JEE Main Preparation Resources for Competitive Exams

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JEE Main MCQ & Objective Questions

The JEE Main exam is a crucial step for students aspiring to enter prestigious engineering colleges in India. It tests not only knowledge but also the ability to apply concepts effectively. Practicing MCQs and objective questions is essential for scoring better, as it helps in familiarizing students with the exam pattern and enhances their problem-solving skills. Engaging with practice questions allows students to identify important questions and strengthen their exam preparation.

What You Will Practise Here

Fundamental concepts of Physics, Chemistry, and Mathematics
Key formulas and their applications in problem-solving
Important definitions and theories relevant to JEE Main
Diagrams and graphical representations for better understanding
Numerical problems and their step-by-step solutions
Previous years' JEE Main questions for real exam experience
Time management strategies while solving MCQs

Exam Relevance

The topics covered in JEE Main are not only significant for the JEE exam but also appear in various CBSE and State Board examinations. Many concepts are shared with the NEET syllabus, making them relevant across multiple competitive exams. Common question patterns include conceptual applications, numerical problems, and theoretical questions that assess a student's understanding of core subjects.

Common Mistakes Students Make

Misinterpreting the question stem, leading to incorrect answers
Neglecting units in numerical problems, which can change the outcome
Overlooking negative marking and not managing time effectively
Relying too heavily on rote memorization instead of understanding concepts
Failing to review and analyze mistakes from practice tests

FAQs

Question: How can I improve my speed in solving JEE Main MCQ questions?
Answer: Regular practice with timed quizzes and focusing on shortcuts can significantly enhance your speed.

Question: Are the JEE Main objective questions similar to previous years' papers?
Answer: Yes, many questions are based on previous years' patterns, so practicing them can be beneficial.

Question: What is the best way to approach JEE Main practice questions?
Answer: Start with understanding the concepts, then attempt practice questions, and finally review your answers to learn from mistakes.

Now is the time to take charge of your preparation! Dive into solving JEE Main MCQs and practice questions to test your understanding and boost your confidence for the exam.

Chemistry Syllabus (JEE Main) Mathematics Syllabus (JEE Main) Physics Syllabus (JEE Main)

Q. What is the relationship between enthalpy and internal energy?

A. H = U + PV
B. H = U - PV
C. H = U * PV
D. H = U / PV

Solution

The relationship is given by the equation H = U + PV, where H is enthalpy, U is internal energy, P is pressure, and V is volume.

Correct Answer: A — H = U + PV

Q. What is the relationship between enthalpy change and internal energy change at constant pressure?

A. ΔH = ΔU + PΔV
B. ΔH = ΔU - PΔV
C. ΔH = ΔU
D. ΔH = PΔV

Solution

At constant pressure, the relationship is given by ΔH = ΔU + PΔV.

Correct Answer: A — ΔH = ΔU + PΔV

Q. What is the relationship between entropy and spontaneity of a process?

A. Higher entropy means the process is non-spontaneous.
B. Lower entropy means the process is spontaneous.
C. Higher entropy generally indicates a spontaneous process.
D. Entropy has no relation to spontaneity.

Solution

A higher entropy generally indicates a spontaneous process, as spontaneous processes tend to increase the overall disorder of the system.

Correct Answer: C — Higher entropy generally indicates a spontaneous process.

Q. What is the relationship between entropy and temperature?

A. Entropy increases with decreasing temperature
B. Entropy decreases with increasing temperature
C. Entropy increases with increasing temperature
D. Entropy is independent of temperature

Solution

Entropy generally increases with increasing temperature due to increased molecular motion and disorder.

Correct Answer: C — Entropy increases with increasing temperature

Q. What is the relationship between frequency and wavelength in a wave traveling at a constant speed?

A. Directly proportional
B. Inversely proportional
C. Independent
D. None of the above

Solution

Frequency and wavelength are inversely proportional when the speed of the wave is constant, as given by the equation v = fλ.

Correct Answer: B — Inversely proportional

Q. What is the relationship between frequency and wavelength in a wave?

A. Frequency is directly proportional to wavelength
B. Frequency is inversely proportional to wavelength
C. Frequency is independent of wavelength
D. Frequency equals wavelength

Solution

The relationship is given by the equation v = fλ, where v is the speed of the wave, f is the frequency, and λ is the wavelength. Thus, frequency is inversely proportional to wavelength.

Correct Answer: B — Frequency is inversely proportional to wavelength

Q. What is the relationship between Gibbs Free Energy and spontaneity?

A. ΔG < 0 indicates non-spontaneous reactions.
B. ΔG = 0 indicates spontaneous reactions.
C. ΔG > 0 indicates spontaneous reactions.
D. ΔG < 0 indicates spontaneous reactions.

Solution

A negative ΔG (< 0) indicates that a reaction is spontaneous under the given conditions.

Correct Answer: D — ΔG < 0 indicates spontaneous reactions.

Q. What is the relationship between Gibbs Free Energy and the equilibrium constant (K)?

A. ΔG = -RT ln(K)
B. ΔG = RT ln(K)
C. ΔG = KRT
D. ΔG = K - RT

Solution

The relationship is given by ΔG = -RT ln(K), where R is the gas constant and T is the temperature in Kelvin.

Correct Answer: A — ΔG = -RT ln(K)

Q. What is the relationship between gravitational field strength and gravitational potential?

A. Field strength is the gradient of potential.
B. Field strength is the integral of potential.
C. They are independent.
D. Field strength is the square of potential.

Solution

Gravitational field strength is the negative gradient of gravitational potential.

Correct Answer: A — Field strength is the gradient of potential.

Q. What is the relationship between gravitational potential and gravitational field strength?

A. V = -g * r
B. g = -dV/dr
C. V = g * r
D. g = dV/dr

Solution

The gravitational field strength g is the negative gradient of the gravitational potential V, given by g = -dV/dr.

Correct Answer: B — g = -dV/dr

Q. What is the relationship between heat capacity at constant pressure (Cp) and heat capacity at constant volume (Cv)?

A. Cp = Cv
B. Cp > Cv
C. Cp < Cv
D. Cp = Cv + R

Solution

For an ideal gas, Cp is always greater than Cv because it includes the work done against the atmospheric pressure.

Correct Answer: B — Cp > Cv

Q. What is the relationship between heat capacity at constant pressure (C_p) and at constant volume (C_v)?

A. C_p = C_v
B. C_p > C_v
C. C_p < C_v
D. C_p = 0

Solution

For an ideal gas, the heat capacity at constant pressure (C_p) is greater than the heat capacity at constant volume (C_v).

Correct Answer: B — C_p > C_v

Q. What is the relationship between heat capacity at constant pressure (C_p) and heat capacity at constant volume (C_v)?

A. C_p = C_v
B. C_p > C_v
C. C_p < C_v
D. C_p = 2C_v

Solution

For an ideal gas, C_p is always greater than C_v due to the work done during expansion.

Correct Answer: B — C_p > C_v

Q. What is the relationship between heat capacity at constant volume (Cv) and heat capacity at constant pressure (Cp) for an ideal gas?

A. Cp = Cv
B. Cp = Cv + R
C. Cp = Cv - R
D. Cp = 2Cv

Solution

For an ideal gas, the relationship is given by Cp = Cv + R, where R is the universal gas constant.

Correct Answer: B — Cp = Cv + R

Q. What is the relationship between heat capacity at constant volume (Cv) and heat capacity at constant pressure (Cp)?

A. Cp = Cv
B. Cp = Cv + R
C. Cp = Cv - R
D. Cp = 2Cv

Solution

The relationship is Cp = Cv + R for an ideal gas, where R is the gas constant.

Correct Answer: B — Cp = Cv + R

Q. What is the relationship between Joules and Newton-meters?

A. They are equal
B. Joule is greater
C. Newton-meter is greater
D. They are unrelated

Solution

1 Joule is defined as 1 Newton-meter, so they are equal.

Correct Answer: A — They are equal

Q. What is the relationship between Ka and Kb for a conjugate acid-base pair?

A. Ka + Kb = Kw
B. Ka * Kb = Kw
C. Ka - Kb = Kw
D. Ka / Kb = Kw

Solution

For a conjugate acid-base pair, the relationship is Ka * Kb = Kw, where Kw is the ion product of water.

Correct Answer: B — Ka * Kb = Kw

Q. What is the relationship between Kp and Kc for the reaction aA(g) + bB(g) ⇌ cC(g) + dD(g)?

A. Kp = Kc(RT)^(d+c-b-a)
B. Kp = Kc(RT)^(a+b-c-d)
C. Kp = Kc/(RT)^(d+c-b-a)
D. Kp = Kc/(RT)^(a+b-c-d)

Solution

The relationship between Kp and Kc is given by Kp = Kc(RT)^(Δn), where Δn = (d+c) - (a+b).

Correct Answer: A — Kp = Kc(RT)^(d+c-b-a)

Q. What is the relationship between linear velocity (v) and angular velocity (ω) for a point on a rotating object?

A. v = ωr
B. v = r/ω
C. v = ω/r
D. v = rω²

Solution

The relationship is given by v = ωr, where r is the radius of the rotation.

Correct Answer: A — v = ωr

Q. What is the relationship between mass and weight?

A. Weight = Mass × Acceleration due to gravity
B. Weight = Mass / Acceleration due to gravity
C. Weight = Mass + Acceleration due to gravity
D. Weight = Mass - Acceleration due to gravity

Solution

Weight is defined as the force due to gravity acting on a mass, given by the formula Weight = Mass × g (where g is the acceleration due to gravity).

Correct Answer: A — Weight = Mass × Acceleration due to gravity

Q. What is the relationship between pKa and Ka for a weak acid?

A. pKa = -log(Ka)
B. pKa = log(Ka)
C. pKa = Ka
D. pKa = 1/Ka

Solution

The relationship is given by the equation pKa = -log(Ka), where Ka is the acid dissociation constant.

Correct Answer: A — pKa = -log(Ka)

Q. What is the relationship between pKa and Ka for an acid?

A. pKa = -log(Ka)
B. pKa = log(Ka)
C. pKa = Ka
D. pKa = 1/Ka

Solution

The relationship is given by the formula pKa = -log(Ka), where Ka is the acid dissociation constant.

Correct Answer: A — pKa = -log(Ka)

Q. What is the relationship between pKa and Ka?

A. pKa = -log(Ka)
B. pKa = log(Ka)
C. pKa = Ka
D. pKa = 1/Ka

Solution

The relationship is given by pKa = -log(Ka), where Ka is the acid dissociation constant.

Correct Answer: A — pKa = -log(Ka)

Q. What is the relationship between power, force, and velocity?

A. P = F * v
B. P = F / v
C. P = F + v
D. P = F - v

Solution

Power is defined as the product of force and velocity in the direction of the force, given by P = F * v.

Correct Answer: A — P = F * v

Q. What is the relationship between power, voltage, and current in an electrical circuit?

A. P = V/I
B. P = VI
C. P = V + I
D. P = V - I

Solution

In an electrical circuit, power is given by the formula P = VI, where P is power, V is voltage, and I is current.

Correct Answer: B — P = VI

Q. What is the relationship between pressure and depth in a fluid?

A. Pressure increases with depth
B. Pressure decreases with depth
C. Pressure remains constant
D. Pressure is independent of depth

Solution

In a fluid at rest, pressure increases with depth due to the weight of the fluid above.

Correct Answer: A — Pressure increases with depth

Q. What is the relationship between pressure and temperature for a fixed amount of gas at constant volume?

A. Directly proportional
B. Inversely proportional
C. No relationship
D. Exponential

Solution

Pressure and temperature are directly proportional for a fixed amount of gas at constant volume, as described by Gay-Lussac's law.

Correct Answer: A — Directly proportional

Q. What is the relationship between pressure and temperature in Gay-Lussac's Law?

A. Directly proportional
B. Inversely proportional
C. No relationship
D. Exponential relationship

Solution

Gay-Lussac's Law states that pressure is directly proportional to temperature when volume is constant.

Correct Answer: A — Directly proportional

Q. What is the relationship between pressure and volume in Boyle's Law?

A. P ∝ V
B. PV = constant
C. P + V = constant
D. P/V = constant

Solution

Boyle's Law states that the pressure of a gas is inversely proportional to its volume at constant temperature, which can be expressed as PV = constant.

Correct Answer: B — PV = constant

Q. What is the relationship between pressure and volume of a gas at constant temperature?

A. Directly proportional
B. Inversely proportional
C. Independent
D. Exponential

Solution

According to Boyle's Law, pressure and volume of a gas are inversely proportional at constant temperature.

Correct Answer: B — Inversely proportional

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