Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

The radius of the cylinder whose lateral surface area is $$704\ cm^{2}$$ and height $$8\ cm$$, is ___________.

A
$$6\ cm$$

B
$$4\ cm$$

C
$$8\ cm$$

D
$$14\ cm$$

Solution

Let radius of cylinder be $$r$$.

Lateral surface area$$=2\pi rh$$

$$\Rightarrow 704=2\pi r \times 8$$

$$\Rightarrow r=\dfrac{704}{16\pi}=14$$

$$\therefore $$ Radius of cylinder is $$14$$ $$cm$$.
Question 2
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Find the total surface area of a cylinder with diameter of base $$7\ cm$$ and height $$40\ cm$$.

A
$$1016\ cm^{2}$$

B
$$880\ cm^{2}$$

C
$$1540\ cm^{2}$$

D
$$957\ cm^{2}$$

Solution

we k.n.t total surface area of cylinder $$=2\pi r(r+h) $$.

Given $$d=7\ cm$$ and $$h=40\ cm$$ .

Total surface area $$=2\pi r(r+h)$$
$$=2\pi\times 3.5\times(3.5+40)=956.13$$
$$\therefore T.S.A=957$$ $$cm^2$$.
Question 3
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If the radius of a sphere is doubled, then its volume is increase by __________.

A
$$100\%$$

B
$$200\%$$

C
$$700\%$$

D
$$800\%$$

Solution

Given that radius of the sphere is doubled.
Let the Initial volume of sphere $$V=\dfrac {4}{3}\pi r^3$$
Incresed volume $$V'=\dfrac{4}{3}\pi (2r)^3=8V$$
Thus, Volume is increased by:
$$\dfrac{8V-V}{V}\times 100=700 \%$$
Question 4
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If the radius of a sphere is increased by $$2$$cm, then its surface area increases by $$352cm^2$$. The radius of the sphere before the increase was _________.

A
$$3$$cm

B
$$4$$cm

C
$$5$$cm

D
$$6$$cm

Solution

Let the radius of the sphere before increase$$=r$$.

Given, increased surface area $$=352cm^2$$.

Also, the surface area of the sphere before the increase is $$A=4\pi r^2$$.

After increasing radius by $$2 \ cm$$, the area is increased by $$352 \ cm^2$$.

Applying the above conditions:
$$\therefore 4\pi r^2+352=4\pi (r +2)^2\\$$

$$ \Rightarrow 352=16\pi+16\pi r$$

$$ \Rightarrow 1+r=\dfrac{352}{16\pi }=7$$

$$ \Rightarrow r=6cm$$.

Hence, option $$D$$ is correct.
Question 5
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The largest sphere is cut off from a cube of side $$5\ cm$$. The volume of the sphere will be __________.

A
$$27\pi \ cm^3$$

B
$$30\pi\ cm^3$$

C
$$108\pi\ cm^3$$

D
$$\displaystyle\frac{125\pi}{6}\ cm^3$$

Solution

Diameter of the largest sphere= side of the cube $$=5\ cm $$
$$\therefore $$ volume of the sphere
$$=\dfrac{4}{3}\pi r^3=\dfrac{4}{3}\pi \times (\dfrac{5}{2})^3\\=\dfrac{125\pi}{6}\ cm ^3 $$
Question 6
1 / -0

Calculate the surface area of a sphere whose radius is $$21 cm$$.

A
$$5544\ { cm }^{ 2 }$$

B
$$6622\ { cm }^{ 2 }$$

C
$$7733\ { cm }^{ 2 }$$

D
$$4455\ { cm }^{ 2 }$$

Solution

Given, radius $$r=21 cm$$.

We know, the surface area of a sphere of radius $$r$$ $$ = 4\pi { r }^{ 2 } $$

$$= 4 \times \dfrac {22}{7} \times 21 \times 21 $$

$$= 5544 {cm}^{2} $$.
Therefore, option $$A$$ is correct.
Question 7
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If the volume of a sphere is divided by its surface area, we obtain 27 cm. The radius of the sphere is

A
9 cm

B
81 cm

C
27 cm

D
24 cm

Solution
Question 8
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The total surface area of a sphere is, $$154\ { cm }^{ 2 }$$. Its radius is 3.5 cm. Find its volume.

A
$$169.77\ { cm }^{ 3 }$$

B
$$189.07\ { cm }^{ 3 }$$

C
$$159.27\ { cm }^{ 3 }$$

D
$$179.67\ { cm }^{ 3 }$$

Solution

solution:
volume of sphere is: $$\dfrac{4}{3} \pi r^3$$
V = $$\dfrac{4}{3} \times3.14 \times(3.5)^3$$
$$V = 179.5 cm^3$$
hence the correct opt: D
Question 9
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The volume of a sphere is $$\dfrac {88}{21}\times (14)^{3} cm^{3}$$. The curved surface of the sphere is (Take $$\pi = \dfrac {22}{7}$$).

A
$$2424\ cm^{2}$$

B
$$2446\ cm^{2}$$

C
$$2484\ cm^{2}$$

D
$$2464\ cm^{2}$$

Solution

Given that,
Volume of sphere $$= \dfrac {88}{21} \times (14)^{3}$$
We know that,
Volume of the sphere $$= \dfrac {4}{3} \pi r^{3}$$
$$\dfrac {4}{3}\pi r^{3} = \dfrac {88}{21}\times (14)^{3}$$
$$\Rightarrow \dfrac {4}{3}\times \dfrac {22}{7} \times r^{3} = \dfrac {88}{21} \times (14)^{3}$$
$$r^{3}\Rightarrow (14)^{3} \Rightarrow r = 14$$
We know that,
Curved surface of the sphere $$= 4\pi r^{2}$$
$$= 4\times \dfrac {22}{7}\times 14 \times 14 = 2464\ cm^{2}$$.
Question 10
1 / -0

The number of balls of radius $$1$$ cm that are made from a solid sphere of radius $$4$$ cm

A
$$64$$

B
$$16$$

C
$$12$$

D
$$4$$

Solution

$$No\quad of\quad balls=\cfrac { Volume\quad of\quad solid\quad sphere }{ volume\quad of\quad ball } =\cfrac { \cfrac { 4 }{ 3 } \pi .4\times 4\times 4 }{ \cfrac { 4 }{ 3 } \pi \times 1\times 1\times 1 } =64\quad balls$$

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Re-Attempt Weekly Quiz Competition

Surface Areas and Volumes Test - 46

Result

Surface Areas and Volumes Test - 46

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SHARING IS CARING

Question 1

$$6\ cm$$

$$4\ cm$$

$$8\ cm$$

$$14\ cm$$

Solution

Question 2

$$1016\ cm^{2}$$

$$880\ cm^{2}$$

$$1540\ cm^{2}$$

$$957\ cm^{2}$$

Solution

Question 3

$$100\%$$

$$200\%$$

$$700\%$$

$$800\%$$

Solution

Question 4

$$3$$cm

$$4$$cm

$$5$$cm

$$6$$cm

Solution

Question 5

$$27\pi \ cm^3$$

$$30\pi\ cm^3$$

$$108\pi\ cm^3$$

$$\displaystyle\frac{125\pi}{6}\ cm^3$$

Solution

Question 6

$$5544\ { cm }^{ 2 }$$

$$6622\ { cm }^{ 2 }$$

$$7733\ { cm }^{ 2 }$$

$$4455\ { cm }^{ 2 }$$

Solution

Question 7

9 cm

81 cm

27 cm

24 cm

Solution

Question 8

$$169.77\ { cm }^{ 3 }$$

$$189.07\ { cm }^{ 3 }$$

$$159.27\ { cm }^{ 3 }$$

$$179.67\ { cm }^{ 3 }$$

Solution

Question 9

$$2424\ cm^{2}$$

$$2446\ cm^{2}$$

$$2484\ cm^{2}$$

$$2464\ cm^{2}$$

Solution

Question 10

$$64$$

$$16$$

$$12$$

$$4$$

Solution

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