Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 23

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Weekly Quiz Competition

Question 1
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A solid is in the form of a right circular cone mounted on a solid hemisphere with the same radius is made from a piece of metal. The radius of the hemisphere is $$\displaystyle \frac{1}{3}$$ of the vertical height of the conical part. If the radius of the base of the cone is $$r$$, the volume of the piece of metal is

A
$$\displaystyle \frac{11}{3}\pi\, r^{3}$$

B
$$\displaystyle \frac{8}{3}\pi\, r^{3}$$

C
$$\displaystyle \frac{7}{3}\pi\, r^{3}$$

D
$$\displaystyle \frac{5}{3}\pi\, r^{3}$$

Solution

Volume of the solid $$ = $$ Volume of the hemispherical part $$ + $$ Volume of conical part

Volume of a hemisphere of radius $$'r'$$ $$ = \dfrac { 2 }{ 3 } \pi { r }^{ 3 } $$

Volume of a cone $$ = \dfrac { 1 }{ 3 } \pi { r }^{ 2 }h $$ where $$r$$ is the radius of the base of the cone and $$h$$ is the height.

Given, $$r$$$$ = \dfrac {1}{3} h \Rightarrow h = 3r $$

Hence,
Volume of the solid $$ =\left(\dfrac { 2 }{ 3 } \pi { r }^{ 3 }\right)+ \left(\dfrac { 1 }{ 3 } \times \pi \times { r }^{ 2 }\times 3r\right )$$
$$=\dfrac { 2 }{ 3 } \pi { r }^{ 3 }+\pi r^3$$
$$= \dfrac {5}{3} \pi {r}^{3} $$
Question 2
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A vessel is in the form of a hemispherical bowl, surmounted by a hollow cylinder. The diameter of the hemisphere is $$12\ \text{cm}$$ and the total height of the vessel is $$16\ \text{cm}$$. Find the capacity of the vessel. (Take $$\pi$$ = 22/7). Also find the internal surface area of the vessel.

A
Capacity = $$1584\, \text{cm}^{3};\,$$ Internal Surface Area = $$ 602.88\, \text{cm}^{2}$$

B
Capacity = $$1684\, \text{cm}^{3};\,$$ Internal Surface Area = $$ 612.88\, \text{cm}^{2}$$

C
Capacity = $$1784\, \text{cm}^{3};\,$$ Internal Surface Area = $$ 622.88\, \text{cm}^{2}$$

D
Capacity = $$1884\, \text{cm}^{3};\,$$ Internal Surface Area = $$ 632.88\, \text{cm}^{2}$$

Solution

The diameter of the hemispherical bowl is $$12\ \text{cm}$$
The radius of the hemisphere r=$$\frac{12}{2}=6\ \text{cm}$$
The curved surface area of the bowl=$$2\pi r^2$$
= $$2\times 3.14\times 6\times 6\ \text{cm}$$
= $$226.08\ \text{cm}^2$$
The total height of the vessel is =$$16\ \text{cm}$$
The height of the cylindrical part=$$(16-6)\ \text{cm}$$
=$$10\ \text{cm}$$
The surface area of the cylinder =$$2\pi r h$$
=$$2\times 3.14\times 6\times10\ \text{cm}^2$$
=$$376.8\ \text{cm}^2$$
The surface area of the vessel = Surface area of the cylinder + Surface area of the sphere
=$$(377.14+226.28)\ \text{cm}^2$$
=$$602.88\ \text{cm}^2$$
Question 3
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A solid toy is in the form of a right circular with hemispherical ends. The total height of the solid is 19 cm and diameter of the cylinder is 7 cm. Find the volume and total surface area of the solid. (Take $$\pi = \dfrac{22}{7}$$)

A
Volume = $$681.66\, cm^{3};\,$$ Total Surface Area = $$ 478\, cm^{2}$$

B
Volume = $$681.66\, cm^{3};\,$$ Total Surface Area = $$ 418\, cm^{2}$$

C
Volume = $$641.66\, cm^{3};\,$$ Total Surface Area = $$ 478\, cm^{2}$$

D
Volume = $$641.66\, cm^{3};\,$$ Total Surface Area = $$ 418\, cm^{2}$$

Solution

height of the solid = 19 cm,
and diameter of the cylinder = 7 cm, radius=3.5 cm,
Height of the cylinder =19-7=12 cm,
the volume of the solid=volume of cylinder+volume of 2 hemisphere
$$V=\Pi { r }^{ 2 }h+2*\Pi\frac { 2 }{ 3 } { r }^{ 3 }$$
$$V=\frac { 22 }{ 7 } { \times 3.5 }^{ 2 }\times 12+2*\frac { 22 }{ 7 } \times \frac { 2 }{ 3 } { 3.5 }^{ 3 }$$
$$V=\frac { 22 }{ 7 } { \times 3.5 }^{ 2 }\times 12+2*\frac { 22 }{ 7 } \times \frac { 2 }{ 3 } { 3.5 }^{ 3 }$$
$$V=462+179.67$$
$$V=641.67cm^3$$
total surface area of the solid=
$${ S }_{ a }=2\times 2\Pi { r }^{ 2 }+2\Pi rh$$
$${ S }_{ a }=2\times 2\times \frac { 22 }{ 7 } { 3.5 }^{ 2 }+2\frac { 22 }{ 7 } \times 3.5\times 12$$
$${ S }_{ a }=154+264=418$$
Answer $$641.66\, cm^{3};\, 418\, cm^{2}$$
Question 4
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The diameter of a metallic sphere is $$6$$ cm. The sphere is melted and drawn into a wire of uniform circular cross-section. If the length of the wire is $$36$$ m, find the radius of its cross-section.

A
$$0.8$$ cm

B
$$0.5$$ cm

C
$$0.3$$ cm

D
$$0.1$$ cm

Solution

The wire is in the shape of a cylinder.
Since the sphere is melted and a cylindrical wire is formed, their volumes are equal.
Volume of a sphere $$ = \cfrac { 4 }{ 3 } \pi { r }^{ 3 } $$
As the diameter of the sphere is $$6$$ cm, its radius $$r = 3$$ cm
Volume of a Cylinder $$ = \pi { R }^{2 }h $$
Length of the wire $$ h = 36 m = 3600 cm $$
Hence, Volume of sphere $$ = $$ Volume of the wire
$$ \cfrac { 4 }{ 3 } \pi { r }^{ 3 }=\pi { R }^{ 2 }h $$
$$ {R}^{2} = \cfrac {1}{100} $$
$$ R = \cfrac {1}{10} = 0.1 cm $$
Hence, radius of the cross-section of the wire $$ = 0.1 cm $$
Question 5
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The slant height of the frustrum of a cone is $$4$$ cm. If the perimeters of its circular bases be $$18$$ cm and $$6$$ cm, find the curved surface area of the frustum and also find the cost of painting its total surface at the rate of Rs. $$12.50$$ per $$100 cm^{2}$$.

A
Rs. $$9.58$$

B
Rs. $$10.93$$

C
Rs. $$12.72$$

D
Rs. $$25$$
Question 6
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The radii of the circular ends of the frustum of a right circular cone are $$14$$ cm and $$12$$ cm and its thickness is $$9$$ cm. Find the lateral surface of the frustum. (Take $$\pi$$ $$= 3.14$$)

A
$$783.6\, \text{cm}^{2}$$

B
$$763.6\, \text{cm}^{2}$$

C
$$753.6\, \text{cm}^{2}$$

D
$$733.6\, \text{cm}^{2}$$

Solution

Curved Surface area of a frustum of a cone $$=\pi l({r}_{1}+{ r }_{ 2}) $$, where $$l$$ is the slant height
$$ l = \sqrt { { h }^{ 2 }+ (r_1 - r_2)^2}$$, where $$h$$ is the height, $$ { r }_{ 1 } $$ and $${ r }_{ 2 }$$ are the radii of the lower and upper ends of a frustum of a cone.
So,
$$l = \sqrt { { 9 }^{ 2 }+{ (14 - 12) }^{ 2 } } $$ $$ = \sqrt {85} \ \text{cm}$$
Hence, Curved Surface area of this frustum of a cone $$=3.14 \times \sqrt {85} \times (14 + 12)\ \text{cm}^2 $$
$$= 753.6 \ \text{cm}^{2} $$
Question 7
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A rectangular tank $$28\ m$$ long and $$22\ m$$ wide is required to receive entire water from a full cylindrical tank of internal diameter $$28\ m$$ and depth $$4\ m.$$ Find the least height of the tank that will serve the purpose. $$\bigg($$Take $$\pi \ = \ \, \dfrac{22}{7}\bigg)$$

A
$$4.0 m$$

B
$$4.5 m$$

C
$$5.0m$$

D
$$5.2 m$$

Solution

Volume of water in the rectangular tank will be equal to the volume of water in the cylindrical tank.

Volume of a Cylinder $$ = \pi{ R }^{ 2 }h $$

Radius of the cylindrical tank $$ = \dfrac {Diameter}{2} = 14\ m $$
Volume of a cuboid $$ = l\times b \times h $$
Hence, $$ 28\times 22\times h = \cfrac {22}{7} \times {(14) }^{ 2 }\times 4 $$

$$\Rightarrow \ \ \ \ \ \ \ \ \ \ h=4\ m$$
Hence, the least height of the tank that will serve the purpose is $$ 4\ m. $$
Question 8
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A godown building as shown in figure is made in the form of a cubical base with dimensions $$40m\, \times\, 14 m\, \times\, 4 m$$, surmounted by a half cylindrical curved roof having same length as that of the base. The diameter of the cylinder is $$14m$$. Find the volume of the building and its total outer surface area.

A
Volume = $$5310\, m^{3};\,$$ Outer Surface Area = $$ 1466\, m^{2}$$

B
Volume = $$5310\, m^{3};\,$$ Outer Surface Area = $$ 1486\, m^{2}$$

C
Volume = $$5320\, m^{3};\,$$ Outer Surface Area = $$ 1466\, m^{2}$$

D
Volume = $$5320\, m^{3};\,$$ Outer Surface Area = $$ 1486\, m^{2}$$

Solution

Volume of building$$=$$volume of cuboid+volume of half cylinder
$$ =l\times b\times h+\cfrac { 1 }{ 2 } \pi { r }^{ 2 }h\\ =(40\times 14\times 4)+\cfrac { 1 }{ 2 } \times \cfrac { 22 }{ 7 } \times \cfrac { 14 }{ 2 } \times \cfrac { 14 }{ 2 } \times 40\\ =2240+3080\\ =5320{ m }^{ 3 }$$
total outer surface area$$=2h(l+b)+\cfrac { 1 }{ 2 } \pi { r }^{ 2 }h+\pi { r }^{ 2 }\\ =2\times 4(40+14)+\cfrac { 22 }{ 7 } \times 7\times 40+\cfrac { 22 }{ 7 } \times 7\times 7\\ =8\times 54+22\times 40+22\times 7\\ =1466{ m }^{ 2 }$$
Question 9
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A cone is made from a circular sector of radius 2 units. If total surface area of cone (including base) is $$\displaystyle\pi $$ sq. units, find the radius of base of the cone

A
$$\displaystyle \left ( 3-2\sqrt{2} \right )\pi$$

B
$$\left( 2\sqrt { 2 } -2 \right) \pi $$

C
$$\displaystyle \sqrt{2}\pi$$

D
$$\displaystyle \sqrt{3}\pi$$

Solution

TSA of cone $$=\pi rl+\pi { r }^{ 2 }=\pi r\left( l+r \right) =\pi \quad \left[ Given \right] $$
$$\Rightarrow \quad r\left( l+r \right) =1\Rightarrow r\left( 2+r \right) =1\Rightarrow 2r+{ r }^{ 2 }=1\Rightarrow { r }^{ 2 }+2r+1=2$$

$$\Rightarrow \quad { \left( r+1 \right) }^{ 2 }=2\Rightarrow r+1=\sqrt { 2 } \Rightarrow r=\left( \sqrt { 2 } -1 \right) $$

CSA of cone $$=\pi rl=$$ CSA of sector

$$\Rightarrow \quad CSA\quad of\quad sector=\pi \left( \sqrt { 2 } -1 \right) \times 2=\left( 2\sqrt { 2 } -2 \right) \pi $$
Question 10
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A reservoir is in the shape of a frustum of a right circular cone It is $$8 \ m$$ across at the top and $$4\ m$$ across at the bottom. If it is $$6\ m$$ deep its capacity is

A
$$176 \displaystyle \ m^{3}$$

B
$$196 \displaystyle\ m^{3}$$

C
$$200 \displaystyle \ m^{3}$$

D
$$110 \displaystyle \ m^{3}$$

Solution

Volume of a frustum of a cone $$ = \dfrac { 1 }{ 3 } \pi h({ R }^{ 2 }+{ r }^{ 2 }+Rr)$$
Where $$h$$ is the height and $$R$$ and $$r$$ are the radii of the upper and lower ends of a frustum of a cone.
Height $$h=6\ m$$
Since, the reservoir is $$ 8 \ m $$ across the top, upper radius $$R= 4 \ m $$
Similarly, lower radius $$r = 2 m $$

So, volume of the reservoir $$ = \dfrac { 1 }{ 3 } \times \dfrac {22}{7} \times 6({ 4 }^{ 2 }+{ 2 }^{ 2 }+ 4 \times 2)\ m^3$$

  $$ = \dfrac {22}{7} \times 2(16+4+ 8)\ m^3$$

  $$ = \dfrac {22}{7} \times 2\times28\ m^3$$

$$ = 176\ {m}^{3} $$

Hence, the volume of the reservoir is  $$ 176\ {m}^{3} $$

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Re-Attempt Weekly Quiz Competition

Surface Areas and Volumes Test - 23

Result

Surface Areas and Volumes Test - 23

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SHARING IS CARING

Question 1

$$\displaystyle \frac{11}{3}\pi\, r^{3}$$

$$\displaystyle \frac{8}{3}\pi\, r^{3}$$

$$\displaystyle \frac{7}{3}\pi\, r^{3}$$

$$\displaystyle \frac{5}{3}\pi\, r^{3}$$

Solution

Question 2

Capacity = $$1584\, \text{cm}^{3};\,$$ Internal Surface Area = $$ 602.88\, \text{cm}^{2}$$

Capacity = $$1684\, \text{cm}^{3};\,$$ Internal Surface Area = $$ 612.88\, \text{cm}^{2}$$

Capacity = $$1784\, \text{cm}^{3};\,$$ Internal Surface Area = $$ 622.88\, \text{cm}^{2}$$

Capacity = $$1884\, \text{cm}^{3};\,$$ Internal Surface Area = $$ 632.88\, \text{cm}^{2}$$

Solution

Question 3

Volume = $$681.66\, cm^{3};\,$$ Total Surface Area = $$ 478\, cm^{2}$$

Volume = $$681.66\, cm^{3};\,$$ Total Surface Area = $$ 418\, cm^{2}$$

Volume = $$641.66\, cm^{3};\,$$ Total Surface Area = $$ 478\, cm^{2}$$

Volume = $$641.66\, cm^{3};\,$$ Total Surface Area = $$ 418\, cm^{2}$$

Solution

Question 4

$$0.8$$ cm

$$0.5$$ cm

$$0.3$$ cm

$$0.1$$ cm

Solution

Question 5

Rs. $$9.58$$

Rs. $$10.93$$

Rs. $$12.72$$

Rs. $$25$$

Question 6

$$783.6\, \text{cm}^{2}$$

$$763.6\, \text{cm}^{2}$$

$$753.6\, \text{cm}^{2}$$

$$733.6\, \text{cm}^{2}$$

Solution

Question 7

$$4.0 m$$

$$4.5 m$$

$$5.0m$$

$$5.2 m$$

Solution

Question 8

Volume = $$5310\, m^{3};\,$$ Outer Surface Area = $$ 1466\, m^{2}$$

Volume = $$5310\, m^{3};\,$$ Outer Surface Area = $$ 1486\, m^{2}$$

Volume = $$5320\, m^{3};\,$$ Outer Surface Area = $$ 1466\, m^{2}$$

Volume = $$5320\, m^{3};\,$$ Outer Surface Area = $$ 1486\, m^{2}$$

Solution

Question 9

$$\displaystyle \left ( 3-2\sqrt{2} \right )\pi$$

$$\left( 2\sqrt { 2 } -2 \right) \pi $$

$$\displaystyle \sqrt{2}\pi$$

$$\displaystyle \sqrt{3}\pi$$

Solution

Question 10

$$176 \displaystyle \ m^{3}$$

$$196 \displaystyle\ m^{3}$$

$$200 \displaystyle \ m^{3}$$

$$110 \displaystyle \ m^{3}$$

Solution

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