Q. In order to quadruple the resistance of a uniform wire, a part of its length was uniformly stretched till the final length of the entire wire was 1.5 times the original length, the part of the wire was fraction equal to
A) $1/8$
B) $1/6$
C) $1/10$
D) $1/4$
Explanation
(a) Let $l$ be the original length of wire and $x$ be its length stretched uniformly such that final length is $1.5\ l$
Then $4R = \rho\frac{(l - x)}{A} + \rho\frac{(0.5l + x)}{A^{'}}$ where $A^{'} = \frac{x}{(0.5l + x)}A$
$\therefore 4\rho\frac{l}{A} = \rho\frac{l - x}{A} + \rho\frac{(0.5l + x)^{2}}{xA}$
or $4l = l - x + \frac{1}{4}\frac{l^{2}}{x} + \frac{x^{2}}{x} + \frac{lx}{x}$ or $\frac{x}{l} = \frac{1}{8}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.