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In an electric circuit, a wire of resistance 10 2 is used. If this wire is stretched to a length double of its original value, the current in the circuit would become :
Explanation
The resistance (R) of a conductor is determined by the formula R = ρ(L/A), where ρ is resistivity, L is length, and A is the cross-sectional area. When a wire is stretched, its total volume (V = A × L) remains constant. If the length is doubled (L' = 2L), the cross-sectional area must be halved (A' = A/2) to keep the volume unchanged.
Substituting these new values into the resistance formula: R' = ρ(2L / (A/2)) = 4 × ρ(L/A) = 4R. This shows that stretching the wire to double its length increases its resistance fourfold.
According to Ohm’s Law (V = IR), current (I) is inversely proportional to resistance (I = V/R), assuming the voltage remains constant. Since the resistance has increased to four times its original value, the current will decrease to one-fourth of its original value (I' = V/4R = I/4). Thus, Option C is correct.
SIMILAR QUESTIONS
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:
The resistance of a wire is 10 fi. If it is stretched ten times, the resistance will be
An electric wire of resistance 50 ohm is cut into five equal wires. These wires are then connected in parallel. What is the equivalent resistance of this combination?
Two wires have their lengths, diameters and resistivities, all in the ratio of 1 : 2. If the resistance of the thinner wire is 10 ohms, the resistance of the thicker wire is
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