A premium Physics-only practice page for serious NEET aspirants, built from the Physics section of the uploaded assessment paper with clickable answers, official solutions, question navigation, and NEET-style scoring.
If you really want to master Physics, your preparation should be built around three powerful books: S.C. Verma, Resnick Halliday, and I.E. Irodov. These books form the backbone of strong conceptual Physics. Most high-quality coaching modules, advanced practice sheets, and competitive examination questions are inspired, modified, or developed from the concepts found in these standard books. Therefore, students should not study Physics only for formula memorisation. They should focus on concept clarity, logical understanding, and proper application of ideas. Kumar Sir recommends that every serious NEET aspirant should keep these three books in mind while preparing Physics, because they help develop the depth of thinking required to solve difficult and unfamiliar questions confidently.
Why Strong Physics Preparation Is Now More Important Than Ever
Physics has become one of the strongest rank-deciding subjects in NEET because it rewards clear concepts, disciplined practice, and calm application under time pressure. Students who understand the ideas behind formulas can handle unfamiliar questions more confidently, while students who depend only on memorisation often struggle when a question is framed differently.
A strong Physics foundation improves problem-solving speed, accuracy, and confidence across the full syllabus. It also trains the mind to read data carefully, connect principles, and avoid careless mistakes. For aspirants aiming at top medical colleges, Physics preparation should therefore be regular, concept-based, and guided by high-quality question practice.
Important Message for NEET 2027 and Future Aspirants
Dear NEET Aspirants, your preparation journey should begin with patience, discipline, and honest self-analysis. The students who start early and build concepts step by step usually develop a much deeper command over Physics than those who rush at the end.
For NEET 2027 and future attempts, focus on understanding every law, graph, equation, and diagram from the root. Practice questions regularly, revise mistakes, and do not ignore difficult chapters. With the right guidance and consistent effort, Physics can become your scoring advantage instead of your fear.
Question Index
Q1+4 / -1 / 0
Change in wavelength is Δλ = -λ √2Ψmv2 cosθ. Here symbols have their usual notations (except of Ψ). Ψ can represent
Correct Answer: Option 1
Official PDF Solution: From the dimensional relation in the PDF: [Delta lambda] = [lambda] and cos theta is dimensionless. Therefore Psi/(mv^2) is dimensionless, so [Psi] = ML^2T^-2. Hence Psi represents kinetic energy.
Q2+4 / -1 / 0
You have a ring balanced at the center of the table by three forces F1, F2 and F3. The forces F1 and F2 have components: F1x = 10 N, F1y = -50 N, F2x = -40 N, F2y = 100 N. The force F3 is required to make the ring stationary at the center of the table. Which one of the four vectors in the diagram could represent F3?
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Correct Answer: Option 2
Official PDF Solution: For equilibrium, F1 + F2 + F3 = 0. Using the given components: 10i - 50j - 40i + F3 + 100j = 0, so F3 = 30i - 50j. This lies in the fourth quadrant, corresponding to vector B.
Q3+4 / -1 / 0
A rocket rises vertically up from the surface of earth so that its distance from the earth's surface is l = ct2 where c is a constant. After 10 sec the rocket has travelled 2 km. Determine its speed (in m/s) at that moment.
Correct Answer: Option 4
Official PDF Solution: Given l = ct^2 and l = 2000 m at t = 10 s, c = 20 m/s^2. Speed v = dl/dt = 2ct = 2 x 20 x 10 = 400 m/s.
Q4+4 / -1 / 0
If block A shown in figure is moving with speed v as observed by a person on ground then the magnitude of velocity of A with respect to block B is (the wedge is fixed)
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Correct Answer: Option 3
Official PDF Solution: The official solution uses relative velocity components: v_AB = sqrt(v^2 + v^2 + 2v^2 cos 60 deg) = v sqrt(3).
Q5+4 / -1 / 0
A block starts motion at t = 0 with constant acceleration a on a horizontal plane. After t0 second, the acceleration remains same in magnitude but reverses its direction. After what time from start does it come back to initial point?
Correct Answer: Option 3
Official PDF Solution: The velocity becomes zero at 2t0. The displacement during that time is s = 2 x (1/2) a t0^2 = a t0^2. In return, the same distance is covered with reversed acceleration, giving an additional time sqrt(2)t0. Total time = 2t0 + sqrt(2)t0 = (2 + sqrt(2))t0.
Q6+4 / -1 / 0
A cannon ball is thrown with an initial velocity of 40 m/s at an angle of 53° above the horizontal from ground. After some time, the ball is seen to travel at an angle of 37° below horizontal. Find the time elapsed (in second) till then.
Correct Answer: Option 3
Official PDF Solution: The PDF uses tan 37 deg = vy/24, giving vy = 18 m/s downward. With vertical motion, -18 = 32 - 10t, so 10t = 50 and t = 5 s.
Q7+4 / -1 / 0
Figure shows two swimmers starting from point A and B on opposite banks. They started at same instant with a constant velocity. Both of them start to swim in a direction parallel to line AB always. The river flows towards east.
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Correct Answer: Option 3
Official PDF Solution: The relative velocity between A and B remains along the line joining them. Therefore they collide, but the river flow shifts the collision point to the east of line AB.
Q8+4 / -1 / 0
A horizontal force F = 2 N is applied on the system shown in the figure. All surfaces are smooth. All pulleys and string are ideal. Mass of A and B each is 1 kg. What is the acceleration of B with respect to A?
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Correct Answer: Option 3
Official PDF Solution: Official result: Answer (3). From the pulley constraint for the given smooth system, the acceleration of B with respect to A is 2 m/s^2.
Q9+4 / -1 / 0
The system shown in figure is released from rest with spring at natural length. The spring starts extending as soon as it is released. (Neglect friction and masses of pulley, string and spring)
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Correct Answer: Option 4
Official PDF Solution: Just after release, the acceleration of mass M is g and the acceleration of mass m is zero. Hence the spring starts extending. This happens for any value of M.
Q10+4 / -1 / 0
A particle is moving on a circle of radius 1 m and its speed is changing as v = 2t. The magnitude of the acceleration of particle at t = 1 sec is
Correct Answer: Option 4
Official PDF Solution: At t = 1 s, v = 2 m/s. Tangential acceleration dv/dt = 2 m/s^2 and centripetal acceleration v^2/R = 4 m/s^2. Resultant acceleration = sqrt(2^2 + 4^2) = 2sqrt(5) m/s^2.
Q11+4 / -1 / 0
Two blocks of mass 2 kg and 3 kg are kept on level ground, having coefficient of static friction 0.3 and 0.4 respectively from ground. The blocks are connected by a spring of constant 100 N/m. What can be a possible extension in the spring so that both the blocks are in equilibrium? (g = 10 m/s2)
Correct Answer: Option 2
Official PDF Solution: For limiting equilibrium, kx = mu1 m1 g and kx = mu2 m2 g. The possible range includes x = 6 cm; hence 6 cm can be a possible extension.
Q12+4 / -1 / 0
A particle moving along the x axis is acted upon by a single force F = F0e-kx, where F0 and k are constants. The particle is released from rest at x = 0. It will attain a maximum kinetic energy of
Correct Answer: Option 1
Official PDF Solution: Maximum kinetic energy equals work done by the force from x = 0 to infinity: W = integral F0 e^(-kx) dx = F0/k.
Q13+4 / -1 / 0
A ball hits a smooth surface and rebounds with the same speed, as diagramed below. The changes in the components of the velocity of the ball are
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Correct Answer: Option 2
Official PDF Solution: The surface is smooth, so the x-component of velocity remains unchanged: Delta vx = 0. The y-component reverses direction, so Delta vy > 0.
Q14+4 / -1 / 0
A block of mass m placed on a smooth horizontal surface is attached to a spring and is held at rest by a force P as shown. Suddenly the force P changes its direction opposite to the previous one. Find the ratio l2/l1, where l2 is the maximum extension in the spring and l1 is the initial compression.
Kumar Sir +91 9958461445
Correct Answer: Option 1
Official PDF Solution: Initially P = kl1. Using energy after reversing P: P(l1 + l2) = (1/2)k l2^2 - (1/2)k l1^2. Substitution gives 3l1^2 + 2l1l2 - l2^2 = 0, so l2 = 3l1.
Q15+4 / -1 / 0
A ball A moving with certain velocity along positive x-axis collides with a stationary ball B. After collision their directions of motion make angles α and β with the x-axis. The possible values of α and β are
Correct Answer: Option 1
Official PDF Solution: Linear momentum of the system along the y-axis must remain zero, while along the x-axis it remains non-zero. The possible values are alpha = 30 deg and beta = -45 deg.
Q16+4 / -1 / 0
A metal sheet 14 cm x 2 cm of uniform thickness is cut into two pieces of width 1 cm each. The two pieces are joined and laid along XY plane as shown. The centre of mass has the coordinates.
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Correct Answer: Option 4
Official PDF Solution: Using centre of mass coordinates from the two pieces: y_c = (m x 7 + m x 1/2)/(2m) = 15/4 cm and x_c = (m x 8 + m x 0.5)/(2m) = 17/4 cm.
Q17+4 / -1 / 0
When the speed of a rear wheel-drive car is increasing on a horizontal road the direction of the frictional force on the tyres is
Correct Answer: Option 4
Official PDF Solution: For a rear wheel-drive car speeding up, friction is backward on the front tyres and forward on the rear tyres.
Q18+4 / -1 / 0
A horizontal rod of mass 'M' and length 'L' is tied to two vertical strings symmetrically as shown in the figure. One of the strings at end Q is cut at t = 0 and the rod starts rotating about the other end P. Then
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Correct Answer: Option 4
Official PDF Solution: Taking torque about P: Mg(L/2) = (ML^2/3) alpha, so alpha = 3g/(2L). The centre of mass acceleration is alpha L/2 = 3g/4 downward. Hence both statements (1) and (3) are correct.
Q19+4 / -1 / 0
L shaped narrow tube containing liquid of density ρ accelerating horizontally with acceleration g. The pressure at A will be (atmospheric pressure is P0)
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Correct Answer: Option 1
Official PDF Solution: For the accelerating liquid, PA a - PB a = (2ha)rho g and PB = rho gh. Hence PA - rho gh = 2rho gh, so PA = 3rho gh above atmospheric contribution as given in the option.
Q20+4 / -1 / 0
A metallic vessel of negligible heat capacity contains 0.5 kg of water (of specific heat capacity 4200 J kg-1 K-1). It is heated from 15°C to 85°C by 750 W immersion heater in 4 minutes. The average loss of heat in watts to the surroundings from the vessel during this time is nearly
Correct Answer: Option 3
Official PDF Solution: Heat supplied = 750 x 4 x 60 = 18 x 10^4 J. Heat gained by water = 0.5 x 4200 x 70 = 14.7 x 10^4 J. Heat loss = 3.3 x 10^4 J, so power loss = 3.3 x 10^4 / 240 = 137.5 W, nearly 138 W.
Q21+4 / -1 / 0
10 litre of water per second is lifted from well through 20 m and delivered with a velocity of 10 ms-1, then the power of the motor is
Correct Answer: Option 2
Official PDF Solution: Power = total energy/time = (mgh + 1/2 mv^2)/t = 10 x 10 x 20 + (1/2) x 10 x 10^2 = 2500 W = 2.5 kW.
Q22+4 / -1 / 0
A snapshot of a travelling wave is shown in figure-1 and a snapshot of a standing wave is shown in figure-2. (Particles are at their respective extremes. The amplitude, frequency, and the speed of both the waves are the same. The string in both cases is identical). If the energy per unit length of system at point A in figure-1 is x1 and that in figure-2 is x2, what is the value of x1/x2?
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Correct Answer: Option 1
Official PDF Solution: In figure 1, energy at A is both potential and kinetic. In figure 2, energy at A is only potential. Therefore x1/x2 = 2:1.
Q23+4 / -1 / 0
Two point sound sources S1 and S2 are both of same power and send out sound waves in the same phase. The wavelength of both the waves is 48/5 m. The intensity due to S2 alone at D is 25 W/m2. The resultant intensity at D is I. Find I (in W/m2).
Kumar Sir +91 9958461445
Correct Answer: Option 3
Official PDF Solution: Using interference intensity: I = I1 + I2 + 2sqrt(I1I2) cos(Delta theta). With I1 = 16 and I2 = 25, the PDF obtains I = 41 + 20 = 61 W/m^2.
Q24+4 / -1 / 0
The young's modulus of material of a thin ring shaped elastic body is Y. The mass of ring is m, area of cross section is A, its initial radius is R. Ring is a little elongated, then left alone. At what time will ring circumference be same as it was initially? (Neglect loss of energy)
Correct Answer: Option 1
Official PDF Solution: The official solution finds SHM of the ring radius variation: T = 2pi/omega, and the circumference first becomes the original value after T/4. This gives t = sqrt(pi mR/(8YA)).
Q25+4 / -1 / 0
Two masses are executing SHM under influence of a vertical spring as shown with an amplitude of 2 cm and angular frequency of 5 rad/s. When the extension of spring is maximum, suddenly the string breaks and 3 kg mass falls off. What is the maximum velocity of the 1 kg mass (in cm/s) during subsequent motion?
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Correct Answer: Option 1
Official PDF Solution: Initial extension x_i = A + 4g/k = 42 cm. With k = m omega^2 = 4 x 5^2 = 100 N/m, the new equilibrium extension for 1 kg is x0' = g/k = 10 cm. New amplitude = 42 - 10 = 32 cm and omega_new = 10 rad/s. Hence vmax = 320 cm/s.
Q26+4 / -1 / 0
Consider two infinitely long parallel uniformly charged non conducting wires with linear charged density λ and -2λ as shown. A charge hangs in equilibrium by means of an insulating chargeless thread between the wires. The mass of the charge is given by
Kumar Sir +91 9958461445
Correct Answer: Option 1
Official PDF Solution: The charge is in equilibrium, so the net electric force balances its weight. The official calculation gives m = 40 lambda q/(9 pi epsilon0 l g).
Q27+4 / -1 / 0
In a double slit experiment, two parallel slits are illuminated first by light of wavelength 400 nm and then by light of unknown wavelength. The fourth order dark fringe resulting from the known wavelength of light falls in the same place on the screen as the second order bright fringe from the unknown wavelength. The value of unknown wavelength of the light is
Correct Answer: Option 2
Official PDF Solution: For dark fringes, y_n = (n + 1/2)D lambda/d. Given the fourth order dark fringe for 400 nm coincides with the second order bright fringe of the unknown light, lambda = (7/4) x 400 = 700 nm.
Q28+4 / -1 / 0
Three charges are placed (fixed) at the vertices of an isosceles right angle triangle ABC with point charge +Q, +2Q and -Q placed at points A, B and C respectively. The magnitude of electric field strength at point D (mid point of AC) is √η KQ/a2. Find η.
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Correct Answer: Option 1
Official PDF Solution: Adding the electric field components at D from +Q, +2Q and -Q gives E = (32KQ/a^2) in magnitude form used in the question. Therefore eta = 32.
Q29+4 / -1 / 0
Consider a grounded conductor with a spherical cavity in it. A point charge Q is placed in it. The electric potential at centre of cavity is _______ volts. Given K = 9 x 109 Nm2/C2, a = 1 m, b = 1/2 m, Q = 1 x 10-9 C.
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Correct Answer: Option 4
Official PDF Solution: Using the expression for the grounded conductor cavity, V = KQ(1/b - 1/a) = 9 x 10^9 x 10^-9 x (2 - 1) = 9 V.
Q30+4 / -1 / 0
2 µC charge is shifted from P to Q. If work done by external force in this process is 160 x 10-6 joule, then potential difference between points P and Q i.e. VP - VQ is
Correct Answer: Option 4
Official PDF Solution: Work by external force is W_ext = -q(VP - VQ). Thus 160 x 10^-6 = 2 x 10^-6(VQ - VP), so VQ - VP = 80 V and VP - VQ = -80 V.
Q31+4 / -1 / 0
A hollow conducting sphere of inner radius R and outer radius 2R is given a charge Q as shown in the figure, then the
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Correct Answer: Option 4
Official PDF Solution: Charge resides on the outer surface of the conductor, so the potential inside the conducting region and cavity is constant. Hence potentials at A, B, C and O are the same.
Q32+4 / -1 / 0
Two light bulbs shown in the circuit have ratings A (24 V, 48 W) and B (24 V and 36 W) as shown. When the switch is closed
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Correct Answer: Option 4
Official PDF Solution: For bulbs, P = V^2/R, and PB < PA, so RB > RA. Initially VA = 24 x 3/7 < 12 V and VB = 24 x 4/7 > 12 V. After closing the switch, VA = VB = 12 V; hence A becomes brighter and B dimmer.
Q33+4 / -1 / 0
Three copper rods are subjected to different potential difference. Compare the drift speed of electrons through them. Assume that all 3 are at the same temperature.
Length
Diameter
Potential difference
a.
L
3d
V
b.
2L
d
2V
c.
3L
2d
2V
Correct Answer: Option 1
Official PDF Solution: Drift speed is proportional to E = V/L. For the three rods, v1 = V/L, v2 = 2V/(2L) = V/L, and v3 = 2V/(3L). Thus v1 = v2 > v3.
Q34+4 / -1 / 0
On quadrupling (making four times) the resistance 'R' connected to a battery the power dissipated in the resistance surprisingly does not change. What is the internal resistance of the battery in Ohm? (Take R = 10 Ω)
Correct Answer: Option 3
Official PDF Solution: For unchanged power when R becomes nR, the official relation gives r^2 = nR^2. With n = 4 and R = 10 ohm, r = sqrt(4)R = 20 ohm.
Q35+4 / -1 / 0
In the given circuit, ratio of charge on capacitor 2C and capacitor C in steady state is
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Correct Answer: Option 2
Official PDF Solution: In steady state, the potential difference across each capacitor is the same, V/2. Since charge Q = CV, the ratio of charge on 2C to C is 2:1.
Q36+4 / -1 / 0
Two particles having the same specific charge enter a uniform magnetic field with the same speed but at angles of 30° and 60° with the field. Let a, b and c be the ratios of their pitches, radii and periods of their helical paths respectively, then
Correct Answer: Option 4
Official PDF Solution: For a uniform helical path, T = 2pi m/qB, pitch = 2pi mv cos theta/qB, and R = mv sin theta/qB. Comparing 30 deg and 60 deg gives ab = c.
Q37+4 / -1 / 0
Bohr model is applied to singly ionized helium He+. Consider the lines that lie in the visible region of the spectrum. Which of the following transitions can give a photon of visible light?
Correct Answer: Option 2
Official PDF Solution: For He+, Z = 2 and 1/lambda = RZ^2(1/n^2 - 1/m^2). Visible hydrogen-like lines correspond here to transitions ending at n = 4, including m = 7 to n = 4.
Q38+4 / -1 / 0
From a cylinder of radius R, a cylinder of radius R/2 is removed, as shown. Current flowing in the remaining cylinder is I. Magnetic field strength is
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Correct Answer: Option 4
Official PDF Solution: The remaining conductor is treated as a superposition of a full cylinder and an oppositely directed removed cylinder. The PDF obtains B at A = mu0 I/(3pi R) and B at B = mu0 I/(3pi R), so both statements 2 and 3 are correct.
Q39+4 / -1 / 0
A uniform magnetic field exists in region given by B = 3î + 4ĵ + 5k̂. A rod of length 5 m which is placed along y-axis is moved along x-axis with constant speed 1 m/sec. Then induced e.m.f. in the rod will be
Correct Answer: Option 2
Official PDF Solution: Motional emf e = (v x B).l. With v along i-hat and B = 3i + 4j + 5k, dotting with a 5 m rod along j gives e = 25 V.
Q40+4 / -1 / 0
A transparent cylinder has its right half polished so as to act as a mirror. A paraxial light ray is incident from left, that is parallel to principal axis, exits parallel to the incident ray as shown. The refractive index n of the material of the cylinder is
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Correct Answer: Option 4
Official PDF Solution: Using refraction at the spherical surface, n2/v - n1/u = (n2 - n1)/R. With the ray emerging parallel as shown, the PDF gives n = 2.
Q41+4 / -1 / 0
A glass hemisphere of refractive index 4/3 and of radius 4 cm is placed on a plane mirror. A point object is placed at distance 'd' on axis of this sphere as shown. If the final image is at infinity, then find the value of 'd' in cm.
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Correct Answer: Option 1
Official PDF Solution: The object first forms an image at I by refraction at the plane surface with AI = (4/3)d. That image acts for the curved surface and mirror sequence. Solving the final infinity condition gives d = (3/4)R = 3 cm.
Q42+4 / -1 / 0
All electrons ejected from a surface by incident light of wavelength 200 nm can be stopped before travelling 1 m in the direction of uniform electric field of 4 N/C. The work function of the surface is
Correct Answer: Option 4
Official PDF Solution: Stopping work by the electric field gives (1/2)mvmax^2 = eEd = 4 eV. Photon energy for 200 nm is 1240/200 = 6.2 eV. Therefore work function phi0 = 6.2 - 4 = 2.2 eV.
Q43+4 / -1 / 0
The rate of disintegration of a radioactive substance falls from 800 decay/min to 100 decay/min in 6 hours. The half-life of the radioactive substance is
Correct Answer: Option 2
Official PDF Solution: Activity follows A = A0 e^(-lambda t). Since 100 = 800e^(-lambda x 6 h), the activity falls by 1/8 = (1/2)^3 in 6 hours. Thus the half-life is 2 hours.
Q44+4 / -1 / 0
When a battery is connected to a P-type semiconductor with a metallic wire, the current in the semiconductor (predominantly), inside the metallic wire and that inside the battery respectively due to
Correct Answer: Option 1
Official PDF Solution: The main charge carriers are holes in the P-type semiconductor, electrons in the metallic wire, and ions inside the battery.
Q45+4 / -1 / 0
A parallel plate capacitor with plate area A and separation between the plates d, is charged by a constant current i. Consider a plane surface of area A/2 parallel to the plates and drawn symmetrically between the plates, the displacement current through this area will be
Correct Answer: Option 2
Official PDF Solution: If the capacitor charge is Q, E = Q/(epsilon0 A). Through an area A/2, flux is phi_E = Q/(2epsilon0). Differentiating gives displacement current I/2.
