Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

The number of balls of radius 1 cm that can be made from a sphere of radius 10 cm will be

A
1000

B
10000

C
100000

D
100

Solution

Given the radius of ball is 1 cm and radius of sphere is 10cm
Then volume of sphere=$$\frac{4}{3}\pi (10^{3})=\frac{4000}{3}\pi cm^{3}$$
And volume of ball=$$\frac{4}{3}\pi (1^{3})=\frac{4}{3}\pi cm^{3}$$
Then number of ball made=$$\frac{\frac{4000}{3}\pi }{\frac{4}{3}\pi }=1000$$
Question 2
1 / -0

The total surface area of a hemisphere of radius r is given by

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

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

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

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

Solution

The surface area of a sphere of radius r is $$4\pi r^{2}$$. Half of this is $$2\pi r^{2}$$.
If you have a hemispherical object then it has a base which is a circle of radius r.
The area of a circle of radius r is $$\pi r^{2}$$ and thus if the hemisphere is meant to include the base then the surface area is $$2\pi r^{2}+\pi r^{2}=3\pi r^{2}$$.
Question 3
1 / -0

A square sheet of side $$28$$ cm is folded into a cylinder by joining its two sides. Find the base area of the cylinder thus formed (in $$\displaystyle cm^{2}$$).

A
$$\displaystyle \frac{524}{11}$$

B
$$\displaystyle \frac{575}{11}$$

C
$$\displaystyle \frac{629}{11}$$

D
$$\displaystyle \frac{686}{11}$$

Solution

As the square sheet is folded to form a cylinder, circumference of the base of the cylinder will be equal to $$ 28 $$ cm
Let $$R$$ be the radius of the cylinder
Circumference of base of cylinder $$ = 2 \pi R $$
So, $$ 2 \times \dfrac {22}{7} \times R = 28 $$
$$ \Rightarrow R = \dfrac {49}{11} $$
And area of the base of the cylinder $$ = \pi{R}^{2} $$
$$= \dfrac {22}{7} \times \dfrac {49}{11} \times \dfrac {49}{11} $$
$$ = \dfrac {686}{11} $$ sq.cm
Question 4
1 / -0

Three spheres of radii 6 cm, 8 cm and 10 cm are melted to form a sphere of radius

A
11 cm

B
13 cm

C
24 cm

D
none

Solution

Given the three sphere whose radii are 6 cm , 8 cm and 10 cm melted and made a new sphere
Then volume of fist sphere=$$\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi (6)^{3}=\frac{4\pi }{3}216 cm^{3}$$
And volume of second sphere=$$\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi (8)^{3}=\frac{4\pi }{3}512 cm^{3}$$
And volume of third sphere=$$\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi (10)^{3}=\frac{4\pi }{3}1000 cm^{3}$$
Then total volume of new sphere=$$\frac{4\pi }{3}216+\frac{4\pi }{3}512+\frac{4\pi }{3}1000=\frac{4\pi }{3}1728cm^{3}$$
Let the radius of new sphere is R
$$\frac{4}{3}\pi R^{3}=\frac{4\pi }{3}1728cm^{3}\Rightarrow R^{3}=1728\Rightarrow R=12 $$ cm
Question 5
1 / -0

The radius of a sphere is increased by $$50 \%$$, then the increase in surface area of a sphere is:

A
$$200 \%$$

B
$$150 \%$$

C
$$125 \%$$

D
$$50 \%$$

Solution

Let the radius of sphere is $$r$$.
Then surface area $$=4\pi r^{2}$$.
If radius increased by $$50\%$$ new radius $$=r\times \dfrac{150}{100}=\dfrac{3r}{2}$$.
So surface area of new sphere $$=4\pi (\dfrac{3r}{2})^{2}=4\times \dfrac{9}{4}\pi r^{2}=9\pi r^{2}$$.
Then increased in surface area of sphere $$=9\pi r^{2}-4\pi r^{2}=5\pi r^{2}$$.
So $$\%$$ increased in $$S.A=\dfrac{5\pi r^{2}}{4\pi r^{2}}=\dfrac{5}{4}=\dfrac{5}{4}\times\dfrac{100}{100}=\dfrac{5\times25}{100}=125\%$$.
Therefore, option $$C$$ is correct.
Question 6
1 / -0

If the volume in $$m^3$$ and the surface area in $$m^2$$ of a sphere are numerically equal, then the radius of the sphere in m is

A
4

B
2

C
3.5

D
3

Solution

Let the radius of the sphere be $$r$$.

Since, the volume and surface area of the sphere are equal.

Therefore,
$$4\pi r^2=\dfrac{4}{3}\pi r^3$$
$$r=3\ m$$

Hence, this is the answer.
Question 7
1 / -0

A cuboidal container has dimensions of $$20$$ cm $$\times$$ $$18$$ cm $$\times$$ $$16$$ cm. Find the maximum number of syrup bottles whose contents can be emptied into the container, if each bottle contains $$24$$ $$\displaystyle cm^{3}$$ of syrup.

A
$$120$$

B
$$180$$

C
$$240$$

D
$$270$$

Solution

As the container is cuboidal, its volume $$=$$ length $$\times$$ breadth $$\times $$ height

So, amount of syrup contained in it $$ = (20 \times 18 \times 16)\ cm^3 $$

Given: Each syrup bottle can contain $$ 24\ {cm}^{3} $$ of syrup

The number of bottles that can be filled $$ = \dfrac {20 \times 18 \times 16 }{24} = 240 $$
Question 8
1 / -0

Find the surface area of the sphere whose radius is $$35 \text{ cm}$$.

A
$$15400 \text{ cm}^{2}$$

B
$$1540 \text{ cm}^{2}$$

C
$$154 \text{ cm}^{2}$$

D
$$15.400 \text{ cm}^{2}$$

Solution

Given $$\Rightarrow$$ Radius of the sphere, $$r=35 \text{ cm}$$.

We know, Surface area of the sphere is equal to $$4\pi r^{2}.$$ So,
$$\begin{aligned}{}A &= 4\pi {r^2}\\ &= 4 \times \frac{{22}}{7} \times 35 \times 35\\& = 15400\text{ cm}^2\end{aligned}$$

Therefore, option $$A$$ is correct.
Question 9
1 / -0

The volume of a cube whose edge measures 12 m is ............ times the volume of a cuboid of dimensions 8 m $$\times$$ 6 m $$\times $$ 4 m

A
6

B
7

C
5

D
9

Solution

Volume of the cube $$= 12 \times 12 \times 12 = 1,728 $$ cu m
Volume of the cuboid $$= 8 \times 6 \times 4 = 192 $$ cu m
$$\displaystyle \frac{1728}{192} = 9 times$$
Question 10
1 / -0

Length, breadth and height of a room are $$12$$ ft, $$5$$ ft and $$8$$ ft respectively. Find the volume of the room.

A
$$\displaystyle 440{ ft }^{ 3 }$$

B
$$\displaystyle 480{ ft }^{ 3 }$$

C
$$\displaystyle 400{ ft }^{ 3 }$$

D
$$\displaystyle 500{ ft }^{ 3 }$$

Solution

Length of the room $$= 12$$ ft
Breadth of the room $$= 5$$ ft
Height of the room $$= 8$$ ft
Therefore, volume of the room $$=$$ length $$\times$$ breadth $$\times$$ height
$$\displaystyle =12\times 5\times 8{ ft }^{ 3 }$$
$$\displaystyle =480$$ $${ ft }^{ 3 }$$

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Re-Attempt Weekly Quiz Competition

Surface Areas and Volumes Test - 33

Result

Surface Areas and Volumes Test - 33

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SHARING IS CARING

Question 1

1000

10000

100000

100

Solution

Question 2

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

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

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

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

Solution

Question 3

$$\displaystyle \frac{524}{11}$$

$$\displaystyle \frac{575}{11}$$

$$\displaystyle \frac{629}{11}$$

$$\displaystyle \frac{686}{11}$$

Solution

Question 4

11 cm

13 cm

24 cm

none

Solution

Question 5

$$200 \%$$

$$150 \%$$

$$125 \%$$

$$50 \%$$

Solution

Question 6

4

2

3.5

3

Solution

Question 7

$$120$$

$$180$$

$$240$$

$$270$$

Solution

Question 8

$$15400 \text{ cm}^{2}$$

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

$$154 \text{ cm}^{2}$$

$$15.400 \text{ cm}^{2}$$

Solution

Question 9

6

7

5

9

Solution

Question 10

$$\displaystyle 440{ ft }^{ 3 }$$

$$\displaystyle 480{ ft }^{ 3 }$$

$$\displaystyle 400{ ft }^{ 3 }$$

$$\displaystyle 500{ ft }^{ 3 }$$

Solution

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