\times 2}{\sqrt{\pi}}$$

$$\therefore$$ $$r=\dfrac{24}{\sqrt{\pi}}$$

The diameter of cylinder $$=2r=\dfrac{48}{\sqrt{\pi}}$$
Question 9
1 / -0

The area of the floor of a room is $$15{m}^{2}$$. If its height is $$4m$$, then the volume of the air contained in the room is

A
$$60{dm}^{3}$$

B
$$600{dm}^{3}$$

C
$$6000{xm}^{3}$$

D
$$60000{dm}^{3}$$

Solution

The area of the floor $$(A)=15\,m^2$$
Height of the room $$(h)=4\, m$$
We have to find the volume of the air in the room
So, capacity of the room to contain air,
$$V=A\times h$$
$$=15\times 4$$
$$=60\,m^3$$
$$=60\times (10\,dm)^3$$ [ $$1m=10\,dm$$ ]
$$=60\times 1000\,dm^3$$
$$=60,000\,dm^3$$
Volume of the air contained in the room is $$60,000\,dm^3.$$
Question 10
1 / -0

The largest sphere is carved out of a cube of side 10.5 cm. Find the volume of the sphere.

A
$$ 606.375\,cm^{3}$$

B
$$ 66.3\,cm^{3}$$

C
$$ 60.375\,cm^{3}$$

D
$$ 66.375\,cm^{3}$$

Solution

We have,

Diameter of the largest sphere$$ = 10.5 cm\Rightarrow 2r = 10.5 cm \Rightarrow r = 5.25\, cm$$

$$ \therefore $$ Volume of the sphere$$ = \dfrac{4}{3} \times \dfrac{22}{7} \times \left ( 5.25 \right )^{3} \,cm^{3}$$

$$ = \dfrac{4}{3} \times \dfrac{22}{7} \times \left ( \dfrac{21}{4} \right )^{3} cm^{3} = 4 \times 22 \times \dfrac{21^2}{64} cm^3 $$

$$ = \dfrac{11 \times 441}{ 8 } \,cm^3 = 606.375\, cm^3 $$

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Re-Attempt Weekly Quiz Competition

Surface Areas and Volumes Test - 49

Result

Surface Areas and Volumes Test - 49

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SHARING IS CARING

Question 1

$$460 \ cm^2$$

$$490 \ cm^2$$

$$360 \ cm^2$$

$$230 \ cm^2$$

Solution

Question 2

$$10.1848$$ $$m^3$$

$$0.001858$$ $$m^3$$

$$0.001848$$ $$m^3$$

$$18.48$$ $$m^3$$

Solution

Question 3

1370 $$\text{cm}^3$$

1800 $$\text{cm}^3$$

1437 $$\text{cm}^3$$

1734 $$\text{cm}^3$$

Solution

Question 4

$$5226.66 \ cm^3$$

$$4281.33 \ cm^3$$

$$1437.33 \ cm^3$$

$$2382.66 \ cm^3$$

Solution

Question 5

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 6

$$2\pi r(r+h)$$

$$\pi r(r+2h)$$

$$\pi r(2r+h)$$

$$2\pi r^2+h$$

Solution

Question 7

$$40\pi$$

$$80\pi$$

$$160\pi$$

$$200\pi$$

Solution

Question 8

$$\dfrac{24}{\sqrt{\pi}}$$

$$\dfrac{48}{\sqrt{\pi}}$$

$$\dfrac{32}{\sqrt{\pi}}$$

$$\dfrac{48}{\pi}$$

Solution

Question 9

$$60{dm}^{3}$$

$$600{dm}^{3}$$

$$6000{xm}^{3}$$

$$60000{dm}^{3}$$

Solution

Question 10

$$ 606.375\,cm^{3}$$

$$ 66.3\,cm^{3}$$

$$ 60.375\,cm^{3}$$

$$ 66.375\,cm^{3}$$

Solution

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