Dear Students, this NEET Physics assessment paper has been prepared for students who genuinely want to test, strengthen, and improve their Physics preparation. This is not an ordinary practice sheet. It has been designed to check concept clarity, calculation accuracy, formula selection, and the ability to solve Physics questions under real exam pressure.
This paper has been prepared and solved by an experienced Physics Tutor in Golf Links - New Delhi. The quality of this paper is excellent because the questions are selected and solved in a systematic, conceptual, and exam-focused manner. Students should attempt this paper with patience, discipline, and full concentration.
If students are searching for Physics Tutor, NEET Physics Tutor, or Physics Tutor in Golf Links - New Delhi and they are unable to solve these questions properly, they should contact Kumar Sir for one-to-one online Physics classes.
This paper should be attempted only after revising the important Class 11 and Class 12 Physics formulas. First revise the formula bank, then solve the complete question paper under timed conditions. Do not open the solution immediately. First think, calculate, choose your answer, and then compare it with the official solution. Every mistake should be treated as a learning point.
Important Formula Revision for NEET Physics: Class 11 and Class 12
Before starting this paper, revise the important formulas from Class 11 and Class 12 Physics. NEET Physics often tests whether a student can select the correct formula, apply it correctly, and avoid calculation mistakes. Many students remember formulas but still lose marks because they do not know where and how to apply them. This formula bank is added to help students quickly revise the major concepts before attempting the paper.
Units and Dimensions
Dimensional formula: write every physical quantity as MaLbTcAdKemolfcdg.
Principle of homogeneity: dimensions of LHS and RHS must be identical; used in formula checking and option elimination.
Percentage error: ΔZ/Z = aΔA/A + bΔB/B for Z = AaBb; used in numerical error questions.
Significant figures: final answer follows the least precise measurement in multiplication/division.
Vectors
|A + B| = √(A2 + B2 + 2AB cosθ); used in resultant vector problems.
A.B = AB cosθ; dot product is used in work, flux, and projection questions.
|A x B| = AB sinθ; cross product is used in torque, angular momentum, and magnetic force.
Projection of A on B = A cosθ; useful in components and inclined-plane applications.
Kinematics
v = u + at, s = ut + (1/2)at2, v2 = u2 + 2as; used when acceleration is constant.
x = x0 + vt for uniform motion; useful in relative motion.
R = u2 sin2θ/g, H = u2 sin2θ/(2g), T = 2u sinθ/g; projectile formulas.
a = dv/dt, v = dx/dt; used in graph and calculus-based questions.
Laws of Motion
F = ma; central equation for force and acceleration.
fs,max = μsN, fk = μkN; used in friction numericals.
Impulse J = ∫F dt = Δp; direction of impulse is direction of change in momentum.
Tension and normal reaction are constraint forces; draw a free-body diagram before calculation.
Work, Energy and Power
W = F s cosθ; work by constant force.
K = (1/2)mv2, U = mgh, U = (1/2)kx2; energy conservation applications.
Power P = dW/dt = Fv; used in engine, pump, and motion questions.
Work-energy theorem: Wnet = ΔK; useful when time is not given.
Circular Motion
ac = v2/r = ω2r; centripetal acceleration.
Fc = mv2/r; used in banking, loop, and tension questions.
v = rω, T = 2π/ω; relation between angular and linear motion.
tanθ = v2/(rg); banked road without friction.
Centre of Mass
Rcm = (Σmiri)/Σmi; used in particle systems.
Vcm = Ptotal/M; useful in explosions and collisions.
m1r1 = m2r2 about COM for two-particle systems.
External force changes COM motion; internal forces cannot change the motion of COM.
Logic gates: AND gives 1 only for A = 1 and B = 1; OR gives 1 if any input is 1; NOT inverts input.
One Year of Serious Tapasya for NEET Physics
Dear students, NEET preparation is not casual work. You may not be able to study 18 to 20 hours every day, but a serious aspirant can still study with discipline for long focused hours. During this one-year tapasya, reduce distractions, sleep only as much as genuinely required, and use every day with purpose. This does not mean studying blindly; it means studying intelligently, revising formulas, solving NEET papers, checking mistakes, and improving every week.
This paper should be solved like a real exam. Sit with a timer, attempt every question honestly, and do not open the solution before trying properly. If you are living in Golf Links - New Delhi and searching for a Physics Tutor for NEET, IB, ICSE, IIT-JEE, CBSE, IGCSE, AP Physics or any serious Physics preparation, contact Kumar Sir for one-to-one online Physics guidance. Kumar Sir helps students understand concepts deeply, solve difficult numericals, and build confidence for competitive exams.
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 student learns the formula, studies the condition behind it, practises mixed questions, and analyses every mistake honestly.
Important Message for NEET 2027, 2028, 2029, 2030 and Future Aspirants
Future NEET aspirants must prepare seriously for online-style or changing exam patterns, where question variation and concept application may become more important. Students should practise papers under timed conditions, revise formulas regularly, analyse mistakes, and strengthen weak chapters before they become scoring problems in the final exam.
Why Study Physics with Kumar Sir?
Kumar Sir provides personalised one-to-one online Physics classes. He clears each and every concept, explains difficult topics in simple language, and helps students prepare for NEET, CBSE, JEE, IB, ICSE, IGCSE, AP Physics and other exams. His teaching style focuses on conceptual clarity, numerical practice, doubt-solving, and exam-oriented preparation. If you are struggling in Physics or aiming for high marks in NEET Physics, Kumar Sir can guide you step by step and help you build confidence through disciplined practice.
Physics Guidance for Serious Students
If you are searching for a Physics Tutor in Golf Links - New Delhi for NEET, IB, ICSE, IIT-JEE, CBSE, IGCSE, AP Physics or any advanced Physics preparation, contact Kumar Sir. Kumar Sir explains Physics topics in a very clear, step-by-step, and exam-oriented way.
The dimensional formula of intensity of magnetisation is
Official solution: I = M/V = NiA/V = [A][L2]/[L3] = [L-1A].
Question 2+4 / -1
A particle at rest starts moving in straight line with acceleration a = 10 m/s2, then moves with constant velocity and then decelerates to rest at 10 m/s2. Total time of journey is 25 s. Average velocity during this time interval is 40 m/s. The time interval for which the car moved with constant velocity is
Official solution: Assume the car accelerates and retards for time t each, and moves with constant velocity for (25 - 2t). Total displacement/time = average velocity. [10t2 + 10t(25 - 2t)]/25 = 40, so t2 - 25t + 100 = 0. Taking t = 5 s, constant-velocity time = 25 - 2t = 15 s.
Question 3+4 / -1
A particle moves along z-axis such that its z-coordinate as a function of time t is given as z = 5 + 2t - t2. If v is speed, vector v is velocity and vector a is acceleration then which of the following graphs is/are correct?
Kumar Sir +91 9958461445
Kumar Sir +91 9958461445
Kumar Sir +91 9958461445
Kumar Sir +91 9958461445
Official solution: z = 5 + 2t - t2. Therefore dz/dt = 2 - 2t, so velocity changes linearly with time. Also a = dv/dt = -2 m/s2, a constant negative acceleration. Hence the correct set is I, II and IV.
Question 4+4 / -1
A bomb is to be dropped on a target by a jet flying with horizontal velocity 60 km/hr at an altitude of 490 m. How far before the target should the bomb be released (g = 9.8 m/s2)?
Official solution: R = v√(2H/g). With v = 60 km/hr = 50/3 m/s and H = 490 m, time of fall = √(980/9.8) = 10 s. Hence R = (50/3) x 10 = 500/3 m.
Question 5+4 / -1
Two particles start moving from same position on a circle of radius 20 m with speeds 18π m/s and 14π m/s respectively in the same direction. The time after which they will collide again is
Official solution: t = 2πR/(v2 - v1) = 2π x 20 / (18π - 14π) = 10 s.
Question 6+4 / -1
An impulse is imparted to a moving object at 60° to the velocity vector. The angle between the impulse vector and the change in momentum vector is
Official solution: dP = F dt. Impulse and change in momentum are in the same direction, so θ = 0°.
Question 7+4 / -1
A particle is moving on a circle of radius R. If the centripetal force is inversely proportional to R then linear speed is proportional to
Official solution: Given F ∝ 1/R. Since F = mv2/R, mv2/R = K/R. Therefore v is constant, so v ∝ R0.
Question 8+4 / -1
A bomb of mass 5 kg at rest, explodes into two pieces of masses 4 kg and 1 kg. The smaller mass starts at a speed of 20 m/s. The total energy of the two masses just after the explosion is
Official solution: By conservation of linear momentum, 4v + 1 x 20 = 0, so v = -5 m/s for the 4 kg mass. Total kinetic energy = (1/2)(1)(20)2 + (1/2)(4)(5)2 = 200 + 50 = 250 J.
Question 9+4 / -1
A water pump of 0.5 hp fills a bucket in 2 min. If the bucket is to be filled in half the time then the power of water pump should be
Official solution: The PDF gives Answer (1) and uses the relation Power ∝ v3.
Question 10+4 / -1
For a body under pure rolling the fraction of total energy which is purely rotational is 1/2. The rolling body may be
Official solution: Krot/Ktotal = (mK2v2/R2) / [mv2 + mK2v2/R2] = 1/2, so K2/R2 = 1. This is true for a ring and a hollow cylinder.
Question 11+4 / -1
A hollow sphere starts pure translatory motion on a rough surface with initial speed v0. The speed of its centre of mass when it just starts pure rolling is
Official solution: Using angular momentum about the contact point, mv0R = mvR + (2/3)mR2(v/R). Hence v = 3v0/5.
Question 12+4 / -1
The minimum time period an artificial satellite can have around the earth is approximately
Official solution: T = 2π√(R/g) = 84.6 min approximately.
Question 13+4 / -1
A solid cylindrical rod has length L, cross-sectional area A and Young's modulus Y. The spring constant of the rod is
Official solution: Y = FL/(A l). Therefore F = (YA/L)l = kl, so spring constant k = YA/L.
Question 14+4 / -1
Blood pressure rises in old age due to arteries becoming narrow. This phenomenon could be explained on the basis of
Official solution: Poiseuille's equation, Q = dV/dt = πpr4/(8ηl), explains the strong dependence of flow on radius.
Question 15+4 / -1
The pressure at the bottom of a water tank is 5P, where P is atmospheric pressure. If half of the water is drained out then the pressure at the bottom is
Official solution: Gauge pressure due to water is 5P - P = 4P. After half the water is drained, gauge pressure becomes 2P. Total pressure = P + 2P = 3P.
Question 16+4 / -1
In Siberia air is flowing at -10°C over a lake 20 m deep. It is given that the coefficient of thermal conductivity of water is 1.5 times that of ice. If the temperature of the bottom of the lake is 4°C then maximum thickness of ice formed is
Official solution: Equating heat conduction through ice and water gives 1 x 10/t = 1.5 x 4/(20 - t). Thus 200 - 10t = 6t and t = 200/16 = 25/2 m.
Question 17+4 / -1
A cup of tea cools from 70°C to 60°C in 5 minutes and to 50°C in next 7 minutes. Assuming Newton's law of cooling, the temperature of surroundings is
Official solution: (70 - 60)/5 divided by (60 - 50)/7 = (65 - T)/(55 - T). Solving gives T = 30°C.
Question 18+4 / -1
If NP and NC be the number of holes and conduction electrons in an extrinsic semiconductor then,
Official solution: If pentavalent impurity is doped then NC > NP. If trivalent impurity is doped then NP > NC.
Question 19+4 / -1
At 27°C the average kinetic energy of an ideal gas molecules is 5 x 10-21 J. Its average kinetic energy at 327°C will be
Official solution: E = (3/2)kT, so E ∝ T. Therefore E2 = (600/300) x 5 x 10-21 = 10-20 J.
Question 20+4 / -1
A wooden cylinder of mass m and cross-sectional area A is floating in a liquid of density ρ. It is depressed slightly and then left to do SHM. Time period for small oscillations will be
Official solution: In equilibrium, mg - Alρg = 0. When depressed by y, restoring force F = -Aρgy. Hence k = Aρg and T = 2π√(m/(Aρg)).
Question 21+4 / -1
A particle is subjected to three SHMs: X1 = 3 sinωt, X2 = 4 cosωt, X3 = 6 sin(ωt + π). Where X is in metre. The resultant amplitude of particle is
Official solution: Resultant amplitude = √[(3 - 6)2 + 42] = 5 m.
Question 22+4 / -1
In a resonance tube experiment for finding speed of sound in air the first and second resonance occurs at length 15 cm and 48 cm. The end correction at the open end is
Official solution: (l2 + e)/(l1 + e) = 3. Therefore (48 + e)/(15 + e) = 3, giving 48 + e = 45 + 3e and e = 1.5 cm.
Question 23+4 / -1
Energy of an electron in an excited hydrogen atom is -3.40 eV. Its angular momentum will be
Official solution: E = -13.6/n2 eV. Given -3.4 = -13.6/n2, so n = 2. Angular momentum L = nh/(2π) = h/π.
Question 24+4 / -1
Two short-electric dipoles kept at distance d apart, apply force 8F on each other. If the separation between them is doubled then the force between them will become approximately
Official solution: Force between short dipoles varies as 1/r4. Doubling the separation makes force 8F/16 = F/2.
Question 25+4 / -1
If on the x-axis electric potential increases uniformly from 30 V to 90 V between x = -3 m to x = +3 m then magnitude of electric field at origin may be
Official solution: ΔV/Δr = 60/6 = 10 V/m. The PDF states E ≥ ΔV/Δr, so E ≥ 10 V/m. Therefore any of the above may be possible.
Question 26+4 / -1
The refractive angle of prism is 60° and refractive index of material of prism is √3. The angle of minimum deviation is
Official solution: μ = sin[(A + δmin)/2]/sin(A/2). With A = 60° and μ = √3, sin[(60 + δmin)/2] = √3/2. Thus (60 + δmin)/2 = 60° and δmin = 60°.
Question 27+4 / -1
A potentiometer wire has length 2 m. Supply voltage Vs = 100 volt and R = 50 Ω. An emf E = 10 volt is balanced against 40 cm. The resistance per unit length of the potentiometer wire is
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Official solution: Assume resistance per unit length is R'. E = Kl. Therefore 10 = [100/(2R' + 50)]R' x 0.4. This gives 20R' + 500 = 40R', hence R' = 500/20 = 25 Ω/m.
Question 28+4 / -1
In a region an electron is moving in a circle. What can be the possible field there?
Official solution: A time-varying electric field produces a magnetic field, and a uniform magnetic field can make a charged particle move in a circle.
Question 29+4 / -1
A circular ring of radius R carries a current I in it. The magnetic field at the centre is B0. The field B on the axis at a distance x from the centre is B0/8. Here x is equal to
Official solution: B = μ0NIR2/[2(R2 + x2)3/2] and B0 = μ0NI/(2R). Setting B = B0/8 gives x = √3 R.
Question 30+4 / -1
If magnetic monopoles exists then Gauss's law in magnetism can be written as
Official solution: If magnetic monopoles exist, Gauss's law in magnetism becomes ∮B.dS = μ0mnet.
Question 31+4 / -1
Like poles of two bar magnets are tied together and the time period of their angular SHM in horizontal plane is K/2 second. When their unlike poles are tied together for similar SHM then the time period is K/√2 second. The ratio of magnetic moments of the bar magnets is
Official solution: From the given time periods, (μ1 + μ2)/(μ1 - μ2) = (2/√2)2 = 2. Hence μ1/μ2 = 3.
Question 32+4 / -1
The energy stored in the capacitor at steady state shown is
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Official solution: U = (1/2)CV2 = (1/2)(4 μF)(10 V)2 = 200 μJ.
Question 33+4 / -1
In the circuit shown the reading of ammeter just after the key is closed is
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Official solution: Just after the key is closed, i = 20/(5 + 5) = 2 A.
Question 34+4 / -1
At the centre of a large circular ring of radius R lies a small circular coplanar ring of radius r. The mutual induction of larger ring with respect to the smaller ring is
Official solution: Flux through the smaller ring due to current in the larger ring is (μ0I/2R)πr2 = MI. Hence M = μ0πr2/(2R).
Question 35+4 / -1
In a LC oscillation the value of inductance L = 0.4 mH and the maximum current is 20 A. If the capacitance C = 4 μF then its maximum voltage rating is
Official solution: (1/2)CV2 = (1/2)LI2. Therefore V = I√(L/C) = 20√(0.4 mH / 4 μF) = 200 V.
Question 36+4 / -1
In a series RLC circuit, R = 50 Ω, L = 180 mH and C = 8 μF. What is that frequency (in rad/s) which drives the current in phase with the voltage?
Official solution: At resonance, ω = 1/√(LC) = 1/√(180 x 10-3 x 8 x 10-6) = 104/12 rad/s.
Question 37+4 / -1
A point source of light of power P is kept above a table at a distance R. If C is speed of light and ε0 is permittivity then amplitude of electric field at distance R is
Official solution: P/(4πR2C) = (1/2)ε0E02. Hence E0 = (1/R)√[P/(2πCε0)].
Question 38+4 / -1
On a plane mirror a glass slab of μ = 1.6 and thickness 8 cm is kept. An object is kept 10 cm above the upper surface of the glass slab. How much behind the mirror is the image formed?
Official solution: Apparent calculation from the PDF: (10 + 8/1.6) - (8 - 8/1.6) = 12 cm.
Question 39+4 / -1
Parallel rays are falling on a convex lens of focal length 20 cm. A concave lens is kept behind the convex lens at 8 cm such that the rays passing through this concave lens again become parallel. The focal length of the concave lens is
Official solution: The virtual object is at the focus of the concave lens. Therefore fconcave = fconvex - d = 20 - 8 = 12 cm.
Question 40+4 / -1
Appearance of colours on thin soap film is due to
Official solution: Appearance of few colours in soap film is due to interference of light.
Question 41+4 / -1
Sunlight is passed through a polariser. Now the polariser is rotated through 360°. If the light passing through the polariser is observed then its intensity
Official solution: Sunlight is partially polarised, so on rotating the polariser, the observed intensity varies between maximum and minimum twice in a rotation.
Question 42+4 / -1
The radius of gyration of disc of mass and radius R about an axis which is at distance 2R from its centre and in the plane of the disc will be
Official solution: I = mR2/4 + m(2R)2 = 17mR2/4. Since I = mk2, k = (√17/2)R.
Question 43+4 / -1
A nucleus of mass M emits γ-rays of energy E. The recoil energy of the nucleus is
Official solution: Apply momentum conservation: 0 = Mv + E/C, so v = E/(MC). Recoil energy of nucleus = (1/2)Mv2 = E2/(2MC2).
Question 44+4 / -1
A radioactive sample can decay through two independent modes with half-lives 30 min and 60 min. If initially N0 nuclei are present, the number left after 60 min is
Official solution from the supplied PDF: Half-life of mixture = (30 x 60)/(30 + 60) = 20 min. Number of half lives = 3. Therefore N = N0(1/2)3 = N0/8.
Question 45+4 / -1
The logic-gate combination represented by the truth table in the official solution is equivalent to
Official solution from the supplied PDF: Truth table: A=0, B=0 gives Y=0; A=1, B=0 gives Y=0; A=0, B=1 gives Y=0; A=1, B=1 gives Y=1. This is an AND gate.
