1
Dimension formula of Planck's constant is
Official PDF solution
E = hν
ML2T-2 = hT-1
h = [ML2T-1]
ML2T-2 = hT-1
h = [ML2T-1]
Kumar Physics Classes
Phone / WhatsApp: +91 9958461445
Email: kumarsirphysics@gmail.com
Website: kumarphysicsclasses.com
Physics Section | Q1 to Q45
NEET PHYSICS TUTOR DOUBT 33
A focused premium practice page for Class XI and Class XII NEET Physics, built for careful revision, instant checking, official solutions, and exam-style marking.
+4 / -1
NEET marking: correct +4, wrong -1, unattempted 0.
Premium Physics Practice for Serious NEET Preparation
This page is designed to help aspirants revise the Physics section with attention, accuracy, and confidence. Each question includes four clickable options, answer checking, correct answer reveal, official solution support, and final NEET-style score calculation.
At Kumar Physics Classes, every session is shaped around concept clarity, disciplined practice, and doubt removal. Students preparing for NEET need more than memorised formulas; they need the ability to read a question calmly, identify the principle, avoid traps, and finish with confidence. This practice page gives students a compact way to revise important Physics ideas while staying connected with Kumar Sir for deeper guidance, structured mentoring, and regular support. Consistent practice, honest correction, and the right teacher can turn Physics into a scoring strength.
Concept Growth Message
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.
Why Strong Physics Preparation Is Now More Important Than Ever
Physics rewards students who can connect ideas, not merely remember results. A strong preparation plan helps an aspirant understand laws, graphs, circuits, mechanics, optics, waves, and modern physics as one connected subject. This matters because high-quality questions often test reasoning through familiar concepts placed in unfamiliar situations.
When students practise with discipline and review their mistakes honestly, Physics becomes predictable and enjoyable. They begin to see patterns, choose the right formula faster, and avoid wasting time on avoidable confusion. That confidence can make a major difference in the final score.
Important Message for NEET 2027 and Future Aspirants
Students preparing for NEET 2027 and future attempts should start building Physics slowly and seriously. Do not wait for pressure to become the teacher. Begin with concepts, revise formulas with meaning, solve selected questions, and analyse every wrong answer. A small but regular effort today becomes a strong advantage later.
Build fundamentals.
Understand why a formula works before depending on it.
Understand why a formula works before depending on it.
Correct mistakes early.
Every error is useful when it teaches the next step clearly.
Every error is useful when it teaches the next step clearly.
Stay consistent.
Daily practice creates calm speed and examination confidence.
Daily practice creates calm speed and examination confidence.
Question Index
Not attempted
Selected
Correct
Wrong
2
If the maximum percentage error in the measurement of length of thread of simple pendulum and acceleration due to gravity is 2% and 4% respectively, then maximum percentage error in time period of simple pendulum is
Official PDF solution
T = 2π√(ℓ/g)
ΔT/T × 100 = (1/2)(Δℓ/ℓ) × 100 + (1/2)(Δg/g) × 100
= 1% + 2% = 3%
ΔT/T × 100 = (1/2)(Δℓ/ℓ) × 100 + (1/2)(Δg/g) × 100
= 1% + 2% = 3%
3
A block of mass m is sliding down on a fixed smooth inclined plane as shown in figure. The acceleration of the block is (Take g = 10 m/s2)
Kumar Sir +91 9958461445
Official PDF solution
a = g sinθ = 10 × 3/5 = 6 m/s2
4
The potential at the junction A will be
Kumar Sir +91 9958461445
Official PDF solution
(10 - VA)/C + (20 - VA)/C + (30 - VA)/C + (40 - VA)/C = 0
4VA = 100
VA = 25 V
4VA = 100
VA = 25 V
5
In the situation shown in figure, the potential difference between P and Q is
Kumar Sir +91 9958461445
Official PDF solution
i = 20/8 = 2.5 A
VP + 4 - 2.5 × 4 = VQ
VP - VQ = 6 V
VP + 4 - 2.5 × 4 = VQ
VP - VQ = 6 V
6
A straight wire of length L carrying current I is turned into a circular loop of radius r. The magnitude of its magnetic moment will be
Official PDF solution
2πr = L
M = IA = I × πr2 = I × π × [L/(2π)]2
M = IL2/4π
M = IA = I × πr2 = I × π × [L/(2π)]2
M = IL2/4π
7
An electron of charge e moves in a circular orbit of radius r with frequency ν. The magnetic field at the center of orbit is
Official PDF solution
B = μ0I/2r
= (μ0/2r) × (q/t) = (μ0/2r) × (e/t)
B = μ0eν/2r
= (μ0/2r) × (q/t) = (μ0/2r) × (e/t)
B = μ0eν/2r
8
Two thin convex lens each of focal length 20 cm are placed in contact coaxially. If a point object is placed at a distance of 20 cm from the lens combination, then the image formed from the combination is at a distance of
Official PDF solution
Focal length of combination = 10 cm
1/10 = 1/v + 1/20
1/v = 1/10 - 1/20 = 1/20
v = 20 cm
1/10 = 1/v + 1/20
1/v = 1/10 - 1/20 = 1/20
v = 20 cm
9
Two circular coils of one turn have radii in ratio 1 : 2. If the magnetic field at their respective centers are equal then the ratio of the current in the coil is
Official PDF solution
B1 = B2
μ0i1/2r1 = μ0i2/2r2
i1/i2 = r1/r2 = 1/2
μ0i1/2r1 = μ0i2/2r2
i1/i2 = r1/r2 = 1/2
10
Three rods of same thermal conductivity are joined like an English alphabet y as shown in figure. The temperature at the junction O is
Kumar Sir +91 9958461445
Official PDF solution
KA(20 - T)/2L + KA(40 - T)/L = KA(T - 0)/2L
On solving, T = 25°C
On solving, T = 25°C
11
Electromagnetic waves do not transport
Official PDF solution
The PDF states: Answer (2). Electromagnetic waves transport energy, momentum and information, but not charge.
12
Two periodic coherent waves of intensities I1 and I2 pass through a region at the same time in the same direction. The sum of the minimum and maximum intensities is
Official PDF solution
Imax = I1 + I2 + 2√(I1I2)
Imin = I1 + I2 - 2√(I1I2)
Imax + Imin = 2(I1 + I2)
Imin = I1 + I2 - 2√(I1I2)
Imax + Imin = 2(I1 + I2)
13
In Young's double slit experiment, the distance between two slits is halved and the distance between the screen and slit is made three times. Then width of the fringe becomes
Official PDF solution
Use β = λD/d
β' = λ × 3D / (d/2) = 6β
β' = λ × 3D / (d/2) = 6β
14
The de Broglie wavelength of an electron in the first Bohr orbit is equal to
Official PDF solution
λ = h/mv
According to Bohr theory, mvr = h/2π
Therefore h/mv = λ = 2πr
According to Bohr theory, mvr = h/2π
Therefore h/mv = λ = 2πr
15
The logic symbol shown in figure represents
Kumar Sir +91 9958461445
Official PDF solution
Y = overline(overline(A · B)) = A · B
16
For the given position time relation t = αx2 + βx, where α and β are constants, the retardation of the particle is [v = instantaneous velocity]
Official PDF solution
As t = αx2 + βx
dt/dx = 2αx + β
v = 1/(2αx + β) = (2αx + β)-1
dv/dt = a = -2αv3
Retardation = 2αv3
dt/dx = 2αx + β
v = 1/(2αx + β) = (2αx + β)-1
dv/dt = a = -2αv3
Retardation = 2αv3
17
A man can throw a stone up to maximum distance x on ground. The maximum height to which it will rise is
Official PDF solution
For maximum range θ = 45°
R = X = u2/g
H = u2sin245°/2g = u2/4g = X/4
R = X = u2/g
H = u2sin245°/2g = u2/4g = X/4
18
A spring of spring constant 500 N/m is stretched initially by 5 cm from unstretched position. Then work required to stretch it further by another 5 cm is
Official PDF solution
w = (1/2)k[x22 - x12]
= (1/2) × 500[100 - 25] × 10-4
= 1.875 J
= (1/2) × 500[100 - 25] × 10-4
= 1.875 J
19
The bob of a simple pendulum having length 2 m, is displaced from mean position to an angular position θ from vertical. If it is released, then velocity of bob at lowest position will be
Official PDF solution
mgℓ(1 - cosθ) = (1/2)mv2
v = √(2gℓ(1 - cosθ)) = 2√(g(1 - cosθ))
v = √(2gℓ(1 - cosθ)) = 2√(g(1 - cosθ))
20
The potential energy of a point object of mass 1 kg free to move along the x-axis is given by U(x) = (x4/4 - x2/2) J. The total mechanical energy of the particle is 2 J. Then the maximum speed (in m/s) is
Official PDF solution
U = x4/4 - x2/2
dU/dx = x3 - x
For Umin, dU/dx = 0, so x = ±1
Umin = 1/4 - 1/2 = -1/4
K + (-1/4) = 2
(1/2)mv2 = 2.25
v2 = 9/2, so v = 3/√2
dU/dx = x3 - x
For Umin, dU/dx = 0, so x = ±1
Umin = 1/4 - 1/2 = -1/4
K + (-1/4) = 2
(1/2)mv2 = 2.25
v2 = 9/2, so v = 3/√2
21
The initial angular acceleration of the rod, when the string A is just cut is
Kumar Sir +91 9958461445
Official PDF solution
When string is cut:
mgL/4 = (mL2/12 + mL2/16)α
Therefore α = 12g/7L
mgL/4 = (mL2/12 + mL2/16)α
Therefore α = 12g/7L
22
If the radius of earth contracts to half of its present value, then length of the day will be
Official PDF solution
I1ω1 = I2ω2
(2/5)mR2(2π/24) = (2/5)M(R/2)2(2π/T)
On solving, T = 6 h
(2/5)mR2(2π/24) = (2/5)M(R/2)2(2π/T)
On solving, T = 6 h
23
A uniform rod of length l, hinged at lower end is free to rotate in the vertical plane. If the rod is held vertically in beginning and released, then angular acceleration of the rod at the instant when it makes an angle of 45° with the horizontal for first time is
Official PDF solution
τ = Iα
mgℓ/(2√2) = (mℓ2/3)α
Therefore α = 3g/(2√2ℓ)
mgℓ/(2√2) = (mℓ2/3)α
Therefore α = 3g/(2√2ℓ)
24
An object of mass 2 kg is fired vertically upward from the surface of earth and reaches up to height equal to radius of earth. The initial speed of object is (g = 10 m/s2)
Official PDF solution
Use (1/2)mv2 = mgh/(1 + h/R)
v = √(gR) = √(10 × 6.4 × 106) = 8 km/s
v = √(gR) = √(10 × 6.4 × 106) = 8 km/s
25
For a satellite moving in an orbit around the planet, the ratio of magnitude of potential energy to kinetic energy is
Official PDF solution
|P.E|/K.E = |-GMem/Re| / [(1/2)GMem/Re] = 2/1
26
A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will
Official PDF solution
The PDF states: Answer (3). The force between the plates remains unchanged.
27
Two capillary tubes of same length l but radii r1 and r2 are fitted in parallel to the bottom of a vessel. What should be the radius r of the single tube of same length, that can replace the two tubes, so that the rate of flow is same as before.
Official PDF solution
πpr4/8ηℓ = πpr14/8ηℓ + πpr24/8ηℓ
r4 = r14 + r24
r4 = r14 + r24
28
A wire suspended vertically from one of its end is stretched by attaching a block of mass 200 kg to the lower end. The weight stretches the wire by 1 mm. Then the elastic potential energy stored in the wire is g = 10 m/s2
Official PDF solution
U = (1/2)Fℓ = (1/2) × 200 × 10 × 1 × 10-3 = 1 J
29
Equation of SHM is given as x = 3 sin 20πt + 4 cos 20πt, where x is in cm and t in seconds. The amplitude of SHM is
Official PDF solution
A = √(32 + 42) = 5 cm
30
A simple pendulum performs simple harmonic motion about x = 0 with an amplitude A and time period T. The speed of the pendulum at x = A/2 will be
Official PDF solution
u = ω√(A2 - A2/4) = (√3/2)Aω
u = πA√3/T
u = πA√3/T
31
Velocity of sound in a mixture of gases is 320 m/s. Pressure of mixture is now made 27 times keeping the temperature same. Velocity of sound in the given mixture will now be
Official PDF solution
vsound = √(γRT/M), which is independent of pressure.
32
Angle subtended by the vector 4î + 3ĵ + 12k̂ with y-axis is
Official PDF solution
cosθ = 3/√(42 + 32 + 122) = 3/13
θ = cos-1(3/13)
θ = cos-1(3/13)
33
A projectile is fired at angle 45° with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is
Official PDF solution
tanφ = 2h/R = [2u2/4g] / [u2/g] = 1/2
Therefore φ = sin-1(1/√5)
Therefore φ = sin-1(1/√5)
34
A Carnot engine operates with source at 127°C and sink at 27°C. If the source supplies 40 kJ of heat energy, then work done by the engine is
Official PDF solution
η = 1 - 300/400 = 1/4 = w/Q1
w = Q1/4 = 10 kJ
w = Q1/4 = 10 kJ
35
Standing waves are produced in a 10 m long stretched string fixed at both the ends. If the string vibrates in 5 segments and wave velocity is 20 m/s, then frequency is
Official PDF solution
n = 5u/2ℓ = (5 × 20)/(2 × 10) = 5 Hz
36
A police car with a siren of frequency 8 kHz is moving with uniform velocity 36 km/hr towards a tall building which reflects the sound waves. The speed of sound in air is 320 m/s. The frequency of siren observed by the car driver is (Approximately)
Official PDF solution
n' = n[(v + v0)/(v - vs)]
= 8[330/310] ≃ 8.5 kHz
= 8[330/310] ≃ 8.5 kHz
37
n small drops of same size are charged to potential V volt each. If they coalesce to form a single large drop, then its potential will be
Official PDF solution
V' = KQ/R = KnCV/(n1/3r) = n2/3V
38
In a potentiometer, when the cell of emf E is connected null point is obtained at 1.2 m. When the resistor of resistance 10 Ω is connected across it, the balance point decreases by 40 cm. The internal resistance of cell is
Official PDF solution
r = R(ℓ1/ℓ2 - 1)
= 10[120/80 - 1] = 5 Ω
= 10[120/80 - 1] = 5 Ω
39
Two cells with same emf E and different internal resistance r1 and r2 are connected in series to an external resistance R. The value of R so that the potential difference across the first cell be zero is
Official PDF solution
V = E - ir1 = 0
i = E/r1
i = 2E/(R + r1 + r2) = E/r1
R + r1 + r2 = 2r1
R = r1 - r2
i = E/r1
i = 2E/(R + r1 + r2) = E/r1
R + r1 + r2 = 2r1
R = r1 - r2
40
In a given system of masses, the work done by gravity during the fourth second after the system is released from rest, will be (g = 10 m/s2)
Kumar Sir +91 9958461445
Official PDF solution
a = [(3 - 2)/(3 + 2)]g = g/5 = 2 m/s2
S4th = (2/2)(2 × 4 - 1) = 7 m
w = 3 × 10 × 7 - 2 × 10 × 7 = 70 J
S4th = (2/2)(2 × 4 - 1) = 7 m
w = 3 × 10 × 7 - 2 × 10 × 7 = 70 J
41
A transformer of efficiency 90% draws an input power of 4 kW. An electrical appliance connected across the secondary draws a current of 6 A. The impedance of the appliance is
Official PDF solution
Efficiency, η = output/input = 0.9
Output = 0.9 × 4 = 3.6 kW
3.6 × 103 = 36R
R = 100 Ω
Output = 0.9 × 4 = 3.6 kW
3.6 × 103 = 36R
R = 100 Ω
42
The ratio of the intensity at the centre of a bright fringe to the intensity at a point one quarter of the fringe width from the centre is
Official PDF solution
Phase difference at quarter of fringe width = π/2
IC/I = [a2 + a2 + 2a2cos0] / [a2 + a2 + 2a2cos(π/2)]
= 4a2/2a2 = 2
IC/I = [a2 + a2 + 2a2cos0] / [a2 + a2 + 2a2cos(π/2)]
= 4a2/2a2 = 2
43
The de-Broglie wavelength of electron having energy 10 keV is
Official PDF solution
λ = 12.27/√V = 12.27/√(10 × 103) = 0.12 Å
44
An equiconvex lens of refractive index 1.5 has power 4 D in air. Its power in water is (R.I. of water = 4/3)
Official PDF solution
Pa = 1/f = (1.5 - 1)[1/R1 - 1/R2] = 4
(1/R1 - 1/R2) = 8
Pw = (9/8 - 1)[1/R1 - 1/R2] = (1/8) × 8 = 1.0 D
(1/R1 - 1/R2) = 8
Pw = (9/8 - 1)[1/R1 - 1/R2] = (1/8) × 8 = 1.0 D
45
A: An iron needle placed carefully on the surface of water may float, whereas a solid ball of same material will always sink
R: The buoyancy on an object depends both on the material and shape of object.
Official PDF solution
Buoyancy depends on shape of body and density of the fluid.
Final Result
Use the button after attempting the paper to calculate NEET marking: Correct Answer +4, Wrong Answer -1, Unattempted 0.