Q. Certain wire has resistance of $10\Omega$. If its is stretched by 1/10th of its length, then its resistance is nearly
A) $9\ \Omega$
B) $10\ \Omega$
C) $11\ \Omega$
D) $12\ \Omega$
Explanation
(d) $R = \frac{\rho l}{A} = 10$
Now length $l_{1} = l + l/10 = 11\ l/10$
$\therefore$ New area $A_{1} = Al/l_{1}$
$\therefore$ New resistance,
$R_{1} = \frac{\rho l_{1}}{A_{1}} = \rho(11/10)/(10/11)A$
$= \frac{121}{100}\frac{\rho l}{A} = \frac{121}{100} \times 10 = 12.1\ \Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.