MCQ Questions for Class 9 Maths Surface Areas And Volumes Quiz 10

Which of the following is a unit of Volume?

0%
Meters per second
0%
Cubic millimeters
0%
Litres
0%
Both B and C

Explanation

Volume can be cubic millimeters as well as litres

$1lt=1000000$ cubic millimeters
Both B anc C is correct

If the radius of a sphere is half of the radius of another sphere, then their respective volumes are in the ratio

0%
$1:8$
0%
$2:1$
0%
$1:2$
0%
$8:1$

Explanation

Let the radius of sphere be

$r$ and radius of other sphere be

$\dfrac{r}{2}.$

We know that the volume of the sphere

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

Therefore,

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

$=\dfrac{1}{8}\times \dfrac{\dfrac{4}{3}\pi r^3}{\dfrac{4}{3}\pi r^3}$

$=\dfrac{1}{8}\times 1$

$=\dfrac{1}{8}$

Hence, the ratio is

$1:8$

A bottle of drink holds

$250$

$\text{cm}^3$ . How many litres will

$24$ bottles hold?

0%
$600$ litres
0%
$60$ litress
0%
$6$ litres
0%
$6000$ litres

Explanation

$1 \ \text{cm}^3=0.001 \ \text{L}$

$250 \ \text{cm}^3=250 \times 0.001=0.25 \ \text{L}$
So,

$1$ bottle will hold

$0.25 \ \text{L}$
Therefore,

$24 \times 0.25=6 \ \text{L}$

How many liters is

$50 m^3$ ?

0%
$50$
0%
$500$
0%
$50000$
0%
$5$

Explanation

$1m^3=1000 L$

$50m^3=50000L$

How many

$\text{m}^3$ are

$7,500$ litres?

0%
$750 \ \text{m}^3$
0%
$75 \ \text{m^}3$
0%
$7.5 \ \text{m}^3$
0%
$100 \ \text{m}^3$

Explanation

We know

$1 \ \text{m}^3=1000$ L

$1$ L

$=\dfrac{1}{1000} \ \text{m}^3$

$7500$ L

$=\dfrac{7500}{1000}=7.5 \ \text{m}^3$

Option C is correct.

A water tank holds $50,000,000$ liters of water. How many cubic meters is that?

0%
$50,000,000\ m^3$
0%
$5,000,000\ m^3$
0%
$500,000\ m^3$
0%
$50,000\ m^3$

Explanation

$1\ m^3=1000\ L$

$1\ L=\dfrac{1}{1000}\ m^3$

$50000000\ L=\dfrac{50000000}{1000}\ m^3$

$=50000\ m^3$

Hence, option

$D$ is correct.

What is the volume in cubic centimetres of a

$0.6$ L bottle of shampoo?

0%
$6000 cm^3$
0%
$60 cm^3$
0%
$400 cm^3$
0%
$600 cm^3$

Explanation

$1L=1000cm^3$

$0.6L=0.6 \times 1000=600cm^3$

How many

$cm^3$ is 1.2litres?

0%
$1,200 cm^3$
0%
$1,200,00 cm^3$
0%
$120 cm^3$
0%
$12,000 cm^3$

Explanation

$1cm^3=\dfrac{1}{1000}=0.001L$

$1L=1000cm^3$

$1.2L=1.2 \times 1000=1200cm^3$

How many spherical bullets can be made out of a cube of lead whose edge measures

$22$ cm, each bullet being

$2$ cm in diameter?

0%
$1347$
0%
$2541$
0%
$2662$
0%
$5324$

Explanation

Volume of cube,

$V=(side)^3=(22^3 )cm ^3$

Volume of one spherical bullets,

$v=\dfrac{4}{3}\pi r^3=\dfrac{4}{3}\pi (1^3)cm^3$

Number of spherical bullets

$=\dfrac{V}{v}=\dfrac{22^3\times 3}{4\times \pi \times1 ^3}=2541$

If the radius of a sphere is increased by

$2$ cm, then its surface area increases by

$352cm^2$ . The radius of the sphere before the increase was _________.

0%
$3$ cm
0%
$4$ cm
0%
$5$ cm
0%
$6$ cm

Explanation

Let the radius of the sphere before increase

$=r$ .

Given, increased surface area

$=352cm^2$ .

Also, the surface area of the sphere before the increase is

$A=4\pi r^2$ .

After increasing radius by

$2 \ cm$ , the area is increased by

$352 \ cm^2$ .

Applying the above conditions:

$\therefore 4\pi r^2+352=4\pi (r +2)^2\\$

$\Rightarrow 352=16\pi+16\pi r$

$\Rightarrow 1+r=\dfrac{352}{16\pi }=7$

$\Rightarrow r=6cm$ .

Hence, option

$D$ is correct.

How many liters of water can be stored in a hemisphere of radius 10.5 dm?

0%
3465 lit
0%
2425.5 lit
0%
2465.5 lit
0%
3425 lit

Explanation

solution:

volume of hemisphere =

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

V =

$\dfrac{2}{3} \pi (10.5)^3$

V =

$\dfrac{2}{3} 3.14 \times 1157.6$

V = 2425.5 lit

hence the correct opt: B

Calculate the surface area of a sphere whose radius is

$21 cm$ .

0%
$5544\ { cm }^{ 2 }$
0%
$6622\ { cm }^{ 2 }$
0%
$7733\ { cm }^{ 2 }$
0%
$4455\ { cm }^{ 2 }$

Explanation

Given, radius $r=21 cm$ .

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

$= 4 \times \dfrac {22}{7} \times 21 \times 21$

$= 5544 {cm}^{2}$ .

Therefore, option

$A$ is correct.

If the volume of a sphere is divided by its surface area, we obtain 27 cm. The radius of the sphere is

0%
9 cm
0%
81 cm
0%
27 cm
0%
24 cm

If the radius of a sphere is doubled, then its volume is increase by __________.

0%
$100\%$
0%
$200\%$
0%
$700\%$
0%
$800\%$

Explanation

Given that radius of the sphere is doubled.

Let the Initial volume of sphere

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

Incresed volume

$V'=\dfrac{4}{3}\pi (2r)^3=8V$

Thus, Volume is increased by:

$\dfrac{8V-V}{V}\times 100=700 \%$

The largest sphere is cut off from a cube of side

$5\ cm$ . The volume of the sphere will be __________.

0%
$27\pi \ cm^3$
0%
$30\pi\ cm^3$
0%
$108\pi\ cm^3$
0%
$\displaystyle\frac{125\pi}{6}\ cm^3$

Explanation

Diameter of the largest sphere= side of the cube

$=5\ cm$

$\therefore$ volume of the sphere

$=\dfrac{4}{3}\pi r^3=\dfrac{4}{3}\pi \times (\dfrac{5}{2})^3\\=\dfrac{125\pi}{6}\ cm ^3$

The number of balls of radius

$1$ cm that are made from a solid sphere of radius

$4$ cm

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

Explanation

$No\quad of\quad balls=\cfrac { Volume\quad of\quad solid\quad sphere }{ volume\quad of\quad ball } =\cfrac { \cfrac { 4 }{ 3 } \pi .4\times 4\times 4 }{ \cfrac { 4 }{ 3 } \pi \times 1\times 1\times 1 } =64\quad balls$

The volume of a sphere is

$\dfrac {88}{21}\times (14)^{3} cm^{3}$ . The curved surface of the sphere is (Take

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

0%
$2424\ cm^{2}$
0%
$2446\ cm^{2}$
0%
$2484\ cm^{2}$
0%
$2464\ cm^{2}$

Explanation

Given that,
Volume of sphere

$= \dfrac {88}{21} \times (14)^{3}$
We know that,
Volume of the sphere

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

$\dfrac {4}{3}\pi r^{3} = \dfrac {88}{21}\times (14)^{3}$

$\Rightarrow \dfrac {4}{3}\times \dfrac {22}{7} \times r^{3} = \dfrac {88}{21} \times (14)^{3}$

$r^{3}\Rightarrow (14)^{3} \Rightarrow r = 14$
We know that,
Curved surface of the sphere

$= 4\pi r^{2}$

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

If the volume and surface area of a sphere are numerically the same, then its diameter is

0%
$6$ units
0%
$8$ units
0%
$10$ units
0%
$12$ units

Explanation

Let radius of sphere be 'r', then

Volume of sphere=surface area of sphere

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

$\therefore$ Diameter

$=2\times 3=6$ units

The radii of two sphere are in the ratio

$3:4$ . The ratio of their surface area is:

0%
$3:16$
0%
$4:16$
0%
$6:16$
0%
$9:16$

Explanation

Given, the ratio of radii of two spheres is

$3:4$ .

We know, the surface area of a sphere

$=4 \pi r^2$ .

Let

$r_1$ and

$r_2$ be the radii of the two sphere,

i.e.

$r_1 : r_2=\dfrac{r_1}{r_2}=3:4=\dfrac{3}{4}$ .

Then, the areas of the spheres be

$4 \pi r_1^2$ and

$4 \pi r_2^2$ .

Therefore, the ratio of their surface area

$=\dfrac{4 \pi r_1^2}{4\pi r_2^2}=\dfrac{r_1^2}{r_2^2}=(\dfrac{r_1}{r_2})^2=(\dfrac{3}{4})^2=\dfrac{9}{16}=9:16$ .

Hence, option

$D$ is correct.

The total surface area of a sphere is,

$154\ { cm }^{ 2 }$ . Its radius is 3.5 cm. Find its volume.

0%
$169.77\ { cm }^{ 3 }$
0%
$189.07\ { cm }^{ 3 }$
0%
$159.27\ { cm }^{ 3 }$
0%
$179.67\ { cm }^{ 3 }$

Explanation

solution:

volume of sphere is:

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

V =

$\dfrac{4}{3} \times3.14 \times(3.5)^3$

$V = 179.5 cm^3$

hence the correct opt: D

The total surface area of a closed cylindrical petrol storage tank whose diameter

$4.2$ m and height

$4.5$ m.

0%
$87.12$ sq. m
0%
$78.21$ sq. m
0%
$69.23$ sq. m
0%
$65.45$ sq. m

Explanation

Total surface area of cylindrical tank

$=2\pi r(r+h) \\ = 2\times \cfrac{22}{7}\times \cfrac{4.2}{2}\left(\cfrac{4.2}{2}+4.5\right) \\ =13.2\times 6.6 = 87.12 cm^2$

The volume of a sphere of radius

$\pi$ cm is ________ cc.

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

Explanation

The volume of a sphere with radius

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

$\therefore$ The volume of a sphere with radius

$\pi \ cm$ .

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

$= \dfrac {4\pi^{4}}{3}(cm)^{3}$ .

A closed water tank has an internal size of

$10m \times 15m \times 20m$ . It needs to be lined with waterproofing cement on all its internal sides. At the rate of Rs.

$250\ \text {per} \ m^2$ , the total cost of lining will be Rs:

0%
$237, 500$
0%
$325,000$
0%
$400,000$
0%
$475,000$

Explanation

Total surface area of water tank

$=2(lb+bh+hl)=2(10\times 15+15\times 20+20\times 10)$

$= 2\times (150+300+200)=2\times 650$

$=1300m^2$

Cost of lining

$=Rs (250\times 1300)=Rs\; 325000$

Hence option

$'B'$ is the answer.

If the curved surface area of a hemisphere is 2772

$cm^2$ , then its diameter is -

0%
42 cm
0%
21 cm
0%
20 cm
0%
40 cm

Explanation

$CSA = 2\pi r^{2} = 2772\,cm^{2}$

$2\times \dfrac{22}{7}\times r^{2} = 2772$

$r^{2} = \dfrac{126\times 7}{2}$

$= 63\times 7$

$r^{2} = 3^{2}\times 7^{2}$

$r = 21\,cm$

Since

$d$ =

$2r$

$= 42\,cm$

The volume of a sphere whose diameter is

$14 ~cm$ is

0%
$5226.66 \ cm^3$
0%
$4281.33 \ cm^3$
0%
$1437.33 \ cm^3$
0%
$2382.66 \ cm^3$

A building has

$25$ cylindrical shaped poles. Each has a radius of

$28$ cm and a height of

$4$ cm. Find the cost of painting curved surface of all poles at the rate of Rs.

$8$ per

$m^2$ .

0%
$7.23$ Rs
0%
$8$ Rs
0%
$14.08$ Rs
0%
$56.32$ Rs

Explanation

Given,

Radius of each pillar= 28 cm

Height of each pillar=4 cm

So, as we all know

CSA of a cylinder=

$2 \dfrac {22} {7} rh$

CSA of one pillar

$= 2 \dfrac {22} {7} (28) (4)$

$=704 \ cm^2$ =

$704 \times 10^{-4} \ m^2$

$=704 \times 10^{-4} \times (25)$

$=17600 \times 10^{-4} \ m^2$

Since, cost of painting 1 sq m= Rs. 8

∴Cost of 1.76 sq m =Rs. 8 ( 1.76)

= Rs. 14.08

From a solid sphere of radius

$R$ , a concentric solid sphere of radius

$\dfrac{R}{2}$ is removed. The total surface area increases by

0%
$0\%$
0%
$25\%$
0%
$50\%$
0%
$10\%$

Explanation

Given,

From a solid sphere of radius

$R$ , a concentric solid sphere of radius

$\dfrac{R}{2}$ is removed.

Initial surface area of the sphere

$\left( {A}_{1} \right) = 4 \pi {R}^{2}$

Now, the new surface area of sphere

$\left( {A}_{2} \right) = 4 \pi {R}^{2} + 4 \pi {\left( \cfrac{R}{2} \right)}^{2}$

$=4 \pi {R}^{2} + \pi {R}^{2}$

$= 5 \pi {R}^{2}$ .

$\therefore$ Increase in area is

$= \cfrac{{A}_{2} - {A}_{1}}{{A}_{1}} \times 100\%$

$= \cfrac{5 \pi {R}^{2} - 4 \pi {R}^{2}}{4 \pi {R}^{2}} \times 100\%$

$= \cfrac{\pi {R}^{2}}{4 \pi {R}^{2}} \times 100\%$

$= 25 \%$ .

Hence, the area will be increased by

$25 \%$ .

Therefore, option

$B$ is correct.

Area of the four walls of a big room measuring

$25\ m \times 18\ m\times 10\ m$ is

0%
$1000\ m^{2}$
0%
$940\ m^{2}$
0%
$860\ m^{2}$
0%
None of these

A sphere has volume

$36\pi\ cm^{3}$ , find the radius of the sphere

0%
$4cm$
0%
$3cm$
0%
$6cm$
0%
$8cm$

Explanation

Volume of sphere

$=\dfrac{4}{3}\pi r^3=36\pi \ c m^3$

$\implies r^3=27cm^3$

Hence,

$r=3cm$

Which of the following is the formula for the volume of the cubiod

0%
$V=lbh$
0%
$V=lh$
0%
$V=lb$
0%
$V=l^2bh$

Explanation

We know,

Volume of cuboid =

$length \times breadth \times height$

Therefore,

$V=lbh$

Hence, option A is correct.

CBSE Questions for Class 9 Maths Surface Areas And Volumes Quiz 10 - 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 10 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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