Mensuration Test

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Mensuration Test - 17

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Weekly Quiz Competition

Question 1
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 2
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 3
1 / -0

If the perimeter of one face of a cube is $$20\: cm$$, then its surface area is

A
$$120\: cm^2$$

B
$$150\: cm^2$$

C
$$125\: cm^2$$

D
$$400\: cm^2$$

Solution

Perimeter of one face of a cube is $$4a=20$$
$$\Rightarrow a=5$$ $$cm$$
Now the surface area will be $$6a^2$$
$$=6\times 5\times 5$$
$$=150\: cm^2$$
Hence, option $$B$$ is correct.
Question 4
1 / -0

The ratio of the height of a circular cylinder to the diameter of its base is $$1 : 2$$, then the ratio of the areas of its curved surface to the sum of the areas of its two ends is

A
$$1 : 1$$

B
$$1 : 2$$

C
$$2 : 1$$

D
$$1 : 3$$

Solution

Given,
Height : Diameter of cylinder $$= 1:2$$
Height : Radius of cylinder $$= 1:1$$
Curved surface area $$= 2 \pi r h$$
Areas of two ends $$= 2\pi r^2$$
$$\therefore$$ their Ratio = $$\displaystyle \frac{2\pi rh}{2\pi r^2}= 1:1$$
Question 5
1 / -0

If the radius of the base of a cylinder is $$2$$ cm and its height $$7$$ cm, then its curved surface is

A
$$44 cm^2$$

B
$$22 cm^2$$

C
$$88 cm^2$$

D
$$56 cm^2$$

Solution

Given, radius of cylinder $$= 2$$ cm and height of cylinder $$= 7$$ cm
$$\therefore $$ curved surface area of cylinder $$= 2\pi rh $$
$$= 2 \pi \times 2\times 7$$
$$= 88 cm^2$$
Question 6
1 / -0

If the curved surface of a cylinder be doubled the area of the ends, then the ratio of its height and radius is

A
$$2 : 3$$

B
$$1 : 1$$

C
$$2 : 1$$

D
$$1 : 2$$

Solution

Let the height and radius of the ends of a cylinder be $$h$$ and $$r$$ respectively.
Now Curved surface $$= 2 \times$$ Area of the ends
$$\therefore 2\pi rh = 2\times (\pi r^2+\pi r^2)$$
$$\therefore \dfrac{h}{r} = \dfrac{2}{1}$$
$$\therefore h : r :: 2 :1$$
Question 7
1 / -0

If the curved surface of a right circular cylinder is $$1760 cm^2$$ and its radius is 21 cm, then its height is $$\dfrac {80}{3} cm.$$

A
True

B
False

C
Data insufficient

D
Ambiguous

Solution

Formula for curved surface of a cylinder $$=2\pi rh$$
given radius $$=21\ cm$$
$$2(\pi)rh=1760$$
$$2 \times \dfrac{22}{7} \times 21 \times h=1760$$
$$2\times 22\times 3 \times h=1760$$
$$h=\dfrac{1760}{44 \times 3}$$
$$h=\dfrac{40}{3}\ cm$$
Hence the given statement is false.
Question 8
1 / -0

A sphere has the same curved surface as the total surface area of cylinder of height 4 cm and diameter of base 8 cm. The radius of the sphere is

A
$$2\ cm$$

B
$$3 \ cm$$

C
$$4\ cm$$

D
$$6\ cm$$

Solution

Given that,
The curved surface area of a sphere is same as the total surface area of a cylinder.
The height of the cylinder is $$4\ cm$$ and the diameter of its base is $$8\ cm$$

To find out,
The radius of the sphere.

Let the radius of the sphere be $$x\ cm$$.
We know that, total surface area of cylinder $$= 2\pi rh + 2\pi r^2$$
We have $$r=\frac d2=4\ cm$$ and $$h=4\ cm$$
$$\therefore \ $$ Total surface area of the cylinder $$= 2 \pi (4)(4) + 2\pi (4)^2$$
$$= 32 \pi + 32 \pi $$
$$= 64 \pi $$

We also know that, curved surface area of a sphere $$= 4\pi r^2 $$
Hence, $$4\pi x^2= 64 \pi$$
$$\Rightarrow x^2=16$$
$$\therefore \ x = 4\ cm$$

Hence, the radius of the sphere is $$4\ cm$$.
Question 9
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 10
1 / -0

A tent is in the form of a right circular cylinder, surmounted by a cone. The diameter of the cylinder is $$24$$ m. The height of the cylindrical portion is $$11$$ m, while the vertex of the cone is $$16$$ m above the ground.
The curved surface area of the cylindrical portion is

A
$$(246 \pi)m^2$$

B
$$(264 \pi)m^2$$

C
$$(426 \pi)m^2$$

D
$$(462 \pi)m^2$$

Solution

Given, diameter $$=24$$ m
Therefore, radius $$=\dfrac {d}{2}=\dfrac {24}{2}=12$$ m
and height given is $$11$$ m
Therefore, curved surface area of the cylinder $$=2(\pi)rh$$
$$=2\times (\pi)\times 12\times 11$$
$$=264(\pi)$$

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Re-Attempt Weekly Quiz Competition

Mensuration Test - 17

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Mensuration Test - 17

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SHARING IS CARING

Question 1

1

2

3

0

Solution

Question 2

halved

doubled

same

four times

Solution

Question 3

$$120\: cm^2$$

$$150\: cm^2$$

$$125\: cm^2$$

$$400\: cm^2$$

Solution

Question 4

$$1 : 1$$

$$1 : 2$$

$$2 : 1$$

$$1 : 3$$

Solution

Question 5

$$44 cm^2$$

$$22 cm^2$$

$$88 cm^2$$

$$56 cm^2$$

Solution

Question 6

$$2 : 3$$

$$1 : 1$$

$$2 : 1$$

$$1 : 2$$

Solution

Question 7

True

False

Data insufficient

Ambiguous

Solution

Question 8

$$2\ cm$$

$$3 \ cm$$

$$4\ cm$$

$$6\ cm$$

Solution

Question 9

$$12\ cm^3$$

$$6\ cm^3$$

$$4\ cm^3$$

$$3\ cm^3$$

Solution

Question 10

$$(246 \pi)m^2$$

$$(264 \pi)m^2$$

$$(426 \pi)m^2$$

$$(462 \pi)m^2$$

Solution

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