A focused 45-question NEET Physics practice page with clickable options, official answer matching, PDF-based diagram crops, solutions, and NEET style scoring.
Premium Physics Practice for Serious NEET Preparation
This assessment is designed for students who want to check their Physics preparation with discipline and honesty. Attempt each question before opening the answer or solution, and treat the paper as a timed online-style practice test.
The paper covers important Class 11 and Class 12 Physics ideas, including mechanics, heat, waves, electricity, magnetism, optics and modern physics. The aim is not only to know the answer, but to understand the reasoning behind it.
Why Strong Physics Preparation Is Now More Important Than Ever
Physics is becoming increasingly decisive for NEET aspirants because it rewards conceptual clarity, numerical discipline and the ability to connect ideas across chapters. Students who depend only on memorised formulas often struggle when the same concept is presented in a new format.
A strong Physics foundation helps you read questions faster, identify the correct principle, avoid calculation traps and manage exam pressure with confidence. Regular practice with properly matched solutions is one of the best ways to build that accuracy.
Use this paper as a diagnostic test. Mark the questions that feel slow, revise the related theory, and then reattempt similar problems until the method becomes natural.
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.
You should attempt this paper in the same way you would attempt an online test. Sit with full concentration, use a timer, read each question carefully, select your answer honestly, and then check the solution only after attempting the question yourself.
Before solving these questions, revise the relevant Physics notes properly. You can search on Google for "Physics Notes by Kumar Sir" and study the notes carefully. After reading the concepts, attempt this paper seriously. You will find that these questions help you understand your real preparation level.
If you are searching for a Physics Tutor and still finding Physics difficult to understand, you may contact Kumar Sir for one-to-one online Physics guidance.
Dimensional formula for (h2ν2/mc2) / (1 + hν/mc2) is, where h is Planck's constant, ν is frequency of light, m is mass and c is speed of light.
Correct Answer: Option 3
Official Solution
hν/mc2 is dimensionless. Therefore the required dimensional formula is that of h2ν2/mc2, which is energy: [ML2T-2].
Question 2
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Area of a square plate is (100 ± 2) m2. Its side is
Correct Answer: Option 1
Official Solution
A = l2, so l = 10 m. Since 2Δl/l = ΔA/A, Δl = (2 × 10)/(2 × 100) = 0.1 m. Side with error = (10 ± 0.1) m.
Question 3
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If a = (î + 2ĵ + 2k̂) and b = (2î + ĵ + 2k̂), the projection of b on a is
Correct Answer: Option 1
Official Solution
Projection of b on a = (b · â)â. Here |a| = 3 and b · a = 8. Hence projection = 8/9 (î + 2ĵ + 2k̂).
Question 4
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Figure shows position(x)-time(t) graph for two boys going from their homes to school. Which statement is true about their relative velocity, when both are in motion?
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Correct Answer: Option 4
Official Solution
Velocity of both boys is different and constant. Therefore their relative velocity is non-zero and constant.
Question 5
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A body is moving down a long inclined plane of inclination θ. The coefficient of friction between the body and inclined plane varies as μ = 0.5x, where x is distance travelled down the plane. The body will have maximum velocity when it has travelled a distance x given by (all quantities are in SI units)
Correct Answer: Option 1
Official Solution
When velocity is maximum, acceleration is zero. So g sinθ - 0.5x g cosθ = 0. Hence x = sinθ/(0.5 cosθ) = 2 tanθ.
Question 6
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The moment of inertia of a square sheet of side l and mass per unit area μ about an axis passing through one corner and perpendicular to its plane is
Correct Answer: Option 4
Official Solution
Mass of sheet M = μA = μl2. Moment of inertia about the corner = Ml2/6 + Ml2/2 = 2μl4/3.
Question 7
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A particle of mass m strikes elastically with a disc of same mass with speed v as shown in figure. All surfaces of contact are smooth. The velocity of disc just after collision is (take tangential line as +x-axis and radial line toward C as +y-axis at the point of impact).
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Correct Answer: Option 1
Official Solution
Conserve linear momentum along the normal. Since particle and disc have equal mass, momentum is exchanged along the normal. vD = v cosθ ĵ. From the figure sinθ = 1/2, so θ = 30°. Thus vD = (√3/2)v ĵ.
Question 8
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A pendulum consists of a wooden bob of mass m and a string of length l. A bullet of mass m1 is fired towards the pendulum with velocity v1. The bullet comes out of the bob with speed v1/4 and the bob just completes the vertical circle. The velocity v1 is
Correct Answer: Option 3
Official Solution
Using conservation of linear momentum: m1v1 = m√(5gl) + m1v1/4. Hence (3/4)m1v1 = m√(5gl), so v1 = (4m/3m1)√(5gl).
Question 9
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A body of mass M and radius R, rolling on a smooth horizontal surface with velocity v, rolls up an inclined plane up to a vertical height 3v2/4g. The shape of body is
Correct Answer: Option 1
Official Solution
Apply conservation of energy: KT + KR = Mgh. Thus (1/2)Mv2(1 + K2/R2) = (3/4)Mv2, giving K2/R2 = 1/2, which corresponds to a disk.
Question 10
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The escape velocity from a spherical planet is ve. The escape velocity corresponding to another planet of twice the radius and half the mean density is
Correct Answer: Option 1
Official Solution
Escape velocity ve = √(2GM/R) and, for a sphere, ve ∝ R√ρ. With radius 2R and density ρ/2, the new escape velocity is ve√2.
Question 11
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A glass beaker having mass 390 g and an interior volume of 500 cm3 floats just submerged in water when it is half filled with water. The density of material of beaker is approximately
Correct Answer: Option 2
Official Solution
Total floating mass = 390 + 250 = 640 g, so outer volume = 640 cm3. Volume of beaker material = 640 - 500 = 140 cm3. Density = 390/140 = 2.8 g/cm3.
Question 12
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A cylindrical vessel filled with water up to a height of 2 m stands on a horizontal plane. The side wall has a small circular hole touching the bottom. The velocity of water coming out from the hole is (g = 10 m/s2)
Correct Answer: Option 4
Official Solution
Using Torricelli's theorem, v = √(2gh) = √(2 × 10 × 2) = 2√10 m/s.
Question 13
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A ring is cut from a platinum tube of internal diameter 8.5 cm and external diameter 8.7 cm. It is supported horizontally on water surface. If an extra force of 4.0 g wt is required to pull it away from water surface, then surface tension of water is (g = 10 m/s2)
Correct Answer: Option 2
Official Solution
Force due to surface tension F = T(2πr1 + 2πr2). Therefore T = (4 × 10-3 × 10)/(3.14 × (8.5 + 8.7) × 10-2) = 7.4 × 10-2 N/m.
Question 14
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The temperature of 200 g of water is to be raised from 24°C to 90°C by adding steam at 100°C. The mass of steam required for this purpose is
Correct Answer: Option 2
Official Solution
Heat lost by steam = heat gained by water. m × 540 + m × 1(100 - 90) = 200 × 1(90 - 24). Thus 550m = 13200, so m = 24 g.
Question 15
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If λm denotes the wavelength at which the radiative emission from a black body at temperature T K is maximum, then
0.014 kg of nitrogen is enclosed in a vessel at a temperature of 27°C. The amount of heat transferred to the gas to double the r.m.s. speed of molecules is nearly
Correct Answer: Option 1
Official Solution
The vessel keeps volume constant, so Q = nCvΔT. Since vrms ∝ √T, doubling speed makes T2 = 1200 K from T1 = 300 K. Q = (1/2) × 5 × 2 × 900 = 2250 calorie.
Question 17
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If 90 cal of heat is required to raise the temperature of 2 mole of an ideal gas at constant pressure from 30°C to 35°C, the degree of freedom of the gas molecule is
Correct Answer: Option 2
Official Solution
ΔW = nRΔT = 2 × 2 × 5 = 20 cal. ΔU = Q - ΔW = 90 - 20 = 70 cal. From γ = Cp/Cv = 9/7 and γ = 1 + 2/f, the degree of freedom f = 7.
Question 18
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The temperature inside and outside a refrigerator are 273 K and 303 K respectively. Assume that refrigerator cycle is reversible. The heat rejected outside for every joule of work is approximately
Correct Answer: Option 2
Official Solution
For the reversible refrigerator, Q2/W = T2/(T1 - T2) = 273/30 ≈ 9. Heat rejected outside = Q2 + W ≈ 9 + 1 = 10 J.
Question 19
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Assume a sitar string is plucked at about 1/6 of its length from one end. The most prominent harmonic will be
Correct Answer: Option 3
Official Solution
The official solution states that the most prominent harmonic is the third harmonic.
Question 20
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21 tuning forks are arranged such that every fork gives 5 beats per second with the next. The last fork has a frequency that is double of the first. Frequency of 11th fork in Hz is
Correct Answer: Option 3
Official Solution
For 21 forks there are 20 intervals. 2n - n = 20 × 5, so n = 100. Therefore n11 = n1 + 10 × 5 = 150 Hz.
Question 21
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When an oscillator completes 50 oscillations, its amplitude reduces to 1/3 times of the initial value. The amplitude, when it completes 100 oscillations, will be
Correct Answer: Option 2
Official Solution
A = A0e-bt/2m. After 50 oscillations A/A0 = 1/3. After 100 oscillations, the factor is squared: A/A0 = (1/3)2 = 1/9.
Question 22
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A particle having charge equal to that of electron and mass 1.6 × 10-30 kg is projected with initial speed v at an angle of 45° with the lower plate as shown. The plates are sufficiently long and have separation 2 cm. Electric field between the plates is 0.5 × 103 V/m upward. The maximum value of velocity of particle not to hit the upper plate is
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Correct Answer: Option 2
Official Solution
For the particle just not to hit the upper plate, vy becomes zero at the upper plate. Use v2 = u2 + 2as: 0 = (u/√2)2 - 2(qE/m)(2 × 10-2). This gives u2 = 4 × 1012 and u = 2 × 106 m/s.
Question 23
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In the given circuit diagram E = 5 V, r = 1 Ω, R1 = R3 = 1 Ω, R2 = 4 Ω and C = 3 μF. The numerical value of charge on each capacitor is
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Correct Answer: Option 3
Official Solution
Current i = E/(R2 + r) = 5/(4 + 1) = 1 A. Terminal voltage V = E - ir = 5 - 1 = 4 V. Effective capacitance is 1.5 μF, so q = CeffV = 1.5 × 4 = 6 μC.
Question 24
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The capacitance of capacitor of plate area A1 and A2 (A1 < A2) at a distance d as shown in the figures is
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Correct Answer: Option 2
Official Solution
Only the overlapping effective plate area contributes. Since A1 < A2, effective area = A1. Hence C = ε0A1/d.
Question 25
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As the switch S is closed in the circuit shown in figure, the current passed through point A is
Kumar Sir +91 9958461445
Correct Answer: Option 1
Official Solution
Applying KVL in POQ: 20 - 2i - 4(i - i1) = 5, so 6i - 4i1 = 15. Applying KVL in POE: 20 - 2i - 2i1 = 0, so i + i1 = 10. Solving gives i1 = 4.5 A.
Question 26
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Resistance of a conductor at temperature t°C is R = R0(1 + at + bt2). Here R0 is resistance at 0°C. The temperature coefficient of resistance at temperature t°C is
Correct Answer: Option 1
Official Solution
Temperature coefficient α = (1/R)(dR/dt). With R = R0(1 + at + bt2), dR/dt = R0(a + 2bt). Hence α = (a + 2bt)/(1 + at + bt2).
Question 27
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A potentiometer wire, 10 m long, has a resistance of 40 Ω. It is connected in series with a resistance box and a 2 V storage cell. If the potential gradient along the wire is 0.1 mV/cm, the resistance unplugged in the box is
Correct Answer: Option 2
Official Solution
Potential gradient K = 0.1 × 10-3/10-2 = 0.01 V/m. Since K = i(R/L), 0.01 = 4i, so i = 0.25 × 10-2 A. Also i = 2/(40 + R). Hence 40 + R = 800 and R = 760 Ω.
Question 28
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If E denotes electric field in a uniform conductor and vd the drift velocity of free electrons in the conductor, then which of the following graph is correct?
Correct Answer: Option 1
Official Solution
Drift velocity vd = eEτ/m. Therefore vd ∝ E, so the correct graph is the straight-line proportional graph.
Question 29
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A conducting circular loop of same material is placed as shown in figure. The area of cross section of part APC is A and of AQC is A/3. The magnetic field at centre O is
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Correct Answer: Option 1
Official Solution
The resistances of the two paths become equal, so current divides equally: i1 = i2 = i/2. The magnetic fields due to the two arcs are opposite and unequal, giving net B = μ0i/8R.
Question 30
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An electron and proton having same kinetic energy enter into a magnetic field perpendicular to it, then
Correct Answer: Option 2
Official Solution
Radius of path r = √(2Km)/(qB), so r ∝ √m for equal kinetic energy and equal magnitude of charge. Proton has greater mass, hence larger radius and less curved path.
Question 31
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Figure shows a square current carrying loop ABCD of side 2 m and current 1/2 A. The magnetic moment M of the loop is
Kumar Sir +91 9958461445
Correct Answer: Option 1
Official Solution
Magnetic moment M = iA vector. From the loop geometry, M = (1/2)[4cos60° î - 4sin60° k̂] = (î - √3 k̂) A m2.
Question 32
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An ammeter is obtained by shunting a 30 Ω galvanometer with a 30 Ω resistance. The additional shunt required to double the range is
Correct Answer: Option 1
Official Solution
Initially, the 30 Ω galvanometer and 30 Ω shunt share current equally. For double range, (2i - i/2)S = (i/2) × 30, so S = 10 Ω. Since 10 Ω is made by 30 Ω in parallel with additional shunt S', 1/10 = 1/30 + 1/S', giving S' = 15 Ω.
Question 33
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The coercivity of a bar magnet is 120 A/m. It is to be demagnetised by placing it inside a solenoid having number of turns per unit length 60 m-1. The current flowing through the solenoid is
Correct Answer: Option 3
Official Solution
For a solenoid, H = ni. Therefore i = H/n = 120/60 = 2.0 A.
Question 34
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A conducting rod of length l is hinged at one end and is free to rotate in a vertical plane. A uniform magnetic field B exists perpendicular to the plane of rotation. The rod is released from horizontal position. The potential difference between the two ends of rod when rod makes an angle θ with horizontal is proportional to
Correct Answer: Option 4
Official Solution
Loss of potential energy equals gain in rotational kinetic energy: mgl sinθ = (1/2)Iω2. This gives ω ∝ √(sinθ). Motional emf ΔV = (1/2)Bωl2, so ΔV ∝ (sinθ)1/2.
Question 35
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A capacitor, an inductor and an electric bulb are connected in series to an AC supply of variable frequency. As the frequency of supply is increased gradually, the electric bulb is found to be
Correct Answer: Option 3
Official Solution
At resonant frequency power is maximum; after resonance it starts decreasing. Therefore brightness first increases, reaches a maximum, and then decreases.
Question 36
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White light is incident on the interface of glass and air as shown in figure. If green light is just totally internally reflected, then the emerging rays in air contain
Kumar Sir +91 9958461445
Correct Answer: Option 1
Official Solution
Since μ ∝ 1/λ and sin C ∝ λ, the incidence angle is the critical angle for green. It is less than critical for yellow, orange and red, so those colours refract into air.
Question 37
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Rays from a lens are converging towards a point P as shown in figure. How much thick glass plate having refractive index 1.6 must be located between lens and point P, so image is formed at P'?
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Correct Answer: Option 1
Official Solution
Shift Δx = t(1 - 1/μ). Given Δx = 0.3 cm and μ = 1.6, 0.3 = t(1 - 1/1.6). Therefore t = 0.8 cm.
Question 38
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Refractive index of material of prism is μ = √2 and the angle of prism is 60°. The angle of incidence for minimum deviation is
Correct Answer: Option 2
Official Solution
For minimum deviation, r = A/2 = 30°. Using μ = sin i/sin r, √2 = sin i/sin30°. Hence i = 45°.
Question 39
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The diameter of objective lens of a telescope is 0.61 m. The wavelength of light used is 5000 Å. The resolving power of telescope is
The maximum number of possible interference maxima for slit separation equal to 2.5 times the wavelength in Young's double slit experiment is
Correct Answer: Option 2
Official Solution
For maximum, d sinθ = nλ. With d = 2.5λ, sinθ = n/2.5. Allowed integral n values are -2, -1, 0, 1, 2, so five maxima are obtained.
Question 41
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The threshold frequency for a certain metal is ν0. When light of frequency 2ν0 is incident on it, the maximum velocity of photoelectron is 4 × 106 m/s. If the frequency of incident radiation is increased to 5ν0, then the maximum velocity of photoelectrons will be
In the Bohr model of hydrogen atom, let R, v and E represent radius of orbit, speed of electron and energy of electron respectively. Which of the following quantities is proportional to quantum number n?
Correct Answer: Option 4
Official Solution
Angular momentum mvR = nh/(2π). Hence vR ∝ n.
Question 43
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Two radioactive materials X1 and X2 contain same number of nuclei. If 6λ s-1 and 4λ s-1 are the decay constants of X1 and X2 respectively, the ratio of number of nuclei undecayed of X1 to that of X2 will be 1/e after a time
Correct Answer: Option 1
Official Solution
N1 = N0e-6λt and N2 = N0e-4λt. Therefore N1/N2 = e-2λt = e-1. Hence 2λt = 1 and t = 1/(2λ).
Question 44
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In the circuit shown in the figure, the current through ideal diode is
Kumar Sir +91 9958461445
Correct Answer: Option 4
Official Solution
For an ideal diode, forward resistance is zero. No current passes through the 20 Ω branch. Thus i = E/R = 2/80 = 0.025 A = 25 mA.
Question 45
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To get an output y = 1 from the circuit shown, the input A, B and C must be respectively
Kumar Sir +91 9958461445
Correct Answer: Option 3
Official Solution
From the logic circuit, Y = (A + B) · C. To get Y = 1, C must be 1 and one or both of A and B should be 1. Therefore A, B, C = 1, 0, 1 works.
