This is a complete Class 11 Physics paper for NEET aspirants. The topics covered include the complete syllabus of Class XI Physics, making this paper highly useful for students who want to strengthen their full Class 11 base before entering the final NEET race. A strong Class 11 foundation is essential because many NEET questions directly or indirectly depend on conceptual clarity from these chapters.
This assessment has been prepared in a systematic and exam-oriented way by a Physics Tutor in Indiranagar – Bengaluru. The questions are selected to improve concept application, calculation ability, speed, confidence and exam discipline.
First revise the complete Class 11 formula bank, then solve the paper honestly under timed conditions, and only after that check the official solutions.
Important Formula Revision for NEET Physics: Complete Class 11 Physics
Before starting this paper, revise the important formulas of complete Class 11 Physics. NEET Physics often tests whether a student can select the correct formula, understand the condition in which it applies, and use it correctly in numerical and conceptual questions. This formula bank is added to help students revise quickly before attempting the paper.
Physical World
Physics laws use models, approximations and symmetry.
Least count = one main-scale division - one vernier-scale division.
Used in dimensional and error questions.
Motion in a Straight Line
v = u + at
s = ut + (1/2)at2
v2 = u2 + 2as
Average velocity = total displacement / total time.
Used in numerical questions.
Motion in a Plane
R = u2sin2θ/g
H = u2sin2θ/2g
T = 2u sinθ/g
Relative velocity: vAB = vA - vB.
Used in projectile questions.
Laws of Motion
F = ma
fmax = μsN, fk = μkN
On incline: mg sinθ, mg cosθ
Centripetal force = mv2/r.
Used in friction and circular motion.
Work, Energy and Power
W = F s cosθ
K = (1/2)mv2
Wnet = ΔK
P = dW/dt = Fv.
Used in power and energy questions.
System of Particles and Rotation
xcm = Σmixi/Σmi
τ = rF sinθ
L = Iω
Krot = (1/2)Iω2
Rolling: v = Rω
Used in rotational motion numericals.
Moment of Inertia
Ring about centre: MR2
Disc/cylinder: (1/2)MR2
Solid sphere: (2/5)MR2
Parallel axis: I = Icm + Md2.
Used in rotation and rolling.
Gravitation
F = Gm1m2/r2
g = GM/R2
vorb = √(GM/r)
vesc = √(2GM/R)
Used in orbit and field questions.
Mechanical Properties of Solids
Stress = F/A
Strain = ΔL/L
Y = stress/strain
Elastic energy = (1/2)FΔL = stress2 volume / 2Y.
Used in elasticity questions.
Mechanical Properties of Fluids
P = P0 + ρgh
Fb = ρVg
A1v1 = A2v2
P + (1/2)ρv2 + ρgh = constant.
Used in Bernoulli and buoyancy.
Surface Tension and Viscosity
Excess pressure: 2T/R (drop), 4T/R (soap bubble)
F = ηA dv/dy
vterminal = 2r2(ρ - σ)g/9η
Used in fluid layer questions.
Thermal Properties
Q = mcΔT
Q = mL
ΔL = αLΔT
ΔA = 2αAΔT
ΔV = γVΔT.
Used in heat and expansion.
Heat Transfer
dQ/dt = kAΔT/L
H = εσAT4
Newton cooling: dT/dt ∝ -(T - Ts).
Used in conduction and cooling.
Thermodynamics
ΔQ = ΔU + W
Wisothermal = nRT ln(V2/V1)
PVγ = constant
Wadiabatic = (P1V1 - P2V2)/(γ - 1).
Used in thermodynamics process questions.
Kinetic Theory
PV = nRT
vrms = √(3RT/M)
vavg = √(8RT/πM)
vmp = √(2RT/M)
Kavg = (3/2)kT.
Used in gas speed questions.
Oscillations
x = A sin(ωt + φ)
v = ω√(A2 - x2)
a = -ω2x
T = 2π√(m/k)
Tpendulum = 2π√(l/g).
Used in SHM questions.
Waves
v = fλ
vstring = √(T/μ)
Beat frequency = |f1 - f2|
Doppler: f' = f(v ± vo)/(v ∓ vs).
Used in SHM and wave questions.
Strengthen Class 11 Physics for NEET
Class 11 Physics is the backbone of NEET Physics. If the student becomes strong in motion, force, work, energy, rotation, gravitation, properties of matter, heat, thermodynamics, kinetic theory, oscillations and waves, then Class 12 preparation also becomes stronger. Treat this paper like a serious self-test. Attempt every question carefully and do not open the solution immediately. Think first, calculate properly, and then compare your answer with the official solution. Every mistake should become a learning opportunity.
If a student living in Indiranagar – Bengaluru is searching for a Physics Tutor for NEET, CBSE, IIT-JEE, IB, ICSE, IGCSE or AP Physics, they may contact Kumar Sir for one-to-one guidance.
Why Strong Physics Preparation Is Now More Important Than Ever
NEET Physics is becoming more conceptual and competitive. Students must build conceptual clarity, calculation accuracy, speed and the ability to solve unfamiliar problems. Memorising formulas is not enough; students must understand when, where and how to apply them. A strong teacher can help students connect formulas with physical meaning and avoid common mistakes in pressure, rotation, waves, heat and mechanics.
Important Message for NEET 2027, 2028, 2029, 2030 and Future Aspirants
Future NEET aspirants must prepare seriously and strengthen their Class 11 Physics foundation early. Question patterns may continue to become more application-based and concept-driven. Students should practise papers under timed conditions, revise formulas regularly, analyse mistakes honestly and strengthen weak chapters before they become long-term gaps.
Why Study Physics with Kumar Sir?
Kumar Sir provides personalised one-to-one online Physics classes. He explains each concept step by step, clears doubts properly and teaches difficult topics in a simple exam-oriented manner. His teaching is highly useful for NEET, CBSE, IIT-JEE, IB, ICSE, IGCSE, AP Physics and other serious academic or competitive exams. He focuses on conceptual clarity, numerical practice, revision and confidence building. If students are struggling with complete Class 11 Physics or want stronger preparation, Kumar Sir can guide them personally with disciplined study planning and focused doubt clearing.
The 9th division of vernier scale coincides with 6th division of main scale in a vernier calliper. If one main scale division is 0.01 mm, then the least count of instrument is
Official PDF solution: 9 VSD = 6 MSD, so 1 VSD = 2/3 MSD. With 1 MSD = 0.01 mm, least count = 1 MSD - 1 VSD = 0.01 - (2/3)(0.01) = 0.0033 mm.
Question 2+4 / -1
The position of a particle moving along x-axis varies with time t as 4x + t2 = 4. At what position is the particle at rest?
Official PDF solution: Differentiate 4x + t2 = 4 with respect to time: 4v + 2t = 0. At rest v = 0, hence t = 0. Substituting in the original equation gives x = 1.
Question 3+4 / -1
In covering a certain distance, the time decreases by 20%, if speed (uniform) increases by 20 m/s. Find the original speed.
Official PDF solution: Let original speed be u and time be t. Distance s = ut. New time is 0.8t and new speed is u + 20, so (u + 20)(0.8t) = ut. Therefore 4u + 80 = 5u and u = 80 m/s.
Question 4+4 / -1
Figure shows a quarter circle. A particle moves from B to O with 5 m/s, from O to A with 7 m/s. With what speed should the particle move from A to B so that the average speed for the total journey is 6 m/s?
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Official PDF solution: Use average speed = total distance / total time. Total distance is R + R + πR/2 and total time is R/5 + R/7 + (πR/2)/v. Setting the average speed equal to 6 and solving gives v = 210/(35 - 4/π) m/s.
Question 5+4 / -1
The relation between acceleration and velocity for a body moving in straight line is a ∝ v3. Choose the incorrect option. (x is position and t is time)
Official PDF solution: Using a = dv/dt ∝ v3 gives 1/v2 ∝ t. Also using a = v dv/dx gives the position relation used in the official solution. The option x ∝ 1/v2 is the incorrect relation.
Question 6+4 / -1
A man running at speed of 5 km/h finds the rain hitting his head vertically, but he has to hold the umbrella at 30° with vertical while at rest. The speed of rain is
Official PDF solution: From the velocity triangle, VR cos 60° = 5 km/h. Hence VR = 10 km/h.
Question 7+4 / -1
A particle moves in x-y plane according to the law v = 3y2 î + 2x ĵ. The equation of trajectory is
Official PDF solution: Here dx/dt = 3y2 and dy/dt = 2x. Thus dx/dy = 3y2/(2x), so 2x dx = 3y2 dy. Integrating gives x2 - y3 = constant.
Question 8+4 / -1
A block has been placed on an inclined plane. The slope angle θ of the plane is such that the block slides down the plane at a constant speed. The coefficient of kinetic friction is equal to
Official PDF solution: For constant speed, net force along the plane is zero: mg sinθ = μ mg cosθ. Hence μ = tanθ.
Question 9+4 / -1
A block of mass m is constrained to move inside a ring of radius R fixed on a horizontal smooth table. At time t = 0, the block moves along the inside of the ring with velocity v0. The coefficient of friction between the block and the ring is μ. The speed of the block at time t is
Official PDF solution: Normal reaction N = mv2/R, so the limiting friction gives retardation dv/dt = -μv2/R. Integrating from v0 to v gives v = v0/(1 + μv0t/R).
Question 10+4 / -1
A block of mass 2 kg is pulled along a horizontal surface by applying a force at an angle 30° with the horizontal. If the block travels with a uniform velocity and has a displacement 115 cm and the coefficient of friction is 0.15, then the work done by the applied force is
Official PDF solution: For uniform motion, F cosθ = μ(mg - F sinθ). The work done is W = Fd cosθ = μmgd cosθ/(cosθ + μsinθ). Substituting m = 2 kg, d = 1.15 m, θ = 30°, μ = 0.15 gives W = 3.2 J.
Question 11+4 / -1
A body of radius R and radius of gyration K (about axis of rotation) is rolling without slipping. The fraction of its kinetic energy in the form of rotational kinetic energy is
Official PDF solution: For rolling, rotational kinetic energy is (1/2)Iω2 = (1/2)mK2v2/R2. Dividing by total kinetic energy gives K2/(R2 + K2).
Question 12+4 / -1
A motor of power P1 is used to deliver water at a certain speed through a given horizontal pipe. To increase the speed of water through the same pipe three times, the power of the motor is increased to P2. The ratio of P2 to P1 is
Official PDF solution: For the same pipe, power varies as v3. Tripling the speed needs 33 = 27 times the power, so P2 : P1 = 27 : 1.
Question 13+4 / -1
If linear density of a rod of length 4 m varies as λ = 4 + x, then the position of the centre of gravity of the rod is
Official PDF solution: xcm = ∫x dm / ∫dm = ∫04x(4 + x)dx / ∫04(4 + x)dx = 20/9 m.
Question 14+4 / -1
A cannon ball is fired with a velocity of 100 m/s at an angle of 30° with the horizontal. At the highest point it explodes into three equal fragments, one going vertically upwards with velocity 50 m/s and the second falling vertically downwards with velocity 50 m/s. The third fragment will be moving with a velocity of
Official PDF solution: At the highest point, horizontal velocity of the original ball is 100 cos30° = 50√3 m/s. The two vertical fragments have equal and opposite vertical momenta, so the third fragment carries the horizontal momentum: v = 150√3 = about 260 m/s horizontally.
Question 15+4 / -1
A and B are two loops made from the same wire. The radii of A and B are r1 and r2, and their moments of inertia about same axes are I1 and I2. If I1/I2 = 1/4, then r2/r1 is
Official PDF solution: For loops made from the same wire, mass is proportional to radius. Moment of inertia I = mr2, so I ∝ r3. Thus I1/I2 = (r1/r2)3 = 1/4, giving r2/r1 = 22/3.
Question 16+4 / -1
A wheel having moment of inertia 3 kg-m2 about its vertical axis rotates at 30 rpm. The torque which can stop the wheel's rotation in one minute would be
Official PDF solution: Initial angular speed is ω = 2π(30)/60 = π rad/s. Angular retardation = π/60 rad/s2. Torque = Iα = 3(π/60) = π/20 Nm.
Question 17+4 / -1
The ratio of the density and pressure of a fixed mass of an ideal gas is 5 at 10°C. This ratio at 110°C is
Official PDF solution: For a fixed mass of ideal gas, ρ/P ∝ 1/T. Therefore (ρ/P)110°C = 5 × 283/383 = 3.69.
Question 18+4 / -1
A mixture of n1 moles of diatomic gas and n2 moles of monatomic gas has Cp/Cv = γ = 3/2, then
Official PDF solution: Using the mixture heat-capacity relation from the official solution, the condition γ = 3/2 gives n1 = n2.
Question 19+4 / -1
The velocities of four gas molecules are 1 km/s, 2 km/s, 6 km/s and 8 km/s respectively. The rms velocity is
Official PDF solution: vrms = √[(12 + 22 + 62 + 82)/4] = √(105/4).
Question 20+4 / -1
The graph of pressure and temperature for an ideal gas is as shown in figure. The given process is
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Official PDF solution: For an ideal gas, PV = nRT, so P = (nR/V)T. Since the graph shows P ∝ T, volume is constant and the process is isochoric.
Question 21+4 / -1
A particle of mass 0.2 kg executes SHM under a force F = -20x N. If speed of particle at mean position is 12 m/s, then the amplitude of oscillations is
Official PDF solution: Using conservation of energy, (1/2)kA2 = (1/2)mv2. Here k = 20 N/m, m = 0.2 kg and v = 12 m/s, so A = v√(m/k) = 1.2 m.
Question 22+4 / -1
Which of the following statement is incorrect?
Official PDF solution: Speed of sound in a gas is v = √(γRT/M), so it depends on temperature. Therefore statement (3) is incorrect.
Question 23+4 / -1
If two SHMs are represented by y1 = 4 sin(3πt + π/3) and y2 = 4[sin(3πt) + √3 cos(3πt)], then the ratio of their amplitudes is
Official PDF solution: Amplitude of y1 is 4. For y2, amplitude = √[42 + (4√3)2] = 8. Ratio = 4 : 8 = 1 : 2.
Question 24+4 / -1
If a spring of force constant K is cut into two parts, such that one part is thrice in length of the other part, then the force constants of each part are
Official PDF solution: Let the two lengths be l1 and l2, with l1 = 3l2. Then l1 = 3l/4 and l2 = l/4. Since spring constant is inversely proportional to length, k1 = 4K/3 and k2 = 4K.
Question 25+4 / -1
Moment of inertia of a ring about its diametric axis is I. Two such rings are welded as shown in figure. The new moment of inertia about XX' axis is
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Official PDF solution: For one ring, I = mR2/2 about a diametric axis. For the welded pair about XX', use the parallel axis theorem for both rings: Inew = 2(I + mR2) = 6I.
Question 26+4 / -1
Two wires are in unison. If the tension in one of the wires is increased by 2%, 5 beats are produced per second. The initial frequency of each wire is
Official PDF solution: Frequency of a stretched wire is proportional to √T, so percentage change in frequency is half the percentage change in tension, i.e. 1%. Since 1% of f is 5, f = 500 Hz.
Question 27+4 / -1
A string is hanging from a rigid support. A transverse pulse is excited at its free end. The speed at which the pulse travels a distance x is proportional to
Official PDF solution: At distance x from the free end, tension T = (M/L)xg and mass per unit length is M/L. Hence v = √(T/μ) = √(gx), so v ∝ √x.
Question 28+4 / -1
When a constant power is delivered to an object initially at rest, then the graph between velocity and position of the object is (motion of object is along x-axis)
Official PDF solution: For constant power, P = Fv = m v dv/dt. This gives v ∝ t1/2 and x ∝ t3/2, hence v ∝ x1/3. The correct graph is option (4).
Question 29+4 / -1
In an experiment, sonometer wire vibrates in 4 loops when a 50 g weight is placed in the pan of weight 15 g. To make the wire vibrate in 6 loops the mass that has to be removed from the pan is approximately, frequency remaining constant
Official PDF solution: For constant frequency, P√T is constant for the loop relation used in the official solution. With initial mass 50 + 15 = 65 g, the required mass becomes about 28.8 g. Mass removed = 65 - 28.8 = 36.2 g, approximately 36 g.
Question 30+4 / -1
A string is rigidly tied at two ends and its equation of vibration is y = cos 2πt sin 2πx. The minimum length of string is
Official PDF solution: Comparing with the standard stationary wave form gives angular frequency ω = 2π and wave number k = 2π. Thus wavelength λ = 1 m and the minimum length of a string fixed at both ends is λ/2 = 0.5 m.
Question 31+4 / -1
Two particles each of mass m are moving in a circle of radius r with speed v each around the third mass M as shown in figure. The value of v is
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Official PDF solution: Centripetal force on one mass is provided by attraction due to M and the other m: GmM/r2 + Gm2/(2r)2 = mv2/r. Therefore v = √[(G/r)(M + m/4)].
Question 32+4 / -1
In the following figure, if mass M is in equilibrium, then the ratio of m1 and m2 is (Neglect Earth's gravity and m1 and m2 are fixed)
Kumar Physics Classes +91 9958461445
Official PDF solution: For equilibrium, T sin37° = GMm1/d2 and T cos37° = GMm2/d2. Dividing gives m1/m2 = tan37° = 3/4.
Question 33+4 / -1
The molar heat capacity of a certain substance varies with temperature T as c = a + bT where a = 27.2 J mol-1K-1 and b = 4 × 10-3 J mol-1K-2. The heat necessary to change the temperature of 2 moles of this substance from 27°C to 427°C is nearly
Official PDF solution: Use Ti = 300 K and Tf = 700 K. Q = n∫c dT = 2∫300700(27.2 + 4 × 10-3T)dT = 23,360 J.
Question 34+4 / -1
The angular speed with which earth should spin about its axis, such that a person feels (5/9)th times his weight at equator as compared to his weight at pole, is
Official PDF solution: At the equator, effective gravity is g - Rω2. Given g - Rω2 = 5g/9, so Rω2 = 4g/9 and ω = (2/3)√(g/R).
Question 35+4 / -1
Coefficient of linear expansion of a substance along three mutually perpendicular directions are 1 × 10-5/°C, 2 × 10-5/°C and 3 × 10-5/°C. The coefficient of cubical expansion of the substance is
Official PDF solution: For anisotropic expansion, cubical expansion coefficient γ = α1 + α2 + α3 = 6 × 10-5/°C.
Question 36+4 / -1
Breaking stress of a steel wire of radius 1 mm and length 1 m is 1012 N/m2. Breaking stress of another steel wire of radius 2 mm and length 0.5 m will be
Official PDF solution: Breaking stress is a material property and is independent of length and area at a given temperature. Hence it remains 1012 N/m2.
Question 37+4 / -1
Two identical rods are connected in series and 100 joule heat is transferred in 10 seconds. Find the time taken to transfer the same amount of heat if the rods are connected in parallel, keeping the temperature of ends same as before.
Official PDF solution: In series, thermal resistance is 2R. In parallel, equivalent thermal resistance is R/2. For the same heat and same temperature difference, time is proportional to thermal resistance, so the time becomes one-fourth of 10 s = 2.5 s.
Question 38+4 / -1
Figure shown below is a graph of a monoatomic gas which undergoes a cyclic process. If work done by gas in process BC is x, the heat absorbed in process CA is
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Official PDF solution: Process BC is isobaric, so WBC = nR(T0 - 2T0) = -nRT0 = x. Process CA is isothermal, W = nRT0 ln(P0/2P0) = -nRT0ln2. Hence heat absorbed in CA = x ln2.
Question 39+4 / -1
Liquid is filled in a container. Find the value of acceleration a for which point x is exposed to air as shown in the figure.
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Official PDF solution: From the free-surface geometry in the official solution, tanθ = a/g = 4L/3L. Therefore a = 4g/3.
Question 40+4 / -1
A container is filled with liquid upto height H as shown in figure. There are 2 holes in the container such that water coming out from them have same range. If the range of water coming out of the holes is half of maximum range, then the distance x between these holes is
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Official PDF solution: For range R = H/2, use R = 2√[(H - h)h]. This gives h = H/2 ± √3H/4. The distance between the two holes is x = h2 - h1 = √3H/2.
Question 41+4 / -1
Bernoulli's equation is valid for
Official PDF solution: Bernoulli theorem is valid for ideal flow of liquids or gases: non-viscous, incompressible and streamlined flow. Hence all of these.
Question 42+4 / -1
A small block of mass m is attached at the bottom end of an elastic massless rod of length L, area A and Young's modulus Y. Elastic energy stored in the rod is
Official PDF solution: Elastic energy stored = (stress)2(volume)/(2Y). With stress = mg/A and volume = AL, energy = (mg)2L/(2AY).
Question 43+4 / -1
If the velocity of water layers in river is given by v = ky2, where y is the height from the river bed, the force required to keep a square plate of area A with constant velocity at height h is [η = coefficient of viscosity]
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Official PDF solution: Given v = ky2, dv/dy = 2ky. Viscous force F = ηA(dv/dy). At y = h, F = 2ηAkh.
Question 44+4 / -1
A car starts from rest at time t = 0 and its a-t graph is shown in figure. A coin is initially at rest on the floor of the car. At t = 1 s the coin begins to slip and it stops slipping at t = 3 s. The coefficient of static friction between the floor and the coin is (g = 10 m/s2)
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Official PDF solution: At the limiting condition for slipping, dv/dt = μg. From the graph the required acceleration is 4 m/s2. Therefore 4 = μg and μ = 0.4.
Question 45+4 / -1
A solid sphere of mass m and radius R is on a rough surface as shown in figure. Two forces having magnitude 2F and F are applied as shown. If the sphere performs pure rolling along right direction, then frictional force acting on the sphere is
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Official PDF solution: The official solution uses the rolling-friction relation f = F[(h/R) - (k2/R2)]/[1 + k2/R2]. For the shown force arrangement on a solid sphere, the sign is forward, so friction acts in the forward direction.
Final Result
Total Questions45
Attempted Questions0
Correct Answers0
Wrong Answers0
Unattempted Questions45
Positive Marks0
Negative Marks0
Final Score0
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