The length of a wire having a resistance of 15 ohms is stretched by 20%. If the volume of the wire remains constant after stretching and its cross-section remains uniform throughout, determine the new resistance of the wire
WBBSE
Class X
Physical Science
Current Electricity
02 Mark
2026
Initial resistance (R₁) = 15 Ω
Let initial length = L₁
After stretching by 20%,
Final Length (L₂) = L₁ + 20% of L₁ = 1.2L₁
Since the volume of the wire remains constant,
A₁L₁ = A₂L₂
Therefore,
A₂ = \frac{\text{A₁L₁}}{\text{L₂}}
= \frac{\text{A₁}}{1.2}
We know,
R = \frac{\text{ρL}}{\text{A}}
Therefore, \frac{\text{R₂}}{\text{R₁}}=\frac{\text{L₂A₁}}{\text{L₁A₂}}
= 1.2 × 1.2
= 1.44
R₂ = 1.44 × 15
= 21.6 Ω
Hence, the new resistance of the wire is 21.6 Ω.
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