Q1
The dimensional formula of magnetic dipole moment is
Solution: Official solution: Unit of magnetic dipole moment is ampere-metre2. Therefore the dimensional formula is [L2A].
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NEET PHYSICS TUTOR DOUBT 42
Premium NEET Physics practice paper for focused assessment, concept revision, formula application and exam-style numerical practice. This page includes Physics questions Q1 to Q45 with clickable options, answers, official solutions and NEET marking.
Why Strong Physics Preparation Is Now More Important Than Ever
Physics can strongly influence a NEET rank because it rewards clarity, speed and disciplined formula application. Modern NEET aspirants need more than memorisation: they must connect concepts with numericals, recognise hidden assumptions, and avoid calculation traps under time pressure. A structured paper like this helps students revise, attempt, check and improve in one focused sitting.
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Important Message for NEET 2027, 2028, 2029, 2030 and Future Aspirants
Future NEET aspirants should build Physics slowly, chapter by chapter, with constant practice in concepts, doubt-solving, numericals and formula selection. Early preparation gives students enough time to correct weak areas before they become habits. Every serious aspirant should revise Class 11 and Class 12 Physics carefully, practise mixed questions regularly, and compare every wrong answer with the official method.
Important Formula Revision: all physics of class 11 and 12
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 Gravity, Solids, Fluids, Thermal Properties and other Physics topics in a very clear, step-by-step and exam-oriented way.
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Email: kumarsirphysics@gmail.com
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Important Practice for kinetic theory, oscillation, wave
These chapters often look formula-based, but NEET questions test interpretation, sign convention, graph reading and quick numerical judgement. Revise the formulas below before starting the paper.
Important Formula Revision for NEET Physics: kinetic theory, oscillation, wave
Kinetic Theoryp = (1/3)ρcrms2 and PV = (1/3)Nm crms2
RMS Speedcrms = √(3RT/M) = √(3kT/m)
Average Speedcavg = √(8RT/πM)
Most Probable Speedcmp = √(2RT/M)
Molecular Energy<K> = (3/2)kT, U = (f/2)nRT
Ideal GasPV = nRT = NkT, pVγ = constant
Mean Free Pathλ = 1/(√2πd2n)
SHM Displacementx = A sin(ωt + φ), ω = 2π/T
SHM Velocityv = ω√(A2 - x2), vmax = ωA
SHM Accelerationa = -ω2x, amax = ω2A
Spring OscillatorT = 2π√(m/k), E = (1/2)kA2
Simple PendulumT = 2π√(l/g), geff changes in lift fields
Wave Equationy = A sin(kx - ωt + φ), k = 2π/λ
Wave Speedv = fλ = ω/k
String Wavev = √(T/μ)
Sound Wavev = √(γP/ρ)
Beatsfbeat = |f1 - f2|
Pipes And Stringsf = nv/(2L) for open pipe/string, f = (2n - 1)v/(4L) for closed pipe
Doppler Effectf' = f[(v +/- vo)/(v -/+ vs)]
IntensityI ∝ A2ω2, β = 10 log(I/I0)
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
Q2
The velocity-time (v-t) graph of a particle moving along x-axis is as shown. Choose the correct statement.
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Solution: Official solution: Area under the velocity-time graph gives displacement. Since the area keeps increasing with time, displacement continuously increases.
Q3
For a projectile motion in x-y plane on some planet, the position coordinates as a function of time t are given as x = 4t and y = 4t - 5t2, where x, y are in metre and time t is in second. Assume x is horizontal and y is vertical. The horizontal range of the projectile is
Solution: Official solution: y = 0 gives t = 4/5 s. Hence range R = x = 4t = 16/5 m.
Q4
For a particle moving on a circle of radius 2 m, the distance covered (s) as a function of time (t) is given as s = t2 (s is in metre and t is in second). The acceleration of the particle at t = 1 second is
Solution: Official solution: v = ds/dt = 2t, so at t = 1 s, v = 2 m/s. Tangential acceleration is 2 m/s2 and centripetal acceleration is v2/r = 2 m/s2. Resultant acceleration = 2√2 m/s2.
Q5
A water pump of 1 horse power is required to flow water through a cylindrical pipe line at 1 m/s. If flow rate of water is to be doubled then the power of new water pump required is
Solution: Official solution: Power required for flow is proportional to v3. If speed is doubled, power becomes 23 = 8 times, so required power is 8 HP.
Q6
A hollow sphere of mass 2 kg and radius 2 cm is rolling down a rough incline of inclination θ = 30°. The acceleration of its centre down the incline is (g = 10 m/s2)
Solution: Official solution: a = g sinθ/(Icm/MR2 + 1). For a hollow sphere Icm = (2/3)MR2, so a = 5/(5/3) = 3 m/s2.
Q7
In a large rectangular water tank, water is coming out of a small hole at the bottom of the tank. It takes almost 5 minute to empty half of the tank. How much approximate further time will be needed to empty the tank completely?
Solution: Official solution: For draining under gravity, x/5 = 1/(√2 - 1). This gives x approximately 12 minute for the further time.
Q8
A steel scale (calibrated at 25°C) has been employed to measure the length of a copper pipe at 25°C. However, the reading has been taken at 65°C. If thermal coefficient of linear expansion of steel and copper are 3 × 10-5 per °C and 2 × 10-5 per °C respectively then percentage error in reading taken is
Solution: Official solution: Percentage error = (αsteel - αcopper)Δθ × 100 = 10-5 × 40 × 100 = 0.04%.
Q9
A thermodynamic cyclic process abcd has two isothermals, one isobaric and one isochoric process as shown on pressure-volume (P-V) diagram. The same cyclic process on a volume-temperature (V-T) diagram is best represented by
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Solution: Official solution: b to a is isobaric, a to d is isothermal, d to c is isochoric, and c to b is isothermal. The matching V-T graph is option 1.
Q10
From a uniform circular plate of radius R and area A, a small circular portion of area A/31 is cut out at a distance 3R/4 from the centre O as shown. The centre of mass of leftover disc is at a distance x from O, where x is
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Solution: Official solution: Treat the removed part as negative mass. x = [(A/31)(3R/4)]/(31 - 1) in equivalent ratio, giving x = R/40.
Q11
Four organ pipes of equal length have their overtones as shown. Frequency ratio of sound waves in them is
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Solution: Official solution: From the shown modes, f1 : f2 : f3 : f4 = 8 : 7 : 6 : 5.
Q12
In which of the following motions, the time T will have approximate value of 84.6 minute? a. Time period T of a satellite revolving very close to the surface of the earth. b. Time period T of rotation of the earth so that a body kept on equator feels weightlessness. c. Time period T of oscillation of an infinitely large simple pendulum. d. Time period T of oscillation of a body along a complete tunnel through a chord of the earth.
Solution: Official solution: This is a standard result. All four listed motions have approximate time period 84.6 minute.
Q13
A plastic cylinder of mass M and cross-sectional area A is undergoing vertical SHM while floating in a liquid of density d. The time period of oscillation can be increased by increasing
Solution: Official solution: T = 2π√(M/dAg). Therefore T increases when M increases.
Q14
A parallel plate capacitor with dielectric is charged completely and then the battery is disconnected. Now the dielectric is pulled out. Due to such pulling the
Solution: Official solution: After disconnecting the battery, charge remains same. Pulling out the dielectric decreases capacitance. Since U = q2/2C, energy stored increases.
Q15
The electric potential V as a function of position (x, y, z) in a space is given by V = xy2z3 volt. The electric field (in N/C) at point P(1, 2, -1) m will be given as
Solution: Official solution: E = -∇V. At P(1, 2, -1), Ex = 4, Ey = 4 and Ez = -12. Hence E = 4î + 4ĵ - 12k̂.
Q16
A regular hexagonal loop of side a is formed by a uniform wire. A current flows into it as shown. The magnetic field at the centre C of the hexagon is
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Solution: Official solution: The magnetic field contributions of the current-carrying wire sections cancel at the centre. Net magnetic field is zero.
Q17
The magnetic dip angle at magnetic equator of the earth is
Solution: Official solution: At the magnetic equator, B = BH, so B = B cosθ. Therefore θ = 0°.
Q18
A metallic slider of length L is moving on wire frame with resistance R (kept in uniform magnetic field B) at a speed v as shown in figure. Force F applied by external agent in pulling this slider is
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Solution: Official solution: Induced current i = BLv/R. Magnetic force F = iLB = B2L2v/R.
Q19
In the circuit shown, the power consumed by the bulb B1 long time after the key K is closed is (inductor is ideal)
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Solution: Official solution: Long after closing the key, the ideal inductor behaves like a short. Current i = 10/5 = 2 A through B1. Power P = i2R = 22 × 5 = 20 W.
Q20
In the situation shown the distance between the object O and its image is 24 cm. The object is 6 cm above water level and the beaker containing water upto 8 cm is kept on a plane mirror. Refractive index of the water is
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Solution: Official solution: 2[6 + 8/μ] = 24. Solving gives μ = 4/3.
Q21
In a demonstration of dispersion without deviation phenomenon of thin prism combination, a prism X is chosen with refracting angle 5° and refractive index 1.40. It is combined with another prism Y of material refractive index 1.50. The refracting angle of prism Y is
Solution: Official solution: For no deviation, (1.40 - 1)5° = (1.50 - 1)A. Hence A = 4°.
Q22
Originally the Young's double slit experiment was performed in white light in which higher order spectra in the fringe pattern overlap each other. Assume that wavelength of blue light is 520 nm and that of red light is 780 nm. The value of x for which xth bright red band coincides with (x + 1)th blue band is
Solution: Official solution: (x + 1)520 = x(780). Therefore x = 520/260 = 2.
Q23
Assuming Bohr's model of hydrogen atom, the wavelength of α-line of Balmer series is λ. The wavelength of β-line of Lyman series is
Solution: Official solution: Using the Rydberg formula for Balmer-α and Lyman-β, the required wavelength is 5λ/32.
Q24
Which of the following current-voltage (I-V) graph is characteristics of solar cell?
Solution: Official solution: The I-V characteristic curve of a solar cell lies in the fourth quadrant. Hence option 4 is correct.
Q25
For a particle moving on a straight line, the variation of its acceleration a as a function of time t is as shown. Variation of its position x with time t is best represented by
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Solution: Official solution: Since a is proportional to t, velocity varies as t2 and position varies as t3. The matching x-t graph is option 4.
Q26
From a suitable height two projectiles are thrown simultaneously in horizontally opposite direction with equal speed 20 m/s. Then the time after which the projectiles will be moving perpendicular to each other is (g = 10 m/s2)
Solution: Official solution: For perpendicular velocities, v1.v2 = 0. This gives (gt)2 = 20 × 20, so t = 20/10 = 2 s.
Q27
In the situation shown, reading of the spring balance will be
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Solution: Official solution: Acceleration a = (60 - 15 - 10)/(2 + 3) = 4 m/s2. For the 3 kg block, 60 - T - 15 = 3 × 4, hence T = 33 N.
Q28
A carrom coin of mass 5 g moving with speed 2 m/s strikes elastically with the board wall at 45°. If it remains in contact with the wall for 0.02 second, then the average force acting on the coin is approximately
Solution: Official solution: Average force F = mΔv/t = (5 × 10-3 × 2 × √2)/0.02 = 1/√2 N.
Q29
A circular disc of mass 1 kg and radius 20 cm is rolling down a string without slipping as shown in figure. If acceleration due to gravity at that place is 9 m/s2, then the linear acceleration of the centre of mass of the disc is
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Solution: Official solution: For the disc, mg - T = ma and T = ma/2. Therefore a = 2g/3 = 6 m/s2.
Q30
A spherical raindrop of radius R is moving down in air with terminal velocity 2 m/s. If it clubs with another identical raindrop, then the larger drop will have the terminal velocity
Solution: Official solution: Terminal velocity is proportional to R2. After coalescence, new radius is 21/3R. Hence V' = 2 × 22/3 = 25/3 m/s.
Q31
In the following question, a statement of assertion (A) is followed by a statement of reason (R). A: Pascal's law is the working principle of hydraulic lift. R: Pressure = Thrust/Area.
Solution: Official solution: Assertion and reason are both true, but the stated pressure formula is not the correct explanation of Pascal's law in a hydraulic lift.
Q32
Three metallic rods of identical lengths and cross-section area are welded together to form the letter Y of English alphabet. If their conductivities are in the ratio k : 2k : 3k and their peripheral temperatures are 60°C, 51°C and 20°C respectively as shown, then the junction temperature will be
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Solution: Official solution: T = (60k + 51 × 2k + 20 × 3k)/(k + 2k + 3k) = 37°C.
Q33
A liquid cools from 71°C to 69°C in two minutes. It cools from 41°C to 39°C in 5 minutes. If Newton's law of cooling is obeyed by the liquid, then temperature of surroundings will be
Solution: Official solution: (71 - 69)/2 is proportional to (70 - T), and (41 - 39)/5 is proportional to (40 - T). Solving gives T = 20°C.
Q34
In general in which of the following cases, the speed of mechanical waves is maximum?
Solution: Official solution: Longitudinal waves generally have maximum speed in solids.
Q35
A particle is undergoing simple harmonic motion in a straight line with a time period of 3 second. Starting from the mean position, the 3/8 oscillation will be completed in
Solution: Official solution: Distance covered = 4A × 3/8 = 3A/2. Starting from mean position, t = T/4 + T/6 = 5T/12 = 1.25 s.
Q36
Reading of the ideal ammeter in the circuit shown below is (R = 5 Ω)
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Solution: Official solution: Equivalent resistance gives current I = 10 volt/(5 + 5)Ω = 1 A.
Q37
The potential difference between points x and y in the circuit shown below is
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Solution: Official solution: From potential division, Vx = 40 volt and Vy = 20 volt. Therefore Vx - Vy = 20 volt.
Q38
A point charge q is rotating in a circle of radius R with angular frequency ω. The magnetic field created by it at the centre of the circle will be
Solution: Official solution: Equivalent current I = qf = qω/2π. Magnetic field at centre B = μ0I/(2R) = μ0qω/(4πR).
Q39
Two inductor coils L1 = 18 mH and L2 = 8 mH are perfectly coupled. If a current of 2 A is reversed in first coil within 2 millisecond, then the EMF induced in the second coil will be
Solution: Official solution: For perfect coupling, M = √(L1L2) = 12 mH. Current change is 4 A in 2 ms. EMF = M di/dt = 24 volt.
Q40
An electromagnetic wave has its electric field pointing towards negative z direction and the magnetic field at that instant is pointing towards positive y direction. The direction of propagation of EM wave is
Solution: Official solution: Direction of propagation is along E × B. Here (-z) × (+y) = +x.
Q41
A glass slab is kept over three coloured dots viz., blue, yellow and red. If the coloured dots are viewed from the top then the dot appearing closest to the eye will be
Solution: Official solution: The refractive index is minimum for blue in the given reasoning, so the blue dot appears closest to the eye.
Q42
Power factor is maximum in an LCR series circuit when
Solution: Official solution: cosφ = R/Z = R/√[R2 + (XL - XC)2]. It is maximum when XL = XC.
Q43
In a photoelectric experiment, the stopping potential (V) versus applied frequency (ν) graph for three different metals A, B, C are as shown. The metal with maximum threshold wavelength is
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Solution: Official solution: The threshold frequency is smallest for metal A. Since threshold wavelength is c/ν0, A has maximum threshold wavelength.
Q44
If binding energy of nuclei P, Q, R are 112 MeV, 106 MeV and 230 MeV respectively then energy released in the reaction P + Q → R is
Solution: Official solution: Energy released = 230 - (112 + 106) = 12 MeV.
Q45
The symbolic representation of the network of various logic gates shown is equivalent to
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Solution: Official solution: Simplifying the logic-gate network gives NAND gate.
Final Result
Total Questions45
Attempted Questions0
Correct Answers0
Wrong Answers0
Unattempted Questions45
Positive Marks0
Negative Marks0
Final Score0