Units and Dimensions
Dimensional formula: [MaLbTc]; principle of homogeneity; percentage error: ΔZ/Z = aΔA/A + bΔB/B.
NEET PHYSICS TUTOR DOUBT 48
Kumar Physics Classes
Phone / WhatsApp: +91 9958461445
Email: kumarsirphysics@gmail.com
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Dear Students,
This question paper has been specially prepared for serious NEET Physics aspirants. The solved questions have been prepared and explained by an experienced Physics Tutor in Mumbai in a clear, systematic, and exam-oriented manner. These questions are not meant for casual practice. They are designed to test your actual understanding of Class 11 and Class 12 Physics concepts.
Before attempting this paper, first revise the important formulas carefully. Once the formulas are fresh in your mind, attempt the complete question paper honestly under timed conditions. Read every question carefully, select your answer, and then compare it with the official answer and solution.
The purpose of this page is not only to show correct answers but to help students understand how to approach NEET-level Physics questions. If you are able to solve these questions properly, your confidence in Physics will improve. If you are unable to solve them, it means you should revise the concepts again and take proper guidance.
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. Physics questions in NEET often involve 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 x B| = AB sinθ, resultant: R = sqrt(A2 + B2 + 2AB cosθ).
Kinematics
v = u + at, s = ut + 1/2 at2, v2 = u2 + 2as, projectile range R = u2sin2θ/g.
Laws of Motion
F = ma, friction f ≤ μN, banking tanθ = v2/rg, impulse J = Δp.
Work, Energy and Power
W = Fs cosθ, K = 1/2 mv2, P = W/t = Fv, spring energy 1/2 kx2.
Circular Motion
v = ωr, ac = v2/r = ω2r, Fc = mv2/r.
Centre of Mass
Rcm = Σmiri/Σmi, momentum conservation: Σp = constant.
Rotational Motion
τ = Iα, L = Iω, Krot = 1/2 Iω2, rolling v = ωR.
Gravitation
F = Gm1m2/r2, g = GM/R2, escape speed ve = sqrt(2GM/R).
Solids
Stress = F/A, strain = ΔL/L, Young modulus Y = stress/strain, elastic energy 1/2 stress x strain x volume.
Fluids
P = P0 + ρgh, continuity Av = constant, Bernoulli P + 1/2ρv2 + ρgh = constant.
Thermal Properties
Q = mcΔT, ΔL = αLΔT, ΔA = 2αAΔT, ΔV = 3αVΔT.
Thermodynamics
ΔQ = ΔU + W, isothermal W = nRT ln(V2/V1), Carnot efficiency η = 1 - T2/T1.
Kinetic Theory
PV = nRT, P = 1/3 ρcrms2, crms = sqrt(3RT/M), K = 3/2 kT.
Oscillations
x = A sin(ωt + φ), T = 2πsqrt(m/k), pendulum T = 2πsqrt(l/g).
Waves
v = fλ, string wave speed v = sqrt(T/μ), Doppler approximation f' = f(v +/- vo)/(v -/+ vs).
Electrostatics
F = kq1q2/r2, E = F/q, point charge field E = kq/r2, potential V = kq/r.
Capacitance
C = Q/V, parallel plate C = ε0A/d, energy U = 1/2 CV2 = Q2/2C.
Current Electricity
V = IR, R = ρl/A, P = VI = I2R, Wheatstone balance P/Q = R/S.
Moving Charges and Magnetism
F = qvB sinθ, r = mv/qB, wire force F = BIl sinθ, B = μ0I/2πr.
Magnetism and Matter
τ = MB sinθ, U = -MB cosθ, bar magnet axial field B = μ02M/4πr3.
Electromagnetic Induction
e = -dΦ/dt, motional emf e = Blv, inductor emf e = -L dI/dt, energy 1/2 LI2.
Alternating Current
XL = ωL, XC = 1/ωC, Z = sqrt(R2 + (XL-XC)2), cosφ = R/Z.
Electromagnetic Waves
c = 1/sqrt(μ0ε0), E/B = c, propagation along E x B.
Ray Optics
Mirror/lens: 1/f = 1/v - 1/u, power P = 1/f(m), lens maker 1/f = (μ-1)(1/R1-1/R2).
Wave Optics
YDSE fringe width β = λD/d, single slit minima a sinθ = nλ, resolving power depends on aperture and wavelength.
Dual Nature
Photon energy E = hν, photoelectric hν = φ + Kmax, de Broglie λ = h/p.
Atoms
Bohr radius rn = n2r1/Z, energy En = -13.6Z2/n2 eV.
Nuclei
Activity A = λN, decay N = N0e-λt, half-life T1/2 = 0.693/λ, binding energy Δmc2.
Semiconductor Electronics
Diode forward bias conducts, reverse bias blocks; transistor current gain β = IC/IB; logic: AND, OR, NOT basics.
This question paper has been designed to help serious NEET aspirants practise Physics in a focused and exam-oriented way. 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.
If you want to solve these questions properly, first revise the formula section, then attempt the complete question paper from beginning to end. Do not open the solution immediately. First think, calculate, choose your option, and then check the official explanation. This habit will improve your confidence, speed, and accuracy in NEET Physics.
Physics is the only subject that can strongly influence your NEET rank if your concepts are clear. 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.
If you still cannot understand these concepts properly, study one-to-one with Kumar Sir through personalised online Physics classes.
Question Index
Q1
The number of significant digits in 7000.50 are
Official Solution:
Number of significant digit in 7000.50 is 6.
Q2
The acceleration of a particle moving along straight line decreases linearly from a value of 15 m/s2 at t = 0. It has zero value at t = 5 s. At what time the particle acquires its initial speed?
Official Solution:
Let a = mt + c. At t = 0, a = 15, so c = 15. At t = 5, a = 0, so m = -3. Thus a = -3t + 15. At t = 10 s, the net change in velocity is zero, so the velocity again becomes equal to the initial velocity.
Q3
If the frequency of Kα, Kβ and Lα X-rays of a material are γ1, γ2 and γ3 respectively, then correct relation is
Official Solution:
EKβ = EKα + ELα. Therefore γ2 = γ3 + γ1, so γ3 = γ2 - γ1.
Q4
A ball is projected upward at an angle θ with the horizontal. The graph between kinetic energy and horizontal displacement will be a
Official Solution:
v2 = (u sinθ - gt)2 + (u cosθ)2 and x = u cosθ t. Substituting t = x/(u cosθ) in kinetic energy gives a quadratic equation in x; therefore variation is parabolic.
Q5
When the frame of reference is rotated
Official Solution:
On rotating frame of reference, vector will not change but component of vector will change.
Q6
A fireman slides down a rope with a maximum acceleration of g/4 without breaking it. The ratio of breaking load to its weight is
Official Solution:
mg - T = mg/4, hence T = 3mg/4. Therefore T/mg = 3/4.
Q7
A block rests on a rough horizontal plane whose inclination θ with the horizontal can be varied. The graph which best represents the variation of frictional force F with θ.
Official Solution:
If block is at rest then f = mg sinθ. It increases until the limiting case θ = tan-1(μ). After this angle, friction is kinetic friction f = μN = μmg cosθ.
Q8
A ball falls on the ground and rebounds elastically along the same line. Then
Official Solution:
Fact.
Q9
An elastic spring has length x, when it has tension of 4 N and length y, when tension is 5 N. The tension in the spring, when its length is (5y - 4x) will be
Official Solution:
kx = 4 N and ky = 5 N. Hence 9kx = 16 N, 5ky = 25 N, and (5y - 4x)k = 9 N.
Q10
A particle is suspended from a fixed point by a light inextensible thread of length l. With what velocity, the particle must be projected horizontally from its lowest point so that it leaves the circular path at an angle 120° with vertical?
Official Solution:
At leaving point, mg cos60° = mv2/l. Using conservation of energy between the lowest point and leaving point gives v02 = 7gl/2. Therefore v0 = sqrt(7gl/2).
Q11
A ball x is projected vertically up. Another similar ball y is projected at an angle θ. They rise to same vertical height. If the ratio of their initial kinetic energy is 1 : 2 then the angle of projection is
Official Solution:
From equal height, vx2 = vy2 sin2θ. Given kinetic energy ratio vx2/vy2 = 1/2. Thus sin2θ = 1/2 and θ = 45°.
Q12
If the universal gravitational constant decreases uniformly with time, then the path of earth around sun will be
Official Solution:
E = -GMm/2r. On decreasing G, radius r will increase with time and path becomes spiral.
Q13
A satellite of mass m moves along an elliptical path around earth. Its areal velocity is
Official Solution:
dA/dt = L/2m = mvr/2m = vr/2. It is independent of mass.
Q14
A hollow sphere of mass M and radius R is in pure rolling on the horizontal rough surface. The ratio of rotational kinetic energy to the total kinetic energy of the hollow sphere will be
Official Solution:
For hollow sphere, I = 2MR2/3. Therefore KR/KT = (2/3)/(1 + 2/3) = 2/5.
Q15
A piston with small block on top of it is undergoing vertical S.H.M. If the amplitude at which the block separate from the piston is 40/π2. The time period of oscillation is
Official Solution:
At separation, g = Aω2. With A = 40/π2, ω2 = π2/4, so ω = π/2. Thus T = 2π/ω = 4 s.
Q16
A simple pendulum is oscillating with time period T. Now charge q is given to the bob and similar charge is kept at the point of suspension, as shown in figure then
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Official Solution:
Here net electrostatic force on the system is zero, therefore there will be no change in g and time period remains same.
Q17
A man of mass 50 kg is standing on a plank of mass 150 kg placed on a horizontal smooth floor. If the man walks from end A to B, then displacement of plank is
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Official Solution:
(L - x) x 50 - 150x = 0. With L = 12 m, 50 x 12 - 200x = 0, hence x = 3 m left.
Q18
Pulley and string connecting masses 2 kg and 3 kg are massless. Initially both blocks are at rest. If they start moving at t = 0, then the velocity of the 2 kg at t = 2 s is (g = 10 m/s2)
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Official Solution:
a = (3g - 2g)/5 = 2 m/s2. Since v = u + at, v = 2 x 2 = 4 m/s.
Q19
Dimensional formula of e4 / (ε02h2) is (e = charge of electron, ε0 = permittivity of free space, h = Planck's constant)
Official Solution:
Using the given relation, e4/(h2ε02) = v2. Hence dimensional formula is M0L2T-2.
Q20
A tunnel is dug along the diameter of earth and a block is released from the surface. The speed of the block at the centre of the earth is
Official Solution:
By conservation of energy, -GMm/R = -3GMm/2R + 1/2 mv2. Therefore v = sqrt(gR).
Q21
A plate having a circular hole is heated and allowed freely to expand. Mark the incorrect statement
Official Solution:
Plate expands freely, therefore there will be no stress.
Q22
The ratio of thermal conductivity of two rods is 1 : 2. Their cross-sectional area is same. If the ratio of their thermal resistance is in the ratio of 4 : 1 then the ratio of their respective lengths will be
Official Solution:
R = l/KA. Therefore R1/R2 = (l1/l2)(k2/k1)(A2/A1). With data, l1/l2 = 2 : 1.
Q23
The rms value of current for the following current time graph will be
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Official Solution:
Irms = sqrt(∫i2dT / ∫dT). For the given square waveform, Irms = 3 A.
Q24
Two thin convex lenses of same material and focal length 10 cm and 20 cm are used coaxially such that chromatic aberration is absent in the image produced jointly by them. The separation between lenses is
Official Solution:
d = (f1 + f2)/2 = (10 + 20)/2 = 15 cm.
Q25
The process not involved in the formation of rainbow is
Official Solution:
Total internal reflection is not involved in the formation of rainbow.
Q26
A charge is placed in front of a conducting sphere as shown in the figure. The potential at the centre of the sphere due to induced charges is [K = 1/(4πε0)]
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Official Solution:
For the conducting sphere, induced charge contribution at the centre gives qind = 0. Hence Vind = qind/(4πε0R) = 0.
Q27
An uncharged spherical solid conductor has a cavity as shown in the figure. A positive charge is placed at the centre. If the charge is displaced towards right then
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Official Solution:
Distribution of charge on outer surface will not be affected.
Q28
A conducting rod PQ is rotated with angular velocity ω in a uniform magnetic field B as shown in the figure. If OP = l and VOQ = (9/2)Bωl2 then length of rod PQ is
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Official Solution:
For rotating rod segment, VOQ = Bωx2/2. Given Bωx2/2 = (9/2)Bωl2, so x = 3l. Therefore PQ = 3l + l = 4l.
Q29
A proton is accelerated by a potential difference of 8.2 x 104 V. Its de-Broglie wavelength is
Official Solution:
eV = 1/2 mv2, so p = sqrt(2emV). Thus λ = h/sqrt(2emV). Substitution gives λ = 1.0 x 10-13 m.
Q30
The time taken for a radioactive substance to decay to 1/16th of its initial value is 400 μs. Its half life is
Official Solution:
N = N0e-λt. For N/N0 = 1/16, λ x 400 μs = ln16 = 4ln2. Hence T1/2 = ln2/λ = 100 μs.
Q31
Density of nucleus of matter is nearly
Official Solution:
ρ = mA / ((4/3)πR03A) = 3m/(4πR03), which is nearly 1017 kg/m3.
Q32
In the adjoining figure, transistor is used as a common emitter amplifier. If VCE = 6 V and VBE is neglected, then its current gain is
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Official Solution:
iB = (10 - VBE)/RB = 10/(250 x 103) = 40 x 10-6 A. iC = (10 - VCE)/RC = 4 x 10-3 A. Therefore β = iC/iB = 100.
Q33
Which of the following is reverse biased?
Official Solution:
In option (3), P is at lower potential than N, so the diode is reverse biased.
Q34
The frequency range of 1 x 109 Hz to 3 x 1011 Hz corresponds for
Official Solution:
Fact.
Q35
When a body is placed over liquids, it will sink down if
Official Solution:
A body sinks when its gravitational force is more than the upthrust of the liquid.
Q36
An isotropic material has coefficient of linear expansion α1, α2 and α3 along x, y and z axis respectively. If γ is coefficient of cubical expansion of its material, then
Official Solution:
Fact. For expansion along three perpendicular directions, γ = α1 + α2 + α3.
Q37
An ideal gas undergoes a thermodynamic cycle as shown in figure. Which of the following graphs represents the same cycle?
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Official Solution:
A to B is isobaric process and temperature is increasing, therefore volume is increasing. B to C is isochoric process and temperature is decreasing. C to A is isothermal process; pressure is increasing, so volume decreases. Hence graph (1) represents the same cycle.
Q38
The distance up to which defective eye of a person see is 50 cm. The power of lens he should use to correct the vision is
Official Solution:
For correction, v = -50 cm and final image should be at infinity. Thus 1/(-50) = 1/f, so f = -50 cm = -0.5 m. Hence P = 1/f = -2 D.
Q39
An ideal gas is compressed isothermally V to V' and then expanded adiabatically to its original volume V. Then the net work done is
Official Solution:
Work done in isothermal process is more than work done in adiabatic process. Hence the net work done is negative.
Q40
If E and B are the electric and magnetic field vectors of e.m waves, then the direction of propagation of e.m wave is along
Official Solution:
Fact. The direction of propagation of an electromagnetic wave is along E x B.
Q41
Power factor in series L.C.R. circuit at resonance is
Official Solution:
cosφ = R/Z = R/sqrt(R2 + (XL - XC)2). At resonance, XL = XC, hence cosφ = 1.
Q42
An uncharged capacitor having capacitance C is connected across a battery of emf V. Now the battery is disconnected and reconnected across same battery but with reverse polarity, then the ratio of total thermal energy produced to the total energy supplied by the battery is
Official Solution:
As given in solution, total work supplied W = 3CV2. Total thermal energy H = 1/2 CV2 + 3/2 CV2 = 2CV2. Therefore H/W = 2/3.
Q43
A supersonic Jet is moving with a certain velocity, such that vertex of mach cone is 60°. The ratio of the velocity of Jet to sound is
Official Solution:
Vp/Vs = cosecθ = cosec 60° = 2 as used in the official solution.
Q44
A plane mirror can form
Official Solution:
A plane mirror can form real or virtual image.
Q45
The magnifying power of a microscope decreases when its length is
Official Solution:
M = -L/fo(1 + D/fe). Magnifying power is proportional to tube length L, so it decreases when the length is decreased.