Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

Find the volume and surface area of a sphere of radius $$4.2$$ cm. $$\displaystyle \left [ \pi =\frac{22}{7} \right ]$$

A
$$380.46 \,\text{cm}^3,\:221.76 \,\text{cm}^2$$

B
$$320.46 \,\text{cm}^3,\:221.76 \,\text{cm}^2$$

C
$$310.46 \,\text{cm}^3,\:221.76 \,\text{cm}^2$$

D
$$370.46 \,\text{cm}^3,\:221.76 \,\text{cm}^2$$

Solution

Given, radius of sphere $$=4.2$$ cm
Volume of sphere $$ = \dfrac { 4 }{ 3 } \pi { r }^{ 3 } $$
$$= \dfrac {4}{3} \times \dfrac {22}{7} \times 4.2 \times 4.2 \times 4.2 = 310.46 \ \text{cm}^{3} $$
Surface area of a sphere of radius '$$r$$' $$ = 4\pi { r }^{ 2 } = 4 \times \dfrac {22}{7} \times 4.2 \times 4.2 = 221.76 \ \text{cm}^{2} $$
Question 2
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The total surface area of a solid right cylinder of radius $$r$$ and height $$h$$ is $$2\pi r(h+r)$$

A
True

B
False

C
Either

D
Neither

Solution

The curved surface area for a cylinder is $$2(\pi)rh$$
Total surface area = curved surface area +area of the two circles$$= 2(\pi)rh+2(\pi)r^2$$
$$=2(\pi)r(r+h)$$
Question 3
1 / -0

The volume of a solid is the measurement of the portion of the space occupied by it.
State True or False.

A
True

B
False

C
Data Insufficient

D
None of the above

Solution

The volume of a solid is defined as the space occupied by the solid shape in 3-dimensional region.

For example: Let us say that there is a glass of water; The amount of water the glass can hold is called volume of the glass.
Question 4
1 / -0

The volume of a sphere of diameter $$d$$ is:

A
$$\dfrac {\pi d^3}{3}$$

B
$$\dfrac {\pi d^3}{6}$$

C
$$\dfrac {2\pi d^3}{3}$$

D
$$\dfrac {\pi d^3}{4}$$

Solution

Volume of sphere $$=\cfrac 43(\pi)r^3=\cfrac 43(\pi)\left(\cfrac d2\right)^3 = \cfrac {\pi d^3}{6}$$
Question 5
1 / -0

The ratio of the volumes of two spheres is $$8 : 27$$. The ratio of their radii is

A
$$3 : 2$$

B
$$2 : 3$$

C
$$4 : 3$$

D
$$2 : 9$$

Solution

Given, ratio of volume of spheres $$= 8 : 27$$
Volume of sphere $$= \displaystyle \frac{4}{3} \pi r^3$$
Let $$R$$ be the radius of one sphere and $$r$$ be the radius of another sphere.
$$\therefore $$ Ratio $$=\dfrac{\frac{4}{3} \pi R^3}{\frac{4}{3} \pi r^3}$$ $$= 8:27$$
$$\therefore \dfrac{R^3}{r^3}$$ $$= 8:27$$
$$\therefore \dfrac{R}{r} = 2: 3$$
Question 6
1 / -0

Curved surface area of hemisphere of diameter $$2 r$$ is :

A
$$2\pi r^{2}$$

B
$$3\pi r^{2}$$

C
$$4\pi r^{2}$$

D
$$8\pi r^{2}$$

Solution

Diameter $$=2r$$
Radius $$=r$$
$$\therefore $$ Curved surface area of hemisphere $$=2\pi { r }^{ 2 }$$
Question 7
1 / -0

If the diameter of the sphere is doubled, the surface area of the resultant sphere becomes $$x$$ times that of the original one. Then, $$x$$ would be:

A
$$2$$

B
$$3$$

C
$$4$$

D
$$8$$

Solution

Let $$S_1$$ be the original surface area of the sphere and $$S_2$$ be the surface area with double the diameter.
Doubling the diameter means doubling the radius.
Let $$r_1$$ and $$r_2$$ corresponding radii.
We know that $$S = 4 \pi r^2$$ and $$r_2=2r_1$$ ....(Given)
Hence, $$\cfrac {S_1}{S_2} = \cfrac {4 \pi r_1^2}{4 \pi r_2^2}$$
$$\Rightarrow \cfrac {S_1}{S_2} = \cfrac {r_1^2}{(2r_1)^2}$$
$$\Rightarrow S_2 = 4S_1$$.

In complying with the question above we can say that $$x$$ is equivalent to $$4$$.
So, doubling radius will increase $$S$$ by a factor of $$4$$.
That is, option $$C$$ is correct.
Question 8
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If the curved surface area of a cylinder is $$1760$$ sq.cm and its base radius is $$14$$ cm, then its height is:

A
$$10$$ cm

B
$$15$$ cm

C
$$20$$ cm

D
$$40$$ cm

Solution

Given, curved surface varea of cone $$=1760$$ sq.cm and radius $$=14$$ cm
Curved surface area of a cylinder of radius "$$R$$" and height "$$h$$" $$ = 2\pi Rh$$
Hence, curved surface area of the given cylinder $$ = 2\times \dfrac {22}{7} \times 14\times h = 1760 $$sq.cm
$$ \therefore h = 20 $$ cm
Question 9
1 / -0

If radius of a sphere is doubled, how many times its volume will be affected?

A
2 times

B
4 times

C
6 times

D
8 times

Solution

$$\upsilon = \displaystyle \frac{4}{3} \pi r^3$$
Now new radius: $$r' = 2r$$
New volume: $$\upsilon' = \displaystyle \frac{4}{3} \pi (2r)^3$$
$$= 8 \displaystyle \left ( \frac{4}{3} \pi r^3 \right )$$
$$\upsilon' = 8 \upsilon$$.
Question 10
1 / -0

Find the volume of a sphere whose radius is $$7\ cm$$

A
$$1437\cfrac{1}{3}{cm}^{3}$$

B
$$1437\cfrac{1}{4}{cm}^{3}$$

C
$$1437\cfrac{2}{3}{cm}^{3}$$

D
$$1437\cfrac{1}{2}{cm}^{3}$$

Solution

radius of sphere$$=7cm$$
volume of sphere$$=\cfrac { 4 }{ 3 } \pi { r }^{ 3 }=\cfrac { 4 }{ 3 } \times \cfrac { 22 }{ 7 } \times 7\times 7\times 7=\cfrac { 4312 }{ 3 } cm\\ =1437\cfrac { 1 }{ 3 } { cm }^{ 3 }$$

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Re-Attempt Weekly Quiz Competition

Surface Areas and Volumes Test - 16

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Surface Areas and Volumes Test - 16

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Question 1

$$380.46 \,\text{cm}^3,\:221.76 \,\text{cm}^2$$

$$320.46 \,\text{cm}^3,\:221.76 \,\text{cm}^2$$

$$310.46 \,\text{cm}^3,\:221.76 \,\text{cm}^2$$

$$370.46 \,\text{cm}^3,\:221.76 \,\text{cm}^2$$

Solution

Question 2

True

False

Either

Neither

Solution

Question 3

True

False

Data Insufficient

None of the above

Solution

Question 4

$$\dfrac {\pi d^3}{3}$$

$$\dfrac {\pi d^3}{6}$$

$$\dfrac {2\pi d^3}{3}$$

$$\dfrac {\pi d^3}{4}$$

Solution

Question 5

$$3 : 2$$

$$2 : 3$$

$$4 : 3$$

$$2 : 9$$

Solution

Question 6

$$2\pi r^{2}$$

$$3\pi r^{2}$$

$$4\pi r^{2}$$

$$8\pi r^{2}$$

Solution

Question 7

$$2$$

$$3$$

$$4$$

$$8$$

Solution

Question 8

$$10$$ cm

$$15$$ cm

$$20$$ cm

$$40$$ cm

Solution

Question 9

2 times

4 times

6 times

8 times

Solution

Question 10

$$1437\cfrac{1}{3}{cm}^{3}$$

$$1437\cfrac{1}{4}{cm}^{3}$$

$$1437\cfrac{2}{3}{cm}^{3}$$

$$1437\cfrac{1}{2}{cm}^{3}$$

Solution

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