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

A brick whose length, breadth and height are

$5m, 6m$ , and

$7m$ respectively. Find the surface area of the brick.

0%
$214\ m^2$
0%
$202\ m^2$
0%
$201\ m^2$
0%
$210\ m^2$

Explanation

Surface area of the cuboid having length, breadth and height as

$l, b$ and

$h$ respectively is given as

$2 (lb + bh + hl)$

Brick has the shape of cuboid. Let

$S$ be the require surface area.

$S=2 (5 \times 6 + 6 \times 7 + 7 \times 5)$

$2 (30 + 42 + 35)$

$214 m^2$

The radius of a hemisphere is

$10\ cm$ . Calculate it's volume.

0%
$3125.53$ $cm^3$
0%
$2095.24$ $cm^3$
0%
$3005.35$ $cm^3$
0%
$2672.18$ $cm^3$

Explanation

Given,

radius of the hemisphere

$=10\ cm$

Volume of hemisphere

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

$=\dfrac{2}{3}\times\dfrac{22}{7}\times 10^3\ cm^3$

$=\dfrac{2}{3}\times\dfrac{22}{7}\times 1000\ cm^3$

$=2095.24\ cm^3$

Hence, the volume of the hemisphere is

$2095.24\ cm^3$ .

What is the volume of the hemisphere with radius

$21$ cm?

0%
$16404$ $cm^3$
0%
$17404$ $cm^3$
0%
$18404$ $cm^3$
0%
$19404$ $cm^3$

Explanation

Given, radius of hemisphere

$=21$ cm

Volume of the hemisphere

$=$

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

$=$

$\dfrac{2}{3}\times \dfrac{22}{7} \times 21^3$

$=$

$19404$

$cm^3$

Calculate the surface area of hemisphere having the radius of

$1.4$ cm.

0%
$1.232\ cm^2$
0%
$12.32\ cm^2$
0%
$123.2\ cm^2$
0%
$1232\ cm^2$

Explanation

Formula for surface area of the hemisphere with radius

$r$ is

$S=2\pi r^2$
Given , radius

$= 1.4$ cm

$S=2\times \cfrac{22}{7}\times 1.4^2$

$=12.32\ cm^2$

The largest possible sphere is carved out of a wooden solid cube of side

$7$ cm. Find the volume of the wood left.

$\left[\text{Use } \displaystyle \pi =\dfrac { 22 }{ 7 }\right]$

0%
$163.3\ cm^3$
0%
$164\ cm^3$
0%
$165\ cm^3$
0%
$170\ cm^3$

Explanation

For such sphere its diameter should be equal to the side of sphere.

Diameter of sphere $= 7$ cm

So, radius $= 3.5$ cm

Volume of the sphere carved out $= \dfrac{4}{3}\pi r^3$

$=\displaystyle \frac{4}{3}\times \frac{22}{7}\times 3.5^{3}$

$= 179.7 cm^3$

$\text{Volume of the cube of edge} 'a' \text{units} =a^3 \\= 7^3\\=343$

$\text{Volume of the wooden left} = 343 – 179.7 \\= 163.3\ cm^3$

The diagram shows the cross section of six identical marbles touching each other on a horizontal surface.
If the volume of a mabrle is

$\dfrac{9 \pi}{2} cm^3$ , calculate the length of PQ, in cm.

0%
9
0%
27
0%
18
0%
22

Explanation

Given: Volume of one marble

$=\dfrac{9\pi}{2}\ cm^3$

Let the radius of each marble be

$r$ cm.

Then volume of each marble

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

$\dfrac{9 \pi}{2} =\dfrac{4}{3} \pi r^3$

$r^3=\dfrac{9\pi}{2}\times \dfrac{3}{4}\times \dfrac{1}{\pi}$

$r^3=\dfrac{27}{8}$

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

Hence diameter of each marble

$=2\times \dfrac32=3\ cm$

We know that

$PQ = 6 \times$ diameter of each marble

$= 6 \times 3 = 18\ cm$

Calculate the total surface area of the right circular cylinder shown in the above figure (in square units).

0%
$\displaystyle 300\pi$
0%
$\displaystyle 400\pi$
0%
$\displaystyle 500\pi$
0%
$\displaystyle 600\pi$

Explanation

From the given figure, we can see that:

The diameter of the cylinder

$(d)=20$ and

The height of the cylinder

$(h)=20$

Hence, the radius of the cylinder will be

$r=\dfrac{20}{2}=10$

We know that, the total surface area of a cylinder is

$2\pi r^2+2\pi r h$

Here,

$r=10,\ h=20$

$\therefore \$ The total surface area

$=2\pi (10)^{2}+2\pi(10\times20)$

$=2\pi(100+200)$

$=\pi(2\times 300)$

$=600\pi\ square\ units$

Hence, the total surface area of the given cylinder is

$600\pi \ square\ units$ .

A closed metallic cylindrical box is 1.25 m high and its base radius is 35 cm. If the sheet metal costs Rs. 80 per m

$^2$ , the cost of the material used in the box is

0%
Rs. 281.60
0%
Rs. 290
0%
Rs. 340.50
0%
Rs. 500

Explanation

T.S.A. of cyl

$= 2\pi rh+2\pi r^{2} = 2\pi \times \left ( \dfrac{35}{100} \right )\times 1.25+2\pi \left ( \dfrac{35}{100} \right )^{2}$

$\Rightarrow TSA = 3.52\,m^{2}$ & as cost of

$1\,m^{2}$ is Rs 80

$\Rightarrow$ Total cost

$= 3.52\times 80 = Rs\,281.6$

If the total surface area of a solid hemisphere is

$462$

$\displaystyle { cm }^{ 2 }$ , find its volume.
Note: Take

$\displaystyle \pi =\frac { 22 }{ 7 }$

0%
$716$
0%
$718.67$
0%
$720.87$
0%
$840$

Explanation

Total surface area of the hemisphere $= 462 {cm}^2$

Total surface area of the hemisphere $= 2\pi r^2$

$\Rightarrow 462 = 3\pi r^2$

$\Rightarrow r = 7$ cm

Volume of hemisphere $= \dfrac{2}{3}\pi r^3$

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

$V = 718.67 {cm}^3$

The addition of curved surface and circular surfaces are called

0%
area
0%
perimeter
0%
lateral surface area
0%
total surface area

The curved surface area of a right circular cone of radius

$11.3$ cm is

$355$ cm

$^2$ . What is its slant height?

0%
$8$ cm
0%
$9$ cm
0%
$10$ cm
0%
$11$ cm

Explanation

Given, radius

$r=11.3 \text{ cm}$

and curved surface area

$=355 \text{ cm}^2$ .

We know, curved surface area of a cone

$=\pi rl$

$11.3 \pi l= 355$

$l = \dfrac{355}{\pi \times 11.3}$

${l = 10\text{ cm}}$

Therefore, option

$C$ is correct.

Find the surface area of a cube in square inches, whose each edge measures

$3$ inches.

0%
9
0%
18
0%
27
0%
36
0%
54

Explanation

Surface area of cube

$=6a^2$

Where '

$a$ ' is the edge of the cube

Surface area

$=$

$6\times (3)^2$

$\Rightarrow 6\times 9=54$ square inches

What is the surface area of a cylinder that has a top and bottom if the height is

$8.5$ units and the diameter is

$3$ units?

0%
$18\pi$ square units
0%
$30\pi$ square units
0%
$32\pi$ square units
0%
$36\pi$ square units
0%
$54\pi$ square units

Explanation

Given: height

$= 8.5$ , diameter

$= 3$

$\therefore$ radius

$=$

$\dfrac{D}{2}=\dfrac{3}{2}=1.5$
Surface area

$=$

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

$=2\pi\times1.5\times 8.5+2\pi\times 1.5^2$

$=2\pi(12.75+2.25)$

$=2\pi(15)$

$=30\pi$ sq.units

Each edge in a cube is

$5\ cm$ . What is the surface area in square cm?

0%
$25$
0%
$30$
0%
$100$
0%
$125$
0%
$150$

Explanation

Surface area of a cube is

$6a^2$ where

$a$ is an edge of the cube.

The surface area of the cube with edge

$5$ cm is given by:

$A=6a^{ 2 }=6\times (5)^{ 2 }=6\times 25=150cm^{ 2 }$

There are two cuboidal boxes as shown in the given figure. Does Box 1 require the less amount of material to make ?

0%
True
0%
False

Explanation

Total surface area of Box 1

$=2(lb + bh + hl)$

$= 2(60 \times 40 + 40 \times 50 + 60 \times 50)$

$= 2 (2400 + 2000 + 3000)$

$= 2 (7400)$

$= 14800$ sq. units

Total surface area of Box 2

$= 2(lb + bh + hl)$

$= 2 (50 \times 50 + 50 \times 50 + 50 \times 50)$

$= 2(2500 \times 3)$

$= 6 \times 2500$

$= 15000$ sq. units

$(TSA)_A < (TSA)_B$

$\therefore$ Box A requires less material

The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

0%
$5:16$
0%
$1:16$
0%
$3:16$
0%
None of these

Explanation

Let the diameter of earth be

$D$ & of moon be

$d$ .

$\therefore$ According to the statement,

$d=\dfrac { D }{ 4 }$ .

Then,

$\dfrac{d}{2}=\dfrac { D }{ 2\times4 }$ .

$\Rightarrow r=\dfrac { R }{ 4 }$

$\Rightarrow \dfrac { r }{ R } =\dfrac { 1 }{ 4 }$ .

$\therefore$ The ratio of their surface areas is

$= \dfrac { 4\pi { r }^{ 2 } }{ 4\pi { R }^{ 2 } } ={ \left( \dfrac { r }{ R } \right) }^{ 2 }={ \left( \dfrac { 1 }{ 4 } \right) }^{ 2 }=\dfrac { 1 }{ 16 } =1:16$ .

Hence, option

$B$ is correct.

The diameter of a sphere is decreased by

$25$ %. By what percent does its curved surface area decrease?

0%
$53.85 \%$
0%
$34.85 \%$
0%
$45.65 \%$
0%
$43.75 \%$

Explanation

$\textbf{Step 1: Calculating}$

$\text{Let the diameter of the sphere be } r$

$\implies \text{Surface area}_1=\pi d^2$

$\textbf{Step 2: Calculating}$

$\text{The diameter of the sphere changes to }r'$

$\implies d'=d-0.25d=0.75d$

$\implies \text{Surface area}=\pi d'^2$

$\implies \text{Surface area}_2=\pi (0.75d)^2=0.5625\pi d^2$

$\implies \text{% Change in surface area}=\dfrac{S.A_2-S.A_1}{S.S_1}\times 100=\dfrac{(0.5625-1)\pi d^2}{(1)\pi d^2}=-43.75$

$\textbf{Hence, The surface area decreases by 43.75%}$

Curved surface area of solid sphere is

$24{ cm }^{ 2 }$ . If the sphere is divided into two hemispheres, then the total surface area of one of the hemispheres is :

0%
$12{ cm }^{ 2 }$
0%
$8{ cm }^{ 2 }$
0%
$16{ cm }^{ 2 }$
0%
$18{ cm }^{ 2 }$

Explanation

$\text{Curved surface area of solid sphere}=4\pi r^2$

$\Rightarrow 4\pi r^2=24$

$\Rightarrow \pi r^2=6$ ...(i)
Now, total surface area of hemisphere

$=3\pi r^2$

$\therefore \text{Total surface area of hemisphere}=3(6)=18cm^2$ (from (i))
Option D is correct.

If the radius of a sphere is

$2\ cm$ , then the curved surface area of the sphere is equal to:

0%
$8\pi \ cm^{2}$
0%
$16 \ cm^{2}$
0%
$12\pi \ cm^{2}$
0%
$16\pi \ cm^{2}$

Explanation

Curved surface area of sphere

$=4\pi r^2$ .
Here,

$r=2 cm$ .

$\therefore$ Curved surface area of sphere

$=4\pi (2)^2=16\pi cm^2$ .
Therefore, option

$D$ is correct.

The total surface area of a solid hemisphere of diameter

$2$ cm is equal to

0%
$12{cm}^{2}$
0%
$12\pi {cm}^{2}$
0%
$4\pi {cm}^{2}$
0%
$3\pi {cm}^{2}$

Explanation

Total surface area of hemisphere

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

Given, diameter

$=2$ cm

Therefore, radius

$=1$ cm.

Therefore, total surface area is

$3\pi r^2$

$=3\times \pi\times (1)^2$

$=3\pi$ sq.cm.

The curved surface area of a right circular cylinder of radius

$1\ cm$ and height

$1\ cm$ is equal to

0%
$\pi$ ${cm}^{2}$
0%
$2\pi$ ${cm}^{2}$
0%
$3\pi$ ${cm}^{2}$
0%
$2$ ${cm}^{2}$

Explanation

We have,

$h=1\ cm, r=1\ cm$

We know that curved surface area of the cylinder

$=2\pi r h$

$=2\pi \times 1 \times 1$

$=2\pi\ cm^2$

Hence, this is the answer.

The curved surface area of a right circular cylinder whose radius is

$a$ units and height is

$b$ units, is equal to

0%
$\pi {a}^{2}b$ sq.cm
0%
$2\pi ab$ sq.cm
0%
$2\pi$ sq.cm
0%
$2$ sq.cm

Explanation

We have,

$r=a\ units, h=b\ units$

We know that the curved surface area

$=2\pi r h$

So,

$=2\pi a b\ sq.\ units$

Hence, this is the answer.

The inner curved surface area of a hemispherical dome of a building needs to be painted. If the circumference of the base is

$17.6\ m$ , find the cost of painting it at the rate of Rs.

$5$ per sq.m

0%
Rs. $246.40$
0%
Rs. $24.40$
0%
Rs. $26.40$
0%
None of these

Explanation

We have,

Circumference

$=17.6\ m$

Circumference of circle

$= 2\pi r$

$17.6 = 2\times \dfrac{22}{7}\times r$

$44r = 17.6\times 7$

$r = \dfrac{123.2}{44}=2.8\ m$

Now, Curved surface area of the hemisphere

$= 2\pi r^2$

$=2\times \dfrac{22}{7}\times 2.8\times 2.8$

$49.28\ m^2$

Cost of painting

$= Rs.\ 5\ per\ m^2$

Total cost of painting

$= 49.28\times 5$

$= Rs.\ 246.4$

Hence, the cost of painting is

$= Rs.\ 246.4$

If the surface area of a sphere is

$36\pi {cm}^{2}$ , then the volume of the sphere is equal to

0%
$12\pi {cm}^{3}$
0%
$36\pi {cm}^{3}$
0%
$72\pi {cm}^{3}$
0%
$108\pi {cm}^{3}$

Explanation

We have,

$surface\ area=36\pi\ cm^2$

We know that the surface area of the sphere

$=4\pi r^2$

So,

$4\pi r^2=36\pi$

$r^2=9$

$r=3\ cm$

We know that the volume of the sphere

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

Therefore,

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

$=4\times \pi \times 9$

$=36\pi\ cm^3$

Hence, this is the answer.

If the volume of a sphere is

$\cfrac{9}{16}\pi$

$cu.cm$ , then its radius is

0%
$\cfrac{4}{3}cm$
0%
$\cfrac{3}{4}cm$
0%
$\cfrac{3}{2}cm$
0%
$\cfrac{2}{3}cm$

Explanation

We have,

$V=\dfrac{9\pi}{16}\ cm^3$

We know that the volume of the sphere

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

So,

$\dfrac{4}{3}\pi\times r^3=\dfrac{9\pi}{16}$

$\dfrac{4}{3} r^3=\dfrac{9}{16}$

$r^3=\dfrac{27}{64}$

$r=\dfrac{3}{4}\ cm$

Hence, this is the answer.

The surface area of two spheres are in the ratio of

$9:25$ . Then their volumes are in the ratio

0%
$81:625$
0%
$729:15625$
0%
$27:75$
0%
$27:125$

Explanation

Given,

The ratio of surface area of two sphere

$=\dfrac{9}{25}$

We know that the surface area of the sphere

$=4\pi r^2$

So,

$\dfrac{4\pi r_1^2}{4\pi r_2^2}=\dfrac{9}{25}$

$\dfrac{r_1^2}{r_2^2}=\dfrac{9}{25}$

$\dfrac{r_1}{r_2}=\dfrac{3}{5}$

We know that the volume of the sphere

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

Therefore,

$=\dfrac{\dfrac{4}{3}\pi r_1^3}{\dfrac{4}{3}\pi r_2^3}$

$=\dfrac{3^3}{5^3}$

$=\dfrac{27}{125}$

Hence, this is the answer.

The cub (cubical) volume of a hemisphere, having diameter

$1\text{ cm}$ , will be .............

$\text{ cm }^{ 3 }$

0%
$\cfrac { \pi }{ 6 }$
0%
$\cfrac { \pi }{ 12 }$
0%
$\cfrac {2 \pi }{ 3 }$
0%
$\cfrac {4 \pi }{ 3 }$

Explanation

Here diameter of the hemisphere is given as

$1\ cm$

$\therefore$

$r=\cfrac{1}{2}$
Volume of the hemisphere

$=\cfrac { 2 }{ 3 } \pi { r }^{ 3 }=\cfrac { 2 }{ 3 } \times \pi \times \cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } =\cfrac { \pi }{ 12 }$

Thus, the answer is

$\dfrac{\pi}{12}$ .

If the total surface area of a solid hemisphere is

$12\pi \ {cm}^{2}$ , then its curved surface are is equal to

0%
$6\pi \ {cm}^{2}$
0%
$24\pi \ {cm}^{2}$
0%
$36\pi \ {cm}^{2}$
0%
$8\pi \ {cm}^{2}$

Explanation

Given, the total surface area of the hemisphere

$=12\pi\ cm^2$ .

We know that the total surface area of a hemisphere

$=3\pi r^2$ .

Hence,

$3\pi r^2=12\pi$

$\Rightarrow r^2=4$

$\therefore \ r=2\ cm$

Now, we know that the curved surface area of the hemisphere

$=2\pi r^2$ .

Hence, curved surface area

$=2\pi \times 2^2$

$\Rightarrow 8\pi\ cm^2$

Hence, the curved surface area of the solid hemisphere is

$8\pi \ cm^2$ .

The total surface area of a solid hemisphere whose radius is

$a$ units, is equal to

0%
$2\pi {a}^{2}$ sq. units
0%
$3\pi {a}^{2}$ sq. units
0%
$3\pi {a}^{}$ sq. units
0%
$3 {a}^{2}$ sq. units

Explanation

We have,

$r=a\ units$

We know that the total surface area of the hemisphere

$=3\pi r^2$

So,

$=3\pi a^2\ sq.\ units$

Hence, this is the answer.

How many litres of water does this fish tank will hold ?

0%
$2.4$ L
0%
$24$ L
0%
$4.8$ L
0%
$100$ L

Explanation

Volume

$=l \times b \times h=24 \times 20 \times 50=24000 \ \text{cm}^3$

$1L = 1000 \ \text{cm}^3$
So,

$24$ L

$=24000 \ \text{cm}^3$

Thus

$24$ litres of water the fish tank will hold.

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

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Practice Class 9 Maths Quiz Questions and Answers

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