📊 PYQ Analysis — Last 10 Years
Previous Year Question Patterns
10 years of CBSE, NEET, JEE Main, and JEE Advanced questions analyzed. Know the patterns. Predict what's coming.
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JEE Main Questions (10 yrs)
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NEET Questions (10 yrs)
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JEE Advanced Qs (10 yrs)
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Most-Repeated Topic Types
📊 JEE Main — EMI Questions Per Year (2015–2024)
📈 NEET — EMI Questions Per Year (2015–2024)
🍕 Topic-Wise Weightage (All Exams Combined)
🕸 Difficulty Radar — By Exam & Topic
📐 Topic-Wise Weightage Analysis
JEE Main — Topic Weightage
NEET — Topic Weightage
🔬 CBSE Pattern Summary
CBSE allocates 8–10 marks to EMI annually. High-priority: (1) Derivations — Faraday's law, solenoid inductance, AC generator EMF = 5 marks each. (2) Numerical applications — 2-3 mark short answers. (3) Application MCQs from eddy currents and transformers.
State Faraday's laws of electromagnetic induction. A circular coil of 200 turns, radius 0.1 m is placed in a uniform magnetic field of 0.5 T. If the field reverses direction in 0.05 s, calculate the average induced EMF.
💡 ΔΦ = 2BA (field reversal doubles the flux change). Answer = N×2BA/t = 200×2×0.5×π×0.01/0.05 = 1256.6 V
Derive an expression for the self-inductance of a long solenoid of N turns, length l, and cross-sectional area A. A solenoid of 500 turns, length 0.4 m, area 4 cm² carries 3A. Find self-inductance and energy stored.
💡 Standard derivation: L = μ₀N²A/l. Then U = ½LI². Practice this derivation — appears almost every year.
With a labelled diagram, explain the principle, construction, and working of an AC generator. Derive the expression for instantaneous EMF produced.
💡 Derivation: Φ = NBAcos(ωt), ε = −dΦ/dt = NBAω sin(ωt). Include diagram of rotating coil in B field with slip rings and brushes.
Define eddy currents. Give two disadvantages and two applications. Why are transformer cores laminated?
💡 Lamination reduces area of eddy current loops → higher resistance → less I²R loss. Standard marks question.
Draw a labelled diagram of a step-up transformer. Derive the expression relating turns ratio with voltage ratio. Why is the primary current greater when the secondary is loaded?
💡 N₂/N₁ = V₂/V₁. When load is connected, I₂ increases → power conservation (V₁I₁ = V₂I₂) requires I₁ to increase.
🔬 NEET Pattern Summary
NEET asks 1–3 EMI questions per year. 60% are conceptual/directional (Lenz's Law, application recognition). 30% are formula-based numerical. 10% are higher-order application (AC generator principle, eddy current applications). Rarely asks derivations — focus on concepts and quick calculations.
A conducting loop is held stationary in a non-uniform magnetic field. The induced EMF in the loop is: (A) Maximum (B) Zero (C) Proportional to field strength (D) Depends on resistance
💡 Answer: B — Zero. Non-uniform but STATIC field → flux through loop is constant → dΦ/dt = 0 → EMF = 0. Classic NEET trap.
Which of the following is NOT an application of eddy currents? (A) Induction furnace (B) Speedometer (C) Galvanometer damping (D) Capacitor charging
💡 Answer: D. Capacitor charging has nothing to do with eddy currents. Others all use the heating or braking effect of eddy currents.
Two coils have mutual inductance of 0.005 H. If the current in the primary changes from 0 to 10 A in 0.01 s, what is the magnitude of EMF induced in the secondary?
💡 ε = M × dI/dt = 0.005 × (10/0.01) = 0.005 × 1000 = 5 V
The induced charge in a coil is independent of: (A) Number of turns (B) Resistance of circuit (C) Change in flux (D) Time of flux change
💡 Answer: D — Time. q = NΔΦ/R — no time dependence. The charge depends on total flux change, not how fast it changes. This is the ballistic galvanometer principle.
A coil of 100 turns and resistance 50 Ω has an initial flux of 0.1 Wb. If flux is reduced to zero in 0.25 s, find the charge through the circuit.
💡 q = NΔΦ/R = 100 × 0.1 / 50 = 0.2 C. Note: time plays no role in charge calculation!
🔬 JEE Main Pattern Summary
JEE Main consistently asks 2–4 EMI questions. Top patterns: (1) Rail/rod problems with multi-step calculations, (2) Graph-based — Φ-t graph, find EMF, (3) Rotating rod EMF = ½BωL², (4) Energy in inductor, (5) LC oscillation frequency. 2023–2024 trend: increasing graph interpretation questions.
A rod of length 1.5 m rotates about one end with angular velocity 2π rad/s in a uniform B = 0.4 T field perpendicular to the plane of rotation. Find the EMF induced between the ends of the rod.
💡 ε = ½BωL² = ½ × 0.4 × 2π × (1.5)² = ½ × 0.4 × 2π × 2.25 = 0.9π ≈ 2.83 V
The magnetic flux through a coil varies with time as Φ = 5t³ − 10t + 4 (Wb). What is the EMF induced at t = 2s? The coil has 10 turns.
💡 dΦ/dt = 15t² − 10. At t=2: dΦ/dt = 60 − 10 = 50 Wb/s. ε = −N × dΦ/dt = −10 × 50 = −500 V (magnitude = 500 V)
A rod (mass 0.5 kg, length 2 m) on rails (resistance 4 Ω) in B = 0.5 T moves at terminal velocity. Find the terminal velocity if external force applied is 2 N.
💡 At terminal: F = B²L²v_T/R → v_T = FR/B²L² = 2×4/(0.25×4) = 8/1 = 8 m/s
Two identical inductors L are connected in series with mutual inductance M between them (aiding). The effective inductance is: (A) 2L (B) 2L+M (C) 2L+2M (D) 2(L+M)
💡 Answer: C — L_eff = L₁ + L₂ ± 2M. For aiding configuration: + sign → L_eff = L + L + 2M = 2L + 2M = 2(L+M). Note: Option D is the same as C if simplified.
In an LC circuit, the inductance L = 1 mH and capacitance C = 1 μF. The maximum current is 2 A. Find the maximum charge on the capacitor.
💡 Energy: ½LI²_max = ½CV²_max = q²_max/2C → q_max = I_max √(LC) = 2 × √(10⁻³ × 10⁻⁶) = 2 × 10⁻⁴·⁵ = 2×10⁻⁴·⁵ ≈ 6.32 × 10⁻⁵ C
🔬 JEE Advanced Pattern Summary
JEE Advanced tests EMI at multi-concept depth. Common types: (1) Rail problems with variable force/friction + dynamics, (2) Rotating disk/rod with current distribution, (3) Mutual inductance between non-coaxial configurations, (4) LC circuits with EMI derivation. Typically 1–2 questions per year, each multi-part and heavily analytical.
A conducting rod of length L and mass m slides on two vertical conducting rails (separation L) connected by resistance R at top. Magnetic field B is horizontal. Rod starts from rest. Write the equation of motion and find the velocity as a function of time.
💡 ma = mg − B²L²v/R → m(dv/dt) + (B²L²/R)v = mg. Solve: v(t) = v_T(1 − e^(−t/τ)) where v_T = mgR/B²L² and τ = mR/B²L² (time constant)
A conducting disk of radius R rotates with angular velocity ω in a uniform magnetic field B perpendicular to the disk. Find the EMF between the center and the rim, and the current if the disk resistance is ρ (resistivity), thickness t.
💡 EMF = ½BωR². This is analogous to rotating rod: integrate dε = B(rω)dr from 0 to R → ε = ½BωR². Current calculation requires computing disk resistance (like resistors in parallel).
A long solenoid (radius a, n₁ turns/length) has a smaller circular coil (radius b, N₂ turns) placed coaxially inside it. Find M. Now the coil is tilted by 60°. Find new M.
💡 Original: M = μ₀n₁N₂πb². When tilted by 60°: effective area = πb² cos60° = πb²/2. New M = μ₀n₁N₂πb²/2 = M_original/2.
A rod (resistance R, mass m, length L) on horizontal rails in vertical B field is pushed and released. Show that the velocity decays exponentially. Find the total heat generated in the circuit after infinite time.
💡 ma = −B²L²v/R → exponential decay: v(t) = v₀e^(−B²L²t/mR). Total heat = initial KE = ½mv₀² (energy conservation — all KE converts to heat). This is elegant application of energy conservation.
🔁 Patterns That Repeat Every Year
Pattern 1: Rail Problem (All Exams)
Setup: Rod on parallel rails in B field. Resistance may be in rails, rod, or external.
Trap: Forget rod's own resistance. Calculate total R first, then I = BLv/R.
Extension types: (a) constant velocity → find required force, (b) released from rest → find terminal velocity, (c) exponential velocity equation for JEE Adv.
Trap: Forget rod's own resistance. Calculate total R first, then I = BLv/R.
Extension types: (a) constant velocity → find required force, (b) released from rest → find terminal velocity, (c) exponential velocity equation for JEE Adv.
Pattern 2: "Find direction of induced current" (CBSE & NEET)
Appears in NEET 7 out of 10 years. CBSE almost every year.
Use 4-step method: (1) identify flux change direction, (2) apply Lenz's Law, (3) find induced B direction, (4) right-hand rule for current.
Setups: magnet moving toward/away from loop, expanding/shrinking loop in B field, loop entering/leaving B field region.
Use 4-step method: (1) identify flux change direction, (2) apply Lenz's Law, (3) find induced B direction, (4) right-hand rule for current.
Setups: magnet moving toward/away from loop, expanding/shrinking loop in B field, loop entering/leaving B field region.
Pattern 3: Self-Inductance Derivation (CBSE)
5-mark question appears in CBSE Board almost every alternate year. Steps: B inside solenoid → flux per turn → total flux linkage → L = NΦ/I → L = μ₀N²A/l. Practice writing this derivation cleanly in exam conditions.
Pattern 4: Rotating Rod/Disk EMF (JEE Main)
Appears in JEE Main 6 out of last 10 years. Formula ε = ½BωL² derived by integration. Sometimes the rod is a sector of a disk — same formula with R instead of L. Sometimes B is not perpendicular — need B cosα component.
Pattern 5: Induced Charge is Independent of Time (NEET)
NEET asks this concept directly. q = NΔΦ/R. No time variable. Tests whether student knows the ballistic galvanometer principle. The answer is always "time" when asked "which quantity does induced charge NOT depend on."
Pattern 6: EMF from Φ = f(t) Equation (JEE Main)
Flux given as polynomial/trig function of time. Find EMF at specific t. Process: differentiate Φ, multiply by N, take magnitude. If Φ is linear → EMF constant. If quadratic → EMF varies linearly. If sinusoidal → EMF is cosinusoidal.
Pattern 7: AC Generator Derivation (CBSE)
5-mark question in CBSE. With diagram of rotating coil in B field. Derive ε = NBAω sin(ωt). State the positions of maximum and zero EMF. Relate to applications in power stations.