Complete Physics practice page from the uploaded assessment paper, covering only Physics questions Q1 to Q45 with clickable options, official answer mapping, solutions, NEET marking and final scoring.
Physics Q1-Q45Correct +4Wrong -1Full-width Elementor HTML
Physics can be the subject that carries a serious NEET aspirant across the finish line. When concepts, formula application and numericals are revised with discipline, marks improve quickly and confidence becomes stable. If Physics is still creating doubt, confusion or fear, contact Kumar Sir for clear, step-by-step and exam-oriented guidance.
Why Strong Physics Preparation Is Now More Important Than Ever
NEET Physics rewards clarity, speed and accuracy. Students who revise formulas but also understand when and how to apply them can solve numerical questions with far fewer mistakes. A strong Physics score can separate a prepared aspirant from the crowd because every correct calculation directly improves rank.
Important Message for NEET 2027 ,2028,2029,2030 and Future Aspirants
Future NEET aspirants should start Physics early and practise consistently. Class 11 and Class 12 chapters are connected, so weak basics in motion, energy, waves, electricity or modern physics can affect later topics. Build habits of formula revision, timed practice, error analysis and solution reading from the beginning.
Kumar Sir Physics Guidance
If you are searching for a Physics Tutor for NEET and still facing difficulty in concepts, doubt-solving, numericals, or formula application, you may contact Kumar Sir. Kumar Sir explains Physics topics in a very clear, step-by-step and exam-oriented way.
Important Formula Revision: all physics of class 11 and 12
Before starting the question paper, revise the most important formulas from Class 11 and Class 12 Physics. The quick cards below are designed for fast NEET revision, especially for calculation-heavy areas where students often know the formula but lose marks in application.
Important Formula Revision for NEET Physics: all physics neet
Focus chapters for immediate revision: kinetic theory, oscillation and wave. All formulas are shown in bold styling so students can scan them quickly before solving the paper.
Kinematics
v = u + at s = ut + (1/2)at2 v2 = u2 + 2as Range = u2sin2θ/g
Laws, Work, Power
F = ma p = mv W = Fscosθ K = (1/2)mv2 P = dW/dt = Fv
Rotation
τ = Iα L = Iω Krot = (1/2)Iω2 v = Rω
Gravitation
F = GMm/r2 g = GM/R2 V = -GM/r ve = √(2GM/R)
Properties of Matter
Y = stress/strain P = F/A v = √(Y/ρ) Surface energy = TΔA
F = -kx x = A sin(ωt + φ) vmax = Aω T = 2π√(m/k) Tpendulum = 2π√(l/g)
Waves
v = fλ y = A sin(ωt - kx) vstring = √(T/μ) fbeat = |f1 - f2| β = 10log(I/I0)
Electrostatics
F = kq1q2/r2 E = F/q V = kq/r C = Q/V U = (1/2)CV2
Current Electricity
V = IR R = ρL/A P = VI = I2R vd = I/neA ε = V + Ir
Magnetism, EMI, AC
F = qvBsinθ F = BILsinθ Φ = BAcosθ ε = -dΦ/dt Z = √[R2 + (XL-XC)2]
Optics
1/f = 1/v - 1/u m = h'/h = v/u μ = sin i/sin r β = λD/d Power = 1/f
Modern Physics
E = hf = hc/λ eVs = hf - φ rn = r0A1/3 N = N0e-λt
Semiconductors
IE = IB + IC β = IC/IB α = IC/IE gm = ΔIC/ΔVBE
Students preparing for NEET must revise all physics of class 11 and class 12 very carefully because these chapters often involve calculation, formula selection and conceptual confusion. Many students know the formulas but make mistakes while applying them in numerical questions. This question sheet has been designed to help students practise these topics in a focused and exam-oriented way. Every future NEET aspirant should attempt these questions seriously, check the solution carefully, and identify weak areas before the final exam.
Question Index
Status: cream = not attempted, gold = selected, green = correct, red = wrong
Q1+4 / -1
The dimensional formula of electrical conductivity is
Correct Answer: Option 2
Official solution: V/I = resistance, and conductivity is the reciprocal of resistivity. Using the dimensional relation gives σ = M-1L-3T3A2.
Q2+4 / -1
The speed-time (v-t) graph for a particle in straight line motion is as shown. Average speed from time t = 0 to t = 4 s
Kumar Sir +91 9958461445
Correct Answer: Option 2
Official solution: Total distance from 0 to 4 s is 50 m. Therefore vav = 50/4 = 12.5 m/s.
Q3+4 / -1
The position of a particle along a straight line is denoted by the equation x = 6 + 12t - t3. Here x is in metre and t is in second. The magnitude of acceleration of the particle when it is at rest is
Correct Answer: Option 2
Official solution: v = 12 - 3t2. At rest, v = 0, so t = 2 s. Also a = -6t, hence |a| = 12 m/s2.
Q4+4 / -1
An object of mass 3 kg is at rest. If a force F = 3t2î + 9ĵ (N) acts on the object, then velocity of the object at t = 2 second (in m/s) is
Correct Answer: Option 4
Official solution: Acceleration is a = t2î + 3ĵ. Integrating, v = (t3/3)î + 3tĵ. At t = 2 s, v = (8/3)î + 6ĵ.
Q5+4 / -1
A bullet of mass 5 gram leaves a rifle of mass 20 kg with a speed of 1000 m/s and strikes a wall at same level with a speed of 500 m/s at a distance of 50 m from rifle. The work done by bullet in overcoming the air resistance is
Correct Answer: Option 1
Official solution: Work lost against air resistance equals loss in kinetic energy: W = (1/2)(5 x 10-3)[10002 - 5002] = 1875 J.
Q6+4 / -1
On the surface of a planet of mass M and radius R, the acceleration due to gravity is g. If we go on another planet of same density but half the radius, then on the new planet the acceleration due to gravity
Correct Answer: Option 4
Official solution: For same density, g = (4π/3)ρGR, so g is directly proportional to R. If radius becomes R/2, gravity decreases by g/2.
Q7+4 / -1
Masses are kept at the corners of a regular pentagon as shown. Distance between centroid and any vertex of the pentagon is R. If another point mass m0 is kept at the centroid of the pentagon then force acting on m0 due to masses is
Kumar Sir +91 9958461445
Correct Answer: Option 1
Official solution: Vector addition of forces at the centroid gives a resultant equal in magnitude to GMm0/R2 along the required direction.
Q8+4 / -1
From a uniform circular disc of mass M and radius R, a small circular disc of radius R/3 has been cut out as shown. The centre of mass of the original disc shifts by a distance
Kumar Sir +91 9958461445
Correct Answer: Option 4
Official solution: Treat the removed disc as negative mass. X = [(-M/9)(2R/3)]/(M - M/9) = R/12.
Q9+4 / -1
A thin rod of mass M and length L is bent at its mid point at 110°. Moment of inertia of this bent rod about an axis perpendicular to the plane and passing through mid-point as shown is
Kumar Sir +91 9958461445
Correct Answer: Option 2
Official solution: Each half contributes about the bending point. Adding both halves gives I = ML2/24 + ML2/24 = ML2/12.
Q10+4 / -1
A solid cubical block of length 10 cm is put in liquid A of density 1.2 g/cm3 floating over another liquid B of density 7.2 g/cm3. If the density of cubical block is 3.6 g/cm3, then the ratio of length of block immersed in liquid A to liquid B is
Correct Answer: Option 3
Official solution: If x is the fraction in liquid A, then 1.2x + 7.2(1 - x) = 3.6. Thus x = 3/5 and 1 - x = 2/5, so the ratio is 3 : 2.
Q11+4 / -1
Two isochoric lines for volume V1 and V2 are shown on pressure-temperature (P-T) diagram below. It can be concluded that
Kumar Sir +91 9958461445
Correct Answer: Option 1
Official solution: From PV = nRT, slope of P-T graph is nR/V. The steeper line has smaller volume, hence V1 < V2.
Q12+4 / -1
An ideal monatomic gas at 27°C is compressed suddenly to 8/27 of its original volume adiabatically. The rise in temperature of the gas is
Correct Answer: Option 2
Official solution: For adiabatic compression of a monatomic gas, TVγ-1 = constant. Thus T = 675 K and rise = 675 - 300 = 375 K.
Q13+4 / -1
If the length of simple pendulum is increased by 1.44%, then its time period will
Correct Answer: Option 4
Official solution: ΔT/T = (1/2)(ΔL/L). Therefore increase in time period = 0.72%.
Q14+4 / -1
A standing wave having 3 nodes and 2 antinodes is formed in a string of length one metre. The wavelength of standing wave is
Correct Answer: Option 3
Official solution: With 3 nodes and 2 antinodes, the string contains two loops, so the length equals one wavelength. Hence λ = 1 m.
Q15+4 / -1
The equation of a plane progressive wave is given by y = 5 sin[2π(t/10 - x/18) + π/4], where x and y are in cm and t is in second. Now which of the following is true?
Correct Answer: Option 4
Official solution: Comparing with the standard wave equation gives frequency = 0.1 cps, wavelength = 18 cm = 0.18 m and velocity = 18/10 = 1.8 cm/s. Therefore all statements are true.
Q16+4 / -1
Two concentric conducting spheres of radii R and r (R < r) have similar charges with equal surface charge density σ. The electric potential at their common centre is
Correct Answer: Option 2
Official solution: The potentials due to the two charged conducting spheres add at the common centre, giving V = σ(R + r)/ε0.
Q17+4 / -1
Four equal resistors R each are connected to form a square ABCD, one in each side. If an electric potential V is applied between points A and B then current flowing out of B is
Correct Answer: Option 2
Official solution: Between A and B, one branch is R and the other is 3R in parallel. Hence Req = 3R/4 and I = V/Req = 4V/3R.
Q18+4 / -1
The ratio of the radii of the nuclei of 2713Al and 12552Te is
Correct Answer: Option 2
Official solution: Nuclear radius R ∝ A1/3. Therefore ratio = 271/3 : 1251/3 = 3 : 5.
Q19+4 / -1
In the circuit shown, if the power consumed in 6 Ω resistor is 24 W then the power consumed in 4 Ω resistor is
Kumar Sir +91 9958461445
Correct Answer: Option 4
Official solution: Current through 6 Ω is 2 A, so voltage across the branch is 36 V. Current in the 4 Ω-5 Ω branch is 4 A. Hence power in 4 Ω is I2R = 42 x 4 = 64 W.
Q20+4 / -1
In the circuit shown the reading of ammeter long time after the key is closed is
Kumar Sir +91 9958461445
Correct Answer: Option 3
Official solution: After a long time, the inductor behaves like a short circuit. Therefore the current is I = 10 V / 10 Ω = 1 A.
Q21+4 / -1
The truth table given below belongs to X Y Z 0 0 0 0 1 1 1 0 1 1 1 0
Correct Answer: Option 4
Official solution: The output is 1 only when the inputs are different and 0 when they are same. This is the truth table of an XOR gate.
Q22+4 / -1
The typical I-V characteristic graph for photo-diode is
Correct Answer: Option 3
Official solution: A photodiode is normally operated under reverse bias, and its characteristic corresponds to the graph shown in option 3.
Q23+4 / -1
For an equilateral prism of material refractive index μ = √2, the angle of minimum deviation is
Correct Answer: Option 2
Official solution: For a prism at minimum deviation, μ = sin[(A + δm)/2] / sin(A/2). With A = 60° and μ = √2, δm = 30°.
Q24+4 / -1
A rocket is fired vertically up with net acceleration of 5 m/s2. After 10 second the fuel got exhausted. Maximum height from the ground upto which the rocket goes is (Take g = 10 m/s2)
Correct Answer: Option 3
Official solution: During powered motion, velocity after 10 s is 50 m/s and height is 250 m. Extra height after fuel exhaustion is 502/(2g) = 125 m. Total height = 375 m.
Q25+4 / -1
A projectile is thrown up with initial speed u making angle θ with the horizontal. It reaches the maximum height H. If maximum height is to be doubled then initial speed of throw for same angle θ is
Correct Answer: Option 4
Official solution: H = u2sin2θ/2g. For the same angle, height is proportional to u2, so doubled height requires speed √2u.
Q26+4 / -1
In the arrangement shown, the coefficient of friction between the block B (mass = 10 kg) and the horizontal floor is μ = 0.4. If the mass of the hanging block A is 10 kg then the acceleration of the blocks is (pulley and string are ideal and g = 10 m/s2)
Kumar Sir +91 9958461445
Correct Answer: Option 3
Official solution: Driving force = 10 x 10 - 10 x 10 x 0.4 = 60 N. Total mass = 20 kg. Hence a = 60/20 = 3 m/s2.
Q27+4 / -1
A block of mass m1 moving with speed u undergoes head-on elastic collision with another mass m2 at rest. The fraction of energy transferred by m1 to m2 is
Correct Answer: Option 4
Official solution: For a head-on elastic collision with the second mass initially at rest, the transferred energy fraction is 4m1m2/(m1 + m2)2.
Q28+4 / -1
A block of mass m falls vertically on a horizontal floor of coefficient of restitution e = 0.8. The fraction of energy lost during collision is
Correct Answer: Option 4
Official solution: Energy fraction retained is e2 = 0.64. Therefore energy fraction lost = 1 - 0.64 = 0.36.
Q29+4 / -1
A boy of mass 40 kg is standing at one end of a floating boat of mass 200 kg. The boat is 6 m long. If the boy moves to the other end of the boat then the distance moved by the boat with respect to the fixed ground is
Correct Answer: Option 4
Official solution: With no external horizontal force, centre of mass remains fixed. Using m1Δx1 = m2Δx2 and total relative motion 6 m gives boat displacement 1 m.
Q30+4 / -1
A uniform wooden plank of mass 30 kg is kept on two supports X and Y as shown. The length of the plank is L. Reaction force applied by Y as shown is (g = 10 m/s2)
Kumar Sir +91 9958461445
Correct Answer: Option 3
Official solution: Taking moments gives 300 x (L/2) = RY x (3L/4). Therefore RY = 200 N.
Q31+4 / -1
A hollow sphere of mass m and radius R is kept on a rough horizontal surface. It is hit hard so that it starts sliding with initial speed v. Its velocity of centre of mass when it starts pure rolling is
Correct Answer: Option 2
Official solution: Using angular impulse and rolling condition for a hollow sphere gives final centre of mass velocity v' = 3v/5.
Q32+4 / -1
An ideal gas undergoes a thermodynamic process a to b as shown on pressure-volume (P-V) diagram. The work done by the gas during this process
Kumar Sir +91 9958461445
Correct Answer: Option 3
Official solution: During the process, volume is increasing continuously; hence the work done by the gas continuously increases.
Q33+4 / -1
The capacitance of each capacitor C in the network shown is 60 μF. The equivalent capacitance between points a and b is
Kumar Sir +91 9958461445
Correct Answer: Option 2
Official solution: The network reduces to 60/3 + 60 = 80 μF between points a and b.
Q34+4 / -1
A cylindrical metal rod of length L and cross-sectional area S is joined to an EMF source. The drift speed of electrons in the rod is v. If this rod is replaced by another rod of length 2L and cross-sectional area 2S then drift speed of electrons in the new rod will be
Correct Answer: Option 1
Official solution: Drift speed is proportional to current density. For the same EMF, current density is inversely proportional to length. Doubling length makes drift speed v/2.
Q35+4 / -1
If e, m and h represent electronic charge, mass of electron and Planck's constant respectively then Bohr's magneton is represented as
Correct Answer: Option 3
Official solution: Bohr magneton is μB = eh/(4πm).
Q36+4 / -1
Magnetic susceptibility is small and negative for
Correct Answer: Option 1
Official solution: Diamagnetic substances have small negative magnetic susceptibility.
Q37+4 / -1
Two induction coils of self induction 3 mH and 6 mH are connected in series. Their mutual induction is absent. The equivalent inductance is
Correct Answer: Option 2
Official solution: With mutual induction absent and coils in series, L = L1 + L2 = 3 + 6 = 9 mH.
Q38+4 / -1
In the circuit shown, resistance/reactance of R, L and C are 40 Ω, 80 Ω and 50 Ω respectively. The supply emf is 250 V, 50 Hz AC. The RMS current flowing in the circuit is
Kumar Sir +91 9958461445
Correct Answer: Option 2
Official solution: Net reactance = 80 - 50 = 30 Ω. Therefore Z = √(402 + 302) = 50 Ω and I = 250/50 = 5 A.
Q39+4 / -1
An astronomical telescope has an objective of focal length 100 cm and eye piece of focal length 5 cm. Maximum magnification of this telescope is
Correct Answer: Option 4
Official solution: Maximum magnification m = (f0/fe)(1 + fe/D). With f0 = 100 cm, fe = 5 cm and D = 25 cm, m = 24.
Q40+4 / -1
In a Young's double slit experiment in front of one of the slits a thin transparent sheet of thickness t and refractive index μ is inserted. The entire fringe pattern will be shifted by a distance (d - separation between slits; D - distance of screen from the slits and λ - wavelength of light used)
Correct Answer: Option 3
Official solution: Path difference due to the sheet is (μ - 1)t. Fringe shift is x = D(μ - 1)t/d.
Q41+4 / -1
In a photoelectric experiment the stopping potential V is plotted against applied frequency f for three different materials A, B and C as shown. Which of the following materials has largest threshold wavelength?
Kumar Sir +91 9958461445
Correct Answer: Option 1
Official solution: The material with the smallest threshold frequency has the largest threshold wavelength. From the graph, A has the smallest threshold frequency, so A has the largest threshold wavelength.
Q42+4 / -1
If potential energy of first excited state of hydrogen atom is taken zero then kinetic energy of ground state electron in the hydrogen atom is
Correct Answer: Option 1
Official solution: Shifting the zero of potential energy does not change kinetic energy. Ground state kinetic energy of hydrogen is 13.6 eV.
Q43+4 / -1
A radioactive nuclei of mass M emits gamma ray of wavelength λ. The recoil energy of the nuclei is (symbols have usual meanings)
Correct Answer: Option 3
Official solution: Photon momentum is p = h/λ. Recoil energy is E = p2/2M = h2/(2Mλ2).
Q44+4 / -1
Two radioactive substances X and Y have decay constants 3λ and λ respectively. Initially they have same number of nuclei. The ratio of number of nuclei of X to those of Y will be (1/e)3 after a time interval
Correct Answer: Option 1
Official solution: NX/NY = e-3λt/e-λt = e-2λt. Setting this equal to e-3 gives t = 3/(2λ).
Q45+4 / -1
The input resistance of a CE amplifier is 5 kΩ and the load resistance is 20 kΩ. If the current gain is 100 then mutual conductance of the transistor used is
Correct Answer: Option 3
Official solution: gm = Iout/(IinRin) = 100/(5 x 103) = 0.02 Ω-1 = 20 x 10-3 Ω-1.
Final NEET Score
Kumar Physics Classes
For NEET Physics concepts, doubt-solving, numerical practice and formula application, contact Kumar Sir.