NEET Physics Tutor Doubt 7 - Very Special NEET Physics Questions
These concept-rich current electricity and magnetism questions are built for serious NEET revision with typed questions, instant checking, official answers, and detailed solutions.
This lightweight version removes the heavy snapshot images and keeps the same quiz flow with typed questions, option checking, answer reveal, and detailed solutions.
The important circuits, graphs, and field diagrams are retained as small original WebP snapshot assets so visual accuracy matches the source.
Inspired by Trusted Physics Traditions
H.C. Verma style conceptual clarity
I.E. Irodov level thinking
Resnick Halliday style fundamentals
Previous year NEET Physics patterns
Kota coaching institute problem-solving style
Question 1 of 45
NEET Physics Practice Question 1
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Question 1. A set of n equal resistors, each of value R, are connected in parallel to a battery of emf E and internal resistance R. The current drawn is I. Now the n resistors are connected in series to the same battery. If the current becomes I/9, the value of n is
3
18
9
6
Correct answer: Option (3) — 9
Solution:
For the parallel combination, the external resistance is R/n, so I = E/(R/n + R).
For the series combination, the external resistance is nR, so I/9 = E/(nR + R).
Dividing the two expressions gives 9 = (n + 1)/(1/n + 1) = n.
Therefore, the correct option is (3).
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Question 2 of 45
NEET Physics Practice Question 2
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Question 2. The resistance of a wire is R ohm. It is melted and then stretched to three times its original length. The new resistance is
3R
9R
R3
R9
Correct answer: Option (2) — 9R
Solution:
On melting and stretching, the volume remains constant: lA = 3lA1, hence A1 = A/3.
The new resistance is R1 = ρ(3l)/(A/3) = 9ρl/A = 9R.
Therefore, the correct option is (2).
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Question 3 of 45
NEET Physics Practice Question 3
+4 / −1
Question 3. The potential difference (VP − VQ) between points P and Q in the circuit shown is
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4 V
−2 V
+2 V
Zero
Correct answer: Option (4) — Zero
Solution:
Moving from P to Q along the current direction, the drops/rises are: 1 × 2 − 6 + 1 × 2 + 2.
Thus VP − VQ = 2 − 6 + 2 + 2 = 0.
Therefore, the correct option is (4).
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Question 4 of 45
NEET Physics Practice Question 4
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Question 4. Two metal wires of identical dimensions are connected in parallel. Their conductivities are σ and 2σ respectively. The effective conductivity of the combination is
3σ2
σ2
2σ3
σ
Correct answer: Option (1) — 3σ2
Solution:
The resistances are R1 = l/(σA) and R2 = l/(2σA).
For the equivalent wire of length l and total area 2A, Req = l/(σeff2A).
Using conductances: 2σeffA/l = σA/l + 2σA/l, so σeff = 3σ/2.
Therefore, the correct option is (1).
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Question 5 of 45
NEET Physics Practice Question 5
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Question 5. A uniform wire of resistance 20 Ω has resistance per unit length 1 Ω/m and is bent in the adjoining circular form. If the equivalent resistance between M and N is 1.8 Ω, the length of the shorter section is
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2 m
5 m
18 m
1.8 m
Correct answer: Option (1) — 2 m
Solution:
Let the resistance of the shorter section between M and N be x. The longer section has resistance 20 − x.
The two sections are in parallel, so Req = x(20 − x)/20 = 1.8.
This gives x = 2 Ω for the shorter section. Since resistance per unit length is 1 Ω/m, the length is 2 m.
Therefore, the correct option is (1).
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Question 6 of 45
NEET Physics Practice Question 6
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Question 6. When a battery is connected across a 16 Ω resistor, the voltage across the resistor is 12 V. When the same battery is connected across a 10 Ω resistor, the voltage across it is 11 V. The internal resistance of the battery is
107 Ω
207 Ω
257 Ω
157 Ω
Correct answer: Option (2) — 207 Ω
Solution:
For the 16 Ω resistor, E = 12 + (12/16)r.
For the 10 Ω resistor, E = 11 + (11/10)r.
Equating the two expressions: 12 + 3r/4 = 11 + 11r/10, which gives r = 20/7 Ω.
Therefore, the correct option is (2).
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Question 7 of 45
NEET Physics Practice Question 7
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Question 7. In a meter bridge set-up with null deflection in the galvanometer as shown, the value of the unknown resistor R is
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80 Ω
100 Ω
160 Ω
120 Ω
Correct answer: Option (3) — 160 Ω
Solution:
For a balanced meter bridge, Rleft/l = Rright/(100 − l).
Here 40/20 = R/80, hence R = 160 Ω.
Therefore, the correct option is (3).
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Question 8 of 45
NEET Physics Practice Question 8
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Question 8. In the potentiometer circuit shown, the balance length AP = 70 cm when switch S is open. When S is closed and R = 6 Ω, the balance length AP′ = 60 cm. The internal resistance of the cell C′ is
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2 Ω
1.5 Ω
1 Ω
0.5 Ω
Correct answer: Option (3) — 1 Ω
Solution:
For a cell tested by potentiometer, r = R(l1/l2 − 1).
Thus r = 6(70/60 − 1) = 6(1/6) = 1 Ω.
Therefore, the correct option is (3).
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Question 9 of 45
NEET Physics Practice Question 9
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Question 9. When 115 V is applied across a wire that is 10 m long and has a radius of 0.30 mm, the current density is 1.4 × 104 A/m2. The resistivity of the wire is
2.0 × 104 Ωm
4.1 × 10−4 Ωm
8.2 × 10−4 Ωm
2.0 × 10−3 Ωm
Correct answer: Option (3) — 8.2 × 10−4 Ωm
Solution:
Using J = I/A and R = ρl/A, the potential difference is V = IR = Jρl.
Question 10. Two electric lamps of 60 watt each are connected in parallel. The power consumed by the combination will be, assuming the same source voltage,
80 watt
120 watt
60 watt
30 watt
Correct answer: Option (2) — 120 watt
Solution:
In parallel, each lamp gets the same rated source voltage and therefore consumes 60 W.
The total power is the sum of powers: 60 W + 60 W = 120 W.
Therefore, the correct option is (2).
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Question 11 of 45
NEET Physics Practice Question 11
+4 / −1
Question 11. In the circuit shown, the reading of the voltmeter is
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2 V
2.5 V
3 V
5 V
Correct answer: Option (1) — 2 V
Solution:
The 100 Ω resistor is in parallel with the 200 Ω voltmeter, giving Rp = (100 × 200)/(100 + 200) = 200/3 Ω.
The total resistance is 100 + 200/3 = 500/3 Ω, so current from the 5 V source is 5/(500/3) = 3/100 A.
The voltmeter reads the potential across the parallel part: V = (3/100)(200/3) = 2 V.
Therefore, the correct option is (1).
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Question 12 of 45
NEET Physics Practice Question 12
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Question 12. For the balanced bridge shown in the figure, the current through the 2 Ω resistance is
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227 A
137 A
167 A
207 A
Correct answer: Option (4) — 207 A
Solution:
The bridge is balanced because 4/10 = 2/5, so no current flows through the galvanometer.
The top branch has resistance 4 + 2 = 6 Ω, and the bottom branch has resistance 10 + 5 = 15 Ω.
With total current 4 A, current in the top branch is 4 × 15/(6 + 15) = 20/7 A. This is the current through the 2 Ω resistor.
Therefore, the correct option is (4).
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Question 13 of 45
NEET Physics Practice Question 13
+4 / −1
Question 13. The V-I graph for a conductor at two temperatures T1 and T2 is shown. The relation between the temperatures is
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T1 > T2
T1 < T2
T1 = T2
T1 = 1/T2
Correct answer: Option (1) — T1 > T2
Solution:
In a V-I graph, the slope V/I gives resistance.
The line for T1 is steeper, so the resistance at T1 is greater.
For a metallic conductor, resistance increases with temperature. Hence T1 > T2.
Therefore, the correct option is (1).
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Question 14 of 45
NEET Physics Practice Question 14
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Question 14. In the network shown, the potential difference between A and B is
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2 V
6 V
4 V
1 V
Correct answer: Option (3) — 4 V
Solution:
Both branches have total resistance 10 Ω, so the 4 A current splits equally: 2 A in each branch.
From the left junction to A, the drop is 2 × 4 = 8 V. From the left junction to B, the drop is 2 × 6 = 12 V.
Thus VA − VB = 12 − 8 = 4 V.
Therefore, the correct option is (3).
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Question 15 of 45
NEET Physics Practice Question 15
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Question 15. A potentiometer wire of length 100 cm has resistance 10 Ω. It is connected in series with a 5 Ω resistance and an accumulator of emf 6 V. A source of 2.4 V is balanced against length L of the potentiometer wire. The value of L is
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40 cm
30 cm
50 cm
60 cm
Correct answer: Option (4) — 60 cm
Solution:
The current through the potentiometer wire is I = 6/(5 + 10) = 0.4 A.
The potential drop across the 100 cm potentiometer wire is 0.4 × 10 = 4 V, so the potential gradient is 4/100 = 0.04 V/cm.
For balance, 2.4 = 0.04L, hence L = 60 cm.
Therefore, the correct option is (4).
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Question 16 of 45
NEET Physics Practice Question 16
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Question 16. In a metre bridge, the resistance in the left gap and the right gap are in the ratio 2 : 3. The balance point from the left end is
60 cm
50 cm
40 cm
20 cm
Correct answer: Option (3) — 40 cm
Solution:
At balance, the ratio of gap resistances equals the ratio of bridge-wire lengths.
Therefore l : (100 − l) = 2 : 3, giving l = 40 cm from the left end.
Therefore, the correct option is (3).
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Question 17 of 45
NEET Physics Practice Question 17
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Question 17. A resistor R and a 10 Ω resistor are connected in parallel across a 20 V source. If the total power consumed is 50 W, the value of R is
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20 Ω
10 Ω
30 Ω
40 Ω
Correct answer: Option (4) — 40 Ω
Solution:
The equivalent resistance is Req = 10R/(10 + R).
Using P = V2/Req, 50 = 400(10 + R)/(10R).
Thus 40 + 4R = 5R, giving R = 40 Ω.
Therefore, the correct option is (4).
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Question 18 of 45
NEET Physics Practice Question 18
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Question 18. Consider the statements: (A) Kirchhoff’s junction law follows from conservation of energy. (B) Kirchhoff’s loop law follows from conservation of charge. Which option is correct?
Both statements are correct
Both statements are incorrect
A is correct but B is incorrect
A is incorrect but B is correct
Correct answer: Option (2) — Both statements are incorrect
Solution:
Kirchhoff’s junction law follows from conservation of charge, not conservation of energy.
Kirchhoff’s loop law follows from conservation of energy, not conservation of charge.
Both given statements are therefore incorrect.
Therefore, the correct option is (2).
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Question 19 of 45
NEET Physics Practice Question 19
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Question 19. Two batteries of emf 2 V and 3 V with internal resistances 1 Ω and 2 Ω respectively are connected in parallel. The effective emf is
53 V
73 V
43 V
5 V
Correct answer: Option (2) — 73 V
Solution:
For two cells in parallel, Eeq = (E1r2 + E2r1)/(r1 + r2).
Thus Eeq = (2 × 2 + 3 × 1)/(1 + 2) = 7/3 V.
Therefore, the correct option is (2).
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Question 20 of 45
NEET Physics Practice Question 20
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Question 20. The rate of flow of charge through a metallic conductor of non-uniform cross-section is uniform. Which quantity remains constant along the conductor?
Current
Current density
Electric potential
Electric field
Correct answer: Option (1) — Current
Solution:
The rate of flow of charge is current, I = dq/dt, and it remains the same through every cross-section in steady state.
Current density changes with area because J = I/A.
Therefore, the correct option is (1).
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Question 21 of 45
NEET Physics Practice Question 21
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Question 21. When the temperature of a conductor is increased, its resistance increases because
relaxation time increases
mass of electron increases
electron density decreases
relaxation time decreases
Correct answer: Option (4) — relaxation time decreases
Solution:
In a metal, increasing temperature increases lattice vibrations and electron collisions.
The relaxation time τ decreases, and since ρ = m/(ne2τ), resistivity and resistance increase.
Therefore, the correct option is (4).
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Question 22 of 45
NEET Physics Practice Question 22
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Question 22. In the circuit shown, the ammeter reading is
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0.4 A
0.8 A
0.2 A
0.6 A
Correct answer: Option (2) — 0.8 A
Solution:
The three 6 Ω branches are in parallel, giving equivalent resistance 2 Ω.
This is in series with the 3 Ω branch containing the cell and ammeter, so total resistance is 3 + 2 = 5 Ω.
The ammeter current is I = 4/5 = 0.8 A.
Therefore, the correct option is (2).
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Question 23 of 45
NEET Physics Practice Question 23
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Question 23. Three parallel straight wires are perpendicular to the plane of paper and carry equal currents I in the same direction. They are placed at the vertices of an equilateral triangle of side d. The force per unit length on wire A is
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μ0I2√2 πd
√3 μ0I22πd
μ0I22πd
μ0I2πd
Correct answer: Option (2) — √3 μ0I22πd
Solution:
The force per unit length on A due to either B or C is F0 = μ0I2/(2πd).
The two equal forces make an angle of 60° with each other.
The resultant is √3F0 = √3 μ0I2/(2πd).
Therefore, the correct option is (2).
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Question 24 of 45
NEET Physics Practice Question 24
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Question 24. A wire carrying current is bent into a circular loop of one turn and the magnetic field at its centre is B. If the same wire is bent into a circular coil of four turns, the magnetic field at the centre becomes
4B
B4
16B
B8
Correct answer: Option (3) — 16B
Solution:
For the original one-turn loop, the wire length is 2πR and field at the centre is B = μ0I/(2R).
For n turns using the same wire length, n2πr = 2πR, so r = R/n.
The new field is Bn = nμ0I/(2r) = n2B. For n = 4, Bn = 16B.
Therefore, the correct option is (3).
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Question 25 of 45
NEET Physics Practice Question 25
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Question 25. A charged particle enters a uniform magnetic field. Its kinetic energy
decreases
increases
becomes zero
remains constant
Correct answer: Option (4) — remains constant
Solution:
The magnetic force is q(v × B), which is always perpendicular to velocity.
Since the magnetic force does no work, the speed and kinetic energy remain constant.
Therefore, the correct option is (4).
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Question 26 of 45
NEET Physics Practice Question 26
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Question 26. An arc of a circle of radius R subtends an angle π/3 at the centre and carries current I. The magnetic field at the centre is
μ0I12R
μ0I6R
μ0I4R
μ0I3R
Correct answer: Option (1) — μ0I12R
Solution:
Magnetic field at the centre due to a circular arc is B = (μ0I/2R)(φ/2π).
For φ = π/3, B = (μ0I/2R)(1/6) = μ0I/(12R).
Therefore, the correct option is (1).
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Question 27 of 45
NEET Physics Practice Question 27
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Question 27. A wire carrying current 2I and another wire carrying current 3I in the opposite direction produce magnetic field B at the midpoint as shown. What will be the field at the midpoint when the 2I wire is switched off?
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3B5
2B5
2B3
4B3
Correct answer: Option (1) — 3B5
Solution:
Initially the magnetic fields at the midpoint are in the same direction, so B = 2μ0I/(2πd) + 3μ0I/(2πd) = 5μ0I/(2πd).
When the 2I wire is switched off, the field is B1 = 3μ0I/(2πd).
Therefore B1/B = 3/5.
Therefore, the correct option is (1).
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Question 28 of 45
NEET Physics Practice Question 28
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Question 28. An electric field 1200 ĵ V/m and a magnetic field 0.20 k̂ Wb/m2 act on a moving electron. The electron will move with uniform speed, without deflection, when its speed is
3.0 × 106 m/s
6.0 × 103 m/s
4.0 × 107 m/s
2.0 × 108 m/s
Correct answer: Option (2) — 6.0 × 103 m/s
Solution:
For no deflection, electric and magnetic forces must balance: qE = qvB.
Thus v = E/B = 1200/0.20 = 6.0 × 103 m/s.
Therefore, the correct option is (2).
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Question 29 of 45
NEET Physics Practice Question 29
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Question 29. A long hollow copper tube carries a uniform current I. Which statement is true?
Magnetic field B is zero at all points inside the tube.
Magnetic field B is zero only at the point on the axis of the tube.
Magnetic field is maximum at points on the axis of the tube.
Magnetic field is zero at any point outside the tube.
Correct answer: Option (1) — Magnetic field B is zero at all points inside the tube.
Solution:
For an Amperian loop lying inside the hollow region, the enclosed current is zero.
By Ampere’s law, ∮B·dl = μ0Ienc = 0, so B is zero everywhere inside the hollow tube.
Therefore, the correct option is (1).
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Question 30 of 45
NEET Physics Practice Question 30
+4 / −1
Question 30. For the current-carrying wire shown, the magnetic field induction at point O is
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μ0I4R
μ0I4R(1/π + 1)
μ0I4R(1 − 1/π)
Zero
Correct answer: Option (2) — μ0I4R(1/π + 1)
Solution:
The magnetic field at O has contribution from the semicircular arc and from the straight part.
For the semicircular arc, B1 = μ0I/(4R).
For the straight wire contribution, B2 = μ0I/(4πR).
Hence B = μ0I/(4R)(1 + 1/π).
Therefore, the correct option is (2).
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Question 31 of 45
NEET Physics Practice Question 31
+4 / −1
Question 31. The magnetic force on a charged particle of charge −4 μC moving in a magnetic field of 1 T in the y-direction with velocity (3 î + 4 ĵ) × 106 m/s is
6 N in y-direction
12 N in negative z-direction
8 N in z-direction
4 N in negative y-direction
Correct answer: Option (2) — 12 N in negative z-direction
Solution:
The force is F = q(v × B).
Here v × B = [(3 î + 4 ĵ) × 106] × ĵ = 3 × 106 k̂.
Multiplying by q = −4 × 10−6 C gives F = −12 k̂ N.
Therefore, the correct option is (2).
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Question 32 of 45
NEET Physics Practice Question 32
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Question 32. A wire of length 4 m carrying a current of 2 A is bent into a circle. Its magnetic moment is
4π A m2
4π A m2
8π A m2
2π A m2
Correct answer: Option (3) — 8π A m2
Solution:
The circumference is 2πR = 4, so R = 2/π.
Area A = πR2 = π(2/π)2 = 4/π.
Magnetic moment M = IA = 2 × 4/π = 8/π A m2.
Therefore, the correct option is (3).
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Question 33 of 45
NEET Physics Practice Question 33
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Question 33. The correct graph between magnetic moment M of a current-carrying loop and its radius r is
For a current loop, magnetic moment M = NIA = NIπr2.
For fixed N and I, M ∝ r2, so the graph is an increasing parabola.
Therefore, the correct option is (3).
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Question 34 of 45
NEET Physics Practice Question 34
+4 / −1
Question 34. Two α-particles have velocities in the ratio 4 : 3 and enter a magnetic field. If they move in different circular paths, the ratio of their radii is
1 : 2
4 : 3
16 : 9
9 : 16
Correct answer: Option (2) — 4 : 3
Solution:
The radius of a charged particle in a magnetic field is r = mv/(qB).
For identical α-particles in the same magnetic field, r ∝ v.
Therefore the ratio of radii is 4 : 3.
Therefore, the correct option is (2).
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Question 35 of 45
NEET Physics Practice Question 35
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Question 35. A 4 cm long wire carrying a current of 5 A is placed inside a solenoid perpendicular to its axis. If the magnetic field inside the solenoid is 0.4 T, the magnetic force on the wire is
4 × 10−2 N
0.4 N
8 × 10−2 N
0.8 N
Correct answer: Option (3) — 8 × 10−2 N
Solution:
Magnetic force on a current-carrying wire is F = ILB sinθ.
Here θ = 90°, so F = 5 × 4 × 10−2 × 0.4 = 8 × 10−2 N.
Therefore, the correct option is (3).
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Question 36 of 45
NEET Physics Practice Question 36
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Question 36. A charged particle moves in a transverse magnetic field. If its velocity is doubled, the radius of its circular path becomes
doubled
halved
increased by four times
unchanged
Correct answer: Option (1) — doubled
Solution:
The radius is r = mv/(qB).
With m, q, and B constant, r is directly proportional to v. Therefore doubling velocity doubles the radius.
Therefore, the correct option is (1).
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Question 37 of 45
NEET Physics Practice Question 37
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Question 37. A voltmeter measuring 4 V is constructed with a galvanometer of resistance 16 Ω and full-scale current 2 mA. The value of series resistance R is
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898 Ω
1984 Ω
2280 Ω
2598 Ω
Correct answer: Option (2) — 1984 Ω
Solution:
For full-scale deflection at 4 V, total resistance required is V/Ig = 4/(2 × 10−3) = 2000 Ω.
The galvanometer already has 16 Ω resistance, so the required series resistance is 2000 − 16 = 1984 Ω.
Therefore, the correct option is (2).
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Question 38 of 45
NEET Physics Practice Question 38
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Question 38. If the current sensitivity of a galvanometer is doubled by increasing the number of turns, its voltage sensitivity will be
doubled
halved
four times
unchanged
Correct answer: Option (4) — unchanged
Solution:
Voltage sensitivity = current sensitivity / galvanometer resistance.
Increasing the number of turns doubles current sensitivity, but it also doubles the resistance of the galvanometer.
The ratio therefore remains unchanged.
Therefore, the correct option is (4).
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Question 39 of 45
NEET Physics Practice Question 39
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Question 39. A current-carrying loop is placed in a uniform magnetic field. The torque on the loop does not depend upon
area of loop
value of current
magnetic field
none of these
Correct answer: Option (4) — none of these
Solution:
Torque on a current loop is τ = NIAB sinθ.
It depends on the area of the loop, the current, and the magnetic field. Hence none of the listed quantities is independent of torque.
Therefore, the correct option is (4).
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Question 40 of 45
NEET Physics Practice Question 40
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Question 40. The power supplied by the magnetic force on a moving charged particle is
always zero
never zero
zero only when v ⟂ B
zero only when v ∥ B
Correct answer: Option (1) — always zero
Solution:
The magnetic force F = q(v × B) is always perpendicular to velocity.
Power is P = F·v, so the power supplied by magnetic force is always zero.
Therefore, the correct option is (1).
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Question 41 of 45
NEET Physics Practice Question 41
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Question 41. A thin disc of radius R and mass M has charge q uniformly distributed on it. It rotates with angular velocity ω. The magnetic moment of the disc about the axis of rotation will be
qR2ω4
qR2ω2
3qR2ω4
qR2ω3
Correct answer: Option (1) — qR2ω4
Solution:
For a uniformly charged rotating body, magnetic moment μ and angular momentum L are related by μ = qL/(2M).
For a thin disc, L = Iω = (MR2/2)ω.
Thus μ = q(MR2ω/2)/(2M) = qR2ω/4.
Therefore, the correct option is (1).
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Question 42 of 45
NEET Physics Practice Question 42
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Question 42. The magnetic field at point P due to the current-carrying circular coil shown, of radius a with P at axial distance √3a, is
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μ0I8a
μ0I4a
μ0I16a
3μ0I4a
Correct answer: Option (3) — μ0I16a
Solution:
Magnetic field on the axis of a circular coil is B = μ0I a2/[2(x2 + a2)3/2].
Here x = √3a, so x2 + a2 = 3a2 + a2 = 4a2.
Therefore B = μ0I a2/[2(4a2)3/2] = μ0I/(16a).
Therefore, the correct option is (3).
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Question 43 of 45
NEET Physics Practice Question 43
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Question 43. Ampere’s circuital law can be derived from
Ohm’s law
Biot-Savart’s law
Kirchhoff’s law
Gauss’s law
Correct answer: Option (2) — Biot-Savart’s law
Solution:
Ampere’s circuital law for steady currents can be derived from the Biot-Savart law.
Therefore, the correct option is (2).
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Question 44 of 45
NEET Physics Practice Question 44
+4 / −1
Question 44. A proton enters a region of uniform magnetic field 1 T with velocity u at an angle 37° with B. If P is the pitch of the helical path followed, the radius of the helix is
Pπ
3P8π
2P3π
3P4π
Correct answer: Option (2) — 3P8π
Solution:
Pitch of the helix is P = (2πmv cosθ)/(qB).
Radius is r = (mv sinθ)/(qB).
Thus P/r = 2π cotθ. With θ = 37°, cot37° = 4/3, so P/r = 8π/3.
Therefore r = 3P/(8π).
Therefore, the correct option is (2).
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Question 45 of 45
NEET Physics Practice Question 45
+4 / −1
Question 45. A long solenoid has n turns per unit length and current I ampere is flowing through it. The magnetic field at the end of the solenoid will be
8μ0nI
2μ0n2I
μ0nI2
μ0n2I
Correct answer: Option (3) — μ0nI2
Solution:
The magnetic field well inside a long solenoid is B = μ0nI.
At the end of a long solenoid, the field is half of the central value: Bend = μ0nI/2.
Therefore, the correct option is (3).
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