General Knowledge (NDA)

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General Knowledge (NDA)

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General Knowledge (NDA) MCQ & Objective Questions

General Knowledge (NDA) is a crucial component of various competitive exams, including the NDA entrance test. Mastering this subject not only enhances your awareness of current affairs but also significantly boosts your exam scores. Practicing MCQs and objective questions helps you identify important topics and improves your ability to tackle exam questions effectively. By engaging with practice questions, you can solidify your understanding and prepare thoroughly for your upcoming exams.

What You Will Practise Here

Current Affairs: National and International Events
History: Key Dates and Important Figures
Geography: Physical and Political Maps
Indian Polity: Constitution and Governance
Economics: Basic Concepts and Terminology
Science: Fundamental Principles and Discoveries
General Awareness: Sports, Awards, and Recognitions

Exam Relevance

General Knowledge is a significant section in various exams such as CBSE, State Boards, NEET, and JEE. In these assessments, you can expect questions that test your understanding of historical events, geographical facts, and current affairs. Common question patterns include multiple-choice questions that require you to select the correct answer from a set of options, making it essential to be well-prepared with important General Knowledge (NDA) questions for exams.

Common Mistakes Students Make

Overlooking current affairs, assuming they are less important.
Confusing similar historical events or dates.
Neglecting to revise basic geographical concepts.
Misunderstanding key terms in Indian polity and governance.
Failing to connect economic principles with real-world scenarios.

FAQs

Question: What are the best ways to prepare for General Knowledge (NDA)?
Answer: Regularly read newspapers, follow current affairs, and practice MCQs to enhance your knowledge.

Question: How can I improve my speed in answering General Knowledge questions?
Answer: Time yourself while practicing and focus on understanding concepts rather than rote memorization.

Start your journey towards mastering General Knowledge (NDA) today! Solve practice MCQs and test your understanding to ensure you are well-prepared for your exams. Every question you tackle brings you one step closer to success!

Biology Chemistry Current Affairs Geography History & Civics Physics

Q. A 1 kg block of metal at 100°C is placed in 2 kg of water at 20°C. If the final temperature is 30°C, what is the specific heat capacity of the metal? (2023)

A. 0.5 J/kg°C
B. 1 J/kg°C
C. 2 J/kg°C
D. 4 J/kg°C

Solution

Using conservation of energy: m1c1(T1 - Tf) = m2c2(Tf - T2), we find c1 = 2 J/kg°C.

Correct Answer: C — 2 J/kg°C

Q. A 10 kg mass is lifted to a height of 5 m. What is the potential energy gained? (Take g = 10 m/s²)

A. 50 J
B. 100 J
C. 150 J
D. 200 J

Solution

Potential energy PE = mgh = 10 kg * 10 m/s² * 5 m = 500 J.

Correct Answer: B — 100 J

Q. A 10 kg object is dropped from a height of 5 m. What is its speed just before it hits the ground? (Use g = 9.8 m/s²) (2021)

A. 10 m/s
B. 15 m/s
C. 20 m/s
D. 25 m/s

Solution

Using energy conservation, Potential Energy = Kinetic Energy at the ground. mgh = (1/2)mv². v = sqrt(2gh) = sqrt(2 × 9.8 m/s² × 5 m) ≈ 10 m/s.

Correct Answer: B — 15 m/s

Q. A 10 kg object is lifted to a height of 5 m. What is the potential energy gained by the object? (Take g = 10 m/s²)

A. 50 J
B. 100 J
C. 150 J
D. 200 J

Solution

Potential energy PE = mgh = 10 kg × 10 m/s² × 5 m = 500 J.

Correct Answer: B — 100 J

Q. A 10 kg object is lifted to a height of 5 m. What is the potential energy gained? (Take g = 10 m/s²)

A. 50 J
B. 100 J
C. 150 J
D. 200 J

Solution

Potential energy PE = mgh = 10 kg * 10 m/s² * 5 m = 500 J.

Correct Answer: D — 200 J

Q. A 10 kg object is lifted to a height of 5 m. What is the potential energy gained? (Take g = 9.8 m/s²) (2021)

A. 490 J
B. 980 J
C. 390 J
D. 590 J

Solution

Potential Energy (PE) = mgh = 10 kg × 9.8 m/s² × 5 m = 490 J.

Correct Answer: B — 980 J

Q. A 10 kg object is lifted to a height of 5 m. What is the work done against gravity? (Take g = 10 m/s²)

A. 50 J
B. 100 J
C. 150 J
D. 200 J

Solution

Work done W = mgh = 10 kg × 10 m/s² × 5 m = 500 J.

Correct Answer: B — 100 J

Q. A 10 kg object is moving with a speed of 2 m/s. What is its momentum?

A. 20 kg·m/s
B. 10 kg·m/s
C. 5 kg·m/s
D. 15 kg·m/s

Solution

Momentum (p) = mass × velocity = 10 kg × 2 m/s = 20 kg·m/s.

Correct Answer: A — 20 kg·m/s

Q. A 10 kg object is moving with a speed of 5 m/s. What is its momentum? (2022)

A. 50 kg m/s
B. 10 kg m/s
C. 25 kg m/s
D. 15 kg m/s

Solution

Momentum = mass × velocity = 10 kg × 5 m/s = 50 kg m/s.

Correct Answer: A — 50 kg m/s

Q. A 10 kg object is moving with a velocity of 5 m/s. What is its kinetic energy?

A. 125 J
B. 50 J
C. 100 J
D. 75 J

Solution

Kinetic energy KE = (1/2)mv² = (1/2)(10)(5²) = 125 J.

Correct Answer: A — 125 J

Q. A 100 g piece of metal at 150°C is placed in 200 g of water at 25°C. What is the final temperature of the system assuming no heat loss? (2019)

A. 30°C
B. 40°C
C. 50°C
D. 60°C

Solution

Using the principle of conservation of energy, we can calculate the final temperature to be 50°C.

Correct Answer: C — 50°C

Q. A 2 kg object is dropped from a height of 10 m. What is its speed just before it hits the ground? (Use g = 9.8 m/s²) (2021)

A. 14 m/s
B. 20 m/s
C. 10 m/s
D. 15 m/s

Solution

Using energy conservation, PE = KE: mgh = 0.5mv². Solving gives v = √(2gh) = √(2 × 9.8 m/s² × 10 m) = 14 m/s.

Correct Answer: A — 14 m/s

Q. A 2 kg object is heated from 20°C to 80°C. If the specific heat capacity is 0.5 J/g°C, how much heat is absorbed? (2019)

A. 60 kJ
B. 80 kJ
C. 100 kJ
D. 120 kJ

Solution

Heat absorbed (Q) = mcΔT = (2000 g)(0.5 J/g°C)(60°C) = 60,000 J = 60 kJ.

Correct Answer: B — 80 kJ

Q. A 2 kg piece of copper (specific heat = 0.385 J/g°C) is heated from 25°C to 75°C. How much heat is absorbed? (2022)

A. 2000 J
B. 3000 J
C. 4000 J
D. 5000 J

Solution

Q = mcΔT = 2000 g * 0.385 J/g°C * (75°C - 25°C) = 38500 J.

Correct Answer: B — 3000 J

Q. A 2 kg piece of iron at 150°C is placed in 1 kg of water at 20°C. What will be the final temperature of the system? (Specific heat of iron = 0.45 J/g°C, water = 4.18 J/g°C) (2023)

A. 25°C
B. 30°C
C. 35°C
D. 40°C

Solution

Using heat transfer equations, we find the final temperature to be approximately 35°C.

Correct Answer: C — 35°C

Q. A 5 kg block of metal is heated from 20°C to 100°C. If the specific heat capacity of the metal is 0.5 J/g°C, how much heat is required? (2023)

A. 2000 J
B. 2500 J
C. 3000 J
D. 3500 J

Solution

Heat (Q) = m * c * ΔT = 5000 g * 0.5 J/g°C * (100°C - 20°C) = 5000 * 0.5 * 80 = 200000 J = 2000 J.

Correct Answer: C — 3000 J

Q. A 5 kg object is dropped from a height of 10 m. What is its potential energy at the top? (Take g = 9.8 m/s²) (2020)

A. 49 J
B. 98 J
C. 39 J
D. 59 J

Solution

Potential Energy (PE) = mgh = 5 kg × 9.8 m/s² × 10 m = 490 J.

Correct Answer: B — 98 J

Q. A 5 kg object is dropped from a height of 10 m. What is its speed just before it hits the ground? (Use g = 9.8 m/s²) (2021)

A. 14 m/s
B. 20 m/s
C. 10 m/s
D. 15 m/s

Solution

Using energy conservation, Potential Energy = Kinetic Energy. mgh = 1/2 mv². v = √(2gh) = √(2 × 9.8 m/s² × 10 m) = 14 m/s.

Correct Answer: A — 14 m/s

Q. A 5 kg object is dropped from a height of 20 m. What is its velocity just before it hits the ground? (Take g = 10 m/s²)

A. 10 m/s
B. 20 m/s
C. 15 m/s
D. 25 m/s

Solution

Using the formula v² = u² + 2gh, where u = 0, we have v = √(0 + 2*10*20) = √400 = 20 m/s.

Correct Answer: B — 20 m/s

Q. A 5 kg object is lifted to a height of 10 m. What is the potential energy gained by the object? (Take g = 10 m/s²)

A. 50 J
B. 100 J
C. 150 J
D. 200 J

Solution

Potential energy PE = mgh = 5 kg * 10 m/s² * 10 m = 500 J.

Correct Answer: B — 100 J

Q. A 50 kg object is moving with a velocity of 10 m/s. What is its kinetic energy? (2022)

A. 2500 J
B. 500 J
C. 1000 J
D. 2000 J

Solution

Kinetic energy KE = (1/2)mv² = (1/2)(50)(10²) = 2500 J.

Correct Answer: A — 2500 J

Q. A balanced diet consists of 50% carbohydrates, 30% proteins, and 20% fats. If a person consumes 2400 calories, how many calories are from carbohydrates?

A. 1000 calories
B. 1200 calories
C. 1400 calories
D. 1600 calories

Solution

50% of 2400 calories = 0.50 * 2400 = 1200 calories.

Correct Answer: B — 1200 calories

Q. A balanced diet should consist of what percentage of carbohydrates, proteins, and fats respectively? (2021)

A. 50%, 30%, 20%
B. 60%, 25%, 15%
C. 40%, 40%, 20%
D. 70%, 20%, 10%

Solution

A balanced diet typically consists of 60% carbohydrates, 25% proteins, and 15% fats.

Correct Answer: B — 60%, 25%, 15%

Q. A ball is dropped from a height of 80 m. How long will it take to reach the ground? (Take g = 10 m/s²) (2021)

A. 4 s
B. 5 s
C. 6 s
D. 8 s

Solution

Using the formula h = (1/2)gt², we have 80 = (1/2)(10)t². Solving gives t² = 16, so t = 4 s.

Correct Answer: B — 5 s

Q. A ball is thrown upwards with a speed of 15 m/s. How long will it take to reach the highest point? (Take g = 10 m/s²) (2023)

A. 1.5 s
B. 2 s
C. 3 s
D. 4 s

Solution

Time to reach highest point = u/g = 15/10 = 1.5 s.

Correct Answer: B — 2 s

Q. A basketball player has a free throw success rate of 75%. If he takes 20 free throws, how many successful throws can be expected? (2019)

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

Solution

Expected successful throws = 75% of 20 = 0.75 * 20 = 15.

Correct Answer: A — 15

Q. A basketball player has a free throw success rate of 75%. If he takes 20 free throws, how many successful throws can he expect? (2019)

A. 10
B. 15
C. 12
D. 18

Solution

Expected successful throws = 75% of 20 = 0.75 * 20 = 15.

Correct Answer: B — 15

Q. A basketball player has a shooting accuracy of 75%. If he takes 20 shots, how many shots can he be expected to make? (2019)

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

Solution

Expected successful shots = 75% of 20 = 0.75 * 20 = 15.

Correct Answer: A — 15

Q. A basketball player has a shooting accuracy of 75%. If he takes 40 shots in a game, how many shots can he be expected to make? (2019)

A. 25
B. 30
C. 35
D. 20

Solution

Expected successful shots = 75% of 40 = 0.75 * 40 = 30 shots.

Correct Answer: B — 30

Q. A beam of light passes from air into a medium with a refractive index of 1.5. If the angle of incidence is 30 degrees, what is the angle of refraction? (2022)

A. 19.1 degrees
B. 20 degrees
C. 25 degrees
D. 30 degrees

Solution

Using Snell's law, n1 * sin(i) = n2 * sin(r). Thus, sin(r) = (1 * sin(30)) / 1.5 = 0.333, giving r ≈ 19.1 degrees.

Correct Answer: A — 19.1 degrees

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