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Test: Surface Area & Volumes- 1 - Class 9 MCQ


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25 Questions MCQ Test - Test: Surface Area & Volumes- 1

Test: Surface Area & Volumes- 1 for Class 9 2024 is part of Class 9 preparation. The Test: Surface Area & Volumes- 1 questions and answers have been prepared according to the Class 9 exam syllabus.The Test: Surface Area & Volumes- 1 MCQs are made for Class 9 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Surface Area & Volumes- 1 below.
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Test: Surface Area & Volumes- 1 - Question 1

A rectangular sand box is 5 m wide and 2 m long. How many cubic metres of sand are needed to fill the box upto a depth of 10 cm ?

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 1

To calculate the volume of sand needed to fill the rectangular sandbox, we can use the formula:Volume = Length × Width × DepthGiven:
Length = 2 m
Width = 5 m
Depth = 10 cm = 0.1 mSubstituting the values into the formula:Volume = 2 m × 5 m × 0.1 m
Volume = 1 cubic meterTherefore, the correct answer is A: 1 cubic meter.

Test: Surface Area & Volumes- 1 - Question 2

A beam 9 m long, 40 cm wide and 20 cm deep is made up of iron which weighs 50 kg per cubic metre.

The weight of the beam is :

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 2

To find the weight of the iron beam, we need to calculate its volume and then multiply it by the weight of 1 cubic meter of iron, which is given as 50 kg.Given:
Length of beam (l) = 9 m
Breadth (b) = 40 cm = 0.4 m
Height (h) = 20 cm = 0.2 m
Weight of 1 cubic meter of iron = 50 kgVolume of the beam = l × b × h = 9 m × 0.4 m × 0.2 m = 0.72 cubic metersWeight of the beam = Volume × Weight of 1 cubic meter of iron = 0.72 cubic meters × 50 kg/cubic meter = 36 kg

Therefore, the weight of the iron beam is 36 kg, which corresponds to option B.

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Test: Surface Area & Volumes- 1 - Question 3

The maximum length of a pencil that can be kept in a rectangular box of dimensions 8 cm × 6 cm × 2 cm, is:

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 3

To find the maximum length of a pencil that can be kept in a rectangular box of dimensions 8 cm × 6 cm × 2 cm, we need to find the length of the diagonal of the box, which will be the maximum length of the pencil that can fit inside the box.Using the Pythagorean theorem, the length of the diagonal of the box can be found as:Diagonal = √(length² + breadth² + height²)
Diagonal = √(8² + 6² + 2²)
Diagonal = √(64 + 36 + 4)
Diagonal = √104
Diagonal = 2√26Therefore, the maximum length of a pencil that can be kept in the rectangular box is 2√26 cm, which corresponds to option C.

Test: Surface Area & Volumes- 1 - Question 4

How many bricks, each measuring 25 cm × 12.5 cm × 7.5 cm will be needed to construct a wall 15 m long, 1.8 m high and 37.5 cm thick?

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 4

Test: Surface Area & Volumes- 1 - Question 5

A wooden box of dimensions 8 m × 7 m × 6 m is to carry rectangular boxes of dimensions 8 cm × 7 cm × 6 cm. The maximum number of boxes that can be carried in the wooden box, is :

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 5

Test: Surface Area & Volumes- 1 - Question 6

The surface area of a cube of side 27 cm is :

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 6

The surface area of a cube of side 27 cm can be found using the formula:Surface area of a cube = 6a^2, where a is the length of the side of the cube.Given:
Length of side of the cube = 27 cmSubstituting the value into the formula:Surface area of the cube = 6 × 27^2
Surface area of the cube = 6 × 729
Surface area of the cube = 4374 cm^2Therefore, the surface area of the cube of side 27 cm is 4374 cm^2, which corresponds to option C

Test: Surface Area & Volumes- 1 - Question 7

The perimeter of one face of a cube is 40 cm. The volume of the cube (in cm3) is :

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 7

Given that the perimeter of one face of a cube is 40 cm. Let each side of the cube be a cm. Each face of a cube is a square. Therefore, the perimeter of one face of the cube is 4a. We can write:4a = 40 cm
a = 10 cmTherefore, the length of each side of the cube is 10 cm. The volume of the cube is given by the formula V = a^3, where a is the length of each side of the cube. Substituting the value of a, we get:V = 10^3 = 1000 cm^3Therefore, the volume of the cube is 1000 cm^3, which corresponds to option B

Test: Surface Area & Volumes- 1 - Question 8

The volume of a cube with surface area 384 sq. cm, is :

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 8

The volume of a cube with a surface area of 384 sq. cm is 512 cm3, which corresponds to option D.

Test: Surface Area & Volumes- 1 - Question 9

The length of the longest rod that can fit in a cubical vessel of side 10 cm, is

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 9

The length of the longest rod that can fit in a cubical vessel of side 10 cm is equal to the length of the diagonal of the cube. Using the Pythagorean theorem, the length of the diagonal of the cube can be found as:Diagonal = √(length² + breadth² + height²)
Diagonal = √(10² + 10² + 10²)
Diagonal = √300
Diagonal = 10√3 cmTherefore, the length of the longest rod that can fit in a cubical vessel of side 10 cm is 10√3 cm, which corresponds to option C.

Test: Surface Area & Volumes- 1 - Question 10

If the length of diagonal of a cube is 4√3  cm, then the length of its edge is :

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 10

Test: Surface Area & Volumes- 1 - Question 11

If the diameter of the base of a cylindrical pillar is 4 m and its height is 21 m, then the cost of construction of the pillar at Rs. 1.50 per cubic metre is :

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 11

Test: Surface Area & Volumes- 1 - Question 12

The volume of the cylinder whose height is 14 cm and diameter of base 4 cm, is :

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 12

Test: Surface Area & Volumes- 1 - Question 13

Find the weight of a cylindrical metal structure with base radius 10.5 cm and height 6 m where the weight of 1 cm3 of metal is 5 grams.(use π = 3.14)

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 13

Given:

Radius = 10.5 cm

Height = 6 m

Weight of 1 cm3 = 5 grams

Formula used:

Volume of cylinder = πr2h

1 m = 100 cm

⇒ 6 m = 600 cm

1 kg = 1000 gram

Calculation:

According to the question

Volume of cylinder = πr2h

⇒ (3.14 × 10.5 × 10.5 × 600) cm3

⇒ 207711 cm3

Now, Weight of 1 cm3​ = 5 grams

So, Weight of 207,711 cm3 = 5 × 207711

⇒ 1,038,555 grams

⇒ 1038.555 kg ~ 1038.5 kg

Test: Surface Area & Volumes- 1 - Question 14

If the diameter of a cylinder is 28 cm and its height is 20 cm, then total surface area (in cm2) is :

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 14

Given a cylinder with

Radius = 28 / 2 = 14 cm

Height = 20 cm

ii.) Total surface area = 2πr ( r + h )

= 2 x 22 / 7 x 14cm ( 14 + 20 )cm

= 2991.99cm2
=2992cm2
(b) 2992cm2 is the correct option.

Test: Surface Area & Volumes- 1 - Question 15

If the curved surface area of a cylinder is 1760 sq. cm and its base radius is 14 cm, then its height is :

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 15

Test: Surface Area & Volumes- 1 - Question 16

The volume (in cm3) of a right circular cone of height 12 cm and base radius 6 cm, is :

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 16

Test: Surface Area & Volumes- 1 - Question 17

If the radius of the base of a right circular cone is 3r and its height is equal to the radius of the base, then its volume is :

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 17

Test: Surface Area & Volumes- 1 - Question 18

The lateral surface area (in cm2) of a cone with height 3 cm and radius 4 cm, is :

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 18

The lateral surface area of a cone with height 3 cm and radius 4 cm can be found using the formula:
Lateral surface area of a cone = πrl, where r is the radius of the base of the cone and l is the slant height of the cone.
Given:
Height of the cone = 3 cm
Radius of the base of the cone = 4 cm
Using the Pythagorean theorem, the slant height of the cone can be found as:
l = √(r^2 + h^2) = √(4^2 + 3^2) = √(16 + 9) = √25 = 5 cm
Substituting the values into the formula:
Lateral surface area of the cone = πrl = π × 4 × 5 = 20π cm^2
Approximating π to 22/7, we get:
Lateral surface area of the cone = 20 × 22/7 = 440/7 cm^2
Therefore, the lateral surface area of the cone with height 3 cm and radius 4 cm is 440/7 cm^2, which corresponds to option A

Test: Surface Area & Volumes- 1 - Question 19

A cone of height 8 m has a curved surface area 188.4 2. The radius of the base is : (Take π = 3.14) :

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 19

Test: Surface Area & Volumes- 1 - Question 20

Find the length of canvas 1.1 m wide required to build a conical tent of height 14 m and the floor area 346.5 m2, is

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 20

Solution

Let r= radius of the base.
Given area = 346.5

π×r2 =346.5

r2 = 346.5×722 = 4414

r = 212

l = √r2+h2=√4414+142

l = 352 m

The area of canvas = surface area of the cone =πrl = π×212×352 m2 = 33×352 m2

Given length of canvas = 1.1 m
Let width = x
Area of canvas = 1.1x
So, 1.1 x=33×352
x=33×352×1.1 = 525 m

Test: Surface Area & Volumes- 1 - Question 21

The volume of a sphere is 38808 cu.cm. The curved surface area of the sphere (in cm2) is :

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 21

Test: Surface Area & Volumes- 1 - Question 22

The volume of a spherical shell whose internal and external diameters are 8 cm and 10 cm respectively (in cubic cm) is:

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 22

Test: Surface Area & Volumes- 1 - Question 23

If the ratio of volumes of two spheres is 1 : 8, then the ratio of their surface areas is :

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 23

Test: Surface Area & Volumes- 1 - Question 24

If a hemi-spherical dome has an inner diameter of 28 m, then its volume (in m3) is :

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 24

Test: Surface Area & Volumes- 1 - Question 25

Three solid spherical beads of radii 3 cm, 4 cm and 5 cm are melted into a spherical bead. Its radius is :

Detailed Solution for Test: Surface Area & Volumes- 1 - Question 25

Let three spheres are S1, S2 & S3
having radii r₁ = 3cm, r₂ = 4cm & r₃ = 5cm respectively.
Let the radius of new Big sphere S is R.
A/Q,
Volume of new Sphere  S = Sum of volumes three Spheres S1, S2 & S3

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