IOE entrance Physics · Chapter 5
Electricity and Magnetism
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71 questions in 7 syllabus topics · 28 are 2-mark questions.
5.1 Electrostatics and capacitors
12 questions
1. Two point charges repel each other with a force F. If the magnitude of each charge is doubled and the distance between them is halved, the new force is
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Use Coulomb's law F = kq₁q₂/r² and scale each factor.
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Answer: C. 16F
F ∝ q₁q₂/r². Doubling both charges multiplies F by 4; halving r multiplies it by 4 again, so F' = 16F.
2. Which of the following is NOT a unit of electric field intensity?
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Electric field = force per unit charge = potential gradient.
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Answer: D. J/C
J/C is the volt, a unit of potential. N/C = V/m = kg m s⁻²/(A s) = kg m s⁻³ A⁻¹ are all units of electric field.
3. A point charge q is placed at the centre of a cube. The electric flux through any one face of the cube is
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Total flux from Gauss's law, then use symmetry.
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Answer: C. q/(6ε₀)
By Gauss's law the total flux through the closed cube is q/ε₀; by symmetry it is shared equally among the 6 faces, giving q/(6ε₀).
4. A hollow conducting sphere of radius R carries a charge Q. The electric field at a point inside it at distance r (r < R) from the centre is
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Apply Gauss's law to a sphere drawn inside the shell.
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Answer: B. zero
A Gaussian sphere of radius r < R encloses no charge (all charge resides on the outer surface), so E = 0 inside.
5. The electric potential in a region is V = 2x² − 3y² volt (x, y in metre). The magnitude of the electric field at the point (1 m, 1 m) is
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E = −(potential gradient); take partial derivatives along x and y.
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Answer: A. 2√13 V/m
Ex = −∂V/∂x = −4x = −4 V/m and Ey = −∂V/∂y = 6y = 6 V/m at (1, 1). |E| = √(16 + 36) = √52 = 2√13 V/m.
6. Capacitors of 3 μF and 6 μF are connected in series across a 90 V supply. The potential difference across the 3 μF capacitor is
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In series the charge is the same on each capacitor; V = Q/C.
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Answer: D. 60 V
C_eq = (3×6)/(3+6) = 2 μF, so Q = 2 μF × 90 V = 180 μC on each. V on 3 μF = 180/3 = 60 V (and 30 V on the 6 μF).
7. A parallel-plate air capacitor has capacitance C₀. A dielectric slab of dielectric constant 4 and thickness equal to half the plate separation is inserted between the plates (parallel to them). The new capacitance is
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Treat the air gap and slab as two capacitors in series, or use C = ε₀A/(d − t + t/K).
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Answer: C. 1.6C₀
C = ε₀A/(d − t + t/K) = ε₀A/(d/2 + d/8) = ε₀A/(5d/8) = 8C₀/5 = 1.6C₀. (2.5C₀ would be for a slab filling half the area.)
8. A 10 μF capacitor is charged to a potential difference of 100 V. The energy stored in it is
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Use U = ½CV².
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Answer: D. 0.05 J
U = ½CV² = ½ × 10×10⁻⁶ × (100)² = 0.05 J.
9. A capacitor of capacitance C charged to potential V is disconnected from the battery and joined in parallel to an identical uncharged capacitor. The energy lost in the process is
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Charge is conserved; find the common potential, then compare energies.
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Answer: D. ¼CV²
Initial energy ½CV². Charge CV is shared over 2C, so common potential V/2; final energy = ½(2C)(V/2)² = ¼CV². Loss = ½CV² − ¼CV² = ¼CV² (as heat/radiation).
10. The work done in moving a charge from one point to another on an equipotential surface is
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Work depends only on the potential difference between the end points.
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Answer: C. zero
W = q(V_B − V_A) and V_B = V_A on an equipotential surface, so W = 0.
11. Point charges +9q and +4q are fixed 50 cm apart. The point on the line joining them where the net electric field is zero lies at a distance of
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Equate the two field magnitudes and take square roots.
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Answer: D. 30 cm from the +9q charge
Between like charges: 9q/x² = 4q/(50 − x)² ⇒ 3/x = 2/(50 − x) ⇒ 150 − 3x = 2x ⇒ x = 30 cm from +9q.
12. A slab of dielectric constant K is placed in a uniform external electric field E₀ (field normal to the slab faces). The electric field inside the slab is
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Think of what the dielectric constant does to the field.
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Answer: C. E₀/K
Polarisation produces bound surface charges whose field opposes E₀, reducing the field inside the dielectric to E₀/K.
5.2 DC circuits
13 questions
13. If you are given three equal resistors, how many different combinations of these resistors are possible?
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List series, parallel and the two mixed arrangements.
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Answer: C. 4
Using all three: all in series (3R), all in parallel (R/3), two in series with the third in parallel (2R/3), and two in parallel with the third in series (3R/2): four combinations.
14. A wire of resistance R is stretched uniformly so that its length becomes double. Its new resistance is
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Keep the volume constant: A × L = constant.
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Answer: D. 4R
Volume is constant, so doubling L halves A. R = ρL/A becomes ρ(2L)/(A/2) = 4R.
15. The resistivity of a metallic wire depends on
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Resistance depends on shape; resistivity does not.
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Answer: C. the material of the wire and its temperature
Resistivity is a property of the material and varies with temperature; it does not depend on the dimensions of the wire or the current.
16. A cell sends a current of 1 A through a 4 Ω resistor and a current of 0.5 A through a 9 Ω resistor. The emf and internal resistance of the cell are
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Write E = I(R + r) for both cases and solve simultaneously.
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Answer: D. 5 V and 1 Ω
E = 1(4 + r) and E = 0.5(9 + r). So 4 + r = 4.5 + 0.5r ⇒ r = 1 Ω and E = 5 V.
17. A cell of emf 2 V and internal resistance 0.5 Ω is connected to an external resistance of 3.5 Ω. The terminal potential difference of the cell is
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V = E − Ir.
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Answer: D. 1.75 V
I = E/(R + r) = 2/4 = 0.5 A. Terminal p.d. = IR = 0.5 × 3.5 = 1.75 V (= E − Ir).
18. The resistance of the filament of a bulb rated 100 W, 220 V is
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Use P = V²/R with the rated values.
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Answer: C. 484 Ω
R = V²/P = (220)²/100 = 48400/100 = 484 Ω.
19. Two bulbs rated 100 W, 220 V and 60 W, 220 V are connected in series across a 220 V supply. Which statement is correct?
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Same current in series; compare I²R.
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Answer: D. The 60 W bulb glows brighter
R = V²/P, so the 60 W bulb has the larger resistance. In series the current is the same, so power I²R is greater in the 60 W bulb.
20. A galvanometer of resistance 99 Ω gives full-scale deflection for 1 mA. To convert it into an ammeter of range 0–100 mA, one should connect
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A shunt carries the extra current: I_g G = (I − I_g)S.
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Answer: C. a 1 Ω resistance in parallel
Shunt S = I_g G/(I − I_g) = (1 × 99)/(100 − 1) = 1 Ω, connected in parallel with the galvanometer.
21. A galvanometer of resistance 50 Ω gives full-scale deflection for a current of 2 mA. The resistance needed to convert it into a voltmeter of range 0–10 V is
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A voltmeter needs a high resistance in series: V = I_g(G + R).
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Answer: A. 4950 Ω in series
Total resistance = V/I_g = 10/(2×10⁻³) = 5000 Ω. Series resistance = 5000 − 50 = 4950 Ω.
22. In a balanced Wheatstone bridge, the four arms are P = 10 Ω, Q = 20 Ω, R = 15 Ω and S. The value of S is
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Use the balance condition P/Q = R/S.
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Answer: B. 30 Ω
Balance condition P/Q = R/S ⇒ S = RQ/P = 15 × 20/10 = 30 Ω.
23. Kirchhoff's loop (voltage) law is a consequence of the conservation of
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Think of the work done on a charge taken round a closed loop.
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Answer: D. energy
Going round a closed loop, a charge returns to the same potential, so the net work done (sum of emfs minus IR drops) is zero — conservation of energy. The junction law follows from conservation of charge.
24. A battery of emf 12 V and internal resistance 2 Ω is connected to a variable external resistor. The maximum power that can be delivered to the external resistor is
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Maximum power transfer theorem: R = r.
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Answer: B. 18 W
Maximum power transfer occurs when R = r = 2 Ω. Then I = 12/4 = 3 A and P = I²R = 9 × 2 = 18 W (= E²/4r).
25. Two cells, one of emf 4 V and the other of emf 2 V, each of internal resistance 1 Ω, are connected in parallel (positive to positive) across an external resistor of 1.5 Ω. The current through the external resistor is
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Apply Kirchhoff's laws, or replace the two cells by one equivalent cell.
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Answer: A. 1.5 A
Equivalent emf = (E₁/r₁ + E₂/r₂)/(1/r₁ + 1/r₂) = (4 + 2)/2 = 3 V and r_eq = 0.5 Ω. I = 3/(1.5 + 0.5) = 1.5 A. Check by Kirchhoff: V = 2.25 V, cell currents 1.75 A and −0.25 A, sum 1.5 A.
5.3 Thermoelectric effect
6 questions
26. The production of an emf in a circuit of two dissimilar metals when their two junctions are kept at different temperatures is called the
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Which effect is the basis of a thermocouple?
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Answer: B. Seebeck effect
This is the Seebeck effect, the working principle of the thermocouple.
27. Which of the following statements about the Peltier effect is correct?
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Peltier effect is a reversible junction effect.
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Answer: B. Heat absorbed or evolved at a junction is proportional to the current and reverses when the current is reversed
Peltier heat = πIt at a junction of two dissimilar conductors; it is reversible. I²R heating is Joule's, and the effect along a single conductor with a temperature gradient is the Thomson effect.
28. The thermo-emf of a thermocouple with its cold junction at 0 °C is E = αθ + ½βθ², where α = 40 μV/°C and β = −0.08 μV/°C². The neutral and inversion temperatures are respectively
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Neutral: dE/dθ = 0; inversion: E = 0.
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Answer: B. 500 °C and 1000 °C
Neutral temperature: dE/dθ = α + βθ = 0 ⇒ θn = −α/β = 40/0.08 = 500 °C. Inversion: E = 0 ⇒ θi = −2α/β = 1000 °C.
29. The cold junction of a thermocouple is kept at 20 °C and its neutral temperature is 270 °C. Its inversion temperature is
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θn lies midway between θc and θi.
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Answer: B. 520 °C
The neutral temperature is the mean of the cold-junction and inversion temperatures: θn = (θc + θi)/2 ⇒ θi = 2 × 270 − 20 = 520 °C.
30. A thermopile is
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Many small thermo-emfs add up when joined in series.
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Answer: B. a number of thermocouples connected in series, used to detect thermal radiation
Connecting many thermocouples in series adds their small emfs, making a sensitive detector of radiant heat.
31. At the neutral temperature of a thermocouple, the thermoelectric power (dE/dθ) is
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Thermo-emf is maximum at the neutral temperature.
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Answer: A. zero
At the neutral temperature the thermo-emf is maximum, so its slope dE/dθ (the thermoelectric power) is zero.
5.4 Magnetic effect of current
12 questions
32. A charged particle moves parallel to a uniform magnetic field. The magnetic force on it is
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Use F = qvB sin θ.
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Answer: D. zero
F = qvB sin θ and θ = 0°, so F = 0.
33. A proton and an α-particle having the same kinetic energy enter a uniform magnetic field perpendicular to it. The ratio of the radius of the α-particle's path to that of the proton is
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Write r in terms of kinetic energy: r = √(2mK)/(qB).
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Answer: B. 1 : 1
r = mv/(qB) = √(2mK)/(qB). For α: √(2·4m·K)/(2eB) = √(2mK)/(eB), the same as for the proton. Ratio 1 : 1.
34. The period of revolution of a charged particle moving perpendicular to a uniform magnetic field is
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Find r = mv/qB, then T = 2πr/v.
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Answer: A. independent of its speed
T = 2πr/v = 2πm/(qB); the radius grows in proportion to v, so T does not depend on speed (the basis of the cyclotron).
35. A straight wire 0.5 m long carrying 4 A is placed in a uniform magnetic field of 0.2 T, making an angle of 30° with the field. The force on the wire is
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F = BIL sin θ.
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Answer: C. 0.2 N
F = BIL sin θ = 0.2 × 4 × 0.5 × sin 30° = 0.4 × 0.5 = 0.2 N.
36. A rectangular coil of 100 turns and area 20 cm² carries 0.5 A in a uniform magnetic field of 0.4 T. The torque on the coil when its plane is parallel to the field is
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τ = NIAB sin θ, θ measured from the normal to the coil.
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Answer: A. 0.04 N m
τ = NIAB sin θ, where θ is between the field and the normal to the coil. Plane parallel to B ⇒ θ = 90°. τ = 100 × 0.5 × 20×10⁻⁴ × 0.4 = 0.04 N m (maximum).
37. The Hall effect is used to determine
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What does the polarity and size of the Hall voltage tell you?
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Answer: D. the sign and number density of charge carriers in a conductor
The polarity of the Hall voltage gives the sign of the carriers, and V_H = BI/(net) gives their number density n.
38. A copper strip of thickness 1 mm carries a current of 8 A in a magnetic field of 1 T perpendicular to the strip. If the free-electron density is 10²⁹ m⁻³, the Hall voltage is
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V_H = BI/(n e t), where t is the thickness along B.
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Answer: C. 0.5 μV
V_H = BI/(net) = (1 × 8)/(10²⁹ × 1.6×10⁻¹⁹ × 10⁻³) = 8/(1.6×10⁷) = 5×10⁻⁷ V = 0.5 μV.
39. A circular loop of radius 10 cm carries a current of 5 A. The magnetic field at its centre is (μ₀ = 4π×10⁻⁷ T m/A)
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Field at the centre of a circular coil: B = μ₀I/(2R).
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Answer: A. π×10⁻⁵ T
From the Biot–Savart law, B = μ₀I/(2R) = 4π×10⁻⁷ × 5/(2 × 0.1) = π×10⁻⁵ T.
40. A solenoid 0.5 m long has 500 turns and carries a current of 2 A. The magnetic field well inside it is (μ₀ = 4π×10⁻⁷ T m/A)
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B = μ₀nI with n = turns per unit length.
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Answer: D. 8π×10⁻⁴ T
n = 500/0.5 = 1000 turns/m. By Ampere's law, B = μ₀nI = 4π×10⁻⁷ × 1000 × 2 = 8π×10⁻⁴ T.
41. A long straight solid cylindrical wire of radius R carries a current I distributed uniformly over its cross-section. The ratio of the magnetic field at distance R/2 from the axis to that at distance 2R from the axis is
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Inside B ∝ r; outside B ∝ 1/r.
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Answer: A. 1 : 1
Ampere's law: inside, B = μ₀Ir/(2πR²) ⇒ B(R/2) = μ₀I/(4πR). Outside, B = μ₀I/(2πr) ⇒ B(2R) = μ₀I/(4πR). Ratio 1 : 1.
42. Two long parallel straight wires carry currents in the same direction. The wires
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Use the right-hand rule for the field of one wire and the force on the other.
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Answer: B. attract each other
Each wire lies in the magnetic field of the other; F = IL × B gives a force towards the other wire when the currents are parallel.
43. Two long parallel wires 10 cm apart carry currents of 10 A and 20 A. The force per unit length between them is (μ₀ = 4π×10⁻⁷ T m/A)
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F/L = μ₀I₁I₂/(2πd).
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Answer: A. 4×10⁻⁴ N/m
F/L = μ₀I₁I₂/(2πd) = 2×10⁻⁷ × 10 × 20/0.1 = 4×10⁻⁴ N/m.
5.5 Magnetic properties of matter
8 questions
44. At a place, the horizontal component of the earth's magnetic field is 0.2 G and the angle of dip is 60°. The total intensity of the earth's field there is
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H = B cos δ, where δ is the angle of dip.
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Answer: B. 0.4 G
H = B cos δ ⇒ B = H/cos 60° = 0.2/0.5 = 0.4 G. (0.35 G is the vertical component H tan 60°.)
45. A substance having a small negative magnetic susceptibility is
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Which kind of material is weakly repelled by a magnet?
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Answer: A. diamagnetic
Diamagnetic materials are weakly magnetised opposite to the applied field, so χ is small and negative.
46. According to Curie's law, the magnetic susceptibility of a paramagnetic substance is
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Higher temperature means more random dipole orientation.
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Answer: C. inversely proportional to its absolute temperature
Curie's law: χ = C/T. Thermal agitation disturbs the alignment of atomic dipoles, so χ falls as T rises.
47. A material of magnetic susceptibility 999 is placed in a magnetising field H = 1000 A/m. The magnetic flux density inside it is (μ₀ = 4π×10⁻⁷ T m/A)
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μr = 1 + χ and B = μ₀μrH.
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Answer: C. 0.4π T ≈ 1.26 T
μr = 1 + χ = 1000. B = μ₀μrH = 4π×10⁻⁷ × 1000 × 1000 = 0.4π ≈ 1.26 T. (1.26×10⁻³ T is μ₀H, the field without the material.)
48. The area enclosed by the hysteresis loop (B–H curve) of a ferromagnetic material represents
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Think of ∮H dB.
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Answer: B. the energy lost as heat per unit volume per cycle
The loop area equals the work done per unit volume in taking the material through one cycle of magnetisation, which appears as heat.
49. The core of a transformer or an electromagnet should be made of a material having
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The core is magnetised and demagnetised repeatedly; losses must be small.
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Answer: A. high permeability and low coercivity
High permeability gives a strong field for a small current; low coercivity (narrow hysteresis loop) gives small energy loss — soft iron is used.
50. Which of the following statements is NOT correct?
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Recall the sign of χ for each type of material.
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Answer: D. A diamagnetic substance is weakly attracted towards a strong magnet
Diamagnetic substances are weakly repelled (move from stronger to weaker field). The other three statements are true.
51. The apparent angles of dip observed in two mutually perpendicular vertical planes at a place are both 45°. The true angle of dip at the place is
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Use cot²δ = cot²δ₁ + cot²δ₂.
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Answer: A. tan⁻¹(1/√2)
cot²δ = cot²δ₁ + cot²δ₂ = 1 + 1 = 2 ⇒ cot δ = √2 ⇒ δ = tan⁻¹(1/√2) ≈ 35°.
5.6 Electromagnetic induction
10 questions
52. Lenz's law is a consequence of the law of conservation of
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Ask what would happen if the induced current helped the change.
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Answer: A. energy
If the induced current aided the change producing it, energy would be created from nothing; opposing the change means work must be done, conserving energy.
53. The magnetic flux through a coil varies as φ = (5t² − 4t + 1) mWb, where t is in seconds. The magnitude of the induced emf at t = 2 s is
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Faraday's law: e = −dφ/dt.
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Answer: C. 16 mV
|e| = dφ/dt = (10t − 4) mV = 10 × 2 − 4 = 16 mV. (13 mV is the flux value, not its rate of change.)
54. A conducting rod of length 50 cm slides at a constant speed of 4 m/s on rails, perpendicular to a uniform magnetic field of 0.5 T. The total resistance of the circuit is 2 Ω. The external force needed to keep the rod moving at constant speed is
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Find the motional emf BLv, the current, then F = BIL.
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Answer: A. 0.125 N
e = BLv = 0.5 × 0.5 × 4 = 1 V; I = 1/2 = 0.5 A. Retarding force F = BIL = 0.5 × 0.5 × 0.5 = 0.125 N, which the external force must balance.
55. An AC generator has a coil of 100 turns, each of area 0.02 m², rotating at an angular speed of 50 rad/s in a uniform magnetic field of 0.5 T. The peak emf generated is
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e = NABω sin ωt.
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Answer: D. 50 V
e₀ = NABω = 100 × 0.02 × 0.5 × 50 = 50 V. (35.4 V is the rms value.)
56. The self-inductance of a long solenoid is L. If the number of turns is doubled keeping its length and area of cross-section the same, the self-inductance becomes
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Write L = μ₀N²A/l.
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Answer: B. 4L
L = μ₀N²A/l, so L ∝ N². Doubling N makes it 4L.
57. An emf of 10 V is induced in a coil when the current in it changes at the rate of 50 A/s. The energy stored in the coil when it carries a steady current of 4 A is
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First find L from e = L dI/dt, then U = ½LI².
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Answer: B. 1.6 J
L = e/(dI/dt) = 10/50 = 0.2 H. U = ½LI² = ½ × 0.2 × 16 = 1.6 J.
58. When the current in a primary coil falls uniformly from 5 A to zero in 0.02 s, an emf of 25 V is induced in a nearby secondary coil. The mutual inductance of the pair is
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e₂ = M dI₁/dt.
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Answer: A. 0.1 H
e = M dI/dt ⇒ M = 25/(5/0.02) = 25/250 = 0.1 H.
59. An ideal transformer has a turns ratio Ns/Np = 10. If 220 V is applied to the primary and the primary current is 5 A, the secondary voltage and current are
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Voltage scales with turns ratio; power in = power out.
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Answer: C. 2200 V and 0.5 A
Vs = Vp(Ns/Np) = 2200 V. For an ideal transformer VpIp = VsIs ⇒ Is = 220 × 5/2200 = 0.5 A.
60. A step-down transformer of efficiency 90% converts 220 V to 22 V and supplies a current of 18 A to a load. The current drawn from the primary is
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Efficiency = output power/input power.
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Answer: C. 2 A
Output = 22 × 18 = 396 W. Input = 396/0.9 = 440 W. Primary current = 440/220 = 2 A.
61. A metal rod of length 1 m rotates about one of its ends with an angular velocity of 20 rad/s in a plane perpendicular to a uniform magnetic field of 0.5 T. The emf induced between its ends is
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Integrate BvdR with v = ωr, or use the average speed ωl/2.
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Answer: C. 5 V
e = ½Bωl² = ½ × 0.5 × 20 × 1² = 5 V (the average speed of the rod is ωl/2).
5.7 Alternating currents
10 questions
62. An alternating voltage is given by V = 311 sin(100πt) volt. Its rms value and frequency are
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Compare with V = V₀ sin 2πft.
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Answer: B. 220 V and 50 Hz
V_rms = V₀/√2 = 311/1.414 ≈ 220 V. ω = 100π = 2πf ⇒ f = 50 Hz.
63. The inductive reactance of a coil of inductance (1/π) H connected to a 50 Hz AC supply is
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X_L = ωL = 2πfL.
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Answer: C. 100 Ω
X_L = 2πfL = 2π × 50 × (1/π) = 100 Ω.
64. In a purely capacitive AC circuit,
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Current is the rate of change of charge, q = CV.
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Answer: D. the current leads the voltage by 90°
For a capacitor i = C dV/dt; with V = V₀ sin ωt, i = ωCV₀ cos ωt, which leads V by π/2.
65. A series LCR circuit has R = 30 Ω, X_L = 80 Ω and X_C = 40 Ω, and is connected to a 200 V (rms) AC supply. The average power consumed is
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Only the resistor consumes power: P = I_rms²R.
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Answer: D. 480 W
Z = √(R² + (X_L − X_C)²) = √(900 + 1600) = 50 Ω. I = 200/50 = 4 A. P = I²R = 16 × 30 = 480 W (power factor 0.6).
66. A series LCR circuit has L = 0.1 H, C = 10 μF and R = 10 Ω. Its resonant angular frequency and quality factor are
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ω₀ = 1/√(LC); Q = ω₀L/R.
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Answer: C. 1000 rad/s and 10
ω₀ = 1/√(LC) = 1/√(0.1 × 10⁻⁵) = 1/10⁻³ = 1000 rad/s. Q = ω₀L/R = 1000 × 0.1/10 = 10.
67. Which of the following is NOT true for a series LCR circuit at resonance?
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At resonance X_L = X_C.
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Answer: A. The impedance is maximum
At resonance X_L = X_C, so Z = R is minimum, the current is maximum and the power factor cos φ = 1.
68. The average power consumed over a complete cycle by a pure inductor connected to an AC source is
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P = V_rms I_rms cos φ.
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Answer: D. zero
The phase difference is 90°, so the power factor cos 90° = 0 and P = V_rms I_rms cos φ = 0 (wattless current).
69. In a series LCR circuit connected to an AC source, the rms voltages across R, L and C are 40 V, 90 V and 60 V respectively. The rms voltage of the source is
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V_L and V_C are opposite in phase; V_R is at 90° to both.
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Answer: C. 50 V
Using phasors, V = √(V_R² + (V_L − V_C)²) = √(40² + 30²) = √2500 = 50 V. The voltages cannot simply be added because they are out of phase.
70. The rms value of an alternating current is equal to that steady current which
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rms is defined by the heating effect.
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Answer: C. produces the same heat in a given resistor in the same time
The rms (virtual) value is defined through the heating effect: I_rms²R t equals the heat produced by the AC in the same resistor in the same time.
71. A resistor of 10 Ω and an inductor of reactance 10√3 Ω are connected in series to a 220 V (rms) AC supply. The power factor of the circuit and the rms current are
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cos φ = R/Z.
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Answer: A. 0.5 and 11 A
Z = √(10² + (10√3)²) = √400 = 20 Ω. I = 220/20 = 11 A. Power factor = R/Z = 10/20 = 0.5.