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 $\frac{3}{2}$ 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

The correct option is B $\frac{1}{8}$

Solution:

Let $R,\rho ,L$ and $A$ be the original wire’s resistance, resistivity, length and area of cross-section.

Suppose, a part of wire’s length equal to $l_1$ is stretched uniformly to make the resistance of the new wire $4R$. Consequently, the area of cross section of the stretched portion will also change such that the volume of the stretched portion remains unchanged in the process. Therefore, if $l_1^{\prime }$ and $A^{\prime }$ are the length and area of cross section of the stretched out part of the wire, then

$l_1^{\prime }\times A^{\prime }=l_1\times A\dots \dots \dots \dots ..\left(1\right)$

Also, $(L-l_1)+l_1^{\prime }=1.5L\dots \dots \dots \dots \dots ..\left(2\right)$

Now, in the new configuration of the wire, we have,

Resistance due to the unstretched part, $R_{unstretched}=\rho \frac{(L-l_1)}{A}$

Resistance due to the stretched part, $R_{stretched}=\rho \frac{l_1^{\prime }}{A^{\prime }}=\rho \frac{(l_1^{\prime }{)}^2}{l_{1}A}$ (from (1))

By the problem, $4R=R_{unstretched}+R_{stretched}$

which implies

$4\rho \frac{L}{A}=\rho \frac{(L-l_1)}{A}+\rho \frac{(l_1^{\prime }{)}^2}{l_{1}A}⟹3L=\frac{(l_1^{\prime }{)}^2}{l_1}-l_1............\left(3\right)$

But, $l_1^{\prime }–l_1=0.5L$ from (2).

Substituting this in equation (3) above for $l_1$­ and simplifying, we get,

$2L.l_1^{\prime }–1.25L^2=0$

$⟹l_1^{\prime }=0.625L$

$l_1=0.125L=\frac{1}{8}L$ (using equation (2) again for $l_1$).