Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

If radius of sphere is doubled, what is the ratio of volume of original sphere to that of second?

A
$$1 : 4$$

B
$$8 : 1$$

C
$$1 : 8$$

D
None

Solution

Given, radius of a sphere is doubled, i.e. $$\displaystyle r=2r$$
Volume of sphere $$\displaystyle =\frac { 4 }{ 3 } \pi { r }^{ 3 }$$ (original)
Vol of new sphere $$\displaystyle =\frac { 4 }{ 3 } \pi { \left( 2r \right) }^{ 3 }=\frac { 4 }{ 3 } \pi { \left( 8 \right) r }^{ 3 }$$
Volume of original sphere / vol. of second sphere $$\displaystyle =\frac { { 4 }/{ 3 }\pi { r }^{ 3 } }{ { 4 }/{ 3 }\pi { r }^{ 3 }\left( 8 \right) } =\frac { 1 }{ 8 } $$
Question 2
1 / -0

Fill in the blank:
____________ is the measure of the amount of space inside of a solid figure, like a cube, ball, cylinder or pyramid.

A
Mass

B
Volume

C
Capacity

D
Density

Solution

Volume is the measure of the amount of space inside of a solid figure, like a cube, ball, cylinder or pyramid.
Question 3
1 / -0

The volume of the global hemisphere is $$\displaystyle 19404{ \ in }^{ 3 }$$. Find its diameter.

A
$$21$$ in

B
$$42$$ in

C
$$10.5$$ in

D
$$9$$ in

Solution

Volume of the hemisphere $$\displaystyle =\frac { 2 }{ 3 } \pi { r }^{ 3 }$$
$$\displaystyle 19404=\frac { 2 }{ 3 } \times \frac { 22 }{ 7 } \times { r }^{ 3 }$$
$$\displaystyle 19404\times 3\times \dfrac {7}{( 2\times 22 )} ={ r }^{ 3 }$$
$$\displaystyle \dfrac {407484}{44}={ r }^{ 3 }$$
$$\displaystyle { r }^{ 3 }=9261$$
$$\displaystyle r=\sqrt [ 3 ]{ 9261 } $$
$$\displaystyle r=21$$ in
Diameter $$= 2 \times $$ radius
Diameter $$=2 \times 21$$
Diameter $$=$$ $$42$$ in
Question 4
1 / -0

The area of the hemisphere is $$\displaystyle 100{ m }^{ 2 }$$. Find its volume. (use $$\displaystyle \pi =3.14$$).

A
$$\displaystyle 132.97{ m }^{ 3 }$$

B
$$\displaystyle 102.97{ m }^{ 3 }$$

C
$$\displaystyle 132.97m$$

D
$$\displaystyle 132.97{ cm }^{ 3 }$$

Solution

Area of hemisphere $$=\displaystyle 2\pi { r }^{ 2 }$$
$$\Rightarrow \displaystyle 100=2\times 3.14\times { r }^{ 2 }$$
$$\Rightarrow \displaystyle \dfrac {100}{( 2\times 3.14 ) }={ r }^{ 2 }$$
$$\Rightarrow \displaystyle R=\sqrt { 15.923 } $$
$$\Rightarrow \displaystyle R=3.99$$ m
Volume of the hemisphere $$=$$ $$\displaystyle \frac { 2 }{ 3 } \pi { r }^{ 3 }$$
$$\displaystyle =\frac { 2 }{ 3 } \times 3.14\times 3.99\times 3.99\times 3.99$$
$$\displaystyle =132.97{ m }^{ 3 }$$
Question 5
1 / -0

Fill in the blank: The volume of solid objects is measured in _____ units.

A
Cubic

B
Square

C
Meter

D
Liter

Solution

The volume of solid objects is measured in cubic units.
Question 6
1 / -0

It cost Rs $$4020$$ to paint the inner curved surface area of hemisphere of radius $$8 m$$. If it is painted at rate of Rs.$$10\: per\: m^2$$. Find inner curved surface.

A
$$\displaystyle 402{ m }^{ 2 }$$

B
$$\displaystyle 400{ m }^{ 2 }$$

C
$$\displaystyle 200{ m }^{ 2 }$$

D
$$\displaystyle 201{ m }^{ 2 }$$

Solution

CSA of hemisphere $$\displaystyle =2\pi { r }^{ 2 }$$
$$\displaystyle =2\pi { \left( 8 \right) }^{ 2 }=2\pi \times 64=402$$
Question 7
1 / -0

If lateral surface area of a cylinder of height 10 cm is 100 cm $$^{ 2 }$$, find radius of its base.

A
$$\displaystyle 5\pi $$

B
$$\displaystyle \frac {5}{\pi}$$

C
$$\displaystyle 6\pi $$

D
$$\displaystyle \frac { \pi }{ 5 } $$

Solution

Given that, the lateral surface area of a cylinder of height $$10\ cm$$ is $$100\ cm^2$$.
To find out: The radius of the base of the cylinder.

We know that, lateral surface area of a cylinder is $$\displaystyle 2\pi rh$$
$$\therefore \ 2\pi rh=100\\$$
$$\Rightarrow 2\pi r (10)=100\\$$
$$\Rightarrow 20\pi r =100\\$$
$$\Rightarrow r =\dfrac{100}{20\pi}\\$$
$$\therefore \ \displaystyle r=\dfrac { 5 }{ \pi }$$

Hence, the radius of the base of the given cylinder is $$\dfrac { 5 }{ \pi }$$.
Question 8
1 / -0

A brick whose length, breadth and height are $$5m, 6m$$, and $$7m$$ respectively. Find the surface area of the brick.

A
$$214\ m^2$$

B
$$202\ m^2$$

C
$$201\ m^2$$

D
$$210\ m^2$$

Solution

Surface area of the cuboid having length, breadth and height as $$l, b$$ and $$h$$ respectively is given as $$2 (lb + bh + hl)$$

Brick has the shape of cuboid. Let $$S$$ be the require surface area.

$$S=2 (5 \times 6 + 6 \times 7 + 7 \times 5)$$

$$2 (30 + 42 + 35)$$

$$214 m^2$$
Question 9
1 / -0

A swimming pool is $$200$$ft long and $$14$$ ft wide and its depth being $$20$$ ft respectively. Find the volume of the pool.

A
$$55,000 ft^{3}$$

B
$$54,000 ft^{3}$$

C
$$56,000 ft^{3}$$

D
$$57,000 ft^{3}$$

Solution

Volume of the pool $$= l \times b \times h$$
$$= 200 \times 14 \times 20$$
$$= 56,000 ft^{3}$$
Question 10
1 / -0

A swimming pool is 100 m long and 15 m wide its deep and 24 m height respectively. Find the capacity of the pool.

A
36,000 m

B
36,000 $$m^3$$

C
36,000 $$m^2$$

D
36,000 $$cm$$

Solution

Volume of the pool = $$l \times b \times h$$
= $$100 \times 15 \times 24$$
= $$36000 m^3$$

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Re-Attempt Weekly Quiz Competition

Surface Areas and Volumes Test - 41

Result

Surface Areas and Volumes Test - 41

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SHARING IS CARING

Question 1

$$1 : 4$$

$$8 : 1$$

$$1 : 8$$

None

Solution

Question 2

Mass

Volume

Capacity

Density

Solution

Question 3

$$21$$ in

$$42$$ in

$$10.5$$ in

$$9$$ in

Solution

Question 4

$$\displaystyle 132.97{ m }^{ 3 }$$

$$\displaystyle 102.97{ m }^{ 3 }$$

$$\displaystyle 132.97m$$

$$\displaystyle 132.97{ cm }^{ 3 }$$

Solution

Question 5

Cubic

Square

Meter

Liter

Solution

Question 6

$$\displaystyle 402{ m }^{ 2 }$$

$$\displaystyle 400{ m }^{ 2 }$$

$$\displaystyle 200{ m }^{ 2 }$$

$$\displaystyle 201{ m }^{ 2 }$$

Solution

Question 7

$$\displaystyle 5\pi $$

$$\displaystyle \frac {5}{\pi}$$

$$\displaystyle 6\pi $$

$$\displaystyle \frac { \pi }{ 5 } $$

Solution

Question 8

$$214\ m^2$$

$$202\ m^2$$

$$201\ m^2$$

$$210\ m^2$$

Solution

Question 9

$$55,000 ft^{3}$$

$$54,000 ft^{3}$$

$$56,000 ft^{3}$$

$$57,000 ft^{3}$$

Solution

Question 10

36,000 m

36,000 $$m^3$$

36,000 $$m^2$$

36,000 $$cm$$

Solution

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