Question

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. $\frac{1}{8}$
• B. $\frac{1}{6}$
• C. $\frac{1}{10}$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

$\frac{1}{8}$

Let $l$ be the original length of wire and $x$ be its length stretched uniformly such that final length is $1.5l$

Then $4R=\rho \frac{(l-x)}{A}+\rho \frac{(0.5l+x)}{A^{\prime }}$ where $A^{\prime }=\frac{x}{(0.5l+x)}A$

$\therefore 4\rho \frac{l}{A}=\rho \frac{l-x}{A}+\rho \frac{(0.5l+x{)}^2}{xA}$

or $4l=l-x+\frac{1}{4}\frac{l^2}{x}+\frac{x^2}{x}+\frac{lx}{x}$

or $\frac{x}{l}=\frac{1}{8}$