Question

In order to increase the resistance of a given wire of uniform cross section to four times its initial value a fraction of its length is stretched uniformly till the full length of the wire becomes $\dfrac{3}{2}$ times the original length. What is the value of the fraction?

Answer 1

In the question it is given to us that a particular fraction of the wire is only stretched. Hence the effective resistance after stretching will be a sum of the stretched portion and the un stretched portion of the wire. When the wire is stretched the volume remains constant though the dimensions change. Hence substituting the value of the new resistance such that the resistance becomes 4 times the initial value will give the value of the fraction of wire that is stretched
Formula used:
$R=\dfrac{\rho l}{a}$

Complete answer:

Let us say that there is a wire of length ‘l’, as shown in the figure i.e. before stretching area of cross section ‘a’ and coefficient of resistivity as $\rho$ than the resistance of the wire is given by,
$R=\dfrac{\rho l}{a}$
Let us the fraction of the wire to be stretched is ‘x’. Hence the length of the wire to be stretched is ‘xl’ which implies that the un stretched length is $l-xl$. Hence the resistance of the wire of the un stretched portion is,
${{R}_{1}}=\dfrac{\rho (l-xl)}{a}$
In the question it is also given that the total length of the wire when the fraction is stretched is $\dfrac{3}{2}l$ . Hence from the figure above the length (L) of the stretched wire is,
$\begin{aligned} & L+(l-xl)=\dfrac{3}{2}l \\\ & \therefore L=\dfrac{1}{2}l+xl \\\ \end{aligned}$
The volume of the stretched portion of the wire with area of cross section ‘a’ is the same as that fraction of the wire that was to be stretched with area of cross section ‘A’. Hence we get
$\begin{aligned} & A\times \left( \dfrac{1}{2}l+xl \right)=a\times xl \\\ & \Rightarrow A=\dfrac{a\times x}{\dfrac{2x+1}{2}} \\\ & \therefore A=\dfrac{2ax}{2x+1} \\\ \end{aligned}$
Therefore the resistance of the wire that is stretched is,
$\begin{aligned} & {{R}_{2}}=\dfrac{\rho L}{A} \\\ & \Rightarrow {{R}_{2}}=\dfrac{\rho \left( \dfrac{1}{2}l+xl \right)}{\dfrac{2ax}{2x+1}} \\\ & \therefore {{R}_{2}}=\dfrac{\rho {{\left( 2x+1 \right)}^{2}}l}{4ax} \\\ \end{aligned}$
The wire after stretching can be divided into two parts one which is not stretched and the other fraction that is stretched. These two can be considered to be in series. In the question it is given that the effective resistance of the stretched wire is equal to the 4 times that of the wire without stretching. Hence using this information we get,
$\begin{aligned} & {{R}_{1}}+{{R}_{2}}=4R \\\ & \Rightarrow \dfrac{\rho (l-xl)}{a}+\dfrac{\rho {{\left( 2x+1 \right)}^{2}}l}{4ax}=4\dfrac{\rho l}{a} \\\ & \Rightarrow 1-x+\dfrac{{{\left( 2x+1 \right)}^{2}}}{4x}=4 \\\ & \Rightarrow 1-x+\dfrac{4{{x}^{2}}+4x+1}{4x}=4 \\\ & \Rightarrow 1-x+x+1+\dfrac{1}{4x}=4 \\\ & \Rightarrow \dfrac{1}{4x}=2 \\\ & \therefore x=\dfrac{1}{8} \\\ \end{aligned}$
Therefore the fraction of the wire which gets stretched is $\dfrac{1}{8}$ of the original length.

Note:
The wire is stretched uniformly. Therefore we can imply that the stretched wire can be taken in series with that of the un stretched wire. If the wire is not uniformly stretched, then we need to add the sections of the wires that are stretched uniformly and then take them in series of effective resistance respectively.