A copper wire of length L and cross-sectional area A has resistance R. If the wire is stretched to double its length while maintaining the same volume, what is the new resistance?
Answer: B
When stretched to double length, volume remains constant. New area = A/2. R' = ρL'/A' = ρ(2L)/(A/2) = 4ρL/A = 4R
Q.322Easy
In a circuit, three resistors of 2Ω, 3Ω, and 6Ω are connected in parallel. The equivalent resistance is:
A battery of EMF 12V and internal resistance 2Ω is connected to an external resistance of 4Ω. The current flowing through the circuit is:
Answer: A
I = E/(R + r) = 12/(4 + 2) = 612 = 2A
Q.324Easy
The resistivity of a material depends on:
Answer: C
Resistivity is an intrinsic property depending only on the material type and temperature, not on geometry
Q.325Medium
A wire of resistance 5Ω is bent into a square loop. What is the equivalent resistance between two adjacent corners?
Answer: A
Each side has resistance 1.25Ω. Between adjacent corners: one path has 1.25Ω, parallel path has 3.75Ω. R_eq = (1.25 × 3.75)/(1.25 + 3.75) = 1.25Ω
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Q.326Medium
In a Wheatstone bridge at balance, if P = 10Ω, Q = 20Ω, and R = 15Ω, then S equals:
Answer: A
At balance: P/Q = R/S. Therefore S = (Q × R)/P = (20 × 15)/10 = 30Ω
Q.327Easy
A light bulb rated 60W, 120V is connected to a 120V supply. The resistance of the bulb is:
Answer: B
R = V²/P = (120)²/60 = 6014400 = 240Ω
Q.328Medium
Two resistors R₁ and R₂ are connected in series with a battery. If R₁ = 2R₂ and the voltage across R₁ is 8V, the voltage across R₂ is:
Answer: A
In series, current is same. V₁/V₂ = R₁/R₂ = 2R₂/R₂ = 2. Therefore V₂ = V₁/2 = 28 = 4V
Q.329Medium
In a potentiometer experiment, the balancing length for a cell of EMF E is 75 cm. If another cell of EMF E/2 is used, the balancing length would be:
Answer: B
EMF is proportional to balancing length. If EMF becomes E/2, balancing length becomes 275 = 37.5 cm
Q.330Medium
The EMF of a cell is 2V and its internal resistance is 0.5Ω. When connected to external resistance, the terminal voltage is 1.8V. The current in the circuit is:
Answer: A
E - I×r = V_terminal. 2 - I×0.5 = 1.8. I×0.5 = 0.2. I = 0.4A
Q.331Easy
Three identical cells, each of EMF 1.5V and internal resistance 1Ω, are connected in series. The total EMF and internal resistance are:
Answer: C
In series: Total EMF = 1.5 + 1.5 + 1.5 = 4.5V. Total internal resistance = 1 + 1 + 1 = 3Ω
Q.332Easy
A rheostat is used in a circuit to:
Answer: B
A rheostat is a variable resistor used to control and vary the current in a circuit
Q.333Medium
When a conductor is heated, its resistance increases because:
Answer: C
Heating increases atomic vibrations, leading to increased collisions between electrons and atoms, thus increasing resistance
Q.334Medium
A copper wire and an aluminum wire of same length and cross-sectional area carry the same current. Which has greater voltage drop?
Answer: B
Aluminum has higher resistivity than copper. Since V = I×R and resistivity differs, aluminum wire has greater voltage drop
Q.335Easy
In a parallel combination of resistors, the voltage across each resistor is:
Answer: B
In parallel combination, all resistors are connected across same two points, hence voltage across each is same
Q.336Hard
A heating element of resistance R is connected to a battery of EMF E and internal resistance r. Maximum power is dissipated in R when:
Answer: B
By maximum power transfer theorem, maximum power is transferred to external load when load resistance equals internal resistance
Q.337Easy
In a circuit with a battery of 10V and total resistance 5Ω, the power dissipated is:
Answer: C
P = V²/R = (10)²/5 = 5100 = 20W
Q.338Medium
Two resistors of 4Ω and 6Ω are connected in series with a 10V battery. The current through the circuit and power dissipated are respectively:
Answer: C
Total R = 4 + 6 = 10Ω. I = V/R = 1010 = 1A. P = V×I = 10×1 = 10W
Q.339Easy
In a circuit, if the current increases when voltage increases, the resistance must:
Answer: C
From Ohm's law: I = V/R. If I increases with increasing V while R stays constant, then the relationship is proportional and follows Ohm's law
Q.340Medium
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
