Q. The resistance of a 10 m long wire is $10\ \Omega$. Its length is increased by 25% by stretching the wire uniformly. The resistance of wire will change to (approximately)
A) $12.5\ \Omega$
B) $14.5\ \Omega$
C) $15.6\ \Omega$
D) $16.6\ \Omega$
Explanation
(c) Given, $l_{1} = l + \frac{25}{100}l = \frac{5l}{4}$. Since volume or wire remains unchanged on increasing length, hence
$Al - A_{1} \times \frac{5l}{4\ or\ A_{1}} = 4A/5$
Given, $R = 10 = \rho l/A,\ and$
$R_{1} = \frac{\rho l_{1}}{A_{1}} = \frac{\rho 5l/4}{4A/5} = \frac{25\rho l}{16A}$
$or\ R_{1} = \frac{25}{16} \times 10 = \frac{250}{16} = 15.6\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.