Major Competitive Exams

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Major Competitive Exams

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Major Competitive Exams MCQ & Objective Questions

Major Competitive Exams play a crucial role in shaping the academic and professional futures of students in India. These exams not only assess knowledge but also test problem-solving skills and time management. Practicing MCQs and objective questions is essential for scoring better, as they help in familiarizing students with the exam format and identifying important questions that frequently appear in tests.

What You Will Practise Here

Key concepts and theories related to major subjects
Important formulas and their applications
Definitions of critical terms and terminologies
Diagrams and illustrations to enhance understanding
Practice questions that mirror actual exam patterns
Strategies for solving objective questions efficiently
Time management techniques for competitive exams

Exam Relevance

The topics covered under Major Competitive Exams are integral to various examinations such as CBSE, State Boards, NEET, and JEE. Students can expect to encounter a mix of conceptual and application-based questions that require a solid understanding of the subjects. Common question patterns include multiple-choice questions that test both knowledge and analytical skills, making it essential to be well-prepared with practice MCQs.

Common Mistakes Students Make

Rushing through questions without reading them carefully
Overlooking the negative marking scheme in MCQs
Confusing similar concepts or terms
Neglecting to review previous years’ question papers
Failing to manage time effectively during the exam

FAQs

Question: How can I improve my performance in Major Competitive Exams?
Answer: Regular practice of MCQs and understanding key concepts will significantly enhance your performance.

Question: What types of questions should I focus on for these exams?
Answer: Concentrate on important Major Competitive Exams questions that frequently appear in past papers and mock tests.

Question: Are there specific strategies for tackling objective questions?
Answer: Yes, practicing under timed conditions and reviewing mistakes can help develop effective strategies.

Start your journey towards success by solving practice MCQs today! Test your understanding and build confidence for your upcoming exams. Remember, consistent practice is the key to mastering Major Competitive Exams!

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Q. Calculate the moment of inertia of a hollow sphere of mass M and radius R about an axis through its center.

A. 2/5 MR^2
B. 3/5 MR^2
C. 2/3 MR^2
D. MR^2

Solution

The moment of inertia of a hollow sphere about an axis through its center is I = 2/5 MR^2.

Correct Answer: B — 3/5 MR^2

Q. Calculate the perimeter of a square with side length 4 cm. (2015)

A. 16 cm
B. 12 cm
C. 8 cm
D. 20 cm

Solution

Perimeter = 4 × side = 4 × 4 cm = 16 cm.

Correct Answer: A — 16 cm

Q. Calculate the perimeter of a square with side length 6 cm. (2015)

A. 24 cm
B. 20 cm
C. 18 cm
D. 30 cm

Solution

Perimeter = 4 × side = 4 × 6 = 24 cm.

Correct Answer: A — 24 cm

Q. Calculate the pH of a 0.01 M solution of NaHCO3. (2023)

A. 8.3
B. 9.0
C. 7.5
D. 8.0

Solution

NaHCO3 is a weak base. The pH can be calculated using the formula pH = 7 + 0.5(pKa - log[C]). pKa of HCO3- is about 10.3, so pH ≈ 8.3.

Correct Answer: A — 8.3

Q. Calculate the pH of a 0.05 M NH4Cl solution (Kb for NH3 = 1.8 x 10^-5).

A. 4.75
B. 5.25
C. 5.75
D. 6.25

Solution

Using the formula for weak bases, pH = 14 - 0.5(pKb - logC) = 14 - 0.5(4.74 - log(0.05)) = 5.25.

Correct Answer: B — 5.25

Q. Calculate the pH of a 0.1 M acetic acid solution (Ka = 1.8 x 10^-5).

A. 2.87
B. 3.87
C. 4.87
D. 5.87

Solution

Using the formula for weak acids, pH = 0.5(pKa - log[C]), where pKa = -log(1.8 x 10^-5) ≈ 4.74. Thus, pH = 0.5(4.74 - log(0.1)) = 3.87.

Correct Answer: B — 3.87

Q. Calculate the pH of a 0.1 M NaOH solution.

A. 12
B. 13
C. 14
D. 11

Solution

pOH = -log[OH-] = -log(0.1) = 1, thus pH = 14 - pOH = 14 - 1 = 13.

Correct Answer: C — 14

Q. Calculate the pH of a 0.2 M solution of KOH.

A. 12
B. 13
C. 14
D. 11

Solution

pOH = -log(0.2) = 0.7, thus pH = 14 - 0.7 = 13.3.

Correct Answer: B — 13

Q. Calculate the pH of a buffer solution containing 0.1 M acetic acid and 0.1 M sodium acetate.

A. 4.76
B. 5.76
C. 6.76
D. 7.76

Solution

Using Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]); pKa of acetic acid = 4.76, so pH = 4.76 + log(1) = 4.76

Correct Answer: B — 5.76

Q. Calculate the pH of a solution that is 0.1 M in acetic acid (Ka = 1.8 x 10^-5).

A. 2.87
B. 3.87
C. 4.87
D. 5.87

Solution

Using the formula for weak acids, pH = 0.5(pKa - logC) = 0.5(4.74 - log(0.1)) = 3.87.

Correct Answer: B — 3.87

Q. Calculate the range of the data set: 12, 15, 22, 30, 5.

A. 25
B. 27
C. 30
D. 22

Solution

Range = Maximum - Minimum = 30 - 5 = 25.

Correct Answer: A — 25

Q. Calculate the range of the data set: 4, 8, 15, 16, 23, 42.

A. 38
B. 36
C. 34
D. 32

Solution

Range = Maximum - Minimum = 42 - 4 = 38.

Correct Answer: A — 38

Q. Calculate the range of the data set: 8, 12, 15, 20, 22.

A. 10
B. 12
C. 14
D. 15

Solution

Range = Max - Min = 22 - 8 = 14.

Correct Answer: A — 10

Q. Calculate the range of the following data set: 12, 15, 20, 22, 30.

A. 10
B. 12
C. 18
D. 20

Solution

Range = Maximum - Minimum = 30 - 12 = 18.

Correct Answer: C — 18

Q. Calculate the range of the following data set: 15, 22, 8, 19, 30.

A. 7
B. 15
C. 22
D. 30

Solution

Range = max - min = 30 - 8 = 22.

Correct Answer: D — 30

Q. Calculate the range of the following data set: 4, 8, 15, 16, 23, 42.

A. 38
B. 36
C. 34
D. 32

Solution

Range = Maximum - Minimum = 42 - 4 = 38.

Correct Answer: A — 38

Q. Calculate the range of the following data set: 8, 12, 15, 7, 10.

A. 5
B. 8
C. 10
D. 12

Solution

Range = Maximum - Minimum = 15 - 7 = 8.

Correct Answer: A — 5

Q. Calculate the RMS speed of a gas with molar mass 0.028 kg/mol at 300 K. (R = 8.314 J/(mol K))

A. 500 m/s
B. 600 m/s
C. 700 m/s
D. 800 m/s

Solution

Using v_rms = sqrt(3RT/M), we find v_rms = sqrt(3 * 8.314 * 300 / 0.028) = 600 m/s.

Correct Answer: B — 600 m/s

Q. Calculate the scalar product of A = (1, 1, 1) and B = (2, 2, 2).

A. 3
B. 4
C. 5
D. 6

Solution

A · B = 1*2 + 1*2 + 1*2 = 2 + 2 + 2 = 6.

Correct Answer: D — 6

Q. Calculate the scalar product of the vectors (1, 0, 0) and (0, 1, 0).

A. 0
B. 1
C. 2
D. 3

Solution

Scalar product = 1*0 + 0*1 + 0*0 = 0.

Correct Answer: A — 0

Q. Calculate the scalar product of the vectors (1, 2, 3) and (4, 5, 6).

A. 32
B. 33
C. 34
D. 35

Solution

Scalar product = 1*4 + 2*5 + 3*6 = 4 + 10 + 18 = 32.

Correct Answer: A — 32

Q. Calculate the scalar product of the vectors (2, 3, 4) and (4, 3, 2).

A. 28
B. 29
C. 30
D. 31

Solution

Scalar product = 2*4 + 3*3 + 4*2 = 8 + 9 + 8 = 25.

Correct Answer: A — 28

Q. Calculate the scalar product of the vectors (3, 0, -3) and (1, 2, 1).

A. 0
B. 1
C. 2
D. 3

Solution

Scalar product = 3*1 + 0*2 + (-3)*1 = 3 + 0 - 3 = 0.

Correct Answer: A — 0

Q. Calculate the scalar product of the vectors A = (1, 2, 3) and B = (4, 5, 6).

A. 32
B. 30
C. 28
D. 26

Solution

A · B = 1*4 + 2*5 + 3*6 = 4 + 10 + 18 = 32.

Correct Answer: B — 30

Q. Calculate the scalar product of the vectors A = (4, -1, 2) and B = (2, 3, 1).

A. 10
B. 8
C. 6
D. 12

Solution

A · B = 4*2 + (-1)*3 + 2*1 = 8 - 3 + 2 = 7.

Correct Answer: A — 10

Q. Calculate the scalar product of the vectors K = (0, 1, 2) and L = (3, 4, 5).

A. 10
B. 11
C. 12
D. 13

Solution

K · L = 0*3 + 1*4 + 2*5 = 0 + 4 + 10 = 14.

Correct Answer: A — 10

Q. Calculate the surface area of a cylinder with a radius of 2 cm and a height of 10 cm.

A. 75.40 cm²
B. 62.83 cm²
C. 40.00 cm²
D. 50.27 cm²

Solution

Surface Area = 2πr(h + r) = 2π(2)(10 + 2) = 2π(2)(12) = 48π ≈ 150.80 cm².

Correct Answer: A — 75.40 cm²

Q. Calculate the surface area of a cylinder with a radius of 3 cm and a height of 10 cm.

A. 60 cm²
B. 62.83 cm²
C. 94.25 cm²
D. 100.00 cm²

Solution

Surface Area = 2πr(h + r) = 2π(3)(10 + 3) = 2π(3)(13) = 78π ≈ 245.04 cm²

Correct Answer: C — 94.25 cm²

Q. Calculate the surface area of a cylinder with a radius of 5 cm and a height of 10 cm.

A. 314.16 cm²
B. 250.00 cm²
C. 200.00 cm²
D. 150.00 cm²

Solution

Surface Area = 2πr(h + r) = 2π(5)(10 + 5) = 2π(5)(15) = 150π ≈ 471.24 cm²

Correct Answer: A — 314.16 cm²

Q. Calculate the surface area of a sphere with a radius of 6 cm.

A. 113.10 cm²
B. 150.80 cm²
C. 452.39 cm²
D. 226.20 cm²

Solution

Surface Area = 4πr² = 4π(6)² = 144π ≈ 452.39 cm².

Correct Answer: C — 452.39 cm²

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