This NEET Physics assessment paper is based on the first major half of Class 12 Physics: Electrostatics, Current Electricity, Moving Charges and Magnetism, Magnetism and Matter, Electromagnetic Induction, Alternating Current and Electromagnetic Waves. These chapters are extremely important for NEET, CBSE, ICSE and other board-level Physics preparation because they test concept clarity, circuit understanding, field concepts, magnetic force, induction, AC analysis and electromagnetic wave fundamentals.
This paper has been prepared in a systematic and exam-oriented way by a Physics Tutor in Koramangala – Bengaluru. The questions are selected to help students improve concept application, numerical accuracy, speed, confidence and exam discipline.
Class 12 Physics becomes scoring only when concepts are clear and formulas are applied correctly. Electrostatics, Current Electricity, Magnetism, EMI and AC are not chapters for blind memorisation. They require patience, circuit sense, field understanding, diagram practice and step-by-step numerical thinking.
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 be treated as a learning opportunity.
Important Formula Revision for NEET Physics: Class 12 Electrostatics to EM Waves
Before starting this paper, revise the important formulas from Electrostatics, Current Electricity, Magnetism, EMI, AC and EM Waves. NEET Physics often checks 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.
Electrostatics
Coulomb's lawF = kq1q2/r2used in force numericals
Electric fieldE = F/qused in field definition questions
Field due to point chargeE = kq/r2used in point charge problems
Potential due to point chargeV = kq/rused in potential numericals
Field and potentialE = -dV/drused in graph-based questions
Potential energyU = kq1q2/rused in charge systems
Electric fluxΦ = E·A = EAcosθused in Gauss law questions
Gauss law∮E·dA = qenclosed/ε0used in symmetry problems
Infinite line chargeE = λ/(2πε0r)used in field comparison
Infinite sheetE = σ/(2ε0)used in sheet charge cases
Outside conductorE = σ/ε0used in conductor questions
CapacitanceC = Q/Vused in capacitor numericals
Parallel plate capacitorC = ε0A/dused in plate questions
With dielectricC = Kε0A/dused in dielectric cases
Capacitor energyU = 1/2 CV2 = 1/2 QV = Q2/2Cused in energy questions
Series capacitance1/C = 1/C1 + 1/C2 + ...used in series networks
Transformer voltageVs/Vp = Ns/Npused in transformer ratios
Transformer currentIs/Ip = Np/Nsused in ideal transformers
Electromagnetic Waves
Speed of EM wavec = 1/√(μ0ε0)used in EM wave conceptual questions
Field relationE0/B0 = cused in wave field ratios
Electric energy densityuE = 1/2 ε0E2used in energy density
Magnetic energy densityuB = B2/(2μ0)used in energy density
Total energy densityu = uE + uBused in EM energy
Speed in mediumv = c/nused in refractive medium
Wavelength-frequencyc = fλused in wave numericals
Question 1NEET Physics
The unit of permittivity of free space is
Official Solution: Answer (4). SI unit of ε is Farad per meter, so the correct unit is F m-1.
Question 2NEET Physics
The electrostatic field lines
Official Solution: Answer (4). Electrostatic lines of force do not form closed loops, as the electrostatic field is a conservative field.
Question 3NEET Physics
A closed solid conductor of irregular shape is given some charge. Which of the following statement is correct?
Official Solution: Answer (1). For a charged conductor, the electric field inside is zero and all points of the conductor are at the same potential, but surface charge density can vary on an irregular surface. Changing the shape changes the capacitance and hence the potential.
Question 4NEET Physics
The electric field decreases most rapidly, with distance for
Official Solution: Answer (4). Electric field due to a short dipole varies as E ∝ 1/r3. For a point charge E ∝ 1/r2, for a line charge E ∝ 1/r, and for a large sheet E ∝ r0.
Question 5NEET Physics
Choose the incorrect statement.
Official Solution: Answer (4). When the Gaussian surface encloses no charge, the electric flux through that surface is zero. Electric field over the Gaussian surface need not be zero at every point.
Question 6NEET Physics
Eight identical spherical liquid drops each charged to potential of 2 volt. They coalesce to make one big spherical liquid drop, whose potential is
Official Solution: Answer (3). V' = n2/3V = 82/3 × 2 volt = 4 × 2 volt = 8 volt.
Question 7NEET Physics
Three capacitors C1 = 3 µF, C2 = 6 µF and C3 = 12 µF are joined in series. This series combination is connected to a 14 volt source. The P.D across the plates of capacitor C2 is
Official Solution: Answer (2). Let Cs be the effective capacitance. 1/Cs = 1/3 + 1/6 + 1/12, so Cs = 12/7 µF. Charge on C2 = charge on Cs; C2V2 = CsV. Therefore V2 = [(12/7) × 14]/6 = 4 V.
Question 8NEET Physics
A uniform electric field of 50√2 V m-1 exists in XY plane, making an angle of 45° with positive X axis as shown in below figure. The potential difference [VA - VB] is
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Official Solution: Answer (2). VA - VB = -∫BA Ē · dr̄ = Ē · (r̄B - r̄A). Using Ē = 50√2[(î + ĵ)/√2] and r̄B - r̄A = -4î - 2ĵ, the result is -300 volt.
Question 9NEET Physics
In the circuit shown below, C1 = 3 µF; C2 = 6 µF; C3 = 6 µF; C4 = 12 µF and C5 = 9 µF. The total energy stored in these five capacitors is
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Official Solution: Answer (2). The Wheatstone bridge is in the balanced condition, so Ceff = 6 µF. Total energy stored U = 1/2 CeffV2 = 1/2 × 6 µF × 22 = 12 µJ.
Question 10NEET Physics
A point charge Q is located at a distance R/2 from the centre of an uncharged conducting spherical shell as shown in below figure. The electric field at a distance 3R from the point charge is
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Official Solution: Answer (3). The field at point P due to charge Q within the shell and the charge induced on the inner surface of the shell is zero. The field at P is only due to the induced charge on the outer surface of shell. Hence E = (1/4πε0) Q/(3R + R/2)2 = (1/4πε0) Q/(7R/2)2.
Question 11NEET Physics
In the circuit shown, R1 = 10 Ω; R2 = 20 Ω; R3 = 30 Ω and R4 = 20 Ω. Potential difference between b and d (Vb - Vd) is
In the circuit shown, the effective resistance between the points a and b is
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Official Solution: Answer (4). By symmetry, the circuit reduces to three parallel branches of 6 Ω, 12 Ω and 12 Ω. Therefore the effective resistance between a and b is Rab = 3 Ω.
Question 13NEET Physics
For the circuit shown below, when the switch S is closed, then the total amount of charge that flows from point a to point b through the switch is
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Official Solution: Answer (3). Before switch S is closed, charge on each of the 12 µF and 4 µF capacitors is [(12 × 4)/(12 + 4)] × 9 = 27 µC. The current through 6 Ω and 3 Ω resistors is 9/(6 + 3) = 1 A, so P.D. across 6 Ω is 6 V and across 3 Ω is 3 V. After closing S, charge on 4 µF capacitor is 4 × 3 = 12 µC and charge on 12 µF capacitor is 12 × 6 = 72 µC. Charge flowed from a to b through S is (+72) - 12 = 60 µC.
Question 14NEET Physics
In the circuit shown, the effective resistance between the points a and b is
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Official Solution: Answer (4). By symmetry, currents through R1 and R4 are same and currents through R2 and R3 are same. From the loop equation, 12i = 36i1, so i = 3i1. Then Va - Vb = 18i1 + 6(i - i1) = 10i. Hence Reff = 10 Ω.
Question 15NEET Physics
In the infinite ladder network shown below, if R = 1.235 Ω, then the effective resistance between the points a and b (approximately) is (√5 = 2.235)
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Official Solution: Answer (3). Let Re be the effective resistance between a and b. The self-similar ladder gives Rab = R + RRe/(R + Re). Solving gives Re2 - RRe - R2 = 0, so Re = [(√5 + 1)/2]R. With R = 1.235 = √5 - 1, Re = [(√5 + 1)/2](√5 - 1) = 2 Ω.
Question 16NEET Physics
The current through the 2 Ω resistor in the circuit shown below is
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Official Solution: Answer (4). εnet = (ε1/r1 - ε2/r2)/(1/r1 + 1/r2) = (10 - 5)/(1 + 1) = 5/2. Current i = (5/2)/(2 + 1/2) = 1 A. Since b is at higher potential than a, current flows from b to a.
Question 17NEET Physics
In the circuit shown below, if the tap key K is pressed at time t = 0, then which of the following statements is false?
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Official Solution: Answer (3). At t = 0, the capacitor offers zero opposition, so i = 4/(500 + 500) = 4 mA. At t = ∞, the capacitor offers infinite opposition, so i = 4/(500 + 1000 + 500) = 2 mA. P.D. across 1000 Ω is 2 × 10-3 × 1000 = 2 V. Therefore charge on 6 µF capacitor = 6 × 2 = 12 µC. Statement (3) is false.
Question 18NEET Physics
The current flowing through the 3 Ω resistor, in the circuit shown below, is
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Official Solution: Answer (4). The branch ab containing the 3 Ω resistor is not a part of the closed circuit. If current flows in this branch, Kirchhoff's first law will be violated. Therefore no current flows through the 3 Ω resistor.
Question 19NEET Physics
The equivalent resistance between A and B is
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Official Solution: Answer (4). The network reduces to a top branch R in parallel with a bottom branch 3R. Thus RAB = R ∥ 3R = 3R/4.
Question 20NEET Physics
The resistance of the resistor shown below is
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Official Solution: Answer (1). First band violet = 7, second band green = 5, third band orange gives multiplier 103, and last band gold gives tolerance ±5%. Therefore R = [75 × 103 Ω] ± 5%.
Question 21NEET Physics
An electron revolves in a circular orbit of radius 0.5 Å with a frequency of 5 × 1016 rev/s. The current in that circular orbit is
Official Solution: Answer (2). i = q/T = fq = fe = 5 × 1016 × 1.6 × 10-19 = 8 × 10-3 A = 8 mA.
Question 22NEET Physics
Which one of the following statements concerning Ohm's law is true?
Official Solution: Answer (4). Ohm's law is true when the resistivity of the material is independent of the applied electric field.
Question 23NEET Physics
A steady current flows in a metal wire of non-uniform cross-section. Which quantity remains constant along the length of the wire?
Official Solution: Answer (4). Mobility μ = vd/E = (J/ne)/(J/σ) = σ/(ne), which is constant for the same metal.
Question 24NEET Physics
As shown in diagram, a current carrying conductor is subjected to steady uniform magnetic field. Then
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Official Solution: Answer (2). The straight wire experiences force only and the circular wire experiences torque only. Overall, the conductor experiences both torque as well as force.
Question 25NEET Physics
The magnetic dipole moment of current loop as shown in figure is
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Official Solution: Answer (1). Radius of the circular path r = L/2. Magnetic moment is M̄ = IL2(-ĵ) + IπL2/4(-î) = -IL2ĵ - (IπL2/8)î.
Question 26NEET Physics
A circular flexible current loop of radius R carrying current i is placed in an inward magnetic field B. If we spin the loop clockwise with angular speed ω, then tension in string (assume the mass of the loop is m)
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Official Solution: Answer (2). Due to outward centrifugal force, tension in the wire becomes (BiR + mRω2/2π), so it is more than iBR.
Question 27NEET Physics
An electron with mass m, velocity v and charge e describing half a revolution in a circle of radius r in a magnetic field B, will experience change in energy equal to
Official Solution: Answer (4). The direction of magnetic force remains perpendicular to direction of velocity, so no work is done on the charged particle and its kinetic energy does not change.
Question 28NEET Physics
Two non-interacting inductors L1 = 2 mH and L2 = 5 mH are connected in parallel then respective ratio of
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Official Solution: Answer (1). In parallel, emf across both inductors is same. Therefore dφ1/dt = dφ2/dt, or φ1 = φ2.
Question 29NEET Physics
A conducting rod PQ is rotated in a magnetic field B about an axis passing through point O as shown in figure. Then potential difference between P and Q is (ω: angular speed)
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Official Solution: Answer (4). The induced emf is e = ∫ab Bωx dx = (Bω/2)(b2 - a2).
Question 30NEET Physics
A magnet is suspended in the magnetic meridian with an untwisted wire. The upper end of wire is rotated through 181° to deflect magnet by 37° from magnetic meridian. Now this magnet is replaced by another magnet and upper end of wire has to be rotated by 273° to deflect magnet by 53° from magnetic meridian. The ratio of magnetic moment of the two magnets respectively is
Official Solution: Answer (2). Twisting torque = deflecting torque. Case I: k(181 - 37) = M1H sin37°, so 144k = M1H(3/5). Case II: k(273 - 53) = M2H sin53°, so 220k = M2H(4/5). Dividing gives M1/M2 = (4/3)(144/220) = 48/55.
Question 31NEET Physics
A coil in the shape of square is placed in a variable magnetic field, which varies at the rate of dB/dt as shown in figure. The magnitude of emf between points a and d along path abcd will be
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Official Solution: Answer (4). If side length is a, then L = a/√2 and a = L√2. The emf between a and d along path abcd is (2L2 dB/dt)(3/4) = (3L2/2)dB/dt.
Question 32NEET Physics
The network shown in the figure is a part of a complete circuit. If at a certain instant, the current i = 1 A and potential at point A and B are equal, then the value of |di/dt| is
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Official Solution: Answer (1). Let i be increasing. VA - 1 × 2 + 6 - 10 × 10-3(di/dt) = VB. Since VA = VB, di/dt = 400 A/s.
Question 33NEET Physics
The resistance of three parts of a circular loop is shown in the figure. The magnetic field at the centre O is
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Official Solution: Answer (1). Since current in ab and ac are same, potential at b and c are same; hence current in bc is zero. The magnetic field at the centre cancels out, so net field is zero.
Question 34NEET Physics
If an electromagnetic wave propagating through vacuum is described by Ey = E0 sin(kx - ωt), Bz = B0 sin(kx - ωt), then
Official Solution: Answer (1). E0/B0 = c = ω/k. Therefore E0k = B0ω.
Question 35NEET Physics
Relation between electric field and magnetic field in electromagnetic wave is E0 = cB0 (where c = speed of light in vacuum). Ratio of energy density in electric to the magnetic fields is
Official Solution: Answer (4). For an electromagnetic wave, the electric and magnetic energy densities are equal, so UE = UB and the ratio is 1 : 1.
Question 36NEET Physics
If a current I = I0 sin(ωt - π/2) flows in an AC circuit across which an AC potential of V = V0sinωt has been applied, then the average power consumption P in the circuit will be
Official Solution: Answer (4). The phase difference is π/2. Average power P = (V0I0/2)cos(π/2) = 0.
Question 37NEET Physics
An AC source is rated 220 V, 50 Hz. The average voltage calculated in time interval of 0.01 s
Official Solution: Answer (1). The value of average voltage depends on the interval chosen; over a suitably chosen 0.01 s interval it may be zero.
Question 38NEET Physics
A capacitor of capacitance C is connected to a V volt potential. Battery is then disconnected and inductor of inductance L is connected across the capacitor, so that LC oscillations are set up. The maximum current in the coil is
Official Solution: Answer (1). Maximum current in LC oscillation is imax = q0ω = CV × 1/√(LC) = V√(C/L).
Question 39NEET Physics
The inductance of a closed packed coil of 400 turns is 8 mH. A current of 5 mA is passed through it. The magnetic flux per turn is (μ0 = Permeability of free space)
A coil having n turns and resistance R is connected with a galvanometer of resistance 4R. The combination is moved in time t such that magnetic flux φ1 per turn changes to flux φ2 per turn. The average induced current in the circuit is
Official Solution: Answer (2). e = -Δφ/Δt = -n(φ2 - φ1)/t. Total resistance is R + 4R = 5R, so i = -n(φ2 - φ1)/(5Rt).
Question 41NEET Physics
A massless square loop of side l carrying current i is suspended from a pulley as shown in the arrangement. Lower edge of the loop is in transverse magnetic field B. The minimum current i required in the loop to just move the block of mass m up the inclined plane is
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Official Solution: Answer (3). T = mg sinθ + μmg cosθ. The magnetic force gives 2T = Bil. Therefore i = (2mg sinθ + 2μmg cosθ)/(Bl).
Question 42NEET Physics
A conducting loop is placed in a magnetic field of strength B perpendicular to its plane. Radius of loop is r, current in the loop is i and linear mass density of loop is m. Speed of any transverse wave in the loop will be
Official Solution: Answer (1). F = 2T and 2irB = 2T, so T = Bir. Wave speed v = √(T/m) = √(Bir/m).
Question 43NEET Physics
A particle of charge q and mass m starts moving from origin under the action of an electric field Ē = E0î and B̄ = B0î with velocity v = v0ĵ. The time after which its speed becomes 2v0 is
Official Solution: Answer (4). Magnetic force does no work. Displacement by the electric field is s = (1/2)(qE0/m)t2î. Using work-energy theorem, qE0s = (1/2)m[(2v0)2 - v02], giving t = √3 mv0/(qE0).
Question 44NEET Physics
A conducting circular loop of radius r carries a constant current i. It is placed in uniform magnetic field B̄, such that B̄ is perpendicular to plane of the loop. Find the magnetic force acting on the loop.
Official Solution: Answer (3). Net force on a current carrying loop in a uniform magnetic field is zero.
Question 45NEET Physics
If current is passed in a spring, then it
Official Solution: Answer (1). Two current carrying wires with parallel current always attract each other. Adjacent turns of the spring attract, so the spring gets compressed.
Final Result
Use this after a sincere timed attempt. The result follows NEET marking: correct +4, wrong -1, unattempted 0.
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; a student must understand when, where, and how to apply each relation in a fresh situation.
A disciplined student should revise formulas, solve mixed questions, practise diagrams and circuits, and analyse every wrong answer. This is where guidance from a One-to-one Physics Tutor in Bengaluru can make preparation more focused and accountable.
Important Message for NEET 2027, 2028, 2029, 2030 and Future Aspirants
Future NEET aspirants must prepare seriously and strengthen their Class 12 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 pressure points.
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 and 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 Class 12 Physics or want stronger preparation, Kumar Sir can guide them personally with a structured learning plan and regular doubt clearing.
This Class 12 half-syllabus paper has been prepared in the style of a serious guided assessment by a Physics Tutor in Koramangala – Bengaluru, and students should solve it sincerely for best results.
