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 :

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.