Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 17

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Weekly Quiz Competition

Question 1
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The slant height of a frustum of a flower pot is 2 mm and the perimeters of its circular ends are 12 mm and 4 mm. Find the curved surface area of the flower pot.

A
13.7 $$mm^2$$

B
14.7 $$mm^2$$

C
15.7 $$mm^2$$

D
16.7 $$mm^2$$

Solution

Radius, $$R = perimeter/2 \pi = 12/2 \times 3.14 \Rightarrow 1.9 mm$$
$$r = perimeter/2 \pi = 4/2 \times 3.14 = 0.6 mm$$
Curved surface area = $$ \pi (R + r)s$$
$$= 3.14 \times (1.9 + 0.6) \times 2$$
$$= 3.14 \times 2.5 \times 2$$
$$= 15.7 mm^2$$
Question 2
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If the radii of the circular ends of a bucket are $$14$$ and $$6$$ m whose height is $$6$$ m. What is the slant height of the bucket ?

A
$$10$$

B
$$20$$

C
$$30$$

D
$$40$$

Solution

Given: $$R=14\space\mathrm{m}, r=6\space\mathrm{m}$$ and $$h=6\space\mathrm{m}$$
To find: Slant height, $$l$$
Solution: We know, $$l=\sqrt{(R-r)^2+h^2}$$
$$=\sqrt{(14-6)^2+6^2}$$
$$=\sqrt{8^2+6^2}$$
$$=\sqrt{64+36}$$
$$=\sqrt{100}$$
$$=10$$
$$\therefore$$ Slant height $$(l)=10\space\mathrm{m}$$
So, $$\text{A}$$ is the correct option.
Question 3
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The perimeter of the ends of a frustum of a cone are $$44cm$$ and $$8.4\pi cm$$. If the depth is $$14cm$$, then find its volume.

A
$$1408.6{cm}^{3}$$

B
$$1486{cm}^{3}$$

C
$$1406{cm}^{3}$$

D
None of these

Solution

Let the bottom and top radius be $$R$$ and $$r$$ respectively
Perimeter of the ends of the frustum cone
$$2\pi R=44\ cm\\\Rightarrow R=7\ cm$$
$$2\pi r=8.4\pi\ cm\\\Rightarrow r=4.2\ cm$$

$$depth(h)=14\ cm$$

We know that the volume of the frustum cone
$$=\dfrac{1}{3}\pi (R^2+r^2+R r)h$$

So,
$$=\dfrac{1}{3}\times \dfrac{22}{7} (7^2+(4.2)^2+7\times 4.2)\times 14$$

$$=\dfrac{1}{3}\times 22 \times (49+17.64+29.4)\times 2$$

$$=\dfrac{44}{3} \times (96.04)$$

$$=1408.6\ cm^3$$
Question 4
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A friction clutch is in the form of a frustum of cone. The radius of the ends bring $$16$$ cm and $$10$$ cm. Find its curved surface area. The slant height of the friction clutch is $$12$$ cm.

A
$$302$$ $$\pi cm^2$$

B
$$312$$ $$cm^2$$

C
$$318$$ $$cm^2$$

D
$$312$$ $$\pi cm^2$$

Solution

Curved surface area = $$\pi ( R+ r) s$$
$$= \pi (16+ 10) \times 12$$
$$= 26 \pi \times 12$$
$$= 312 \pi cm^2$$
Question 5
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The slant height of a frustum of a drum is 12 ft and the perimeters of its circular ends are 10 ft and 2 ft. Find the curved surface area of the drum.(Use $$\pi$$ = 3.14)

A
719.8912 $$ft^2$$

B
709.8912 $$ft^2$$

C
729.8912 $$ft^2$$

D
739.8912 $$ft^2$$

Solution

Radius, $$R = perimeter/2 \pi = 10/2 x 3.14 \Rightarrow 15.7 ft$$
$$r = perimeter/2 \pi = 2/2 x 3.14 = 3.14 ft$$
Curved surface area = $$ \pi (R + r)s$$
$$= 3.14 \times (15.7 + 3.14) x 12$$
$$= 3.14 \times 18.84 \times 12$$
$$= 709.8912 ft^2$$
Question 6
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Find the capacity of a glass which is in the shape of frustum of height $$14$$cm and diameters of both circular ends are $$4$$cm and $$2$$cm.

A
$$\displaystyle\frac{308}{3}cm^3$$

B
$$\displaystyle\frac{298}{21}cm^3$$

C
$$\displaystyle 112cm^2$$

D
$$\displaystyle\frac{298}{21}cm^2$$

Solution

We know that volume of frustum $$\displaystyle v=\frac{\pi h}{3}(r^2_1+r^2_2+r_1r_2)$$
Given: $$r_1 = \dfrac{2}{2} = 1cm$$, $$r_2 = \dfrac{4}{2} = 2cm$$ and $$h=14cm$$

$$\displaystyle v=\frac{22}{7}\times \frac{14}{3}(1^2+2^2+1\times 2)$$

$$\displaystyle =\frac{44}{3}(7)=\frac{308}{3}cm^3$$.
Question 7
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Volume of the figure $$C$$ is :

A
Same as volume of $$A$$

B
Same as volume of $$B$$

C
Sum of volumes of $$A$$ and $$B$$

D
Difference of volumes of $$A$$ and $$B$$

Solution

$$C$$ is a shape formed by combining both $$A$$ and $$B$$.
Hence, volume of $$C$$ is the sum of volumes of $$A$$ and $$B$$.
Question 8
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A capsule shape is formed by combining which of the following solids?

A
$$I$$ and $$II$$

B
$$II$$ and $$III$$

C
$$I$$ and $$III$$

D
$$I$$, $$II$$ and $$III$$

Solution

A capsule is a shape containing curved surfaces and no flat surfaces .
By joining $$I,II,III$$ we will get the shape of a capsule.
Hence, the answer is $$I,II$$ and $$III$$.
Question 9
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The formula for finding the total surface area of a cylinder having cone shaped lid at both the ends, will be

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

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

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

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

Solution

Total Surface Area of cylindrical part $$=2\pi rh$$
Surface Area of two cones $$=\pi rl+\pi rl=2\pi rl\quad $$
Then, Total Surface Area $$=2\pi rh+2\pi rl=2\pi r(h+l)$$
Question 10
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The slant height of a cone is $$13\ cm$$ and radius is $$5\ cm$$. then its height is-

A
$$5\ cm$$

B
$$22\ cm$$

C
$$12\ cm$$

D
$$18\ cm$$

Solution

As given in this diagram
Here the height will be perpendicular to the base
So it will form a right angled triangle we can apply Pythagoras theorem in this right angled triangle
$$ S = 13\ cm $$ and $$ r= 5\ cm $$
$$ S^{2} = h^{2} + r^{2} $$
$$13^{2} = h^{2} + 5^{2} $$
Hence $$h =12\ cm$$

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Surface Areas and Volumes Test - 17

Result

Surface Areas and Volumes Test - 17

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Question 1

13.7 $$mm^2$$

14.7 $$mm^2$$

15.7 $$mm^2$$

16.7 $$mm^2$$

Solution

Question 2

$$10$$

$$20$$

$$30$$

$$40$$

Solution

Question 3

$$1408.6{cm}^{3}$$

$$1486{cm}^{3}$$

$$1406{cm}^{3}$$

None of these

Solution

Question 4

$$302$$ $$\pi cm^2$$

$$312$$ $$cm^2$$

$$318$$ $$cm^2$$

$$312$$ $$\pi cm^2$$

Solution

Question 5

719.8912 $$ft^2$$

709.8912 $$ft^2$$

729.8912 $$ft^2$$

739.8912 $$ft^2$$

Solution

Question 6

$$\displaystyle\frac{308}{3}cm^3$$

$$\displaystyle\frac{298}{21}cm^3$$

$$\displaystyle 112cm^2$$

$$\displaystyle\frac{298}{21}cm^2$$

Solution

Question 7

Same as volume of $$A$$

Same as volume of $$B$$

Sum of volumes of $$A$$ and $$B$$

Difference of volumes of $$A$$ and $$B$$

Solution

Question 8

$$I$$ and $$II$$

$$II$$ and $$III$$

$$I$$ and $$III$$

$$I$$, $$II$$ and $$III$$

Solution

Question 9

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

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

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

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

Solution

Question 10

$$5\ cm$$

$$22\ cm$$

$$12\ cm$$

$$18\ cm$$

Solution

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