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 projectile is thrown with a speed of 20 m/s at an angle of 45 degrees. What is the vertical component of the velocity?

A. 10 m/s
B. 14.14 m/s
C. 20 m/s
D. 28.28 m/s

Solution

Vertical component (u_y) = u * sin(θ) = 20 * (√2/2) = 14.14 m/s.

Correct Answer: B — 14.14 m/s

Q. A proton moves in a magnetic field and experiences a force. If the velocity of the proton is doubled, what happens to the magnetic force?

A. It doubles
B. It halves
C. It remains the same
D. It quadruples

Solution

The magnetic force is proportional to the velocity of the charge, so if the velocity is doubled, the magnetic force also doubles.

Correct Answer: A — It doubles

Q. A public administration course has 40 students, and 25% of them are from rural areas. How many students are from rural areas?

A. 5
B. 10
C. 15
D. 20

Solution

Number of rural students = 25% of 40 = 0.25 * 40 = 10.

Correct Answer: C — 15

Q. A public administration reform is expected to reduce costs by 10%. If the current costs are $800,000, what will be the new costs?

A. $720,000
B. $740,000
C. $760,000
D. $780,000

Solution

New costs = Current costs - (10% of Current costs) = 800,000 - (0.10 * 800,000) = 720,000.

Correct Answer: A — $720,000

Q. A public health campaign costs $750,000 and reaches 150,000 people. What is the cost per person reached?

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

Solution

Cost per person = $750,000 / 150,000 = $5.

Correct Answer: C — $5

Q. A public policy aims to reduce waste by 30%. If the current waste level is 500 tons, what is the target waste level?

A. 350 tons
B. 400 tons
C. 450 tons
D. 500 tons

Solution

Target waste level = 500 tons - (30% of 500 tons) = 500 - 150 = 350 tons.

Correct Answer: A — 350 tons

Q. A public policy initiative aims to reduce traffic congestion by 15%. If the current congestion level is measured at 80 units, what will be the target congestion level after the initiative?

A. 60 units
B. 65 units
C. 68 units
D. 70 units

Solution

Target congestion level = 80 units - (15% of 80) = 80 - 12 = 68 units.

Correct Answer: A — 60 units

Q. A public policy initiative costs $1,200,000. If it is funded by 3 different agencies equally, how much does each agency contribute?

A. $300,000
B. $400,000
C. $500,000
D. $600,000

Solution

Each agency contributes = Total cost / Number of agencies = 1,200,000 / 3 = 400,000.

Correct Answer: A — $300,000

Q. A public policy initiative is expected to create 1,200 jobs. If each job costs $10,000 to create, what is the total cost of the initiative?

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

Solution

Total cost = 1,200 jobs * $10,000/job = $12 million.

Correct Answer: B — $12 million

Q. A public policy is implemented to reduce corruption by 25%. If the initial corruption level was estimated at $1,000,000, what is the new estimated corruption level?

A. $500,000
B. $750,000
C. $800,000
D. $900,000

Solution

25% of $1,000,000 is $250,000. Therefore, $1,000,000 - $250,000 = $750,000.

Correct Answer: B — $750,000

Q. A quiz competition awards 4 prizes: 1st place gets $500, 2nd place gets $300, 3rd place gets $200, and 4th place gets $100. What is the total amount awarded?

A. $1000
B. $1100
C. $1200
D. $1300

Solution

Total amount = 500 + 300 + 200 + 100 = $1100.

Correct Answer: B — $1100

Q. A radar system can detect objects up to 100 km away. If an object is detected at 75 km, how far is it from the radar? (2022)

A. 25 km
B. 50 km
C. 75 km
D. 100 km

Solution

Distance from Radar = Maximum Range - Detected Distance = 100 km - 75 km = 25 km

Correct Answer: A — 25 km

Q. A radar system can detect objects up to 100 km away. If it scans an area of 50 km radius, what is the area it can cover? (2023)

A. 7850 km²
B. 1500 km²
C. 2000 km²
D. 2500 km²

Solution

Area = π × r² = π × (50 km)² ≈ 7850 km²

Correct Answer: A — 7850 km²

Q. A radar system can detect objects up to 200 km away. If it scans an area in a circular pattern, what is the area it can cover? (2023)

A. 12560 km²
B. 31415 km²
C. 40000 km²
D. 50000 km²

Solution

Area = πr² = π(200 km)² = 125600 km² (approx 31415 km²)

Correct Answer: B — 31415 km²

Q. A radar system can detect objects up to 200 km away. If it scans at a speed of 20 km/h, how long will it take to complete a full scan? (2022)

A. 8 hours
B. 10 hours
C. 12 hours
D. 15 hours

Solution

Time = Distance / Speed = 200 km / 20 km/h = 10 hours

Correct Answer: B — 10 hours

Q. A radar system can detect objects up to a distance of 100 km. If it scans at a speed of 20 km/h, how long will it take to complete a full scan? (2023)

A. 4 hours
B. 5 hours
C. 6 hours
D. 7 hours

Solution

Time = Distance / Speed = 100 km / 20 km/h = 5 hours

Correct Answer: A — 4 hours

Q. A radar system can detect objects up to a distance of 15 km. What is this distance in meters? (2022)

A. 1000 m
B. 5000 m
C. 10000 m
D. 15000 m

Solution

Distance in meters = 15 km × 1000 m/km = 15000 m

Correct Answer: D — 15000 m

Q. A radar system can detect objects up to a distance of 50 km. If it scans an area of 20 km radius, what is the area it can cover? (2023)

A. 1256 km²
B. 314 km²
C. 400 km²
D. 500 km²

Solution

Area = π × r² = π × (20 km)² ≈ 1256 km²

Correct Answer: A — 1256 km²

Q. A ray of light in a medium with n=1.33 strikes the boundary with air at an angle of 45°. What is the behavior of the ray?

A. Total internal reflection
B. Partial reflection and refraction
C. Complete absorption
D. Total refraction

Solution

Since 45° < θc ≈ 48.6°, the ray will partially reflect and refract.

Correct Answer: B — Partial reflection and refraction

Q. A ray of light in glass (n=1.5) strikes the glass-air interface at an angle of 30°. What will be the angle of refraction in air?

A. 60°
B. 30°
C. 45°
D. 90°

Solution

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1.5, θ1 = 30°, n2 = 1.0. Thus, sin(θ2) = (1.5 * sin(30°))/1.0 = 0.75, giving θ2 ≈ 60°.

Correct Answer: A — 60°

Q. A ray of light passes from air into glass at an angle of 45 degrees. What is the angle of refraction if the refractive index of glass is 1.5? (2022)

A. 30 degrees
B. 45 degrees
C. 60 degrees
D. 90 degrees

Solution

Using Snell's law, n1 * sin(theta1) = n2 * sin(theta2). Here, n1 = 1 (air), theta1 = 45 degrees, n2 = 1.5 (glass). Thus, sin(theta2) = (1 * sin(45))/1.5 = 0.471, giving theta2 ≈ 30 degrees.

Correct Answer: A — 30 degrees

Q. A ray of light passes from air into glass at an angle of incidence of 30 degrees. If the refractive index of glass is 1.5, what is the angle of refraction?

A. 20 degrees
B. 30 degrees
C. 18.4 degrees
D. 22 degrees

Solution

Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1 (air), i = 30 degrees, n2 = 1.5 (glass). Thus, sin(r) = (1 * sin(30))/1.5 = 0.333, giving r ≈ 18.4 degrees.

Correct Answer: C — 18.4 degrees

Q. A ray of light passes from air into glass at an angle of incidence of 45 degrees. If the refractive index of glass is 1.5, what is the angle of refraction?

A. 30 degrees
B. 45 degrees
C. 60 degrees
D. 90 degrees

Solution

Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1 (air), n2 = 1.5 (glass), i = 45 degrees. Solving gives r = 30 degrees.

Correct Answer: A — 30 degrees

Q. A ray of light passes from air into glass with a refractive index of 1.5. If the angle of incidence is 30 degrees, what is the angle of refraction?

A. 20 degrees
B. 30 degrees
C. 18.4 degrees
D. 22 degrees

Solution

Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1, n2 = 1.5, i = 30 degrees. Solving gives r ≈ 18.4 degrees.

Correct Answer: C — 18.4 degrees

Q. A ray of light passes from air into water at an angle of incidence of 30 degrees. What is the angle of refraction?

A. 22 degrees
B. 30 degrees
C. 45 degrees
D. 18 degrees

Solution

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1 (air), n2 = 1.33 (water), θ1 = 30 degrees. Thus, sin(θ2) = (1 * sin(30))/1.33 = 0.375. Therefore, θ2 = sin^(-1)(0.375) which is approximately 22 degrees.

Correct Answer: A — 22 degrees

Q. A ray of light passes from air into water at an angle of incidence of 30 degrees. What is the angle of refraction in water (n = 1.33)?

A. 22.5 degrees
B. 30 degrees
C. 40 degrees
D. 45 degrees

Solution

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, sin(30) = 0.5, so 1 * 0.5 = 1.33 * sin(θ2). Solving gives θ2 ≈ 22.5 degrees.

Correct Answer: A — 22.5 degrees

Q. A ray of light passes from air into water at an angle of incidence of 30 degrees. What is the angle of refraction in water? (Refractive index of water = 1.33)

A. 22.5 degrees
B. 30 degrees
C. 40 degrees
D. 20 degrees

Solution

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2), we find θ2 = sin^(-1)(sin(30 degrees)/1.33) = 22.5 degrees.

Correct Answer: A — 22.5 degrees

Q. A ray of light passes from air into water. If the angle of incidence is 30 degrees, what is the angle of refraction?

A. 22 degrees
B. 30 degrees
C. 45 degrees
D. 18 degrees

Solution

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1 (air), n2 = 1.33 (water). Solving gives θ2 ≈ 22 degrees.

Correct Answer: A — 22 degrees

Q. A ray of light passes from air into water. If the angle of incidence is 30 degrees, what is the angle of refraction in water (n = 1.33)?

A. 22.5 degrees
B. 30 degrees
C. 22 degrees
D. 40 degrees

Solution

Using Snell's law: n1 * sin(θ1) = n2 * sin(θ2). Here, sin(30) = 0.5, so sin(θ2) = (1 * 0.5) / 1.33 = 0.375. Thus, θ2 = sin^(-1)(0.375) ≈ 22.5 degrees.

Correct Answer: A — 22.5 degrees

Q. A ray of light passes from air into water. If the angle of incidence is 30 degrees, what is the angle of refraction in water? (Refractive index of water = 1.33)

A. 22.1 degrees
B. 30 degrees
C. 36.9 degrees
D. 45 degrees

Solution

Using Snell's law, n1*sin(θ1) = n2*sin(θ2), we find θ2 = 22.1 degrees.

Correct Answer: A — 22.1 degrees

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