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A metal wire of length l and diameter d has a resistance R. What would be the resistance of another wire of the same metal and of the same length but having double the diameter?
Explanation
The resistance (R) of a metallic conductor is directly proportional to its length (l) and inversely proportional to its cross-sectional area (A), expressed as R = ρ(l/A) [2]. For a wire with a circular cross-section, the area is calculated as A = π(d/2)² or A = πd²/4, where d is the diameter. This implies that resistance is inversely proportional to the square of the diameter (R ∝ 1/d²). When the diameter of the wire is doubled (2d) while keeping the length and material constant, the new cross-sectional area increases by a factor of four (2² = 4). Since resistance is inversely proportional to this area, the new resistance becomes one-fourth of the original value (R/4). Therefore, doubling the diameter reduces the opposition to electron flow by providing four times the space, resulting in a resistance of R/4.
Sources
- [1] Science , class X (NCERT 2025 ed.) > Chapter 11: Electricity > Activity 11.3 > p. 178
- [2] Science , class X (NCERT 2025 ed.) > Chapter 11: Electricity > What you have learnt > p. 192
SIMILAR QUESTIONS
Let us consider a copper wire having radius r and length l. Let its resistance be R. If the radius of another copper wire is 2r and the length is l/2 then the resistance of this wire will be
A copper wire of radius r and length l has a resistance of R. A second copper wire with radius 2r and length l is taken and the two wires are joined in a parallel combination. The resultant resistance of the parallel combination of the two wires will be:
A) R/5
B) 5R/4
C) R/2
D) 4R/5
The resistance of a wire of length / and area of cross-section a is x ohm. If the wire is stretched to double its length, its resistance would become:
Two metallic wires made from copper have the same length, but the radius of wire 1 is half that of wire 2. The resistance of wire 1 is R. If both the wires are joined together in series, the total resistance becomes:
- (A) 2R
- (B) 1.25R
- (C) 0.5R
- (D) 1.33R