Kumar Physics Classes
Phone / WhatsApp: +91 9958461445

NEET PHYSICS TUTOR DOUBT 41

Class 11 + Class 12 NEET Physics Practice Paper

This premium practice page brings together the Physics section of the assessment paper in a focused quiz format. Attempt every question honestly, review your mistakes with the official solution, and use the score analysis to understand where your preparation needs more attention.

The design is kept lightweight, mobile responsive, and suitable for direct use inside a single Elementor HTML widget.

45 Physics Questions NEET marking: +4 for correct, -1 for wrong, 0 for unattempted. Select an answer, submit, and then open the official solution.

Why Strong Physics Preparation Is Now More Important Than Ever

Physics has become a decisive subject for serious NEET aspirants because it tests clarity, speed, accuracy, and calm thinking under pressure. A student who understands concepts deeply can solve new questions even when the language or diagram looks unfamiliar. Strong Physics preparation also improves overall exam discipline because it builds the habit of logical problem solving. Regular practice, formula revision, error correction, and concept-based doubt solving are now essential parts of a high-quality NEET strategy.

Important Message for NEET 2027 and Future Aspirants

Students preparing for NEET 2027 and future medical entrance exams should start building Physics slowly and consistently. Do not wait for the syllabus to feel complete before solving questions. Learn a concept, revise the formula, solve examples, and immediately test yourself with exam-level problems. Every small improvement in Physics adds up over time. If you stay disciplined from the beginning, the subject becomes manageable, scoring, and even enjoyable.

Concepts Grow Like Savings

Just as money slowly accumulates through hard work and discipline, Physics understanding also grows step by step as concepts become clearer. Every solved question, every corrected mistake, and every understood concept adds value to a student's preparation. The main purpose of these questions is to help students build real conceptual clarity, so that Physics does not remain a burden but becomes a subject they can understand, enjoy, and solve with confidence.

Before You Attempt This Paper: Revise the Formula Bank First

Dear Students, creating and researching a good Physics paper requires a lot of effort. Every question is selected with a purpose, so you should not attempt this paper casually. Before starting the paper, make sure that you revise the important formulas properly. At least the major Class 11 and Class 12 Physics formulas should be fresh in your mind.

This is an attempt to provide a useful formula bank before the question paper so that students can quickly revise the important concepts and then attempt the questions with better confidence. Read the formulas, understand where they apply, and then solve the questions honestly.

Physics becomes easier when formulas are connected with concepts. Do not only memorise formulas; try to understand the meaning behind them. This will help you solve NEET, CBSE, JEE, IB, IGCSE, AP Physics and other Physics exam questions with better accuracy.

Class 11 + Class 12 Physics Formula Bank

Units and Dimensions

  • v = s/t
  • F = ma
  • [E] = ML2T-2
  • [P] = ML-1T-2

Vectors

  • |A + B| = √(A2 + B2 + 2AB cosθ)
  • A · B = AB cosθ
  • |A x B| = AB sinθ

Kinematics

  • v = u + at
  • s = ut + 1/2 at2
  • v2 = u2 + 2as
  • Range = u2 sin2θ/g

Laws of Motion

  • F = ma
  • fmax = μsN
  • fk = μkN
  • Impulse = Δp

Work, Energy and Power

  • W = F s cosθ
  • K = 1/2 mv2
  • P = W/t = Fv
  • Wcon = -ΔU

Circular Motion

  • ac = v2/r = ω2r
  • Fc = mv2/r
  • v = rω

Centre of Mass

  • xcm = Σmixi/Σmi
  • Vcm = Σmivi/M
  • Fext = Macm

Rotational Motion

  • τ = rF sinθ
  • L = Iω
  • K = 1/2 Iω2
  • Irod, centre = ML2/12

Gravitation

  • F = GMm/r2
  • g = GM/R2
  • V = -GM/r
  • ve = √(2GM/R)

Mechanical Properties of Solids

  • Stress = F/A
  • Strain = ΔL/L
  • Y = stress/strain
  • U = 1/2 FΔL

Mechanical Properties of Fluids

  • P = F/A
  • P = P0 + ρgh
  • A1v1 = A2v2
  • P + 1/2ρv2 + ρgh = constant

Thermal Properties of Matter

  • Q = mcΔT
  • Q = mL
  • ΔL = αLΔT
  • H = kAΔT/L

Thermodynamics

  • ΔQ = ΔU + W
  • PV = nRT
  • W = nRΔT
  • η = 1 - T2/T1

Kinetic Theory of Gases

  • PV = 1/3 mNvrms2
  • vrms = √(3RT/M)
  • Kavg = 3/2 kT
  • λ = RT/(√2 πd2NAP)

Oscillations

  • x = A sin(ωt + φ)
  • T = 2π√(m/k)
  • T = 2π√(l/g)
  • E = 1/2 kA2

Waves

  • v = fλ
  • v = √(T/μ)
  • I ∝ A2
  • Beat frequency = |f1 - f2|

Electrostatics

  • F = kq1q2/r2
  • E = kq/r2
  • V = kq/r
  • U = kq1q2/r

Capacitance

  • C = Q/V
  • C = ε0A/d
  • U = 1/2 CV2
  • Cseries-1 = Σ1/C

Current Electricity

  • I = Q/t
  • V = IR
  • R = ρL/A
  • P = VI = I2R

Moving Charges and Magnetism

  • F = qvB sinθ
  • F = BIL sinθ
  • B = μ0I/(2πr)
  • r = mv/(qB)

Magnetism and Matter

  • τ = MB sinθ
  • U = -MB cosθ
  • Baxial = μ02M/(4πr3)

Electromagnetic Induction

  • ε = -dφ/dt
  • φ = BA cosθ
  • L = Nφ/I
  • U = 1/2 LI2

Alternating Current

  • V = V0 sinωt
  • XL = ωL
  • XC = 1/ωC
  • Z = √(R2 + (XL-XC)2)

Electromagnetic Waves

  • c = 1/√(μ0ε0)
  • c = fλ
  • E0/B0 = c

Ray Optics

  • 1/f = 1/v - 1/u
  • m = h'/h = v/u
  • Lens maker: 1/f = (μ-1)(1/R1 - 1/R2)

Wave Optics

  • β = λD/d
  • Path difference = d sinθ
  • a sinθ = nλ
  • I = 4I0 cos2(φ/2)

Dual Nature of Matter

  • E = hν
  • Kmax = hν - φ
  • λ = h/p
  • λ = h/√(2mK)

Atoms

  • En = -13.6/n2 eV
  • rn = n2a0
  • hν = Ei - Ef

Nuclei

  • N = N0e-λt
  • T1/2 = 0.693/λ
  • E = Δmc2
  • R = R0A1/3

Semiconductor Electronics

  • I = I0(eV/ηVT - 1)
  • β = IC/IB
  • α = IC/IE
  • Gain = output/input

Personalised Online Physics Guidance

If you are searching for a Physics Tutor and not getting proper guidance, you may contact Kumar Sir for personalised one-to-one online Physics classes. Whether you are searching for a Physics Tutor in India or anywhere in the world, Kumar Sir can guide you online with concept clarity, doubt-solving, numerical practice, and exam-oriented preparation.

Kumar Physics Classes
Phone / WhatsApp: +91 9958461445
Email: kumarsirphysics@gmail.com
Website: kumarphysicsclasses.com

Question Index
Question 1+4 / -1
The dimensional formula of expression √LC is (where C is capacitance and L is inductance).
Correct Answer: Option 1
[√LC] = 1/[ω] = [T]. Therefore the correct dimensional formula is [M0L0T1].
Question 2+4 / -1
A body is thrown in downward direction from a certain height. The velocity (v) of body varies with position (x) as:
Correct Answer: Option 1
v2 = u2 + 2gx. Therefore the v-x graph is parabolic.
Question 3+4 / -1
Forces are acting at a point as shown in the given figure. The net force is:
Question 3 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 3
The resultant components are F1 and F2 in perpendicular directions. Hence net force = √(F12 + F22).
Question 4+4 / -1
A uniform bar of length 2 m and mass 3 kg is as shown in the figure. Tension at a point in the bar at 1 m from one end is:
Question 4 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 4
For the half bar, T - 6 = 1.5[(12 - 6)/3]. This gives T = 9 N.
Question 5+4 / -1
Work done to carry bob upto angle 60° with the vertical very slowly from initial vertical position is (Given g = 10 m s-2):
Question 5 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 3
W = ΔP.E. = mgl(1 - cosθ) = 1 x 10 x 1 x (1 - 1/2) = 5 J.
Question 6+4 / -1
Four rods each of mass m and length l are arranged in form of square as shown in the figure. The moment of inertia about a side of the square is:
Question 6 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 1
I = 2 x (ml2/3) + ml2 = (5/3)ml2.
Question 7+4 / -1
Which of the following graph represents the variation of gravitational potential (V) with distance (r) from the center of a solid sphere?
Correct Answer: Option 3
Inside the sphere, V = -GM(3R2 - r2)/(2R3). Outside, V = -GM/r. Hence the correct graph is option 3.
Question 8+4 / -1
Ideal fluid is flowing through tube as shown in the figure. If PA and PB are pressure at point A and B respectively, then:
Question 8 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 2
Using Bernoulli's theorem, P + 1/2ρv2 = constant. As vB < vA, PB > PA.
Question 9+4 / -1
Two ends of a uniform conductor of length l are at temperature difference 100°C. The temperature at point P at distance l/4 from one end is about (assuming heat transfers by conduction only):
Question 9 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 1
θ = [100 x (R/4) + 0 x (3R/4)]/R = 25°C.
Question 10+4 / -1
How mean free path (λ) of a gas at constant pressure depends on temperature (T) of the gas?
Correct Answer: Option 2
λ = RT/(√2πd2NP). At constant pressure, λ ∝ T, so the graph is a straight line.
Question 11+4 / -1
The phase difference between two SHMs y1 = 10 sin(20t + π/6) and y2 = 20 cos(20t - π/6) is:
Correct Answer: Option 1
Write y2 = 20 sin(20t - π/6 + π/2). Thus φ = π/6 - (-π/6 + π/2) = π/6.
Question 12+4 / -1
Which of the following statement is true about stationary waves?
Correct Answer: Option 4
Nodes are stationary points while all points between two consecutive nodes oscillate with same frequency. Hence all the listed statements are correct.
Question 13+4 / -1
A point charge is kept at the centre of a conducting shell. The surface charge density is:
Question 13 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 4
Charge density is more at a point closer to a point charge. Here the charge is at the centre, so all shown points are equally distant and the density is same.
Question 14+4 / -1
Electric field due to point charge depends on distance r as:
Correct Answer: Option 4
Electric field due to a point charge is E = KQ/r2. Therefore E ∝ 1/r2.
Question 15+4 / -1
Which of the following is not true for equipotential surfaces?
Correct Answer: Option 3
Equipotential surfaces are denser in strong field and rarer in weak field. They never intersect and electric field lines touch them normally. Work between different equipotential surfaces is not zero.
Question 16+4 / -1
In the given circuit, the reading of voltmeter shown in the figure is:
Question 16 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 1
Rex = (600 x 1200)/(600 + 1200) + 300 = 700 Ω. V = (7 x 400)/700 = 4 V.
Question 17+4 / -1
In the given figure, current carrying conductor will:
Question 17 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 2
The force acting on the conductor creates a clockwise turning tendency. So it will turn clockwise.
Question 18+4 / -1
The magnetic induction at the centre O, if the wire carries current I, is:
Question 18 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 2
For the arc, B = μ0Iθ/(4πR) = μ0I(5π)/(4π x 3R) = 5μ0I/(12R).
Question 19+4 / -1
In the given figure, if magnitude of current I is decreasing continuously in cable then induced current in the loop is:
Question 19 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 3
By Lenz's rule, e = -dφ/dt. The induced current in the loop is clockwise.
Question 20+4 / -1
A very small metallic ring of radius r is kept coplanar at the centre of a square wire of dimension l. The mutual inductance between the loops is:
Question 20 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 2
B = 2√2μ0I/(πl), A = πr2. Thus φ = 2√2μ0Ir2/l and M = 2√2μ0r2/l.
Question 21+4 / -1
In the given circuit, the current through resistor R3, at instant K plugged in is:
Question 21 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 1
At t = 0, inductor provides open circuit whereas capacitor provides short circuit. Therefore current through R3 is zero.
Question 22+4 / -1
The minimum required size of plane mirror to visualize complete image of an object is:
Correct Answer: Option 4
According to the official solution, the minimum size of mirror depends on distance of the object and the observer from the mirror.
Question 23+4 / -1
When YDSE is carried from air to water then choose incorrect statement:
Correct Answer: Option 3
Shift is in proportion to distance of fringe from the center of screen. Therefore shift of all fringes is not same.
Question 24+4 / -1
How much percent of nuclei remain undecayed after one mean life?
Correct Answer: Option 2
N = N0e-λt. After one mean life, N = N0/e = 0.37N0 = 37%.
Question 25+4 / -1
If α is the current gain in common base transistor and β is the current gain in common emitter transistor then:
Correct Answer: Option 3
α = IC/IE and β = IC/IB. As IE > IB, α < β.
Question 26+4 / -1
Two resistors of resistance R1 = (300 ± 6) Ω and R2 = (600 ± 3) Ω are connected in series combination. The maximum percentage error in resistance in series combination is:
Correct Answer: Option 2
R = R1 + R2 = (900 ± 9) Ω. Therefore (ΔR/R) x 100 = 1%.
Question 27+4 / -1
The position (x) of body changes with time (t) as shown in the graph. The average speed of the body in 10 s is:
Question 27 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 2
Average speed = Distance/Time = (5 + 2 + 4)/10 = 1.1 m s-1.
Question 28+4 / -1
The ratio of kinetic energy of a projectile at the point of projection to the maximum height is 4 : 1. The ratio of maximum height to the range is:
Correct Answer: Option 3
KE/KEmax height = 4 = 1/cos2θ, so cosθ = 1/2 and θ = 60°. H/R = tanθ/4 = √3/4.
Question 29+4 / -1
In the given figure strings and pulley are ideal, the reading of spring balance S is:
Question 29 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 3
a = [(2m - m)/(3m)]g = g/3. For the block, mg - T = ma, so T = m(g - a) = m(g - g/3) = 2mg/3.
Question 30+4 / -1
Work done by conservative forces acting on body is always equal to:
Correct Answer: Option 1
For conservative forces, Wconservative = -ΔU. Hence it is the negative of change in potential energy.
Question 31+4 / -1
Three point masses each of mass m are connected by 3 light rigid rod and system is rotating about one rod as shown in the given figure. Moment of inertia of the system is:
Question 31 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 3
The mass away from the axis is at distance l√3/2. Therefore I = m(l√3/2)2 = (3/4)ml2.
Question 32+4 / -1
If at a small height h above the surface of earth acceleration due to gravity decreases by 0.2%, then at depth h below the surface of earth, the acceleration due to gravity will:
Correct Answer: Option 4
At height h, g' = g(1 - 2h/R), so percentage decrease = 2h/R = 0.2%. At depth h, g' = g(1 - h/R), so percentage decrease = h/R = 0.1%.
Question 33+4 / -1
A capillary tube is dipped into the liquid of surface tension S. The rise or fall of a liquid in a capillary tube is given as (where symbols have their usual meanings):
Correct Answer: Option 3
Height of liquid column in a capillary tube is h = 2S cosθ/(rρg).
Question 34+4 / -1
Assertion (A): Power radiated by a body depends only on temperature of body not on properties of its surface.
Reason (R): Good absorbers are good emitters.
Correct Answer: Option 4
According to Stefan's law, P = σeAsT4. Power depends on surface, area and temperature. The reason is true, so option 4 is correct.
Question 35+4 / -1
The pressure P versus volume V of an ideal gas varies as shown in the given figure. Which of the following graph may represent the variation of pressure (P) with temperature (T)?
Question 35 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 3
From the given P-V variation, temperature first increases and then decreases. Hence the correct P-T graph is option 3.
Question 36+4 / -1
A proton and deuteron are accelerated by same potential difference and enter in region of uniform electric field in direction perpendicular to it. The deflection of proton is:
Correct Answer: Option 1
y = qEx2/(2mv2). Since qV = 1/2mv2, y = Ex2/(4V), independent of particle mass and charge ratio here.
Question 37+4 / -1
In the given circuit diagram, the potential difference across 12 μF capacitor is:
Question 37 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 1
V = 18 x [3/(3 + 6)] x [6/(6 + 12)] = 2 V.
Question 38+4 / -1
In the given circuit diagram, the potential difference between A and B is:
Question 38 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 2
VA - VB = (6 x 2 + 3 x 1)/(2 + 1) = 5 V.
Question 39+4 / -1
Infinite long cable is bent as shown in the given figure. The magnetic induction at point O is:
Question 39 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 1
B = 2[μ0I/(4πa)(sin45° + sin90°)] = μ0I(√2 + 1)/(2√2πa).
Question 40+4 / -1
A linear magnet of magnetic moment M is bent in arc as shown in the given figure. New magnetic moment is:
Question 40 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 3
M = ml. For the arc, l = 2πr/6 = πr/3, so r = 3l/π. Hence M' = mr = 3M/π.
Question 41+4 / -1
The current I changes with time t in an inductor as shown in the given figure. The potential difference (e) across the inductor is:
Question 41 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 4
For 0 to t, e = -L(dI/dt) = +L dI/dt. For t to 2t, e = 0. For 2t to 3t, e = -L dI/dt. This matches option 4.
Question 42+4 / -1
For observer P current in loop is clockwise, the moment K is plugged. The induced current in loop observed is:
Question 42 diagram
Kumar Sir +91 9958461445
Correct Answer: Option 2
By Lenz's rule, current in coil B for observer P is anti-clockwise.
Question 43+4 / -1
An electric bulb rated 100 W, 40 V has to be operated across 50 V, 50 Hz A.C. supply. What is the value of inductance of inductor in series with the bulb?
Correct Answer: Option 2
I = 100 W/40 V = 2.5 A. Using V2 = VR2 + VL2, solving gives L = 0.12/π H.
Question 44+4 / -1
Real image can be obtained from:
Correct Answer: Option 4
A concave mirror can form a real image directly. Also, a virtual object can form real image in convex mirror and plane mirror. Hence all of these.
Question 45+4 / -1
Let ν1 be the frequency of first line of Lyman series and ν2 be the frequency of first line of Balmer series, then the frequency of second line of Lyman series is:
Correct Answer: Option 2
ν1 = (E2 - E1)/h and ν2 = (E3 - E2)/h. The second Lyman line is (E3 - E1)/h = ν1 + ν2.

Final Result

Use the button after attempting the paper. Selected but unsubmitted answers are still counted in the final score.

Scroll to Top