MCQ Questions for Class 9 Maths Surface Areas And Volumes Quiz 6

If the surface area of a sphere is

$324\pi cm^2$ then its volume is

0%
$950 \pi cm^3$
0%
$972 \pi cm^3$
0%
$975 \pi cm^3$
0%
$980 \pi cm^3$

Explanation

Consider the given surface area of sphere $=324\pi \,c{{m}^{3}}$

Let, radius of sphere =r

We know that,

Surface area of the sphere $=4\pi {{r}^{2}}\,c{{m}^{3}}$

Now,

$4\pi {{r}^{2}}=324\pi$

${{r}^{2}}=81$

$r=9\,cm$

$\dfrac{4}{3}\pi {{r}^{3}}=4851$

${{r}^{3}}=\dfrac{3\times 4851\times 7}{4\times 22}$

Now, volume of the sphere,

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

$=\dfrac{4}{3}\pi \times {{9}^{3}}c{{m}^{3}}$

$=972\pi c{{m}^{3}}$

Hence, this is the answer.

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

0%
1000
0%
10000
0%
100000
0%
100

Explanation

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=

$\frac{\frac{4000}{3}\pi }{\frac{4}{3}\pi }=1000$

The volume of a sphere of diameter 2p cm is given by

0%
$\pi p^2 cm^3$
0%
$\pi p^3 cm^3$
0%
$4\pi p^3 cm^3$
0%
$\frac {4}{3}\pi p^3 cm^3$

Explanation

Given, the diameter of the sphere $=2p\,cm$

radius = $2p/2$ = $p$

We know that the volume of the sphere

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

$=\dfrac{4}{3}\pi {{\left( 2p \right)}^{3}}$

$=\dfrac{4}{3}\pi {8{p}^{3}}\,c{{m}^{3}}$

$=\dfrac{32}{3}\pi {{p}^{3}}\,c{{m}^{3}}$

Hence, this is the answer.

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

0%
11 cm
0%
13 cm
0%
24 cm
0%
none

Explanation

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

The radius of a solid sphere is

$r cm$ . It is bisected, then the total surface area of the two pieces obtained is:

0%
$8\pi r^2 cm^2$
0%
$4\pi r^2 cm^2$
0%
$5\pi r^2 cm^2$
0%
$6\pi r^2 cm^2$

Explanation

Given, the radius of the sphere $=r\,cm$ .

If the sphere is divided into two equal parts, then the resulting parts are hemispheres with radius $=r\,cm$ .

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

$=2\times 3\pi {{r}^{2}}=6\pi {{r}^{2}}\,c{{m}^{2}}$ .

Hence, option $D$ is correct.

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

0%
4
0%
2
0%
3.5
0%
3

Explanation

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.

The radius of a sphere is increased by

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

0%
$200 \%$
0%
$150 \%$
0%
$125 \%$
0%
$50 \%$

Explanation

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

$=9\pi r^{2}-4\pi r^{2}=5\pi r^{2}$ .

$\%$ 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.

Find the surface area of the sphere whose radius is

$35 \text{ cm}$ .

0%
$15400 \text{ cm}^{2}$
0%
$1540 \text{ cm}^{2}$
0%
$154 \text{ cm}^{2}$
0%
$15.400 \text{ cm}^{2}$

Explanation

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.

If the circumference of the inner edge of a hemispherical bowl is

$\displaystyle \frac{132}{7}$ cm then what is the capacity?

0%
$\displaystyle 12\pi cm^{3}$
0%
$\displaystyle 18\pi cm^{3}$
0%
$\displaystyle 24\pi cm^{3}$
0%
$\displaystyle 36\pi cm^{3}$

Explanation

Given the circumference of the inner edge of a hemispherical bowl is

$\frac{132}{7}$ cm

Then

$2\pi r=\frac{132}{7}$

$\Rightarrow r=\frac{132}{7}\times \frac{7}{44}$

$\Rightarrow r=3$ cm

Then capacity =volume of hemispherical =

$\frac{2}{3}\pi (3)^{3}=\frac{2}{3}\times 27\pi =18\pi cm^{3}$

What is the volume of the cuboid of the brick? (Given : length

$= 23$ cm, breadth

$= 10$ cm, height

$= 12$ cm)

0%
$\displaystyle 2765{ cm }^{ 3 }$
0%
$\displaystyle 2760{ cm }^{ 3 }$
0%
$\displaystyle 2340{ cm }^{ 3 }$
0%
$\displaystyle 2901{ cm }^{ 3 }$

Explanation

$Volume\ of\ cuboid=length\times breadth\times height$

$length=223\ cm$

$breadth=10\ cm$

$height=12\ cm$

$volume=23\times 10\times 12=2760\ cm^3$

A plastic box

$3.5$ m long,

$2$ m wide and

$20$ m deep is to be made. Find its volume.

0%
$\displaystyle 160\ { m }^{ 3 }$
0%
$\displaystyle 70\ { m }^{ 3 }$
0%
$\displaystyle 140\ { m }^{ 3 }$
0%
$\displaystyle 50\ { m }^{ 3 }$

Explanation

Volume of the plastic box

$=$ length

$\times$ breadth

$\times$ height
Therefore, length

$= 2$ m, breadth

$= 20$ m, height

$= 3.5$ m
Therefore, volume

$=$

$\displaystyle 2\times 20\times 3.5=140{ m }^{ 3 }$

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

$\displaystyle cm^{2}$ ).

0%
$\displaystyle \frac{524}{11}$
0%
$\displaystyle \frac{575}{11}$
0%
$\displaystyle \frac{629}{11}$
0%
$\displaystyle \frac{686}{11}$

Explanation

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

If the side of cube is

$2\text{ m}$ , then surface area of cube is

0%
$24\:\text{m}^2$
0%
$124\:\text{m}^2$
0%
$24\:\text{m}$
0%
$124\:\text{m}$

Explanation

Given, side of a cube

$=2\text{ m}$

Let

$'a'$ be the length of the side of the cube.

Surface area of the cube

$=$

$6 \times a^2$

$= 6 \times 2^2$

$=6 \times 4$

$=24 \text{ m}^2$

So, option A is correct.

$s$ represents the side, then the formula of surface area of cube is

0%
$6(s)^2$
0%
$3(s)^2$
0%
$5(s)^2$
0%
$16(s)^2$

Explanation

Side of the cube

$=$

$s$

Area of

$1$ surface of cube

$=s \times s = s^2$

A cube has

$6$ surfaces.

$\therefore \mbox{Total surface area of cube = }$

$6 \times \mbox{Area of 1 surface}$

$=6s^2$

So, option A is correct.

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.

0%
$120$
0%
$180$
0%
$240$
0%
$270$

Explanation

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

$= \dfrac {20 \times 18 \times 16 }{24} = 240$

Capacity is not measured in ..........

0%
Litres
0%
Millilitres
0%
Cubic centimetres
0%
Centimetres

Kyle bought 2 snack boxes. Box A has a measure of

$2\ in\times 3\ in\times 5 \ in.$ Box B has a measure of

$11 \ in\times 4\ in \times 3\ in.$ Find the surface area of the boxes and their difference.

0%
$\displaystyle 116{ in }^{ 2 }$
0%
$\displaystyle 201{ in }^{ 2 }$
0%
$\displaystyle 279{ in }^{ 2 }$
0%
$\displaystyle 198{in }^{ 2 }$

Explanation

Surface area of rectangle is 2 x (Length x width + Length x height + Width x height)

Here Box A has length

$= 2$ in, height

$= 3$ in and width

$= 5$ in

Therefore, the surface area of Box A is:

$A=2((2\times 5)+(2\times 3)+(3\times 5))=2(10+6+15)=2\times 31=62$ in

$^2$

Now, Box B has length

$= 11$ in, height

$= 4$ in and width

$= 3$ in

Therefore, the surface area of Box B is:

$A=2((11\times 4)+(11\times 3)+(4\times 3))=2(44+33+12)=2\times 89=178$ in

$^2$

Hence, the difference between the surface area of box A and box B is

$178-62=116$ in

$^2$ .

What is the surface area of a television

$20$ inches long,

$15$ inches wide and

$5$ inches high?

0%
$\displaystyle 980{\ in }^{ 2 }$
0%
$\displaystyle 950{ \ in }^{ 2 }$
0%
$\displaystyle 915{\ in }^{ 2 }$
0%
$\displaystyle 925{\ in }^{ 2 }$

Explanation

Surface area

$= 2 ($ Length

$\times$ width

$+$ Length

$\times$ height

$+$ Width

$\times$ height

$)$
Length

$= 20$ in; Height

$= 5$ in; Width

$= 15$ in

$\displaystyle = 2\times \left( 20\times 15+20\times 5+15\times 5 \right)$

$\displaystyle = 2\times \left( 300+100+75 \right)$

$\displaystyle = 2\times 475$

$\displaystyle = 950{\ in }^{ 2 }$

Length, breadth and height of a room are

$12$ ft,

$5$ ft and

$8$ ft respectively. Find the volume of the room.

0%
$\displaystyle 440{ ft }^{ 3 }$
0%
$\displaystyle 480{ ft }^{ 3 }$
0%
$\displaystyle 400{ ft }^{ 3 }$
0%
$\displaystyle 500{ ft }^{ 3 }$

Explanation

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

$\displaystyle =12\times 5\times 8{ ft }^{ 3 }$

$\displaystyle =480$

${ ft }^{ 3 }$

A parcel has to be packed with length

$1$ m, width

$0.1$ m and height

$0.05$ m. Find the volume of the parcel in cubic cm.

0%
$\displaystyle { 6050\ cm }^{ 3 }$
0%
$\displaystyle { 7000\ cm }^{ 3 }$
0%
$\displaystyle { 5000\ cm }^{ 3 }$
0%
$\displaystyle { 1300\ cm }^{ 3 }$

Explanation

Given dimensions of the parcel are, length

$= 1$ m or

$100$ cm, width

$= 0.1$ m or

$10$ cm and height

$= 0.05$ m or

$5$ cm
Therefore, volume of the parcel

$=$ length

$\times$ width

$\times$ height

$\displaystyle =\left( 100\times 10\times 5 \right) { cm }^{ 3 }$

$\displaystyle =5000{\ cm }^{ 3 }$

Find the total surface area of a cuboid given

$l =10\ cm,$

$h = 4\ cm$ and

$w = 13\ cm$ .

0%
$111{ \ cm }^{ 2 }$
0%
$444{ \ cm }^{ 2 }$
0%
$222{\ cm }^{ 2 }$
0%
$333{ \ cm }^{ 2 }$

Explanation

$\begin{aligned}{}{\text{Area}} &= 2(lw + wh + hl)\\ &= 2 \left( {10 \times 13 + 13 \times 4 + 4 \times 10} \right)\\& = 2 \left( {130 + 52 + 40} \right)\\& = 2 \times 222\\ &= 444\ c{m^2}\end{aligned}$

Find the surface area of a cuboid with dimensions

$4 \times 2.5 \times 2$ (in inches)

0%
$\displaystyle 46{ \text{ in.} }^{ 2 }$
0%
$\displaystyle 12{\text{ in.} }^{ 2 }$
0%
$\displaystyle 26\text{ in.}^{ 2 }$
0%
$\displaystyle 14{\text{ in.} }^{ 2 }$

Explanation

According to the given dimensions let,

$l=4\text{ in.}$

$w=2.5\text{ in.}$

$h=2\text{ in.}$

By using the formula of total surface area of a cuboid we get,

$\begin{aligned}{}{\rm{Area}} &= 2(lw + wh + hl)\\ &= 2\left( {4 \times 2.5 + 2.5 \times 2 + 4 \times 2} \right)\\ &= 2\left( {10 + 5 + 8} \right)\\ &= 2 \times 23\\ &= 46\text{ in.}^2\end{aligned}$

Find the volume of a cuboid whose length is

$100\ m,$ breadth is

$0.2\ m$ and height is

$0.15\ m.$

0%
$3\ { m }^{ 3 }$
0%
$1.25\ { m }^{ 3 }$
0%
$4\ { m }^{ 3 }$
0%
$5\ { m }^{ 3 }$

Explanation

Volume of the cuboid whose length is

$l,$ breadth is

$b$ and height is

$h$ is given by,

$V=l\times b\times h$

Given:

$l= 100\ m$

$b= 0.2\ m$

$h= 0.15\ m$

So,

$V=100\times 0.2\times 0.15$

$=20\times 0.15$

$=3\ m^3$

Hence, option

$A$ is correct.

Find the volume of the cuboid having length, breadth and height as

$12$ cm,

$15$ cm and

$30$ cm.

0%
$\displaystyle 6000{ cm }^{ 3 }$
0%
$\displaystyle 5400{ cm }^{ 3 }$
0%
$\displaystyle 4200{ cm }^{ 3 }$
0%
$\displaystyle 5000{ cm }^{ 3 }$

Explanation

Volume of the cuboid

$=$ length

$\times$ breadth

$\times$ height
Therefore, length

$= 12$ cm, breadth

$= 15$ cm, height

$= 30$ cm
Volume of the cuboid

$\displaystyle \\ = 12\times 15\times 30=5400{ cm }^{ 3 }\\$

Find the surface area of a cuboid:

0%
$\displaystyle 110$ unit $^{ 2 }$
0%
$\displaystyle 120$ unit $^{ 2 }$
0%
$\displaystyle 130$ unit $^{ 2 }$
0%
$\displaystyle 140$ unit $^{ 2 }$

Explanation

Surface area

$= 2 ($ Length

$\times$ width

$+$ Length

$\times$ height

$+$ Width

$\times$ height

$)$

Length

$= 5$ units; Height

$= 5$ units; Width

$= 4$ units

$\displaystyle =2\times \left( 5\times 4+5\times 5+4\times 5 \right)$

$\displaystyle =2\times \left( 20+25+20 \right)$

$\displaystyle =2\times 65$

$\displaystyle= 130$ unit

$^{ 2 }$

Michael has a box measured

$2.1$ meters in length,

$1.9$ meters in height and

$2.5$ meters in width. What is the surface area of his box can hold?

0%
$\displaystyle 18.21{ m }^{ 2 }$
0%
$\displaystyle 20.21{ m }^{ 2 }$
0%
$\displaystyle 27.98{ m }^{ 2 }$
0%
$\displaystyle 18.08{ m }^{ 2 }$

Explanation

Surface area

$=$

$2 \times$

$($ Length

$\times$ width

$+$ Length

$\times$ height

$+$ Width

$\times$ height

$)$
Length

$= 2.1$ m; Height

$= 1.9$ m; Width

$= 2.5$ m

$\displaystyle =2\times \left( 5.25+3.99+4.75 \right)$

$\displaystyle =2\times 13.99$

$\displaystyle =27.98$

${ m }^{ 2 }$

Find the total surface area of the given cuboid.

0%
$\displaystyle 130{ m }^{ 2 }$
0%
$\displaystyle 131{ m }^{ 2 }$
0%
$\displaystyle 132{ m }^{ 2 }$
0%
$\displaystyle 133{ m }^{ 2 }$

Explanation

Given, length

$(l)=8\ m$

breadth

$(b)=5\ m$

height

$(h)=2\ m$

Surface area of the cuboid

$\displaystyle =2\times \left( lb+lh+bh \right)$

$\displaystyle =2\times \left( 8\times 5+8\times 2+5\times 2 \right) \ m^2$

$\displaystyle =2\times \left( 40+16+10 \right) \ m^2$

$\displaystyle =2\times 66\ m^2$

$\displaystyle =132\ { m }^{ 2 }$

Hence, total surface area is

$132\ { m }^{ 2 }$

The edge of the cube measures

$9$ cm. Find its surface area.

0%
$\displaystyle 418{ cm }^{ 2 }$
0%
$\displaystyle 446{ cm }^{ 2 }$
0%
$\displaystyle 486{ cm }^{ 2 }$
0%
$\displaystyle 441{ cm }^{ 2 }$

Explanation

Given, edge

$= 9 \text{ cm}$

Total surface area of a cube

$=6\times \text{side}^2$

$= 6\times 9\times 9$

$= 486 \text{ cm}^2$

Find the surface area of the cuboid:

0%
$\displaystyle 300.1{ mm }^{ 2 }$
0%
$\displaystyle 320.5{ mm }^{ 2 }$
0%
$\displaystyle 316.4{ mm }^{ 2 }$
0%
$\displaystyle 305.2{ mm }^{ 2 }$

Explanation

Surface area of the cuboid

$=$

$\displaystyle 2\times \left( lw+lh+hw \right)$

$\displaystyle =2\times \left( 8\times 2.1+8\times 14+2.1\times 14 \right)$

$\displaystyle =2\times \left( 16.8+112+29.4 \right)$

$\displaystyle =2\times 158.2$

$\displaystyle =316.4$

${ mm }^{ 2 }$

Calculate the total surface area of the cuboid:

0%
$\displaystyle 171\ { cm }^{ 2 }$
0%
$\displaystyle 172\ { cm }^{ 2 }$
0%
$\displaystyle 173\ { cm }^{ 2 }$
0%
$\displaystyle 174\ { cm }^{ 2 }$

Explanation

Total surface area of the cuboid

$\displaystyle 2\times \left( lw+lh+hw \right)$

$\displaystyle =2\times \left( 10\times 1+10\times 7+7\times 1 \right)$

$\displaystyle =2\times \left( 10+70+7 \right)$

$\displaystyle =2\times 87$

$=\displaystyle 174{ cm }^{ 2 }$

So, option D is correct.

CBSE Questions for Class 9 Maths Surface Areas And Volumes Quiz 6 - 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 6 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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