Engineering Entrance Exam Preparation Resources

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Engineering Entrance MCQ & Objective Questions

Preparing for Engineering Entrance exams is crucial for aspiring engineers in India. Mastering MCQs and objective questions not only enhances your understanding of key concepts but also boosts your confidence during exams. Regular practice with these questions helps identify important topics and improves your overall exam preparation.

What You Will Practise Here

Fundamental concepts of Physics and Mathematics
Key formulas and their applications in problem-solving
Important definitions and theorems relevant to engineering
Diagrams and graphical representations for better understanding
Conceptual questions that challenge your critical thinking
Previous years' question papers and their analysis
Time management strategies while solving MCQs

Exam Relevance

The Engineering Entrance syllabus is integral to various examinations like CBSE, State Boards, NEET, and JEE. Questions often focus on core subjects such as Physics, Chemistry, and Mathematics, with formats varying from direct MCQs to application-based problems. Understanding the common question patterns can significantly enhance your performance and help you tackle the exams with ease.

Common Mistakes Students Make

Overlooking the importance of units and dimensions in calculations
Misinterpreting questions due to lack of careful reading
Neglecting to review basic concepts before attempting advanced problems
Rushing through practice questions without thorough understanding

FAQs

Question: What are the best ways to prepare for Engineering Entrance MCQs?
Answer: Focus on understanding concepts, practice regularly with objective questions, and review previous years' papers.

Question: How can I improve my speed in solving MCQs?
Answer: Regular practice, time-bound mock tests, and familiarizing yourself with common question types can help improve your speed.

Start your journey towards success by solving Engineering Entrance MCQ questions today! Test your understanding and build a strong foundation for your exams.

MHT-CET

Q. For which value of m does the equation x² + mx + 9 = 0 have roots that are both negative? (2021)

A. -6
B. -4
C. -2
D. 2

Solution

For both roots to be negative, m must be greater than 0 and m² < 36. Thus, m must be in the range (-6, 0). The suitable value is -4.

Correct Answer: B — -4

Q. For which value of m does the equation x² - mx + 9 = 0 have roots 3 and 3? (2023)

A. 6
B. 9
C. 3
D. 0

Solution

The sum of the roots is 3 + 3 = 6, hence m = 6.

Correct Answer: A — 6

Q. For which value of p does the equation x² + px + 4 = 0 have roots 2 and -2? (2022)

A. 0
B. 4
C. -4
D. 2

Solution

Using the sum of roots: 2 + (-2) = -p, hence p = 0.

Correct Answer: C — -4

Q. For which value of p does the equation x² + px + 4 = 0 have roots that are both negative? (2022)

A. -8
B. -6
C. -4
D. -2

Solution

For both roots to be negative, p must be greater than 0 and p² > 16. Thus, p < -4.

Correct Answer: C — -4

Q. For which value of p does the equation x² + px + 9 = 0 have roots that are both negative? (2021)

A. -6
B. -4
C. -3
D. -2

Solution

For both roots to be negative, p must be positive and p² > 4*9. Thus, p > 6, so p = -4 is valid.

Correct Answer: B — -4

Q. For which value of p does the equation x² - px + 9 = 0 have roots 3 and 3? (2021)

A. 6
B. 3
C. 9
D. 0

Solution

The sum of the roots is 3 + 3 = 6, hence p = 6.

Correct Answer: A — 6

Q. How does gravitational potential energy (U) change when the distance from the center of the Earth increases? (2020)

A. It increases
B. It decreases
C. It remains constant
D. It becomes zero

Solution

Gravitational potential energy increases as the distance from the center of the Earth increases.

Correct Answer: A — It increases

Q. How does the gravitational force between two objects change if the mass of one object is halved?

A. It doubles
B. It halves
C. It remains the same
D. It becomes zero

Solution

Gravitational force is directly proportional to the product of the masses. Halving one mass halves the force.

Correct Answer: B — It halves

Q. How many different 4-digit PINs can be formed using the digits 0-9 if digits cannot be repeated?

A. 5040
B. 10000
C. 9000
D. 7200

Solution

The first digit can be any of 10 digits, the second can be any of 9, the third can be any of 8, and the fourth can be any of 7. Total = 10 * 9 * 8 * 7 = 5040.

Correct Answer: A — 5040

Q. How many different 4-digit PINs can be formed using the digits 0-9 if repetition is allowed? (2020)

A. 10000
B. 1000
C. 100
D. 1000

Solution

Each digit can be any of the 10 digits (0-9), so the total number of PINs is 10^4 = 10000.

Correct Answer: A — 10000

Q. How many different 4-digit PINs can be formed using the digits 0-9 without repetition?

A. 5040
B. 9000
C. 10000
D. 1000

Solution

The number of 4-digit PINs is P(10, 4) = 5040.

Correct Answer: A — 5040

Q. How many different ways can 2 out of 5 different fruits be chosen?

A. 10
B. 15
C. 20
D. 25

Solution

The number of ways to choose 2 from 5 is C(5, 2) = 10.

Correct Answer: A — 10

Q. How many different ways can 3 boys and 2 girls be seated in a row?

A. 30
B. 60
C. 120
D. 180

Solution

The number of ways to arrange 5 people is 5! = 120.

Correct Answer: C — 120

Q. How many different ways can 4 different colored balls be arranged in a row? (2020)

A. 12
B. 24
C. 36
D. 48

Solution

The number of arrangements of 4 different colored balls is 4! = 24.

Correct Answer: B — 24

Q. How many different ways can 4 prizes be distributed among 10 students? (2020)

A. 5040
B. 10000
C. 2100
D. 120

Solution

The number of ways to distribute 4 prizes among 10 students is 10P4 = 10! / (10-4)! = 5040.

Correct Answer: A — 5040

Q. How many different ways can 6 people be seated in a row? (2020)

A. 720
B. 600
C. 360
D. 480

Solution

The number of arrangements of 6 people is 6! = 720.

Correct Answer: A — 720

Q. How many different ways can the letters of the word 'BOOK' be arranged? (2014)

A. 12
B. 24
C. 36
D. 48

Solution

The number of arrangements of the letters in 'BOOK' is 4! / 2! = 12.

Correct Answer: B — 24

Q. How many different ways can the letters of the word 'MATH' be arranged?

A. 12
B. 24
C. 36
D. 48

Solution

The number of arrangements of the letters in 'MATH' is 4! = 24.

Correct Answer: B — 24

Q. How many grams are in 0.5 moles of glucose (C6H12O6)? (2023)

A. 90 g
B. 180 g
C. 45 g
D. 60 g

Solution

Molar mass of C6H12O6 = 6*12 + 12*1 + 6*16 = 180 g/mol. Mass = 0.5 moles x 180 g/mol = 90 g.

Correct Answer: A — 90 g

Q. How many grams are in 4 moles of potassium chloride (KCl)? (2022)

A. 74 g
B. 148 g
C. 296 g
D. 37 g

Solution

Molar mass of KCl = 39 + 35.5 = 74.5 g/mol. Mass = 4 moles x 74.5 g/mol = 298 g.

Correct Answer: C — 296 g

Q. How many grams are in 4 moles of sulfuric acid (H2SO4)? (2023)

A. 196 g
B. 98 g
C. 392 g
D. 294 g

Solution

Molar mass of H2SO4 = 2*1 + 32 + 4*16 = 98 g/mol. Mass = 4 moles x 98 g/mol = 392 g.

Correct Answer: C — 392 g

Q. How many grams of KCl are needed to prepare 0.5 M solution in 500 mL of water? (2023)

A. 14.9 g
B. 7.45 g
C. 29.8 g
D. 3.73 g

Solution

Molarity = moles/volume(L). Moles = 0.5 mol/L * 0.5 L = 0.25 mol. Mass = moles * molar mass = 0.25 mol * 74.5 g/mol = 18.625 g.

Correct Answer: B — 7.45 g

Q. How many grams of KCl are needed to prepare 250 mL of a 0.5 M solution? (2023) 2023

A. 7.45 g
B. 12.5 g
C. 9.25 g
D. 5.25 g

Solution

Molar mass of KCl = 74.55 g/mol. Grams = Molarity * Volume (L) * Molar mass = 0.5 M * 0.25 L * 74.55 g/mol = 9.31 g.

Correct Answer: A — 7.45 g

Q. How many grams of KCl are needed to prepare 250 mL of a 2 M solution? (2023)

A. 37.25 g
B. 50 g
C. 75 g
D. 25 g

Solution

Molar mass of KCl = 74.55 g/mol. Grams = moles × molar mass = 0.5 moles × 74.55 g/mol = 37.25 g.

Correct Answer: A — 37.25 g

Q. How many grams of KCl are needed to prepare 250 mL of a 2 M solution? (Molar mass of KCl = 74.5 g/mol) (2023)

A. 37.25 g
B. 74.5 g
C. 18.625 g
D. 9.25 g

Solution

Grams = moles × molar mass = (2 moles/L × 0.25 L) × 74.5 g/mol = 37.25 g.

Correct Answer: A — 37.25 g

Q. How many liters of a 0.5 M NaOH solution are needed to obtain 1 mole of NaOH? (2023)

A. 1 L
B. 2 L
C. 0.5 L
D. 0.25 L

Solution

Volume = moles / molarity = 1 mole / 0.5 M = 2 L.

Correct Answer: A — 1 L

Q. How many liters of a 0.5 M solution contain 2 moles of solute? (2023)

A. 2 L
B. 4 L
C. 1 L
D. 0.5 L

Solution

Volume = moles / molarity = 2 moles / 0.5 M = 4 L

Correct Answer: B — 4 L

Q. How many liters of a 3 M solution can be made from 6 moles of solute? (2023)

A. 1 L
B. 2 L
C. 3 L
D. 4 L

Solution

Volume = moles / molarity = 6 moles / 3 M = 2 L.

Correct Answer: B — 2 L

Q. How many molecules are in 0.1 moles of oxygen gas (O2)? (2021)

A. 6.022 x 10^22
B. 6.022 x 10^23
C. 1.2044 x 10^23
D. 3.011 x 10^23

Solution

Number of molecules = moles x Avogadro's number = 0.1 moles x 6.022 x 10^23 molecules/mole = 6.022 x 10^22 molecules.

Correct Answer: A — 6.022 x 10^22

Q. How many molecules are in 0.25 moles of oxygen gas (O2)? (2022)

A. 3.011 x 10^23
B. 6.022 x 10^22
C. 1.2044 x 10^23
D. 1.5055 x 10^23

Solution

Number of molecules = moles x Avogadro's number = 0.25 moles x 6.022 x 10^23 molecules/mole = 1.5055 x 10^23 molecules.

Correct Answer: B — 6.022 x 10^22

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