# Free Surface Areas and Volumes 02 Practice Test - 10th Grade

### Question 1

A container made up of a metal sheet is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of the milk which can completely fill the container at the rate of Rs. 15 per litre and the cost of the metal sheet used, if it costs Rs. 5 per 100 cm2. (Take π= 3·14)

Rs. 156.75 , Rs. 92.45

Rs. 155.75, Rs. 95.63

Rs. 156.75 , Rs 97.97

Rs. 150, Rs. 100

#### SOLUTION

Solution :C

R = 20 cm, r = 8cm, h = 16 cm

l = √h2+(R−r)2=√256+144cm=20cm

Volume of container = 13πh(R2+r2+Rr)

=13×(3.14)×16(400+64+160)cm3

=13×3.14×16×624cm3

=3.14×16×208cm3

=10449.93cm3

Therefore, the quantity of milk in the container = 10449.921000 = 10.45 litersCost of milk at the rate of Rs.15 per liters = Rs.{10.45 × 15 } = Rs. 156.75

Surface area of the metal sheet used to make the container=πl(R+r)+πr2=π[l(R+r)+r2]

=3.14×[20×(20+8)+82]cm2

=3.14×[20×28+64]cm2

=3.14×624cm2=1959.36cm2

Therefore, the cost of the metal sheet at rate of Rs.5 per 100 cm2

= Rs. 1959.36 × 5100 = Rs.97.97 approx.

### Question 2

A cubical block of side 7 cm is surmounted by a hemisphere. Find the surface area of the solid(in cm2 ).

332.5 cm2

346.8 cm2

312.5 cm2

320 cm2

#### SOLUTION

Solution :A

The hemisphere surmounts the cube, the maximum diameter the hemisphere can have = side of cube as shown in the figure below. The hemisphere will just touch the sides of top face of the cube.

Therefore, maximum diameter Hemisphere can have = 7 cm

⇒ Radius of Hemisphere = r = 72=3.5 cm

Surface area of solid = Surface area of 5 Faces of cube + Surface area of Hemisphere + Surface area left uncovered on the top face of the cube (shown in blue)=5l2+2.π.r2+ Area of Square ABCD - Area of inscribed circle in square ABCD

=(5×7×7)+(2×227×3.5×3.5)+(49−(227×3.5×3.5))

=245+77+(49−38.5)=322+10.5=332.5 cm2

∴ The surface area of combined solid = 332.5 cm2

### Question 3

The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum(in cm2).

56cm2

40cm2

52cm2

48cm2

#### SOLUTION

Solution :D

Slant height of a frustum of a cone =

l = 4 cm

Perimeter of its first circular end =

18 cm

Perimeter of its second circular end =

6 cm

Curved Surface Area of frustum of the cone

=π.(r1+r2)l

=(π.r1+π.r2)l

=(2π.r1+2π.r2)l2

=(18+6)42=48cm2

### Question 4

Shanta runs an industry in a shed which is in the shape of a cuboid surmounted by a half cylinder (see Fig.). If the base of the shed is of dimension 7 m × 15 m, and the height of the cuboidal portion is 8 m, find the volume of air that the shed can hold. Further, suppose the machinery in the shed occupies a total space of 300 m3, and there are 20 workers, each of whom occupy about 0.08 m3 space on an average. Then, how much air is in the shed (in m3)? (Take π=227)

900

827.15

800

845.3

#### SOLUTION

Solution :B

The volume of air inside the shed (when there are no people or machinery) is given by the

volume of air inside the cuboid and inside the half cylinder, taken together.

Now, the length, breadth and height of the cuboid are 15 m, 7 m and 8 m, respectively.

Also, the diameter of the half cylinder is 7 m and its height is 15 m.

So, the required volume = volume of the cuboid +12 volume of the cylinder

= [15×7×8+12×227×72×72×15]m3=1128.75 m3

Next, the total space occupied by the machinery = 300 m3

And the total space occupied by the workers = 20×0.08 m3=1.6 m3

Therefore, the volume of the air, when there are machinery and workers

=1128.75−(300.00+1.60)=827.15 m3

### Question 5

A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

10

30

25

20

#### SOLUTION

Solution :A

Let number of cones which can be filled = n

Diameter of cylinder = d = 12 cm

⇒ Radius of cylinder =

r=d2=122=6 cmHeight of cylinder = h = 15 cm

Volume of cylinder = π.r2.h=(π×62×15)=540π cm3

Diameter of cone = d1=6 cm

Radius of cone = r1=d12=62=3 cm

Height of cone = h1=12 cm

Volume of cone =13π(r1)2h1=13×π×32×12=36π cm3

Radius of hemispherical top of the cone = r1=3 cm

Volume of hemisphere top =23π(r1)3=23×π×33=18π cm3

According to given condition we have:

n × ( Volume of Cone + Volume of Hemispherical top ) = volume of cylinder

⇒ n × (36 π+18π)=540π

⇒ n × (54 π)=540π

n=54054=10

### Question 6

A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If, water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled(in minutes)?

80

100

150

200

#### SOLUTION

Solution :B

Given that diameter of pipe, d = 20 cm

Then, radius of pipe,

r=202=10 cm=0.1 m

Speed of water flowing through pipe = 3 km/h = 3000 m/h

Pipe is in the form of a cylinder.

So, volume of water flowing through pipe in 1 hour

=πr2h=π×0.1×0.1×3000=30π m3Let the cylindrical tank be filled in x hours.

Then, volume of water flowing through pipe in x hours

=30xπ m3Also it is given that,

diameter of cylindrical tank, d1=10 m

⇒ radius of cylindrical tank, r1=102=5 m

Height of cylindrical tank, h1=2 mVolume of cylindrical tank

=π(r1)2h1=π×5×5×2=50π m3

Now, volume of water flowing through pipe in x hours = volume of cylindrical tank⇒30xπ=50π

⇒x=5030=53 hours

i.e., x=5×603 minutes =100 minutes

∴ The tank will be filled in 100 minutes.

### Question 7

A toy in shape of a hemisphere of radius 14 cm is surmounted by a cone of height 22 cm. Find the approximate volume of the toy.

10266.66 cm3

17266.67 cm3

15266.67 cm3

12266.67 cm3

#### SOLUTION

Solution :A

The volume of the toy = volume of hemisphere + volume of a cone

V=23πr3+13πr2h =23×227(14)3+13×227(14)2(22) =23×22×2×14×14+13×22×2×14×22 =13[2×22×2×14×14+22×2×14×22] =13[17248+13552]=13×30800V=10266.66 cm3

∴ Volume of the toy = 10266.66 cm3

### Question 8

A rectangular paper is folded into a cylinder. The length and breadth of the paper are L and B respectively. The surface area of the cylinder is

#### SOLUTION

Solution :The cylinder is a 3-D object. Here the 3-D surface of the cylinder can be unrolled into a 2-D rectangular sheet of paper. The area of the 3-D surface and that of the 2-D plane are the same. It is clear that the area of rectangle = Length × Breadth= LB

### Question 9

A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameter of its two circular ends are 4 cm and 2 cm. The capacity of the glass is 10223 cm3.

True

False

#### SOLUTION

Solution :A

Height of frustum of cone = h = 14 cm

Radius of one end = r1=1 cm

Radius of the other end = r2=2 cm

Volume of frustum of cone =

13 π.h((r1)2+(r2)2+(r1)(r2))=13×227×14(12+22+(1)(2))

=13×227×14×7=3083=10223 cm3

### Question 10

In order to paint a solid frustum, which area would be taken into consideration?

Curved surface area

Curved surface area + area of the smaller circle.

curved surface area + area of the larger circle.

Curved surface area + area of both the circular ends.

#### SOLUTION

Solution :D

If the object is solid, then all the surfaces have to be painted. It consists of no hollow faces.

Therefore the area to be painted is,

curved surface area of frustum + area of both the circular ends.