neet physics tutor doubt 6: 45 Premium NEET Physics Questions Selected by Kumar Sir
neet physics tutor doubt 6 Previous Year NEET Physics questions, NCERT-based conceptual questions and trap-focused practice for serious NEET aspirants.
Question 1. A particle moves such that its acceleration is given by a = - p(x - c), where p and c are positive constants. Time period of oscillations of particles is
2π√c
2π 1√(p + c)
2π√p
2π 1√(p - c)
Correct answer: (3) 2π√p
Solution:
Given a = -p(x - c). Comparing with SHM form a = -ω2(x - c), we get ω2 = p.
So ω = √p and T = 2πω = 2π√p.
Therefore, the correct option is (3).
Question 2Unattempted
Question 2. The ratio of lengths, radii and Young moduli of two wires of material A and B are x, y and z respectively. The ratio of increase in their lengths is
Question 3. A wall is made of two layers P and Q of same thickness but different materials. The thermal conductivity of P is thrice that of Q. In steady state, temperature difference across the wall is 36°C. The temperature difference across layer P is
10°C
12°C
9°C
15°C
Correct answer: (3) 9°C
Solution:
In steady state, heat current through both layers is same.
kP(36 - T) = kQT. Since kP = 3kQ, 3(36 - T) = T.
Thus 108 - 3T = T, so T = 27°C. Temperature drop across layer P is 36°C - 27°C = 9°C.
Therefore, the correct option is (3).
Question 4Unattempted
Question 4. A gas mixture consists of 2 moles of oxygen and 4 moles of helium at absolute temperature T. Neglecting all vibrational modes, the total internal energy of system is
11 RT
14 RT
9 RT
15 RT
Correct answer: (1) 11 RT
Solution:
Oxygen is diatomic with 5 active degrees of freedom, so UO2 = 2 × (5/2)RT = 5RT.
Helium is monoatomic with 3 degrees of freedom, so UHe = 4 × (3/2)RT = 6RT.
Total internal energy = 5RT + 6RT = 11RT.
Therefore, the correct option is (1).
Question 5Unattempted
Question 5. Two persons A and B standing at ends of a plank of length 5 m and mass 80 kg swap their positions. The plank is placed on a smooth horizontal surface and masses of A and B are 60 kg and 40 kg respectively. What is displacement of A with respect to the horizontal surface?
609 m
409 m
507 m
556 m
Correct answer: (2) 409 m
Solution:
External horizontal force is zero, so the centre of mass does not shift. Let the plank move left by x.
Taking displacement of the centre of mass zero: 60(5 - x) - 40(5 + x) + 80(-x) = 0.
300 - 60x - 200 - 40x - 80x = 0, so x = 100180 = 59 m.
Displacement of A with respect to ground = 5 - 59 = 409 m.
Therefore, the correct option is (2).
Question 6Unattempted
Question 6. A particle of mass 4 kg is thrown from ground with speed of 50 m/s at an angle of 53° from horizontal. Torque on it about point of projection at time t (t < time of flight) is
(1200 t) Nm
(550 t) Nm
(600 t) Nm
(880 t) Nm
Correct answer: (1) (1200 t) Nm
Solution:
The force is weight mg downward. The perpendicular lever arm about the point of projection is the horizontal displacement x = u cos 53° · t.
Torque magnitude τ = mgx = 4 × 10 × 50 × cos 53° × t = 4 × 10 × 50 × 0.6 × t = 1200t Nm.
Therefore, the correct option is (1).
Question 7Unattempted
Question 7.P-V diagram for an ideal gas is shown in process. At A and B, internal energy of gas is 10 J and 50 J respectively. If heat given to gas during the process BC is 60 J, then internal energy of gas in the state C is
180 J
200 J
250 J
150 J
Correct answer: (3) 250 J
Solution:
For process BC, work done by gas equals area under the line. Since volume decreases from 6 m3 to 2 m3, work is negative.
WBC = -(1/2)(20 + 50)(6 - 2) = -140 J.
By first law, Q = ΔU + W. Hence 60 = ΔU - 140, so ΔU = 200 J.
UC - UB = 200 J, so UC = 50 + 200 = 250 J.
Therefore, the correct option is (3).
Question 8Unattempted
Question 8. Current I in an inductor coil varies with time t, according to the adjoining graph. The graph which correctly shows the variation of voltage in the coil is
Graph (1)
Graph (2)
Graph (3)
Graph (4)
Correct answer: (3) Graph (3)
Solution:
For an inductor, V = -L dI/dt.
From 0 to t1, current has a positive constant slope, so voltage is negative constant. From t1 to t2, slope is zero, so voltage is zero. From t2 to t3, slope is negative constant, so voltage is positive constant.
This matches graph (3).
Therefore, the correct option is (3).
Question 9Unattempted
Question 9. A square metal loop of side 10 cm and resistance 1 Ω is moved with constant velocity partly inside a uniform magnetic field 2 T, directed into the paper as shown. The loop is connected to a network of three resistors R = 3 Ω. If steady current of 2 mA flows in loop, then speed of loop is
1 cm/s
2 cm/s
0.5 cm/s
4 cm/s
Correct answer: (2) 2 cm/s
Solution:
Induced emf in the loop is E = BvL = 2 × v × 0.10 = 0.2v volt.
The resistor network has equivalent resistance R/3 = 1 Ω. Including loop resistance, total resistance is 1 Ω + 1 Ω = 2 Ω.
Current I = E/Req, so 2 × 10-3 = 0.2v/2.
Thus v = 2 × 10-2 m/s = 2 cm/s.
Therefore, the correct option is (2).
Question 10Unattempted
Question 10. The work functions for two metals P and Q are φ1 and φ2 respectively with φ1 > φ2. The graphs between stopping potential (V0) and frequency (ν) of incident radiation for them would look like
Graph (1)
Graph (2)
Graph (3)
Graph (4)
Correct answer: (2) Graph (2)
Solution:
Photoelectric equation gives V0 = (h/e)ν - φ/e.
The slope h/e is same for both metals, so the lines are parallel. The metal with larger work function has larger threshold frequency, so graph for P is shifted to the right of Q.
This is graph (2).
Therefore, the correct option is (2).
Question 11Unattempted
Question 11. An electron with kinetic energy K (eV) collides with hydrogen atom in ground state. The collision is observed to be elastic for
0 < K < ∞
Only 0 < K < 3.4 eV
0 < K < 10.2 eV
0 < K < 13.6 eV
Correct answer: (3) 0 < K < 10.2 eV
Solution:
An inelastic collision can occur only if the hydrogen atom absorbs enough energy to reach its first excited state.
For hydrogen, energy needed from ground state to first excited state is 10.2 eV. If K < 10.2 eV, excitation is impossible, so collision remains elastic.
Therefore, the correct option is (3).
Question 12Unattempted
Question 12. The fraction of the initial activity of a radioactive material which remains after half of a half-life of the radioactive sample is
0.707
0.37
0.63
0.5
Correct answer: (1) 0.707
Solution:
Activity follows A/A0 = (1/2)t/T1/2.
Here t = T1/2/2, so A/A0 = (1/2)1/2 = 1/√2 = 0.707.
Therefore, the correct option is (1).
Question 13Unattempted
Question 13. Electric field in a region is given by E = (2x î + 4y ĵ) V/m where x and y are in metre. A charged particle 1 C is moved from origin to point P(2 m, 4 m), first along the y-axis and then along x-axis. The work done by field is
Zero
36 J
26 J
20 J
Correct answer: (2) 36 J
Solution:
Along y-axis from (0,0) to (0,4), work W1 = ∫4y dy from 0 to 4 = 32 J.
Then along x-axis from (0,4) to (2,4), work W2 = ∫2x dx from 0 to 2 = 4 J.
Total work = 32 + 4 = 36 J.
Therefore, the correct option is (2).
Question 14Unattempted
Question 14. The electric potential V in space as a function of co-ordinates is given by V = 1/x + 1/y + 1/z. The electric field intensity at (2, 2, 2) is given by (in V/m)
î + ĵ + k̂
î + ĵ + k̂2
2î + 2ĵ + 2k̂
î + ĵ + k̂4
Correct answer: (4) î + ĵ + k̂4
Solution:
E = -∇V. Since V = 1/x + 1/y + 1/z, we get Ex = 1/x2, Ey = 1/y2, and Ez = 1/z2.
At (2,2,2), E = (î + ĵ + k̂)/4 V/m.
Therefore, the correct option is (4).
Question 15Unattempted
Question 15. For a satellite to appear stationary to an observer on earth, which of the following is false?
Its sense of rotation must be from west to east
It must be rotating about the Earth's axis
It must be rotating in the equatorial plane
Its time period must be 84.6 min
Correct answer: (4) Its time period must be 84.6 min
Solution:
A geostationary satellite must move west to east in the equatorial plane about Earth's axis, and its period must be 24 hours.
84.6 min is the period of a near-earth satellite, not a geostationary satellite.
Therefore, the correct option is (4).
Question 16Unattempted
Question 16. A uniform wire of resistance 8 Ω is bent into the form of circle of radius r. A specimen of same wire is connected along the diameter of the circle. The equivalent resistance across the ends of this wire is
(48 + π) Ω
(84 + 3π) Ω
(84 + π) Ω
(84 + 2π) Ω
Correct answer: (3) (84 + π) Ω
Solution:
Resistance of circular wire is R = 8 Ω. Each semicircle has resistance R/2.
Resistance of the diameter made from same wire is (R/2πr) × 2r = R/π.
The two semicircles and diameter are in parallel, so 1/R' = 1/(R/2) + 1/(R/2) + 1/(R/π) = 4/R + π/R.
Thus R' = R/(4 + π) = 8/(4 + π) Ω.
Therefore, the correct option is (3).
Question 17Unattempted
Question 17. An electric heater is designed to operate in 200 V mains with output of 1000 W. When it is connected to 50 V source, power output is
62.5 W
250 W
500 W
1000 W
Correct answer: (1) 62.5 W
Solution:
For a fixed resistance heater, P ∝ V2.
P2 = P1(V2/V1)2 = 1000(50/200)2 = 1000/16 = 62.5 W.
Therefore, the correct option is (1).
Question 18Unattempted
Question 18. A capacitor of capacity C1 = 10 μF is charged to 40 V and a second capacitor of capacity C2 = 15 μF is charged to 30 V. If they are connected in parallel, with like charged plates, the charge flow is
20 μC
45 μC
60 μC
75 μC
Correct answer: (3) 60 μC
Solution:
Common potential V = (C1V1 + C2V2)/(C1 + C2).
V = (10×40 + 15×30)/(10+15) = 850/25 = 34 V.
Charge flowing from C1 = C1(40 - 34) = 10 μF × 6 V = 60 μC.
Therefore, the correct option is (3).
Question 19Unattempted
Question 19. A positive charged particle q, mass m is dropped from certain point above ground. A constant horizontal magnetic field of magnitude B is present. If charge falls down by height h, then speed of charge is (neglect effect of air)
√(2gh)
√(2gh + qB/m)
√(qB/m)
√(gh + qB/2m)
Correct answer: (1) √(2gh)
Solution:
Magnetic force is always perpendicular to velocity, so it does no work.
The gain in kinetic energy equals work done by gravity: (1/2)mv2 = mgh.
Thus v = √(2gh).
Therefore, the correct option is (1).
Question 20Unattempted
Question 20. If ratio of time periods of circular motion of two charged particles in magnetic field is 1 : 2, and they have same charge and same speed, then the ratio of their kinetic energies is
1 : 1
1 : 2
2 : 1
1 : √2
Correct answer: (2) 1 : 2
Solution:
For circular motion in magnetic field, T = 2πm/qB. Same charge and same field imply T ∝ m.
So m1 : m2 = 1 : 2. With same speed, K = (1/2)mv2, hence K ∝ m.
Therefore K1 : K2 = 1 : 2.
Therefore, the correct option is (2).
Question 21Unattempted
Question 21. If a magnetic dipole of dipole moment M is rotated through an angle θ with respect to the direction of magnetic field B, then work done by field is
-MB cosθ
MB[1 - cosθ]
MB[sinθ - 1]
MB[cosθ - 1]
Correct answer: (4) MB[cosθ - 1]
Solution:
Potential energy of magnetic dipole is U = -MB cosθ.
Work done by field from θ = 0 to θ is -ΔU = -[-MB cosθ - (-MB)] = MB(cosθ - 1).
Therefore, the correct option is (4).
Question 22Unattempted
Question 22. A plane mirror of length L moves vertically along y-axis as shown in figure with velocity v. A point source S illuminates mirror. The rate at which length of light spot is growing on wall is
Zero
v
2v
4v
Correct answer: (1) Zero
Solution:
For a plane mirror, the image displacement effect doubles the mirror displacement, so both end-points of the light spot on the wall move with speed 2v.
Since both ends move at the same speed, the separation PQ, i.e. the length of the light spot, remains constant.
Hence the rate of growth of the length is zero.
Therefore, the correct option is (1).
Question 23Unattempted
Question 23. A ray of light is incident at angle 60° on a prism kept in air. The angle of refraction of prism, angle of deviation and angle of emergence are equal. The refractive index of prism is
1.414
1.73
1.67
1.33
Correct answer: (2) 1.73
Solution:
Let prism angle A, deviation δ and emergence angle e be equal. Given i = 60°.
δ = i + e - A. Since e = A, δ = i = 60°. Also e = i, so this is minimum deviation condition.
Question 24. Two incoherent waves of intensities I0 and 4I0 superimpose at a place. What is resultant intensity at that point?
4I0
5I0
7I0
3I0
Correct answer: (2) 5I0
Solution:
For incoherent waves, intensities add directly.
I = I0 + 4I0 = 5I0.
Therefore, the correct option is (2).
Question 25Unattempted
Question 25. In YDSE experiment, electron beam is used in place of light which was accelerated through some potential difference. If accelerating potential difference is made four times, then width of fringes will become
2 times
4 times
12 times
14 times
Correct answer: (3) 12 times
Solution:
For electrons, de Broglie wavelength λ = h/√(2meV).
Fringe width β = λD/d, so β ∝ 1/√V.
If V becomes 4V, fringe width becomes 1/2 of the original.
Therefore, the correct option is (3).
Question 26Unattempted
Question 26. A particle moves along x-axis in such a way that distance covered by it in any second exceeds that in the previous second by 2 m. If it covers 1 m distance in the first second, then its distance at the end of n seconds is (in metre)
2n
n2
n3
2n2
Correct answer: (2) n2
Solution:
Distances covered in successive seconds form an AP: 1, 3, 5, ... up to n terms.
Total distance S = n/2[2×1 + (n - 1)2] = n/2(2n) = n2.
Therefore, the correct option is (2).
Question 27Unattempted
Question 27. A boatman finds that he can save 2 seconds in crossing a river by the quickest path, than by the shortest path. If velocity of boat in still water is 13 m/s and speed of flow of water is 5 m/s, then width of river is
189 m
408 m
256 m
312 m
Correct answer: (4) 312 m
Solution:
Quickest path time t1 = d/13.
For shortest path, effective crossing speed is √(132 - 52) = 12 m/s, so t2 = d/12.
Given t2 - t1 = 2, so d/12 - d/13 = 2.
d/(156) = 2, hence d = 312 m.
Therefore, the correct option is (4).
Question 28Unattempted
Question 28. A very small block of weight W is pulled at uniform velocity by a constant force F, through a string AB having length a. The end B is at height b above the horizontal rough surface. The coefficient of friction between block and horizontal surface is
W(a + b)F(a - b)
F√(a2 - b2)Wa - Fb
W√(a2 + b2)F(b - a)
F(a - b)W√(a2 + b2)
Correct answer: (2) F√(a2 - b2)Wa - Fb
Solution:
Let the string make angle θ with the horizontal. Since AB = a and height is b, sinθ = b/a and cosθ = √(a2 - b2)/a.
Vertical equilibrium: N + F sinθ = W, so N = W - F sinθ.
Horizontal equilibrium at uniform velocity: F cosθ = μN.
Thus μ = F cosθ/(W - F sinθ) = F√(a2 - b2)/(Wa - Fb).
Therefore, the correct option is (2).
Question 29Unattempted
Question 29. The magnitude of the force acting along (6î - 2k̂ + 3ĵ), which displaces a particle from (2, 1, 0) to (1, 4, -1) and does work 5 J, is
11 N
7 N
15 N
9 N
Correct answer: (2) 7 N
Solution:
Direction vector is 6î + 3ĵ - 2k̂ with magnitude √(36 + 9 + 4) = 7.
Question 30. Efficiency of a water pump is 80% and it lifts 25 kg water per second to 8 m height. The power consumed by pump is
2.5 kW
2.0 kW
1.6 kW
2.4 kW
Correct answer: (1) 2.5 kW
Solution:
Useful output power = mgh/t = 25 × 10 × 8 = 2000 W.
Efficiency = Pout/Pin = 0.80.
Pin = 2000/0.80 = 2500 W = 2.5 kW.
Therefore, the correct option is (1).
Question 31Unattempted
Question 31. Four particles each having mass m are placed at the four corners of a square having edge length l. The moment of inertia of system about an axis along its one diagonal is
4 ml2
2ml2
ml2
ml22
Correct answer: (3) ml2
Solution:
Two masses lie on the diagonal axis, so their distance from the axis is zero.
The other two masses are each at perpendicular distance l/√2 from the diagonal.
I = 2m(l/√2)2 = 2m(l2/2) = ml2.
Therefore, the correct option is (3).
Question 32Unattempted
Question 32. Select incorrect statement.
For mountains, the position of centre of gravity will be below that of centre of mass of mountain.
Centre of mass of regular bodies is always at its geometrical centre.
The centre of mass is closer to the heavier mass compared to a lighter mass.
Centre of mass may lie outside an object.
Correct answer: (2) Centre of mass of regular bodies is always at its geometrical centre.
Solution:
The centre of mass of a regular body is at its geometrical centre only when mass distribution is uniform.
Therefore the word “always” makes statement (2) incorrect.
Therefore, the correct option is (2).
Question 33Unattempted
Question 33. Within proportional limit, the slope of the stress-strain curve yields
Elastic fatigue
Modulus of elasticity
Poisson's ratio
Working stress
Correct answer: (2) Modulus of elasticity
Solution:
Within proportional limit, stress ∝ strain and slope = stress/strain.
This ratio is Young's modulus or modulus of elasticity.
Therefore, the correct option is (2).
Question 34Unattempted
Question 34. The terminal speed of a spherical rain drop of mass m and radius r, falling through air is proportional to
mr2
mr
r2
mr3
Correct answer: (2) mr
Solution:
For a spherical drop falling through air, terminal speed by Stokes' law is vt = 2r2(ρ - σ)g/(9η).
For drops of the same material, m ∝ r3. Hence r2 ∝ m/r.
So vt ∝ m/r.
Therefore, the correct option is (2).
Question 35Unattempted
Question 35. If velocity of light (3 × 108 m/s), acceleration due to gravity (9.8 m/s2) and density of water (103 kg/m3) are adopted as the fundamental units, then unit of length will be
6.2 × 1016 m
4 × 1017 m
10.5 × 1012 m
9.2 × 1015 m
Correct answer: (4) 9.2 × 1015 m
Solution:
Dimensionally, length can be obtained from velocity and acceleration as L = v2/g.
Unit length = (3 × 108)2/9.8 = 9.2 × 1015 m.
Therefore, the correct option is (4).
Question 36Unattempted
Question 36. A particle executes SHM between two points separated by 20 cm along x-axis. If the force experienced by it be 2 N at displacement 1 cm from mean position, then the maximum force experienced by it is
20 N
40 N
10 N
80 N
Correct answer: (1) 20 N
Solution:
The amplitude is half of total separation, so A = 10 cm.
In SHM, restoring force magnitude F = kx, so F ∝ x.
At x = 1 cm, F = 2 N. At x = 10 cm, Fmax = 10 × 2 = 20 N.
Therefore, the correct option is (1).
Question 37Unattempted
Question 37. If the difference in frequencies corresponding to tones emitted by stretched string vibrating in its 2nd overtone and seventh harmonic is 1600 Hz, then fundamental frequency of string is
200 Hz
400 Hz
800 Hz
600 Hz
Correct answer: (2) 400 Hz
Solution:
Second overtone corresponds to the third harmonic, i.e. 3f.
Seventh harmonic is 7f. Given 7f - 3f = 1600 Hz.
Thus 4f = 1600 Hz, so f = 400 Hz.
Therefore, the correct option is (2).
Question 38Unattempted
Question 38. The emission spectrum of a black body at two different temperatures are as shown by curves x and y. The ratio of the area under the two curves x and y is
1 : 2
2 : 1
1 : 16
16 : 1
Correct answer: (4) 16 : 1
Solution:
Area under a blackbody spectrum is proportional to total emissive power, hence to T4.
By Wien's law, λmT = constant, so Tx/Ty = λy/λx = 8000/4000 = 2.
Area ratio x : y = Tx4 : Ty4 = 24 : 1 = 16 : 1.
Therefore, the correct option is (4).
Question 39Unattempted
Question 39. A diode of Si is used in a circuit as shown in figure. The electric current through cell is
Zero
10-4 A
10-3 A
10-2 A
Correct answer: (1) Zero
Solution:
A silicon diode requires about 0.7 V forward bias to conduct appreciably.
The applied cell voltage is only 0.1 V, so the diode remains off and current through the cell is zero.
Therefore, the correct option is (1).
Question 40Unattempted
Question 40. The Boolean equation for circuit shown in figure is
A.B
A + B
A.B
A.B
Correct answer: (1) A.B
Solution:
The upper gate gives A. The lower gate gives B. These two outputs are ORed.
Y = A + B = A.B by De Morgan's theorem.
Therefore, the correct option is (1).
Question 41Unattempted
Question 41. A transistor operating in common emitter configuration and change in base current from 200 μA to 250 μA produces change in collector current from 10 mA to 15 mA. The current gain is
150
90
50
100
Correct answer: (4) 100
Solution:
Current gain in common emitter configuration is β = ΔIC/ΔIB.
ΔIC = 15 mA - 10 mA = 5 mA. ΔIB = 250 μA - 200 μA = 50 μA.
β = 5×10-3 / 50×10-6 = 100.
Therefore, the correct option is (4).
Question 42Unattempted
Question 42. The intensity of electromagnetic wave is equal to (symbols have their usual meanings)
ε0ErmsBrmsc2
ε0Erms2
ε0ErmsBrms
ε0Brms2c
Correct answer: (1) ε0ErmsBrmsc2
Solution:
Intensity I = uavc = ε0Erms2c.
Since Erms = cBrms, this can be written as I = ε0ErmsBrmsc2.
Therefore, the correct option is (1).
Question 43Unattempted
Question 43. At resonance in a series LCR circuit, if the input voltage is Vin and the quality factor is Q, then the voltage across the inductor is
Vin/Q
Vin
QVin
Q2Vin
Correct answer: (3) QVin
Solution:
At resonance, impedance Z = R, so current I = Vin/R.
Voltage across inductor VL = IXL = (Vin/R)XL.
Since Q = XL/R at resonance, VL = QVin.
Therefore, the correct option is (3).
Question 44Unattempted
Question 44. According to Weiss theory, ferromagnetism arises due to
interaction of spin of electrons of one atom with that of neighbouring atoms
interaction of orbital motion of electrons with the nucleus
random thermal motion of magnetic domains
motion of free electrons through the crystal lattice
Correct answer: (1) interaction of spin of electrons of one atom with that of neighbouring atoms
Solution:
Ferromagnetism arises because neighbouring atomic magnetic moments tend to align through strong exchange interaction, commonly described as interaction of electron spins.
Therefore, the correct option is (1).
Question 45Unattempted
Question 45. A particle of mass M moving with velocity u collides elastically in one dimension with a stationary particle of mass 2M. The impulse on the first particle is
-Mu/3
-2Mu/3
-Mu
-4Mu/3
Correct answer: (4) -4Mu/3
Solution:
For elastic collision in one dimension, velocity of first particle after collision is v1 = [(M1 - M2)/(M1 + M2)]u1 + [2M2/(M1 + M2)]u2.
Here M1 = M, M2 = 2M, u1 = u and u2 = 0. Hence v1 = (M - 2M)/(M + 2M) · u = -u/3.
Impulse on first particle = Mv1 - Mu = -Mu/3 - Mu = -4Mu/3.
Therefore, the correct option is (4).
Kumar Physics Classes
If you are unable to solve these questions or are facing conceptual difficulty in Physics, Kumar Sir provides personalized one-to-one online Physics classes.
NEETJEE MainJEE AdvancedCBSE Class 11CBSE Class 12AP PhysicsIB PhysicsIGCSE Physics