Surface Areas and Volumes Test

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Surface Areas and Volumes Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

Which part in the cylinder is included for total surface area?

A
curved surface area

B
curved surface area and two circular bases

C
one circular bases

D
two circular bases

Solution

Total surface area = Curved surface area + Area of two circular bases.
Question 2
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A ball in the shape of a sphere has a surface area of $$221.76$$ cm$$^2$$. Calculate its diameter.

A
$$8.4cm$$

B
$$7.4cm$$

C
$$9.4cm$$

D
None of these

Solution

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

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

$$\implies$$ $$221.76 cm^2 =4\pi { r }^{ 2 }$$

$$\implies$$ $$ { { r }^{ 2 } }=\cfrac { 221.76 }{ 4\pi } =\dfrac{221.76\times 7}{4\times 22}=17.64$$

$$\implies$$ $$ r=\sqrt { 17.64 } =4.2 cm$$ .
Then, diameter of the sphere, $$d=2r=2\times4.2=8.4 cm$$.

Therefore, option $$A$$ is correct.
Question 3
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Find the radius of a solid sphere that has a volume of $$10$$ m$$^3$$.

A
$$1.336 m$$

B
$$3.336 m$$

C
$$2.336 m$$

D
None of these

Solution

$$ V = \dfrac{4}{3}\pi r^{3} $$
$$ \Rightarrow r = 3 \sqrt{\dfrac{3V}{4\pi }} = \left ( \dfrac{3\times 10}{4\pi} \right )^{1/3} $$
$$ \Rightarrow \boxed{r = 1.336\,m} $$
Question 4
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Find the radius of a hemisphere with the volume of $$20$$ cm$$^3$$. $$(\pi = 3.142)$$

A
$$2.121 \text{ cm}$$

B
$$4.121 \text{ cm}$$

C
$$5.121 \text{ cm}$$

D
None of these

Solution

Volume of hemisphere is $$ V = \dfrac{2}{3}\pi r^{3} $$
$$ \Rightarrow r = \left(\dfrac{3V}{2\pi }\right)^{\dfrac{1}{3}}$$
$$ \Rightarrow r = \left ( \dfrac{3\times 20}{2\pi} \right )^{\dfrac{1}{3}} $$
$$ \Rightarrow r = 2.121\,\text{ cm} $$
Question 5
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Find the volume of the sphere whose radius is $$5\sqrt 3$$ cm.

A
$$2127.72$$ cm$$^3$$

B
$$2720.72$$ cm$$^3$$

C
$$2721.71$$ cm$$^3$$

D
$$2727.27$$ cm$$^3$$

Solution

$$ V = \dfrac{4}{3}\pi r^{3} = \dfrac{4}{3}\pi (5\sqrt{3})^{3} = \boxed{2720.7\,cm^{3}} $$
Question 6
1 / -0

The radius of a spherical balloon increases from $$7$$ cm to $$14$$ cm as air is being pumped into it. Find ratio of surface areas of the balloon in the two cases.

A
$$\dfrac{3}{4}$$

B
$$\dfrac{1}{4}$$

C
$$\dfrac{1}{5}$$

D
None of these

Solution

We know, the surface area of a sphere is $$4\pi r^{2}$$, where $$r$$ is the radius.
Let $$r_1$$ be the original radius and $$r_2$$ be the new radius.
Then, $$r_1=7$$ and $$r_2=14$$.

Therefore, the ratio of the surface area will be:
$$=\dfrac { 4\pi { r }_{ 1 }^{ 2 } }{ 4\pi { r }_{ 2 }^{ 2 } } =\dfrac { { r }_{ 1 }^{ 2 } }{ { r }_{ 2 }^{ 2 } } =\dfrac { { 7 }_{ }^{ 2 } }{ { 14 }_{ }^{ 2 } } =\dfrac { 1 }{ 4 } $$
Hence, the required ratio is $$\dfrac{1}{4}$$.
Therefore, option $$B$$ is correct.
Question 7
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The volume of sphere with radius $$3cm$$ is ................. $${cm}^{3}$$.

A
$$14\pi $$

B
$$18\pi $$

C
$$2\pi $$

D
$$36\pi $$

Solution

Given : Radius$$(r)=3\ cm$$
We know, Volume of a sphere $$=\cfrac { 4 }{ 3 } \pi { r }^{ 3 }$$
$$\implies V=\cfrac { 4 }{ 3 } \times \pi \times 3\times 3\times 3$$
$$\implies V=4\times \pi \times 3\times 3=36\pi $$
Hence, volume of a sphere is $$36\pi$$.
Question 8
1 / -0

The surface area of a solid hemisphere is

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

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

C
$$\dfrac { 4 }{ 3 } \pi { r }^{ 2 }$$

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

Solution

Surface area of a circle $$=\pi r^2$$
Since, the hemisphere is solid, then it will include the base too.
$$\therefore$$ surface area of a solid hemisphere $$=\pi r^2+2\pi r^2=3\pi r^2$$
Hence, option D is correct.
Question 9
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The formula to find the total surface area of a Rs.$$5$$ coin is ..............

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

B
$$\pi r(r+h)$$

C
$$\pi { r }^{ 3 }h$$

D
$$2\pi r(h+r)$$

Solution

A $$5$$ rupee coin is cylindrical in shape having radius $$r$$ and haight $$h$$.
TSA of a cylinder is given by $$2\pi r(r+h)$$
Thus, the total surface area of the cylinder can be calculated by using the formula $$2\pi r(h+r)$$
Hence, option D is correct.
Question 10
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The formula of lateral surface area of cylinder is ________.

A
$$A = 2\pi r h$$

B
$$A = 2\pi r$$

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

D
$$A = 2\pi r^{2}h$$

Solution

Lateral surface area and cylinder $$2\pi rh$$
$$\therefore $$ Part (A) is correct answer.

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Re-Attempt Weekly Quiz Competition

Surface Areas and Volumes Test - 21

Result

Surface Areas and Volumes Test - 21

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SHARING IS CARING

Question 1

curved surface area

curved surface area and two circular bases

one circular bases

two circular bases

Solution

Question 2

$$8.4cm$$

$$7.4cm$$

$$9.4cm$$

None of these

Solution

Question 3

$$1.336 m$$

$$3.336 m$$

$$2.336 m$$

None of these

Solution

Question 4

$$2.121 \text{ cm}$$

$$4.121 \text{ cm}$$

$$5.121 \text{ cm}$$

None of these

Solution

Question 5

$$2127.72$$ cm$$^3$$

$$2720.72$$ cm$$^3$$

$$2721.71$$ cm$$^3$$

$$2727.27$$ cm$$^3$$

Solution

Question 6

$$\dfrac{3}{4}$$

$$\dfrac{1}{4}$$

$$\dfrac{1}{5}$$

None of these

Solution

Question 7

$$14\pi $$

$$18\pi $$

$$2\pi $$

$$36\pi $$

Solution

Question 8

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

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

$$\dfrac { 4 }{ 3 } \pi { r }^{ 2 }$$

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

Solution

Question 9

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

$$\pi r(r+h)$$

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

$$2\pi r(h+r)$$

Solution

Question 10

$$A = 2\pi r h$$

$$A = 2\pi r$$

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

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

Solution

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