A closed cylindrical tank, made of thin iron sheet, had diameter $$8.4$$ m and height $$5.4$$ m. How much metal sheet to the nearest $$\displaystyle m^{2},$$ is used in making this tank, if $$\displaystyle \frac{1}{15}$$ of the sheet actually used was wasted in making the tank ?

$$270$$ $$\displaystyle m^{2}$$

$$272$$ $$\displaystyle m^{2}$$

$$271$$ $$\displaystyle m^{2}$$

$$242$$ $$\displaystyle m^{2}$$

MCQ Questions for Class 9 Maths Surface Areas And Volumes Quiz 3

Given a sphere with the radius

7 c m

$7\ cm$ . Find the volume of this sphere. Take

π

$\pi$ as

3.14

$3.14$ .

0%
$1436.03 \ cm^3$
0%
$87.92 \ cm^3$
0%
$1345.01 \ cm^3$
0%
$100.02 \ cm^3$

Explanation

Volume of sphere =

\frac{4}{3} π r^{3}

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

V = \frac{4}{3} (3.14) (7)^{3}

$V=\dfrac{4}{3} (3.14) (7)^3$

V = \frac{4}{3} \times 3.14 \times 343 = 1436.03 c m^{3}

$V=\dfrac{4}{3} \times 3.14 \times 343=1436.03 \ cm^3$

Calculate the surface area of a sphere with radius

3.2 c m

$3.2 cm$ .

0%
$110.65 \ cm^2$
0%
$128.61 \ cm^2$
0%
$131.54 \ cm^2$
0%
None of these

Explanation

Given, radius of the sphere,

r = 3.2 c m

$r=3.2 cm$ .

Therefore, the surface area of sphere

= 4 π r^{2}

$=4\pi r^2$

= 4 π (3.2)^{2} c m^{2}

$=4\pi (3.2)^2\ cm^2$

= 4 \times 3.14 \times 3.2 \times 3.2 c m^{2}

$=4 \times 3.14 \times 3.2 \times 3.2\ cm^2$

= 128.61 c m^{2}

$=128.61 \ cm^2$ .

Hence, option

B

$B$ is correct.

What is the total surface area of hemisphere?

0%
$4\pi r^{2}$
0%
$3\pi r^{2}$
0%
$\dfrac {2}{3}\pi r^{2}$
0%
$\dfrac {4}{3}\pi r^{2}$

Explanation

Let

r

$r$ be the radius of a hemisphere.

The total surface area of a hemisphere

=

$=$ flat surface area

+

$+$ curved surface area

We know that, the curved surface area of a hemisphere

= 2 π r^{2}

$=2\pi r^2$ and flat surface area

= π r^{2}

$=\pi r^2$

∴

$\therefore \$ Total surface area of a hemisphere

= 2 π r^{2} + π r^{2} = 3 π r^{2}

$=2\pi r^2+ \pi r^2=3\pi r^2$

Hence, the total surface area of hemisphere is

3 π r^{2}

$3\pi r^2$ .

State whether the statement are true (T) or false (F)
Two cylinders with equal volume will always have equal surface areas.

0%
True
0%
False

Explanation

The given statement is False.
Consider two cylinders with the following measures-
e.g.

r_{1} = 2 c m

$r_{1}= 2 cm$ , h= 9 cm, r=3 cm, h=4 cm
for the first cylinder,
Vol. =

π r^{2} h

$\pi r^{2}h$ =

π 2^{2} \times 9

$\pi 2^{2}\times 9$ =

36 π c m^{3}

$36 \pi cm ^{3}$
Again for the second cylinder,
Vol. =

π r^{2} h

$\pi r^{2}h$ =

π 3^{2} \times 4

$\pi 3^{2}\times 4$ =

36 π c m^{3}

$36 \pi cm ^{3}$
Now, surface area of first cylinder =

2 π r h

$2\pi rh$
=

2 π r \times 2 \times 9 = 36 π c m^{2}

$2\pi r \times2\times9 = 36 \pi cm ^{2}$
Surface area of second cylinder =

2 π r h

$2\pi rh$
=

2 π \times 3 \times 4 = 24 π c m 2

$2\pi \times3\times4 = 24 \pi cm{2}$
Therefore, the surface areas are not equal.

The curved surface area of a solid hemisphere of radius r is

0%
$4 \pi r^2$
0%
$2 \pi r^2$
0%
$4/3 \pi r^3$
0%
$3 \pi r^2$

Explanation

The curved surface area of a solid hemisphere of radius r is

2 π r^{2}

$2 \pi r^2$

The surface area of a solid hemisphere with radius

r

$r$ is

0%
$4\pi r^2$
0%
$2\pi r^2$
0%
$3\pi r^2$
0%
$\cfrac{2}{3} \pi r^3$

Explanation

Surface area of a solid sphere with radius

r

$r$ is

4 π r^{2}

$4 \pi r^2$
A solid sphere can be divided into two equal hemispheres with a flat surface and a curved surface.
Curved surface area of a solid hemisphere will be half the surface area of solid sphere.
Hence, curved surface area of solid hemisphere

= \frac{1}{2} * 4 π r^{2} = 2 π r^{2}

$=\dfrac{1}{2}*4 \pi r^2 = 2 \pi r^2$
And, flat surface area of hemisphere will be the area of circle with radius

r

$r$ .
Hence, flat surface area of hemisphere

= π r^{2}

$=\pi r^2$
Total surface area of a solid hemisphere with radius

r

$r$ is,

S = 2 π r^{2} + π r^{2} = 3 π r^{2}

$S=2\pi r^2+\pi r^2 = 3\pi r^2$

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratios of the surface areas of the balloon in the two cases is

0%
1 : 4
0%
1 : 3
0%
2 : 3
0%
2 : 1

Explanation

Total surface area of a Hemisphere

= 2 π r^{2} + π r^{2} = 3 π r^{2}

$=2\pi r^2 + \pi r^2 = 3\pi r^2$

r = 6, S u r f a c e a r e a = 3 π (6)^{2} = 36

$r=6,Surface\ area=3 \pi (6)^2=36$

r = 12, S u r f a c e a r e a = 3 π (12)^{2} = 144

$r=12,Surface\ area=3 \pi (12)^2=144$
Ratio =

36 : 144

$36:144$
=

1 : 4

$1:4$

The radius of a sphere is

2 r

$2r$ , then its volume will be:

0%
$\dfrac{4}{3} \pi r^{3}$
0%
$4 \pi r^{3}$
0%
$\dfrac{8 \pi r^{3}}{3}$
0%
$\dfrac{32}{3} \pi r^{3}$

Explanation

As,

r = 2 r

$r=2r$
Volume of sphere

= \frac{4}{3} π (2 r)^{3} = \frac{32}{3} π r^{3}

$=\frac{4}{3} \pi (2r)^3=\frac{32}{3} \pi r^3$

The amount of water displaced by a solid spherical ball of diameter

4.2

$4.2$ cm, when it is completely immersed in water, is

0%
$45.808$ cm $^3$
0%
$38.808$ cm $^3$
0%
$14.808$ cm $^3$
0%
$58.808$ cm $^3$

Explanation

The amount of water displaced by a solid spherical ball when it is completely immersed in water is equal to its volume.
Volume of a sphere of radius

r

$r$ is

\frac{4}{3} π r^{3}

$\dfrac { 4 }{ 3 } \pi { r }^{ 3 }$
As the diameter of the ball is

4.2

$4.2$ cm, its radius

r = 2.1

$r = 2.1$ cm
Hence, volume of water displaced

= \frac{4}{3} π r^{3} = \frac{4}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1 = 38.808

$= \dfrac { 4 }{ 3 } \pi { r }^{ 3 } = \dfrac { 4 }{ 3 } \times \dfrac { 22 }{ 7 } \times 2.1\times 2.1\times 2.1 = 38.808$ cm

^{3}

$^{ 3 }$

The water for a factory is stored in a hemispherical tank whose internal diameter is

14 m

$14\ m$ . The tank contains

50

$50$ kilo litres of water. Water is pumped into the tank to fill to its capacity. Calculate the volume of water pumped into the tank.

0%
$568.67\ m^3$
0%
$668.67\ m^3$
0%
$648.67\ m^3$
0%
$688.67\ m^3$

Explanation

Radius of hemispherical tank,

r = 7

$r= 7$ m
So volume of water to be pumped

=

$=$ Volume of hemispherical tank

- 50

$-50$ kl

∵ 1 k l = 1 m^{3}

$\because 1\ kl = 1\ m^3$

\Rightarrow \frac{2}{3} π r^{3} - 50 m^{3}

$\Rightarrow \dfrac{2}{3} \pi r^3 - 50\ m^3$

\Rightarrow \frac{2}{3} \times \frac{22}{7} \times {(7)}^{3} - 50 m^{3}

$\Rightarrow \dfrac{2}{3} \times \dfrac{22}{7}\times \left(7\right)^3 - 50\ m^3$

\Rightarrow 718.67 - 50 m^{3}

$\Rightarrow 718.67-50\ m^3$

\Rightarrow 668.67 m^{3}

$\Rightarrow 668.67\ m^3$

Hence, option B is correct.

If two solid hemispheres of same base radius r are joined together along their bases, then curved surface area of this new solid is

0%
$4\pi r^{2}$
0%
$6\pi r^{2}$
0%
$3\pi r^{2}$
0%
$8\pi r^{2}$

Explanation

We know that,

Curved surface area of a hemisphere

= 2 π r^{2}

$=2\pi r^2$

So, when two solids hemispheres of same base radius are joined together along their bases, then

Curved surface area of newly formed solid sphere

= 2 \times 2 π r^{2} = 4 π r^{2}

$=2\times2\pi r^2=4\pi r^2$

Hence, A is the correct option.

In a cylinder, radius is doubled and height is halved, curved surface area will

0%
halved
0%
doubled
0%
same
0%
four times

Explanation

Step 1: Using the given condition find Curved Surface Area of Cylinder

$\textbf{Step 1: Using the given condition find Curved Surface Area of Cylinder}$

Let us assume r is the radius of cylinder and h is the height of cylinder

$\text{Let us assume r is the radius of cylinder and h is the height of cylinder}$

∴ Curved surface area of cylinder is

$\therefore \text{Curved surface area of cylinder is}$

2 π r h

$2 \pi rh$ .

If the radius is doubled and height is halved

$\text{If the radius is doubled and height is halved}$

\Rightarrow R = 2 r, H = \frac{h}{2}

$\Rightarrow R=2r, H=\dfrac{h}{2}$

∴ Curved surface area of the cylinder

$\therefore \text{Curved surface area of the cylinder}$

= 2 π R H

$=2\pi RH$

= 2 π (2 r) (\frac{h}{2})

$=2\pi(2r)\left (\dfrac{h}{2}\right)$

= 2 π r h

$=2\pi rh$

Therefore, the curved surface area of the cylinder will be the same.

$\textbf{Therefore, the curved surface area of the cylinder will be the same.}$

The number of dimensions, a solid has :

0%
1
0%
2
0%
3
0%
0

Explanation

$\textbf{Step -1: Describing the dimensions in solid.}$

$\text{Dimension is a measurement of something in a particular direction.}$

$\text{And as a solid has length, breadth and height.}$

$\textbf{Step -2: Finding the number of dimensions in solid.}$

$\text{So, the number of dimensions in solid}=3.$

$\textbf{Hence, correct option is C.}$

If the surface area of a sphere is

144 π c m^{2}

$144 \pi \ cm^{2}$ , then its radius is:

0%
$6 cm$
0%
$8 cm$
0%
$12 cm$
0%
$10 cm$

Explanation

Let radius of sphere be

r

$r$ cm.

We know, surface area of sphere

= 4 π r^{2}

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

⟹

$\implies$

144 π = 4 π r^{2}

$144\pi =4\pi { r }^{ 2 }$

⟹

$\implies$

r^{2} = \frac{144 π}{4 π} = 36

${ { r }^{ 2 } }=\cfrac { 144\pi }{ 4\pi } =36$

⟹

$\implies$

r = \sqrt{36} = 6 c m

$r=\sqrt { 36 } =6 cm$ .

Therefore, option

A

$A$ is correct.

The curved surface area of a hemisphere is

77 c m^{2}

$77 cm^2$ . The radius of the hemisphere is :

0%
$3.5 cm$
0%
$7cm$
0%
$10.5cm$
0%
$11cm$

Explanation

Le the radius of hemisphere be

^{'} r

$'r$ cm.

Now, cuoved surface area of hemisphere

= 2 π r^{2}

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

\Rightarrow 2 π r^{2} = 77

$\Rightarrow 2\pi { r }^{ 2 }=77$

\Rightarrow r^{2} = \frac{77 \times 7}{2 \times 22} = \frac{7 \times 7}{2 \times 2}

$\Rightarrow { r }^{ 2 }=\dfrac { 77\times 7 }{ 2\times 22 } =\dfrac { 7\times 7 }{ 2\times 2 }$

\Rightarrow r = \sqrt{\frac{49}{21}} = \frac{7}{2}

$\Rightarrow r=\sqrt { \dfrac { 49 }{ 21 } } =\dfrac { 7 }{ 2 }$

\Rightarrow r = 3.5 c m

$\Rightarrow r=3.5cm$

Area of the base of a solid hemisphere is 36

π c m^{2}

$\pi\ cm^2$ . Then its volume is :

0%
$288\pi \ cm^{3}$
0%
$108\pi \ cm^{3}$
0%
$144\pi\ cm^{3}$
0%
$72\pi \ cm^{3}$

Explanation

Let the radius of the hemisphere be

^{'} r^{'}

$'r'$ cm.

Area of base

= 36 π {c m}^{2}

$=36\pi { cm }^{ 2 }$

\Rightarrow π r^{2} = 36 π

$\Rightarrow \pi { r }^{ 2 }=36\pi$

\Rightarrow r^{2} = 36

$\Rightarrow { r }^{ 2 }=36$

\Rightarrow r = 6 c m

$\Rightarrow r=6cm$

∴

$\therefore$ Volume of hemisphere

= \frac{2}{3} π r^{3}

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

= \frac{2}{3} \times π \times 6 \times 6 \times 6

$=\dfrac { 2 }{ 3 } \times \pi \times 6\times 6\times 6$

= 144 π {c m}^{3}

$=144\pi { cm }^{ 3 }$

Given dimensions of a cuboid as

l = 3 c m, b = 2 c m

$l = 3\ cm, b = 2\ cm$ and

h = 1 c m .

$h = 1\ cm.$ Find the volume.

0%
$12\ cm^3$
0%
$6\ cm^3$
0%
$4\ cm^3$
0%
$3\ cm^3$

Explanation

Given:

l = 3 c m

$l=3\ cm$

b = 2 c m

$b=2\ cm$

h = 1 c m

$h=1\ cm$

Volume of cuboid

= l b h

$=lbh$

= 3 \times 2 \times 1

$=3\times 2\times 1$

= 6 {c m}^{3}

$=6 \ { cm }^{ 3 }$

The volume of a sphere of radius

r

$r$ is:

0%
$\dfrac{4}{3}\pi r^{3}$
0%
$2\pi r^{2}$
0%
$\dfrac{2}{3}\pi r^{3}$
0%
$4\pi r^{2}$

Explanation

Step-1:Write the relevant formula of volume of sphere using given radius.

$\textbf{Step-1:Write the relevant formula of volume of sphere using given radius.}$

We have given,

$\text{We have given,}$

Radius = r  unit.

$\text{Radius = r unit.}$

∴

$\therefore$

Volume of a sphere

$\text{Volume of a sphere}$

= \frac{4}{3} π r^{3}

$=\cfrac { 4 }{ 3 } \pi { r }^{ 3 }$

Cubic unit.

$\text{Cubic unit.}$

Hence, option - A

$\textbf{Hence, option - A}$

The ratio of the height of a circular cylinder to the diameter of its base is

1 : 2

$1 : 2$ , then the ratio of the areas of its curved surface to the sum of the areas of its two ends is

0%
$1 : 1$
0%
$1 : 2$
0%
$2 : 1$
0%
$1 : 3$

Explanation

Given,
Height : Diameter of cylinder

= 1 : 2

$= 1:2$
Height : Radius of cylinder

= 1 : 1

$= 1:1$
Curved surface area

= 2 π r h

$= 2 \pi r h$
Areas of two ends

= 2 π r^{2}

$= 2\pi r^2$

∴

$\therefore$ their Ratio =

\frac{2 π r h}{2 π r^{2}} = 1 : 1

$\displaystyle \frac{2\pi rh}{2\pi r^2}= 1:1$

If the perimeter of one face of a cube is

20 c m

$20\: cm$ , then its surface area is

0%
$120\: cm^2$
0%
$150\: cm^2$
0%
$125\: cm^2$
0%
$400\: cm^2$

Explanation

Perimeter of one face of a cube is

4 a = 20

$4a=20$

\Rightarrow a = 5

$\Rightarrow a=5$

c m

$cm$
Now the surface area will be

6 a^{2}

$6a^2$

= 6 \times 5 \times 5

$=6\times 5\times 5$

= 150 c m^{2}

$=150\: cm^2$
Hence, option

B

$B$ is correct.

If the curved surface of a cylinder be doubled the area of the ends, then the ratio of its height and radius is

0%
$2 : 3$
0%
$1 : 1$
0%
$2 : 1$
0%
$1 : 2$

Explanation

Let the height and radius of the ends of a cylinder be

h

$h$ and

r

$r$ respectively.
Now Curved surface

= 2 \times

$= 2 \times$ Area of the ends

∴ 2 π r h = 2 \times (π r^{2} + π r^{2})

$\therefore 2\pi rh = 2\times (\pi r^2+\pi r^2)$

∴ \frac{h}{r} = \frac{2}{1}

$\therefore \dfrac{h}{r} = \dfrac{2}{1}$

∴ h : r :: 2 : 1

$\therefore h : r :: 2 :1$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Explanation

Given, surface area of sphere

= 221.76 c m^{2}

$=221.76cm^2$

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

\Rightarrow 4 (π) r^{2} = 221.76

$\Rightarrow 4(\pi)r^2=221.76$

\Rightarrow r^{2} = \frac{7}{22} \times 221.76 \times \frac{1}{4} = 7 \times (10.08) \times \frac{1}{4}

$\Rightarrow r^2=\dfrac {7}{22} \times 221.76\times \dfrac {1}{4}=7\times (10.08)\times \dfrac {1}{4}$

\Rightarrow r = 4.2 c m

$\Rightarrow r=4.2 cm$
Then, diameter equals

2 r = 2 \times 4.2 = 8.4

$2r=2\times4.2=8.4$ cm, which is true and reason is also correct.

Moreover reason is the correct explanation of assertion.
Hence, option

A

$A$ is correct.

The number of balls of radius 1 cm that can be made from a solid sphere of radius 4 cm is

0%
$64$
0%
$16$
0%
$12$
0%
$4$

Explanation

Radius of sphere,

R = 4

$R = 4$ cm
Radius of ball,

r = 1

$r = 1$ cm
Volume of sphere

V_{1} = \frac{4}{3} π R^{3}

$V_{1}= \dfrac{4}{3} \pi R^3$ =

\frac{4}{3} π (4)^{3}

$\dfrac{4}{3} \pi (4)^3$
Volome of ball

V_{2} = \frac{4}{3} π r^{3}

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

\frac{4}{3} π (1)^{3}

$\dfrac{4}{3} \pi (1)^3$
Number of balls that can made are

= \frac{V_{1}}{V_{2}} = \frac{\frac{4}{3} π (4)^{3}}{\frac{4}{3} π (1)^{3}}

$= \displaystyle \frac{V_{1}}{V_{2}} = \displaystyle \frac{\frac{4}{3} \pi (4)^3}{\frac{4}{3} \pi (1)^3}$

= 4^{3}

$= 4^3$

= 64

$= 64$

A sphere has the same curved surface as the total surface area of cylinder of height 4 cm and diameter of base 8 cm. The radius of the sphere is

0%
$2\ cm$
0%
$3 \ cm$
0%
$4\ cm$
0%
$6\ cm$

Explanation

Given that,

The curved surface area of a sphere is same as the total surface area of a cylinder.

The height of the cylinder is

4 c m

$4\ cm$ and the diameter of its base is

8 c m

$8\ cm$

To find out,

The radius of the sphere.

Let the radius of the sphere be

x c m

$x\ cm$ .

We know that, total surface area of cylinder

= 2 π r h + 2 π r^{2}

$= 2\pi rh + 2\pi r^2$

We have

r = \frac{d}{2} = 4 c m

$r=\frac d2=4\ cm$ and

h = 4 c m

$h=4\ cm$

∴

$\therefore \$ Total surface area of the cylinder

= 2 π (4) (4) + 2 π (4)^{2}

$= 2 \pi (4)(4) + 2\pi (4)^2$

= 32 π + 32 π

$= 32 \pi + 32 \pi$

= 64 π

$= 64 \pi$

We also know that, curved surface area of a sphere

= 4 π r^{2}

$= 4\pi r^2$

Hence,

4 π x^{2} = 64 π

$4\pi x^2= 64 \pi$

\Rightarrow x^{2} = 16

$\Rightarrow x^2=16$

∴ x = 4 c m

$\therefore \ x = 4\ cm$

Hence, the radius of the sphere is

4 c m

$4\ cm$ .

If the diameter of the sphere is doubled, the surface area of the resultant sphere becomes

x

$x$ times that of the original one. Then,

x

$x$ would be:

0%
$2$
0%
$3$
0%
$4$
0%
$8$

Explanation

Let

S_{1}

$S_1$ be the original surface area of the sphere and

S_{2}

$S_2$ be the surface area with double the diameter.

Doubling the diameter means doubling the radius.

Let

r_{1}

$r_1$ and

r_{2}

$r_2$ corresponding radii.

We know that

S = 4 π r^{2}

$S = 4 \pi r^2$ and

r_{2} = 2 r_{1}

$r_2=2r_1$ ....(Given)

Hence,

\frac{S_{1}}{S_{2}} = \frac{4 π r_{1}^{2}}{4 π r_{2}^{2}}

$\cfrac {S_1}{S_2} = \cfrac {4 \pi r_1^2}{4 \pi r_2^2}$

\Rightarrow \frac{S_{1}}{S_{2}} = \frac{r_{1}^{2}}{(2 r_{1})^{2}}

$\Rightarrow \cfrac {S_1}{S_2} = \cfrac {r_1^2}{(2r_1)^2}$

\Rightarrow S_{2} = 4 S_{1}

$\Rightarrow S_2 = 4S_1$ .

In complying with the question above we can say that

x

$x$ is equivalent to

4

$4$ .
So, doubling radius will increase

S

$S$ by a factor of

4

$4$ .

That is, option

C

$C$ is correct.

If the volume and the surface area of a solid sphere are numerically equal, then its radius is ____

0%
$9$ units
0%
$2$ units
0%
$6$ units
0%
$3$ units

Explanation

Volume of sphere

= \frac{4}{3} (π) r^{3}

$=\dfrac {4}{3}(\pi)r^3$
Surface area of sphere

= 4 (π) r^{2}

$=4(\pi)r^2$

In question it is given that volume and surface area of a solid sphere are equal.

∴ \frac{4}{3} (π) r^{3} = 4 (π) r^{2}

$\therefore \dfrac {4}{3}(\pi)r^3=4(\pi)r^2$

∴ r = 3

$\therefore r=3$ units

A tent is in the form of a right circular cylinder, surmounted by a cone. The diameter of the cylinder is

24

$24$ m. The height of the cylindrical portion is

11

$11$ m, while the vertex of the cone is

16

$16$ m above the ground.

The curved surface area of the cylindrical portion is

0%
$(246 \pi)m^2$
0%
$(264 \pi)m^2$
0%
$(426 \pi)m^2$
0%
$(462 \pi)m^2$

Explanation

Given, diameter

= 24

$=24$ m

Therefore, radius

= \frac{d}{2} = \frac{24}{2} = 12

$=\dfrac {d}{2}=\dfrac {24}{2}=12$ m

and height given is

11

$11$ m

Therefore, curved surface area of the cylinder

= 2 (π) r h

$=2(\pi)rh$

= 2 \times (π) \times 12 \times 11

$=2\times (\pi)\times 12\times 11$

= 264 (π)

$=264(\pi)$

A closed cylindrical tank, made of thin iron sheet, had diameter

8.4

$8.4$ m and height

5.4

$5.4$ m. How much metal sheet to the nearest

m^{2},

$\displaystyle m^{2},$ is used in making this tank, if

\frac{1}{15}

$\displaystyle \frac{1}{15}$ of the sheet actually used was wasted in making the tank ?

0%
$272$ $\displaystyle m^{2}$
0%
$271$ $\displaystyle m^{2}$
0%
$242$ $\displaystyle m^{2}$
0%
$270$ $\displaystyle m^{2}$

If the radius of the base of a cylinder is

2

$2$ cm and its height

7

$7$ cm, then its curved surface is

0%
$44 cm^2$
0%
$22 cm^2$
0%
$88 cm^2$
0%
$56 cm^2$

Explanation

Given, radius of cylinder

= 2

$= 2$ cm and height of cylinder

= 7

$= 7$ cm

∴

$\therefore$ curved surface area of cylinder

= 2 π r h

$= 2\pi rh$

= 2 π \times 2 \times 7

$= 2 \pi \times 2\times 7$

= 88 c m^{2}

$= 88 cm^2$

the height of the cylinder correct to one decimal place.

0%
$1.5$ cm
0%
$1.1$ cm
0%
$1.9$ cm
0%
$1.6$ cm

Explanation

Given, a cylinder with diameter

20

$20$ cm and curved surface area

100

$100$ sq. cm.
Radius

= \frac{d}{2}

$=\dfrac {d}{2}$

= 10

$=10$ cm
Curved surface area

= 2 π r h

$=2\pi rh$

\Rightarrow 100 = 2 \times \frac{22}{7} \times 10 \times h

$\Rightarrow 100= 2\times \dfrac { 22 }{ 7 } \times 10\times h$

\Rightarrow h = \frac{100 \times 7}{2 \times 22 \times 10}

$\Rightarrow h= \dfrac { 100\times 7 }{ 2\times 22\times 10 }$

\Rightarrow h = 1.6

$\Rightarrow h= 1.6$ cm
Height of cylinder

= 1.6

$=1.6$ cm

CBSE Questions for Class 9 Maths Surface Areas And Volumes Quiz 3 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Surface Areas And Volumes Quiz 3 - MCQExams.com

0:0:11

Practice Class 9 Maths Quiz Questions and Answers

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