Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 21

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Question 1
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A cone of radius $$r$$ and height '$$3h$$' is cut by two planes parallel to the base at height $$h$$ and $$20 h$$. The volumes of three parts of the cone are in the ratio :

A
$$1 : 7 : 19$$

B
$$1 : 8 : 27$$

C
$$1 : 2 : 3$$

D
$$1 : 4 : 9$$

Solution

By similar triangles,
$$OG : OH = OI = GB : HD : IF = 1 : 2 : 3$$
Hence, volumes of comes
$$OAB, OCD, OEF$$ are in the ratio $$1^3 : 2^3 : 3^3 = 1 : 8 : 27$$
Hence, volumes of cone $$ OAB$$, frustum $$ABCD$$ and frustum $$CDFE$$ are in the ratio $$1 : 7 : 19$$.
Question 2
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If a cubical metal of dimensions 44 cm $$\times$$ 30 cm $$\times$$ 15 cm was melted and cast into a cylinder of height of 28 cm, then what is its radius?

A
$$15$$

B
$$15\sqrt 3$$

C
$$3\sqrt 6$$

D
$$5\sqrt 3$$

Solution

$$\Rightarrow$$ Dimension of cuboid $$=44\,cm\times 30\,cm\times 15\,cm$$

$$\Rightarrow$$ Height of a cylinder $$(H)=28\,cm$$

According to the question,

$$\Rightarrow$$ Volume of a cuboid $$=$$ Volume of cylinder.

$$\Rightarrow$$ $$l\times b\times h=\pi \times r^2\times H$$

$$\Rightarrow$$ $$44\times 30\times 15=\dfrac{22}{7}\times (r^2)\times 28$$

$$\Rightarrow$$ $$44\times 30\times 15=22\times 4\times r^2$$

$$\Rightarrow$$ $$r^2=\dfrac{44\times 30\times 15}{22\times 4}$$

$$\Rightarrow$$ $$r^2=225\,cm^2$$

$$\therefore$$ $$r=15\,cm$$
Question 3
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A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 $$cm^{3}$$. Check whether she is correct, taking the above as the inside measurements, and $$\pi$$ = 3.14

A
Yes

B
No

C
Less Data

D
None of the above

Solution

Radius of cylinder$$=1cm$$

Height of cylinder$$=8cm$$

Radius of sphere $$=8.5$$

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

$$=\pi\times1^2\times8$$

$$=8\pi cm^3$$

Volume of sphere $$=\dfrac{4}{3}\pi r^3$$

$$=\dfrac{4}{3}\pi \times(\dfrac{8.5}{2})^3$$

$$=\dfrac{614125}{6000}\pi cm^3$$

$$\therefore $$ total Volume $$=\left(\dfrac{614125}{6000}+8\right)\pi $$

$$=\left(\dfrac{614125+48000}{6000}\right)\pi $$

$$=\dfrac{662125}{6000}\pi$$

$$=110.35\times3.41$$

$$=346.5cm^3$$

So, $$345cm^3$$ is not correct.
Question 4
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A toy is in the form of a cone of radius $$3.5$$ cm mounted on a hemisphere of same radius. The total height of the is $$15.5$$ cm. Find the total surface area of the toy.

A
$$234.5cm^2$$

B
$$224.5cm^2$$

C
$$214.5cm^2$$

D
$$204.5cm^2$$

Solution

Given,

Radius of a cone is $$3.5$$cm,

Height of a cone is $$15.5-3.5=12$$cm

Slant height of cone is,

$$l=\sqrt{h^2+r^2}$$

$$=\sqrt{12^2+3.5^2}$$

$$=\sqrt{144+12.25}$$

$$=\sqrt{156.25}$$

$$=12.5$$cm

Curved surface area of cone is,

$$=\pi rl$$

$$=\dfrac{22}{7}\times3.5\times12.5$$

$$=137.5cm^2$$

Curved surface area of hemispherical portion is,

$$=2\pi r^2$$

$$=2\times\dfrac{22}{7}\times3.5\times3.5$$

$$=77cm^2$$

Hence,

total surface area,

$$137.5+77=214.5cm^2$$
Question 5
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A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also find the cost of the canvas of the tent at the rate of Rs. 500 per $$m^{2}$$ (note that the base of the tent will not be covered with canvas).

A
$$20,000$$

B
$$21,000$$

C
$$22,000$$

D
$$23,000$$

Solution

Radius of the cylindrical base
$$= 2m$$
Height of the cylindrical is
$$ = 2.1 m$$
Slant Height of conical top is
$$=2.8$$
Curved Surface Area of cylindrical portion is,
$$=2\pi rh$$
$$=2\pi \times 2\times 2.1$$
$$=8.4\pi m^2$$
Curved Surface Area of conical portion is,
$$=\pi rl$$
$$=\pi \times 2\times 2.8$$
$$=5.6\pi m^2$$
Total Curved Surface Area
$$=8.4\pi + 5.6\pi$$
$$=14\times \frac{22}{7}$$
$$=44m^2$$
Therefore,
Cost of canvas=Rate\times Surface Area
$$=500\times 44$$
$$=22000$$
Question 6
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A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of $$\pi$$.

A
$$2\pi \ cm^3$$

B
$$8\pi \ cm^3$$

C
$$7\pi \ cm^3$$

D
$$\pi \ cm^3$$

Solution

Area of solid $$=$$ Area of cone $$+$$ area of cone
Radius $$=1$$cm
Height $$=1$$cm
Volume of Hemisphere
$$=\dfrac{2}{3}\pi r^3$$
$$=\dfrac{2}{3}\pi \times 1^3$$
$$=\dfrac{2}{3}\pi \ cm^2$$
Volume of cone is,
$$\dfrac{1}{3}\pi r^2h$$
$$=\dfrac{1}{3}\pi \times1^2\times 1$$
$$=\dfrac{1}{3}\pi \ cm^2$$
$$\therefore $$ total Volume $$=\dfrac{2}{3}\pi + \dfrac{1}{3}\pi$$
$$=\pi (\dfrac{2+1}{3})$$
$$=\pi \ cm^3$$
Question 7
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A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

A
$$557 cm^2$$

B
$$517 cm^2$$

C
$$537 cm^2$$

D
$$572 cm^2$$

Solution

$$\textbf{Step-1: Apply the relevant formula of curved surface area of a cylinder and hemispherical portion.}$$
$$\text{Radius = 7 cm}$$

$$\text{Height of the cylindrical portion = 13 - 7 = 6 cm}$$

$$\text{Area of a Curved surface of cylindrical portion is,}$$

$$= 2\pi rh$$

$$= 2\times\frac{22}{7}\times 7\times 6$$

$$ = 264cm^2$$

$$\text{Area of a curved of hemispherical portion is}$$

$$ = 2\pi r^2$$

$$ = 2\times\dfrac{22}{7}\times 7\times 7$$

$$ = 308cm^2$$

$$\textbf{Step-2: Find the sum of above results to get the required unknown.}$$

$$\therefore$$ $$\text{Total surface area is}$$ $$=308+264=572 cm^2.$$

$$\textbf{Hence, correct option is D.}$$
Question 8
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A drinking glass in the shape of a frustrum of a cone of height $$14$$ cm.Two diameters of its two circular ends are $$4$$ cm. and $$2$$ cm, then the capacity of glass is:

A
$$102\frac{2}{3}$$ cm$$^3$$

B
$$102\frac{1}{3}$$ cm$$^3$$

C
$$101\frac{2}{3}$$ cm$$^3$$

D
$$101\frac{1}{3}$$ cm$$^3$$

Solution

Volume of frustum $$= \dfrac{1}{3}\pi h({r}_{1}^{2}+{r}_{2}^{2}+{r}_{1}{r}_{2})$$

$$=\dfrac{1}{3}\times \dfrac{22}{7}\times 14({2}^{2}+{1}^{2}+2)$$

$$ =\dfrac{1}{3}\times 22\times 2\times 7=\dfrac{308}{3}=102\dfrac{2}{3}{cm}^{3}$$
Hence, option 'A' is correct.
Question 9
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The dimensions of a solid metal cuboid are $$72\, cm\, \times\, 30\, cm\, \times\, 75\, cm.$$ it is metaled and recast into identical solid metal cubes with edge of each $$6$$ cm. Find the number of cubes formed.
Also, find the cost of polishing the surface of all the cubes formed at the rate Rs. $$150$$ per sq. m.

A
$$427$$ ; Rs. $$2,430$$

B
$$830$$ ; Rs. $$2,430$$

C
$$571$$ ; Rs. $$2,430$$

D
$$750$$ ; Rs. $$2,430$$

Solution

Given a solid metal cuboid with dimension $$72cm\times 30cm\times 75cm$$. It is melted, recast into cube of edge $$6$$ cm.
For forming cube by melting, we have to find volume of cuboid $$=l\times b\times h\\ =72cm\times 30cm\times 75cm\\ =162000\quad { cm }^{ 3 }$$
Volume of the cube to be formed $$={ (side) }^{ 3 }\\ ={ 6 }^{ 3 }\\ =216{ \quad cm }^{ 3 }$$

So, number of cubes formed $$=\dfrac { \text {Volume of cuboid}}{ \text {Volume of cube}} $$ $$=\dfrac { 162000 }{ 216 } =750$$

For polishing the cube we need to find total surface area of cube.
Total surface area of one cube $$=6\times { (side) }^{ 2 }\\ =6\times { 6 }^{ 2 }\\ =216\quad { cm }^{ 2 }$$

Total surface area of the $$750$$ cubes $$=216{ cm }^{ 2 }\times 750\\ =162000{ cm }^{ 2 }\\ =16.2 { m }^{ 2 }$$

Cost of polishing per sq.m is Rs. $$150$$.
Therefore, cost of polishing 16.2 sq m $$=16.2 { m }^{ 2 }\times 150\\ =Rs.2430$$
Question 10
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A medicine capsule is in the shape of a cylinder with two hemisphere struck to each of its ends (see figure). The length of the entire capsule is $$14 mm$$ and the diameter of the capsule is $$5 mm$$. Find its surface area.

A
$$240 mm^2$$

B
$$280 mm^2$$

C
$$220 mm^2$$

D
$$120 mm^2$$

Solution

Height of the cylinder
$$=14-5 =9$$mm
Radius$$=2.5$$mm
Curved Surface Area of Cylinder = $$2\pi rh$$
$$=2\pi \times 2.5\times 9$$
$$=45\pi mm^2$$
Curved Surface Area of two Hemispheres = $$4\pi r^2$$
$$=4\pi \times 2.5^2$$
$$=25\pi mm^2$$
Total Surface Area = $$=45\pi +25\pi$$
$$=70\pi$$
$$=70\times \frac{22}{7}$$
$$=220mm^2$$