Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

A military of height $$8.25\ m$$ is in the form of a right circular cylinder of base diameter $$30\ m$$ and height $$5.5\ m$$ surmounted by a right circular cone of same base radius. Find the length of the canvas use in making the tent, if the breadth of the canvas is $$1.5\ m$$

A
$$500.26\ m$$

B
$$224.26\ m$$

C
$$824.26\ m$$

D
None of these

Solution

Let the length of the canvas of width $$1.5\ m$$ required be $$L.$$

Height of the cone $$h= 8.25 - 5.5\ m$$
$$= 2.75 m$$
Radius of the cone, $$r=15\ m$$
Radius of the cylinder, $$R=15\ m$$

Slant height of the cone will be,
$$l= \sqrt{ h^2 + r^2}$$
$$ = 15.25\ m$$

Surface area of cone $$= πrl $$
$$= 3.14\times 15\times 15.25$$
$$= 718.3$$

Surface area of cylinder $$= 2πrl$$
$$ = 2\times 3.14\times 15\times 5.5$$
$$= 518.1$$

Surface Area of tent $$= 718.3 + 518.1$$
$$ =1236.4\ m^2$$

Area of the canvas used $$=$$ Surface area of the tent
$$L\times 1.5=1236.4$$
$$L= \dfrac{1236.4}{1.5}$$
$$=824.26\ m$$
Question 2
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If the height of a bucker in the shape of frustum of a cone is $$16\ cm$$ and the diameters of its two circumends are $$40\ cm$$ and $$16\ cm$$,then its slant height is

A
$$20\ cm$$

B
$$12\sqrt5\ cm$$

C
$$8\sqrt13\ cm$$

D
$$16\ cm$$

Solution

Let Slant Height be the Hypotenuse of a Imaginary Right angled Triangle whose height will be equal to that of the frustum and its base will be equal to $$\dfrac{40}{2}-\dfrac{16}{2}=20-8=12$$

Hence, According to Pythagoras theorem
$$SH^2=16^2+12^2$$
$$=400$$
$$\Rightarrow SH=\sqrt{400}\ cm=20\ cm$$
Question 3
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$$2$$ cubes each of volume $$64{cm}^{3}$$ are joined end to end. Find the surface area of the resulting cuboid.

A
$$110cm^2$$

B
$$100cm^2$$

C
$$260cm^2$$

D
$$160cm^2$$

Solution

length of cube $$= a$$
$$a^3 = 64$$
$$\Rightarrow a = 4$$
Length of cuboid = $$ 8 = l$$
breadth of cuboid = $$4 = b$$
height of cuboid = $$4 = h$$
Ref. image
Surface area $$\Rightarrow 2(lb + bh + lh)$$
$$= 2 ((8 \times 4) + (4 \times 4) + (8 \times 4))$$
$$= 2 (32 + 16 + 32)$$
$$= 160\ cm^2$$
Question 4
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A hollow iron pipe is $$21\,cm$$ long and its external diameter is $$8\,cm$$. If the thickness of the pipe is $$1\,cm$$ and iron weighs $$8\,g/c{m^3}$$, then the weight of pipe is :

A
$$3.6\,kg$$

B
$$3.696\,kg$$

C
$$36\,kg$$

D
$$36.9\,kg$$

Solution

Volume of iron $$=\pi(R^2-r^2)\times h$$

$$=\pi(4^2-3^2)\times 21$$

$$=\cfrac{22}{7}\times(16-9)\times 21$$

$$=462cm^2$$

Weight $$=\cfrac{8\times 462}{1000}kg=3.696kg$$
Question 5
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12 sphere of the same size are made from melting a solid cylinder of 16 cm diameter and 2 cm height. The diameter of each sphere is:

A
$$\sqrt 3 \ cm$$

B
$$2 \ cm$$

C
$$3 \ cm$$

D
$$4 \ cm$$

Solution

Given that, $$12$$ spheres of same size are made by melting a cylinder of $$16 \ cm$$ diameter and $$2\ cm$$ height.
to find out: The diameter of each sphere.

Since, the spheres are made by melting the cylinder, the volume of $$12$$ spheres $$=$$ the volume of cylinder
We know that, volume of cylinder $$=\pi r^2h$$
and volume of sphere $$=\frac{4}{3}\pi r^3$$

$$\therefore \ 12\times \left(\dfrac{4}{3}\pi r^3\right)=\pi(8)^2\times 2=128\pi$$

$$\Rightarrow r^3=\dfrac{128\times 3}{12\times 4}=8$$

$$\Rightarrow r^3=(2)^3$$

$$\therefore \ r=2\ cm$$

Hence, diameter of each sphere $$=2r=4\ cm$$.
Hence, option D is correct.
Question 6
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A metallic bucket, open at the top, of height $$24\ cm$$ is in the form of the frustum of a cone, the radii of whose lower and upper circular ends are $$7\ cm$$ and $$14\ cm$$ respectively. Find:
(i) the volume of water which can completely fill the bucket.
(ii) the area of metal sheet used to make the bucket. [Use $$\pi=\dfrac{22}{7}$$]

A
$$8622$$

B
$$8624$$

C
$$8660$$

D
$$8824$$

Solution

Volume of water$$=$$Volume of frustum of cone
Given $$h=24$$cm
$$R=14$$cm
$$r=17$$cm
Volume of frustum of cone
$$=\dfrac{\pi h}{3}(R^2+Rr+r^2)$$
$$=\dfrac{22\times 24}{7\times 3}((14)^2+14\times 7+(7)^2)$$
$$=\dfrac{22\times 24}{7\times 3}[196+98+49]$$
$$=\dfrac{22\times 8}{7}[343]$$
$$=22\times 8\times 49$$
$$=8624cm^3$$.
Question 7
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The radius of the base of a cone is $$4\ \text{cm}$$ and the height is $$3\ \text{cm}$$. Find its Curved Surface area. $$\left(\pi=\dfrac{22}{7}\right)$$

A
$$20\ \text{cm}^{2}$$

B
$$20\pi\ \text{cm}^{2}$$

C
$$30\pi\ \text{cm}^{2}$$

D
None of these

Solution

Curved Surface area of cone = $$\pi \, r \,l$$,
where $$r$$ is the radius of the cone,
$$l$$ is the slant height of the cone
By Pythagoras theorem, slant height $$l$$ can be calculated as follows:
$$l=\sqrt{r^2+h^2}$$
$$=\sqrt{4^{2}+3^{2}}$$
$$=\sqrt{25}$$
$$=5$$
Curved Surface Area $$=\pi rl= \dfrac{22}{7}\times 5\times 4\ \text{cm}^2$$
$$=62.85 \, \text{cm}^{2}$$
$$=20 \pi \, \text{cm}^{2}$$
Question 8
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The side of a solid metallic cube is $$50\ \text{cm}$$. it is melted and recast into $$8000$$ similar solid cubical dies. Find the side of each die.

A
$$2.5\ \text{cm}$$

B
$$6\ \text{cm}$$

C
$$1.9\ \text{cm}$$

D
$$5.2\ \text{cm}$$

Solution

According to the statement,
Volume of big cube $$=8000\times $$ Volume of small cube.

Let $$x$$ be the side of small cube,

$$(50)^3=8000(x^3)$$

$$x^3=\dfrac{125000}{8000}$$

$$\implies x=\dfrac{50}{20}=2.5\ \text{cm}$$
Hence, the length of side of each small die will be $$2.5\ \text{cm}$$.
Question 9
1 / -0

The slant height of a cone is fixed at 7 cm. The rate of increase in the volume of the cone corresponding to the rate of increase of 0.3 cm/s in the height when h = 4 cm is

A
$$\dfrac {33\pi}{10}cc/s$$

B
$$\dfrac {3\pi}{10}cc/s$$

C
$$\dfrac {\pi}{5}cc/s$$

D
$$\dfrac {7\pi}{10}cc/s$$

Solution

Using Pythagoras Theorem on the triangle formed, we have,
$${ r^{ 2 } }+{ 4^{ 2 } }={ 7^{ 2 } } \\ { r^{ 2 } }=49-16 \\ { r^{ 2 } }=33$$
Volume of the cone $$=\dfrac { 1 }{ 3 } \pi { r^{ 2 } }h \\ $$

Rate of increase in volume $$=\dfrac { 1 }{ 3 } \pi { r^{ 2 } }\times \left( \text{ rate of } h \right)$$
$$=\dfrac { 1 }{ 3 } \pi\times 33\times 0.3\, \text{ cm/s}$$
$$=\pi\times 3.3\, \, \text{ cm}^{3}/\text{sec}$$
$$=\dfrac{33\pi}{10} \text{cm}^3/\text{sec} $$
Question 10
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A cylindrical can of internal diameter 21 cm contains water. A solid sphere whose diameter is 10.5 cm is lowered into the cylindrical can. The sphere is completely immersed in water. Calculate the rise in water level, assuming that no water overflows.

A
$$1\ cm$$

B
$$2.5\ cm$$

C
$$1.75\ cm$$

D
$$3\ cm$$

Solution

Given that, a solid hemisphere of diameter $$10.5\ cm$$ is lowered into a cylindrical can of diameter $$21\ cm$$, containing water.
To find out: The rise in water level in the can.

Let $$h_1$$ be the rise in level of water.

We know that, the volume of a cylinder is $$\pi r^2h$$
$$\therefore \ $$ Volume of water rise when the sphere is immersed $$=\pi r^2h_1=\pi (10.5)^2h_1~~~~~~-(1)$$

Volume of sphere $$=\dfrac{4}{3}\pi r^3=\dfrac{4}{3}\pi(5.25)^3~~~~~-(2)$$

The volume of the water rise $$=$$ volume of the sphere

Hence, From $$(1)$$ and $$(2)$$, we get
$$ \pi(10.5)^2h_1=\dfrac{4}{3}\pi (5.25)^3\\$$
$$\Rightarrow h_1=\dfrac{4\times (5.25)^3}{3\times (10.5)^2}=\dfrac{578.81}{330.75}\\$$
$$\Rightarrow h_1=1.75\ cm$$

Hence, the rise in level of water is $$1.75\ cm$$.

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Surface Areas and Volumes Test - 47

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Surface Areas and Volumes Test - 47

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Question 1

$$500.26\ m$$

$$224.26\ m$$

$$824.26\ m$$

None of these

Solution

Question 2

$$20\ cm$$

$$12\sqrt5\ cm$$

$$8\sqrt13\ cm$$

$$16\ cm$$

Solution

Question 3

$$110cm^2$$

$$100cm^2$$

$$260cm^2$$

$$160cm^2$$

Solution

Question 4

$$3.6\,kg$$

$$3.696\,kg$$

$$36\,kg$$

$$36.9\,kg$$

Solution

Question 5

$$\sqrt 3 \ cm$$

$$2 \ cm$$

$$3 \ cm$$

$$4 \ cm$$

Solution

Question 6

$$8622$$

$$8624$$

$$8660$$

$$8824$$

Solution

Question 7

$$20\ \text{cm}^{2}$$

$$20\pi\ \text{cm}^{2}$$

$$30\pi\ \text{cm}^{2}$$

None of these

Solution

Question 8

$$2.5\ \text{cm}$$

$$6\ \text{cm}$$

$$1.9\ \text{cm}$$

$$5.2\ \text{cm}$$

Solution

Question 9

$$\dfrac {33\pi}{10}cc/s$$

$$\dfrac {3\pi}{10}cc/s$$

$$\dfrac {\pi}{5}cc/s$$

$$\dfrac {7\pi}{10}cc/s$$

Solution

Question 10

$$1\ cm$$

$$2.5\ cm$$

$$1.75\ cm$$

$$3\ cm$$

Solution

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