Question

In order to increase the resistance of a given wire of uniform cross section to four times its value, a fraction of its length is stretched uniformly till the full length of the wire becomes $1.5$ times the original length. What is the value of this fraction ?

• A. $\frac{1}{4}$
• B. $\frac{1}{8}$
• C. $\frac{1}{16}$
• D. $\frac{1}{6}$

Answer 1

Answer: (B)

Suppose $n(< 1)$ fraction of the length of resistance is stretched to $m$-times.
Then its new length is given as
$(1-n) l+(n l) m =1.5\, l$
$n m-n =0.5\,\,\,...(i)$
Resistance is given as
$R=\frac{\rho l}{A}$
$R=\frac{\rho l}{(V / l)} $
[ $V$ is the volume of wire]
$R=\frac{\rho l^{2}}{V}$
$\Rightarrow R \propto l^{2}$
[As $V$ remain constant]
The final resistance becomes four times so we use
$(1-n) R+(n R) m^{3} =4 R $
$\Rightarrow n m^{2}-n =3\,\,\,...(ii)$
Solving above two equations we get
$n=\frac{1}{8}$