Units and Dimensions
- Dimensional formula: [MaLbTc]
- Percentage error: ΔQ/Q = aΔA/A + bΔB/B
- Significant figures follow least precise measurement
NEET PHYSICS TUTOR DOUBT 50
NEET Physics full-paper assessment with formula revision, answer checking, official solutions and final score.
Dear Students,
This NEET Physics assessment paper has been created for serious aspirants who want to check their real preparation level. These questions are not ordinary practice questions. They are designed to test your conceptual clarity, calculation accuracy, formula application, and ability to solve Physics problems under exam pressure.
These questions have been solved by an experienced Physics Tutor in South Delhi. For NEET, Physics can strongly affect your final rank. In a competitive city like Delhi, students often need at least 36 correct questions overall to remain competitive, and in Physics, a serious aspirant should aim for 160+ marks. To reach that level, students must practise high-quality conceptual papers like this one, analyse mistakes, revise formulas, and improve speed and accuracy.
Before attempting this paper, revise all important Class 11 and Class 12 Physics formulas carefully. Then solve the complete paper honestly under timed conditions. Do not open the solution immediately. First think, calculate, choose your option, and then check the official solution.
If you cannot solve these questions or if your Physics concepts are still not clear, contact Kumar Sir for one-to-one online Physics classes. Kumar Sir explains difficult Physics concepts in a simple, step-by-step, and exam-oriented manner so that students can understand the logic behind every formula and every question.
Important Formula Revision for NEET Physics: Class 11 and Class 12
Dear Students,
Before attempting this paper, revise the important formulas from Class 11 and Class 12 Physics. NEET Physics often involves calculation, formula selection, conceptual clarity, and careful application. Many students know the formulas but make mistakes while applying them in numerical questions. This formula bank is added so that you can quickly revise the major concepts before starting the paper.
Vectors
- A · B = AB cosθ
- A × B = AB sinθ n̂
- Resultant: R = √(A2 + B2 + 2AB cosθ)
Kinematics
- v = u + at
- s = ut + (1/2)at2
- v2 = u2 + 2as
- Projectile range: R = u2 sin2θ/g
Laws of Motion
- F = ma
- Friction: fmax = μN
- Impulse: J = Δp
- Banking: tanθ = v2/rg
Work, Energy and Power
- W = Fscosθ
- K = (1/2)mv2
- P = dW/dt = Fv
- Spring energy: U = (1/2)kx2
Circular Motion
- ac = v2/r = ω2r
- Fc = mv2/r
- v = rω
- T = 2π/ω
Centre of Mass
- Rcm = Σmiri/Σmi
- vcm = Σmivi/M
- Momentum: P = Mvcm
Rotational Motion
- τ = Iα
- L = Iω
- Krot = (1/2)Iω2
- Rolling: v = Rω
Gravitation
- F = GMm/r2
- g = GM/R2
- V = -GM/r
- vescape = √(2GM/R)
Mechanical Properties of Solids
- Stress = F/A
- Strain = ΔL/L
- Y = stress/strain
- Elastic energy density = (1/2) stress × strain
Mechanical Properties of Fluids
- P = P0 + ρgh
- Buoyant force = ρVg
- Continuity: Av = constant
- Bernoulli: P + (1/2)ρv2 + ρgh = constant
Thermal Properties
- Q = mcΔT
- Q = mL
- Linear expansion: ΔL = αLΔT
- Stefan law: P = σeAT4
Thermodynamics
- ΔQ = ΔU + W
- W = nRΔT for isobaric process
- Efficiency: η = 1 - T2/T1
- PV = nRT
Kinetic Theory
- PV = (1/3)MNvrms2/V
- vrms = √(3RT/M)
- Average KE = (3/2)kT
- Mean free path: λ = 1/(√2πd2n)
Oscillations
- SHM: a = -ω2x
- T = 2π√(m/k)
- Pendulum: T = 2π√(l/g)
- Energy: E = (1/2)kA2
Waves
- v = fλ
- String speed: v = √(T/μ)
- Beat frequency = |f1 - f2|
- Doppler: f' = f(v ± vo)/(v ∓ vs)
Electrostatics
- F = kq1q2/r2
- E = F/q
- V = kq/r
- Gauss law: Φ = q/ε0
Capacitance
- C = Q/V
- Parallel plate: C = ε0A/d
- Energy: U = (1/2)CV2
- Series: 1/C = Σ1/Ci
Current Electricity
- I = dq/dt
- V = IR
- R = ρl/A
- Power: P = VI = I2R = V2/R
Moving Charges and Magnetism
- F = q(v × B)
- F = BIl sinθ
- r = mv/qB
- B near wire = μ0I/(2πr)
Magnetism and Matter
- M = m/V
- B = μ0(H + M)
- μr = 1 + χ
- Torque: τ = MB sinθ
Electromagnetic Induction
- ε = -dΦ/dt
- Φ = BA cosθ
- Lenz law gives direction
- Self emf: ε = -L dI/dt
Alternating Current
- XL = ωL
- XC = 1/ωC
- Z = √(R2 + (XL-XC)2)
- Resonance: ω = 1/√(LC)
Electromagnetic Waves
- c = 1/√(μ0ε0)
- c = fλ
- E0/B0 = c
- Energy density shared by E and B fields
Ray Optics
- Mirror: 1/f = 1/v + 1/u
- Lens: 1/f = 1/v - 1/u
- m = h'/h = v/u
- Prism: μ = sin[(A + δm)/2]/sin(A/2)
Wave Optics
- YDSE fringe width: β = λD/d
- Path difference: Δx = d sinθ
- Single slit minima: a sinθ = nλ
- Malus law: I = I0cos2θ
Dual Nature
- E = hν
- p = h/λ
- Photoelectric: Kmax = hν - φ
- de Broglie: λ = h/p
Atoms
- Bohr radius: rn = n2a0/Z
- Energy: En = -13.6Z2/n2 eV
- Angular momentum: mvr = nh/2π
Nuclei
- R = R0A1/3
- Activity: A = λN
- N = N0e-λt
- Half-life: T1/2 = 0.693/λ
Semiconductor Electronics
- Diode current flows in forward bias
- Rectifier converts AC to DC
- Transistor: IE = IB + IC
- Logic: NAND and NOR are universal gates
This question sheet has been designed to help students practise Physics in a focused and exam-oriented way. Future NEET aspirants must revise Class 11 and Class 12 Physics carefully because these chapters involve calculation, conceptual confusion, and tricky formula application. Every serious NEET student should attempt this paper, check the solution carefully, and identify weak areas before the final exam.
These questions are inspired by NCERT concepts, standard Physics books, Resonance-style practice, and the problem-solving approaches followed by leading coaching institutes such as Allen, Aakash, Kota-based institutes, and Delhi-based coaching systems. The purpose is not only to test memory, but to check whether the student can apply concepts correctly under exam pressure.
Why Strong Physics Preparation Is Now More Important Than Ever
NEET Physics is becoming more conceptual, more competitive, and less forgiving of superficial preparation. Students must build conceptual clarity, calculation accuracy, speed, and the ability to solve unfamiliar problems under pressure. Memorising formulas is not enough; a student must understand when, where, and how to apply each formula in a changing question pattern.
Important Message for NEET 2027, 2028, 2029, 2030 and Future Aspirants
Future NEET aspirants must prepare seriously for online-style or changing exam patterns, where question variation and concept application may become more important. Practise papers under timed conditions, revise formulas regularly, analyse every mistake, and strengthen weak chapters before they become scoring gaps.
Question Index
Question 1+4 / -1
Neutrino and antineutrino differ from each other in
Correct Answer: Option 3
Official solution: These particles have no charge but opposite spin. Therefore their angular momenta have different directions.
Question 2+4 / -1
Uncertainty in measurement of mass and radius of a spherical body is 2% and 0.5% respectively. The uncertainty in density is
Correct Answer: Option 4
Official solution: Density d = M/V = M/(43πR3). Hence percentage uncertainty = percentage uncertainty in M + 3 times percentage uncertainty in R = 2% + 3 × 0.5% = 3.5%.
Question 3+4 / -1
A ball is thrown vertically upward with some speed from the ground. Ball passes a point h height above the ground at time 6 s and 10 s respectively. The speed with which ball is thrown up from the ground is (g = 10 m/s2)
Correct Answer: Option 3
Official solution: For a projectile moving vertically under gravity, if it crosses the same height at times t1 and t2, then t1 + t2 equals the time of flight. Thus T = 6 + 10 = 16 s. Since T = 2u/g, u = gT/2 = 10 × 16 / 2 = 80 m/s.
Question 4+4 / -1
A particle starts from rest with constant acceleration. The ratio of magnitude of displacement in nth second to that in n seconds is
Correct Answer: Option 4
Official solution: Starting from rest, displacement in the nth second is Sn = a2(2n - 1), while displacement in n seconds is S = 12an2. The ratio is (2n - 1)/n2 = 2/n - 1/n2.
Question 5+4 / -1
Two projectiles are thrown from the ground at different angles so that their initial vertical components of velocities are equal. Then they have same
Correct Answer: Option 4
Official solution: Time of flight and maximum height depend only on the initial vertical component of velocity. Since the vertical components are equal, the time of flight and maximum height are also equal.
Question 6+4 / -1
A river 500 meter wide is flowing at the rate of 2 m/s. A boat is sailing at a velocity of 10 m/s with respect to water, in the direction perpendicular to river. How far from the point directly opposite to the starting point does the boat reach the opposite bank?
Correct Answer: Option 3
Official solution: The crossing time is width/speed across river = 500/10 = 50 s. During this time the river current drags the boat downstream by 2 × 50 = 100 m.
Question 7+4 / -1
A particle is moving on a circular track of radius r. If speed of the particle is increasing with time then the angle between acceleration (linear) and position vector of the particle is (centre of circle is taken as origin).
Correct Answer: Option 2
Official solution: As speed is increasing, the particle has both radial and tangential acceleration. The resultant acceleration has a component toward the centre and a tangential component, so its angle with the outward position vector is obtuse.
Question 8+4 / -1
A block with mass m is kept on a horizontal rough table at distance R from its centre. If the coefficient of friction between block and surface of turn table is μ. What will be maximum angular speed of the turn table so that block does not slip?
Correct Answer: Option 2
Official solution: Static friction supplies the necessary centripetal force. For limiting condition, mRω2 = μmg, so ωmax = √(μg/R).
Question 9+4 / -1
A particle is attached with massless string. It is displaced from A to B by a force of constant magnitude but its direction is always along instantaneous displacement. The work done by the force F is
Kumar Sir +91 9958461445
Correct Answer: Option 4
Official solution: Since the force is always along the displacement, work equals force multiplied by the path length. The path length is lθ, where θ = π/2. Therefore W = F × l × π/2.
Question 10+4 / -1
A ball of mass m1 is moving with speed u towards another stationary ball of mass m2. After collision ball of mass m1 comes to rest and ball of mass m2 starts moving with speed v. The coefficient of restitution is
Correct Answer: Option 1
Official solution: By conservation of linear momentum, m1u = m2v, so v = m1u/m2. Coefficient of restitution e = velocity of separation / velocity of approach = v/u = m1/m2.
Question 11+4 / -1
A meter stick with uniform linear mass distribution is balanced at its centre. Now a small block of mass 4 kg is put at 16 cm mark of scale. The stick is found to be balanced at 40 cm mark. The mass of meter stick is
Correct Answer: Option 4
Official solution: The centre of mass of the meter stick is at the 50 cm mark. Taking torque about the 40 cm mark, 4g(40 - 16) = Mg(50 - 40). Thus 4 × 24 = 10M, so M = 9.6 kg.
Question 12+4 / -1
A uniform disc of mass m and radius R is resting on a table on its rim. The coefficient of friction between disc and table is μ. The disc is pulled horizontally from its central axis with a force F. What is maximum value of F for which disc rolls without slipping?
Kumar Sir +91 9958461445
Correct Answer: Option 2
Official solution: For translation, F - f = Ma. For rotation about COM, fR = Iα. With pure rolling, a = Rα, and for a disc I = MR2/2. Hence f = Ma/2 and a = 2f/M. Substituting in F - f = Ma gives F = 3f. Since f ≤ μmg, Fmax = 3μmg.
Question 13+4 / -1
A ball is dropped from height h = Re above earth surface. The speed with which ball strikes the surface of earth is
Correct Answer: Option 3
Official solution: Using mechanical energy conservation, v = √[2gRe/(1 + h/Re)]. Given h = Re, this becomes v = √(gRe).
Question 14+4 / -1
A wire is elongated by an external force which is always equal to restoring force developed in the wire material. W work is done by the external force in elongating wire by x unit then energy of wire increases by
Correct Answer: Option 1
Official solution: The external force is equal to the restoring force at every instant, so the work done by the external force is stored as elastic potential energy of the wire. Hence the energy increases by W.
Question 15+4 / -1
Two soap bubbles of radius R1 and R2 combine to form a single soap bubble isothermally. The radius of single soap bubble is
Correct Answer: Option 2
Official solution: For isothermal combination of soap bubbles, P1V1 + P2V2 = PV. Using excess pressure of a soap bubble, the result is R2 = R12 + R22, so R = √(R12 + R22).
Question 16+4 / -1
The bottom of a cylindrical vessel has a circular hole of radius r at depth h below the water level in it. If diameter of the vessel is D, then the speed with which the water level in the vessel drops is
Correct Answer: Option 3
Official solution: Instantaneous velocity of liquid from the hole by Torricelli's theorem is v = √(2gh). From continuity, πr2√(2gh) = π(D/2)2V. Thus V = (4r2/D2)√(2gh) = (r2/D2)√(32gh).
Question 17+4 / -1
Assuming oxygen gas as ideal gas, the equation of state of 10 g of oxygen at temperature T, pressure P which occupy volume V is given by
Correct Answer: Option 4
Official solution: For an ideal gas, PV = nRT. For oxygen gas, molar mass is 32 g mol-1, so n = 10/32. Therefore PV = (10/32)RT.
Question 18+4 / -1
A perfectly black body has absorptivity a equal to
Correct Answer: Option 4
Official solution: For a perfect black body, absorptive power is 1. Hence a = 1.
Question 19+4 / -1
A body cools from 80°C to 70°C in 12 minutes and from 70°C to 60°C in time = t. The value of t is (temperature of surrounding is 40°C)
Correct Answer: Option 2
Official solution: Using Newton's law of cooling, (80 - 70)/12 = k[(80 + 70)/2 - 40] and (70 - 60)/t = k[(70 + 60)/2 - 40]. Dividing the two equations gives t = 12 × 35/25 = 16.8 min.
Question 20+4 / -1
Two tuning forks of frequency 520 Hz and 524 Hz are vibrated together. The time interval between successive maximum and minimum of intensity is
Correct Answer: Option 1
Official solution: Beat frequency = 524 - 520 = 4 Hz. The time from maximum intensity to the next minimum intensity is half the beat period, so t = 1/(2 × 4) = 1/8 s.
Question 21+4 / -1
Frequency of first overtone of closed organ pipe of length 20 cm is equal to frequency of third overtone of open organ pipe. The length of open organ pipe is
Correct Answer: Option 4
Official solution: For a closed pipe, first overtone frequency = 3v/(4lc). For an open pipe, third overtone frequency = 4v/(2lo). Equating them gives 3/lc = 8/lo. With lc = 20 cm, lo = 53.3 cm.
Question 22+4 / -1
An observer is moving with speed 20 m/s towards stationary sound source which is emitting sound of frequency 400 Hz. The frequency of sound heard by observer is approximately (Vsound = 340 m/s)
Correct Answer: Option 3
Official solution: By Doppler effect, n' = n(v + vo)/(v + vs) = 400(340 + 20)/340 = 424 Hz, approximately 425 Hz.
Question 23+4 / -1
If pendulum 1 is set to oscillate, then which of the following four pendulums (A, B, C & D) as shown in figure will oscillate with maximum amplitude under forced vibration?
Kumar Sir +91 9958461445
Correct Answer: Option 3
Official solution: Under forced oscillation, maximum amplitude occurs when the external frequency equals the natural frequency of the system. Pendulum 1 and pendulum C have equal length, so C resonates.
Question 24+4 / -1
Two point charges 4Q and Q are fixed 2 m apart. A third charge q0 is kept between two charges so that net force on it is zero. The separation between charge q0 and 4Q is
Correct Answer: Option 4
Official solution: Let the distance from 4Q be x. For zero net force, k(4Qq0)/x2 = k(Qq0)/(2 - x)2. Hence 2/x = 1/(2 - x), giving x = 4/3 m.
Question 25+4 / -1
In a certain region of space, electric field is along the x-axis throughout. The magnitude of the electric field is however not constant but increases uniformly along positive x-axis at a rate of 105 NC-1m-1. The force experienced by a system having total dipole moment equal to 10-7 C m in negative x-axis direction is
Correct Answer: Option 1
Official solution: For a dipole in non-uniform electric field, F = p(dE/dx). Here p = 10-7 C m and dE/dx = 105 NC-1m-1, so magnitude = 10-2 N. Since the dipole moment is along negative x-axis, force is −10-2 î N.
Question 26+4 / -1
Piece of Aluminium and Silicon both are heated from −10°C to 40°C. The resistivity of
Correct Answer: Option 1
Official solution: Aluminium is a conductor, so its resistivity increases with temperature. Silicon is a semiconductor and has negative temperature coefficient of resistivity, so its resistivity decreases as temperature increases.
Question 27+4 / -1
In the circuit shown in the figure, heat developed across 2 Ω, 4 Ω and 3 Ω resistances are in the ratio of (respectively)
Kumar Sir +91 9958461445
Correct Answer: Option 2
Official solution: In the parallel part, current through 2 Ω is 2I/3 and current through 4 Ω is I/3, while current through the series 3 Ω resistor is I. Since H = I2Rt, the ratio is (2I/3)2×2 : (I/3)2×4 : I2×3 = 8 : 4 : 27.
Question 28+4 / -1
The potential difference between points A and B in the circuit shown in figure is
Kumar Sir +91 9958461445
Correct Answer: Option 4
Official solution: Both branches are in parallel and each branch has resistance 9 Ω, so the 6 A current divides equally as 3 A in each branch. VA = V - 3 × 3 and VB = V - 3 × 6. Therefore VA - VB = 9 V.
Question 29+4 / -1
A potentiometer wire of length 2 m has resistance of 20 Ω. It is connected in series with resistance of 10 Ω and an accumulator of 6 volt with negligible internal resistance. A source of 2.4 volt is balanced against length l of potentiometer wire. The value of l is
Kumar Sir +91 9958461445
Correct Answer: Option 1
Official solution: Current through potentiometer wire I = 6/(20 + 10) = 0.2 A. Potential difference across the potentiometer wire = 0.2 × 20 = 4 V. Potential gradient k = 4/2 = 2 V/m. Hence 2.4 = 2l, so l = 1.2 m.
Question 30+4 / -1
Which of the following is incorrect about a cyclotron?
Correct Answer: Option 2
Official solution: In a cyclotron, the charged particle does not move on any arbitrary path; it moves on a spiral path. Therefore statement (2) is incorrect.
Question 31+4 / -1
A uniform conducting wire of length 30a and resistance of R is wound up as current carrying coil in shape of regular hexagon of sides a. If coil is connected to a voltage source V, then magnetic moment of coil is
Kumar Sir +91 9958461445
Correct Answer: Option 3
Official solution: One hexagonal turn requires wire length 6a, so number of turns = 30a/6a = 5. Current I = V/R. Area of a regular hexagon of side a is 3√3a2/2. Magnetic moment M = NIA = 5(V/R)(3√3a2/2) = 15√3Va2/(2R).
Question 32+4 / -1
A solenoid has core of material with relative permeability 500 and its winding carries a current of 1 ampere. The solenoid has 500 turns per meter. Then intensity of magnetisation of material is nearly
Correct Answer: Option 2
Official solution: H = nI = 500 × 1 = 500 A m-1. Since μr = 1 + χ, χ = 500 - 1 = 499. Magnetisation M = χH = 499 × 500 = 2.495 × 105 A m-1, nearly 2.5 × 105 A m-1.
Question 33+4 / -1
A conducting circular loop is kept in uniform magnetic field B perpendicular to the plane of loop. The radius of ring starts decreasing with constant rate β, then the magnitude of induced emf in the loop (when radius of loop is r) is
Correct Answer: Option 2
Official solution: Magnetic flux φ = Bπr2. Therefore induced emf e = |dφ/dt| = Bπ×2r×|dr/dt|. Given |dr/dt| = β, e = 2πBrβ.
Question 34+4 / -1
An AC source of potential V and frequency f is applied to LCR series circuit as shown. At resonance, reading of
Kumar Sir +91 9958461445
Correct Answer: Option 1
Official solution: At resonance in an LCR series circuit, the inductive reactance equals the capacitive reactance. Therefore the voltage across L and C are equal in magnitude, so Voltmeter 1 = Voltmeter 2.
Question 35+4 / -1
An opaque disc is placed on the surface of pond which has liquid of refractive index 5/3. A source of light is placed 4 m below the surface of liquid. Calculate the minimum diameter of the disc placed symmetrically over source so that light does not come out of the pond.
Correct Answer: Option 3
Official solution: For the limiting ray, sin θc = 1/μ = 3/5, so tan θc = 3/4. With h = 4 m, radius R = h tan θc = 4 × 3/4 = 3 m. Therefore diameter = 2R = 6 m.
Question 36+4 / -1
A small object is placed 20 cm in front of a cube of glass of 10 cm edge and its farther side is silvered. The image is formed 22 cm behind the silver face. The refractive index of the glass cube is
Correct Answer: Option 3
Official solution: The silvered back face behaves like a thick mirror. Its apparent distance from the front face is x = t/μ = 10/μ. Since the image is formed as far behind the mirror as the object is in front of it, 20 + x = 22 + (10 - x). Thus x = 6 cm and 10/μ = 6, giving μ = 1.67.
Question 37+4 / -1
A ray of light of incident at an angle of 60° on one of faces of a prism which has refracting angle equal to 30°. The ray emerging out of prism makes an angle of 30° with incident ray. What is refractive index of prism.
Correct Answer: Option 2
Official solution: Using i + e = A + δ, 60° + e = 30° + 30°, so e = 0°. Also r1 + r2 = A and r2 = 0°, hence r1 = 30°. Refractive index μ = sin60°/sin30° = √3.
Question 38+4 / -1
Which of the following properties of electromagnetic waves does not support particle nature of radiation (electromagnetic wave)?
Correct Answer: Option 3
Official solution: Polarization is a wave character of radiation and therefore does not support the particle nature of electromagnetic radiation.
Question 39+4 / -1
Photons of incident radiation has energy 3.4 eV. If work function of metal surface, irradiated by radiations is 2.9 eV then kinetic energy of the ejected electron may be
Correct Answer: Option 4
Official solution: Maximum kinetic energy = incident photon energy - work function = 3.4 - 2.9 = 0.5 eV. Any emitted electron can have kinetic energy less than or equal to this maximum value. Hence any of these is possible.
Question 40+4 / -1
Radiation of energy 12.1 eV is passed through hydrogen gas, then H-atom may be excited to
Correct Answer: Option 2
Official solution: For hydrogen, -13.6 + 12.1 = -13.6/n2. Thus n2 = 9 and n = 3. Energy level n = 3 corresponds to the second excited state.
Question 41+4 / -1
An optical fibre is used for data transmission at high frequency. If refractive index of core and cladding are μ1 & μ2 respectively then angle of acceptance is
Correct Answer: Option 1
Official solution: For optical fibre, maximum acceptance angle satisfies sin θm = √(μ12 - μ22). Therefore θm = sin-1[√(μ12 - μ22)].
Question 42+4 / -1
If MA and MB are atomic masses of parent and daughter nuclei in β+ decay then mass defect of the reaction is (me is mass of electron)
AZ → BZ-1 + β0+1 + ν
AZ → BZ-1 + β0+1 + ν
Correct Answer: Option 3
Official solution: For β+ decay, using atomic masses requires accounting for the electron masses on both sides. Adding Zme to both sides gives mass defect Δm = MA - MB - 2me.
Question 43+4 / -1
Initial activity of a radioactive sample is A0 and its half life is T. The activity of the sample of a time t = T2 is
Correct Answer: Option 3
Official solution: Activity after time t is A = A0(1/2)t/T. Here t = T/2, so A = A0(1/2)1/2 = A0/√2.
Question 44+4 / -1
The diode used in a circuit shown in figure has constant voltage drop of 0.5 V at all currents and a maximum power rating of 100 mW. What should be value of the resistor R, connected in series with diode, for obtaining maximum current.
Kumar Sir +91 9958461445
Correct Answer: Option 3
Official solution: Maximum current through diode I = P/V = (100 × 10-3)/0.5 = 0.2 A. The supply is 1.5 V and diode drop is 0.5 V, so resistor drop is 1.0 V. Hence R = 1.0/0.2 = 5 Ω.
Question 45+4 / -1
Select the correct match.
(A) A · B = A + B
(B) A + B = A · B
(C) A · B + A · B = A + B
(A) A · B = A + B
(B) A + B = A · B
(C) A · B + A · B = A + B
Correct Answer: Option 3
Official solution: Using De Morgan's theorem, A · B = A + B and A + B = A · B. Statement (C) represents the XOR expression AB + AB, not A + B. Therefore only (A) and (B) are correct.
Final Result
Total Questions45
Attempted Questions0
Correct Answers0
Wrong Answers0
Unattempted Questions45
Positive Marks0
Negative Marks0
Final Score0