Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

A solid in shape of a frustum is 21 cm high. Its radius of top is 10 cm and diameter of bottom is 30 cm. The volume of the solid is

A
10500 $$\displaystyle cm^{3}$$

B
10450 $$\displaystyle cm^{3}$$

C
10000 $$\displaystyle cm^{3}$$

D
None of the above

Solution

Volume of frustum = $$\displaystyle \dfrac{1}{3}\pi h\left ( r_{1}^{2}+r_{1}r_{2}+r_{2}^{2} \right )$$
= $$\displaystyle \dfrac{1}{3}\times \dfrac{22}{7}\times 21\times \left ( 10^{2}+10\times 15+15^{2} \right )cm^{3}$$
= $$\displaystyle \dfrac{1}{3}\times \dfrac{22}{7}\times 21\times 475cm^{3}$$
= 10450 $$\displaystyle cm^{3}$$
Question 2
1 / -0

A spherical iron ball of radius $$9$$ cm is melted and recast into three spherical balls. If the radii of two balls are $$1$$ cm and $$8$$ cm, find the radius of the third ball.

A
$$2\;cm$$

B
$$3\;cm$$

C
$$4\;cm$$

D
$$6\;cm$$

Solution

Let the radius of big ball $$R=9$$ cm

Let the radii of the three balls be $$r_1,\ r_2,\ r_3$$

where $$r_1=1\ cm, r_2=8\ cm$$

We know

Volume of iron ball $$=$$ Sum of volumes of $$3$$ balls

$$\Rightarrow \dfrac { 4 }{ 3 } \pi { R }^{ 3 }=\dfrac { 4 }{ 3 } \pi \left( { r }_{ 1 }^{ 3 }+{ r }_{ 2 }^{ 3 }+{ r }_{ 3 }^{ 3 } \right)$$

$$ \Rightarrow { \left( 9 \right) }^{ 3 }={ 1 }^{ 3 }+{ 8 }^{ 3 }+{ r }_{ 3 }^{ 3 }$$

$$\Rightarrow 729=1+512+{ r }_{ 3 }^{ 3 }$$

$$\Rightarrow 216={ r }_{ 3 }^3$$

$$r_3=\sqrt [ 3 ]{ 216 } =6$$ cm

Therefore radius of third sphere is $$6$$ cm.
Question 3
1 / -0

Given that the volume of a cone is $$\displaystyle 2355cm^{3}$$ and the area of its base is $$\displaystyle 314cm^{2}$$ Its height is

A
25.2 m

B
22.5 m

C
25.5 m

D
52.5 m

Solution

$$\displaystyle V_{cone}=\frac{1}{3}\pi r^{2}h$$
or 2355=$$\displaystyle \frac{1}{3}\times 314h$$
$$\displaystyle \therefore h=22.5 m$$'
Question 4
1 / -0

A metallic sphere of radius $$10.5 cm$$ is melted and then recast into small cones each of radius $$3.5 cm$$ and height $$3 cm$$. The number of such cones is

A
$$63$$

B
$$126$$

C
$$21$$

D
$$130$$

Solution

Radius of sphere $$\displaystyle (r)=10.5cm=\frac {21}{2}cm$$

$$\therefore $$ Volume $$\displaystyle =\frac {4}{3}\pi r^{3}=\frac {4}{3}\pi \times \frac {21}{2}\times \frac {21}{2}\times \frac {21}{2}$$

Let $$n$$ cones are formed
Radius of cone $$\displaystyle (R)=3.5cm=\frac {7}{2}cm$$
Height of cone $$(h)=3cm$$
Volume of $$1$$ cone $$\displaystyle =\frac {1}{3}\pi R^{2}h$$
$$\displaystyle =\frac {1}{3}\pi \times \frac {7}{2}\times \frac {7}{2} \times 3=\frac {49}{4}\pi$$

Volume of $$n$$ cones $$\displaystyle =\frac {49n\pi}{4}$$
Vol. of cylinder $$=$$ Vol. of $$n$$ cones

$$\displaystyle \frac {4\pi}{3}\times \frac {21}{2}\times \frac {21}{2}\times \frac {21}{2} = \frac {49n\pi}{4}$$

$$\therefore n=2 \times 3 \times 21$$
$$n=126$$ cones.
Question 5
1 / -0

Fill in the blank:
The volume of a frustum cone top and bottom area of a circle and height is given. Choose the formula which satisfy the given statement ________________.($$A_1, A_2 , h$$ are area of top circle, area of bottom circle and height of frustum respectively).

A
$$\frac{1}{3} (A_1 + A_2 + \sqrt{A_1 A_2 }) h$$

B
$$\frac{1}{3} (A_1 - A_2 + \sqrt{A_1 A_2 }) h$$

C
$$\frac{1}{3} (A_1 + A_2 - \sqrt{A_1 A_2 }) h$$

D
$$\frac{2}{3} (A_1 + A_2 + \sqrt{A_1 A_2 }) h$$

Solution

Let radius of top & bottom be $${ r }_{ 1 }$$ and $${ r }_{ 2 }$$ & height be h.
Now, Area of top = $$\pi { r }_{ 1 }^{ 2 }={ A }_{ 1 }$$
Area of bottom = $$\pi { r }_{ 2 }^{ 2 }={ A }_{ 2 }$$
$$\therefore \quad $$ Volume of frustum = $$\dfrac { 1 }{ 3 } \left( { A }_{ 1 }+{ A }_{ 2 }+\sqrt { { A }_{ 1 }{ A }_{ 2 } } \right) \times h$$
Question 6
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Find the volume of the frustum cone whose base and top radius is 20 ft and 10 ft respectively. The height of the cone is 300 ft. (Use $$\pi$$ = 3).

A
200,000 ft$$^3$$

B
220,000 ft$$^3$$

C
210,000 ft$$^3$$

D
230,000 ft$$^3$$

Solution

Volume of the frustum cone is V = $$\displaystyle{\frac{\pi h}{3}}(R^2 + Rr + r^2)$$
$$R = 20 ft, r = 10 ft, h = 300 ft$$
$$V = (3 \times 300)/(3) [20^2 + 20 \times 10 + 10^2]$$
$$= 300[400 + 200 + 100]$$
$$= 300[700]$$
$$= 210,000 ft^3$$
Question 7
1 / -0

A cone of height $$7$$ cm. and base radius $$3$$ cm. is carved from a rectangular block of wood of dimensions $$10 cm. \times 5 cm. \times 2$$ cm. The percentage of wood wasted is

A
$$34\%$$

B
$$46\%$$

C
$$54\%$$

D
$$66\%$$

Solution

Volume of block $$=(10\times 5\times 2)cm^3$$
$$=100 cm^3$$
Volume of cone $$=\left (\frac {22}{7}\times 9\times 7\right )cm^3$$
$$=66 cm^3$$
$$\Rightarrow$$ volume of wood wasted $$=(100-66)cm^3=34 cm^3$$
$$\therefore$$ The percentage of wood wasted $$=34$$%
Question 8
1 / -0

Find the volume of the frustum cone whose base and top radius is 11 in and 6 in respectively. The height of the cone is 36 in. (Use $$\pi$$ = 3.14).

A
8,302.64 in$$^3$$

B
8,202.64 in$$^3$$

C
8,402.64 in$$^3$$

D
8,102.64 in$$^3$$

Solution

Volume of the frustum cone is $$V = \displaystyle{\frac{\pi h}{3}}(R^2 + Rr + r^2)$$
$$R = 11 in, r = 6 in, h = 36 in$$
$$V = (3.14 \times 36)/(3)[11^2 + 11 \times 6 + 6^2]$$
$$= 3.14 \times 12[121 + 66 + 36]$$
$$= 37.68[223]$$
$$= 8,402.64\, in^3$$
Question 9
1 / -0

The base and top diameter of a cone is 1.2 mm and 0.5 mm respectively. The height of the cone is 24 mm. What is the volume of frustum of a cone? (Use $$\pi$$ = 3).

A
154.14 mm$$^3$$

B
144.14 mm$$^3$$

C
134.14 mm$$^3$$

D
124.14 mm$$^3$$

Solution

Volume of the frustum cone is $$V = \displaystyle{\frac{\pi h}{3}}(R^2 + Rr + r^2)$$
D = 0.5 m, d = 0.1 m, h = 4 m
Diameter = radius/2
Radius = 2 $$\times$$ diameter
Therefore, R = 2.4 mm, r = 0.25 mm
$$V = (3 \times 24)/(3)[2.4^2 + 2.4 \times 0.25 + 0.25^2]$$
$$=$$ 24[5.76 + 0.6 + 0.0625]
$$=$$ 24[6.4225]
$$=$$ 154.14 mm$$^3$$
Question 10
1 / -0

The volume of a frustum cone is 1,600 $$mm^3$$, whose base and upper area of a circle is 16 and 100 $$mm^2$$. Find the height of the cone.

A
30.76 mm

B
11.25 mm

C
12.25 mm

D
13.25 mm

Solution

Volume of a frustum cone = $$\frac{1}{3} (A_1 + A_2 + \sqrt{A_1 A_2 }) h$$
1,600 = $$\frac{1}{3} (16 + 100 + \sqrt{16 \times 100 }) \times h$$
1,600 = $$\frac{1}{3} (116 + \sqrt{1600 }) \times h$$
1,600 = $$\frac{1}{3} (116 + 40) \times h$$
1,600 = $$\frac{1}{3} \times 156 \times h$$
h = $$(1,600 \times 3)/156 = 30.76 mm$$

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Surface Areas and Volumes Test - 14

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Surface Areas and Volumes Test - 14

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SHARING IS CARING

Question 1

10500 $$\displaystyle cm^{3}$$

10450 $$\displaystyle cm^{3}$$

10000 $$\displaystyle cm^{3}$$

None of the above

Solution

Question 2

$$2\;cm$$

$$3\;cm$$

$$4\;cm$$

$$6\;cm$$

Solution

Question 3

25.2 m

22.5 m

25.5 m

52.5 m

Solution

Question 4

$$63$$

$$126$$

$$21$$

$$130$$

Solution

Question 5

$$\frac{1}{3} (A_1 + A_2 + \sqrt{A_1 A_2 }) h$$

$$\frac{1}{3} (A_1 - A_2 + \sqrt{A_1 A_2 }) h$$

$$\frac{1}{3} (A_1 + A_2 - \sqrt{A_1 A_2 }) h$$

$$\frac{2}{3} (A_1 + A_2 + \sqrt{A_1 A_2 }) h$$

Solution

Question 6

200,000 ft$$^3$$

220,000 ft$$^3$$

210,000 ft$$^3$$

230,000 ft$$^3$$

Solution

Question 7

$$34\%$$

$$46\%$$

$$54\%$$

$$66\%$$

Solution

Question 8

8,302.64 in$$^3$$

8,202.64 in$$^3$$

8,402.64 in$$^3$$

8,102.64 in$$^3$$

Solution

Question 9

154.14 mm$$^3$$

144.14 mm$$^3$$

134.14 mm$$^3$$

124.14 mm$$^3$$

Solution

Question 10

30.76 mm

11.25 mm

12.25 mm

13.25 mm

Solution

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