Current electricity and magnetism together form one of the largest blocks of the JEE Physics syllabus. This set covers resistance, circuits with internal resistance, Kirchhoff's rules, the Wheatstone bridge and potentiometer, electrical power, magnetic fields from currents, forces on charges and wires, galvanometer conversion and motional EMF. Use μ₀ = 4π × 10⁻⁷ T m/A.
How to practise: read each question and tap the option you think is correct. You will see right away whether you got it, along with a short explanation, and your score is tracked at the bottom of the screen. This set has 15 questions.
Topics covered in this set
- Resistance, Ohm's law and resistor combinations
- EMF, internal resistance and Kirchhoff's rules
- Wheatstone bridge and potentiometer
- Electrical power and bulbs in series
- Magnetic field of wires, loops and solenoids
- Galvanometer conversion and motional EMF
JEE Physics MCQs with answers and explanations
Q1. A uniform wire is stretched so that its length doubles while its volume stays the same. Its resistance becomes:
a) Unchanged
b) 2 times
c) 4 times
d) Half
Q2. Resistors of 3 Ω and 6 Ω are in parallel, and this pair is in series with a 4 Ω resistor across an ideal 12 V battery. What current flows from the battery?
a) 1 A
b) 4 A
c) 2 A
d) 3 A
Q3. A cell of EMF 12 V and internal resistance 1 Ω is connected to a 5 Ω resistor. What is the terminal voltage of the cell?
a) 11 V
b) 12 V
c) 10 V
d) 2 V
Q4. A Wheatstone bridge is balanced with P = 10 Ω, Q = 20 Ω and R = 15 Ω. What is the fourth resistance S?
a) 15 Ω
b) 30 Ω
c) 45 Ω
d) 7.5 Ω
Q5. On a potentiometer, a 1.5 V standard cell balances at 250 cm. An unknown cell balances at 300 cm on the same wire. What is the EMF of the unknown cell?
a) 1.25 V
b) 1.5 V
c) 1.8 V
d) 2.0 V
Q6. What is the resistance of a bulb rated 100 W, 220 V?
a) 220 Ω
b) 22,000 Ω
c) 484 Ω
d) 2.2 Ω
Q7. A 100 W and a 60 W bulb, both rated 220 V, are connected in series across 220 V. Which glows brighter?
a) 100 W bulb
b) 60 W bulb
c) Both equally
d) Neither glows
Q8. Kirchhoff's junction rule is a consequence of the conservation of:
a) Energy
b) Mass
c) Charge
d) Momentum
Q9. What is the magnetic field at the centre of a circular loop of radius 10 cm carrying a current of 10 A?
a) 2 × 10⁻⁵ T
b) 4π × 10⁻⁵ T
c) π × 10⁻⁵ T
d) 2π × 10⁻⁵ T
Q10. What is the magnetic field 10 cm from a long straight wire carrying 10 A?
a) 2π × 10⁻⁵ T
b) 2 × 10⁻⁵ T
c) 10⁻⁵ T
d) 2 × 10⁻⁴ T
Q11. A long solenoid has 1,000 turns per metre and carries 2 A. What is the magnetic field inside it?
a) 8π × 10⁻⁴ T
b) 4π × 10⁻⁴ T
c) 2 × 10⁻³ T
d) 8π × 10⁻⁷ T
Q12. A charged particle moves perpendicular to a uniform magnetic field. If its speed is doubled, what happens to the radius and period of its circular path?
a) Radius doubles, period doubles
b) Radius unchanged, period halves
c) Both unchanged
d) Radius doubles, period unchanged
Q13. Two long parallel wires 1 m apart each carry 1 A in the same direction. What is the force per unit length between them?
a) 2 × 10⁻⁷ N/m, repulsive
b) 10⁻⁷ N/m, attractive
c) 4π × 10⁻⁷ N/m, attractive
d) 2 × 10⁻⁷ N/m, attractive
Q14. A galvanometer of resistance 50 Ω gives full-scale deflection at 1 mA. What resistance must be connected in series to convert it into a voltmeter of range 10 V?
a) 10,000 Ω
b) 0.05 Ω
c) 9,950 Ω
d) 10,050 Ω
Q15. A 0.5 m rod moves at 4 m/s perpendicular to a uniform magnetic field of 0.5 T, with the rod also perpendicular to the field. What EMF is induced across its ends?
a) 0.25 V
b) 2 V
c) 4 V
d) 1 V
Quick revision notes
- R = ρL/A; stretching n times at constant volume gives R × n².
- Terminal voltage V = E − Ir; balanced bridge P/Q = R/S; potentiometer E ∝ l.
- P = V²/R; in series the lower-wattage bulb glows brighter.
- Loop B = μ₀I/2R; straight wire μ₀I/2πr; solenoid μ₀nI; parallel wires F/L = μ₀I₁I₂/2πd.
- r = mv/qB, T = 2πm/qB; voltmeter R = V/I_g − G; shunt S = I_gG/(I − I_g); ε = Bvl.
Frequently asked questions
Are these JEE previous year questions?
No. They are original practice questions at the JEE Main level.
Why is a potentiometer preferred over a voltmeter for measuring EMF?
At balance it draws no current from the cell, so there is no drop across the internal resistance and the reading equals the true EMF.
How do I find the direction of the force on a moving charge in a magnetic field?
Use F = q(v × B) with the right-hand rule, and reverse the direction for a negative charge.
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