IMO Practice Test — Electromagnetic Induction and Alternating Current
14 Questions • 15 min • Olympiad level
15:00
Question 1 of 14
A magnet is dropped through a vertical copper pipe. It falls slower than in air because:
air resistance
eddy currents oppose its motion
gravity is weaker
the pipe is heavy
Explanation: Changing flux induces eddy currents that oppose the magnet's fall (Lenz's law).
Question 2 of 14
If the flux through a coil is constant, the induced EMF is:
maximum
zero
half
infinite
Explanation: EMF depends on the rate of change of flux; constant flux gives zero EMF.
Question 3 of 14
Doubling the speed of a rod in a fixed magnetic field changes its motional EMF to:
half
double
the same
four times
Explanation: $\varepsilon = Blv$ is proportional to v, so it doubles.
Question 4 of 14
Increasing the frequency of an AC source increases $X_L$ and:
increases $X_C$
decreases $X_C$
leaves $X_C$ unchanged
makes $X_C$ zero
Explanation: $X_C = 1/\omega C$ falls as frequency rises, while $X_L = \omega L$ rises.
Question 5 of 14
At resonance in a series LCR circuit, the current is in phase with the voltage because:
R is zero
$X_L = X_C$
C is removed
L is removed
Explanation: When reactances cancel, the circuit is purely resistive, so the phase angle is zero.
Question 6 of 14
A 200 ohm resistor and an inductor of reactance 150 ohm in series carry AC. The impedance is:
350 ohm
250 ohm
50 ohm
175 ohm
Explanation: $Z = \sqrt{200^2 + 150^2} = \sqrt{62500} = 250\,\Omega$.
Question 7 of 14
An ideal transformer steps 240 V down to 12 V. The turns ratio $N_p:N_s$ is:
1:20
20:1
12:1
1:12
Explanation: $N_p/N_s = V_p/V_s = 240/12 = 20$, so $N_p:N_s = 20:1$.
Question 8 of 14
If the secondary current of an ideal step-up transformer is 2 A and the turns ratio $N_s:N_p = 4:1$, the primary current is:
0.5 A
8 A
2 A
4 A
Explanation: $I_p = I_s \times N_s/N_p = 2 \times 4 = 8\,\text{A}$ (current ratio is inverse of voltage ratio).
Question 9 of 14
Quadrupling C in an LC circuit changes its oscillation frequency to:
double
half
four times
unchanged
Explanation: $f \propto 1/\sqrt{C}$; quadrupling C halves the frequency.
Question 10 of 14
A choke coil controls AC current in a tube light with little power loss because it has:
high R
high inductance and low resistance
high capacitance
no reactance
Explanation: An inductive choke limits current via reactance, dissipating almost no power (wattless).
Question 11 of 14
Two coils have a mutual inductance of 0.5 H. If the current in one changes at 4 A/s, the EMF in the other is:
2 V
8 V
0.125 V
4.5 V
Explanation: $\varepsilon = M\,dI/dt = 0.5 \times 4 = 2\,\text{V}$.
Question 12 of 14
In an AC circuit the current leads the voltage by 90 degrees. The circuit is purely:
resistive
capacitive
inductive
resonant
Explanation: Current leading by 90 degrees is the signature of a pure capacitor.
Question 13 of 14
The peak value of the 220 V (rms) mains supply is about:
156 V
311 V
220 V
440 V
Explanation: $V_0 = V_{rms}\sqrt2 = 220 \times 1.414 \approx 311\,\text{V}$.
Question 14 of 14
Energy stored in an inductor of 2 H carrying 3 A is:
6 J
9 J
18 J
3 J
Explanation: $U = \frac{1}{2}LI^2 = \frac{1}{2}\times 2\times 9 = 9\,\text{J}$.