NDA Electricity and magnetism previous year questions with solutions

Answer: (a) 0

Method. Equilibrium of a charge between two charges

U(x) = k·1(4/⁠x + 16/⁠(0.18 − x)) is minimum where dU/⁠dx = −⁠4/⁠x² + 16/⁠(0.18−⁠x)² = 0, i.e. 0.18 − x = 2x, x = 0.06 m. The field there is E = k(4/⁠x² − 16/⁠(0.18−⁠x)²) = k(4/⁠0.0036 − 16/⁠0.0144) = 0. So the field is zero (standard reference, Coulomb's law and superposition).

Answer: (c) It enhances field and promotes continuous corona discharge.

Method. Action at points / corona

Charge density and the electric field are highest at a sharp tip; the strong field ionises air around the tip (corona discharge), letting charge leak away continuously and giving the lightning a safe path to earth. The tip enhances, not reduces, the field (standard reference, action at points).

Answer: (c) Both perfect conductor and insulator can be charged

Method. Charging in static conditions

Any material can hold charge: an insulator keeps it where it is placed; a conductor can also be charged (isolated), the charge residing on its surface. So (c) is correct (standard reference, electrostatics of conductors).

Answer: (d) Charges can be created and destroyed in an isolated system

Method. Properties of charge

Charge is additive, conserved and quantised. Net charge of an isolated system can neither be created nor destroyed, so (d) is not a property (standard reference, electric charges).

Answer: (a) The flow of charges between different parts of the cloud

Method. cause of lightning

Charge separation builds up in the cloud, and lightning is the sudden discharge, that is a flow of charge, through the air (within the cloud or to the ground). (c) reverses the polarity (the base of the cloud is negative, the top positive) so it is wrong; (d) describes only the induction preceding a ground stroke, and (b) is not the usual description. Standard reference: Class 8 Science, natural phenomena. Not an NCERT row in this chapter's set.

Options (a), (b) and (d) are each partly defensible; answered the discharge of charge between parts of the cloud (a), not sure of the setter's key.

Answer: (a) [figure (a)]

Method. Lightning conductor shape (figure read from its description)

Charge concentrates on sharp points (high surface charge density), so a pointed rod gives the strongest field and lets charge leak off or discharge the cloud safely. The cone with a sharp point, (a), is best; blunt, flat or spherical ends concentrate field less (standard electrostatics, action at points).

Answer: (c) 1 and 2 only

Method. Coulomb's law sign

F is proportional to q1 q2, which is positive when both charges have the same sign (both positive or both negative: repulsion). Unlike charges give a negative sign (attraction). So 1 and 2 only. Standard result (Coulomb's law).

Answer: (d) at right angles to the conducting surface and outwards from the centre of the sphere

Method. field at a conductor's surface

Electrostatic field lines meet a conductor's surface at right angles (otherwise charges would move along it), and for a positive charge they point outward. Standard result (electrostatics of conductors).

Answer: (a) 1 only

Method. static discharge from aircraft

An aircraft in flight picks up static charge by friction with air and particles; conducting rubber tyres let it leak to earth on landing, so statement 1 is right. Statement 2 is not a standard reason: operating electronics does not accumulate net charge on the airframe. Standard aviation-safety reference (static dischargers/⁠bonding).

Answer: (b) Static electricity

Method. charging by friction or induction

An insulator is charged by friction (static electricity), as with a rubbed rod; a current needs free carriers, and magnetic and gravitational fields do not put charge on it. Standard result (electrostatics).

Answer: (b) Electrons do not flow easily through them.

Method. insulators have almost no free electrons

An insulator has very high resistivity because it has almost no free charge carriers; electrons do not flow easily through it (TP3-A28). It does contain electrons (a is false), need not be crystalline (glass, rubber), and is neutral overall (d false).

Answer: (d) The negative charges are transferred from wool to rod.

Method. charging by friction moves electrons

Only electrons move in rubbing; positive charges (nuclei) stay fixed. The rod becomes negative by gaining electrons, so negative charge is transferred from wool to rod. Standard result (frictional charging).

Answer: (a) E = 0 for a < r < b

Method. Gauss's law and shielding by a conductor

Inside the metal (a < r < b) electrostatic field is zero, so (a) is true. For r < a the field is that of +q at the centre, q/⁠(4 pi eps0 r^2), which depends on r and is not zero, so (b) is false. As printed, (c) is wrong (E = 0 in the metal) and (d) is wrong (the field is not q/⁠4 pi eps0 a). The retyped page prints (c) and (d) without the square; with or without it, (c) and (d) are false and (a) alone is right. The answer does not depend on the defect.

Answer: (c) 188% larger

Method. R = ρL/⁠A at fixed volume

Shorter wire: length L, area A₁, volume V. Longer wire: length 1.2L, volume V/⁠2, so area A₂ = (V/⁠2)/⁠(1.2L) = A₁/⁠2.4 (that is A₁ ÷ 2.4). R₂/⁠R₁ = (1.2L/⁠A₂)/⁠(L/⁠A₁) = 1.2 × 2.4 = 2.88. So R₂ is 188% larger (NCERT C11-A24, C11-A25).

Answer: (b) The steady voltage across the ideal resistor decreases as compared to its initial value.

Method. Filament with positive temperature coefficient

As the metal filament heats, its resistance rises (resistivity of metals increases with temperature, NCERT C11-A27, TP3-A29). Total resistance rises, steady current falls, so the voltage across the fixed resistor (IR) falls: (b). (a) and (c) are the opposite; filament power I²R_f can rise or fall depending on the ratio R_f/⁠R, so (d) is not fixed.

Assumes a metallic filament (positive temperature coefficient).

Answer: (b) 3 : 1

Method. R = L/⁠(σA)

Same length. R ∝ 1/⁠(σA). R_X = R_Z gives σ_X A_X = σ_Z A_Z, i.e. A_X = 3A_Z. Let A_Y = a, A_Z = 2a, A_X = 6a. R_Y: R_X = (1/⁠(2a)): (1/⁠(1.6a)) = 3 : 1 (NCERT C11-A24).

Answer: (c) Same

Method. Resistivity is a material property

Resistivity depends on the material and temperature, not on the wire's length or area. Lengthening the wire changes R, not ρ, so ρ is the same (NCERT C11-A25).

Answer: (d) A voltmeter has a low resistance and an ammeter has a high resistance

Method. Meters

An ammeter is in series and has low resistance; a voltmeter is in parallel and has high resistance (NCERT C11-A07, C11-A14). (d) reverses these, so it is not correct.

Answer: (c) 12 V

Method. Potential difference V = W/⁠Q

V = 24 J / 2 C = 12 V (NCERT C11-A12).

Answer: (d) 9000 C

Method. Charge Q = It

Q = 15 A × (10 × 60 s) = 9000 C(NCERT C11-A03).

Answer: (b) Free electrons

Method. Conduction in a metal

In a metal the valence electrons are free (a sea of free electrons); under a field they drift and constitute the current. Bound electrons stay with their atoms and the ions are fixed in the lattice (NCERT TP3-A05, TP3-A06; C11-A01).

Answer: (c) 0.27 A

Method. Ohm's law, I ∝ V

R = 220/⁠0.5 = 440 Ω. I′ = 120/⁠440 = 0.2727 A ≈ 0.27 A (NCERT C11-A15, C11-A20).

Shortcut. I′ = 0.5 × 120/⁠220 = 0.27 A.

Answer: (c) 2 and 3

Method. Resistance versus resistivity

Resistance R = ρL/⁠A depends on length, area and temperature and is proportional to ρ (3 correct). Resistivity is a property of the material and temperature only: it does not depend on area or length (1 and 4 false), but it does vary with temperature (2 true) (NCERT C11-A24, C11-A25, C11-A27).

Answer: (c) one-fourth of its original value.

Method. Resistance and stretching, R = ρL/⁠A at constant volume

Doubling L halves A (volume fixed), so R′ = ρ(2L)/⁠(A/⁠2) = 4R = 40 Ω. With the same supply voltage I = V/⁠R falls to one-fourth (NCERT C11-A22, C11-A23, C11-A15).

The question assumes the same supply voltage.

Answer: (c) 360 C

Method. Charge Q = It

Q = 0.6 A × (10 × 60 s) = 360 C(NCERT C11-A03).

Answer: (c) 600 C

Method. Q = I t

Q = I t = 1.0 A x 600 s = 600 C(C11-A03).

Answer: (b) Galvanometer

Method. galvanometer detects current

A galvanometer detects the presence (and direction) of a small current (TP4-A63). The others measure refractive index, viscosity and diffraction.

Answer: (d) All homogeneous materials obey Ohm's law irrespective of whether the field is within range or strong.

Method. limits of Ohm's law

Ohm's law holds only over a range of field strength and for suitable materials; it fails at strong fields (TP3-A68, TP3-A73). (a) and (b) agree with the statement of the law and (c) with its field form, so (d) is the one that is not correct.

Answer: (d) R/⁠8

Method. R = rho l / A

R proportional to l/⁠r^2. New wire: l/⁠2 and r to 2r gives (1/⁠2)/⁠4 = 1/⁠8, so R/⁠8 (computed). C11-A24.

Answer: (b) 1.6 × 10^-16 J

Method. energy gained = qV

E = eV = 1.6e-19 C x 1000 V = 1.6e-16 J (computed; 1 keV = 1.6e-16 J). Work-energy for a charge through a potential difference (C11-A12).

Answer: (b) R₁ < R₃ < R₂

Method. slope of I-V graph = 1/⁠R

Slope of I against V is 1/⁠R (C11-A17). Line 1 is steepest (smallest R), line 3 intermediate, line 2 flattest (largest R). So R1 < R3 < R2.

Answer: (c) Semi conductor diode

Method. non-ohmic = V-I not a straight line

A semiconductor diode has a non-linear V-I curve (it conducts one way), so it is non-ohmic (TP3-A27). A copper coil, a heating coil (nichrome) and a rheostat wire are ohmic at fixed temperature.

Answer: (a) The current through it

Method. R = rho L / A

R = rho L/⁠A (C11-A24, TP3-A13) depends on length, area and resistivity. The current through the resistor does not change its resistance (V = IR only defines it, TP3-A72).

Answer: (a) L₁ = 3L and A₁ = 2A

Method. Resistance of combination (figure read from its description)

Two wires (L, A) in parallel give ρL/⁠(2A); in series with a third (L, A) of ρL/⁠A: R = 1.5 ρL/⁠A. New wire: ρL₁/⁠A₁ = 3ρL/⁠(2A), so L₁/⁠A₁ = 3L/⁠(2A). Option (a) gives 3L/⁠2A, which matches; (b) gives L/⁠A, (c) 2L/⁠3A and (d) L/⁠A, none of which match (NCERT C11-A24, C11-A31, C11-A34).

Answer: (b) 13 Ohm

Method. Parallel resistors

1/⁠R = 1/⁠1 + 1/⁠1 + 1/⁠1 = 3, so R = 1/⁠3 Ω (NCERT C11-A34).

Answer: (c) 9.0 Ω

Method. Series then parallel

5 + 7 = 12 Ω in series. 12 ∥ 36: R = 12×36/⁠(12+36) = 432/⁠48 = 9.0 Ω (NCERT C11-A31, C11-A34).

Answer: (d) (figure) two battery cells in series, two bulbs in parallel

Method. Bulb current in a circuit of identical cells (figures read from their description)

Let cell emf E, bulb R. (a) I = E/⁠2R; (b) I = 2E/⁠2R = E/⁠R in each bulb; (c) I = E/⁠R; (d) two cells give 2E across each parallel bulb: I = 2E/⁠R. Brightness ∝ I²R, so (d) is the brightest (NCERT C11-A29, C11-A32, C11-A47).

Assumes ideal cells (negligible internal resistance).

Answer: (b) V = 300 V; I = 5.5 A

Method. Parallel branches (figure read from its description)

V across the 100 Ω branch = 3 × 100 = 300 V. All branches share 300 V: I₂ = 300/⁠200 = 1.5 A, I₃ = 300/⁠300 = 1 A. I = 3 + 1.5 + 1 = 5.5 A. So V = 300 V, I = 5.5 A (NCERT C11-A32, C11-A33).

Answer: (b) 2 + 2⁠√5⁠ Ω

Method. Self-similar infinite ladder (figure read from its description)

The rail wires merge the junctions, so each section is a series R in the upper rail and a shunt R across the rails: the network repeats. Let x be the net resistance; the tail beyond the first section is x again. x = R + R·x/⁠(R + x), so x² − Rx − R² = 0 and x = R(1+⁠√5)/⁠2.⁠ With R = 4 Ω, x = 2(1+⁠√5)⁠ = 2 + 2⁠√5⁠ Ω ≈ 6.47 Ω (checked by solving a 150-section network: 6.4721 Ω) (NCERT C11-A31, C11-A34).

Answer: (c) 3.0 mA

Method. Two batteries with internal resistance in parallel (figure read from its description)

Node equation across R: (1 − V)/⁠0.1 + (5 − V)/⁠0.1 = V/⁠1000, so 60 = V(20 + 0.001), V = 60/⁠20.001 ≈ 2.9999 V. I = V/⁠R ≈ 2.9999/⁠1000 ≈ 3.0 mA (NCERT TP3-A43, TP3-A47).

Assumes ideal emfs 1 V and 5 V each with 0.1 Ω in series, in parallel across R.

Answer: (a) 1 A

Method. Short-circuited branch in a parallel group (figure read from its description)

The 0 Ω branch shorts the two 10 Ω resistors, so the group has zero resistance. Total R = 5 + 0 + 5 = 10 Ω. I = V/⁠R = 10/⁠10 = 1 A (NCERT C11-A15, C11-A31, C11-A57).

Answer: (a) (figure) three 3 Ω resistors in parallel between X and Y

Method. Equivalent resistance (figures read from their description)

(a) 3 ∥ 3 ∥ 3 = 1 Ω; (b) 3 ∥ 3 = 1.5 Ω; (c) 1 + 1 = 2 Ω; (d) 1+1+1 = 3 Ω. Minimum is 1 Ω in (a) (NCERT C11-A31, C11-A34, C11-A59).

Answer: (c) 3R/⁠2

Method. Series and parallel combination (figure read from its description)

Two R in parallel: R/⁠2. In series with the third R: R/⁠2 + R = 3R/⁠2 (NCERT C11-A31, C11-A34).

Answer: (a) 2 ohm

Method. cut and join in parallel

Each of five pieces is 50/⁠5 = 10 ohm (C11-A52). Five 10 ohm in parallel give 10/⁠5 = 2 ohm (computed).

Answer: (a) 3R = 2R₁

Method. parallel resistors with R proportional to length

Same material and thickness means R is proportional to length, so R2 = 2 R1. R = R1 x 2R1/⁠(3R1) = 2R1/⁠3, i.e. 3R = 2R1 (C11-A24, C11-A34).

Answer: (a) one-third of the individual resistance.

Method. equal resistors in parallel

1/⁠R = 3/⁠r, so R = r/⁠3 (C11-A34, C11-A59): one-third of each resistance.

Answer: (b) 240 Ω

Method. parallel resistors, Ohm's law

Total R = V/⁠I = 12 V / 0.1 A = 120 ohm. Two equal resistors in parallel give R/⁠2 = 120, so R = 240 ohm (computed). C11-A34.

Answer: (c) 5 Ω

Method. cut and join in parallel

Each half is 10 ohm (resistance is proportional to length, C11-A52). Two 10 ohm in parallel give 5 ohm.

Answer: (d) Conservation of energy.

Method. Kirchhoff's loop rule

The loop rule says the net energy gained and lost by a charge round a closed loop is zero, i.e. conservation of energy; the junction rule is conservation of charge (TP3-A50, TP3-A51, TP3-A76).

Answer: (b) 157 ohm

Method. parallel then series

1/⁠R = 1/⁠2 + 1/⁠4 + 1/⁠8 = 7/⁠8, so the three give 8/⁠7 ohm; adding the 1 ohm in series gives 15/⁠7 ohm (computed, exact). C11-A31, C11-A34.

Answer: (b) 35R

Method. reduce the network stage by stage

R parallel R = R/⁠2; in series with R gives 3R/⁠2 on the upper path; this is in parallel with the lower R: (3R/⁠2 x R)/⁠(3R/⁠2 + R) = 3R/⁠5 (computed, exact fraction 3/⁠5). Rules C11-A31, C11-A34, C11-A61.

Answer: (a) 5 Ω

Method. two semicircles in parallel

A diameter splits the 20 ohm ring into two equal halves of 10 ohm each, joined in parallel between the ends: R = 10x10/⁠(10+10) = 5 ohm. (Series/⁠parallel rule, C11-A34.)

Answer: (b) 4 : 1

Method. Heat H = V²t/⁠R

Each wire has resistance R. Parallel: R/⁠2; series: 2R. At the same V and t, H ∝ 1/⁠R, so H_par: H_ser = 2R: R/⁠2 = 4 : 1 (NCERT C11-A71).

Answer: (d) ₹8

Method. Power P = I²R and energy cost

P = I²R = 2² × 200 = 800 W. Light share = 20% = 160 W. Energy in 10 h = 0.160 × 10 = 1.6 kWh. Cost = 1.6 × 5 = ₹8 (NCERT C11-A44, C11-A47, C11-A49).

Answer: (d) 1100 s

Method. Frequency of mains

Indian mains is 50 Hz: 50 cycles/⁠s, 2 reversals per cycle = 100 reversals/⁠s, so direction changes every 1/⁠100 s (NCERT C12-A41).

Answer: (c) Nichrome

Method. Material for heating elements

A heating element needs high resistivity and a high melting point without oxidising. Nichrome (Ni–⁠Cr alloy) is used in irons, heaters and toasters; copper has low resistivity and tungsten is for bulb filaments (NCERT C11-A28, C11-A55).

Answer: (d) 20W

Method. P = V²/⁠R at fixed resistance

R = V²/⁠P is fixed, so P ∝ V². P′ = 80 × (110/⁠220)² = 80 × 1/⁠4 = 20 W (NCERT C11-A47).

Answer: (a) 3.6 × 10⁶ J

Method. Definition of the kilowatt-hour

1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J (NCERT C11-A49, the commercial unit).

Answer: (a) 132 W

Method. Power P = VI

P = 220 × 0.6 = 132 W (600 mA = 0.6 A; NCERT C11-A47).

Answer: (b) I²/⁠R

Method. Power formulae

P = VI = I²R = V²/⁠R (NCERT C11-A47). I²/⁠R is not one of them (dimensionally A²/⁠Ω is not W), so (b) cannot represent power (C11-A70).

Answer: (d) Tungsten

Method. Material of bulb filament

The filament must have a very high melting point and not oxidise; tungsten (m.p. about 3400 °C) is used, in an inert-gas filled bulb (NCERT C11-A43).

Answer: (b) there are 3 cycles/⁠s

Method. frequency = cycles per second

Frequency in hertz is the number of complete cycles per second, so 3 Hz is 3 cycles/⁠s (C12-A41, TP6-A52).

Answer: (b) 0.25 A

Method. P = V I

I = P/⁠V = 60/⁠240 = 0.25 A (computed). C11-A47.

Answer: (b) I R²

Method. power formulas

P = VI = I^2 R = V^2/⁠R (C11-A47, C11-A70). I R^2 is not a power: it has the units of ampere x ohm squared. So (b).

Answer: (b) ₹ 6,000

Method. energy = power x time

E = 5 kW x 10 h x 30 = 1500 kWh; cost = 1500 x 4 = Rs 6000 (computed). C11-A49, C11-A50.

Answer: (c) Tungsten

Method. tungsten filament

The filament is tungsten: very high melting point (about 3400 C) and high resistivity, so it glows white-hot without melting (C11-A43, C11-A54).

Answer: (c) increases substantially

Method. short circuit means almost zero resistance

With negligible resistance in the path I = V/⁠R becomes very large, so the current increases substantially (C11-A57, C12-A39).

Answer: (a) 1, 2 and 3

Method. Joule heating H = V^2 t / R = I^2 R t

Heat produced H = I^2 R t = V^2 t / R (C11-A40); with R fixed, it depends on the voltage, the current and the time, so all three statements are correct.

Answer: (b) P = P₁/⁠2

Method. P = V^2/⁠R with R = rho l / A

R proportional to l/⁠r^2. Wire B: 2l and 2r gives 2/⁠4 = 1/⁠2, so R_B = R_A/⁠2. At the same V, P is proportional to 1/⁠R so P1 = 2P, i.e. P = P1/⁠2. (C11-A47, C11-A24.)

Answer: (a) conducting and of low melting point

Method. fuse = conductor of low melting point

A fuse wire must conduct (it is in series in the circuit) and melt at a low temperature when current exceeds the rating (C11-A45, C12-A38).

Answer: (c) Red—live wire, Green—ground wire, Black—neutral wire

Method. domestic wiring colour code

Live is red, neutral is black, earth is green (C12-A32), so red-live, green-ground, black-neutral is option (c).

Answer: (a) Uniform

Method. Field inside a long solenoid

Inside a long current-carrying solenoid the field is uniform and parallel to the axis, B = μ₀nI (NCERT C12-A21, TP4-A45).

Answer: (a) Clockwise in a plane perpendicular to the wire

Method. Right-hand thumb rule (power line)

Current flows east to west; seen from the east end it runs away from the viewer. By the right-hand thumb rule, field lines are circles in a plane perpendicular to the wire, clockwise when the current points away (south at the point below) (NCERT C12-A15, C12-A51).

Answer: (d) is uniform inside the solenoid

Method. Field of a long solenoid

In a long straight solenoid the field inside is uniform (B = μ₀nI), away from the ends; it falls to about half at an end, but the stated property of the interior is uniformity (NCERT C12-A21, TP4-A45).

Option (b) is also true near the ends of a real solenoid; (d) is the standard statement for the field inside.

Answer: (a) 1

Method. Which materials are attracted by a magnet

Plastic and carbon (graphite) are not attracted (carbon is diamagnetic); aluminium is only feebly paramagnetic, not attracted by an ordinary magnet. Stainless steel is an iron-based alloy: the ferritic and martensitic grades are ferromagnetic and are attracted, while the austenitic grade (18/⁠8) is nearly non-magnetic when annealed. So one material can be attracted: option (a) (standard materials reference).

Less sure: depends on which grade of stainless steel is meant; if austenitic only, the answer would be (d). Chosen reading: 'can be attracted' holds for some stainless steels.

Answer: (d) If a soft iron bar is inserted inside the solenoid, the magnetic field remains the same.

Method. Properties of a current-carrying solenoid

Inside a solenoid the field is uniform and grows with current; the solenoid acts as a bar magnet (NCERT C12-A21, C12-A22). A soft iron core raises the field strongly, so (d) 'remains the same' is the incorrect statement.

Answer: (c) Magnetic needle

Method. Detecting a magnetic field

A small magnetic needle (compass) deflects in a magnetic field; ammeter and voltmeter measure current and potential difference, a motor uses the field (NCERT C12-A02).

Answer: (a) inversely on the distance from it

Method. B = mu0 I / 2 pi r

Outside a straight wire B = mu0 I/⁠(2 pi r), inversely proportional to distance (TP4-A35, C12-A14). Direct proportion with distance applies only inside a thick wire (TP4-A41), not at a point outside it.

Answer: (c) the Right-Hand Thumb rule and it would be in the upward direction.

Method. right-hand thumb rule

The direction of the current producing a given field circulation is found by the right-hand thumb rule (C12-A15): fingers curling anticlockwise (seen from above, the usual picture) put the thumb pointing upward. Left-hand rules are for force, not field. Reading assumes the circulation is viewed from above.

Answer: (c) Magnetic field lines are open curves

Method. field lines are closed curves

Magnetic field lines are continuous closed curves (C12-A08), have direction and magnitude (C12-A05) and never cross (C12-A10). So the statement that they are open curves is not correct.

Answer: (d) 1 and 2 only

Method. B in a solenoid = mu0 n I

B = mu0 n I (TP4-A45): it depends on turns per unit length and on current, not on the diameter (for a long solenoid). So 1 and 2 only.

Answer: (a) depends upon the current in the wire

Method. B of a long straight wire = mu0 I / 2 pi r

At a point outside the wire B = mu0 I/⁠(2 pi r) (TP4-A35, C12-A13): it depends on the current and the distance, not on the wire's radius or the surroundings' temperature.

Answer: (c) 0.4 tesla

Method. B at the centre of a coil = mu0 N I / 2R

B is proportional to N/⁠R (TP4-A30, TP4-A32). N to 2N and R to R/⁠2 multiplies B by 2 / (1/⁠2) = 4: 0.1 x 4 = 0.4 T.

Answer: (a) Zero

Method. force on a magnet in a uniform field

In a uniform field the forces on the two poles are equal and opposite, so the net force is zero (there is only a torque). Standard result (magnetic dipole in uniform field; TP4-A55).

No correct option. This question is defective as printed in UPSC's paper.

Method. magnetic field lines are closed loops, also inside the magnet

Two options are false: (a) magnetic lines have no source or sink point (no monopole), they are closed curves (C12-A08, TP4-A36); (d) lines do run inside the magnet from S to N (C12-A07). (b) and (c) are true (C12-A10).

Two options are incorrect statements: (a) magnetic field lines never start or end at a point, they are closed loops (no monopole; NCERT Class 12 ch. 5), and (d) field lines do run inside a bar magnet, from S to N (NCERT Class 10 ch. 13). The question has no single answer; if (a) is read loosely as 'lines spread out from a pole', (d) is the evident intended key.

Answer: (b) Magnetic equator

Method. Dip angle is zero at the magnetic equator

At the magnetic equator the dip is 0 degrees, so the Earth's field is horizontal; it is vertical at the magnetic poles. The magnetic meridian is a plane, not a place. Standard result (Earth's magnetism).

Answer: (b) 2B

Method. B = mu0 n I inside a solenoid

B = mu0 n I (TP4-A45), so with I fixed B is proportional to n: n to 2n gives B to 2B.

Answer: (b) Fleming's left-hand rule

Method. Force on a conductor in a field

Fleming's left-hand rule gives the direction of force F on a current-carrying conductor in a magnetic field (thumb force, forefinger field, middle finger current). The right-hand rule is for induced current (NCERT C12-A27).

Answer: (a) A split-ring type commutator

Method. Commutator of a DC generator

An AC generator has slip rings; to give a unidirectional (DC) output they are replaced by a split-ring commutator, which reverses the connection every half turn (NCERT TP6-A50, C12-A43).

Answer: (b) Positive charges and negative charges will separate out

Method. Force on moving charges, F = qv×B

For v perpendicular to B, F = qvB, directed along q(v×B). Positive and negative charges feel forces in opposite directions, so they separate; neither stops, as the force does no work and only turns the path (NCERT TP4-A08, TP4-A16).

Answer: (d) no deflecting force.

Method. Force on a moving charge, F = qvB sinθ

Velocity is along south, field along north: θ = 180°, sinθ = 0, so F = qvB sin180° = 0. A charge moving parallel or antiparallel to B feels no force (NCERT TP4-A09).

Answer: (c) Faraday’s laws of electromagnetic induction

Method. Principle of the generator

A DC (like an AC) generator turns a coil in a magnetic field; the changing flux induces an emf by Faraday's law of electromagnetic induction (NCERT C12-A43, TP6-A49).

Answer: (c) induced current

Method. Fleming's right-hand rule

Right-hand rule: forefinger = field, thumb = motion of conductor, middle finger = induced current (C12-A43).

Answer: (b) generator

Method. generator produces current

A generator converts mechanical to electrical energy by induction (TP6-A49). A motor does the reverse; an ammeter and a galvanometer only measure.

Answer: (d) into the page

Method. force on a charge F = q v x B (Fleming's left-hand rule)

Take right = +x, up = +y, out of the page = +z. v = +y, B = +x, and y x x = -z (computed with a cross product). The proton is positive, so the force is into the page. Same by Fleming's left-hand rule (C12-A27, TP4-A08). The figure's description fixes the answer.

Answer: (d) increasing voltage

Method. step-up transformer raises voltage

A step-up transformer has more secondary turns, Vs/⁠Vp = Ns/⁠Np > 1, so it raises the voltage (and lowers the current). It cannot raise power: output power cannot exceed input. Standard result (transformer).

Answer: (c) Transformer

Method. Transformer changes ac voltage up or down

A transformer (mutual induction between two coils) steps ac voltage up or down; a generator and a motor convert between mechanical and electrical energy, and a vibrator is not a voltage changer. Standard result (Class 12 Physics, ac devices; induction rows TP6-A37, TP6-A42).