MCQ Questions for Class 9 Maths Surface Areas And Volumes Quiz 4

A building has

$8$ right cylindrical pillars whose cross sectional diameter is

$1$ m and whose height is

$4.2$ m. Find the expenditure to paint these pillars at the rate of Rs.

$24$ per

$m^2$

0%
Rs. $2534.40$
0%
Rs. $2506.13$
0%
Rs. $2610.9$
0%
Rs. $2514.5$

Explanation

Given, diameter

$=1$ m, radius

$=$

$\displaystyle \frac{diameter}{2}=\frac{1}{2}$
height of cylinder

$=4.2$ m

Curved surface area of cylinder

$=2 \pi r h=2 \pi \times \dfrac{1}{2} (4.2)=13.2 m^2$

$\therefore 8 \times$ right circular cylinder

$= 8 \times 13.2= 105.6 m^2$

$\therefore$ cost of painting pillar at the rate of Rs.

$24$ per

$m^2=24 \times 105.6=$ Rs.

$2534.4$

A cylindrical container with internal radius of its base

$10$ cm, contains water up to a height of

$7$ cm. Find the area of the wet surface of the cylinder.

0%
$324.29$ cm $^{2}$
0%
$754.29$ cm $^{2}$
0%
$674.29$ cm $^{2}$
0%
$584.29$ cm $^{2}$

Explanation

Given, A cylindrical container with internal radius 10 cm and water filled up to height 7 cm.

Wet surface area of cylinder

$=$ curved surface area of cylinder up to 7 cm + Area of base of cylinder

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

$=\pi r(2h+r)$

$=\dfrac { 22 }{ 7 } \times 10\times (14+10)\\ =\dfrac { 22 }{ 7 } \times 10\times 24\\ =754.29\quad { cm }^{ 2 }$

Curved surface of right circular cylinder is

$4.4{m}^{2}$ , radius of base is

$0.7\ m,$ then the height is

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

0%
$1$ m
0%
$2$ m
0%
$3$ m
0%
$4$ m

Explanation

Curved surface area of a cylinder of radius " $R$ " and height $"h$ " is $2\pi Rh$ .

Given, curved surface area $=44,4$ $m^2$
Therefore, CSA of the given cylinder $= 2\times \dfrac {22}{7} \times 0.7\times h = 4.4$ sq.cm

$\Rightarrow \dfrac {44}{10}=\dfrac {44}{7}\times \dfrac {7}{10}\times h$

$\Rightarrow h = 1$ m

Hence, option 'A' is correct.

If a solid metallic sphere of radius

$8$ cm is melted and recasted into

$n$ spherical solid balls of radius

$1$ cm, then

$n$ is:

0%
$500$
0%
$510$
0%
$512$
0%
$516$

Explanation

Volume of sphere

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

For solid metallic sphere

$radius (r)=8\ cm$

$\therefore Volume=\dfrac{4}{3}\times 3.14 \times 8^3$

$=\dfrac{4}{3} \times 3.14 \times 512$

$=\dfrac{6430.72}{3}$

$=2143.57\ cm^3$

For spherical solid balls

$radius(R)=1\ cm$

$\therefore Volume=\dfrac{4}{3}\times 3.14 \times 1^3$

$=\dfrac{4}{3} \times 3.14 \times 1$

$=\dfrac{12.56}{3}$

$=4.186\ cm^3$

In order to find the number of balls carved from larger sphere we need to divide the larger ball to small ball

$\therefore \dfrac{2143.57}{4.186}=512.08 \approx 512\ balls$

Find the surface area of a sphere of radius

$10.5$ cm.

0%
$1386\ {cm}^{2}$
0%
$1836\ cm^2$
0%
$1368\ cm^2$
0%
$1683\ cm^2$

Explanation

Given, radius $r=10.5$ cm.

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

$= 4 \times \dfrac {22}{7} \times 10.5 \times 10.5$

$= 1386 {cm}^{2}$ .

Therefore, option $A$ is correct.

The dimensions of a car petrol tank are

$50$ cm

$\times\, 32$ cm

$\times\, 24\,$ cm, which is full of petrol. If car's average consumption is

$15$ km per litre, find the maximum distance that can be covered by the car.

0%
$576$ km
0%
$500$ km
0%
$543$ km
0%
$567$ km

Explanation

To find the capacity of the petrol tank, we have to find the volume of the tank.
Volume of a car petrol tank

$=$

$lbh$
With the given measurements, we get

$V =$

$50$ cm

$\times 32$ cm

$\times 24$ cm

$=$

$38400$

${ cm }^{ 3 }$
As

$1$ litre

$= 1000$

${ cm }^{ 3 }$
So,

$38400$

${ cm }^{ 3 }$

$=$

$\dfrac { 38400 }{ 1000 }$

$= 38.4$ Lts of petrol
The quantity of petrol in the tank is

$38.4$ Lts.
As given, the average consumption of the petrol is

$15$ kms/ Lts.
The maximum distance which can be covered by the car

$= 38.4$

$\times$

$15=576$ kms.

A closed cylindrical tank, made of thin iron-sheet, has diameter

$8.4$ m and height

$5.4$ m. How much metal sheet to the nearest

$m^2$ is used in making this tank, if

$\displaystyle \frac{1}{15}$ of the sheet actually used was wasted in making the tank ?

0%
$272\, m^2$
0%
$270 \,m^2$
0%
$235 \,m^2$
0%
$240 \,m^2$

Explanation

Given diameter

$=8.4$ m
We know, diameter

$=2 \times$ radius

$\therefore$ radius

$=4.2$ m
Also given, height

$=5.4$ m

$\therefore$ total surface area of cylinder

$=2 \pi r(r+h) cm^2$

$=2 (3.14)(4.2)(4.2+5.4)=253.44$

$\dfrac{1}{15}$ of the sheet is used was wasted to making the tank

$=\dfrac{1}{15} \times 253.44=16.896$

$\therefore$ total metal sheet used in making tank

$=253.44+16.896=270.336 m^2$

$=270 m^2$

Find the surface area of a sphere of radius

$14$ cm.

0%
$2464\ {cm}^{2}$
0%
$1446\ cm^2$
0%
$2446\ cm^2$
0%
$2664\ cm^2$

Explanation

We know that the surface area of sphere with radius

$r$ is

$A=4πr^{ 2 }$ .

It is given that the radius of the sphere is

$14 cm$ .

Then,

$A=4πr^{ 2 }=4\times \dfrac { 22 }{ 7 } \times (14)^{ 2 }=4\times \dfrac { 22 }{ 7 } \times 196=2464$

$cm^2$ .

Hence, the surface area of sphere is

$2464$

$cm^2$ .

Therefore, option

$A$ is correct.

Volume and total surface area of a solid hemisphere are equal in magnitude. The volume is expressed in cm

$^3$ and the area is expressed in cm

$^2$ . Find the radius of hemisphere.

0%
$3$ cm
0%
$4$ cm
0%
$4.5$ cm
0%
$5.5$ cm

Explanation

Volume of a hemisphere of radius '

$r$ '

$= \dfrac { 2 }{ 3 } \pi { r }^{ 3 }$
Total Surface Area of a hemisphere of radius '

$r$ '

$= 3\pi { r }^{ 2 }$
According to the question, we have

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

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

$\Rightarrow r = \dfrac {9}{2} = 4.5$
Hence, option 'C' is correct.

Find the surface area of a sphere of radius

$5.6$ cm.

0%
$394.24{cm}^{2}$
0%
$394.42\ cm^2$
0%
$493.24\ cm^2$
0%
$493.42\ cm^2$

Explanation

Given, radius $r=5.6$ cm.

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

$= 4 \times \dfrac {22}{7} \times 5.6 \times 5.6$

$= 394.24 {cm}^{2}$ .

Therefore, option $A$ is correct.

If the curved surface area of a solid right circular cylinder of height

$h$ and radius

$r$ is one-third of its total surface area, then

0%
$\displaystyle h\, =\, \frac{1}{3}\, r$
0%
$\displaystyle h\, =\, \frac{1}{2}\, r$
0%
$h = r$
0%
$h = 2r$

Explanation

We know, curved surface area of a cylinder of radius "

$R$ " and height "

$h$ "

$= 2\pi Rh$
Total surface area of a cylinder of radius "

$r$ " and height "

$h$ "

$= 2\pi r(r + h)$
Given,

$2\pi Rh = \dfrac { 1 }{ 3 } \times 2\pi r(r+h)$

$\therefore h = \dfrac { 1 }{ 3 } \times (r+h)$

$\therefore \dfrac { 2 }{ 3 } h = \dfrac { 1 }{ 3 } r$

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

A hemispherical bowl is made of steel,

$0.25cm$ thick. The inner radius of the bowl is

$5cm$ . Find the outer curved surface area of the bowl.

0%
$173\ {cm}^{2}$
0%
$133\ {cm}^{2}$
0%
$143\ {cm}^{2}$
0%
$273\ {cm}^{2}$

Explanation

$r=5cm$ , thickness of steel sheet

$=0.25cm$

$\Rightarrow$

$R=5cm+0.25cm=5.25cm$

outer curved surface area of the bowl

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

$=2\times \cfrac{22}{7}\times \cfrac{525}{100}\times \cfrac {525}{100}{cm}^{2}=173.25{cm}^{2}$

Find the volume of a sphere whose surface area is

$154{cm}^{2}$ .

0%
$179\cfrac{2}{3}{cm}^{3}$
0%
$179\cfrac{4}{3}{cm}^{3}$
0%
$179\cfrac{5}{3}{cm}^{3}$
0%
$179\cfrac{1}{3}{cm}^{3}$

Explanation

Surface area of a sphere

$= 4\pi { r }^{ 2 } = 154$

$=> 4 \times \cfrac {22}{7}\times { r }^{ 2 } = 154 {cm}^{2}$

$=> { r }^{ 2 } = \cfrac {49}{4}$ $=\cfrac {7}{2} cm$
Volume of a sphere $= \cfrac { 4 }{ 3 } \pi { r }^{ 3 }$

$= \cfrac {4}{3} \times \cfrac {22}{7} \times \cfrac {7}{2} \times \cfrac {7}{2} \times \cfrac {7}{2} = 179 \cfrac {2}{3} {cm}^{3}$

Find the surface area of sphere of diameter

$14$ cm.

0%
$616{cm}^{2}$
0%
$661\ cm^2$
0%
$166\ cm^2$
0%
None of these

Explanation

Given, diameter

$d=14$ cm.

$\therefore$ radius

$= \dfrac {d}{2} = \dfrac {14}{2} = 7$ cm

Surface area of a sphere of radius $r$ $= 4\pi { r }^{ 2 }$

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

$= 616 {cm}^{2}$ .

Therefore, option $A$ is correct.

Find the amount of water displaced by a solid spherical ball of diameter

$28cm$ .

0%
$11498\cfrac{2}{3}{cm}^{3}$
0%
$11498\cfrac{2}{7}{cm}^{3}$
0%
$11498\cfrac{1}{3}{cm}^{3}$
0%
$11498\cfrac{2}{5}{cm}^{3}$

Explanation

Radius of the spherical ball $= \cfrac {28}{2} = 14 cm$

The amount of water it displaces is equal to its volume.

Volume of a spherical ball $= \cfrac { 4 }{ 3 } \pi {r }^{ 3 }$

' $= \cfrac {4}{3} \times \cfrac {22}{7} \times 14 \times 14 \times 14$

$= \cfrac { 34496 }{3} = 11498\cfrac 23 {cm}^{3}$

Find the volume of a sphere whose radius is

$0.63m$

0%
$1.05{m}^{3}$
0%
$3.05{m}^{3}$
0%
$1.15{m}^{3}$
0%
$2.05{m}^{3}$

Explanation

$r=0.63m$
Volume

$=\cfrac{4}{3}\times {22}{7}\times {(0.63)}^{3}{m}^{3}$

$=1.047816{m}^{3}=1.05{m}^{3}$ (approximately)

A capsule of medicine is in the shape of a sphere of diameter

$3.5mm$ . How much medicine (in

${mm}^{3}$ ) is needed to fill this capsule?

0%
$22.458\ {mm}^{3}$
0%
$32.346\ {mm}^{3}$
0%
$23.346\ {mm}^{3}$
0%
$12.346\ {mm}^{3}$

Explanation

$r=\cfrac{3.5}{2}mm$
Capacity of the capsule

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

$=\cfrac{4}{3}\times \cfrac{22}{7}\times \cfrac{3.5}{2}\times \cfrac{3.5}{2}\times \cfrac{3.5}{2}{mm}^{3}$

$=\cfrac{4}{3}\times \cfrac{22}{7}\times \cfrac{7}{4}\times \cfrac{7}{4}\times \cfrac{7}{4}{mm}^{3}=\cfrac{11}{24}\times 49{mm}^{3}$

$=\cfrac{539}{24}{mm}^{3}=22.346{mm}^{3}$

Find the total surface area of a hemisphere of radius

$10$ cm. (Use

$\pi =3.14$ )

0%
$942{cm}^{2}$
0%
$492\ cm^2$
0%
$249\ cm^2$
0%
$924\ cm^2$

Explanation

Given, radius

$r=10$ cm

Total surface area of a hemisphere of radius '

$r$ '

$= 3\pi { r }^{ 2 }$
Hence, total surface area of this hemisphere

$= 3 \times 3.14 \times 10 \times 10$

$= 942$ sq.cm

Find the radius of a sphere whose surface area is

$154{cm}^{2}.$

0%
$3.5$ cm
0%
$4.5$ cm
0%
$4$ cm
0%
$3$ cm

Explanation

Given, surface area of sphere $=154 cm^2$

Surface area of a sphere of radius ' $r$ ' $= 4\pi { r }^{ 2 } = 154$

$\Rightarrow 4 \times \dfrac {22}{7} \times { r }^{ 2 } = 154 {cm}^{2}$

$\Rightarrow { r }^{ 2 } = \dfrac {49}{4}$

$\Rightarrow r = \dfrac {7}{2} = 3.5 cm$ .

Therefore, option $A$ is correct.

It costs Rs.

$2200$ to paint the inner curved surface of a cylindrical vessel

$10\ m$ deep. If the cost of painting is at the rate of Rs.

$20$ per

${m}^{2}$ .Find Inner curved surface area of the vessel.

0%
$110\ {m}^{2}$
0%
$140\ {m}^{2}$
0%
$210\ {m}^{2}$
0%
$120\ {m}^{2}$

Explanation

Total cost of point

$=Rs.2200$
cost of painting

$1{ m }^{ 2 }\quad is\quad Rs.20.$

$\therefore$ total area painted

$=\cfrac { 2200 }{ 20 } =110{ m }^{ 2 }$
therefore inner curved surface area is

$110{ m }^{ 2 }.$

A match box measures

$4\ cm\times 2.4\ cm\times 1.5\ cm$ . What will be the volume of a packet containing

$12$ such boxes?

0%
$172.8\ {cm}^{3}$
0%
$120\ {cm}^{3}$
0%
$280\ {cm}^{3}$
0%
$360\ {cm}^{3}$

Explanation

Volume of 1 match box

$=(4\times 2.4\times 1.5){ cm }^{ 3 }\\ =14.4{ cm }^{ 3 }$
volume of 12 such match box

$=14.4\times 12\\ =172.8{ cm }^{ 3 }$

What is the total surface area of a cube whose side is

$0.5$ cm?

0%
$1.5 cm^2$
0%
$0.25 cm^2$
0%
$0.125 cm^2$
0%
$2.5 cm^2$

Explanation

The surface area of the cube of side '

$a$ ' is given by,

$S=6a^2$
Here,

$a=0.5$ cm

$\therefore$ the total surface area of a cube

$=6(0.5)^2=1.5 cm^2$

If the diameter of a cylinder is

$28$ cm and its height is

$20$ cm, then total surface area (in

${cm}^{2}$ ) is:

0%
$2993$
0%
$2992$
0%
$2292$
0%
$2229$

Explanation

Total surface area of a cylinder of Radius " $R$ " and height " $h$ " $= 2\pi R(R + h)$
Radius of the base of the cylinder $= \dfrac {28}{2} = 14$ cm

Hence, total surface area of the cylinder $=2\times \dfrac { 22 }{ 7 } \times 14(14 + 20)$

$= 2992 cm^2$

Find the quantity of water in liters in a hemispherical bowl of radius

$21\text{ cm}$ . The bowl is completely filled with water. (Take

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

0%
$19.404\text{ liters}$
0%
$16.141\text{ liters}$
0%
$22.254\text{ liters}$
0%
$18.204\text{ liters}$

Explanation

Volume of a hemisphere

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

$= \cfrac {2}{3} \times \cfrac {22}{7} \times 21 \times 21 \times 21$

$= 19404\ {cm}^{3}$

$1\text{ liter}= 1000{cm}^{3}$

Hence, in

$19404\ {cm}^{3} , \cfrac {19404}{1000} = 19.404\text{ liters}$ of water can be filled.

How many bricks, each measuring

$25 cm \times 11.25 cm \times 6 cm$ will be needed to build a wall

$8 m \times 6 m \times 22.5 cm$ ?

0%
$5600$
0%
$6000$
0%
$6400$
0%
$7200$

Explanation

Number of bricks

$=\, \displaystyle \frac{Volume\, of\, wall}{Volume\, of\, 1\, brick}$

$=\, \displaystyle \frac{800\, \times\, 600\, \times\, 22.5}{25\, \times\, 11.25\, \times\, 6}$ ....

$(1m = 100 cm)$

$= 6400.$

Find the volume of a cuboid of length

$20$ cm, breadth

$15$ cm and height

$10$ cm.

0%
$8794$ $cm^3$
0%
$7680$ $cm^3$
0%
$3000$ $cm^3$
0%
$5760$ $cm^3$

Explanation

Given, length of the cuboid

$= 20$ cm
Breadth of the cuboid

$= 15$ cm
Height of the cuboid

$= 10$ cm

$\therefore$ Volume of the cuboid

$=$ length

$\times$ breadth

$\times$ height

$=3000 cm^3$

A cuboid is

$3$ cm high,

$4$ cm wide and

$5$ cm long. What is its volume?

0%
$60 cm^3$
0%
$50 cm^3$
0%
$150 cm^3$
0%
$100 cm^3$

Explanation

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

$length=5\ cm$

$breadth=4\ cm$

$height=3\ cm$

$Volume=5*4*3=60\ cm^3$

Find the volume of a sphere whose surface area is

$55.44{cm}^{2}$ . (Take

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

0%
$38.808\ {cm}^{3}$
0%
$38.008\ {cm}^{3}$
0%
$32.808\ {cm}^{3}$
0%
$28.808\ {cm}^{3}$

Explanation

Let

$r$ cm be the radius of the sphere. Its surface area

$=55.44{cm}^{2}$

$\Rightarrow$

$4\pi {r}^{2}=55.44$

$\Rightarrow$

$4\times \cfrac{22}{7}\times {r}^{2}=55.44$

$\Rightarrow$

${r}^{2}=\cfrac{55.44\times 7}{4\times 22}=4.41={(2.1)}^{2}$

$\Rightarrow$

$r=2.1cm$
now, volume of the sphere

$=\cfrac{4}{3}\pi {r}^{3}=\cfrac{1}{3}(4\pi {r}^{2})\times r=\cfrac{1}{3}\times 55.44\times 2.1{cm}^{3}$

$=55.55\times 0.7{cm}^{3}$
Hence the volume of the sphere

$=38.808{cm}^{3}$

If a cistern is

$3$ metres long,

$2$ metres wide and

$1$ metre deep, its capacity is

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

Explanation

Capacity of cistern

$=\, l\, \times\, b\, \times\, h$

$=\, 3\, m\, \times\, 2\, m\, \times\, 1\, m\, =\, 6\, m^3=\, 6000\, litres$ .

$(\therefore\, 1\, m^3\, =\, 1000\, L)$

Write the curved surface area of a right circular cylinder whose radius is 3 cm and height is 5 cm.

0%
$30\pi \ {cm}^{2}$
0%
$20\pi \ {cm}^{2}$
0%
$35\pi \ {cm}^{2}$
0%
$40\pi \ {cm}^{2}$

Explanation

Radius of cylinder

$=3 \text{ cm}$

Height of cylinder

$=5 \text{ cm}$

Cross sectional area of cylinder

$=2\pi rh=2\times \pi \times 3\times 5=30\pi \text{ cm}^2$

CBSE Questions for Class 9 Maths Surface Areas And Volumes Quiz 4 - 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 4 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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