Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

Find the volume of the frustum cone whose base and top radius is $$10$$ cm and $$2$$ cm respectively. The height of the cone is $$3$$cm. (Use $$\pi =\dfrac{22}{7}$$).

A
$$369.36$$ cm$$^3$$

B
$$379.36$$ cm$$^3$$

C
$$389.36$$ cm$$^3$$

D
$$319.36$$ cm$$^3$$

Solution

Volume of frustum $$=\dfrac { 1 }{ 3 } \pi \left( { R }^{ 2 }+{ r }^{ 2 }+Rr \right) \times h$$

$$=\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \left( { 10 }^{ 2 }+{ 2 }^{ 2 }+10\times 2 \right) \times 3$$

$$=\dfrac { 22 }{ 7 } \left( 100+4+20 \right) =\dfrac { 22\times 124 }{ 7 } $$

$$= 389.71$$ $${ cm }^{ 3 }$$
Question 2
1 / -0

Find the volume of the frustum cone whose base and top radius is $$5$$ mm and $$2$$ mm respectively. The height of the cone is $$15$$mm. (Use $$\pi$$ = 3.14).

A
$$612.3$$ mm$$^3$$

B
$$602.3$$ mm$$^3$$

C
$$622.3$$ mm$$^3$$

D
$$632.3$$ mm$$^3$$

Solution

Volume of frustum $$=\dfrac { 1 }{ 3 } \times \pi \left( { R }^{ 2 }+{ r }^{ 2 }+Rr \right) \times h$$

$$=\dfrac { 1 }{ 3 } \times 3.14({ 5 }^{ 2 }+{ 2 }^{ 2 }+5\times 2)\times 15$$

$$=15.70(25+4+10)=15.70\times 39$$

$$= 612.3$$ $${ mm }^{ 3 }$$
Question 3
1 / -0

Find the volume of a frustum cone, whose base and upper area of a circle is 36 and 100 $$mm^2$$. The height of the cone is 32.5 mm.

A
6,270 $$mm^3$$

B
6,170 $$mm^3$$

C
2,123.33 $$mm^3$$

D
6,470 $$mm^3$$

Solution

Volume of a frustum cone = $$\frac{1}{3} (A_1 + A_2 + \sqrt{A_1 A_2 }) h$$
= $$\frac{1}{3} (36 + 100 + \sqrt{36 \times 100}) \times 32.5$$
= $$\frac{1}{3} (136 + \sqrt{3600}) \times 32.5$$
= $$\frac{1}{3} (136+60) \times 32.5$$
= $$\frac{1}{3} \times 196 \times 32.5$$
= 2,123.33 $$mm^3$$
Question 4
1 / -0

Find the volume of the frustum cone whose base and top radius is $$0.1 cm$$ and $$0.01 cm$$ respectively. The height of the cone is $$3cm$$. (Use $$\pi$$ = 3).

A
$$0.0003 cm$$$$^3$$

B
$$0.0033 cm$$$$^3$$

C
$$0.3333 cm$$$$^3$$

D
$$0.0333 cm$$$$^3$$

Solution

Volume of frustum $$ = \dfrac { 1 }{ 3 } \pi \left( { R }^{ 2 }+{ r }^{ 2 }+Rr \right) \times h$$

$$= \dfrac { 1 }{ 3 } \times 3\left( { \left( 0.1 \right) }^{ 2 }+{ \left( 0.01 \right) }^{ 2 }+\left( 0.1\times 0.01 \right) \right) \times 3$$

$$= 3(0.01+0.0001+0.001)$$

$$= 3\times 0.0111=0.0333{ cm }^{ 3 }$$
Question 5
1 / -0

A dish in the shape of a frustum of a cone has a height of $$6$$cm. Its top and its bottom have radii of $$24$$cm and $$16$$cm respectively. Find its curved surface area. ( in $$\displaystyle cm^{2}$$)

A
$$200 \displaystyle \pi $$

B
$$300 \displaystyle \pi $$

C
$$400 \displaystyle \pi $$

D
$$500 \displaystyle \pi $$

Solution

Curved surface area of a frustum of a cone $$=\pi l({ r }_{ 1 }+{ r }_{ 2 })$$

where $$ l $$ is the slant height of the cone $$ = \sqrt { { h }^{ 2 }+{ ({ r }_{ 1 }-{ r }_{ 2 }) }^{ 2 } } $$
$$ h $$ is the height
$$ { r }_{ 1 } $$ and $${ r }_{ 2 }$$ are the radii of the lower and upper ends of a frustum of a cone.

So, $$l = \sqrt { { 6 }^{ 2 }+{ (24 - 16) }^{ 2 } } $$

$$ l= 10 cm $$

Hence, Curved Surface area of this frustum of a cone $$=\pi \times 10 \times (24 + 16) = 400 \pi {cm}^{2} $$
Question 6
1 / -0

A drum is in the shape of a frustum of a cone. Its top and bottom radii are $$20$$ ft and $$10$$ ft respectively. Its height is $$15$$ ft. It is fully filled with water. This water is emptied into a rectangular tank. The base of the tank has the dimensions $$100$$ ft $$\times 50$$ ft. Find the rise in the height of the water level in the tank.

A
$$1.9$$ ft.

B
$$2.1$$ ft.

C
$$2.2$$ ft.

D
$$2.4$$ ft.

Solution

Let $$h$$ be the height and $$R$$ and $$r$$ be the radii of the lower and upper ends of a frustum of a cone.

Volume of a frustum of a cone $$ = \dfrac { 1 }{ 3 } \pi h({ R }^{ 2 }+{ r }^{ 2 }+Rr) $$

Hence, volume of the drum $$ = \dfrac { 1 }{ 3 } \times \dfrac {22}{7} \times 15({ 20 }^{ 2 }+{ 10 }^{ 2 }+ 20 \times 10) $$

$$= 11,000 \ {ft}^{3} $$

As this water is emptied into a cuboidal tank, volume of water in tank till height $$h = 11,000 {ft}^{3} $$

Volume of a cuboid of length $$(l)$$, breadth $$(b)$$ and height $$(h)$$ $$ = l \times b \times h $$

So, $$ 100 \times 50 \times h =11,000 $$

$$ \Rightarrow h = 2.2\ ft $$
Question 7
1 / -0

Find the volume of a frustum cone, whose base and upper area of a circle is $$40cm^2$$ and $$40$$ $$cm^2$$. The height of the cone is $$12$$ cm.

A
$$380$$ $$cm^3$$

B
$$480$$ $$cm^3$$

C
$$580$$ $$cm^3$$

D
$$680$$ $$cm^3$$

Solution

Volume of frustum $$=\dfrac { 1 }{ 3 } \left( { A }_{ 1 }+{ A }_{ 2 }+\sqrt { { A }_{ 1 }{ A }_{ 2 } } \right) \times h$$
$$=\dfrac { 1 }{ 3 } \times \left( 40+40+\sqrt { 40\times 40 } \right) \times 12$$
$$=\left( 80+40 \right) \times 4=120\times 4$$
$$ = 480 $$$${ cm }^{ 3 }$$
Question 8
1 / -0

The base and top radius of a truncated cone are $$20 mm$$ and $$10 mm$$ respectively. The height of the cone is $$360 mm$$. What is the volume of a truncated cone? (Use $$\pi$$ = 3).

A
$$252,000 mm$$$$^3$$

B
$$242,000 mm$$$$^3$$

C
$$232,000 mm$$$$^3$$

D
$$222,000 mm$$$$^3$$

Solution

Since the truncated cone is in the shape of frustum, so volume of the truncated cone $$V = \displaystyle{\frac{\pi h}{3}}(R^2 + Rr + r^2)$$
Here, we have,
$$R = 20 mm, r = 10 mm, h = 360 mm$$
Therefore,
$$V = \dfrac{(3 \times 360)}{3}[20^2 + 20 \times 10 + 10^2]$$
$$=$$ $$360[400 + 200 + 100]$$
$$=$$ $$360[700]$$
$$=$$ $$252,000 \ mm$$ $$^3$$
Question 9
1 / -0

The volume of a frustum cone is 4,200 $$cm^3$$, whose base and upper area of a circle is 40 and 10 $$cm^2$$. Find the height of the cone.

A
140 cm

B
150 cm

C
180 cm

D
170 cm

Solution

Volume of a frustum cone = $$\frac{1}{3} (A_1 + A_2 + \sqrt{A_1 A_2 }) h$$
4,200 = $$\frac{1}{3} (40 + 10 + \sqrt{40 \times 10}) \times h$$
4,200 = $$\frac{1}{3} (50 + \sqrt{400}) \times h$$
4,200 = $$\frac{1}{3} (50+20) \times h$$
$$4,200 \times 3$$ = $$70 \times h$$
h = 4,200 $$\times$$ 3/70
h = 180 cm
Question 10
1 / -0

The volume of a frustum cone is 2,700 $$in^3$$, whose base and upper area of a circle is 20 and 5 $$in^2$$. Find the height of the cone.

A
221.42 in

B
231.42 in

C
211.42 in

D
201.42 in

Solution

Volume of a frustum cone = $$\frac{1}{3} (A_1 + A_2 + \sqrt{A_1 A_2 }) h$$
2,700 = $$\frac{1}{3} (20 + 5 + \sqrt{20 \times 5}) \times h$$
2,700 = $$\frac{1}{3} (25 + \sqrt{100}) \times h$$
2,700 = $$\frac{1}{3} (25+10) \times h$$
$$2,700 \times 3 = 35 \times h$$
$$h = \frac{8,100}{35}$$
$$h = 231.42\, in$$

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Surface Areas and Volumes Test - 27

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Surface Areas and Volumes Test - 27

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Question 1

$$369.36$$ cm$$^3$$

$$379.36$$ cm$$^3$$

$$389.36$$ cm$$^3$$

$$319.36$$ cm$$^3$$

Solution

Question 2

$$612.3$$ mm$$^3$$

$$602.3$$ mm$$^3$$

$$622.3$$ mm$$^3$$

$$632.3$$ mm$$^3$$

Solution

Question 3

6,270 $$mm^3$$

6,170 $$mm^3$$

2,123.33 $$mm^3$$

6,470 $$mm^3$$

Solution

Question 4

$$0.0003 cm$$$$^3$$

$$0.0033 cm$$$$^3$$

$$0.3333 cm$$$$^3$$

$$0.0333 cm$$$$^3$$

Solution

Question 5

$$200 \displaystyle \pi $$

$$300 \displaystyle \pi $$

$$400 \displaystyle \pi $$

$$500 \displaystyle \pi $$

Solution

Question 6

$$1.9$$ ft.

$$2.1$$ ft.

$$2.2$$ ft.

$$2.4$$ ft.

Solution

Question 7

$$380$$ $$cm^3$$

$$480$$ $$cm^3$$

$$580$$ $$cm^3$$

$$680$$ $$cm^3$$

Solution

Question 8

$$252,000 mm$$$$^3$$

$$242,000 mm$$$$^3$$

$$232,000 mm$$$$^3$$

$$222,000 mm$$$$^3$$

Solution

Question 9

140 cm

150 cm

180 cm

170 cm

Solution

Question 10

221.42 in

231.42 in

211.42 in

201.42 in

Solution

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