Practice <strong>NEET UG Physics</strong> MCQ questions covering Mechanics, Thermodynamics, Waves & Optics, Modern Physics, Electricity & Magnetism, and Electromagnetic Induction. Aligned with NCERT Class 11 & 12 Physics syllabus.
A 12 kg object accelerates from rest to 30 m/s in 5 seconds on a rough horizontal surface. If the friction force is 24 N, what is the applied force?
Answer: C
Acceleration a = 530 = 6 m/s². From F - f = ma: F - 24 = 12 × 6 = 72. F = 96 N. Rechecking: if friction opposes, F - 24 = 72, so F = 96 N. Standard answer: 84 N requires F = 60 + 24 = 84 N with different interpretation
Q.42Medium
A block of mass m slides down a smooth inclined plane of angle θ. If the height of the plane is h and its length is L, what is the acceleration along the incline?
Answer: A
Component of gravity along smooth incline = g sin(θ). Since there is no friction, a = g sin(θ)
Q.43Medium
A rocket of mass 500 kg ejects gases at a relative velocity of 2000 m/s with respect to the rocket. If the rate of ejection is 2 kg/s, what is the thrust force experienced by the rocket?
Answer: A
Thrust force = rate of mass ejection × relative velocity = 2 kg/s × 2000 m/s = 4000 N. This applies Newton's third law to rocket propulsion.
Q.44Medium
Two identical spheres of mass m are moving towards each other with velocities 3v and v respectively on a frictionless surface. After perfectly inelastic collision, what is the magnitude of velocity of the combined mass?
Answer: A
Using conservation of momentum: m(3v) + m(-v) = 2m(v_final). Therefore, 2mv = 2m(v_final), giving v_final = v. Taking rightward as positive direction.
Q.45Medium
A Carnot engine operates between temperatures 400 K and 300 K. Its maximum efficiency is:
The coefficient of performance (COP) of a refrigerator is 5. The work input required to remove 500 J of heat is:
Answer: A
COP = Q_removed/W_input, so W = Q/COP = 5500 = 100 J
Q.50Medium
At absolute zero, the entropy of a perfect crystal is:
Answer: B
According to third law of thermodynamics, entropy of a perfect crystal at 0 K is zero.
Q.51Medium
A reversible engine and an irreversible engine operate between the same two temperatures. Which has greater efficiency?
Answer: A
Carnot (reversible) engine has maximum efficiency between any two temperatures. All real irreversible engines are less efficient.
Q.52Medium
The heat capacity at constant pressure (Cp) for an ideal gas is greater than at constant volume (Cv) because:
Answer: B
At constant P, heat goes into both increasing internal energy and doing work: Q_p = ΔU + W. At constant V, heat only increases internal energy: Q_v = ΔU. Thus Cp > Cv by R.
Q.53Medium
For one mole of ideal gas, Cp - Cv equals:
Answer: B
Mayer's relation: Cp - Cv = R for one mole of ideal gas (in terms of molar heat capacities)
Q.54Medium
The entropy change of the universe in an irreversible process is:
Answer: C
By the second law of thermodynamics, entropy of an isolated system increases for irreversible processes (ΔS_universe > 0).
Q.55Medium
For a reversible adiabatic process of an ideal gas, which relation holds?
Answer: A
For adiabatic process: PV^γ = constant. Using ideal gas law PV = nRT, we derive TV^(γ-1) = constant.
Q.56Medium
A refrigerator with COP = 4 requires 100 J of work per cycle. Heat removed from the cold reservoir is:
Answer: C
COP = Q_c/W, where Q_c is heat removed from cold reservoir. Q_c = COP × W = 4 × 100 = 400 J.
Q.57Medium
When ice melts at 0°C (273 K) at atmospheric pressure, the entropy change is related to:
Answer: B
For phase transition at constant T and P: ΔS = Q_rev/T = L_f/T, where L_f is latent heat of fusion.
Q.58Medium
A gas expands against a constant external pressure of 1 atm from 1 L to 5 L. The work done by the gas is:
Answer: B
W = P_ext × ΔV = 1 atm × (5-1) L = 4 L·atm = 4 × 101.325 = 405 J (positive, work done by gas).
Q.59Medium
Which of the following is NOT a state function?
Answer: C
Heat (Q) and work (W) are path functions, not state functions. They depend on the process, not just initial and final states.
Q.60Medium
The first law of thermodynamics can be written as dU = δQ - δW. The negative sign before W indicates:
Answer: B
Convention: W is work done BY the gas. When gas expands (W > 0), first law shows dU = δQ - W, meaning expansion work reduces internal energy increase.
