A cylindrical wire is stretched to increase its length by 10%. The percentage increase in resistance will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 21%
Given that the wire is stretched so that its length increases by 10%.
Initial length = $L$, Final length = $L' = 1.1L$
Volume remains constant ⇒ $A L = A' L'$
⇒ $A' = \frac{A}{1.1}$
Resistance, $R = \rho \frac{L}{A}$
New resistance, $R' = \rho \frac{L'}{A'} = \rho \frac{1.1L}{A/1.1} = \rho \frac{1.21L}{A} = 1.21R$
Percentage increase in resistance = $(1.21 - 1) \times 100 = 21\%$
∴ Percentage increase in resistance = 21%