Surface Areas and Volumes Test

Result

Surface Areas and Volumes Test - 45

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Weekly Quiz Competition

Question 1
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A container made up of metal sheet is in the form of a frustum cone of height $$20$$ cm with radii of its lower and upper ends $$4$$ cm and $$16$$ cm respectively. The amount of liquid the container can hold is (Use $$\pi = 3$$)

A
$$6720 cm^3$$

B
$$6700 cm^3$$

C
$$6680 cm^3$$

D
$$6660 cm^3$$

Solution

Volume of the frustum cone $$=\cfrac{1}{3}\pi h(R^2+r^2+Rr)$$
$$=\cfrac{1}{3}\pi \times 20(16^2+4^2+64)$$
$$=\cfrac{1}{3}\times 3 \times 20 \times 336$$
$$=6720 cm^3$$
Question 2
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The radii of the ends of a frustum of a cone are $$28\ cm$$ and $$7\ cm$$. The height of the cone is $$40\ cm$$, find its volume.

A
$$32340\ cm^{3}$$

B
$$43120\ cm^{3}$$

C
$$10780\ cm^{3}$$

D
$$2156\ cm^{3}$$

Solution

If $$R$$ and $$r$$ are the top and bottom radii of frustrum then,
$$\text{Volume of frustum} =\dfrac { 1 }{ 3 } \pi h\left[ { R }^{ 2 }+{ r }^{ 2 }+Rr \right] \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times 40\left[ { 28 }^{ 2 }+{ 7 }^{ 2 }+28\times 7 \right] $$

$$=\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times 40\times 1029\\=43120\ { cm }^{ 3 }$$
Question 3
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Calculate the volume of a frustum cone given above:

A
$$450$$

B
$$411$$

C
$$487$$

D
$$467$$

Solution

Volume of the frustum cone = $$\cfrac{1}{3}\pi h(R^2+r^2+Rr)$$
= $$\cfrac{1}{3}\pi \times 15(5^2+1^2+5)$$
= $$\cfrac{1}{3}\pi \times 465$$
= $$\cfrac{1}{3}\times 3.14 \times 465$$
= $$\approx 487$$
Question 4
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The height of the frustum cone is $$24$$ ft. The area of the top and bottom circle of the cone is $$6 ft^2$$ and $$10 ft^2$$ respectively. Find its volume.

A
$$369 ft^3$$

B
$$367 ft^3$$

C
$$368 ft^3$$

D
$$362 ft^3$$

Solution

Volume of the frustum cone is $$V = \cfrac{h}{3} [\pi R^2 + \pi r^2 + \sqrt{(\pi R^2 ) (\pi r^2 ))}]$$
Area of the top cone $$= 10 ft^2$$
Area of the bottom cone $$= 90 ft^2$$
Height $$= 24$$ ft
Volume = $$\cfrac{24}{3} [90 + 10 + \sqrt{10\times 90}]$$
= $$8[16 + \sqrt{900}]$$
= $$8 [46]$$
= $$368 ft^3$$
Question 5
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A bucket of height $$16\ cm$$ made up of metal sheet is in the form of frustum of a right circular cone with radii of its lower and upper ends as $$3\ cm$$ and $$15\ cm$$ respectively. What is the slant height of the bucket?

A
$$15\ cm$$

B
$$20\ cm$$

C
$$5\ cm$$

D
$$12\ cm$$

Solution

Slant height = $$\sqrt { { \left( R-r \right) }^{ 2 }+{ h }^{ 2 } } =\sqrt { { \left( 15-3 \right) }^{ 2 }+{ 16 }^{ 2 } } =\sqrt { 144+256 } $$
= $$\sqrt { 400 } =20cm$$
Question 6
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A wooden toy was made by scooping out a hemisphere of same radius from each end of a solid cylinder. If the height of the cylinder is $$10$$ cm, and its base is of radius $$3.5$$ cm, find the volume of wood in the toy. $$\displaystyle \left[ Use \,\,\, \pi = \frac{22}{7} \right]$$

A
$$425$$

B
$$290$$

C
$$474.83$$

D
$$205.25$$

Solution

Given height of cylinder $$=10$$ cm
Radius of cylinder $$=3.5$$ cm

The volume of toy$$=$$ Volume of cylinder $$-2\times$$Volume of a hemisphere
$$=\pi r^{2}h-\left(2\times \dfrac{2}{3}\pi r^{3}\right)$$
$$= \left[\pi (3.5)^{2}\times 10\right]-\left(2\times\dfrac23\times \pi (3.5)^{3}\right) = 205.251\text{ } cm^{3}$$
Question 7
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Adithi has a set of identical cubes. She makes solids by gluing together $$4$$ of these cubes. When cube faces are glued together, they must coincide. Each of the $$4$$ cubes must have a face that coincides with a face of at least one of the other $$3$$ cubes. One such solid is shown. Find the number of unique solids that Adithi can make using the $$4$$ cubes.

A
$$5$$

B
$$6$$

C
$$7$$

D
$$8$$
Question 8
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The solid brick shown is made of small bricks of side $$1$$. When the large brick is disassembled into its component small bricks, the total surface area of all the small bricks is how much greater than the surface area of the large brick?

A
$$32$$

B
$$40$$

C
$$60$$

D
$$72$$

Solution

A solid brick is made up of small block of side 1.
small block of side 1.
The large brick is made up of $$12$$
small cubes.
So, length of the large brick $$= 3$$
width of the large brick $$=2$$
height of the large brick $$= 2$$
$$\therefore$$ total surface area of large brick
= total surface area of cuboid
$$=2(2(3\times 2+2\times 2+3\times 2)$$
$$=2(6+4+6)$$
$$=2(16)$$
$$=32$$
$$\Rightarrow$$ surface area of large brick $$32$$
Total surface area all $$12$$ smaller
bricks $$=10(6\times 1)$$
$$=72$$
$$\therefore $$ Total surface are of all $$12$$
smaller bricks $$-$$ Total surface area of large brick
$$\Rightarrow72-32=40$$
$$\Rightarrow$$ Total surface area of all the small bricks is $$40$$ greater than the surface area of a large brick.
option B $$40$$ is correct answer.
Question 9
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A bucket which is in the form of a frustum of a cone, has radii $$3$$ and $$5$$ units and vertical height $$6$$ units. How much water can the bucket hold?

A
$$33\pi$$ cubic units

B
$$45\pi$$ cubic units

C
$$48\pi$$ cubic units

D
NONE OF THE ABOVE

Solution
Question 10
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A sphere and a cube have same surface area. What is the ratio of the square of volume of the sphere to the square of volume of the cube?

A
$$\pi:6$$

B
$$1:1$$

C
$$6:\pi$$

D
$$3:\pi$$

Solution