Q1
Correct +4 | Wrong -1 | Unattempted 0
The number of components in which a vector can be resolved into is
Official Solution:
The number of components of a vector are infinite.
Class 11 and Class 12 NEET Physics Assessment
Practice a focused Physics paper built from the uploaded assessment PDF. This page includes Physics questions Q1 to Q45 only, with clickable options, NEET marking, official answer mapping, and solution support for disciplined revision.
45 Physics Questions
NEET Marking: +4 / -1 / 0
Class 11 + Class 12 Revision
Why Strong Physics Preparation Is Now More Important Than Ever
Physics has become one of the clearest differentiators for serious NEET aspirants. A strong Biology score can create a base, but a confident Physics score often decides rank, college options, and the final margin between students with similar preparation levels.
The subject rewards clarity over memorisation. Students who understand formulas, graphs, units, dimensions, circuits, mechanics, optics, modern physics, and thermodynamics can handle unfamiliar questions with confidence. This is why regular timed practice, mistake analysis, and concept-wise doubt clearing are essential.
A premium preparation strategy is simple: learn the concept deeply, solve quality questions honestly, review every wrong response, and repeat until speed and accuracy improve together.
Important Message for NEET 2027 and Future Aspirants
Dear NEET Aspirants,
If you are preparing for Physics, you must understand that the NEET examination pattern in the coming years is expected to become more digital, more competitive, and more concept-oriented. From NEET 2027 onwards, students should prepare themselves for an online-style examination environment where speed, accuracy, conceptual clarity, and independent problem-solving will become extremely important.
From NEET 2027 onwards, the examination is expected to move towards an online Computer-Based Test format. This is a very important change for every NEET aspirant, especially in Physics. In an online examination, students must be prepared for a more dynamic testing environment where different students may face different sets of questions, and different shifts may contain different combinations of concepts and difficulty levels.
This makes Physics even more critical. Unlike Biology, where strong memory can help a lot, Physics demands conceptual clarity, calculation speed, logical thinking, and the ability to apply formulas in unfamiliar situations. If the exam becomes online and shift-based, students cannot depend only on repeated question patterns. They must be ready to solve new and concept-based questions confidently.
For NEET 2027 and future aspirants, the correct strategy is very clear. First, study the chapter properly. Understand the theory, formulas, graphs, diagrams, and standard concepts. Then solve high-quality questions under a timer, exactly like an online test. After attempting the paper, check every mistake carefully and understand why the correct answer is correct.
The Physics papers and question sets provided here are designed to help students build this habit. They include concepts from Class 11 and Class 12 Physics and help students test their real preparation level. If a student can solve these questions honestly and accurately, it means the conceptual foundation is becoming strong.
If you still find Physics difficult or are searching for a good Physics Tutor for NEET, you may contact Kumar Sir for one-to-one online Physics guidance. Kumar Sir helps students strengthen concepts, clear doubts, improve problem-solving, and prepare seriously for the changing NEET pattern.
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Phone / WhatsApp: +91 9958461445
Email: kumarsirphysics@gmail.com
Website: kumarphysicsclasses.com
Question Index
Correct Answer: +4 | Wrong Answer: -1 | Unattempted: 0
Q2
Correct +4 | Wrong -1 | Unattempted 0
A particle is moving with velocity v = (2i + 3j) m/s at an instant of time. If a force F = (4i + 6j + 5k) N acts on the particle at that instant, then the radius of curvature of the particle at the instant is (mass of the particle is 2 kg).
Official Solution:
R = v2/a⊥. Here v2 = 22 + 32 = 13 and the perpendicular acceleration is 5/2 m/s2. Therefore R = 13/(5/2) = 5.2 m.
Q3
Correct +4 | Wrong -1 | Unattempted 0
The quantities A and B are related by the relation m = A/B, where m is the linear momentum and A is the angular impulse. The dimensions of B are of
Official Solution:
[m] = [M L T-1] and [A] = [M L2 T-2]. From m = A/B, [B] = [A]/[m] = [L].
Q4
Correct +4 | Wrong -1 | Unattempted 0
The time period of a simple pendulum measured inside a stationary elevator is 4 s. If the lift starts accelerating downwards with an acceleration 3g/4, the time period will be
Official Solution:
For a pendulum, T = 2π√(l/g). When the lift accelerates downwards by 3g/4, the effective acceleration is g' = g/4. Hence T' = 2π√(l/(g/4)) = 8 s.
Q5
Correct +4 | Wrong -1 | Unattempted 0
Consider the following pulley mass system. The frequency of the vibration of the second overtone of the wire between the pulleys is (Given: mass of the wire between the pulleys: 24 g).
Official Solution:
Tension in the wire is T = 9g = 90 N. For the second overtone, f = 3v/(2l) = (3/(2l))√(T/μ). Using l = 1.5 m and μ = 0.024/1.5 kg m-1, f = 75 Hz.
Q6
Correct +4 | Wrong -1 | Unattempted 0
The moment of inertia of a semicircular plate of radius R and mass M about axis AA' in its plane passing through its centre, as shown in the figure, is (θ = 45°).
Official Solution:
Due to symmetry of the semicircular plate about the shown axis, the required moment of inertia is MR2/4.
Q7
Correct +4 | Wrong -1 | Unattempted 0
A rigid body can be hinged about any point on the x-axis. When it is hinged such that the hinge is at x, the moment of inertia is given by I = x2 - 24x + 48. The x-coordinate of centre of mass is
Official Solution:
For minimum moment of inertia, dI/dx = 0. Thus 2x - 24 = 0, so x = 12.
Q8
Correct +4 | Wrong -1 | Unattempted 0
A uniform hollow sphere starts rolling from a height h on an inclined plane. At some instant t, the ratio of its rotational kinetic energy and the total kinetic energy would be
Official Solution:
For rolling motion, the ratio is (1/2 Iω2)/(1/2 Iω2 + 1/2 mv2). For a hollow sphere, I = (2/3)mR2 and v = Rω, giving the ratio 2 : 5.
Q9
Correct +4 | Wrong -1 | Unattempted 0
A system consists of two identical cubes each of mass m, linked together by a compressed weightless spring of force constant k. The cubes are also connected by a thread which is burnt at a certain moment. The minimum value of initial compression of the spring for which the lower cube will bounce up after the burning of the thread is
Official Solution:
To lift the lower block, kx' = mg. Using energy conservation, (1/2)kx2 = mg(x + x') + (1/2)kx'2. This gives x - x' = 2mg/k. Since x' = mg/k, the initial compression is x = 3mg/k.
Q10
Correct +4 | Wrong -1 | Unattempted 0
Two objects A and B each of mass m are connected by a light inextensible string. They are restricted to move on a frictionless ring of radius R in a vertical plane, as shown. The objects are released from rest at the position shown. Then the acceleration of mass A just after release is nearly (g = 10 m/s2).
Official Solution:
For A, T sin45° = ma1. For B, mg - T sin45° = ma2. Also a1 = a2. Hence a = g/2 = 5 m/s2.
Q11
Correct +4 | Wrong -1 | Unattempted 0
The distance of center of mass of a uniform semicircular disc, of radius 3π m, from its center of curvature is
Official Solution:
For a uniform semicircular disc, Ycom = 4R/(3π). With R = 3π m, Ycom = 4 m.
Q12
Correct +4 | Wrong -1 | Unattempted 0
Two soap bubbles of different radii are connected to each other through a tube as shown. Choose the most appropriate alternative.
Official Solution:
Air flows from the smaller bubble to the larger bubble because the smaller bubble has higher internal pressure.
Q13
Correct +4 | Wrong -1 | Unattempted 0
A car starts from rest travelling with constant acceleration. If distance covered by it in 4th second of its journey is 21 m, what will be the acceleration of car?
Official Solution:
Distance in the nth second is sn = u + (a/2)(2n - 1). Here u = 0, n = 4 and 21 = (a/2)(7). Thus a = 6 m/s2.
Q14
Correct +4 | Wrong -1 | Unattempted 0
A parallel beam of light is incident on a tube containing two convex lenses of focal length 5 cm and 10 cm as shown (the rays are incident on the 5 cm focal length lens first). If the diameter of incident beam is 1.6 cm, find the diameter of final beam.
Official Solution:
Using d1/f1 = d2/f2, 1.6/5 = d2/10. Therefore d2 = 3.2 cm.
Q15
Correct +4 | Wrong -1 | Unattempted 0
A thin rod of length f/2 is placed along the principal axis of a concave mirror of focal length f, such that its image which is real and elongated, just touches the rod. Then the longitudinal magnification of the rod is
Official Solution:
The distance of the other end of the rod from the pole of the mirror is 2f - f/2 = 3f/2. Using the mirror formula, the image of this end is at v = -3f. Thus image length is 3f - 2f = f, and longitudinal magnification is f/(f/2) = 2.
Q16
Correct +4 | Wrong -1 | Unattempted 0
Consider the following two statements:
a. Among a given set of axis for a rigid body, the moment of inertia is always minimum about the one passing through the center of mass.
b. Perpendicular axis theorem for moment of inertia would be applicable for a solid sphere.
The correct statement(s) is/are
Official Solution:
The moment of inertia is minimum about the centre of mass only when the axes being compared are parallel. The perpendicular axis theorem is applicable for planar bodies, not a solid sphere. Hence neither statement is correct.
Q17
Correct +4 | Wrong -1 | Unattempted 0
In the circuit shown in figure the current gain, β = 200 for the transistor. What would be the base resistance RB so that VCE = 2 V? Neglect VBE.
Official Solution:
Collector loop: 10 = IC(103) + 2, so IC = 8 mA. Since β = IC/IB = 200, IB = 4 × 10-5 A. With VBE neglected, RB = 10/IB = 2.5 × 105 Ω.
Q18
Correct +4 | Wrong -1 | Unattempted 0
The angular momentum of an electron in hydrogen atom is 5h/(2π). Here h is Planck's constant. The potential energy of this electron is
Official Solution:
The electron is in the n = 5 state. For hydrogen, potential energy is U = -27.2/n2 eV. Therefore U = -27.2/25 = -1.088 eV.
Q19
Correct +4 | Wrong -1 | Unattempted 0
An open pipe of sufficient length is dipping in water with speed v vertically. If at any instant l is length of tube above water, then the rate at which fundamental frequency of pipe changes is (speed of sound = c).
Official Solution:
At time t, the effective air column length is l - vt. Fundamental frequency is f = c/[4(l - vt)]. Hence df/dt = cv/[4(l - vt)2], and at the instant shown it is cv/(4l2).
Q20
Correct +4 | Wrong -1 | Unattempted 0
In a uniform solid sphere of mass M and radius R, a spherical cavity is created such that the center of the cavity does not coincide with the center of the original solid sphere. The gravitational field inside the cavity will be
Official Solution:
Inside an off-centre spherical cavity made in a uniform solid sphere, the gravitational field is uniform in both magnitude and direction.
Q21
Correct +4 | Wrong -1 | Unattempted 0
A mixture of two gases X and Y is enclosed at constant temperature. The molecular mass of X, which is diatomic, is 16 times that of Y, which is monatomic. The ratio of r.m.s. speed of Y molecules to that of molecules of X is
Official Solution:
vrms = √(3RT/M). Since MX = 16MY, vY/vX = √(MX/MY) = 4.
Q22
Correct +4 | Wrong -1 | Unattempted 0
If there are no heat losses, the heat released by the condensation of x g of steam at 100°C into water at 100°C can be used to convert y g of ice at 0°C into water at 100°C. Then ratio y : x is nearly
Official Solution:
Heat released by steam: 540x. Heat required by ice is 80y + 100y = 180y. Thus 540x = 180y, so y/x = 3.
Q23
Correct +4 | Wrong -1 | Unattempted 0
Three rods made of the same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at 0°C and 90°C respectively. The temperature of the junction of the three rods will be
Official Solution:
Let the junction temperature be x. Heat balance gives (90 - x)/R + (90 - x)/R = x/R. Therefore 180 - 2x = x, and x = 60°C.
Q24
Correct +4 | Wrong -1 | Unattempted 0
A mass at the end of a spring executes harmonic motion about an equilibrium position with an amplitude A. Its speed as it passes through the equilibrium position is 2V. If the spring is given a net extension 4A and released, the speed of the mass passing through the equilibrium position will be
Official Solution:
Given 2V = ωA. With amplitude 4A, the maximum speed is v' = ω(4A) = 8V.
Q25
Correct +4 | Wrong -1 | Unattempted 0
Some equipotential surfaces are shown in the figure. The magnitude and direction of the electric field is
Official Solution:
E = -∂V/∂r = (30 - 20)/(10 sin30° cm) = 200 V/m. The electric field is perpendicular to the equipotential surfaces and makes 120° with the x-axis.
Q26
Correct +4 | Wrong -1 | Unattempted 0
An electron is projected as in the figure with kinetic energy K, at an angle θ = 45° between two charged plates. The magnitude of the electric field (uniform) so that the electron just fails to strike the upper plate should be greater than
Official Solution:
For just failing to strike the upper plate, d = u2 sin245° / [2(qE/m)]. Using K = (1/2)mu2, d = K sin245°/(qE). Therefore E = K/(2ed), and the field should be greater than this value.
Q27
Correct +4 | Wrong -1 | Unattempted 0
In the circuit shown, the ratio I1 : I2 is equal to
Official Solution:
Solving the circuit gives I1 = I2. Therefore I1 : I2 = 1 : 1.
Q28
Correct +4 | Wrong -1 | Unattempted 0
In the ladder network shown, current through the resistor 3 Ω is 4 A. The emf V of the ideal battery is equal to
Official Solution:
From the reduced ladder network calculation shown in the official solution, V = 16 × 7 - 8 × 8 + 16 × 9 = 320 V.
Q29
Correct +4 | Wrong -1 | Unattempted 0
Three rings, each having equal radius R, are placed mutually perpendicular to each other and each having its center at the origin of coordinates system. If current I is flowing through each ring, then the magnitude of magnetic field at the common center is
Official Solution:
The fields due to the three perpendicular rings are mutually perpendicular: (μ0I/2R)i + (μ0I/2R)j + (μ0I/2R)k. The magnitude is therefore √3μ0I/(2R).
Q30
Correct +4 | Wrong -1 | Unattempted 0
A metallic rod of length L = 2 m and mass M = 2 kg is moving under the action of two unequal forces F1 = 10 N and F2 = 20 N (directed opposite to each other) acting at its ends along its length. Ignore gravity and any external magnetic field. If specific charge of electrons is α C/kg, then the potential difference between the ends of the rod in steady state must be
Official Solution:
a = (F2 - F1)/M = (20 - 10)/2 = 5 m/s2. In steady state, eE = mea. Since α = e/me, V/L = a/α. With L = 2 m, V = 10/α.
Q31
Correct +4 | Wrong -1 | Unattempted 0
A ring of resistance 20 Ω, radius 1 m and 50 turns is rotated at a rate 100 revolutions per second about its diameter perpendicular to a uniform magnetic field of induction 10 mT. The amplitude of the current in the loop will be nearly (Take: π2 = 10).
Official Solution:
i0 = NBAω/R. Substituting N = 50, B = 10 × 10-3 T, A = π, ω = 200π, and R = 20Ω gives i0 = 50 A.
Q32
Correct +4 | Wrong -1 | Unattempted 0
Three consecutive flash photographs of a travelling wave on a string are reproduced in the figure. The following observations are made. Which of the following options is correct? (Mass per unit length of the string = 6 g/cm.)
Official Solution:
From the figure, A = 0.5/2 = 0.25 m and λ = 2 m. The wave moves λ/4 = 0.5 m in 0.1 s, so v = 0.5/0.1 = 5 m/s. Therefore f = v/λ = 2.5/s.
Q33
Correct +4 | Wrong -1 | Unattempted 0
In Young's double slit experiment carried out with light of wavelength 6000 Å, the distance between the slits is 0.4 mm and the screen is at 250 cm from the plane of slits. The distance of the fifth order maxima from the centre of the screen will be
Official Solution:
y = 5λD/d. Substitution of λ = 6000 Å, D = 2.5 m, and d = 0.4 mm gives y = 1.87 cm.
Q34
Correct +4 | Wrong -1 | Unattempted 0
In the figure shown, the equivalent capacitance between A and B is
Official Solution:
The bridge is a balanced Wheatstone bridge, so the equivalent capacitance between A and B is 2 F.
Q35
Correct +4 | Wrong -1 | Unattempted 0
Ten tuning forks are arranged in increasing order of frequency in such a way that any two consecutive tuning forks produce 8 beats/s. The highest frequency is four times that of the lowest. Possible highest and lowest frequencies (in Hz) are
Official Solution:
If the lowest frequency is f0, then the highest is f0 + 9 × 8 = f0 + 72. Given highest frequency is 4f0, so f0 + 72 = 4f0. Hence f0 = 24 Hz and highest frequency is 96 Hz.
Q36
Correct +4 | Wrong -1 | Unattempted 0
An electron is orbiting around a nucleus in a circle of radius r. Magnetic moment of electron is proportional to
Official Solution:
μ = iA = eπr2/(2πr/v) = evr/2. For orbital motion, v ∝ 1/√r, hence μ ∝ r0.5.
Q37
Correct +4 | Wrong -1 | Unattempted 0
Energy levels A, B and C of a certain atom correspond to increasing values of energy, i.e. EA < EB < EC. If λ1, λ2 and λ3 are the wavelengths of radiations corresponding to transitions C to B, B to A and C to A respectively, then which of the following relations is correct?
Official Solution:
(EC - EB) + (EB - EA) = EC - EA. Therefore hc/λ1 + hc/λ2 = hc/λ3, so λ3-1 = λ1-1 + λ2-1.
Q38
Correct +4 | Wrong -1 | Unattempted 0
The power factor of the circuit shown in figure is
Official Solution:
Power factor cosφ = R/Z. Here net reactance is XL - XC = 60Ω, so Z = √(802 + 602) = 100Ω. Thus cosφ = 80/100 = 0.8.
Q39
Correct +4 | Wrong -1 | Unattempted 0
In an LR circuit current at t = 0 is 20 A. After 2 second current in the circuit decreased by 2 A. Time constant of the circuit is (in second)
Official Solution:
I = I0e-t/τ. At t = 2 s, 18 = 20e-2/τ. Hence e2/τ = 10/9 and τ = 2/ln(10/9).
Q40
Correct +4 | Wrong -1 | Unattempted 0
A faulty thermometer reads 5°C in melting ice and 99°C in steam. The correct temperature in F, when this faulty thermometer reads 52°C, is
Official Solution:
A 94°C interval on the faulty scale corresponds to 100°C actual. Above 5°C, the reading is 52 - 5 = 47°C faulty. Actual temperature is (100/94) × 47 = 50°C. Therefore TF = 32 + (9/5)50 = 122°F.
Q41
Correct +4 | Wrong -1 | Unattempted 0
First excitation potential of a hypothetical hydrogen-like atom is 15 V. Find third excitation potential of the atom.
Official Solution:
For a hydrogen-like atom, excitation energy to level n is proportional to (1 - 1/n2). First excitation is 15 = K(1 - 1/4) = 3K/4, so K = 20. Third excitation is K(1 - 1/16) = 20 × 15/16 = 18.75 V.
Q42
Correct +4 | Wrong -1 | Unattempted 0
A He+ ion is at rest and is in ground state. A neutron with initial kinetic energy K collides head on with the He+ ion. Among the following, the value of K so that there can be an inelastic collision between these two particles is
Official Solution:
For head-on collision, maximum transferable energy is 4K/5. For excitation, 4K/5 > 40.8 eV, so K > 51 eV. Therefore both 55 eV and 69 eV are possible.
Q43
Correct +4 | Wrong -1 | Unattempted 0
The rms value of electric field at a distance r from a source of electromagnetic wave of power P is (ε: permittivity of free space).
Official Solution:
Intensity at distance r is I = P/(4πr2). Since I = εErms2c, Erms = √[P/(4πr2cε)].
Q44
Correct +4 | Wrong -1 | Unattempted 0
Which among the following represents an OR gate?
Official Solution:
Option (2) represents an OR gate. It shows a NOR operation followed by inversion, which gives the OR output.
Q45
Correct +4 | Wrong -1 | Unattempted 0
The fraction f of radioactive material that has decayed in time t varies with time t. The correct variation is given by the curve
Official Solution:
The decayed fraction is f = 1 - e-λt. It increases with time and approaches a limiting value asymptotically, which corresponds to curve B.