JEE Physics Kinetic Theory MCQs Set 04

Physics Objective Questions and Answers: Kinetic Theory of gases

Explore reliable objective questions for Kinetic Theory of gases tailored for JEE learners. Utilizing these Physics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Download Kinetic Theory of gases MCQs with Answers

Access the complete set of multiple-choice questions for Kinetic Theory of gases below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: Pick the correct statements. (1) The rms translational speed of ideal gas molecules at the same temperature depends on the mass. (2) Since \( \tfrac12mv_{rms}^2=\tfrac32kT \), a single particle has a temperature. (3) If the temperature of an ideal gas is doubled from 100 °C to 200 °C, the average kinetic energy of each particle also doubles. (4) Both the pressure and volume of a monatomic ideal gas can change together without any change in its internal energy.
(a) Only (1) and (2)
(b) Only (2) and (3)
(c) Only (1) and (4)
(d) Only (3) and (4)
Show Answer & Explanation

Answer: (c) Only (1) and (4)

Explanation:
1. (1) true: \( v_{rms}=\sqrt{3kT/m} \). (2) false: temperature is a property of a collection of particles.
2. (3) false: 100 °C to 200 °C is 373 K to 473 K, not a doubling.
3. (4) true: an isothermal process changes P and V while U stays constant.

Teacher's Note:
1. Temperature in the kinetic energy formula must be in kelvin.
2. Internal energy of an ideal gas depends only on temperature.

Question 2: A vessel of volume 5 litre contains 1.4 g of nitrogen and 0.4 g of helium at 1500 K. If 30% of the nitrogen molecules dissociate into atoms and the temperature stays constant (R = 25/3 J/mol K), the gas pressure becomes
(a) \( 3.75\times10^5 \) N/m²
(b) \( 4.125\times10^5 \) N/m²
(c) \( 5.0\times10^5 \) N/m²
(d) \( 2.06\times10^5 \) N/m²
Show Answer & Explanation

Answer: (b) \( 4.125\times10^5 \) N/m²

Explanation:
1. Initial moles: \( N_2=0.05 \), He = 0.1. After dissociation: \( N_2=0.035 \), N atoms = 0.03.
2. Total moles = 0.035 + 0.03 + 0.1 = 0.165.
3. \( P=\dfrac{nRT}{V}=\dfrac{0.165(25/3)(1500)}{5\times10^{-3}}=4.125\times10^5 \) N/m².

Teacher's Note:
1. Each dissociating molecule produces two atoms, so the total number of particles rises.
2. Pressure is proportional to the total number of particles.

Question 3: The temperature at which the rms speed of oxygen molecules equals that of nitrogen molecules at 100 °C is nearly
(a) 426.3 K
(b) 456.3 K
(c) 436.3 K
(d) 446.3 K
Show Answer & Explanation

Answer: (a) 426.3 K

Explanation:
1. \( v_{rms}\propto\sqrt{T/M} \): \( \dfrac{T_{O}}{32}=\dfrac{373}{28} \).
2. \( T_O=373\times\dfrac{32}{28}=426.3 \) K.

Teacher's Note:
1. Heavier molecules need a higher temperature for the same rms speed.
2. Convert 100 °C to 373 K.

Question 4: 2 g of gas A is introduced into an evacuated flask at 25 °C and the pressure is 1 atm. When 3 g of another gas B is then added, the total pressure becomes 1.5 atm. The ratio of molecular weights \( M_A:M_B \) is 1 : n. The value of n is
(a) 2
(b) 1.5
(c) 4
(d) 3
Show Answer & Explanation

Answer: (d) 3

Explanation:
1. Partial pressure of B = 0.5 atm, so \( n_A:n_B=1:0.5=2:1 \) (same V and T).
2. \( \dfrac{2/M_A}{3/M_B}=2 \), so \( \dfrac{M_B}{M_A}=3 \), giving n = 3.

Teacher's Note:
1. Pressure is proportional to the number of moles at fixed V and T.
2. Moles = mass / molecular weight.

Question 5: The mean speed of hydrogen molecules in one sample equals the mean speed of helium molecules in another. The ratio of the temperature of the hydrogen sample to that of the helium sample is
(a) 2 : 1
(b) 1 : 1
(c) 1 : 2
(d) 1 : 4
Show Answer & Explanation

Answer: (c) 1 : 2

Explanation:
1. Mean speed \( \propto\sqrt{T/M} \), so equal speeds need \( T_H/M_H=T_{He}/M_{He} \).
2. \( T_H/T_{He}=M_H/M_{He}=2/4=1/2 \).

Teacher's Note:
1. The lighter gas must be colder for the same mean speed.
2. Use molar masses 2 g/mol for hydrogen and 4 g/mol for helium.

Question 6: For the same cylinder (diameter 4.00 cm, piston 13.0 kg, \( P_{atm}=10^5 \) N/m², g = 10 m/s²) started at 20 °C and height 4.00 cm, weights are added to keep the piston at its initial height while the bath is heated to 100 °C. The added mass is nearly
(a) 13 kg
(b) 5.0 kg
(c) 20 kg
(d) 7.0 kg
Show Answer & Explanation

Answer: (d) 7.0 kg

Explanation:
1. Initial pressure \( P_i=10^5+\dfrac{130}{\pi(0.02)^2}=2.03\times10^5 \) Pa.
2. At constant volume \( P_f=P_i\dfrac{373}{293} \), so \( \Delta P=P_i\dfrac{80}{293}=5.55\times10^4 \) Pa.
3. Added weight = \( \Delta P\cdot A=69.8 \) N, so the added mass ≈ 7.0 kg.

Teacher's Note:
1. With the volume fixed, extra weight raises the pressure in proportion to the temperature.
2. Added mass = ΔP × area / g.

Question 7: 16 g of oxygen at 37 °C is mixed with 14 g of nitrogen at 27 °C (no heat loss). The temperature of the mixture is
(a) 34 °C
(b) 32 °C
(c) 30 °C
(d) 37 °C
Show Answer & Explanation

Answer: (b) 32 °C

Explanation:
1. Both are 0.5 mol of diatomic gas with the same \( C_v \).
2. Energy conservation: \( 0.5(310)+0.5(300)=1.0\,T \), so T = 305 K = 32 °C.

Teacher's Note:
1. Equal moles of gases with equal \( C_v \) give the simple mean of the temperatures.
2. Convert to kelvin before averaging.

Question 8: The pressure of an ideal gas can be written as \( E=\dfrac{3PV}{2} \). Here E stands for
(a) the rotational kinetic energy
(b) the total kinetic energy
(c) none of these
(d) the average translational kinetic energy
Show Answer & Explanation

Answer: (d) the average translational kinetic energy

Explanation:
1. \( PV=\tfrac13Nm\overline{v^2} \), so \( \tfrac32PV=N\cdot\tfrac12m\overline{v^2} \).
2. This is the translational kinetic energy of the gas molecules.

Teacher's Note:
1. Rotation and vibration are not included in the pressure formula.
2. For a monatomic gas this is also the total internal energy.

Question 9: The speeds of three molecules are 3V, 4V and 5V. Their root mean square speed is
(a) 4V
(b) \( \sqrt{\dfrac{50}{3}}\,V \)
(c) \( \sqrt{\dfrac{25}{3}}\,V \)
(d) \( \dfrac{50}{3}V \)
Show Answer & Explanation

Answer: (b) \( \sqrt{\dfrac{50}{3}}\,V \)

Explanation:
1. \( v_{rms}=\sqrt{\dfrac{v_1^2+v_2^2+v_3^2}{3}} \).
2. \( =\sqrt{\dfrac{9V^2+16V^2+25V^2}{3}}=\sqrt{\dfrac{50}{3}}\,V\approx4.08V \).

Teacher's Note:
1. The rms speed is the square root of the mean of the squares, not the mean of the speeds.
2. The mean speed here would be exactly 4V, which differs from the rms value.

Question 10: Two closed identical conducting containers hold gases A and B. Experiment 1: the weights of the containers are 225 g (A) and 160 g (B), and an evacuated container weighs 100 g. Experiment 2: when the two containers are given the same amount of heat, the same temperature rise is recorded, and the pressure rises are 2.5 atm for A and 1.5 atm for B. Molar masses (g): monatomic He 4, Ne 20, Ar 40, Kr 84, Xe 131, Rn 222; diatomic H₂ 2, F₂ 38, N₂ 28, O₂ 32, Cl₂ 71. If the containers were at room temperature (300 K) initially, the initial internal energy of the gas in container A (use R = 2 cal/mol K) is
(a) 2812.5 cal
(b) 1406.25 cal
(c) 1000 cal
(d) 5625 cal
Show Answer & Explanation

Answer: (a) 2812.5 cal

Explanation:
1. n = 3.125 mol of monatomic gas: \( U=n\tfrac32RT \).
2. \( U=3.125\times1.5\times2\times300=2812.5 \) cal.

Teacher's Note:
1. For a monatomic gas \( C_v=\tfrac32R \).
2. R is 2 cal/mol K in this problem.

Question 11: A vessel of volume V is evacuated by a piston air pump. In one stroke the piston is pulled out so that the gas occupies a volume \( V+\Delta V \), and then the volume \( \Delta V \) is expelled, leaving V in the cylinder. The process is isothermal. The number of strokes needed to reduce the pressure to \( 1/\eta \) times the initial value is
(a) \( \dfrac{\ln\eta}{\ln(\Delta V/V)} \)
(b) \( \eta\,\dfrac{V}{\Delta V} \)
(c) \( \dfrac{\ln(1+\Delta V/V)}{\ln\eta} \)
(d) \( \dfrac{\ln\eta}{\ln(1+\Delta V/V)} \)
Show Answer & Explanation

Answer: (d) \( \dfrac{\ln\eta}{\ln(1+\Delta V/V)} \)

Explanation:
1. After one stroke \( P_1(V+\Delta V)=P_0V \), so \( P_1=P_0\dfrac{V}{V+\Delta V} \).
2. After n strokes \( P_n=P_0\left(\dfrac{V}{V+\Delta V}\right)^n=\dfrac{P_0}{\eta} \).
3. \( n=\dfrac{\ln\eta}{\ln(1+\Delta V/V)} \).

Teacher's Note:
1. Each stroke multiplies the pressure by the same factor.
2. Take logarithms to solve for n.

Question 12: A mixture of 2 moles of helium (atomic mass 4 u) and 1 mole of argon (atomic mass 40 u) is kept at 300 K in a container. The ratio \( v_{rms}(\text{helium})/v_{rms}(\text{argon}) \) is
(a) 0.32
(b) 0.45
(c) 3.16
(d) 2.24
Show Answer & Explanation

Answer: (c) 3.16

Explanation:
1. Both gases are at the same temperature, so \( v_{rms}\propto1/\sqrt M \).
2. Ratio \( =\sqrt{40/4}=\sqrt{10}=3.16 \).

Teacher's Note:
1. The number of moles of each gas does not matter.
2. The lighter atoms are faster.

Question 13: A mixture of argon and nitrogen is kept at room temperature, and a collision takes place between an argon molecule and a nitrogen molecule. Which of the following are possible? (1) The kinetic energies of both molecules decrease. (2) The kinetic energies of both molecules increase. (3) The kinetic energy of the argon molecule increases and that of the nitrogen molecule decreases. (4) The kinetic energy of the nitrogen molecule increases and that of the argon molecule decreases.
(a) Only (3) and (4)
(b) Only (1) and (2)
(c) Only (1) and (3)
(d) Only (2) and (4)
Show Answer & Explanation

Answer: (a) Only (3) and (4)

Explanation:
1. In a collision the total kinetic energy of the pair is conserved (elastic collision).
2. So one molecule can gain only what the other loses: the energies cannot both rise or both fall.

Teacher's Note:
1. Energy is transferred between the colliding molecules.
2. On average the two gases reach the same mean kinetic energy.

Question 14: When an ideal gas is compressed isothermally, its pressure increases because
(a) its potential energy decreases
(b) its kinetic energy increases and the molecules move apart
(c) the number of collisions per unit area with the walls increases
(d) the energy of each molecule increases
Show Answer & Explanation

Answer: (c) the number of collisions per unit area with the walls increases

Explanation:
1. At constant temperature the average kinetic energy and speed of a molecule do not change.
2. Fewer volume means more molecules per unit volume, so more collisions per unit area per second on the walls.

Teacher's Note:
1. Pressure is the rate of momentum transfer to the walls.
2. Ideal gas molecules have no potential energy.

Question 15: The air in a room communicates with the outside through the windows, so its pressure stays equal to the atmospheric pressure while the temperature changes between day and night. The internal energy of the air in the room
(a) increases as the temperature rises
(b) decreases as the temperature rises
(c) is proportional to the temperature
(d) remains constant
Show Answer & Explanation

Answer: (d) remains constant

Explanation:
1. \( U=\dfrac f2nRT=\dfrac f2PV \).
2. P is atmospheric and V is the room volume, both fixed, so U is constant (the number of molecules changes as air leaves).

Teacher's Note:
1. As the room warms, some air escapes, which exactly compensates the higher temperature.
2. Use PV in place of nRT.

Question 16: One mole of an ideal gas is kept in a rigid container of negligible heat capacity. When 25 J of heat is supplied, the gas temperature rises by 2 °C (R = 8.3 J/mol K). The gas may be (1) helium; (2) argon; (3) oxygen; (4) carbon dioxide.
(a) Only (1) and (3)
(b) Only (3) and (4)
(c) Only (2) and (4)
(d) Only (1) and (2)
Show Answer & Explanation

Answer: (d) Only (1) and (2)

Explanation:
1. \( C_v=\dfrac{Q}{n\Delta T}=\dfrac{25}{1\times2}=12.5 \) J/mol K ≈ \( \tfrac32R \).
2. \( \tfrac32R \) corresponds to a monatomic gas: helium or argon.

Teacher's Note:
1. At constant volume all the heat goes into internal energy.
2. Diatomic gases have \( C_v\approx\tfrac52R \).

Question 17: Two closed identical conducting containers hold gases A and B. Experiment 1: the weights of the containers are 225 g (A) and 160 g (B), and an evacuated container weighs 100 g. Experiment 2: when the two containers are given the same amount of heat, the same temperature rise is recorded, and the pressure rises are 2.5 atm for A and 1.5 atm for B. Molar masses (g): monatomic He 4, Ne 20, Ar 40, Kr 84, Xe 131, Rn 222; diatomic H₂ 2, F₂ 38, N₂ 28, O₂ 32, Cl₂ 71. The gases in containers A and B are respectively
(a) \( N_2 \), Ne
(b) He, \( H_2 \)
(c) Ar, \( O_2 \)
(d) \( O_2 \), Ar
Show Answer & Explanation

Answer: (c) Ar, \( O_2 \)

Explanation:
1. Masses of gas: A = 125 g, B = 60 g. Same heat and same \( \Delta T \): \( n_A\tfrac32R=n_B\tfrac52R \), so \( n_A/n_B=5/3 \).
2. \( \dfrac{125/M_A}{60/M_B}=\dfrac53 \), so \( M_B/M_A=0.8 \).
3. Monatomic A and diatomic B with ratio 0.8: Ar (40) and \( O_2 \) (32).

Teacher's Note:
1. Use the mole ratio from equal heat and equal temperature rise.
2. Check the pair against the list of molar masses.

Question 18: At 300 K the rms speed of the molecules of a certain diatomic gas is 1930 m/s (R = 25/3 J/mol K). The gas, and the temperature at which its rms speed is double this value, are
(a) hydrogen, 1200 K
(b) nitrogen, 600 K
(c) hydrogen, 600 K
(d) oxygen, 1200 K
Show Answer & Explanation

Answer: (a) hydrogen, 1200 K

Explanation:
1. \( M=\dfrac{3RT}{v^2}=\dfrac{3(25/3)(300)}{1930^2}=2.0\times10^{-3} \) kg/mol, which is hydrogen.
2. Since \( v_{rms}\propto\sqrt T \), doubling the speed needs four times the temperature: 4 × 300 = 1200 K.

Teacher's Note:
1. Molar mass of 2 g/mol identifies H₂.
2. Speed doubles when the temperature becomes four times.

Question 19: Three closed vessels A, B and C are at the same temperature T. A contains only \( O_2 \), B only \( N_2 \) and C a mixture of equal quantities of \( O_2 \) and \( N_2 \). If the average speed of \( O_2 \) in A is \( V_1 \) and of \( N_2 \) in B is \( V_2 \), the average speed of \( O_2 \) molecules in C is
(a) \( V_1 \)
(b) \( (V_1+V_2)/2 \)
(c) \( \sqrt{V_1V_2} \)
(d) \( V_1/2 \)
Show Answer & Explanation

Answer: (a) \( V_1 \)

Explanation:
1. The average speed of a given gas depends only on its temperature and molar mass.
2. The temperature of C is T, so oxygen molecules in C move with the same average speed \( V_1 \) as in A.

Teacher's Note:
1. Molecules of one gas in a mixture are unaffected by the presence of the other gas at the same temperature.
2. Maxwell distributions of different gases are independent.

Question 20: A gas is in a rigid container at pressure \( P_0 \). The mass of each molecule is halved, the number of molecules is unchanged and the rms speed is doubled. The new pressure is
(a) \( P_0/2 \)
(b) \( 4P_0 \)
(c) \( 8P_0 \)
(d) \( 2P_0 \)
Show Answer & Explanation

Answer: (d) \( 2P_0 \)

Explanation:
1. \( P=\dfrac{1}{3}\dfrac NV m\overline{v^2} \).
2. The factor \( m\,v_{rms}^2 \) changes by \( \tfrac12\times4=2 \), so \( P=2P_0 \).

Teacher's Note:
1. Pressure depends on the product of molecular mass and the mean square speed.
2. The volume and number of molecules are unchanged.

Question 21: Air at 3 atm is pumped into a cycle-rickshaw tube of volume 0.004 m³. The tube is punctured and its volume falls to 0.0008 m³, the air inside being then at atmospheric pressure (1 atm = \( 10^5 \) Pa). The temperature stays at 300 K and R = 25/3 J/mol K. The number of moles of air that have leaked out is
(a) 0.48
(b) 0.448
(c) 0.032
(d) 0.512
Show Answer & Explanation

Answer: (b) 0.448

Explanation:
1. Initially \( n_1=\dfrac{PV}{RT}=\dfrac{3\times10^5\times0.004}{(25/3)(300)}=0.48 \) mol.
2. Finally \( n_2=\dfrac{10^5\times0.0008}{2500}=0.032 \) mol.
3. Leaked = 0.48 − 0.032 = 0.448 mol.

Teacher's Note:
1. After the puncture the pressure inside equals atmospheric pressure.
2. Use \( n=PV/RT \) before and after.

Question 22: 0.040 g of helium in a closed container is initially at 100.0 °C. The container is heated until the internal energy has increased by 12 J (R = 25/3 J/mol K). The final temperature is
(a) 196 °C
(b) 112 °C
(c) 96 °C
(d) 292 °C
Show Answer & Explanation

Answer: (a) 196 °C

Explanation:
1. n = 0.040/4 = 0.01 mol.
2. \( \Delta U=n\tfrac32R\Delta T=0.01\times1.5\times\tfrac{25}{3}\Delta T=0.125\Delta T \), so \( \Delta T=96 \) K.
3. Final temperature = 100 + 96 = 196 °C.

Teacher's Note:
1. The container is closed (constant volume), so \( \Delta U=nC_v\Delta T \) with no work.
2. A temperature change in kelvin equals that in °C.

Question 23: The average speed of nitrogen molecules in a gas is v. If the temperature is doubled and the \( N_2 \) molecules dissociate into nitrogen atoms, the average speed becomes
(a) v
(b) \( \sqrt2\,v \)
(c) 4v
(d) 2v
Show Answer & Explanation

Answer: (d) 2v

Explanation:
1. Average speed \( \propto\sqrt{T/M} \).
2. Temperature doubles and the molecular mass halves, so the factor is \( \sqrt{2/(1/2)}=2 \).

Teacher's Note:
1. Dissociation halves the mass of each particle.
2. The two effects multiply inside the square root.

Question 24: Butane burns according to \( 2C_4H_{10}+13O_2\to10H_2O+8CO_2 \). Two moles of butane are mixed with 13 moles of oxygen in a fixed volume at pressure \( P_0 \) and completely react. If all species behave as ideal gases and the final temperature equals the initial one, the final pressure is
(a) \( \dfrac{5}{6}P_0 \)
(b) \( P_0 \)
(c) \( \dfrac{6}{5}P_0 \)
(d) \( \dfrac{18}{13}P_0 \)
Show Answer & Explanation

Answer: (c) \( \dfrac{6}{5}P_0 \)

Explanation:
1. Moles before = 2 + 13 = 15; moles after = 10 + 8 = 18.
2. At constant V and T, \( P\propto n \), so \( P=\dfrac{18}{15}P_0=\dfrac65P_0 \).

Teacher's Note:
1. Count the gas moles on each side of the equation.
2. Water is treated as a gas here because the temperature is high.

Question 25: Two closed identical conducting containers hold gases A and B. Experiment 1: the weights of the containers are 225 g (A) and 160 g (B), and an evacuated container weighs 100 g. Experiment 2: when the two containers are given the same amount of heat, the same temperature rise is recorded, and the pressure rises are 2.5 atm for A and 1.5 atm for B. Molar masses (g): monatomic He 4, Ne 20, Ar 40, Kr 84, Xe 131, Rn 222; diatomic H₂ 2, F₂ 38, N₂ 28, O₂ 32, Cl₂ 71. The types of gas in containers A and B respectively are
(a) monatomic, monatomic
(b) diatomic, diatomic
(c) diatomic, monatomic
(d) monatomic, diatomic
Show Answer & Explanation

Answer: (d) monatomic, diatomic

Explanation:
1. Same volume, same heat and same temperature rise: \( \Delta P=\dfrac{nR\Delta T}{V} \) and \( Q=nC_v\Delta T \), so \( \Delta P\propto\dfrac{1}{C_v} \).
2. \( \dfrac{\Delta P_A}{\Delta P_B}=\dfrac{2.5}{1.5}=\dfrac{C_{vB}}{C_{vA}}=\dfrac{5}{3} \), which is \( \dfrac{5R/2}{3R/2} \).
3. So A is monatomic and B is diatomic.

Teacher's Note:
1. A smaller \( C_v \) gives a larger pressure rise for the same heat.
2. The ratio 5/3 matches the degrees of freedom of diatomic and monatomic gases.

Question 26: Which quantity is the same for all ideal gases at the same temperature?
(a) The kinetic energy of equal mass of gas
(b) The number of molecules in equal mass of gas
(c) The number of molecules in equal moles of gas
(d) The total kinetic energy of equal moles of gas
Show Answer & Explanation

Answer: (c) The number of molecules in equal moles of gas

Explanation:
1. Equal moles contain equal numbers of molecules (Avogadro's number per mole).
2. Total kinetic energy depends on the degrees of freedom, and equal masses contain different numbers of molecules, so the others are not universal.

Teacher's Note:
1. Avogadro's number is the same for all gases.
2. Only the translational energy per molecule is the same at a given temperature.

Free study material for Physics

Download Chapter MCQs: JEE Physics Kinetic Theory of gases

JEE Physics Kinetic Theory of gases Objective Test Questions

Test your conceptual understanding of Kinetic Theory of gases with these targeted multiple-choice questions. Designed in alignment with the latest JEE curriculum for JEE Physics, these problem sets build accuracy and prepare students for objective exams.

NCERT-Aligned Objective Questions and Solutions

Built using the official NCERT book for JEE, these Physics objective sets provide reliable academic guidance. Pair your practice with our recommended NCERT solutions to master optimal problem-solving approaches.

Next Steps in Your Exam Preparation

Wrap up your chapter revision by testing your knowledge against standard question formats. Everything on our platform is provided free of charge.

FAQs

Where can I access latest JEE Physics Kinetic Theory MCQs Set 04?

You can get most exhaustive JEE Physics Kinetic Theory MCQs Set 04 for free on StudiesToday.com. These MCQs for JEE Physics are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Physics JEE material?

Yes, our JEE Physics Kinetic Theory MCQs Set 04 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Physics MCQs help in scoring full marks in JEE exams?

By solving our JEE Physics Kinetic Theory MCQs Set 04, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Physics.

Do you provide answers and explanations for JEE Physics Kinetic Theory MCQs Set 04?

Yes, Physics MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Physics JEE MCQs online?

Yes, you can also access online interactive tests for JEE Physics Kinetic Theory MCQs Set 04 on StudiesToday.com as they provide instant answers and score to help you track your progress in Physics.