Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 24

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Weekly Quiz Competition

Question 1
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A few blocks of wood are used to make the shape of a giraffe as shown below. What is the volume of wood used to make the giraffe?

A
$$\displaystyle 640cm^{2}$$

B
$$\displaystyle 1260cm^{2}$$

C
$$\displaystyle 1360cm^{2}$$

D
$$\displaystyle 1400cm^{2}$$

Solution

Volume of one block $$A = 4\times 24\times 34=160 cm^3$$.

So, volume of 4 blocks of $$A$$ is $$640 cm^3$$.

Length of block $$B$$ is $$18cm$$.

Width of block $$B$$ is $$4 cm$$ and height is $$10cm$$.

So, volume of block $$B$$ is $$18\times 10\times 4 = 720 cm^3$$.

$$\therefore$$ The total volume is $$720 + 640 = 1360 cm^3$$
Question 2
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How many small cubes, each of $$96\, cm^2$$ surface area can be formed from the material obtained by melting a larger cube of $$384\, cm^2$$ surface area ?

A
$$80$$

B
$$5$$

C
$$800$$

D
$$8$$

Solution

Let the side of a bigger cube be $$a\,cm$$ and that of smaller cube be $$b\, cm$$.

Area of bigger cube$$=6a^2=384 cm^2$$ ...Given
$$\therefore$$$$a=8\, cm$$

Area of smaller cube$$=6b^2=96cm^2$$ ...Given
$$\therefore$$$$b=4\, cm$$

Number of cubes$$=\dfrac{\text {Volume of bigger cube}}{\text {Volume of smaller cube}}$$

$$\therefore$$ Number of cubes$$=\dfrac{8\times 8 \times 8}{4 \times 4 \times 4}=8$$
Question 3
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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
$$\displaystyle 132\pi \ cm^{3}$$

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

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

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

Solution

Volume of material $$=$$ Volume of frustum $$-$$ Vol. of cavity

                                 $$=\dfrac { 1 }{ 3 } \pi \left( { R }^{ 2 }+{ r }^{ 2 }+R.r \right) H-\dfrac { 1 }{ 3 } \pi { r }_{ 1 }^{ 2 }h$$

                                 $$=\dfrac { 1 }{ 3 } \pi \left[ { 6 }^{ 2 }+{ 3 }^{ 2 }+6\times 3 \right] \times 8-\dfrac { 1 }{ 3 } \pi \times 6\times 6\times 3$$

                                 $$=\dfrac { 1 }{ 3 } \pi \left[ 63\times 8-108 \right] =\dfrac { \pi }{ 3 } \times \left[ 504-108 \right] $$
                                 $$=\dfrac { \pi }{ 3 } \times 396=132\pi $$$${ cm }^{ 3 }$$
Question 4
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Water flows at the rate of 10 m per min from cylindrical pipe 5 mm in diameter. How long will it take to fill up a conical vessel whose diameter at the base is 40 cm and depth 24 cm ?

A
48 min 15 sec

B
51 min 12 sec

C
52 min 1 sec

D
55 min

Solution

Volume of a cone of Radius "R" and height "h" $$ = \dfrac{1}{3} \pi { R }^{

2 }h $$

Radius of the vessel $$ =\dfrac{40}{2} = 20 cm $$

Hence, volume of the conical vessel $$ = \dfrac{1}{3} \pi \times 20 \times 20 \times 24 {cm}^{3} $$

Radius of pipe $$ = \dfrac {5}{2} mm = 2.5 mm = \dfrac {25}{100} cm $$

Since the pipe is cylindrical in shape, Volume $$ = \pi { (Radius) }^{ 2 } \times height $$

Here, height is considered as the distance travelled.

Hence, Volume of water coming out of pipe in one minute $$\displaystyle = \pi \times {(\frac

{25}{100})}^{2} \times 1000 {cm}^{3} $$

Now, $$\textbf{time taken to fill the cistern}$$

$$\displaystyle=\frac { \textit{Volume of the vessel} }{\textit{ Volume of water coming out of pipe in one minute} } =\frac { \frac { 1 }{ 3 } \pi \times 20\times 20\times 24 }{ \pi \times \frac { 25 }{ 100 } \times \frac { 25 }{ 100 } \times 1000 } =51\frac{1}{5}minutes = 51min 12 sec$$
Question 5
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The height of a cone is $$30 cm$$. A frustum is out off from this cone by a plane parallel to the base of the cone. If the volume of the frustum is $$ \cfrac{1}{27}$$ of the volume of the cone, find the height of the frustum.

A
$$20 cm$$

B
$$12 cm$$

C
$$15 cm$$

D
$$18 cm$$

Solution

Lets consider a cone of Radius $$R$$.
Let a cone of height $$h$$ is cut off from the top of this cone whose base is parallel to the original cone. The radius of the the cone cut off be $$r$$.
Here, $$H=30cm$$
In $$\triangle APC$$ and $$\triangle AQE$$, $$PC\parallel QE$$
Therefore,
$$\triangle APC \sim \triangle AQE$$
$$=>\cfrac{AP}{AQ}=\cfrac{PC}{QE}$$
$$=>\cfrac{h}{H}=\cfrac{r}{R}$$ (I)

Given, Volume of the cone$$ ABC$$$$=\cfrac{1}{27}$$Volume of the cone $$ADE$$
$$=>\cfrac{\text {Volume of the cone ABC}}{\text {Volume of the cone ADE} }=\cfrac{1}{27}$$
$$=>\cfrac { \cfrac { 1 }{ 3 } \pi r^{ 2 }h }{ \cfrac { 1 }{ 3 } \pi R^{ 2 }h } =\cfrac{1}{27}$$
$$=>\left(\cfrac{r}{R}\right)^2\times \cfrac{h}{H}=\cfrac{1}{27}$$
$$=>\left(\cfrac{h}{H}\right)^2\times \cfrac{h}{H}=\cfrac{1}{27}$$ ...(from (i))
$$=>\left(\cfrac{h}{H}\right)^3=\cfrac{1}{27}$$
$$=>\cfrac{h}{H}=\cfrac{1}{3}$$
$$=>h=\cfrac{1}{3}H$$
$$=>h=\cfrac{1}{3}\times 30cm$$
$$=>h=10cm$$
Now, $$PQ=H-h$$
$$=30cm-10cm$$
$$=20cm$$
Hence, the section is cut at the height of $$20 cm$$ from the base.
Question 6
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Two steel sheets each of length $$\displaystyle a_{1}$$ and breadth $$\displaystyle a_{2}$$ are used to prepare the surface of two right circular cylinders-one having volume $$\displaystyle V_{1}$$ and height $$\displaystyle a_{2}$$ and the other having volume $$\displaystyle V_{2}$$ and height $$\displaystyle a_{1}$$. Then

A
$$\displaystyle V_{1}=V_{2}$$

B
$$\displaystyle a_{1}V_{1}=a_{2}V_{2}$$

C
$$\displaystyle a_{2}V_{1}=a_{1}V_{2}$$

D
$$\displaystyle \dfrac{V{_{1}}^{2}}{a_{1}}=\dfrac{V{_{2}}^{2}}{a_{2}}$$

Solution

Let the radius of base of the cylinders be r and R
$$\Rightarrow a_{1} = 2\pi r , a_{2} = 2\pi R$$
$$ \Rightarrow r = \dfrac{a_{1}}{2\pi}, R = \dfrac{a_{2}}{2\pi}$$
Also
$$v_{1} = \pi r^2 a_{2} = \pi \left(\dfrac{a_{1}}{2\pi}\right)^2 a_{2}= \dfrac{a_{1}^2a_{2}}{4\pi}$$
$$v_{2} = \pi R^2 a_{1} = \pi \left(\dfrac{a_{2}}{2\pi}\right)^2 a_{1}= \dfrac{a_{2}^2a_{1}}{4\pi}$$
$$\Rightarrow \dfrac{v_{1}}{v_{2}} = \dfrac{\dfrac{a_{1}^2 a_{2}}{4\pi}}{\dfrac{a_{2}^2 a_{1}}{4\pi}} = \dfrac{a_{1}}{a_{2}}$$
$$\Rightarrow \dfrac{v_{1}}{a_{1}} = \dfrac{v_{2}}{a_{2}} = a_{2}v_{1} = a_{1}v_{2}$$
Question 7
1 / -0

A frustum is made by joining edges of following figure. The volume of the frustum thus formed will be $$\displaystyle (\pi =22/7)$$

A

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

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

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

Solution

ar $${ C }_{ 1 }=3\times 120\times \dfrac { \pi }{ 180 } =2\pi $$, ar $${ C }_{ 2 }=6\times 120\times \dfrac { \pi }{ 180 } =4\pi $$

$$2\pi { r }_{ 1 }=2\pi \Rightarrow { r }_{ 1 }=1,\quad { r }_{ 2 }=2$$

Volume of frustum $$=\dfrac { 1 }{ 3 } \pi { r }_{ 2 }^{ 2 }\times 4\sqrt { 2 } -\dfrac { 1 }{ 3 } \pi { r }_{ 1 }^{ 2 }\times 2\sqrt { 2 } $$

$$=\dfrac { 1 }{ 3 } \left( \pi \times { 2 }^{ 2 }\times 4\sqrt { 2 } -\pi \times { 1 }^{ 2 }\times 2\sqrt { 2 } \right) $$
$$=\dfrac { 1 }{ 3 } \left( 16\sqrt { 2 } \pi -2\sqrt { 2 } \pi \right) =\dfrac { 1 }{ 3 } \times 14\sqrt { 2 } \times \dfrac { 22 }{ 7 } =\dfrac { 44\sqrt { 2 } }{ 3 } $$
Question 8
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A metallic solid cone is melted and cast into a cylinder of the same base as that of the cone. If the height of the cylinder is $$7\;cm$$, what was the height of the cone?

A
$$20\;cm$$

B
$$21\;cm$$

C
$$22\;cm$$

D
$$12\;cm$$

Solution

When a solid of one shape is melted into another shape, the amount of volume remains the same.
Volume of cone $$=$$ Volume of cylinder
$$\dfrac { 1 }{ 3 } \pi { r }_{ 1 }^{ 2 }{ h }_{ 1 }=\pi { r }_{ 2 }^{ 2 }{ h }_{ 2 }$$
$$\Rightarrow \quad \dfrac { 1 }{ 3 } \times { r }^{ 2 }\times { h }_{ 1 }={ r }^{ 2 }\times 7\quad \left[ { r }_{ 1 }={ r }_{ 2 }=r \right ] $$ as same base
$$\Rightarrow \quad { h }_{ 1 }=7\times 3=21cm$$
Therefore, height of cone is $$21$$ cm.
Question 9
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Two solid cylinders are $$12\;cm\;and\;18\;cm$$ in height. Their diameters are $$12\;cm\;and\;16\;cm$$ respectively. Both the cylinders are melted and the material is moulded into a solid right circular cone of height $$33\;cm$$. Find its diameter.

A
$$28\;cm$$

B
$$20\;cm$$

C
$$14\;cm$$

D
$$24\;cm$$

Solution

$$Volume of cone = Sum of volumes of two cylinders$$

$$\Rightarrow \quad \dfrac { 1 }{ 3 } \pi { R }^{ 2 }H=\pi { r }_{ 1 }^{ 2 }{ h }_{ 1 }+\pi { r }_{ 2 }^{ 2 }{ h }_{ 2 }\Rightarrow \dfrac { 1 }{ 3 } \pi \times R^{ 2 }\times 33=\pi \left( 6\times 6\times 12+8\times 8\times 18 \right) $$

$$\Rightarrow \quad 11{ R }^{ 2 }=432+1152\Rightarrow { R }^{ 2 }=144\Rightarrow R=\sqrt { 144 } =12cm$$

Therefore, diameter of cone is $$12$$$$\times $$ $$2$$ $$=$$ $$24$$cm.
Question 10
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The diameter of one of the bases of a truncated cone is $$100\ mm$$. If the diameter of this base is increased by $$21 \%$$ such that it still remains a truncated cone with the height and the other base unchanged, the volume also increases by $$21 \% $$. The radius of the other base is

A
$$65\ mm$$

B
$$55\ mm$$

C
$$45\ mm$$

D
$$35\ mm$$

Solution

Let initially $$2$$ bases have a radii $$50\ mm=5\ cm$$ and $$r$$ , and finally bases have radii $$(1.21 \times 5)$$ and $$r.$$ Also, let the height of the truncated cone be $$h$$ which is fixed.

$$\therefore$$ Ratios of volume $$= \dfrac {V_2} {V_1}$$
$$ = 1.21 $$

$$V_2 = \dfrac {\pi h} {3} ((6.50)^2 + 6.05 r + r^2) $$

$$V_1 = \dfrac {\pi h} {3} (5^2 + 5 r + r^2 ) $$

$$\dfrac {V_2} {V_1} = 1.21$$
$$ \Rightarrow \dfrac {(6.05)^2 + 6.05 r + r^2} {5^2 + 5 r + r^2} = 1.21 $$

$$\Rightarrow 36.6025 + 6.05 r + r^2 = 30.25 + 6.05 r + 1.21 r^2 $$

$$\Rightarrow 21r^2 = 6.3525 $$

$$\Rightarrow r^2 = \dfrac {6.3525} {21} $$

$$\Rightarrow r = \dfrac {11} {2}\ cm$$

$$\Rightarrow r = 55\ mm $$

Hence, option $$B$$ is correct.

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Surface Areas and Volumes Test - 24

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Surface Areas and Volumes Test - 24

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SHARING IS CARING

Question 1

$$\displaystyle 640cm^{2}$$

$$\displaystyle 1260cm^{2}$$

$$\displaystyle 1360cm^{2}$$

$$\displaystyle 1400cm^{2}$$

Solution

Question 2

$$80$$

$$5$$

$$800$$

$$8$$

Solution

Question 3

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

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

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

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

Solution

Question 4

48 min 15 sec

51 min 12 sec

52 min 1 sec

55 min

Solution

Question 5

$$20 cm$$

$$12 cm$$

$$15 cm$$

$$18 cm$$

Solution

Question 6

$$\displaystyle V_{1}=V_{2}$$

$$\displaystyle a_{1}V_{1}=a_{2}V_{2}$$

$$\displaystyle a_{2}V_{1}=a_{1}V_{2}$$

$$\displaystyle \dfrac{V{_{1}}^{2}}{a_{1}}=\dfrac{V{_{2}}^{2}}{a_{2}}$$

Solution

Question 7

(44√2)/3

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

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

$$\displaystyle \dfrac{33\sqrt{2}}{4}$$

Solution

Question 8

$$20\;cm$$

$$21\;cm$$

$$22\;cm$$

$$12\;cm$$

Solution

Question 9

$$28\;cm$$

$$20\;cm$$

$$14\;cm$$

$$24\;cm$$

Solution

Question 10

$$65\ mm$$

$$55\ mm$$

$$45\ mm$$

$$35\ mm$$

Solution

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