Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 20

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Weekly Quiz Competition

Question 1
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A solid sphere of radius $$x$$ cm is melted and recast into a shape of a solid cone of radius $$x$$ cm. The height of the cone is:

A
$$3x$$ cm

B
$$x$$ cm

C
$$4x$$ cm

D
$$2x$$ cm

Solution

Let the height of the cone be $$h$$ $$cm$$
Volume remains constant after the recast.
Therefore, volume of cone $$=$$ volume of sphere so,
$$\cfrac { 1 }{ 3 } \pi { x }^{ 2 }h=\cfrac { 4 }{ 3 } \pi { x }^{ 3 }\\ { x }^{ 2 }h=4{ x }^{ 3 }\\ h=4x$$
Question 2
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A sphere is melted and half of the molten liquid is used to form $$11$$ identical cubes, whereas the remaining half is used to form $$7$$ identical smaller spheres. The ratio of the side of the cube to the radius of the new small sphere is -

A
$$\left (\dfrac {4}{3}\right)^{\frac {1}{3}}$$

B
$$\left (\dfrac {8}{3}\right)^{\frac {1}{3}}$$

C
$$(3)^{\frac {1}{3}}$$

D
$$2$$

Solution

Volume of $$11$$ small cubes $$=$$ Volume of $$7$$ small spheres
$$11 \times a^3=7 \times \dfrac 43 \times (\pi) \times r^3$$

$$(\dfrac ar)^3=7\times \dfrac 43 \times \dfrac {22}{7} \times \dfrac {1}{11}$$

$$(\dfrac ar)^3=\dfrac 83$$

$$\dfrac ar=\left(\dfrac 83\right)^{\frac 13}$$
Question 3
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A metallic cube of edge $$1\ cm$$ is drawn into a wire of diameter $$4\ mm$$, then the length of the wire is:

A
$$\dfrac{10}{\pi }\ cm$$

B
$$10\pi\ cm$$

C
$$\dfrac{25}{\pi }\ cm$$

D
$$1000\ cm$$

Solution

Edge of cube $$=1$$ cm
Volume of cube $${ (1) }^{ 3 }=1 cm^{ 3 }=1000 mm^{ 3 }$$
Diameter of wire $$=4mm$$
Radius of wire $$=2mm $$
Now, volume remains same
Therefore, volume of wire $$=$$ volume of cube
$$\pi { r }^{ 2 }h=1000 mm^{ 3 }\\ \pi \times 2\times 2\times h=1000\\ h=\cfrac { 1000 }{ 4\pi } \\ =\cfrac { 250 }{ \pi } mm\\ h=\cfrac { 25 }{ \pi } cm$$
Question 4
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The edges of three cubes of metal are $$3$$ cm, $$4$$ cm and $$5$$ cm. They are melted and formed in to a single cube. Find the edge of the new cube.

A
$$6$$ cm

B
$$5$$ cm

C
$$4$$ cm

D
$$8$$ cm

Solution

Volume of cubes $$=3^3+4^3+5^3=27+64+125=216$$

Volume of new cube $$=$$ edge$$^3=216 =6^3$$

$$\therefore $$ edge $$=6$$ cm
Question 5
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A cylindrical hut $$4$$ m tall and $$7$$ m in diameter is surmounted by aright circular cone of semi-vertical angle $$45$$ $$^{\circ}$$. The volume of air inside the hut is -

A
$$44.9$$ m$$^3$$

B
$$154$$ m$$^3$$

C
$$198.9$$ m$$^3$$

D
None of these

Solution

Total volume $$= \pi r^2 h + \displaystyle \frac{1}{3} \pi r^2 h'$$
$$= \pi (3.5)^2 \times 4 + \dfrac{1}{3} \pi (3.5)^2 (3.5)^2$$
$$=\pi (3.5)^2 \left [ 4 + \dfrac{3.5}{3} \right ]$$
$$=198.9 \ \text{m}^3$$
Question 6
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A drinking glass is in the shape of a frustum of a cone of height $$14\ cm$$. The diameters of its two circular ends are $$4\ cm$$ and $$2\ cm$$. Find the capacity of the glass.

A
$$102.67\ sq. cm$$

B
$$92.67\ sq. cm$$

C
$$115.21\ sq. cm$$

D
$$80.21\ sq. cm$$

Solution

Since glass is in the form of frustum, therefore,

$$h =14 cm, r_1 = \dfrac{4}{2} = 2cm$$ and $$ r_2 = \dfrac{2}{2} = 1 cm.$$

$$\therefore$$ Capacity of frustum $$=$$ Volume of a frustum

$$\displaystyle =\frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2) = \frac{1}{3} \times \frac{22}{7} \times 14 (2^2 + 1^2 + 2 \times 1)$$

$$\displaystyle = \frac{44}{3} (4+1+2) = \frac{44 \times 7}{3} = \frac{308}{3} = 102.67$$ sq. cm.
Question 7
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STATEMENT - 1: The slant height of frustum of a cone is 4 cm. and perimeters of its circular ends are 18 cm. and 6 cm. then the curved surface area is 48 sq. cm.
STATEMENT - 2: Curved surface area of frustum $$ = \pi (r_1+r_2) $$, where $$r_1$$ and $$r_2$$ are radii of the frustum and $$l$$ is a slant height.

A
Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1

B
Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1

C
Statement - 1 is True, Statement - 2 is False

D
Statement - 1 is False, Statement - 2 is True

Solution

Curved surface area offrustum $$= \pi (r_1 + r_2 ) l, $$ where $$r_1$$

and $$r_2$$ are radii of frustum and $$l$$ is a slant height
$$=\displaystyle \frac{22}{7} \left ( \frac{9}{\pi} + \frac{3}{\pi} \right ) \times 4 = \frac{22}{ 7 \times \pi}(9+3) \times 4$$
$$= \displaystyle \frac{22}{7} \times \frac{7}{22} (12) \times 4 = 48 cm^2$$
Question 8
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The given figure shows a solid formed of a solid cube of side 40 cm and a solid cylinder of radius 20 cm and height 50 cm attached to the cubes as shown. Find the volume and the total surface area of the whole solid [Take $$\pi \, =\, 3.14$$]

A
$$122700\, cm^3$$ and $$15880\ cm^2$$

B
$$126800\, cm^3$$ and $$15880\ cm^2$$

C
$$148900\, cm^3$$ and $$15880\ cm^2$$

D
$$148800\, cm^3$$ and $$15880\ cm^2$$

Solution

Given: Side of the cube $$=40\ cm$$ and radius of the cylinder $$=20\ cm$$ and height of the cylinder $$=50\ cm$$

Volume of cube $$=(side)^{3}$$

$$=(40)^{3}$$

$$=64000\ cm^{3}$$

Volume of cylinder $$=\pi r^{2}h$$

$$=3.14\times (20)^{2}\times 50$$

$$=62800\ cm^{3}$$

So, total volume of solid $$=64000+62800=126800\ cm^{3}$$

Now, surface area of cube $$=6(side)^{2}$$

$$=6(40)^{2}$$

$$=9600\ cm^{2}$$

The surface area of cylinder $$=2\pi rh$$

$$=3.14\times 20\times 50$$

$$=6280\ cm^{2}$$

$$\therefore$$ Total surface area of solid $$=9600+6280=15880\ cm^{2}$$
Question 9
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The given figure shows a solid formed of a solid cube of side $$40 cm$$ and a solid cylinder of radius $$20 cm$$ and height $$50 cm$$ attached to the cube as shown. Find the volume of the whole solid. $$\displaystyle \left [ Take\ \pi=3.14 \right ]$$.

A
$$126800$$

B
$$117800$$

C
$$124400$$

D
None of the above

Solution

Solid cube of side $$=40 cm$$
Solid cylinder of radius $$=20 cm$$ and height $$=50 cm $$
Volume of the whole solid $$=$$ Volume of cube $$+$$ Volume of cylinder
$$V={ a }^{ 3 }+\pi { r }^{ 2 }h$$
$$V={ (40) }^{ 3 }+3.14 \times { (20) }^{ 2 } \times 50$$
$$V=126800{ cm }^{ 3 }$$
Question 10
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A sphere of diameter 12.6 cm is melted and cast into a right circular cone of height 25.2 cm. The diameter base of the cone is

A
$$15.75m$$

B
$$12.6 cm$$

C
$$39.69 cm$$

D
$$69.39 cm$$

Solution

Given:
$$\Rightarrow$$ diameter of sphere$$(d)=12.6\ cm$$

$$\Rightarrow$$ radius $$(r_1)= \dfrac{d}{2} = 6.3\ cm$$

$$\Rightarrow$$ height of the cone$$ \ = 25.2\ cm$$
Let radius of the cone $$= r$$

Volume of sphere $$(V_s)$$ $$=$$ volume of cone $$(V_c)$$

$$\Rightarrow$$ $$\dfrac {4}{3}\pi (r_1)^3=\dfrac {1}{3}\pi (r^2)\times h$$

$$\Rightarrow$$ $$\dfrac {4}{3}\pi (6.3)^3=\dfrac {1}{3}\pi (r^2)\times 25.2$$

$$\Rightarrow$$ $$r^2=\dfrac {4\times 6.3\times (6.3)^2}{25.2}$$

$$\Rightarrow$$ $$r=6.3$$
Diameter $$ \ D=2r=12.6 \ cm$$

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Re-Attempt Weekly Quiz Competition

Surface Areas and Volumes Test - 20

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Surface Areas and Volumes Test - 20

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SHARING IS CARING

Question 1

$$3x$$ cm

$$x$$ cm

$$4x$$ cm

$$2x$$ cm

Solution

Question 2

$$\left (\dfrac {4}{3}\right)^{\frac {1}{3}}$$

$$\left (\dfrac {8}{3}\right)^{\frac {1}{3}}$$

$$(3)^{\frac {1}{3}}$$

$$2$$

Solution

Question 3

$$\dfrac{10}{\pi }\ cm$$

$$10\pi\ cm$$

$$\dfrac{25}{\pi }\ cm$$

$$1000\ cm$$

Solution

Question 4

$$6$$ cm

$$5$$ cm

$$4$$ cm

$$8$$ cm

Solution

Question 5

$$44.9$$ m$$^3$$

$$154$$ m$$^3$$

$$198.9$$ m$$^3$$

None of these

Solution

Question 6

$$102.67\ sq. cm$$

$$92.67\ sq. cm$$

$$115.21\ sq. cm$$

$$80.21\ sq. cm$$

Solution

Question 7

Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1

Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1

Statement - 1 is True, Statement - 2 is False

Statement - 1 is False, Statement - 2 is True

Solution

Question 8

$$122700\, cm^3$$ and $$15880\ cm^2$$

$$126800\, cm^3$$ and $$15880\ cm^2$$

$$148900\, cm^3$$ and $$15880\ cm^2$$

$$148800\, cm^3$$ and $$15880\ cm^2$$

Solution

Question 9

$$126800$$

$$117800$$

$$124400$$

None of the above

Solution

Question 10

$$15.75m$$

$$12.6 cm$$

$$39.69 cm$$

$$69.39 cm$$

Solution

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