Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 46

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Question 1
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If the frustum of a cone has a height of $$6cm$$ and radius of $$5cm$$ and $$9cm$$ respectively; then its cubical volume will be .............. $${cm}^{3}$$.

A
$$320 \pi$$

B
$$151 \pi$$

C
$$302 \pi$$

D
$$98 \pi$$

Solution

Given $$h=6\ cm, {R}_{1}=9\ cm$$ and $${R}_{2}=5\ cm$$
Volume of the frustrum of cone $$=\cfrac { 1 }{ 3 } \pi h\left( { R }_{ 1 }^{ 2 }+{ R }_{ 2 }^{ 2 }+{ R }_{ 1 }{ R }_{ 2 } \right) $$
volume of the frustrum of cone $$=\cfrac { 1 }{ 3 } \times \pi \times 6({ (9) }^{ 2 }+{ (5) }^{ 2 }+(9)(5))=302\pi $$
Question 2
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A solid is in the shape of a cylinder surmounted on a hemisphere. If the diameter and the total height of the solid are $$21\ cm,25.5\ cm$$ respectively, then find its volume.

A
$$7623\ {cm}^{3}$$

B
$$7235\ {cm}^{3}$$

C
$$8263\ {cm}^{3}$$

D
None of these

Solution
Question 3
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Lateral surface area of the frustum of a cone is

A
$$\pi (r_2 - r_1) h$$

B
$$\pi (r_1 + r_2) h$$

C
$$\pi (r_1 - r_2) l$$

D
$$\pi (r_1 + r_2) l$$

Solution

By ratio and proportion:
$$\dfrac{l_1}{r_1}=\dfrac{l}{r_1-r_2}$$
$$\Rightarrow l_1= \dfrac{lr_1}{r_1-r_2}$$

And from the figure we observe that:
$$l_2=l_1-l$$
$$=\dfrac{lr_1}{r_1-r_2}- l$$
$$=\dfrac{lr_1-l(r_1-r_2)}{r_1-r_2}$$
$$= \dfrac{lr_2}{r_1-r_2}$$

The length of arc is the circumference:
$$s_1=2\pi r_1 , s_2= 2\pi r_2$$

The lateral surface area of a cone is given by $$\dfrac12sl,$$ where $$s$$ is the circumference of the base and $$l$$ is the slant height.

Hence, by using the above formula,
$$\begin{aligned}{}A &= \frac{1}{2}{s_1}{l_1} - \frac{1}{2}{s_2}{l_2}\\ &= \frac{1}{2}(2\pi {r_1})\left( {\frac{{l{r_1}}}{{{r_1} - {r_2}}}} \right) - \frac{1}{2}(2\pi {r_2})\left( {\frac{{l{r_2}}}{{{r_1} - {r_2}}}} \right)\\ &= \frac{{\pi lr_1^2}}{{{r_1} - {r_2}}} - \frac{{\pi lr_2^2}}{{{r_1} - {r_2}}}\\ &= \frac{{\pi l(r_1^2 - r_2^2)}}{{{r_1} - {r_2}}} = \frac{{\pi l({r_1} + {r_2})({r_1} - {r_2})}}{{{r_1} - {r_2}}}\\ &= \pi ({r_1} + {r_2})l\end{aligned}$$

Hence, option $$D$$ is the correct answer.
Question 4
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A hollow cylinder of negligible thickness and radius $$r$$ is mounted by a hollow hemisphere of the same radius. The height of the cylinder is $$h$$. If the solid formed is to be painted . What is the total surface area that has to be painted?

A
$$2\pi rh+4\pi { r }^{ 2 }$$

B
$$2\pi rh+\pi { r }^{ 2 }$$

C
$$2\pi rh+3\pi { r }^{ 2 }$$

D
None of the above

Solution

Total surface area painted to be $$=$$ curved surface area of cylinder $$+$$ curved surface area of hemisphere $$+$$ area of base circle of cylinder
$$A=2\pi rh+2\pi { r }^{ 2 }+\pi { r }^{ 2 }\\ A=2\pi rh+3\pi { r }^{ 2 }$$
Question 5
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A bucket is of the shape of a frustum. If the radii of the circular ends are $$28$$ cm and $$7$$ cm and the height is $$45$$ cm. The capacity of the bucket is __________.

A
$$48150 \ \text{cm}^3$$

B
$$48510 \ \text{cm}^3$$

C
$$48105 \ \text{cm}^3$$

D
$$48205 \ \text{cm}^3$$

Solution

Given that,
The bucket is in the shape of a frustum.
Height of the bucket $$= 45$$ cm
Radius of the bigger circular end, $$r_{1} = 28$$ cm
and, radius of the smaller circular end, $$r_{2} = 7$$ cm

Volume of bucket $$= \dfrac{\pi}{3}(r_{1}^{2}+r_{1}r_{2}+r_{2}^{2})h$$

$$= \dfrac{22}{3\times 7}[28^{2}+28\times7+7^{2}]\times45\ cm^3$$

$$= \dfrac{22}{21}[784+196+49]\times45\ cm^3$$

$$= \dfrac{22}{21}\times 1029\times45\ cm^3$$

$$= 48510\ cm^{3}$$

Hence, the volume is $$ 48510\ cm^{3}$$
Question 6
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While calculating the surface area of the figure formed by combining all the three given figures, it includes:

A
Only flat areas

B
Only curved areas

C
Both flat and curved areas

D
None of the areas

Solution

After joining all the given figures we will get a cylinder extended with hemi- spheres on both of its flat faces.
This makes the object curved on all surfaces with no flat surfaces at any part of the body.
So, the total surface area includes only curved area not the flat areas.
Hence, the answer is only curved areas.
Question 7
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A metal sheet $$27$$ cm long, $$8$$ cm broad and $$1$$ cm thick is melted into a cube so that volume is constant. The difference between surface areas of two solids is

A
$$284cm^2$$

B
$$286cm^2$$

C
$$296cm^2$$

D
$$300cm^2$$

Solution

We know that,
Total surface area of cuboid = $$2$$ (lb + bh + hl)
Surface area of given cuboid
= $$2 (27$$ $$\times$$ $$8 + 8$$ $$\times$$ $$1 + 27$$ $$\times$$ 1) = $$502$$ $$cm^2$$
Volume of cube = $$27$$ $$\times$$ $$8 \times 1$$ = $$216cm^3$$
$$\therefore$$ Edge of the cube a = $$\sqrt[3]{216}$$ = $$6$$ cm
We know that,
Total surface area of cube = $$6a^2$$
Surface area of this cube = $$6 \times \, (6)^2$$ = $$216$$ $$cm^2$$
Hence, the difference of surface areas = $$502 - 216 = 286$$ $$cm^2$$
Question 8
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Squares are constructed on the outer side of right angled triangle on each of its three sides. If the lengths of the two sides containing the right angle are $$5$$ cm and $$10$$ cm respectively, then the total area of the regions bounded by the diagram is:

A
$$250$$ sq.cm

B
$$200$$ sq.cm

C
$$300$$ sq.cm

D
$$275$$ sq.cm

Solution

In right $$\triangle PQR$$, using Pythagoras theorem,
$${ PR }^{ 2 }={ PQ }^{ 2 }+{ QR }^{ 2 }\Rightarrow PR=\sqrt { { 10 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 125 } =5\sqrt { 5 } ㎝$$
Required area$$=\cfrac { 1 }{ 2 } \times 5\times 10+{ \left( 5 \right) }^{ 2 }+{ \left( 10 \right) }^{ 2 }+{ \left( 5\sqrt { 5 } \right) }^{ 2 }$$
$$=25+25+100+125=275㎠$$
Question 9
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A tent is of the shape of a right circular cylinder upto a height of $$3$$ metres and then becomes a right circular cone with maximum height of $$13.5$$ metres above the ground. If the radius of the base is $$14$$ metres, the cost of painting the inner side of the tent at the rate of $$Rs. 2$$ per square metre is

A
$$Rs. 2050$$

B
$$Rs. 2060$$

C
$$Rs. 2068$$

D
$$Rs. 2080$$

Solution

$$\textbf{Step -1: Calculating surface area .}$$
$$\text{Curved surface area of cylinder}=2\pi rh$$
$$\implies 2\pi\times 14\times3=84\pi m^2$$
$$\text{Curved surface area of cone}=\pi r\sqrt{r^2+h^2}$$
$$= \pi\times14\times\sqrt{14^2+(13.5-3)^2}$$
$$= \pi \times 14\times\dfrac{35}2m^2$$
$$= 245\pi m^2$$
$$\textbf{Step -2: Calculating total rate .}$$
$$\text{Total area to paint}=(245+84)\pi m^2$$
$$\implies 329\pi=1034m^2$$
$$\text{Cost of painting }=Rs2/m^2$$
$$\implies \text{Total cost}=1034\times2=Rs.2068$$
$$\textbf{Hence Option C is correct.}$$
Question 10
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If the radii of the circular ends of a bucket of height $$40cm$$ are of lengths $$35cm$$ and $$14cm$$, then the volume of the bucket in cubic centimetres, is _________.

A
$$60060$$

B
$$80080$$

C
$$70040$$

D
$$80160$$

Solution

Given:-
Height of bucket (h) $$= 40$$ cm
$${r}_{1} = 35 cm$$
$${r}_{2} = 14 cm$$

To find:- volume of bucket

Solution:-
Volume of bucket $$= \cfrac{\pi}{3} \left({({r}_{1})}^{2} + {r}_{1}{r}_{2} + {({r}_{2})}^{2} \right) \times h$$

$$\text{ Volume of bucket} = \cfrac{22}{7 \times 3} \left({35}^{2} + 35 \times 14 + {14}^{2} \right) \times 40$$
$$ = \cfrac{22}{21} \times 1911 \times 40 $$
$$= 880 \times 91 $$
$$= 80080 {cm}^{3}$$

Hence, the volume of the bucket is $$ 80080\ {cm}^{3}$$.