MCQ Questions for Class 9 Maths Surface Areas And Volumes Quiz 5

Find the volume of a hemisphere of radius

$7\;dm$ .

0%
$708.67\;dm^3$
0%
$818.67\;dm^3$
0%
$717.67\;dm^3$
0%
$718.67\;dm^3$

Explanation

Volume of hemisphere

$=$

$\dfrac { 2 }{ 3 } \pi { r }^{ 3 }=\dfrac { 2 }{ 3 } \times \dfrac { 22 }{ 7 } \times 7\times 7\times 7$

$=$

$718.67$

${ dm }^{ 3 }$

Diameter of a football is

$28\;cm$ . It is made up of small hexagones each of area

$112\;cm^2$ . Find the number of small hexagones used in football.

0%
$21$
0%
$27$
0%
$23$
0%
$22$

Explanation

Given, diameter of football

$=28cm$ .

Then, radius of football

$=14cm$ .

$\therefore$ Area of football

$=4\pi { r }^{ 2 }=4\times \dfrac { 22 }{ 7 } \times 14\times 14=2464{ cm }^{ 2 }$ .

Given, area of each hexagon

$=112{ cm }^{ 2 }$

Number of hexagon

$=\dfrac { 2464 }{ 112 } =22$ .

Hence, option

$D$ is correct.

How many balls, each of radius

$1$ cm, can be made from a solid sphere of lead of radius

$8$ cm?

0%
$512$
0%
$500$
0%
$400$
0%
$345$

Explanation

Volume of the spherical ball of radius 8 cm

$=\cfrac {4}{3}\pi \times 8^3 cm^3$

Also, volume of each smaller spherical ball of radius 1 cm

$=\cfrac {4}{3}\pi \times 1^3cm^3$

Let n be the number of smaller balls that can be made. Then, the volume of the larger ball is equal to the sum of all the volumes of n smaller balls.

Hence,

$\cfrac {4}{3}\pi \times n=\cfrac {4}{3}\pi \times 8^3$

$\Rightarrow n=8^3=512$

Hence, the required number of balls

$=512$ .

The capacity of a water tank that measures

$9$ cm

$\times 3.5$ cm

$\times 7.5$ cm is

0%
$236.25$ c $m^3$
0%
$189$ $cm^3$
0%
$236.25$ $m^3$
0%
$189$ $m^3$

Explanation

Here, length

$= 9$ cm, breadth

$= 3.5$ cm and height

$= 7.5$ cm

$\therefore$ capacity of water tank

$=$ length

$\times$ breadth

$\times$ height

$=9 \times 3.5 \times 7.5 = 236.25$

$cm^3$ .

Diameter of a sphere is

$21$ dm. Find its surface area.

0%
$1286\;dm^2$
0%
$1376\;dm^2$
0%
$1386\;dm^2$
0%
$1380\;dm^2$

Explanation

Given, diameter of a sphere,

$d=21$ dm,

$\therefore$ radius of the sphere,

$r=\dfrac{d}{2}=\dfrac{21}{2}=10.5$ dm
Surface area of sphere

$=4 \pi r^2$

$=4 \pi (10.5)^2=4 \times \dfrac{22}{7} \times (10.5)^2$

$=1386 dm^2$ .

Therefore, option

$C$ is correct.

The area of the curved surface of a sphere is

$5544\;m^2$ . Find the radius of the sphere.

0%
$12$ m
0%
$20$ m
0%
$22$ m
0%
$21$ m

Explanation

Given, area of curved surface of a sphere

$=5544$

$cm^2$

We know, curved surface area of sphere

$=4 \pi r^2$

$\Rightarrow 5544=4 \times \dfrac {22}{7}\times r^2$

$\Rightarrow 7 \times 5544=4 \times 22\times r^2$

$\Rightarrow r^2=441$

$\Rightarrow r=21 m$ .

Therefore, option

$D$ is correct.

A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape It is

$30\ cm$ long,

$25\ cm$ wide, and

$25\ cm$ high The area of glass is:

0%
$5000\ cm^{2}$
0%
$4800\ cm^{2}$
0%
$4250\ cm^{2}$
0%
$4500\ cm^{2}$

Explanation

Since the greenhouse is cuboidal in shape, we need to calculate its total surface area by applying the formula.

Total surface area of a cuboid $= 2(l \times b + b\times h + l \times h)$

Given:-

$l=30\ cm$

$b=25\ cm$

$h=25\ cm$

Hence, the total surface area of the herbarium will be,

$S =2(30 \times 25 + 25 \times 25 + 30 \times 25)$

$= 2(750+625+750)$

$=2(2125)$

$= 4250 \ cm^2$

Find the surface area of the following cubes with length of edge

$4\;cm$

$3.2\;cm$

0%
$96\;cm^2\;;\;61.44\;cm^2$
0%
$90\;cm^2\;;\;61.44\;cm^2$
0%
$96\;cm^2\;;\;60.44\;cm^2$
0%
$92\;cm^2\;;\;61.44\;cm^2$

Explanation

Total surface area

$=6(4^2)=96 cm^2$ ,for edge=4

$cm^2$
Total surface area

$=6(3.2^2)=61.44 cm^2$ , for edge=61.44

$cm^2$

A plastic box

$1.5$ m long,

$1.25$ m wide, and

$65$ cm deep is to be made. It is to be opened at the top. Ignoring the thickness of the plastic, the cost of the sheet for covering it, if a sheet measuring

$1$

$\displaystyle m^{2}$ costs Rs.

$20$ is:

0%
Rs. $100$
0%
Rs. $109$
0%
Rs. $115$
0%
Rs. $110$

Explanation

Length of the box,

$l=1.5m$ . Breadth of box,

$b=1.25m$

Depth of box,

$h=65cm=0.65m$

Area of sheet required

$=$ Area of cuboid

$-$ Area of open top

$=2(lb+bh+hl)-lb$

$=lb+2h(b+l)$

$=1.5\times 1.25+2\times 0.65(1.5+1.25)$

$=5.45m^2$

Cost of sheet of

$5.45m^2=$ Rs.

$(5.45\times 20)=$ Rs.

$109$

The radius of the cylinder whose lateral surface area is

$704\, cm^2$ and height

$8$ cm is

0%
$6$ cm
0%
$4$ cm
0%
$8$ cm
0%
$14$ cm

Explanation

Lateral surface area of a cylinder

$=2 \pi r h$
Given,

$\, 2\, \pi\, rh\, =\, 704\, cm^2$ .

$\therefore\, r\, =\, \displaystyle \frac {704}{2\pi h}\,$

$=\, \displaystyle \frac {704}{2\, \times\, \displaystyle \frac {22}{7}\, \times\, \times\, 8}\,$

$=\, 14$ cm

Savitri had to make a model of a cylindrical kaleidoscope for her science project. She wanted to use chart paper to make the curved surface of the kaleidoscope. What would be the area of chart paper required by her, if she wanted to make a kaleidoscope of length

$25\space cm$ with a

$3.5\space cm$ radius? You may take

$\left (\pi = \dfrac{22}{7}\right )$

0%
$540\space cm^2$
0%
$520\space cm^2$
0%
$550\space cm^2$
0%
$560\space cm^2$

Explanation

Kaleidoscope is of cylindrical shape.

$\therefore$ Area of chart paper required

$=$ CSA of cylinder

$=$

$2\pi rh$

$=$

$2\times \dfrac { 22 }{ 7 } \times 3.5\times 25$

$=$

$550$

${ cm }^{ 2 }$

Three solid spheres of copper, whose radii are

$3$ cm,

$4$ cm and

$5$ cm respestively are melted into a single solid sphere of radius R. The value of R is

0%
$12$ cm
0%
$8$ cm
0%
$4$ cm
0%
$6$ cm

Explanation

$\frac {4}{3}\pi R^3=\frac {4}{3}\pi (3^3+4^3+5^3)\Rightarrow R^3=27+64+125=216\Rightarrow R=6$

A right circular cylinder and a sphere are of equal volumes and their radii are also equal If h is the height of the cylinder and d is the diameter of the sphere then

0%
$\displaystyle \frac{h}{3}=\frac{d}{2}$
0%
$\displaystyle \frac{h}{2}=\frac{d}{3}$
0%
$2h = d$
0%
$h = d$

Explanation

Volume of cylinder = Volume of sphere

$\displaystyle \Rightarrow \pi \left ( \dfrac{d}{2} \right )^{2}h=\dfrac{4}{3}\pi \left ( \dfrac{d}{2} \right )^{3}$

$\displaystyle \Rightarrow h=\dfrac{2}{3}d\Rightarrow \dfrac{h}{2}=\dfrac{d}{3}$

Find the difference between total surface area & curved surface area of a hemisphere of radius

$21\space cm$ .

0%
$1376\space cm^2$
0%
$1386\space cm^2$
0%
$1396\space cm^2$
0%
$1404\space cm^2$

Explanation

Curved Surface Area of a hemisphere of radius 'r'

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

Total Surface Area of a hemisphere of radius 'r'

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

Hence, required difference

$= 3\pi { r }^{ 2 } - 2\pi { r }^{ 2 } = \pi { r }^{ 2 } = \dfrac {22}{7} \times 21 \times 21 = 1386\ {cm}^{2}$

Three solid spheres of a lead are melted into a single solid sphere If the radii of the three spheres be 1 cm, 6 cm and 8 cm respectively Then radius of the new sphere is :

0%
$2 cm$
0%
$3 cm$
0%
$5 cm$
0%
$9 cm$

Explanation

Let r be the radius of the new sphere Then
$\displaystyle \frac{4}{3}\pi r^{3}=\frac{4}{3}\pi \left ( 1 \right )^{3}+\frac{4}{3}\pi \left ( 6 \right )^{3}+\frac{4}{3}\pi \left ( 8 \right )^{3}$

$\displaystyle \frac{4}{3}\pi \left [ \left ( 1 \right )^{3}+\left ( 6 \right )^{3}+\left ( 8 \right )^{3} \right ]=\frac{4}{3}\pi \left [ 1+216+512 \right ]$
or

$\displaystyle r^{3}=729=\left ( 9 \right )^{3}$

$\displaystyle \therefore$ Radius of the new sphere (r) =9 cm

The diameter of the base of a right circular cylinder is

$28$ cm and its height is

$21$ cm Curved surface area of the cylinder is

0%
$1540$ $\displaystyle cm^{2}$
0%
$1648$ $\displaystyle cm^{2}$
0%
$1848$ $\displaystyle cm^{2}$
0%
$1548$ $\displaystyle cm^{2}$

Explanation

$Diameter=28\ cm$

$\Rightarrow r=14\ cm$

and

$h =21 cm$
C.S.A of cylinder

$\displaystyle =2\pi rh=2\times \dfrac{22}{7}\times 14\times 21=1848\ cm^{2}$

If the radius of the base of a cylinder is

$2$ cm and its height

$7$ cm, then what is its curved surface area?

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

Explanation

Given,

$r=2$ cm,

$h=7$ cm

Then curved surface area

$=2 \pi r h$

$=2\times \dfrac{22}{7}\times 2\times 7$

$=88 cm^{2}$

The hollow sphere, in which the circus motorcyclist performs his stunts, has diameter of

$7\ m$ . Find the area available to motorcyclist for riding.

0%
$154\ m^2$
0%
$144\ m^2$
0%
$38.5\ m^2$
0%
$176\ m^2$

Explanation

Surface area of a sphere of radius 'r'

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

Radius of the sphere

$= \dfrac {7}{2} m$

Hence, the area available to motorcyclist for riding $$ = 4 \dfrac {22}{7} \times \dfrac {7}{2} \times \dfrac {7}{2} = 154 {m}^{2} $$

Find the volume of a hemisphere of radius

$6.3 \ cm$ (

$\displaystyle \pi =22/7$ )

0%
$523.9 \ \displaystyle cm^{3}$
0%
$520.91 \ \displaystyle cm^{3}$
0%
$512.91 \ \displaystyle cm^{3}$
0%
$510.91 \ \displaystyle cm^{3}$

Explanation

$\Rightarrow$ Volume of a hemisphere

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

$=\dfrac{2}{3}\times \dfrac{22}{7}\times (6.3)^3$

$=\dfrac{2}{3}\times \dfrac{22}{7}\times 6.3\times 6.3 \times 6.3$

$=2\times 22\times 2.1\times 0.9\times 6.3\\$

$=523.9\,cm^3$

The diameter of a garden roller is

$1.4\space m$ and it is

$2\space m$ long. How much area will it cover in

$5$ revolutions? (Use

$\pi = \dfrac{22}{7}$ )

0%
$40\space m^2$
0%
$41.40\space m^2$
0%
$44\space m^2$
0%
$42.40\space m^2$

Explanation

Since the diameter of the roller is
$0.7 m$ , its radius is $0.35 m$ .

The roller is in the shape of a cylinder.

In one revolution, the roller covers the distance of one curved surface area.

Curved Surface Area of a Cylinder of Radius " $R$ " and height " $h$ " $= 2\pi Rh$

Radius of the roller $= 0.7 m$

Curved Surface Area of the roller of radius $0.35 m = 2 \times \dfrac {22}{7}\times 0.7 \times 2 = 8.8 m^2$

Area covered in $5$ revolutions $= 5 \times 8.8 m^2 = 44 m^2$

If volume and surface area of a sphere are numerically equal then it's radius is

0%
$2\ units$
0%
$3\ units$
0%
$5\ units$
0%
$8\ units$

Explanation

Volume of sphere

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

Surface area of sphere

$=4 \pi r^2$

According to the question

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

$\Rightarrow \dfrac{r}{3}=1$

$\Rightarrow r=3$

The volume of a solid hemisphere of radius

$2\ cm$ is ......

0%
$\displaystyle \frac{352}{21}\ cm^{3}$
0%
$\displaystyle 576\ cm^{3}$
0%
$\displaystyle \frac{376}{9}\ cm^{3}$
0%
$\displaystyle 600\ cm^{3}$

Explanation

Volume of a solid hemisphere

$(V) = \dfrac{2}{3}\times \pi \times r^3$

given radius

$r = 2\ cm$

$V=\dfrac{2}{3}\times \dfrac{22}{7}\times (0.02)^3=\dfrac{352}{21}\ cm^{3}$ '

The surface area of a sphere is

$\displaystyle 5544cm^{2}$ . Its volume is

0%
$\displaystyle 38793cm^{3}$
0%
$\displaystyle 24676cm^{3}$
0%
$\displaystyle 30088cm^{3}$
0%
$\displaystyle 83800cm^{3}$

Explanation

$4\pi r^2=5544$

$r^2=441\\ r=21$

$Volume=\dfrac{4\pi}{3}r^3=21^3=38793\ cm^3$

The surface areas of the two spheres are in the ratio

$1:2$ . The ratio of their volumes is

0%
$1:2$
0%
$\displaystyle 1:\sqrt{2}$
0%
$\displaystyle 1:2\sqrt{2}$
0%
$\displaystyle 1:3\sqrt{2}$

Explanation

Area ratio = 1 : 2 (given)

$\displaystyle \therefore$ radii ratio

$\displaystyle =1\sqrt{2}$
Volume ratio

$\displaystyle =1^{3}:\left ( \sqrt{2} \right )^{3}=1:2\sqrt{2}$

A sphere of diameter

$10$ cm weighs

$44$ kg. The weight of a sphere of the same material whose diameter is

$6$ cm is

0%
$2.64$ kg
0%
$1.584$ kg
0%
$0.9504$ kg
0%
$\displaystyle\frac{4}{3}(0.9504)\:kg$

Explanation

Let the weight of the sphere of diameter

$6$ cm be

$x$ kg and density of the material be

$d$ gm/c.c.

$\displaystyle\therefore\frac{x}{4.4}=\frac{\displaystyle\frac{4}{3\pi}\times3^3\times d}{\displaystyle\frac{4}{3\pi}\times5^3\times d}$

$\displaystyle\implies x=\frac{4.4\times27}{125}=0.9504\:kg$

The dimensions of a hall are 40 m, 25 m and 20 m. If each person requires 200 cubic m, then the number of persons who can be accommodated in the hall are

0%
150
0%
140
0%
120
0%
100

Explanation

Volume of hall

$= 40 \times 25 \times 20 = 20000$ cu. m.
If n is the number of persons which can be accommodated in the hall, then

$\times 200 = 20000$

$\therefore$ n = 100

How many spherical bullets can be made out of a solid cube of lead whose edge measures 44 cm if each bullet has radius 2 cm?

0%
3000
0%
2541
0%
1779
0%
2332

Explanation

Volume of the cube of side 44

$\displaystyle cm^{2}$ =

$\displaystyle 44\times 44\times 44cm^{3}$
Let the number of bullets be N Radius is 2 cm Therefore
Total volume of N spheres=Volume of the cube

$\displaystyle N\cdot \frac{4}{3}\times \frac{22}{7}\times 2\times 2\times 2=44\times 44\times 44$

$\displaystyle N=\frac{44\times 44\times 44\times 3\times 7}{4\times 22\times 2\times 2\times 2}=2541cm$

A solid metal sphere is cut through the center into two equal parts. Find the total surface area of each part if the radius of the sphere is

$3.5 cm$ .

0%
$\displaystyle 115.5cm^{2}$
0%
$\displaystyle 151.5cm^{2}$
0%
$\displaystyle 155cm^{2}$
0%
$\displaystyle 151cm^{2}$

Explanation

Given, radius of the hemisphere

$=3.5 cm$ .

According to the statement, the parts obtained are hemisphere.

Then, the total surface area of each hemisphere

$=3\pi r^{2}$

$=\displaystyle 3\times \frac{22}{7}\times 3.5\times 3.5\\=115.5cm^{2}$ .

Therefore, option

$A$ is correct.

The radii of two spheres are in the ratio 3:5 The ratio of their volumes is

0%
$9:25$
0%
$27:125$
0%
$\displaystyle \sqrt{3}:\sqrt{5}$
0%
$\displaystyle \sqrt[3]{3}:\sqrt[3]{5}$

Explanation

Ratio of radii =

$3:5$

$\displaystyle \therefore$ Ratio of volumes

$=$

$\displaystyle \dfrac{4}{3}\pi r_{1}^{3}:\dfrac{4}{3} \pi r_{2}^{3}$

$=\displaystyle 3^{3}:5^{3}$

$=27:125$

Find the volume of this rectangular prism (cuboid).

$($ Given:

$L = 2$ cm,

$B = 10$ cm,

$H = 30$ cm

$)$

0%
$\displaystyle 600\ { cm }^{ 3 }$
0%
$\displaystyle 760\ { cm }^{ 3 }$
0%
$\displaystyle 42\ { cm }^{ 3 }$
0%
$\displaystyle 500\ { cm }^{ 3 }$

Explanation

Volume of the cuboid

$=$ length

$\times$ breadth

$\times$ height

Therefore, length

$= 2$ cm, breadth

$= 10$ cm, height

$= 30$ cm
Volume of the cuboid

$=$

$\displaystyle 2\times 10\times 30=600{ cm }^{ 3 }$

CBSE Questions for Class 9 Maths Surface Areas And Volumes Quiz 5 - MCQExams.com

0:0:1

Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Surface Areas And Volumes Quiz 5 - MCQExams.com

0:0:1

Practice Class 9 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.