Area of cross section of a wire (A) is twice that of wire (B) (given the wires are of the same length and of same material). Whose resistance will be more?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → wire B
The resistance of a conductor is directly proportional to its length and inversely proportional to its area of cross section:
\( \displaystyle R = \rho\,\frac{l}{A} \)
Both wires have the same length l and the same material (same resistivity ρ). With AA = 2AB:
\( \displaystyle \frac{R_A}{R_B} = \frac{A_B}{A_A} = \frac{1}{2} \;\Rightarrow\; R_B = 2R_A \)
So wire B, the thinner wire, has the higher resistance (twice that of A).
Why the other options are incorrect
Option (1) wire A — Incorrect. A has the larger cross section, so it has the smaller resistance.
Option (3) Both are equal — Incorrect. Resistances would be equal only if the areas were equal.
Option (4) Resistance does not depend on the area of cross section — Incorrect. R depends on A through R = ρl/A; it is the resistivity ρ that does not depend on the dimensions.
NCERT (2026-27 edition):
Conductance of Electrolytic Solutions – “… The electrical resistance of any object is directly proportional to its length, l, and inversely proportional to its area of cross section, A. That is, R ∝ l/A or R = ρ l/A (2.17)”