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. A force of 50 N is applied at an angle of 60 degrees to the lever arm of length 1 m. What is the torque about the pivot?

A. 25 Nm
B. 43.3 Nm
C. 50 Nm
D. 0 Nm

Solution

Torque (τ) = F × r × sin(θ) = 50 N × 1 m × sin(60°) = 50 N × 1 m × (√3/2) ≈ 43.3 Nm.

Correct Answer: B — 43.3 Nm

Q. A force of 50 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?

A. 25 Nm
B. 50 Nm
C. 43.3 Nm
D. 100 Nm

Solution

Torque (τ) = F × r × sin(θ) = 50 N × 2 m × sin(60°) = 50 × 2 × (√3/2) = 43.3 Nm.

Correct Answer: C — 43.3 Nm

Q. A force of 50 N is applied at an angle of 60° to the horizontal. What is the horizontal component of the force? (2020)

A. 25 N
B. 50 N
C. 43.3 N
D. 30 N

Solution

Horizontal component = F × cos(θ) = 50 N × cos(60°) = 50 N × 0.5 = 25 N.

Correct Answer: C — 43.3 N

Q. A force of 50 N is applied to a 25 kg object. What is the acceleration of the object? (2019)

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Using F = ma, a = F/m = 50 N / 25 kg = 2 m/s².

Correct Answer: B — 2 m/s²

Q. A force of 50 N is applied to a 25 kg object. What is the object's acceleration? (2020)

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Using F = ma, a = F/m = 50 N / 25 kg = 2 m/s².

Correct Answer: B — 2 m/s²

Q. A force of 50 N is applied to move an object 4 m. How much work is done?

A. 100 J
B. 150 J
C. 200 J
D. 250 J

Solution

Work = Force * Distance = 50 N * 4 m = 200 J.

Correct Answer: C — 200 J

Q. A forced oscillator has a mass of 3 kg and is driven by a force of 12 N at a frequency of 2 Hz. What is the amplitude of the oscillation if the damping coefficient is 0.1 kg/s?

A. 0.1 m
B. 0.2 m
C. 0.3 m
D. 0.4 m

Solution

Using F = mAω², we find A = F / (mω²) = 12 / (3*(2π*2)²) ≈ 0.2 m.

Correct Answer: B — 0.2 m

Q. A forest area of 1000 hectares absorbs 2.5 tons of CO2 per hectare per year. How much CO2 does it absorb in total in a year?

A. 2500 tons
B. 2000 tons
C. 3000 tons
D. 1500 tons

Solution

Total CO2 absorption = Area * CO2 absorption per hectare = 1000 * 2.5 = 2500 tons.

Correct Answer: A — 2500 tons

Q. A foundation awards $5000 to be distributed among 4 projects in the ratio 1:2:3:4. How much does the third project receive? (2023)

A. $500
B. $1000
C. $1500
D. $2000

Solution

Total parts = 1 + 2 + 3 + 4 = 10. Third project's share = (3/10) * 5000 = $1500.

Correct Answer: C — $1500

Q. A fruit seller has 120 apples. He sells 25% of them. How many apples does he have left?

A. 90
B. 100
C. 80
D. 75

Solution

If the seller sells 25% of 120 apples, he sells 0.25 * 120 = 30 apples. Therefore, the number of apples left is 120 - 30 = 90.

Correct Answer: A — 90

Q. A fruit seller has 120 apples. He sells 30% of them. How many apples does he have left? (2023)

A. 70
B. 80
C. 90
D. 100

Solution

Apples sold = 30% of 120 = 36. Apples left = 120 - 36 = 84.

Correct Answer: B — 80

Q. A fruit seller has apples and oranges in the ratio 5:3. If he has 40 apples, how many oranges does he have?

A. 24
B. 30
C. 32
D. 20

Solution

The ratio of apples to oranges is 5:3. If there are 40 apples, we can set up the proportion: 5/3 = 40/x. Cross-multiplying gives us 5x = 120, so x = 120/5 = 24. Therefore, he has 24 oranges.

Correct Answer: B — 30

Q. A fruit seller has apples and oranges in the ratio of 5:3. If he has 40 apples, how many oranges does he have?

A. 24
B. 30
C. 32
D. 36

Solution

If the ratio of apples to oranges is 5:3, then for every 5 apples, there are 3 oranges. If there are 40 apples, we can set up the proportion: 5/3 = 40/x. Cross-multiplying gives us 5x = 120, so x = 24. Therefore, there are 24 oranges.

Correct Answer: B — 30

Q. A gallery has 120 paintings. If 30% of them are abstract, how many abstract paintings are there?

A. 36
B. 30
C. 24
D. 42

Solution

Number of abstract paintings = 30% of 120 = 30/100 * 120 = 36.

Correct Answer: A — 36

Q. A gardener can plant 100 trees in 5 hours. How many trees can he plant in 3 hours?

A. 50
B. 60
C. 70
D. 80

Solution

The rate is 100 trees / 5 hours = 20 trees/hour. In 3 hours, he can plant 20 * 3 = 60 trees.

Correct Answer: A — 50

Q. A gardener has 36 red roses and 48 yellow roses. He wants to plant them in rows with the same number of each type of rose in each row. What is the maximum number of rows he can plant? (2023)

A. 6
B. 12
C. 18
D. 24

Solution

The HCF of 36 and 48 is 12, which is the maximum number of rows he can plant.

Correct Answer: B — 12

Q. A gardener has 60 red flowers and 90 yellow flowers. What is the largest number of bouquets he can make if each bouquet has the same number of red and yellow flowers? (2023)

A. 15
B. 30
C. 45
D. 60

Solution

The largest number of bouquets is the HCF of 60 and 90, which is 30.

Correct Answer: B — 30

Q. A gardener has two types of plants, one type has a height of 3 feet and the other 5 feet. What is the minimum height at which both types can be tied together? (2023)

A. 15
B. 30
C. 60
D. 45

Solution

The minimum height is the LCM of 3 and 5, which is 15 feet.

Correct Answer: C — 60

Q. A gardener has two types of plants, one type requires watering every 4 days and the other every 6 days. If both types are watered together today, in how many days will they be watered together again? (2023)

A. 12
B. 24
C. 18
D. 30

Solution

The LCM of 4 and 6 is 12. Therefore, they will be watered together again in 12 days.

Correct Answer: A — 12

Q. A gas at 300 K has an RMS speed of 400 m/s. What will be its RMS speed at 600 K?

A. 400 m/s
B. 400 sqrt(2) m/s
C. 800 m/s
D. 200 m/s

Solution

The RMS speed is proportional to the square root of the temperature. Therefore, at 600 K, the RMS speed will be 400 * sqrt(600/300) = 400 * sqrt(2) m/s.

Correct Answer: B — 400 sqrt(2) m/s

Q. A gas at 300 K has an RMS speed of 500 m/s. What will be its RMS speed at 600 K?

A. 500 m/s
B. 707 m/s
C. 1000 m/s
D. 250 m/s

Solution

The RMS speed is proportional to the square root of the temperature. Therefore, v_rms at 600 K = 500 * sqrt(600/300) = 500 * sqrt(2) ≈ 707 m/s.

Correct Answer: B — 707 m/s

Q. A gas expands from 2 L to 5 L at a constant pressure of 1 atm. How much work is done by the gas? (2023)

A. 3 L·atm
B. 5 L·atm
C. 2 L·atm
D. 1 L·atm

Solution

Work done by the gas during expansion at constant pressure is W = PΔV. Here, ΔV = 5 L - 2 L = 3 L, so W = 1 atm * 3 L = 3 L·atm.

Correct Answer: A — 3 L·atm

Q. A gas expands from a volume of 2 m³ to 5 m³ at a constant pressure of 100 kPa. What is the work done by the gas? (2022)

A. 300 kJ
B. 500 kJ
C. 700 kJ
D. 3000 kJ

Solution

Work done (W) = P * ΔV = 100 kPa * (5 m³ - 2 m³) = 100 kPa * 3 m³ = 300 kJ.

Correct Answer: A — 300 kJ

Q. A gas expands from volume V1 to V2 at constant pressure P. What is the work done by the gas? (2022)

A. P(V2 - V1)
B. P(V1 + V2)
C. P(V1 * V2)
D. P(V1 - V2)

Solution

Work done (W) = P * ΔV = P * (V2 - V1).

Correct Answer: A — P(V2 - V1)

Q. A gas expands from volume V1 to V2 at constant temperature. If the initial pressure is P1, what is the final pressure P2? (2020)

A. P1(V1/V2)
B. P1(V2/V1)
C. P1
D. P1/2

Solution

According to Boyle's Law, P1V1 = P2V2, thus P2 = P1(V1/V2).

Correct Answer: A — P1(V1/V2)

Q. A gas expands isothermally at 300 K and absorbs 600 J of heat. What is the work done by the gas? (2023)

A. 600 J
B. 300 J
C. 900 J
D. 0 J

Solution

In an isothermal process, the work done by the gas is equal to the heat absorbed. Therefore, work done = 600 J.

Correct Answer: A — 600 J

Q. A gas expands isothermally at 300 K from a volume of 1 m³ to 2 m³. If the pressure of the gas is 100 kPa, what is the work done by the gas? (2020)

A. 0 kJ
B. 10 kJ
C. 20 kJ
D. 30 kJ

Solution

Work done (W) = P * ΔV = 100 kPa * (2 m³ - 1 m³) = 100 kPa * 1 m³ = 100 kJ = 10 kJ.

Correct Answer: B — 10 kJ

Q. A gas expands isothermally at 300 K from a volume of 1 m³ to 2 m³. If the pressure at the initial state is 100 kPa, what is the work done by the gas?

A. 0 kJ
B. 100 kJ
C. 150 kJ
D. 200 kJ

Solution

Work done (W) = nRT ln(Vf/Vi). Here, nRT = PVi = 100 kPa * 1 m³ = 100 kJ. W = 100 kJ * ln(2) ≈ 69.3 kJ.

Correct Answer: B — 100 kJ

Q. A gas expands isothermally at 300 K from a volume of 1 m³ to 2 m³. If the pressure of the gas is 100 kPa, what is the work done by the gas during expansion? (2019)

A. 20 kJ
B. 30 kJ
C. 40 kJ
D. 50 kJ

Solution

Work done (W) = PΔV = 100 kPa * (2 m³ - 1 m³) = 100 kPa * 1 m³ = 100 kJ = 20 kJ.

Correct Answer: A — 20 kJ

Q. A gas expands isothermally at 300 K from a volume of 1 m³ to 2 m³. What is the work done by the gas?

A. 0 J
B. 300 J
C. 600 J
D. 150 J

Solution

The work done in an isothermal expansion is W = nRT ln(Vf/Vi). Assuming 1 mole of gas, W = 1 * 8.31 * 300 * ln(2) ≈ 600 J.

Correct Answer: C — 600 J

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