neet physics tutor doubt 4 | Kumar Physics Classes
neet physics tutor doubt 4 practice page brings 45 NEET Physics questions from complete Class XI and XII syllabus into clean HTML text, with only diagrams, graphs and circuits cropped from the source PDF.
Test before checking solutions. Each answer panel explains the key idea, and every source-PDF visual is limited to the figure area so question text is not duplicated as an image.
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Two vectors P and Q are of same magnitude and inclined at angle 64°. The angle of vector P - Q with P is
Correct option: (2)
For two equal vectors, the difference vector makes an angle 90° - θ/2 with P. Here θ = 64°, so the angle is 90° - 32° = 58°.
If initial velocity of an object is (2ĵ) m/s and it moves with constant acceleration (î + ĵ) m/s2, then the velocity of particle at the end of 2 s is
Correct option: (1)
Use v = u + at. Thus v = 2ĵ + 2(î + ĵ) = (2î + 4ĵ) m/s.
The mass of a cylindrical body depends on density d of material, diameter D and length L as M ∝ dxDyLz. If maximum percentage errors in density, diameter and length are m%, n% and l% respectively, then maximum percentage error in mass is
Correct option: (2)
For products with powers, percentage errors add with powers: %M = x(%d) + y(%D) + z(%L) = (mx + yn + lz)%.
The diagram shows part of a Vernier scale. Which reading is correct? Least count = 0.1 mm.
Correct option: (2)
Reading = main scale reading + least count × Vernier scale reading = 3.9 mm + (0.01 × 3) mm = 3.93 mm.
A particle moving in positive x direction has initial velocity u. It undergoes retardation αv3, where v is instantaneous velocity and α is a positive constant. The velocity as a function of time t is
Correct option: (3)
dv/dt = -αv3. Integrating from u to v gives 1/v2 = 1/u2 + 2αt, hence v = u/√(1 + 2u2αt).
The horizontal range and maximum height of a projectile are R and H. If a constant horizontal acceleration a = g/4 is imparted due to wind, the new horizontal range will be
Correct option: (2)
Original horizontal distance is R = u cosθ T. Extra distance due to wind is (1/2)(g/4)T2 = H. Therefore R' = R + H.
In the figure, three blocks A, B and C of mass m each have accelerations a1, a2 and a3. F1 and F2 are external forces of magnitude 4mg and 3mg respectively. Choose the correct relation.
Correct option: (3)
From the net force on the moving mass systems: a1 = 3g, a2 = 3g/5 and a3 = 3g/2. Thus a1 > a3 > a2.
A block is placed on a rough horizontal floor and a horizontal force F is applied. The friction f by the floor is measured for different F and a graph is plotted between them. The graph will be
Correct option: (4)
For F < μN, static friction adjusts as f = F. Once F reaches limiting friction, f = μN remains constant.
Two blocks each of mass 5 kg are connected to the ends of a light rod as shown. The rod is rotated about a vertical line of symmetry. It breaks if tension exceeds 500 N. The maximum frequency in rotations per second is
Correct option: (3)
T = m l 4π2f2. With T = 500 N, m = 5 kg and l = 1 m, f2 = 100/(4π2), so f = 5/π rps.
A force F = (40 + 0.5x) acts on a particle in the x direction where F is in N and x in m. The work done from x = 0 to x = 4 m is
Correct option: (4)
W = ∫04(40 + 0.5x)dx = [40x + x2/4]04 = 160 + 4 = 164 J.
A ball hits the floor with speed v making an angle of incidence θ with the normal. If the coefficient of restitution is e, the angle of reflection is
Correct option: (1)
Tangential velocity remains v sinθ while normal velocity after impact becomes ev cosθ. Therefore tanθ' = v sinθ/(ev cosθ) = tanθ/e.
Four particles each of mass 2 kg are kept at the corners of a square of edge 2 m. The moment of inertia about an axis perpendicular to the square and passing through its centre is
Correct option: (3)
Each mass is at distance a/√2 from the centre. I = 4m(a/√2)2 = 2ma2. With m = 2 kg and a = 2 m, I = 16 kg m2.
A uniform solid sphere on a fixed rough horizontal surface is pulled horizontally by force F as shown. If it undergoes pure rolling, the frictional force on the sphere is
Correct option: (4)
For translation, F - f = ma. For rolling torque, fR = (2/5)mR2(a/R), so f = (2/5)ma. Combining gives F = 7f/2, hence f = 2F/7.
If radius of Earth's orbit around the sun is R and time period is T, then mass density of sun is (Rs is radius of sun)
Correct option: (1)
Using mRω2 = GMm/R2 and M = (4/3)πRs3ρs, we get ρs = 3πR3/(GT2Rs3).
If a simple-pendulum based clock is taken deep inside Earth's surface, it
Correct option: (2)
T = 2π√(L/g). At depth, g decreases, so the time period increases. The clock runs slow.
If two wires of same material and length have radii in ratio 1 : 3, then under the same stretching force their elongations are in ratio
Correct option: (3)
Δl = FL/AY. Area is proportional to r2, so elongation is inversely proportional to r2. Ratio = 32 : 12 = 9 : 1.
Water flows through a frictionless tube with varying cross section as shown. The pressure variation P at point x along the axis is best represented by
Correct option: (3)
By continuity, speed is larger where area is smaller. From Bernoulli's principle, pressure is lower where speed is larger. Therefore pressure dips at the narrow sections, matching graph (3).
Reynolds number for which flow of water through a pipe will be turbulent is
Correct option: (4)
Flow becomes turbulent for Reynolds number greater than about 3000, so 4000 is turbulent.
A cube of coefficient of cubical expansion γ1 floats in a liquid of coefficient γ2, with γ2 < γ1, such that 50% of its volume is inside. If temperature increases, the percentage of cube inside the liquid may be
Correct option: (3)
Since γ1 > γ2, the cube volume increases more and its density decreases more than the liquid density. The immersed fraction decreases, so it may become 45%.
The thermodynamic state of an ideal gas varies through different processes as shown. The changes in internal energy for processes A, B and C respectively are
Correct option: (4)
For an ideal gas, internal energy depends on temperature. The shown processes correspond to isobaric compression (ΔU negative), isothermal compression (ΔU = 0), and adiabatic compression (ΔU positive).
Mean free path of O2 molecules in a rigid container is λ. The temperature is doubled. The new mean free path is
Correct option: (1)
In a rigid container number density n is fixed. Since λ = 1/(√2πnd2), mean free path is independent of temperature here.
All surfaces are smooth and springs are ideal. If a block of mass m is given velocity v0 to the left, the time period of the block shown is
Correct option: (3)
The time includes half-oscillation with spring K, free travel across l1 and l2, and half-oscillation with spring 4K. Thus T = (3π/2)√(m/K) + 2(l1 + l2)/v0.
An open and a closed organ pipe have same length. The ratio of frequency of their 4th overtone is
Correct option: (1)
For open pipe, 4th overtone is 5v/2L. For closed pipe, 4th overtone is 9v/4L. Ratio = (5v/2L)/(9v/4L) = 10/9.
A dog is barking at a power of 40π mW. Its intensity level at distance 1 m is
Correct option: (2)
I = P/4πr2 = 40π×10-3/4π = 10-2 W/m2. L = 10log(I/I0) = 100 dB.
A long thin wire is charged with charge per unit length +λ. It is inserted in a hollow sphere of radius 2R. The maximum electric flux coming out of the sphere is
Correct option: (3)
Maximum enclosed length is diameter 4R, so qin,max = 4λR. By Gauss law, flux = 4λR/ε0.
If energy stored by capacitor C2 is 40 μJ, then charge stored by capacitor C1, as shown, is (C = 10 μF).
Correct option: (2)
For C2 = 2C, q2/2(2C) = 40×10-6. This gives q = 40 μC for that branch, so charge on C1 is q/2 = 20 μC.
A charge q is placed on a solid metallic object of irregular shape. Charge q will distribute itself
Correct option: (4)
Charge resides on the outer surface of a conductor. For irregular shape, surface charge density is not uniform and depends on curvature.
A 4 V potentiometer is used to determine internal resistance of a 2.5 V cell. The balance point in open circuit is 75 cm. With a 4 Ω resistor in external circuit, the balance point shifts to 60 cm. The internal resistance is
Correct option: (2)
r = (l1/l2 - 1)R = (75/60 - 1)×4 = (5/4 - 1)×4 = 1 Ω.
The current through the battery connected with the infinite network shown is
Correct option: (1)
For the repeating network, Req = 1 + 2Req/(2 + Req). Solving gives Req = 2 Ω, so i = 10/2 = 5 A.
A proton is accelerated in a cyclotron where magnetic field is 0.5 T. If potential gap is 10 kV, how many revolutions are required to acquire kinetic energy 12 MeV?
Correct option: (1)
In one cycle, proton crosses the gap twice, so energy gained per revolution is 2e×104 V = 2×104 eV. Number = 12×106/(2×104) = 600.
The ratio of magnetic length and geometric length of a bar magnet is about
Correct option: (3)
Magnetic length is approximately 0.84 times the geometric length.
Choose the incorrect statement.
Correct option: (4)
A d.c. ammeter reads average value of current. For i = 1 + 2sinωt, the average is 1 A, not 2 A.
A conducting rod of length L rotates in the x-z plane with angular velocity ω about axis yy' in a uniform magnetic field B as shown. The emf induced between P and Q at the instant shown is
Correct option: (1)
At the instant shown, velocity and magnetic field are parallel or antiparallel for rod elements, so v × B gives no separation along the rod. Hence induced emf is zero.
The primary and secondary coils of an ideal transformer have 100 and 500 turns. Magnetic flux linking primary is φ = 40t + 5. The output voltage across secondary is
Correct option: (3)
Vp = dφ/dt = 40 V. Since Vs/Vp = Ns/Np, Vs = 40×500/100 = 200 V.
The speed of light of an electromagnetic wave in a medium with relative permeability 2 and relative permittivity 8 is
Correct option: (4)
v = c/√(μrεr) = 3×108/√(2×8) = 7.5×107 m/s.
Two identical thin lenses are kept on the same optical axis as shown. The right surface of the right lens is silvered. Focal length of each lens is 20 cm and radius of the silvered surface is 20 cm. The focal length of the combined system is
Correct option: (4)
Two lenses combine to fl,eq = f/2 = 10 cm. With the silvered surface acting as a mirror, 1/f = -2/fl,eq + 1/fm = -2/10 + 1/(-10), so f = -10/3 cm = -3.33 cm.
A longsighted person has minimum distance of distinct vision 75 cm. He wants to reduce it to 25 cm. He should use
Correct option: (2)
Use lens formula with u = -25 cm and v = -75 cm: 1/f = 1/v - 1/u = 2/75, so f = 75/2 cm and P = 100/f = 2.67 D.
In a microscope orange light is used for resolution. Resolving power can be increased by using
Correct option: (4)
Resolving power is proportional to μ/λ. Blue light has smaller wavelength and oil increases refractive index, so both improve resolution.
In Young's double slit experiment, separation between 4th maxima and 3rd minima is n times the fringe width. The value of 2n is
Correct option: (3)
y4 = 4λD/d and y'3 = 5λD/2d. Difference = (4 - 5/2)λD/d = 3β/2, so n = 3/2 and 2n = 3.
For a proton accelerated through V volts, de Broglie wavelength is
Correct option: (2)
λ = h/√(2meV). Substituting proton mass and charge gives λ ≈ 0.286/√V Å.
The absorption transition lines between ground state and nth state of hydrogen atom is 4. The emission transition lines between these energy states will be
Correct option: (2)
Absorption transitions from ground to nth state = n - 1 = 4, so n = 5. Number of emission transitions = n(n - 1)/2 = 5×4/2 = 10.
In one average life
Correct option: (3)
For one average life t = 1/λ, N = N0e-1 = 0.37N0. Decayed nuclei = 0.63N0, more than half.
During β- decay
Correct option: (4)
In β- decay, a neutron converts into a proton, electron and antineutrino: n → p + e- + anti-ν.
The input resistance of a common emitter amplifier is 5 kΩ and current gain is 25. If load resistance is 10 kΩ, transconductance of the transistor is
Correct option: (1)
gm = I0/Vin = Ai/Ri = 25/(5×103) = 5×10-3 Ω-1.
The truth table for the circuit shown is
Correct option: (4)
When both A and B are low, no current flows through the resistor, so Y = 0. If A or B or both become high, current flows and Y becomes high. This is OR-gate truth table.
Use this after attempting questions. Unattempted questions stay neutral.
Solve each question first, choose an option, then open the solution only after committing to an answer.
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