Surface Areas and Volumes Test

Result

Surface Areas and Volumes Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

A solid metal ball of radius 8 cm is melted and cast into smaller balls, each of radius 2 cm. The number of such balls is:

A
8

B
16

C
32

D
64

Solution

Let we cast big ball into $$n$$ number of small balls.

Therefore,
Volume of big ball =n$$\times$$ Volume of small balls

$$\implies \dfrac{4}{3}\pi (8)^3=n(\dfrac{4}{3}\pi (2)^3)$$

$$\implies n=\dfrac{512}{8}=64$$

Hence the number of such balls is $$64$$.
Question 2
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How many coins that are $$2$$mm thick and $$1.5$$cm in diameter should be melted in order to form a right circular cylinder having a base diameter $$6$$cm and height $$8$$cm?

A
$$640$$

B
$$540$$

C
$$740$$

D
$$840$$

Solution
Question 3
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Two cubes each of side $$12 cm,$$ are joined end to end .the surface area of the resulting cuboid is :

A
$${ 1240\quad cm }^{ 2 }$$

B
$${ 1440\quad cm }^{ 2 }$$

C
$${ 2250\quad cm }^{ 2 }$$

D
$${ 4252\quad cm }^{ 2 }$$

Solution

$$ length\,of\,the\,resulting\,cuboid $$
$$ 12+12 $$
$$ =24cm $$
$$ Breadth\,and\,height\,\,remaining\,the\,same $$
$$ surface\,area=2\left( lb+bh+lh \right) $$
$$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\left( 24\times 12+12\times 12+24\times 12 \right) $$
$$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=1440c{{m}^{2}} $$
Question 4
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The height of a cone and the radius of its base are respectively 9 and 3 cm. The cone is cut by a plane parallel to its base soas to divide it into two parts. The volume of frustum of cone is $$44{ cm }^{ 3 }$$, then the radius of upper circular of frustum is $$\left\{ Use\ \pi =\dfrac { 22 }{ 7 } \right\} $$.

A
$$\sqrt [ 3 ]{ 12 } cm$$

B
$$\sqrt [ 3 ]{ 13 } cm$$

C
$$\sqrt [ 3 ]{ 6 } cm$$

D
$$\sqrt [ 3 ]{ 20 } cm$$

Solution
Question 5
1 / -0

The number of iron of length $$21\ m$$ and diameter $$4\ cm$$ each that can be made by recasting $$2.64$$ cubic meters of iron is :

A
$$100$$

B
$$200$$

C
$$300$$

D
$$400$$

Solution
Question 6
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The diameter of a metallic sphere is $$6 cm$$. It was melted to make a wire of diameter $$4 mm$$. Find the length of the wire.

A
$$90mm$$

B
$$90cm$$

C
$$9cm$$

D
$$9m$$

Solution

Volume of metallic sphere = $$\dfrac{4}{3} \times \pi \times {r^3}$$
= $$\dfrac{4}{3} \times \pi \times 6^3$$
Volume of cylindrical wire = $$ \pi \times r^{2} \times h$$

Now,
Volume of metallic sphere = Volume of cylindrical wire
$$\therefore \dfrac{4}{3} \times \pi \times 6^3$$ = $$\pi \times 0.02^2 \times h$$
$$\therefore h = 900 cm=9 m$$
Question 7
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Two cubes have their volumes in the ratio $$1:27$$. The ratio of their surface areas is

A
$$1:3$$

B
$$1:9$$

C
$$1:27$$

D
$$None\ of\ these$$

Solution

Volumes are in the ratio $$1:27$$
Let the sides of the cubes are $$a,b$$,
then
$$\dfrac{1}{27}=\dfrac{1}{3}=\dfrac{a^3}{b^3}$$

$$\dfrac{b}{a}=3$$

Since we know that the surface area of a cube is $$6a^2$$
$$i.e$$ the surface ares will be proportional to the square of sides.

Therefore on squaring $$\dfrac{a^2}{b^2}=\dfrac{1^2}{3^2}$$
$$=\dfrac{1}{9}$$
$$=1:9$$
Question 8
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Three cubes of metal whose edges are $$6\ cm,\ 8\ cm$$, and $$10\ cm$$ respectively are melted from a single cube. The edge of the new cube is

A
$$12\ cm$$

B
$$24\ cm$$

C
$$18\ cm$$

D
$$20\ cm$$

Solution

Let the edge length of the new cube formed is equal to $$a.$$

As the three cubes are melted to form a single cube,
Volume of the new cube $$=$$ Sum of the volume of the three cubes
$$a^3 =6^{3}+ 8^{3}+ 10^{3} $$
$$= 216+512+1000 $$
$$= 1728 $$
$$a=12\ cm $$

Hence, option $$A$$ is correct.
Question 9
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The diameters of the two circular ends of the bucket are $$44 cm$$ and $$24 cm$$. The height of bucket is $$35 cm$$. The capacity of bucket is

A
$$32.7L$$

B
$$33.7L$$

C
$$34.7L$$

D
$$37.7L$$

Solution

Bucket is in the shape of a frustum of a cone,
In below figure, $$r_{1}=\dfrac{24}{2}=12cm$$ and $$r_{2}=\dfrac{44}{2}=22cm,h=35cm$$
The volume of the bucket is given by
$$V=\dfrac{1}{3}\pi h(r_{1}^{2}+r_{2}^{2}+r_{1}r_{2})$$
$$=\dfrac{1}{3}\times\dfrac{22}{7}\times35[12^{2}+22^{2}+12\times22]$$
$$=\dfrac{1\times22\times35}{3\times7}[144+484+264]$$
$$=\dfrac{22\times35\times892}{3\times7}=\dfrac{22\times5\times892}{3}=\dfrac{110\times892}{3}cm^{3}$$
$$=\dfrac{110\times892}{3\times1000}=\dfrac{9812}{300}=32.706L$$
It is close to option A.
Question 10
1 / -0

A surahi is the combination of

A
a sphere and a cylinder

B
a hemisphere and a cylinder

C
two hemispheres

D
a cylinder and a cone

Solution

A surahi is the combination of a sphere and a cylinder(as shown in the figure). It is used to store water or for decoration purpose.