Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

The amount of three-dimensional space occupied by an object is called?

A
Area

B
Pressure

C
Volume

D
Mass

Solution

Volume is the quantity of three-dimensional space enclosed by a closed surface. For example: Cylinder, Cuboid etc
Question 2
1 / -0

If all three parameters of a cuboid are in meters$$(m)$$, then unit of volume of a cuboid is:

A
$$m$$

B
$$m^{2}$$

C
$$m^{3}$$

D
$$cm^{3}$$

Solution

As the volume of a cube is $$length\times length \times length $$
If lengths of all are measured in meters.
Then the volume will be measured in $$m^3$$.
Hence, the answer is $$m^{3}$$.
Question 3
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If the circumference of base of a hemisphere is $$2\pi$$ then its volume is _________ $$cm^3$$.

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

B
$$\dfrac{2\pi}{3}$$

C
$$\dfrac{8\pi}{3}$$

D
$$\dfrac{\pi}{12}$$

Solution

The circumference of base of a hemisphere $$=2\pi$$
$$\therefore 2\pi r=2 \pi$$
$$\therefore r=1$$
Its volume $$= \dfrac{2}{3}\pi (r)^3$$
$$= \dfrac{2}{3}\pi (1)^3 =\dfrac{2}{3}\pi \ cm^3$$
Question 4
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The surface area of a cube of side $$27\ \text{cm}$$ is

A
$$2916\ \text{cm}^{2}$$

B
$$729\ \text{cm}^{2}$$

C
$$4374\ \text{cm}^{2}$$

D
$$19683\ \text{cm}^{2}$$

Solution

Let, the given side of square$$=a=27\ \text{cm}$$
Surface area of cube$$=6a^2$$
$$=6(27)^2\ \text{cm}^2$$
$$=4374\ \text{cm}^2$$
Question 5
1 / -0

Find the surface area of a $$10cm \times 4cm \times 3cm$$ brick:

A
84 sq. cm

B
124 sq.cm

C
164 sq.cm

D
180 sq.cm

Solution

The formula for calculating the surface area of a cuboid of dimensions :

$$ l \times b \times h $$ is :

$$ 2\times (l\times b+b\times h+h\times l) $$

Substituting,

$$l=10cm$$

$$b=4cm$$

$$h=3cm$$

Surface area = $$ 2\times (10\times 4+4\times 3+3\times 10) $$

$$ = 2\times (40 + 12 + 30) $$

$$ = 2\times 82 { cm }^{ 2 } $$

$$ = 164 { cm }^{ 2 } $$

Hence, the answer is option (C)
Question 6
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Given a sphere with the radius $$7\ cm$$. Find the volume of this sphere. Take $$\pi$$ as $$3.14$$.

A
$$1436.03 \ cm^3$$

B
$$87.92 \ cm^3$$

C
$$1345.01 \ cm^3$$

D
$$100.02 \ cm^3$$

Solution

Volume of sphere = $$\dfrac{4}{3} \pi r^3$$

$$V=\dfrac{4}{3} (3.14) (7)^3$$

$$V=\dfrac{4}{3} \times 3.14 \times 343=1436.03 \ cm^3$$
Question 7
1 / -0

The surface area of a solid hemisphere with radius $$r$$ is

A
$$4\pi r^2$$

B
$$2\pi r^2 $$

C
$$3\pi r^2 $$

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

Solution

Surface area of a solid sphere with radius $$r$$ is $$4 \pi r^2$$
A solid sphere can be divided into two equal hemispheres with a flat surface and a curved surface.
Curved surface area of a solid hemisphere will be half the surface area of solid sphere.
Hence, curved surface area of solid hemisphere $$=\dfrac{1}{2}*4 \pi r^2 = 2 \pi r^2$$
And, flat surface area of hemisphere will be the area of circle with radius $$r$$.
Hence, flat surface area of hemisphere $$=\pi r^2$$
Total surface area of a solid hemisphere with radius $$r$$ is, $$S=2\pi r^2+\pi r^2 = 3\pi r^2$$
Question 8
1 / -0

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratios of the surface areas of the balloon in the two cases is

A
1 : 4

B
1 : 3

C
2 : 3

D
2 : 1

Solution

Total surface area of a Hemisphere $$=2\pi r^2 + \pi r^2 = 3\pi r^2$$
$$r=6,Surface\ area=3 \pi (6)^2=36$$
$$r=12,Surface\ area=3 \pi (12)^2=144$$
Ratio = $$36:144$$
=$$1:4$$
Question 9
1 / -0

The water for a factory is stored in a hemispherical tank whose internal diameter is $$14\ m$$. The tank contains $$50$$ kilo litres of water. Water is pumped into the tank to fill to its capacity. Calculate the volume of water pumped into the tank.

A
$$568.67\ m^3$$

B
$$668.67\ m^3$$

C
$$648.67\ m^3$$

D
$$688.67\ m^3$$

Solution

Radius of hemispherical tank, $$ r= 7$$ m
So volume of water to be pumped $$=$$ Volume of hemispherical tank $$-50$$ kl
$$\because 1\ kl = 1\ m^3$$
$$\Rightarrow \dfrac{2}{3} \pi r^3 - 50\ m^3$$
$$\Rightarrow \dfrac{2}{3} \times \dfrac{22}{7}\times \left(7\right)^3 - 50\ m^3$$
$$\Rightarrow 718.67-50\ m^3$$
$$\Rightarrow 668.67\ m^3$$
Hence, option B is correct.
Question 10
1 / -0

The radius of a sphere is $$2r$$, then its volume will be:

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

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

C
$$\dfrac{8 \pi r^{3}}{3}$$

D
$$\dfrac{32}{3} \pi r^{3}$$

Solution

As,$$r=2r$$
Volume of sphere$$=\frac{4}{3} \pi (2r)^3=\frac{32}{3} \pi r^3$$

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Re-Attempt Weekly Quiz Competition

Surface Areas and Volumes Test - 22

Result

Surface Areas and Volumes Test - 22

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SHARING IS CARING

Question 1

Area

Pressure

Volume

Mass

Solution

Question 2

$$m$$

$$m^{2}$$

$$m^{3}$$

$$cm^{3}$$

Solution

Question 3

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

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

$$\dfrac{8\pi}{3}$$

$$\dfrac{\pi}{12}$$

Solution

Question 4

$$2916\ \text{cm}^{2}$$

$$729\ \text{cm}^{2}$$

$$4374\ \text{cm}^{2}$$

$$19683\ \text{cm}^{2}$$

Solution

Question 5

84 sq. cm

124 sq.cm

164 sq.cm

180 sq.cm

Solution

Question 6

$$1436.03 \ cm^3$$

$$87.92 \ cm^3$$

$$1345.01 \ cm^3$$

$$100.02 \ cm^3$$

Solution

Question 7

$$4\pi r^2$$

$$2\pi r^2 $$

$$3\pi r^2 $$

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

Solution

Question 8

1 : 4

1 : 3

2 : 3

2 : 1

Solution

Question 9

$$568.67\ m^3$$

$$668.67\ m^3$$

$$648.67\ m^3$$

$$688.67\ m^3$$

Solution

Question 10

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

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

$$\dfrac{8 \pi r^{3}}{3}$$

$$\dfrac{32}{3} \pi r^{3}$$

Solution

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