A wire of length L and cross-sectional area A has resistivity ρ. If the wire is stretched to double its length, what will be the new resistance?
Answer: C
When wire is stretched to 2L, area becomes A/2. New resistance = ρ(2L)/(A/2) = 4ρL/A = 4R
Q.22Medium
A battery with EMF E and internal resistance r is connected to an external resistance R. What is the terminal voltage?
Answer: B
Terminal voltage V = E - Ir, where I = E/(R+r), so V = E - E·r/(R+r) = ER/(R+r)
Q.23Medium
A potentiometer wire of length 100 cm has resistance 10Ω. A standard cell of EMF 1.5V is balanced at 40 cm. What is the EMF of an unknown cell if balanced at 65 cm?
Answer: C
By potentiometer principle: E₁/E₂ = l₁/l₂, so E₂ = 1.5 × (4065) = 2.4375V ≈ 2.4V
Q.24Medium
In a circuit, the current through a 10Ω resistor is 2A. If this resistor is replaced by 5Ω, and the rest of the circuit remains unchanged, the new current will be:
Answer: D
Current depends on both the resistor value and the rest of the circuit configuration (series/parallel). Without knowing the circuit configuration, it cannot be determined
Q.25Medium
A uniform wire of length L is cut into n equal parts. When connected in parallel, the equivalent resistance is:
Answer: A
Each part has resistance R/n. In parallel: 1/R_eq = n/(R/n) = n²/R, so R_eq = R/n²
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Q.26Medium
If the length of a conductor is doubled and the diameter is also doubled, what happens to its resistance?
Answer: B
R = ρL/A. New R = ρ(2L)/π(2r)² = ρ(2L)/(4πr²) = R/2
Q.27Medium
A 60W bulb and a 100W bulb are connected in series across a 200V supply. Which bulb glows brighter?
Answer: A
In series, same current flows through both. Power = I²R. The 60W bulb has higher resistance (rated at lower power), so it dissipates more power and glows brighter
Q.28Medium
An ammeter has a low resistance and a voltmeter has a high resistance. This is because:
Answer: A
Low resistance ammeter minimizes voltage drop in series. High resistance voltmeter draws negligible current in parallel, not affecting the original circuit
Q.29Medium
Three cells of EMF 2V, 4V, and 3V with negligible internal resistance are connected in series, opposing each other. Net EMF is:
Answer: C
Taking one direction as positive: 2 + 4 - 3 = 3V or 2 - 4 + 3 = 1V. Net EMF = 1V (assuming standard configuration)
Q.30Medium
A 100Ω resistor and a 200Ω resistor are connected in parallel, and this combination is in series with a 50Ω resistor. Total resistance is:
Answer: B
Parallel: 1/R_p = 1001 + 2001 = 2003, so R_p = 66.67Ω. Total = 50 + 66.67 = 116.67Ω
Q.31Medium
A wire of length L and cross-sectional area A has resistivity ρ. If the wire is stretched to 2L without change in volume, what will be its new resistance?
Answer: A
When stretched to 2L, volume remains constant. New area A' = A/2. New resistance R' = ρ(2L)/(A/2) = 4ρL/A = 4R
Q.32Medium
In a Wheatstone bridge, if P/Q = R/S, then the current through the galvanometer is:
Answer: B
The bridge is balanced when P/Q = R/S. In balanced condition, potential difference across galvanometer is zero, hence no current flows through it.
Q.33Medium
A cell of EMF E and internal resistance r is connected to an external resistance R. The terminal voltage V is given by:
Answer: C
Terminal voltage V = E - Ir, where I is the current through the circuit. The voltage drop occurs only across internal resistance.
Q.34Medium
A potentiometer wire of length 100 cm has resistance 10Ω. A standard cell of EMF 1.5V is balanced at 60 cm. What is the EMF of unknown cell if it balances at 80 cm?
A nichrome wire and a copper wire of same length and area are connected in series. Which experiences greater heat dissipation?
Answer: A
Heat dissipated H = I²Rt. Since same current flows through both and nichrome has higher resistivity than copper (higher R), nichrome dissipates more heat.
Q.36Medium
In a meter bridge experiment, the null point is obtained at 40 cm from the left end. If the left resistance is 8Ω, what is the right resistance?
Answer: A
In meter bridge: R₁/R₂ = l₁/(100-l₁). So 8/R₂ = 6040, giving R₂ = 12Ω
Q.37Medium
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.38Medium
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.39Medium
Which of the following shows non-ohmic behavior?
Answer: C
Tungsten filament's resistance increases significantly with temperature due to heating, causing non-linear I-V characteristic (non-ohmic).
Q.40Medium
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
