Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

A material handing bucket is in the shape of the frustum of a right circular cone. The small radius is $$5$$ cm and the large radius is $$10$$ cm. The height of the frustum cone is $$20$$ cm. Find the volume and the total surface area of the bucket.

A
$$\text {Volume}$$ = $$6,159.44 cm^3$$ and TSA = $$2000 cm^2$$

B
$$\text {Volume}$$ = $$3,663.33 cm^3$$ and TSA = $$1,363.2 cm^2$$

C
$$\text {Volume}$$ = $$3,562.34 cm^3$$ and TSA = $$1,262.1 cm^2$$

D
$$\text {Volume} $$ = $$3,461.51 cm^3$$ and TSA = $$1,56.2 cm^2$$

Solution

Volume of the frustum cone = $$\displaystyle \frac{\pi h}{3} [R^2 + Rr + r^2] $$
$$= \displaystyle \frac{\pi \times 20}{3} [{10}^2 + 10 \times 5 + 5^2] $$
$$=\displaystyle \frac{3.14\times 20}{3}[100+50+25]$$
= $$3,663.33 \,cm^3$$

Total surface area $$=\displaystyle \pi (R+r)l+\pi R^2+\pi r^2$$
Slant height, $$l = \sqrt{h^2 +r^2} \Rightarrow \sqrt{20^2 + 5^2} \Rightarrow \sqrt{400 +25}$$
$$l = 20.61 cm$$
$$ \therefore TSA = 3.14 (10 + 5) 20.61 + 3.14 \times 10^2 + 3.14 \times 5^2$$
$$ = 3.14 [309.15 + 100 + 25]$$
$$= 1,313.231 \,cm^2$$
$$ \therefore Volume = 3,663.33 cm^3$$
$$ TSA = 1,363.231 cm^3$$
Question 2
1 / -0

Find the curved surface area of a frustum cone whose larger and smaller radius is 4.5 and 2.4 ft. The slant height is 12.5 ft. (Use $$\pi$$ = 3)

A
258.75 ft$$^2$$

B
248.75 ft$$^2$$

C
238.75 ft$$^2$$

D
228.75 ft$$^2$$

Solution

Curved surface area of a frustum cone = $$\pi$$(r + R)s $$\Rightarrow$$ ?
= 3 $$\times$$ (2.4 + 4.5) $$\times$$ 12.5
= 3 $$\times$$ (6.9) $$\times$$ 12.5
Curved surface area of a frustum cone = 258.75 ft$$^2$$
Question 3
1 / -0

The total surface area of a frustum cone is 3,250 cm$$^2$$. The area of larger and smaller circle is 48 and 36 cm$$^2$$. Find the curved surface area of a cone.

A
3,166 cm$$^2$$

B
3,266 cm$$^2$$

C
3,366 cm$$^2$$

D
3,466 cm$$^2$$

Solution

Curved surface area of a frustum cone = $$\pi$$(r + R)s $$\Rightarrow$$  ?
Total surface area of a frustum of cone = $$\pi$$(r + R)s + $$\pi$$r$$^2$$ + $$\pi$$R$$^2$$ $$\Rightarrow$$  3,250 cm$$^2$$
Smaller circle area = $$\pi$$r$$^2$$ $$\Rightarrow$$  36 cm$$^2$$
Larger circle area = $$\pi$$R$$^2$$ $$\Rightarrow$$  48 cm$$^2$$
Therefore,
SA = $$\pi$$(r + R)s + $$\pi$$r$$^2$$ + $$\pi$$R$$^2$$
3,250 = Curved surface area + 36 + 48
$$3,250 - 36 - 48 = $$Curved surface area
Curved surface area of a frustum cone = 3,166 cm$$^2$$
Question 4
1 / -0

The shape of the wooden block is of the form of frustum of a cone. Determine the volume of a frustum of a cone if the diameters of the ends are 15 cm and 3 cm and its perpendicular height is 18 cm.

A
569.6 $$\pi cm^3$$

B
418.5 $$\pi cm^3$$

C
769.6 $$\pi cm^3$$

D
869.6 $$\pi cm^3$$

Solution

Volume of a frustum cone = $$ \displaystyle \frac{\pi h}{3} (R^2 + Rr + r^2) $$
Diameter = radius $$\times 2$$
Radius, R = 7.5 cm, r = 1.5 cm
$$=\displaystyle \frac{\pi\times 18}{3}(7.5^2+7.5\times 1.5+1.5^2)$$
$$= 6 \pi (56.25 + 11.25 + 2.25 )$$
$$=6 \pi \times 69.75$$
$$=418.5 \pi cm^3$$
Question 5
1 / -0

A bouquet is in the shape of a frustum of a cone of height 15 cm. The radii of its two circular ends are 4 cm and 2 cm. Find the volume of the bouquet.

A
339.6 $$cm^3$$

B
439.6 $$cm^3$$

C
239.6 $$cm^3$$

D
139.6 $$cm^3$$

Solution

Volume of a frustum cone = $$ \displaystyle \frac{\pi h}{3} (R^2 + Rr + r^2) $$
$$=\displaystyle \frac{\pi \times 15}{3} (4^2 + 4 \times 2 + 2^2)$$
$$= 5 \pi (16 + 8 + 4 )$$
$$= 5 \pi \times 28$$
$$= 140 \times 3.14$$
= $$439.6 cm^3$$
Question 6
1 / -0

A loudspeaker diaphragm is in the form of a frustum of a cone. The slant height is $$30\ cm$$. The curved surface area is $$3600\ cm^2$$. The base diameter is $$5\ cm$$. Find the top diameter of the loud speaker.

A
$$46.42\ cm$$

B
$$35.71\ cm$$

C
$$36.42\ cm$$

D
$$26.42\ cm$$

Solution

Curved surface area $$= \pi (R + r) l$$
Diameter $$=r\times 2$$
Radius, $$r = 2.5\ cm, l = 30 cm. CSA = 3600\ cm^2$$
$$3600 = \pi (R + 2.5) \times 30$$
$$\dfrac{3600}{30} = 3.14R + 3.14 \times 2.5$$
$$\dfrac{112.15}{3.14} = R$$
$$R = 35.71\ cm$$
Bottom diameter$$, D = 56.42\ cm$$
Question 7
1 / -0

The shape of the wooden block is of the form of frustum of a cone. Determine the height of a frustum of a cone if the radii of the ends are 23 m and 4 m and its volume is 2,400 $$m^3$$.

A
0.5 m

B
1.5 m

C
2.5 m

D
3.5 m

Solution

Volume of a frustum cone = $$ \displaystyle \frac{\pi h}{3} (R^2 + Rr + r^2) $$
R = 23 m, r = 4 cm, Volume = 2,400 $$m^3$$
$$2,400 =\displaystyle \frac{\pi \times h}{3} (23^2 + 23 \times 4 + 4^2)$$
$$7200= \pi h (529 + 92 + 16 )$$
$$7200/637 \pi = h$$
h = $$3.5 m$$
Question 8
1 / -0

A metallic solid of volume 5 $$m^3$$ is melted and drawn into the form of a wire of radius 12 m. Find the length of the wire.

A
0.1 m

B
0.011 m

C
0.05 m

D
0.045 m

Solution

Let the length of the wire be $$x$$ m
$$\therefore$$ The wire is of the shape of the right circular cylinder.
$$\therefore$$ Volume of the cylinder $$= \pi r^2 h$$
$$5 = \pi \times 12 \times 12 \times x$$
$$\displaystyle 5 = \frac{22}{7} \times 12 \times 12 \times x$$
$$\displaystyle \frac{5 \times 7}{22 \times 12 \times 12} = x$$
The length of the wire is $$0.011$$ m
Question 9
1 / -0

A tent of height 16 m is in the form of right circular cylinder with radius of base 2 m and height 6 m, surmounted by a right circular cone of the same base. Find the cost of the canvas of the tent at the rate of Rs. 100 per m.

A
Rs. 12,947

B
Rs. 13,947

C
Rs. 14,947

D
Rs. 15,947

Solution

Radius of the cylindrical base = 2 m
Height of the cylindrical portion = 6 m
Height of the conical portion = 16 - 6 = 10 m
Slant height = $$\sqrt{h^2
+ r^2}$$
= $$\sqrt{10^2 + 2^2}$$
= $$\sqrt{104}$$
= 10.19 m
Surface area = Curved surface area of the cylindrical portion + Curved surface area of the conical portion
= $$2 \pi rh + \pi r l$$
= $$2 \times 22/7 \times 2 \times 6 + 22/7 \times 2 \times 10.19$$
=75.42 + 64.05
= 139.47 $$m^2$$
Therefore the cost of the canvas of the tent at the rate of $$Rs. 100 \times 139.47 = Rs. 13,947$$
Question 10
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A hemispherical water bottle of internal diameter 48 cm contains some liquid. This liquid is to be filled into cylindrical shaped bottles each of radius 4 cm and h = 12 cm. Find the number of bottles necessary to empty the water bottle.

A
24

B
34

C
20

D
14

Solution

Volume of the $$x$$ bottle = Volume of the hemisphere
$$\pi r^2 h x = 1/2 \times 4/3 \times \pi R^3$$

Inner radius R of the bottle $$= 48/2 = 24$$ cm

$$ \displaystyle x = \frac{2/3 \times 24 \times 24 \times 24}{4 \times 4 \times 12} = \frac{13824}{576} $$

$$x = 24$$

$$\therefore$$ Number of bottles $$= 24$$

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Surface Areas and Volumes Test - 41

Result

Surface Areas and Volumes Test - 41

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SHARING IS CARING

Question 1

$$\text {Volume}$$ = $$6,159.44 cm^3$$ and TSA = $$2000 cm^2$$

$$\text {Volume}$$ = $$3,663.33 cm^3$$ and TSA = $$1,363.2 cm^2$$

$$\text {Volume}$$ = $$3,562.34 cm^3$$ and TSA = $$1,262.1 cm^2$$

$$\text {Volume} $$ = $$3,461.51 cm^3$$ and TSA = $$1,56.2 cm^2$$

Solution

Question 2

258.75 ft$$^2$$

248.75 ft$$^2$$

238.75 ft$$^2$$

228.75 ft$$^2$$

Solution

Question 3

3,166 cm$$^2$$

3,266 cm$$^2$$

3,366 cm$$^2$$

3,466 cm$$^2$$

Solution

Question 4

569.6 $$\pi cm^3$$

418.5 $$\pi cm^3$$

769.6 $$\pi cm^3$$

869.6 $$\pi cm^3$$

Solution

Question 5

339.6 $$cm^3$$

439.6 $$cm^3$$

239.6 $$cm^3$$

139.6 $$cm^3$$

Solution

Question 6

$$46.42\ cm$$

$$35.71\ cm$$

$$36.42\ cm$$

$$26.42\ cm$$

Solution

Question 7

0.5 m

1.5 m

2.5 m

3.5 m

Solution

Question 8

0.1 m

0.011 m

0.05 m

0.045 m

Solution

Question 9

Rs. 12,947

Rs. 13,947

Rs. 14,947

Rs. 15,947

Solution

Question 10

24

34

20

14

Solution

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