Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 25

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Weekly Quiz Competition

Question 1
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The adjacent diagram represents a frustum of a cone whose bottom and top radii are 6 cm and 3 cm. Its height is 8 cm. There is a conical cavity to a height of 3 cm at the bottom. The amount of material in the solid is-

A
$$132 \displaystyle \pi cm^{3}$$

B
$$168 \displaystyle \pi cm^{3}$$

C
$$100 \displaystyle \pi cm^{3}$$

D
$$135 \displaystyle \pi cm^{3}$$

Solution

Volume of material in solid $$=$$ frustum vol. $$-$$ cone voloume
$$=\dfrac { 1 }{ 3 } \pi \left( { R }^{ 2 }+{ r }^{ 2 }+Rr \right) h-\dfrac { 1 }{ 3 } \pi { R }^{ 2 }h$$
$$=\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times 8\left( { 6 }^{ 2 }+{ 3 }^{ 2 }+6\times 3 \right) -\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times 6\times 6\times 3$$
$$=\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \left( 8\times 63-108 \right) =\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times 396$$
$$=132\pi { cm }^{ 3 }$$
Question 2
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A bucket is in a shape of a frustum, $$40\;cm$$ in diameter at the top and $$28\;cm$$ in diameter at the bottom. Find the cost of the tin sheet used in making the bucket, if the cost of tin is $$Rs.\; 1.50$$ per sq dm.

A
$$Rs.\;44.25$$

B
$$Rs.\ 42.25$$

C
$$Rs.\ 40.25$$

D
$$Rs.\ 45.25$$

Solution

Given:-
$$r_1=20\ cm, r_2=14\ cm$$ and $$h=21\ cm$$

If $$l$$ is the slant height of the bucket then,
$$\begin{aligned}{}l &= \sqrt {{{({r_1} - {r_2})}^2} + {h^2}} \\ &= \sqrt {{{(20 - 14)}^2} + {{21}^2}} \\ &= \sqrt {{6^2} + {{21}^2}} \\ &= \sqrt {36 + 441} \\ &= \sqrt {477} \\ &= 21.84\;cm\end{aligned}$$

Total surface area of the bucket (which is open at the top)
$$A=\pi l(r_1+r_2)+\pi r_2^2$$
$$=\pi [(r_1+r_2)l+r_2^2]$$
$$=\cfrac {22}{7}[(20+14)21.84+14^2]$$
$$=2949.76\ cm^2$$

$$1\ dm^2=100\; cm^2$$

$$A = \dfrac{{2949.7}}{{100}}$$
$$ = 29.497\ dm^2$$

Required cost of the tin sheet at the rate of $$Rs.\ 1.50$$ per sq. dm is

$$\text{Cost}={1.50\times 2949.76}$$
$$=Rs. 44.25$$
Question 3
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A cylinder of radius $$6\ \text{cm}$$ and height $$h$$ cm is filled with ice cream. The ice cream is then distributed among $$10$$ children in identical cones having hemispherical top. The radius of the base of the cone is $$3\ \text{cm}$$ and its height is $$12\ \text{cm}$$. Then the height $$h$$ of the cylinder must be

A
$$\dfrac{100}{7}\ \text{cm}$$

B
$$18\ \text{cm}$$

C
$$15\ \text{cm}$$

D
$$\dfrac{200}{11}\ \text{cm}$$

Solution

Cone with ice cream is shown in the image
Volume of ice cream $$=$$ Volume of cone $$+$$ Volume of hemispherical top
$$\implies $$ Volume of ice cream$$=\cfrac { \pi { r }^{ 2 }h }{ 3 } +\cfrac { 2 }{ 3 } \pi { r }^{ 3 }$$ $$[\because$$ radius is same for both cone and hemispherical top $$]$$
$$=\pi\left(\dfrac{3\times 3 \times 12}{3}+\dfrac{2\times 3 \times 3\times 3}{3}\right)$$
$$=\left( 36+18 \right) \pi \\=54\times \pi\ cm^3$$

Total volume of $$10$$ cones $$=540\pi ~cm^3$$
Now,
Volume of cylinder$$=\pi { R }^{ 2 }h$$
$$=6\times 6 \times \pi h\ cm^3\\=36\pi h~cm^3$$
As per the question,
Volume of cylinder $$=$$ volume of $$10$$ cones
$$\implies 36\pi h=540\pi$$
$$\implies h=\cfrac { 540\pi }{ 36\pi } =15\ cm$$
$$\therefore$$ Height of the cylinder $$=15$$ cm
Question 4
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A solid sphere of radius 6 cm is melted and recast into small spherical balls each of diameter 1.2 cm. Find the number of balls, thus obtained.

A
$$100$$

B
$$1000$$

C
$$2000$$

D
None of these

Solution

Given diameter of the spherical ball $$=1.2 \text{ cm}$$
So, the radius is $$=\dfrac{1.2}{2}=0.6 \text{ cm}$$

Volume of the large sphere $$=\dfrac { 4 }{ 3 } \pi { R }^{ 3 }=\dfrac { 4 }{ 3 } \times \dfrac { 22 }{ 7 } \times 6\times 6\times 6$$

Volume of one spherical ball $$=\dfrac { 4 }{ 3 } \pi { r }^{ 3 }=\dfrac { 4 }{ 3 } \times \dfrac { 22 }{ 7 } \times 0.6\times 0.6\times 0.6$$

Number of small balls is given by dividing the volume of the large sphere with the volume of one spherical ball, i.e.,

Number of small balls $$=\dfrac { \dfrac { 4 }{ 3 } \times \dfrac { 22 }{ 7 } \times 6\times 6\times 6 }{ \dfrac { 4 }{ 3 } \times \dfrac { 22 }{ 7 } \times 0.6\times 0.6\times 0.6 } =1000$$
Question 5
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A solid consists of a circular cylinder with an exact fitting right circular cone placed at the top If the height of the cone is h and the total volume of the solid is 3 times the volume of the cone, then the height of the circular cylinder is

A
2h

B
$$\frac {2h}{3}$$

C
$$\frac {3h}{2}$$

D
4h

Solution

Let the height of the cylinder be H.
As the circular cylinder and cone have an exact fitting, the radius of the cone and cylinder will be same.

Given, Total volume of the solid = 3 $$ \times $$ Volume of cone

$$\implies$$ Volume of cylinder + Volume of cone = 3 $$\times $$ Volume of cone

$$\implies \pi { r }^{ 2 }H + \dfrac { 1 }{ 3 } \pi { r }^{ 2 }h = 3(\dfrac { 1 }{ 3 } \pi { r }^{ 2 }h) $$

$$ \implies H + \dfrac { 1 }{ 3 } h = h $$

$$ H = \dfrac{2}{3} h $$
Question 6
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Milk World Ltd is a government-owned milk distribution centre. Earlier they used to distribute milk in bottles as shown in the figure for Rs 8 per bottle. Now they have decided to reduce the price per bottle by 50 paise by reducing the size of the bottle. However, they do not want their old customers to notice this change in the size of the bottle so they plan to reduce the circumference of the opening of the bottle without changing the height of the bottle. What is the new circumference of the opening?

A
$$\displaystyle 6\pi $$ cm

B
$$\displaystyle 8\pi $$ cm

C
$$\displaystyle 3.5\pi $$ cm

D
None of these

Solution

Volume of bottle $$=\pi { R }^{ 2 }H+\dfrac { 1 }{ 3 } \pi h\left( { R }^{ 2 }+{ r }^{ 2 }+R.r \right) $$
$$=\pi \times 6\times 6\times 4+\dfrac { 1 }{ 3 } \pi \times 3\left( { 6 }^{ 2 }+{ 4 }^{ 2 }+6\times 4 \right) $$
$$=144\pi +76\pi =220\pi { cm }^{ 3 }$$
% reduction in price $$=\dfrac { 0.50 }{ 8 } \times 100=\dfrac { 25 }{ 4 } $$%
$$\therefore $$ Reduction in frustum volume $$=76\pi -\dfrac { 25 }{ 4 } $$% of $$220\pi $$
$$=76\pi -\dfrac { 25 }{ 400 } \times 220\pi $$
$$=76\pi -\dfrac { 55\pi }{ 4 } $$
$$=\dfrac { 249\pi }{ 4 } $$
$$\therefore \quad \dfrac { 1 }{ 3 } \pi \times 3\left( { 6 }^{ 2 }+{ r }^{ 2 }+6r \right) =\dfrac { 249\pi }{ 4 } $$
$$\Rightarrow \quad 36+{ r }^{ 2 }+6r=\dfrac { 249 }{ 4 } \Rightarrow 144+4{ r }^{ 2 }+24r=249$$
$$\Rightarrow \quad { 4r }^{ 2 }+24r-105=0$$
The value of $$r$$ is $$2.94cm$$.
Question 7
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A frustum is made by joining edges of following figure. The volume of the frustum thus formed will be ($$\displaystyle \pi =22/7$$)

A
$$\displaystyle \frac{44\sqrt{2}}{3}$$

B
$$\displaystyle 11\sqrt{2}$$

C
$$\displaystyle \frac{11\sqrt{2}}{2}$$

D
$$\displaystyle \frac{33\sqrt{2}}{4}$$

Solution

$$ar{ C }_{ 1 }=3\times { 120 }^{ 0 }\times \dfrac { \pi }{ { 180 }^{ 0 } } =2\pi \Rightarrow 2\pi { r }_{ 1 }=2\pi \Rightarrow { r }_{ 1 }=1$$
$$ar{ C }_{ 2 }=6\times { 120 }^{ 0 }\times \dfrac { \pi }{ { 180 }^{ 0 } } =4\pi \Rightarrow 2\pi { r }_{ 2 }=4\pi \Rightarrow { r }_{ 2 }=2$$
$$\therefore \quad OA=\sqrt { 9-1 } =2\sqrt { 2 } ,\quad OC=\sqrt { 36-4 } =4\sqrt { 2 } $$
$$\therefore $$ Volume of frustum $$=\dfrac { 1 }{ 3 } \pi \left( { R }^{ 2 }+{ r }^{ 2 }+Rr \right) h=\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \left( { \left( 2 \right) }^{ 2 }+{ \left( 1 \right) }^{ 2 }+2\times 1 \right) 2\sqrt { 2 } $$
$$=\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times 7\times 2\sqrt { 2 } $$
$$=\dfrac { 44\sqrt { 2 } }{ 3 } $$
Question 8
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A circus tent is cylindrical up to a height of 3 m and conical above it. If the diameter of the base is 12 m and the slant height of the conical part is 10 m, then total area of canvas just exceeds

A
$$200 \ m^2$$

B
$$250 \ m^2$$

C
$$300 \ m^2$$

D
$$350 \ m^2$$

Solution

Given that,
A circus tent is cylindrical up to a height of $$3\ m$$ and conical above it.
The diameter of the base is $$12\ m$$ and the slant height of the conical part is $$10\ m$$.
To find out,
The approximate area of the canvas.

The total area of the canvas is formed by the curved surface area of the cylindrical part and the curved surface area of the conical part.
We know that, the curved surface of a cylinder is $$2 \pi rh$$
And, curved surface area of a cone is $$\pi rl$$

Hence, total area of canvas $$=2\pi r h+\pi rl$$
Here, $$r=\dfrac{12}{2}=6\ m,\ h=3\ m$$ and $$l=10\ m$$
$$\therefore \ $$ Total area $$=\pi (2\times 6\times 3+6\times 10)$$
$$\dfrac{22}{7}\times (36+60)\quad \quad \left[\because \ \pi=\dfrac{22}{7}\right]\\$$
$$\dfrac{22\times 96}{7}\\$$
$$=301.71\ m^2$$

Hence, the total area of canvas just exceeds $$300\ m^2$$.
Question 9
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Volume of a bucket 84 cm high, diameter of top as 40 cm and radius of bottom as 10 cm is

A
30.4 1t

B
61.6 1t

C
30.8 1t

D
60 1t

Solution

Volume of frustum
$$\displaystyle \dfrac{1}{3}\times \dfrac{22}{7}\times 84\times \left ( 20^{2}+20\times 10+10^{2} \right )$$
$$\displaystyle=\dfrac{1}{2}\times \dfrac{22}{7}\times 84\times 700cm^{3}$$
$$\displaystyle =61600cm^{3}$$
=61.6lt
Question 10
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An object is in the shape of a frustum 50 cm high, Area of its top and bottom are 81 $$\displaystyle m^{2}$$ and 36 $$\displaystyle m^{2}$$. Its volume is

A
2800 $$\displaystyle cm^{3}$$

B
2750 $$\displaystyle cm^{3}$$

C
2850 $$\displaystyle cm^{3}$$

D
2700 $$\displaystyle cm^{3}$$

Solution

$$\displaystyle V=\dfrac{50}{3}\left ( 81+36+\sqrt{81\times 36} \right )cm^{3}$$
$$\displaystyle=\dfrac{50}{3}\times 171cm^{3}=2850cm^{3}$$