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. For a number to be divisible by 10, which of the following must be true?

A. It must end in 0
B. It must be a two-digit number
C. It must be a prime number
D. It must be even

Solution

A number is divisible by 10 if it ends in 0.

Correct Answer: A — It must end in 0

Q. For a number to be divisible by 11, which of the following must be true?

A. The difference between the sum of the digits in odd positions and the sum of the digits in even positions must be 0 or divisible by 11
B. The number must be even
C. The number must end in 1
D. The sum of the digits must be divisible by 11

Solution

A number is divisible by 11 if the difference between the sum of the digits in odd positions and the sum of the digits in even positions is 0 or divisible by 11.

Correct Answer: A — The difference between the sum of the digits in odd positions and the sum of the digits in even positions must be 0 or divisible by 11

Q. For a number to be divisible by 8, what must be true about its last three digits?

A. They must be divisible by 8
B. They must be even
C. They must be a multiple of 10
D. They must be prime

Solution

A number is divisible by 8 if the number formed by its last three digits is divisible by 8.

Correct Answer: A — They must be divisible by 8

Q. For a point charge, the electric field varies with distance r as?

A. 1/r
B. 1/r²
C. 1/r³
D. 1/r⁴

Solution

The electric field due to a point charge varies as E = kQ/r², where k is a constant.

Correct Answer: B — 1/r²

Q. For a process at constant volume, which of the following is true? (2023)

A. Work done is zero
B. Heat added equals change in internal energy
C. Both A and B
D. None of the above

Solution

At constant volume, no work is done (W=0), and the heat added equals the change in internal energy (Q=ΔU).

Correct Answer: C — Both A and B

Q. For a process to be reversible, it must be:

A. Fast
B. Quasi-static
C. Adiabatic
D. Isochoric

Solution

A reversible process must be quasi-static, meaning it occurs infinitely slowly.

Correct Answer: B — Quasi-static

Q. For a process with ΔH = 200 kJ and ΔS = 0.5 kJ/K, what is ΔG at 400 K?

A. 200 kJ
B. 180 kJ
C. 220 kJ
D. 160 kJ

Solution

ΔG = ΔH - TΔS = 200 kJ - 400 K * 0.5 kJ/K = 200 kJ - 200 kJ = 0 kJ.

Correct Answer: B — 180 kJ

Q. For a reaction A → B, if the rate of formation of B is 0.5 mol/L/s, what is the rate of disappearance of A?

A. 0.5 mol/L/s
B. 1.0 mol/L/s
C. 0.25 mol/L/s
D. 2.0 mol/L/s

Solution

The rate of disappearance of A is equal to the rate of formation of B, multiplied by the stoichiometric coefficients. Here, it is 1.0 mol/L/s.

Correct Answer: B — 1.0 mol/L/s

Q. For a reaction A → B, if the rate of reaction doubles when the concentration of A is doubled, what is the order of the reaction with respect to A?

A. Zero order
B. First order
C. Second order
D. Third order

Solution

If doubling the concentration of A doubles the rate, the reaction is first order with respect to A.

Correct Answer: B — First order

Q. For a reaction at constant temperature and pressure, which of the following is true?

A. ΔG = ΔH + TΔS
B. ΔG = ΔH - TΔS
C. ΔG = TΔS - ΔH
D. ΔG = ΔS - ΔH

Solution

The correct relationship at constant temperature and pressure is ΔG = ΔH - TΔS.

Correct Answer: B — ΔG = ΔH - TΔS

Q. For a reaction at equilibrium, if the concentration of products increases, what will happen to the equilibrium position?

A. Shift to the left
B. Shift to the right
C. No change
D. Depends on temperature

Solution

According to Le Chatelier's principle, if the concentration of products increases, the equilibrium will shift to the left.

Correct Answer: A — Shift to the left

Q. For a reaction at equilibrium, the change in Gibbs free energy (ΔG) is equal to:

A. ΔH - TΔS
B. 0
C. ΔS - TΔH
D. ΔH + TΔS

Solution

At equilibrium, the change in Gibbs free energy (ΔG) is zero, indicating that the system is at maximum entropy.

Correct Answer: B — 0

Q. For a reaction at standard conditions, if ΔG° is negative, what can be said about the equilibrium constant (K)?

A. K < 1
B. K = 1
C. K > 1
D. K is undefined

Solution

If ΔG° is negative, the equilibrium constant K is greater than 1.

Correct Answer: C — K > 1

Q. For a reaction at standard conditions, if ΔG° is positive, what can be said about the reaction?

A. The reaction is spontaneous in the forward direction.
B. The reaction is spontaneous in the reverse direction.
C. The reaction is at equilibrium.
D. The reaction is impossible.

Solution

A positive ΔG° indicates that the reaction is non-spontaneous in the forward direction, thus spontaneous in the reverse.

Correct Answer: B — The reaction is spontaneous in the reverse direction.

Q. For a reaction at standard conditions, if ΔG° is positive, what does it imply?

A. The reaction is spontaneous in the forward direction.
B. The reaction is at equilibrium.
C. The reaction is non-spontaneous in the forward direction.
D. The reaction will proceed rapidly.

Solution

A positive ΔG° indicates that the reaction is non-spontaneous in the forward direction under standard conditions.

Correct Answer: C — The reaction is non-spontaneous in the forward direction.

Q. For a reaction at standard conditions, if ΔG° is positive, what does it indicate?

A. The reaction is spontaneous in the forward direction.
B. The reaction is non-spontaneous in the forward direction.
C. The reaction is at equilibrium.
D. The reaction is spontaneous in the reverse direction.

Solution

A positive ΔG° indicates that the reaction is non-spontaneous in the forward direction.

Correct Answer: B — The reaction is non-spontaneous in the forward direction.

Q. For a reaction with a rate constant k, what is the relationship between the rate of reaction and the concentration of reactants for a first-order reaction?

A. Rate = k[A]^2
B. Rate = k[A]
C. Rate = k[A]^3
D. Rate = k[A]^0

Solution

For a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant, given by Rate = k[A].

Correct Answer: B — Rate = k[A]

Q. For a reaction with a rate constant of 0.02 M⁻¹s⁻¹ and initial concentration of 0.5 M, what is the time taken to reach 0.25 M in a second-order reaction? (2023)

A. 25 s
B. 50 s
C. 10 s
D. 20 s

Solution

Using t = 1 / (k[A₀]) * (1/[A] - 1/[A₀]), t = 1 / (0.02 * 0.5) * (1/0.25 - 1/0.5) = 25 s.

Correct Answer: A — 25 s

Q. For a reaction with an activation energy of 50 kJ/mol, what is the rate constant at 300 K if R = 8.314 J/(mol·K)? (2022)

A. 0.001 M/s
B. 0.01 M/s
C. 0.1 M/s
D. 1 M/s

Solution

Using the Arrhenius equation, k = Ae^(-Ea/RT). Calculate k using the given values.

Correct Answer: C — 0.1 M/s

Q. For a reaction with an activation energy of 50 kJ/mol, what will happen to the rate if the temperature is increased by 20°C? (2022)

A. Rate decreases
B. Rate remains the same
C. Rate increases significantly
D. Rate increases slightly

Solution

Increasing the temperature generally increases the rate of reaction significantly due to higher kinetic energy.

Correct Answer: C — Rate increases significantly

Q. For a reaction with ΔH = 100 kJ and ΔS = 200 J/K, at what temperature will the reaction become spontaneous?

A. 500 K
B. 250 K
C. 200 K
D. 100 K

Solution

To find the temperature at which the reaction becomes spontaneous, set ΔG = 0: 0 = ΔH - TΔS. Thus, T = ΔH/ΔS = (100,000 J)/(200 J/K) = 500 K.

Correct Answer: A — 500 K

Q. For a reaction with ΔH = 100 kJ/mol and ΔS = 200 J/mol·K, at what temperature will the reaction become spontaneous?

A. 500 K
B. 250 K
C. 200 K
D. 100 K

Solution

To find the temperature at which the reaction becomes spontaneous, set ΔG = 0: 0 = ΔH - TΔS, thus T = ΔH/ΔS = 100,000 J / 200 J/K = 500 K.

Correct Answer: A — 500 K

Q. For a reaction with ΔH = 50 kJ/mol and ΔS = 100 J/mol·K, at what temperature will the reaction become spontaneous?

A. 500 K
B. 250 K
C. 1000 K
D. 200 K

Solution

To find the temperature at which the reaction becomes spontaneous, set ΔG = 0: 0 = ΔH - TΔS, thus T = ΔH/ΔS = (50,000 J/mol) / (100 J/mol·K) = 500 K.

Correct Answer: A — 500 K

Q. For a rectangular plate of mass M and dimensions a x b, what is the moment of inertia about an axis through its center and parallel to side a?

A. 1/12 Mb^2
B. 1/3 Mb^2
C. 1/4 Mb^2
D. 1/6 Mb^2

Solution

The moment of inertia of a rectangular plate about an axis through its center parallel to side a is I = 1/12 Mb^2.

Correct Answer: A — 1/12 Mb^2

Q. For a reversible process, the change in entropy is given by which of the following?

A. ΔS = Q/T
B. ΔS = W/T
C. ΔS = Q + W
D. ΔS = 0

Solution

For a reversible process, the change in entropy is given by ΔS = Q/T, where Q is the heat exchanged and T is the temperature.

Correct Answer: A — ΔS = Q/T

Q. For a reversible process, the change in entropy of the system is equal to the heat absorbed divided by the temperature. This is expressed as:

A. ΔS = Q/T
B. ΔS = T/Q
C. ΔS = Q + T
D. ΔS = Q - T

Solution

For a reversible process, the change in entropy (ΔS) is given by ΔS = Q/T, where Q is the heat absorbed and T is the temperature.

Correct Answer: A — ΔS = Q/T

Q. For a reversible process, the change in entropy of the system is equal to the heat absorbed divided by the temperature. What is the formula?

A. ΔS = Q/T
B. ΔS = T/Q
C. ΔS = Q*T
D. ΔS = Q + T

Solution

The change in entropy (ΔS) for a reversible process is given by ΔS = Q/T, where Q is the heat absorbed and T is the temperature.

Correct Answer: A — ΔS = Q/T

Q. For a reversible process, the change in entropy of the universe is:

A. Zero
B. Positive
C. Negative
D. Undefined

Solution

For a reversible process, the change in entropy of the universe is zero, as the system and surroundings are in equilibrium.

Correct Answer: A — Zero

Q. For a reversible process, the efficiency of a Carnot engine is given by which formula?

A. 1 - (T2/T1)
B. T1/T2
C. T2/T1
D. 1 - (T1/T2)

Solution

The efficiency of a Carnot engine is given by η = 1 - (T2/T1), where T1 is the temperature of the hot reservoir and T2 is the temperature of the cold reservoir.

Correct Answer: A — 1 - (T2/T1)

Q. For a satellite in a circular orbit, which of the following is true about its kinetic and potential energy?

A. K.E. = P.E.
B. K.E. > P.E.
C. K.E. < P.E.
D. K.E. = 0

Solution

For a satellite in a circular orbit, the kinetic energy is less than the potential energy, as K.E. = -1/2 P.E.

Correct Answer: C — K.E. < P.E.

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