JEE Advanced style original question
Advanced 01: Multiple Correct Boundary Reasoning
Difficulty: Moderate | Type: Multiple-correct
Question: JEE Advanced style original question. A closed insulated box contains three conductors with charges +4e, -9e and +2e. During a process, two electrons move from the second conductor to the first. Which statements are correct?
- Charge of the first conductor becomes +2e.
- Charge of the second conductor becomes -7e.
- Net charge of the box remains -3e.
- Charge conservation is violated because individual charges changed.
Complete solution: Moving two electrons to the first conductor adds -2e to +4e, giving +2e. The second conductor loses -2e, so -9e becomes -7e. The total remains +2e - 7e + 2e = -3e.
Final answer: A, B and C
Key concept tested: Internal transfer changes individual charges but not total charge.
JEE Advanced style original question
Advanced 02: Integer Charge Constraint
Difficulty: Moderate | Type: Integer-type
Question: JEE Advanced style original question. The least number of electrons that must be removed from a neutral body so that its charge exceeds +1.0 × 10-18 C is N. Find N.
Complete solution: Need Ne > 1.0 × 10-18. N > 1.0/1.6 × 10 = 6.25. The least integer greater than 6.25 is 7.
Final answer: 7
Key concept tested: Quantization and strict inequality.
JEE Advanced style original question
Advanced 03: Assertion Reason On Invariance
Difficulty: Conceptual | Type: Assertion-reason
Question: JEE Advanced style original question. Assertion: A relativistic electron has the same electric charge as an electron at rest. Reason: Electric charge is invariant under change of inertial frame.
- Both assertion and reason are true, and reason explains assertion.
- Both are true, but reason does not explain assertion.
- Assertion is true and reason is false.
- Assertion is false and reason is true.
Complete solution: The charge of a particle is independent of speed and observer frame. Therefore, the reason is true and directly explains the assertion.
Final answer: A
Key concept tested: Charge invariance.
JEE Advanced style original question
Advanced 04: Matching Charging Methods
Difficulty: Moderate | Type: Matching-type
Question: JEE Advanced style original question. Match the charging process with the best description: P Friction, Q Conduction, R Induction, S Pair production.
- P-electron transfer by rubbing, Q-contact charge sharing, R-separation without contact, S-creation of +e and -e pair
- P-contact sharing, Q-rubbing, R-annihilation, S-grounding only
- P-neutralization, Q-no contact, R-current heating, S-two electrons appear
- P-mass transfer, Q-proton flow, R-charge destroyed, S-net charge +e
Complete solution: Friction transfers electrons during rubbing. Conduction needs contact. Induction separates charge without direct contact. Pair production creates equal opposite charges, conserving net charge.
Final answer: A
Key concept tested: Recognizing mechanisms of charging and conservation.
JEE Advanced style original question
Advanced 05: Paragraph On Three Spheres
Difficulty: Difficult | Type: Paragraph-based
Question: JEE Advanced style original question. Three identical conducting spheres A, B and C carry +12 nC, -6 nC and 0 respectively. A and B are touched and separated; then B and C are touched and separated. Find the final charges.
Complete solution: After A and B touch, total +6 nC is shared equally, so A = +3 nC and B = +3 nC. Then B and C touch: total +3 nC is shared equally, so B = +1.5 nC and C = +1.5 nC. A remains +3 nC.
Final answer: A = +3 nC, B = +1.5 nC, C = +1.5 nC
Key concept tested: Stepwise charge redistribution on identical conductors.
JEE Advanced style original question
Advanced 06: Multiple Correct Quantization
Difficulty: Difficult | Type: Multiple-correct
Question: JEE Advanced style original question. Which exact charges are possible for an isolated body?
- -4.8 × 10-19 C
- +6.0 × 10-19 C
- +1.6 × 10-18 C
- -2.0 × 10-19 C
Complete solution: Divide each magnitude by 1.6 × 10-19 C. The n values are 3, 3.75, 10 and 1.25. Only integer n values are possible.
Final answer: A and C
Key concept tested: Allowed and forbidden charge values.
JEE Advanced style original question
Advanced 07: System Boundary With Escaping Particle
Difficulty: Difficult | Type: Multi-step numerical
Question: JEE Advanced style original question. An isolated container initially has net charge +8 nC. A small particle of charge q escapes from an inner compartment into a region outside the chosen subsystem but still inside the container. The chosen subsystem charge becomes +11 nC. Find q and the charge of the whole container.
Complete solution: For the chosen subsystem, Qafter = Qbefore - q. Thus +11 = +8 - q, so q = -3 nC. The whole container is isolated, so its total charge remains +8 nC.
Final answer: q = -3 nC; whole container remains +8 nC
Key concept tested: Conservation depends on the system boundary chosen.
JEE Advanced style original question
Advanced 08: Pair Production Constraint
Difficulty: Conceptual | Type: Multiple-correct
Question: JEE Advanced style original question. A neutral photon produces particles inside a chamber. Which final charge combinations are allowed by charge conservation alone?
- electron + positron
- proton + antiproton
- electron only
- two positrons and one electron
Complete solution: The initial photon has zero charge. Electron plus positron gives -e + e = 0. Proton plus antiproton gives +e - e = 0. Electron only gives -e, and two positrons plus one electron gives +e.
Final answer: A and B
Key concept tested: Charge conservation in particle creation.
JEE Advanced style original question
Advanced 09: Induced Neutral Conductor
Difficulty: Moderate | Type: Assertion-reason
Question: JEE Advanced style original question. Assertion: A neutral isolated conductor placed near a positive rod may attract the rod. Reason: Charges inside the conductor redistribute, but the conductor's net charge remains zero.
- Both are true, and the reason explains the assertion.
- Both are true, but the reason does not explain the assertion.
- Assertion is true and reason is false.
- Assertion is false and reason is true.
Complete solution: The positive rod draws electrons slightly closer inside the conductor. The near side becomes effectively negative and attraction dominates, while total charge remains zero.
Final answer: A
Key concept tested: Induction and neutrality.
JEE Advanced style original question
Advanced 10: Charge Versus Mass Comparison
Difficulty: Conceptual | Type: Multiple-correct
Question: JEE Advanced style original question. Choose correct comparisons between electric charge and mass in school-level physics.
- Charge can be positive or negative; mass is taken as positive.
- Charge adds algebraically; mass adds arithmetically.
- Electric force can attract or repel; gravitational force is attractive.
- Charge of a particle changes with its speed in the same way as kinetic energy.
Complete solution: The first three statements are standard differences. The fourth is false because charge is invariant and does not change with speed.
Final answer: A, B and C
Key concept tested: Charge and mass distinctions.
JEE Advanced style original question
Advanced 11: Integer Type From Final Neutrality
Difficulty: Moderate | Type: Integer-type
Question: JEE Advanced style original question. A body has charge -9.6 × 10-19 C. How many electrons must be removed from it to make it neutral?
Complete solution: The body has excess electrons. n = 9.6/1.6 = 6, so removing 6 electrons makes the net charge zero.
Final answer: 6
Key concept tested: Electron excess and neutralization.
JEE Advanced style original question
Advanced 12: Possible Final Configurations
Difficulty: Difficult | Type: Multiple-correct
Question: JEE Advanced style original question. An isolated system has total charge +5e. Which final configurations of three bodies are possible?
- +2e, +4e, -1e
- +3e, +3e, 0
- +6e, -2e, +1e
- +2.5e, +2.5e, 0
Complete solution: Final totals must be +5e and each body charge must be an integer multiple of e. A totals +5e. B totals +6e. C totals +5e. D uses fractional charges, so it is not allowed.
Final answer: A and C
Key concept tested: Combining conservation with quantization.
JEE Advanced style original question
Advanced 13: Two Stage Electron Movement
Difficulty: Difficult | Type: Multi-step numerical
Question: JEE Advanced style original question. Bodies A and B initially have +7e and -3e. First, 4 electrons move from B to A. Then 2 electrons move from A to B. Find final charges.
Complete solution: After 4 electrons move B to A, A gains -4e and becomes +3e; B loses -4e and becomes +1e. Then 2 electrons move A to B, so A loses -2e and becomes +5e; B gains -2e and becomes -1e.
Final answer: A = +5e, B = -1e
Key concept tested: Tracking electron movement with signs.
JEE Advanced style original question
Advanced 14: Annihilation Accounting
Difficulty: Conceptual | Type: Paragraph-based
Question: JEE Advanced style original question. A chamber contains 5 electrons and 3 positrons. All possible electron-positron pairs annihilate into photons. Find the remaining net charge and particles.
Complete solution: Three electrons annihilate with three positrons, making photons of zero charge. Two electrons are left. Net charge before was -5e + 3e = -2e, and after it is also -2e.
Final answer: Two electrons remain; net charge = -2e
Key concept tested: Annihilation and charge conservation.
JEE Advanced style original question
Advanced 15: Millikan Type Remainder
Difficulty: Difficult | Type: Integer-type
Question: JEE Advanced style original question. Oil drops are found with charges 8.0, 12.8, 19.2 and 24.0 in units of 10-19 C. Taking e = 1.6 × 10-19 C, find the smallest integer n among these drops.
Complete solution: The n values are 8.0/1.6 = 5, 12.8/1.6 = 8, 19.2/1.6 = 12 and 24.0/1.6 = 15. The smallest integer n is 5.
Final answer: 5
Key concept tested: Millikan-type quantization pattern.