Area And Volume - 3D Test

Result

Area And Volume - 3D Test - 8

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A metallic right circular cone of height 18 cm and base radius 7 cm is melted into a cuboid whose two sides measure 11 cm and 6 cm. What is the length of the third side of the cuboid?

A
7 cm

B
6 cm

C
14 cm

D
10 cm

Solution

According to the given condition, Volume of cone = Volume of cuboid

Let the third side of the cuboid measure x cm.

x = 14 cm
Question 2
1 / -0

A cylindrical vessel of diameter 4 cm is partly filled with water. In the vessel, 300 lead balls are dropped. The rise in the water level is 2.7 cm. The diameter of each ball is

A
0.3 cm

B
0.6 cm

C
0.2 cm

D
none of these

Solution

Increase in the Volume of cylinder = Volume of 300 lead balls
Let radius of lead ball = r cm
solving r = 0.3
diameter = 0.6 cm
Question 3
1 / -0

A hemisphere of inner diameter 9 cm is completely filled with a liquid. This is to be poured into cylindrical bottles of diameter 3 cm and height 2 cm. The number of bottles required is

A
54

B
14

C
27

D
15

Solution

According to given condition
Volume of hemisphere = Volume of n cylinderical bottles

Solving
we get n = 13.5
so closest option is 14
Question 4
1 / -0

From a circle of radius 15 cm, a sector with 288⁰ angle is cut out and its bounding radii are bent so as to form a cone. What is its approximate volume?

A
1268 cm³

B
1071.3 cm³

C
1358 cm³

D
None of these

Solution

' r ' is the radius of the base of cone . ' s ' is the slant length that is the radius (R=15cm) of the sector with 288° angle
The sectoral arc becomes the perimeter of the base circle.
2πr = (288/360)[2πR]
=> r/R = 288/360
r = 15 X (288/360)
= 12 cm.
And h² = s² - r² . . . {Pythagoras theorem}
= 15² - 12² = 9²
h = 9.Volume = cm³ (approx)
Question 5
1 / -0

If one cubic centimetre of cast iron weighs 21 g, then the mass of a cast iron pipe of length 1 m with a bore of 3 cm and thickness of 1 cm is

A
21 kg

B
24.2 kg

C
26.4 kg

D
18.6 kg

Solution

Mass = gm = 26.4kg
Question 6
1 / -0

Water flows through a cylindrical pipe of inner diameter 7 cm at 192.5 litres per minute. Find the rate of flow (in kilometres per hour). [1 litre = 1 dm³]

A
2 km/hr

B
0.75 km/hr

C
4 km/hr

D
5 km/hr

Solution

Volume of water that flows through pipe in 1 hr =192.5 x 60 = 11550 litres
11550 litres =11.55 m³
Let rate of flow = rm/hr

r = 750 m
= 0.75 km/hr
Question 7
1 / -0

A hemispherical bowl is made up of steel of 0.25 cm thickness. The inner radius of the bowl is 5 cm. The volume of steel used is

A
42.15 cm³

B
41.52 cm³

C
41.24 cm³

D
none of these

Solution

Required Volume = cm³
Question 8
1 / -0

If the radii of the circular ends of a conical bucket are 28 cm and 7 cm and the height is 21 cm, then the capacity of the bucket is

A
25638 cm³

B
22638 cm³

C
29638 cm³

D
none of these

Solution

Volume of Conical bucket =
cm³
Question 9
1 / -0

What is the radius of a cylinder whose lateral surface area is 704 cm² and height is 4 cm?

A
16 cm

B
14 cm

C
18 cm

D
28 cm

Solution

r = 28 cm
Question 10
1 / -0

A toy is of the shape of a cone over a hemisphere. The radius of the hemisphere is 3 cm. The total height of the toy is 7 cm. What is the total area of the toy?

A
103.6 sq. cm

B
21.45 sq. cm

C
215.4 sq. cm

D
None of these

Solution

Height of toy = 7cm
radius of hemisphere = radius of cone =3cm
height of conical portion =7-3 = 4cm
slant height of conical part = 5cm
(using pythagorus theorem)
Total surface area of toy = curved surface area of cone +curved surface area of hemisphere =
3.14 x 3 x 5 + (2/3) x 3.14 x 27 = 103.6cm²
Question 11
1 / -0

A cone is divided into two parts by drawing a plane through the mid-point of its axis, parallel to its base. Compare the volumes of the two parts.

A
1 : 7

B
1 : 4

C
2 : 5

D
1 : 8

Solution

Let radius of bigger cone = r
height = hAE= EC
And triangle ADE and triangle ABC are similar
So

So radius of smaller cone = r/2
height of smaller cone = h/2
ratio of volumes of both cones =1/8, so ratio of volumes of both parts by cutting bigger cone =
Question 12
1 / -0

A toy is made in the form of a hemisphere surmounted by a right circular cone, whose circular base coincides with the plane surface of the hemisphere. The base radius of the cone is 3.5 m and its volume is 2/3 of the volume of the hemisphere. Calculate the total height of conical part.

A
4.67 m

B
5.23 m

C
6.34 m

D
4.33 m

Solution

Let height of cone = h
Radius of hemisphere = Radius of cone = r
According to given condition
Volume of cone = 2/3(Volume of hemisphere)
Solving we get
h = 4r/3 =14/3 = 4.66
Question 13
1 / -0

The diameter of a copper sphere is 8 cm. It is beaten and drawn into a wire of diameter 0.2 cm. The length of the wire is

A
36 cm

B
360 cm

C
3600 cm

D
none of these

Solution

Volume of Sphere = Volume of cylinderical wire
= 8533.33 cm
Question 14
1 / -0

The radius of a cylinder is doubled but its lateral surface area is unchanged. Then, how will its height be affected?

A
It becomes doubled.

B
It becomes halved.

C
It becomes tripled.

D
It remains constant.

Solution

Lateral Surface area of cylinder =
If the radius of cylinder is doubled then its height is halved to make the lateral surface area same.
Question 15
1 / -0

A cylindrical rod, whose height is 16 times of its radius, is melted and recast into spherical balls of same radius. The number of balls will be

A
14

B
16

C
12

D
18

Solution

Let radius of cylinder = radius of each spherical ball = r
height of cylinder =16r
According to given condition
where n is the number of spherical balls made by melting the cylinder
Solving we get
n =12
Question 16
1 / -0

A hollow sphere of internal and external diameters 4 cm and 8 cm respectively, is melted into a cone of base diameter 16 cm. Find the height of the cone.

A
3.5 cm

B
7 cm

C
14 cm

D
None of these

Solution

Volume of Sphere = Volume of Cone
where h is the height of the cone
Solving we get
h = 3.5 cm
Question 17
1 / -0

A hollow garden roller, 63 cm wide with a girth of 440 cm is made of iron of thickness 4 cm. The volume of the iron used is

A
56,372 cm³

B
107712 cm³

C
54,982 cm³

D
57,636 cm³

Solution

Circumference = girth = 440cm
Outer radius =70 cm
inner radius = 66 cm
Height = 63 cm
Volume = 22/7(70²- 66²) x 63 = 107712 cm³
Question 18
1 / -0

A cylinder, a hemisphere and a cone stand on the same base and have equal heights. The ratio of the areas of their curved surfaces is

A
2 : 2 : 1

B
: : 1

C
_{: 3 : 1}

D
1 : 3 :

Solution

Let r and h be the radius and height of the cylinder, hemisphere and cone.
The height of the hemisphere is the same as its radius.
So, r = h
The cylinder, hemisphere and cone are made on same base, so the radius is same in all. So, r = h for the cylinder as well as the cone.
Ratio of the curved surface areas of the cylinder, hemisphere and the cone standing on the same base and having equal heights =

Slant height of the cone,

Solving, we get the required ratio =
Question 19
1 / -0

The radius of a sphere is increased by k%. Its surface area increases by

A
k%

B
k²%

C

D

Solution

Let original radius = r
new radius =r+kr/100
percent increase in surface area =
%
Question 20
1 / -0

Vertical and horizontal cross-sections of a right circular cylinder are always ______ and ______ respectively.

A
rectangle, square

B
rectangle, circle

C
square, circle

D
rectangle, ellipse

Solution

Option (2) is correct.