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MPTET Varg 2 Math Mock Test - 5 - MPTET MCQ


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30 Questions MCQ Test - MPTET Varg 2 Math Mock Test - 5

MPTET Varg 2 Math Mock Test - 5 for MPTET 2025 is part of MPTET preparation. The MPTET Varg 2 Math Mock Test - 5 questions and answers have been prepared according to the MPTET exam syllabus.The MPTET Varg 2 Math Mock Test - 5 MCQs are made for MPTET 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MPTET Varg 2 Math Mock Test - 5 below.
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MPTET Varg 2 Math Mock Test - 5 - Question 1

Match the following

Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 1

Concept:

Energy

  • The capacity of doing work is called energy.
  • SI unit of energy is the joule.
  • Various forms of energy are heat energy, light energy, nuclear energy, electrical energy, mechanical energy, etc.
  • Energy can be converted from one form to another.
  • This happens through various processes or using certain appliances.
  • For example, when a bulb glows by electricity, electrical energy is converted into light energy.

Explanation:

Windmill

  • A windmill has large blades just as a fan.
  • It wind rotates the blades attached to the turbine.
  • The turbine rotates and electrical energy is produced.
  • Energy to the motion is kinetic energy.
  • So, the windmill converts the kinetic energy of blades into electrical energy.

Electric cell

  • Chemical reactions take place in the cell.
  • This makes a potential difference between the two terminals.
  • When connected with the conductor, current flows.
  • In this way, chemical energy is converted into electrical energy.

Electric bulb

  • An electric bulb converts the electrical energy into light and heat energy.
  • A filament present in the bulb burns and light energy is produced out of it.

Microphone

  • A microphone receives the sound signal and converts it into electrical signals.
  • In the microphone, there is a diaphragm.
  • It vibrates and the diaphragm makes the coil attached to it vibrate.
  • A permanent magnet near the coil induces a current in the coil by the process of electromagnetic induction.
  • So, sound energy is converted into electrical energy.

Conclusion:


The correct sequence is hence a) - (ii), (b) - (iv), (c) - (i), (d) - (iii)

MPTET Varg 2 Math Mock Test - 5 - Question 2

If log 27 = 1.432, then the value of log 9 = ?

Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 2

Concept:

Logarithms:

  • If ab = x, then we say that loga x = b.
  • log an = n log a.

Calculation:

We know that 27 = 33 and 9 = 32.

Changing the given logarithms to log 3, we get:

log 27 = 1.432

⇒ log 33 = 1.432

Using log an = n log a, we get:

⇒ 3(log 3) = 1.432

⇒ log 3 = 0.4773

Now, log 9 = log 32 = 2(log 3) = 2 × 0.4773 = 0.9546.

MPTET Varg 2 Math Mock Test - 5 - Question 3

The total surface area of a solid hemisphere is 1848 cm2. What is the length of the diameter of the flat surface of the hemisphere. [Use π = 22/7]

Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 3

Given:

The total surface area of a solid hemisphere is 1848 cm2.

Concept used:

Total surface area of a hemisphere = 3πR2

where R is the radius of the hemisphere

Diameter = 2 × Radius

Calculation:

Let the radius of the hemisphere be R.

According to the concept,

3πR2 = 1848

⇒ R2 = 1848/3π

⇒ R2 = 1848/(3 × 22/7)

⇒ R2 = (1848 × 7)/66

⇒ R2 = 196

⇒ R2 = 142

⇒ R = 14 (∵ Radius can't be negative)

Now, the diameter of the flat surface of the hemisphere

⇒ 2 × 14

⇒ 28 cm

∴ The diameter of the flat surface of the hemisphere is 28 cm.

MPTET Varg 2 Math Mock Test - 5 - Question 4
Find the volume of a cylinder (in cm3) if its area of the base is 1,386 cm2 and its height is 30 cm.
Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 4

Given:

Area of base (circle) of the cylinder = 1386 cm2

Height (h) of the cylinder = 30 cm

Formula used:

Cylinder volume = π x (radius)² x height

Area of the circle = π x (radius)² [Radius (r)]

Calculation:

Area of the circle = π x (radius)²

⇒ 1386 = π x (r)² …......(1)

Volume of the cylinder = π x (r)² x height …....(2)

From (1) and (2) we get,

Volume = 1386 x h

Volume = 1386 x 30 = 41580 cm3

∴ The volume of a cylinder (in cm3) is 41580 cm3.

MPTET Varg 2 Math Mock Test - 5 - Question 5
The angles of a triangle are in Arithmetic Progression and the greatest is three times at least. Then find the greatest angle.
Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 5

Given:

The angles of a triangle are in Arithmetic Progression.

The greatest angle is three times at least angle.

Concept Used:

In a triangle,

∠A + ∠B + ∠C = 180°

Where,

∠A, ∠B and ∠C = Angle of the triangle

Calculation:

Let the angle ∠A, ∠B and ∠C be an arithmetic progression

Let the greatest and least angle be ∠C and ∠A

According to the question, we have

∠C = 3∠A

The ∠A, ∠B and ∠C are an arithmetic progression

⇒ ∠B = 2∠A

Now,

∠A + ∠B + ∠C = 180°

⇒ ∠A + 2∠A + 3∠A = 180°

⇒ 6∠A = 180°

⇒ ∠A = 30°

The greatest angle = ∠C

⇒ ∠C = 3 × 30°

⇒ ∠C = 90°

∴ The greatest angle of the triangle is 90°.

MPTET Varg 2 Math Mock Test - 5 - Question 6
Soda bottles are made of thick glass so that they can withstand the ______.
Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 6

Key Points

  • Soda bottles are made of thick glass so that they can withstand the pressure in summer.
  • Soda bottles are generally made of 'Polyethylene Terephthalate', commonly known as PET plastic, or glass using the blow molding process.
  • Usually, the plastic bottles used to hold potable water, soda, and other drinks are made from PET material because that is both strong and light so that they can withstand pressure in summer.
  • *To be able to withstand internal gas pressure, the bottle is made from thick, heavy glass
MPTET Varg 2 Math Mock Test - 5 - Question 7
If the language of mathematics reveals imperfect meanings then
Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 7

It is known that language can either help or hinder learning. If language is used correctly and with clarity, it helps in thinking but if it reveals imperfect meanings it creates a misunderstanding. Since mathematics deals in abstractions and itself is a way of thinking, it creates a dependent relationship between the notions and the language used to describe them.

  • Mathematical language facilitates thinking by complementing ordinary language. Consider how a child gets the notion of a circle. A child handles, manipulates, and observes the shapes of objects like a wheel, bangles, the ring, etc. He may experiment with a model or may stand in a circle while playing.

Important Points

  • Also, mathematics is itself a language; it has its own symbols and rules for correct usage. In spoken language, a usage indicates what words mean, in mathematics, careful defining sharpens word meanings. Mathematical language is clear, concise, consistent, and cogent. Pupils who get the idea and describe it incorrect language are less confused than pupils who memorize terms representing ideas that remain as strange as the terms themselves.

Thus from the above-mentioned points, it is clear that if the language of mathematics reveals imperfect meanings then it creates a misunderstanding.

MPTET Varg 2 Math Mock Test - 5 - Question 8

Find the surface area (in dm2) of a cuboid whose length is 4.2 m, breadth is 30 dm and height is 180 cm.

Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 8

Given:

Length (l) = 4.2 m = 42 dm

Breadth (b) = 30 dm

Height (h) = 180 cm = 18 dm

Formula used:

Surface Area (SA) of Cuboid = 2 [(l × b) + (b × h) + (h × l)]

Calculations:

According to the question,

SA = 2 [(42 × 30) + (30 × 18) + (18 × 42)]

⇒ SA = 2 [(1260) + (540) + (756)]

⇒ SA = 2 [2256]

⇒ SA = 2 × 2556 = 5112 dm2

∴ The surface area (in dm2) of a cuboid is 5112 dm2.

MPTET Varg 2 Math Mock Test - 5 - Question 9
The area of a rectangle whose length and width are in the ratio 9 : 5 is given as 180 cm2. Find the perimeter of the rectangle.
Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 9

Given:

Area of rectangle = 180 cm²

Ratio of length (l) and breadth (b) = 9/5

Formula used:

Perimeter of a Rectangle (P) = 2 × (l + b)

Area of a Rectangle = l × b.

Calculations:

According to the question,

l/b = 9/5

l = 9b/5

Area of rectangle = l × b

180 = (9b/5) × b

900 = 9b2

b = √100 = 10 cm

Then, length = (9 × 10)/5 = 18 cm

The perimeter of the rectangle:

⇒ P = 2 × (l + b) = 2 × (18 + 10)

⇒ P = 2 × (28) = 56 cm

∴The Perimeter of the Rectangle is 56 cm.

MPTET Varg 2 Math Mock Test - 5 - Question 10
The volume of a cuboid is 864 cu. cm. If the area of ​​the base is 108 sq. cm, then its height is:
Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 10

Given:

Volume of cuboid = 864 cu. cm

Area of ​​the base is 108 sq. cm

Formula Used:
Volume of Cuboid = Length × Breadth × Height

Since, Length and Breadth makes the base of cuboid,

Area of Base = Length × Breadth

Volume of Cuboid = Area of Base × Height

Calculation:

Area of Base × Height = 864

⇒ 108 × Height = 864

⇒ Height = 864 ÷ 108 = 8 cm

Hence, Height of cuboid is 8 cm.

MPTET Varg 2 Math Mock Test - 5 - Question 11
If x and 2x are supplementary angles, then the value of 2x.
Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 11

Concept:

Supplementary Angles: Pair of angles that always sum up to 180°.

Calculation:

According to the concept discussed above

⇒ x + 2x = 180°

3x = 180°

x = 180°/3 = 60°

∴ 2x = 2 × 60° = 120°

MPTET Varg 2 Math Mock Test - 5 - Question 12

The total surface area of right circular cone is 360π m2. If the radius is 10 m, then find the height of the right circular cone. (in m)

Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 12

Given:

The total surface area of the right circular cone is 360π m2.

The radius is 10 m.

Formula used:

(1) A = πrl + πr2

Where,

A, is the total surface area of the right circular cone

l, is the slant height of the right circular cone

r, is the radius of the right circular cone

(2) l2 = r2 + h2

Where,

l, is the slant height of the right circular cone

r, is the radius of the right circular cone

h, is the height of the right circular cone

Calculation:

According to the question, the required figure is:

Let 'l' be the slant height of the right circular cone.

According to the question,

⇒ A = πrl + πr2

⇒ 360π = [(π × 10 × l) + (π × 102)]

⇒ 360 = [10l + 100]

⇒ 36 = l + 10

⇒ l = 26 m

Now,

⇒ l2 = r2 + h2

⇒ 262 = 102 + h2

⇒ 676 = 100 + h2

⇒ h2 = 576

⇒ h = 24 m

∴ The height of the right-circular cone is 24 m.

MPTET Varg 2 Math Mock Test - 5 - Question 13

In Fig, ABC is a right triangle and right angled at B such that ∠BCA = 2 ∠BAC then

Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 13

According to the question,

∠BCA = 2 ∠BAC

so

∠BCA = 60°

∠BAC = 30°

So from sine rule,

AC/sin90° = BC/sin30°

AC/1 = BC/(1/2)

AC = 2 BC

MPTET Varg 2 Math Mock Test - 5 - Question 14
A table is bought for Rs. 800 and is sold at 40% loss. What is the selling price of the table?
Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 14

Given:

C.P of a table = Rs. 800

Loss % = 40%

Formula Used:

S.P = {(100 - Loss %)/100} × C.P

Where,

S.P = Selling price

C.P = Cost price

Calculation:

According to the question, we have

S.P = {(100 - 40)/100) × 800

⇒ S.P = (60/100) × 800

⇒ S.P = Rs. 480

∴ The selling price of the table is Rs. 480.

MPTET Varg 2 Math Mock Test - 5 - Question 15

Find the Total surface area of the hemisphere (in cm2) whose radius is 51cm and π = 3.14.

Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 15

Given:

Radius of hemisphere = 51 cm.

Formula used:

T.S.A of hemisphere = 3π r2

Calculation:

∴ The answer is "24501.42".

MPTET Varg 2 Math Mock Test - 5 - Question 16
In a week, the weight of a bag of tea were 350 kg, 280 kg, 270 kg, 360 kg, 310 kg, 300 kg. The range (in kg) is:
Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 16

Given:

The weight of a bag of tea was 350 kg, 280 kg, 270 kg, 360 kg, 310 kg, 300 kg

Concept used:

Range = Highest term - lowest term

Calculation:

Highest term = 360 kg

Lowest term = 270 kg

Range = 360 - 270 = 90 kg

∴ Required range is 90

MPTET Varg 2 Math Mock Test - 5 - Question 17

The difference between the compound and the simple interests for 3 years at the rate of 20% per annum is ₹432. What is the principal lent?

Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 17

Given:

The difference between the Compound (CI) and the Simple interests (SI) is ₹432.

Time (T) = 3 years

The Rate (R) of interest = 20% per annum

Concept used:

Compound Interest = P(1 + R%)T

Simple Interest = (P × R × T)/100

Calculations:

According to the question,

CI - SI = Rs. 432

⇒ 432 = [P(1 + 20%)3 - 1] - [(P × 20 × 3)/100]

⇒ 432 = P[(1 + 20%)3 - 1] - P[(20 × 3)/100]

⇒ 432 = P[(1 + 0.20)3] - P[6/10]

⇒ 432 = P[(1.20)3 - 1] - P[3/5]

⇒ 432 = P[1.728 - 1 - 3/5]

432 = P[(3.64 - 3)/5]

⇒ 432 = P[0.64/5]

⇒ P = 432/0.128

⇒ P = Rs. 3375

∴ The principal lent was Rs. 3375.

Shortcut Trick
20% = 20/100 = 1/5

Time = 3 years

Hence assume that the Principle is 53 = 125 Rs.

Simple Interest for 3 years = 75 Rs.

Compound Interest for 3 years = 91 Rs.

⇒ 91 - 75 = 16

Given that,

⇒ 16 = 432

⇒ 1 = 432/16

The principle, 125

⇒ 125 × 432/16 = 3375 Rs.

∴ The principal lent was Rs. 3375.

MPTET Varg 2 Math Mock Test - 5 - Question 18

The altitude of an equilateral triangle is 12 cm. What is the perimeter of the triangle?

Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 18

Given

Altitude of triangle = 12 cm

Concept

Altitude of an equilateral triangle is a median of an equilateral triangle.

Length of altitude of equilateral triangle = √3/2 × side of the triangle

Calculation

Let the side of an equilateral triangle is x cm.

12 = √3/2 × x

⇒ 24/√3 cm

Perimeter of triangle = 3(side)

⇒ 3 × (24/√3)

⇒ 24√3 cm.

Alternate Method
Given:

Height of the equilateral triangle = 12 cm

Formula used:

The perimeter (P) of a triangle is = Sum of all the sides

Concept used:

Pythagorean Theorem = (Hypotenuse)2 = (Perpendicular)2 + (Base)2

⇒ sin 60° = √3/2

Calculations:

Considering the height of the triangle as perpendicular then,

⇒ sinθ = (Perpendicular)/(Hypotenuse)

⇒ sin 60° = 12/H

⇒ H = (12 × 2)/√3 = 24/√3

Now the perimeter of the triangle,

⇒ P = 3 × Side

⇒ P = 3 × 24/√3 = 24√ 3

∴ The perimeter of the triangle is 24√3.

MPTET Varg 2 Math Mock Test - 5 - Question 19
The length of each of a pair of opposite sides of a parallelogram is 16 cm and the perpendicular distance between these two sides is 10 cm. What is the area (in cm) of the parallelogram?
Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 19

Given:

The length of each a pair of opposite sides of a parallelogram is 16 cm

The perpendicular distance between these two sides is 10 cm

Concept used:

Area of a parallelogram = Base × Height (perpendicular distance between parallel sides)

Calculation:

Area of the parallelogram

⇒ 16 × 10

⇒ 160 cm2

∴ The area of the parallelogram is 160 cm2.

MPTET Varg 2 Math Mock Test - 5 - Question 20

The radius of the base of a right circular cone is increased by 15% keeping the height fixed. The volume of the cone will be increased by:

Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 20

Given:

The radius of the base of a right circular cone is increased by 15% keeping the height fixed.

Concept used:

Volume of cone = 1/3πr2h

Calculation:

Let the radius be r

After an increase in radius, it would be 1.15r.

Volume of new cone = 1/3π(1.15r)2h

And, Volume of old cone = 1/3πr2h

⇒ % Change in volume = [Increase in volume/old volume] × 100%

⇒ 32.25%

∴ The volume of the cone will be increased by 32.25%.

MPTET Varg 2 Math Mock Test - 5 - Question 21

If the radius of the cone is increased by 10% and the height of the cone is increased by 20%, find the percentage increase in the volume of the cone.

Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 21

Shortcut Trick

If the radius of the cone is increased by x% and the height of the cone is increased by y%,

The percentage increase in the volume of the cone =

According to the question,

x = 10

y = 20

Therefore,

The percentage increase in the volume of the cone,

⇒ ΔV = (20 + 20 + 5 + 0.2) × 100

⇒ ΔV = 45.2%

∴ The required answer is 45.2%

MPTET Varg 2 Math Mock Test - 5 - Question 22

A teacher uses the exploratory approach, use manipulatives and involvement of students in discussion while giving the concepts of mathematics. She uses this strategy to

Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 22

According to NCF 2005, the main goal of mathematics education at the primary level is the development of children's ability in mathematization. Basically, it means that children should learn to think about any situation using the language of mathematics. They can use the tools and techniques of mathematics in real life.

Important Point

  • In Mathematics there is a certain way of thinking, people connect the mathematical concepts with real-life which help them think more logically and practically.
  • The National Curriculum Framework 2005 is among the four National Curriculum Frameworks approved by the National Council of Educational Research and Training NCERT in India.
  • At the Primary level, children learn from concrete objects and visualization processes.
  • It must be ambitious, coherent (logical and consistent) and should focus on principles of mathematics.

Hence, we conclude that a 'certain way of thinking and reasoning' means adopting the exploratory approach, use of manipulatives, connecting concepts to real life, involving students in discussions.

MPTET Varg 2 Math Mock Test - 5 - Question 23

The cubes of side length 10 cm, 12 cm and 15 cm all three melted as casted into a single cube. Find the side length of the casted cube. (in cm)

Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 23

Shortcut Trick
The cubes of side lengths x cm, y cm and z cm three melted as cast into a single cube.

The side length of the casted cube =

According to the question,

The side length of the first cube, x = 10 cm

The side length of the second cube, y = 12 cm

The side length of the third cube, z = 15 cm

Therefore,

The side length of the casted cube =

∴ The required answer is 18.27 cm

Alternate Method

Given:

The side length of the first cube = 10 cm

The side length of the second cube = 12 cm

The side length of the third cube = 15 cm

Formula used:

Volume of cube = a3

where,

a = side length of the cube

Calculation:

The volume of the first cube = 103 = 1000 cm3

The volume of the second cube = 123 = 1728 cm3

The volume of the third cube = 153 = 3375 cm3

Since all three cubes are melted and cast into a single cube,

Therefore,

The volume of casted cube = 1000 + 1728 + 3375 = 6103 cm3

Let A be the side length of the casted cube.

⇒ A3 = 6103

⇒ A ≈ 18.27 cm

∴ The required answer is 18.27 cm.

Alternate Method 

The volume of the first cube = 103 = 1000 cm3

The volume of the second cube = 123 = 1728 cm3

The volume of the third cube = 153 = 3375 cm3

Since all three cubes are melted and cast into a single cube,

Therefore,

The volume of casted cube = 1000 + 1728 + 3375 = 6103 cm3

Now, we can also solve through option.

Apply the formula for the volume of cube in each value of side length of the cube.

Option 1: 223 = 10648 cm3

Option 2: 403 = 64000 cm3

Option 3: 183 = 5832 cm3

Option 4: 103 = 1000 cm3

In comparing we get, option B gives the nearest answer.

∴ The required answer is 18.27 cm.

MPTET Varg 2 Math Mock Test - 5 - Question 24

Which of the following is not the unit of work?

Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 24

The correct answer is Joule/second.

  • Work is defined as the component of force along the direction of motion times the displacement of the object acting under a certain motion.
  • Joule - The SI unit of work is the joule (J).
  • Erg - It is the unit of work in the CGS system.
    • Erg is a unit of work equal to 10−7 joules.
  • Newton metre (Nm) - Newton meter is the unit of work in terms of force and displacement.
  • Units of work done is
MPTET Varg 2 Math Mock Test - 5 - Question 25

Find the radius of the semi-circle whose perimeter is 72 cm. (Take π = 22/7)

Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 25

Shortcut Trick
The perimeter of the semi-circle with radius r is given by

πr + 2r = 36r/7

According to the question,

⇒ 72 = 36r/7

⇒ r = 14 cm

Therefore, '14 cm' is the required answer.

Alternate Method
Given:

Perimeter of a semi-circle = 72 cm

Formula used:

The perimeter of semicircle = πr + 2r

Calculation:

According to the question,

⇒ 72 = πr + 2r

⇒ 72 = (π + 2) × r

⇒ 72 = (22/7 + 2) × r

⇒ 72 = 36r/7 × r

⇒ r = 14 cm

Therefore, '14 cm' is the required answer.

Mistake Points
The perimeter of a semicircle is the sum of half of the

circumference of the circle and its diameter i.e. (2r + r).

Don't take it equal to πr.

MPTET Varg 2 Math Mock Test - 5 - Question 26

What will be the difference between mean and median of the given data?

21, 11, 27, 8, 5, 12, 7, 23, 3, 14, 9, 19
Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 26

Mean = (21 + 11 + 27 + 8 + 5 + 12 + 7 + 23 + 3 + 14 + 9 + 19) / 12

= 159/12 = 13.25

Arranging the observations in ascending order,

3, 5, 7, 8, 9, 11, 12, 14, 19, 21, 23, 27

Since the number of observations is even,

Median = (6th term + 7th term) / 2 = (11 + 12)/2 = 11.5

Difference between mean and median of the given data = 13.25 - 11.5 = 1.75

MPTET Varg 2 Math Mock Test - 5 - Question 27
The area of the floor of a cubical room is 192 m2. The length of the longest rod that can be kept in that room is :
Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 27

Given:

Area of the floor of a cubical room = 192 m2

Formula used:

Diagonal of the cube = √3 × side of cube

Floor area = side2 (floor of the cube is in the shape of a square)

Calculation:

According to the question,

The floor area of the cube = side2

⇒ 192 = side2

⇒ side of cube = √192 = 8√ 3 m

Diagonal = √3 × side of the cube

⇒ √3 × 8 × √3

⇒ 8 × 3 = 24 m

The length of the longest rod that can be kept in that room is 24 m.

MPTET Varg 2 Math Mock Test - 5 - Question 28
The energy that will be ideally radiated by a 100 kW transmitter in 1 hour is:
Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 28

Concept:

The amount of electrical energy transferred to an appliance depends on its power and the duration of time it is used.

Electrical energy is measured in kilowatt-hours, kWh. One unit is 1 kWh.

Electrical energy transferred is given as,

E = P × t

Where, P is the power in kilowatts (kW), T is the time in hours (h)

Calculation:

Given,

Power, P = 100 kW

Time, t = 1 hour

∴ E = P × t = 100 × 1 = 100 kWh

E = 100 × 3.6 × 106 J = 36 × 107 J (∵ 1 kWh = 3.6 × 106 J)

Hence, option (2) is the correct answer.

MPTET Varg 2 Math Mock Test - 5 - Question 29
______ is used in buildings and in manufacturing processes to prevent heat loss or heat gain.
Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 29

The correct answer is Insulation.

Key Points

  • Insulation is a general term used to describe material that creates barriers to the transmission of electricity, heat, moisture, shock, or sound between insulated surfaces of adjacent bodies. These materials could be insulating heat, cold, electricity, sound, or radiation.
  • To reduce heat loss or gain, insulation is utilised in buildings and manufacturing processes.
  • Although its major goal is to save money, it also allows for more precise temperature management and worker protection.
  • It reduces rusting caused by moisture on cold surfaces. These materials are porous and have a lot of latent air cells.
  • The building industry uses thermal insulation to decrease heat loss or gain through the home envelope (external walls, windows, roofs, foundation, etc.).
  • Thermal insulation provides thermal comfort inside homes by maintaining a comfortable temperature.

Important Points

  • The flow of a fluid (liquid or gas) between locations of differing temperatures is known as convection.
  • Heat energy is transferred through collisions between surrounding atoms or molecules in the process of conduction.
  • Solidification, commonly known as freezing, is a phase transition in which a liquid becomes a solid when its temperature is decreased below its freezing point.
MPTET Varg 2 Math Mock Test - 5 - Question 30
How many matchboxes of size 4 cm × 3 cm × 1.5 cm can be packed in a cardboard box of size 30 cm × 30 cm × 20 cm?
Detailed Solution for MPTET Varg 2 Math Mock Test - 5 - Question 30

Given:

Dimension of matchboxes = 4 cm × 3 cm × 1.5 cm

Dimension of cardboard box = 30 cm × 30 cm × 20 cm

Formula:

Volume of cuboid = l × b × h

Calculation:

Let number of matchboxes be x.

According to the question

4 × 3 × 1.5 × x = 30 × 30 × 20

⇒ × = (30 × 30 × 20)/(4 × 3 × 1.5)

⇒ × = 1000
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