These essential MCQ questions are selected from the most important topics in CBSE Class 11 Physics 2026 as per the latest NCERT syllabus. Sections include Laws of Motion, Work Energy & Power, Rotational Motion, Gravitation, Properties of Matter, Thermodynamics, Kinetic Theory of Gases, and Waves & Oscillations. These topics also form the direct foundation for JEE Main and NEET. For truly unlimited daily MCQ practice, visit Vooo AI Education.
⚡ CBSE Class 11 Physics
1A body is said to be in equilibrium when the net force acting on it is:
Answer: C — Zero
A body is in translational equilibrium when the vector sum of all forces acting on it is zero (ΣF = 0). For rotational equilibrium, the net torque must also be zero (Στ = 0). Static equilibrium: body at rest. Dynamic equilibrium: body moving at constant velocity. Newton's First Law: a body continues in its state of rest or uniform motion unless acted on by a net external force. When ΣF = 0, acceleration = 0 (from F = ma), so velocity remains constant.
A body is in translational equilibrium when the vector sum of all forces acting on it is zero (ΣF = 0). For rotational equilibrium, the net torque must also be zero (Στ = 0). Static equilibrium: body at rest. Dynamic equilibrium: body moving at constant velocity. Newton's First Law: a body continues in its state of rest or uniform motion unless acted on by a net external force. When ΣF = 0, acceleration = 0 (from F = ma), so velocity remains constant.
2The work done by a force F⃗ over displacement d⃗ when they are perpendicular to each other is:
Answer: C — Zero
Work W = F⃗·d⃗ = Fd cosθ. When force and displacement are perpendicular, θ = 90°, cos90° = 0, so W = 0. Examples: a person carrying a bag walking horizontally (gravity is perpendicular to motion); a satellite in circular orbit (gravity is perpendicular to velocity). Work is positive when force and displacement are in the same direction (θ = 0°), negative when opposite (θ = 180°). Work is a scalar quantity measured in Joules.
Work W = F⃗·d⃗ = Fd cosθ. When force and displacement are perpendicular, θ = 90°, cos90° = 0, so W = 0. Examples: a person carrying a bag walking horizontally (gravity is perpendicular to motion); a satellite in circular orbit (gravity is perpendicular to velocity). Work is positive when force and displacement are in the same direction (θ = 0°), negative when opposite (θ = 180°). Work is a scalar quantity measured in Joules.
3The escape velocity from Earth's surface (radius R, mass M) is given by:
Answer: B — √(2GM/R)
Escape velocity is the minimum velocity needed to escape Earth's gravitational field: vₑ = √(2GM/R) = √(2gR) ≈ 11.2 km/s for Earth. Derivation: set kinetic energy equal to gravitational potential energy: (1/2)mv² = GMm/R → v = √(2GM/R). Orbital velocity (first cosmic velocity): v₀ = √(GM/R) ≈ 7.9 km/s. Note: vₑ = √2 × v₀. Escape velocity does not depend on the mass of the escaping object — both a feather and a rocket need the same escape velocity.
Escape velocity is the minimum velocity needed to escape Earth's gravitational field: vₑ = √(2GM/R) = √(2gR) ≈ 11.2 km/s for Earth. Derivation: set kinetic energy equal to gravitational potential energy: (1/2)mv² = GMm/R → v = √(2GM/R). Orbital velocity (first cosmic velocity): v₀ = √(GM/R) ≈ 7.9 km/s. Note: vₑ = √2 × v₀. Escape velocity does not depend on the mass of the escaping object — both a feather and a rocket need the same escape velocity.
4In simple harmonic motion, the restoring force is:
Answer: B — Proportional to displacement and directed opposite to it
In SHM: F = -kx, where x is displacement from equilibrium and k is force constant. The negative sign shows the restoring force always opposes the displacement. Acceleration: a = -ω²x. The motion is periodic with time period T = 2π√(m/k) for a spring-mass system and T = 2π√(L/g) for a simple pendulum. Maximum velocity occurs at equilibrium (x=0); maximum acceleration occurs at extreme positions (x = ±A). Energy in SHM oscillates between kinetic and potential.
In SHM: F = -kx, where x is displacement from equilibrium and k is force constant. The negative sign shows the restoring force always opposes the displacement. Acceleration: a = -ω²x. The motion is periodic with time period T = 2π√(m/k) for a spring-mass system and T = 2π√(L/g) for a simple pendulum. Maximum velocity occurs at equilibrium (x=0); maximum acceleration occurs at extreme positions (x = ±A). Energy in SHM oscillates between kinetic and potential.
5According to the first law of thermodynamics, for an adiabatic process:
Answer: C — ΔU = -W
First Law of Thermodynamics: ΔU = Q - W (change in internal energy = heat absorbed - work done by system). In an adiabatic process, no heat exchange occurs: Q = 0. Therefore ΔU = -W. If the gas does positive work (expansion), internal energy decreases and temperature falls. Isothermal: ΔT = 0, so ΔU = 0 and Q = W. Isochoric (constant volume): W = 0, so ΔU = Q. Isobaric (constant pressure): W = PΔV. These four processes are fundamental in Class 11 thermodynamics.
First Law of Thermodynamics: ΔU = Q - W (change in internal energy = heat absorbed - work done by system). In an adiabatic process, no heat exchange occurs: Q = 0. Therefore ΔU = -W. If the gas does positive work (expansion), internal energy decreases and temperature falls. Isothermal: ΔT = 0, so ΔU = 0 and Q = W. Isochoric (constant volume): W = 0, so ΔU = Q. Isobaric (constant pressure): W = PΔV. These four processes are fundamental in Class 11 thermodynamics.
6The moment of inertia of a thin uniform rod of mass M and length L about an axis through its centre and perpendicular to its length is:
Answer: B — ML²/12
Moment of inertia of a thin rod about its centre = ML²/12. About one end = ML²/3 (using parallel axis theorem: ML²/12 + M(L/2)² = ML²/12 + ML²/4 = ML²/3). Moment of inertia is the rotational analogue of mass: τ = Iα (analogous to F = ma). Other standard values: solid sphere = 2MR²/5; hollow sphere = 2MR²/3; solid cylinder = MR²/2; ring = MR². The parallel axis theorem: I = Icm + Md².
Moment of inertia of a thin rod about its centre = ML²/12. About one end = ML²/3 (using parallel axis theorem: ML²/12 + M(L/2)² = ML²/12 + ML²/4 = ML²/3). Moment of inertia is the rotational analogue of mass: τ = Iα (analogous to F = ma). Other standard values: solid sphere = 2MR²/5; hollow sphere = 2MR²/3; solid cylinder = MR²/2; ring = MR². The parallel axis theorem: I = Icm + Md².
7At which condition does a real gas behave most like an ideal gas?
Answer: C — Low pressure and high temperature
At low pressure, gas molecules are far apart (intermolecular forces negligible). At high temperature, molecules have high kinetic energy (molecular volume negligible compared to container volume). Both conditions bring real gas behaviour close to ideal gas (PV = nRT). At high pressure, molecules are close — intermolecular forces matter. At low temperature, molecules slow down — they can liquefy. The van der Waals equation corrects for intermolecular forces (a) and molecular volume (b): (P + a/V²)(V - b) = nRT.
At low pressure, gas molecules are far apart (intermolecular forces negligible). At high temperature, molecules have high kinetic energy (molecular volume negligible compared to container volume). Both conditions bring real gas behaviour close to ideal gas (PV = nRT). At high pressure, molecules are close — intermolecular forces matter. At low temperature, molecules slow down — they can liquefy. The van der Waals equation corrects for intermolecular forces (a) and molecular volume (b): (P + a/V²)(V - b) = nRT.
8A wave travelling in a medium has frequency f and wavelength λ. Its speed v is:
Answer: C — v = fλ
The wave equation: v = fλ, where v is wave speed (m/s), f is frequency (Hz), and λ is wavelength (m). Also: f = 1/T where T is time period. So v = λ/T. Angular frequency ω = 2πf; wave number k = 2π/λ; v = ω/k. Speed of sound in air ≈ 340 m/s at room temperature. Speed of light in vacuum = 3 × 10⁸ m/s. The speed of a wave in a medium depends on the properties of the medium, not on frequency or amplitude. Higher frequency = shorter wavelength at the same speed.
The wave equation: v = fλ, where v is wave speed (m/s), f is frequency (Hz), and λ is wavelength (m). Also: f = 1/T where T is time period. So v = λ/T. Angular frequency ω = 2πf; wave number k = 2π/λ; v = ω/k. Speed of sound in air ≈ 340 m/s at room temperature. Speed of light in vacuum = 3 × 10⁸ m/s. The speed of a wave in a medium depends on the properties of the medium, not on frequency or amplitude. Higher frequency = shorter wavelength at the same speed.
9The dimensional formula of pressure is:
Answer: B — [ML⁻¹T⁻²]
Pressure = Force/Area = [MLT⁻²]/[L²] = [ML⁻¹T⁻²]. Dimensional analysis is a powerful tool: it can check equation correctness, derive relations between quantities, and convert units. Common dimensional formulas: Force = [MLT⁻²]; Energy = [ML²T⁻²]; Power = [ML²T⁻³]; Momentum = [MLT⁻¹]; Angular momentum = [ML²T⁻¹]. Dimensional analysis cannot determine dimensionless constants (like 1/2 in KE = 1/2mv²) or distinguish between scalars and vectors of the same dimensions.
Pressure = Force/Area = [MLT⁻²]/[L²] = [ML⁻¹T⁻²]. Dimensional analysis is a powerful tool: it can check equation correctness, derive relations between quantities, and convert units. Common dimensional formulas: Force = [MLT⁻²]; Energy = [ML²T⁻²]; Power = [ML²T⁻³]; Momentum = [MLT⁻¹]; Angular momentum = [ML²T⁻¹]. Dimensional analysis cannot determine dimensionless constants (like 1/2 in KE = 1/2mv²) or distinguish between scalars and vectors of the same dimensions.
10Kepler's second law (law of equal areas) is a consequence of conservation of:
Answer: C — Angular momentum
Kepler's second law: a planet sweeps equal areas in equal times as it orbits the Sun. This is a direct consequence of conservation of angular momentum (L = mvr = constant). Since no torque acts (gravitational force is central — directed towards the Sun), angular momentum is conserved. When a planet is closer to the Sun (perihelion), it moves faster; farther away (aphelion), it moves slower. Kepler's three laws: (1) elliptical orbits, (2) equal areas, (3) T² ∝ a³ (a = semi-major axis).
Kepler's second law: a planet sweeps equal areas in equal times as it orbits the Sun. This is a direct consequence of conservation of angular momentum (L = mvr = constant). Since no torque acts (gravitational force is central — directed towards the Sun), angular momentum is conserved. When a planet is closer to the Sun (perihelion), it moves faster; farther away (aphelion), it moves slower. Kepler's three laws: (1) elliptical orbits, (2) equal areas, (3) T² ∝ a³ (a = semi-major axis).
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