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

Given a sphere with the radius

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

$\pi$ as

$3.14$ .

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

Explanation

Volume of sphere =

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

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

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

Calculate the surface area of a sphere with radius

$3.2 cm$ .

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,

$r=3.2 cm$ .

Therefore, the surface area of sphere

$=4\pi r^2$

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

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

$=128.61 \ cm^2$ .

Hence, option

$B$ 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

$r$ 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

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

$=\pi r^2$

$\therefore \$ Total surface area of a hemisphere

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

Hence, the total surface area of hemisphere is

$3\pi r^2$ .

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.

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

$\pi r^{2}h$ =

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

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

$\pi r^{2}h$ =

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

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

$2\pi rh$
=

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

$2\pi rh$
=

$2\pi \times3\times4 = 24 \pi cm{2}$
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

$2 \pi r^2$

The surface area of a solid hemisphere with radius

$r$ 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

$r$ is

$4 \pi r^2$
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

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

$r$ .
Hence, flat surface area of hemisphere

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

$r$ is,

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

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

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

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

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

$36:144$
=

$1:4$

The radius of a sphere is

$2r$ , 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,

$r=2r$
Volume of sphere

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

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

$4.2$ 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

$r$ is

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

$4.2$ cm, its radius

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

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

$^{ 3 }$

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

$14\ m$ . The tank contains

$50$ 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,

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

$=$ Volume of hemispherical tank

$-50$ kl

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

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

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

$\Rightarrow 718.67-50\ m^3$

$\Rightarrow 668.67\ m^3$

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

$=2\pi r^2$

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

Curved surface area of newly formed solid sphere

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

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

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

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

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

$2 \pi rh$ .

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

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

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

$=2\pi RH$

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

$=2\pi rh$

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

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

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

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

Explanation

Let radius of sphere be

$r$ cm.

We know, surface area of sphere

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

$\implies$

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

$\implies$

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

$\implies$

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

Therefore, option

$A$ is correct.

The curved surface area of a hemisphere is

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

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

Explanation

Le the radius of hemisphere be

$'r$ cm.

Now, cuoved surface area of hemisphere

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

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

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

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

$\Rightarrow r=3.5cm$

Area of the base of a solid hemisphere is 36

$\pi\ cm^2$ . 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

$'r'$ cm.

Area of base

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

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

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

$\Rightarrow r=6cm$

$\therefore$ Volume of hemisphere

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

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

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

Given dimensions of a cuboid as

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

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

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

Explanation

Given:

$l=3\ cm$

$b=2\ cm$

$h=1\ cm$

Volume of cuboid

$=lbh$

$=3\times 2\times 1$

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

The volume of a sphere of radius

$r$ 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

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

$\text{We have given,}$

$\text{Radius = r unit.}$

$\therefore$

$\text{Volume of a sphere}$

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

$\text{Cubic unit.}$

$\textbf{Hence, option - A}$

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

$1 : 2$ , 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

$= 1:2$
Height : Radius of cylinder

$= 1:1$
Curved surface area

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

$= 2\pi r^2$

$\therefore$ their Ratio =

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

If the perimeter of one face of a cube is

$20\: cm$ , 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

$4a=20$

$\Rightarrow a=5$

$cm$
Now the surface area will be

$6a^2$

$=6\times 5\times 5$

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

$B$ 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

$h$ and

$r$ respectively.
Now Curved surface

$= 2 \times$ Area of the ends

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

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

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

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

$=221.76cm^2$

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

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

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

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

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

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

$A$ 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,

$R = 4$ cm
Radius of ball,

$r = 1$ cm
Volume of sphere

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

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

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

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

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

$= 4^3$

$= 64$

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

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

$8\ cm$

To find out,

The radius of the sphere.

Let the radius of the sphere be

$x\ cm$ .

We know that, total surface area of cylinder

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

We have

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

$h=4\ cm$

$\therefore \$ Total surface area of the cylinder

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

$= 32 \pi + 32 \pi$

$= 64 \pi$

We also know that, curved surface area of a sphere

$= 4\pi r^2$

Hence,

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

$\Rightarrow x^2=16$

$\therefore \ x = 4\ cm$

Hence, the radius of the sphere is

$4\ cm$ .

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:

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

Explanation

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.

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

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

$=4(\pi)r^2$

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

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

$\therefore r=3$ units

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

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

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

$16$ 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

$=24$ m

Therefore, radius

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

and height given is

$11$ m

Therefore, curved surface area of the cylinder

$=2(\pi)rh$

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

$=264(\pi)$

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 ?

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

$2$ cm and its height

$7$ 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

$= 2$ cm and height of cylinder

$= 7$ cm

$\therefore$ curved surface area of cylinder

$= 2\pi rh$

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

$= 88 cm^2$

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

$20$ cm and curved surface area

$100$ sq. cm.
Radius

$=\dfrac {d}{2}$

$=10$ cm
Curved surface area

$=2\pi rh$

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

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

$\Rightarrow h= 1.6$ cm
Height of cylinder

$=1.6$ 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

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Practice Class 9 Maths Quiz Questions and Answers

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