A battery of EMF 10V supplies current to a circuit. If the voltage across external resistance is 9V, what is the internal resistance if external resistance is 90Ω?
Answer: B
V = E - Ir. 9 = 10 - I×r. Current I = V/R = 909 = 0.1A. So 9 = 10 - 0.1r, giving r = 10Ω
Q.82Medium
In a circuit with capacitors, which quantity is continuous across the capacitor?
Answer: C
In DC circuits, current is continuous (same) through all series elements including capacitors in steady state. However, in AC circuits, current flows through capacitors.
Q.83Hard
The length of a conductor increases by 10% when stretched. Assuming volume remains constant, the resistance increases by approximately:
Answer: C
R = ρL/A. If L increases by 10% and volume constant, A decreases by ~9.1%. New R = 1.1R₀/0.91 ≈ 1.21R₀, so 21% increase.
Q.84Hard
The equivalent resistance of an infinite ladder network of 1Ω resistors (each rung) is:
Tungsten filament's resistance increases significantly with temperature due to heating, causing non-linear I-V characteristic (non-ohmic).
Q.88Medium
In Joule heating, if voltage is doubled and resistance is halved, the power dissipated becomes:
Answer: C
P = V²/R. New P = (2V)²/(R/2) = 4V²×2/R = 8(V²/R) = 8P₀
Q.89Hard
A superconductor exhibits zero resistance below its critical temperature because:
Answer: B
BCS theory explains superconductivity: below critical temperature, electrons form Cooper pairs with no scattering, resulting in zero resistance.
Q.90Medium
A wire of length L and cross-sectional area A has resistance R. If the wire is stretched to 1.5 times its original length without change in volume, what will be its new resistance?
Answer: A
When stretched, length becomes 1.5L. Volume remains constant (LA = A'L'), so A' = A/1.5. New resistance R' = ρ(1.5L)/(A/1.5) = 2.25ρL/A = 2.25R
Q.91Medium
A heating element rated 1000W, 220V is connected to a 110V supply. The heat produced becomes:
Answer: A
Resistance of element R = V²/P = 220²/1000 = 48.4Ω (constant). At 110V: P' = V'²/R = 110²/48.4 ≈ 250W. Power varies with square of voltage.
Q.92Hard
In a Wheatstone bridge, arms P, Q, R, and S have resistances 10Ω, 15Ω, 20Ω, and 30Ω respectively. A galvanometer is connected between junctions of P-Q and R-S. The galvanometer reading will be:
Answer: A
For balanced bridge: P/Q = R/S. Check: 1510 = 3020 → 32 = 32. The bridge is balanced, so no current flows through galvanometer (zero reading).
