Question

a given wire having restistance R is streched so as to reduce its diameter to half of its previous value. wat will be the new resistance?

Answer 1

The key concept of such type of numerical is that the volume of the wire will remain constant.
$$\begin{gathered} Theresis\tan ceofthewireisgivenas: \\ R=\rho \frac{l}{A} \\ Thevolumeofthewire, \\ V=lA=l\pi {\left(\frac{d}{2}\right)}^2 \\ Nowthewireisstretchedsuchthatthediametergetsreducedtohalf: \\ V=l\text{'}A\text{'} \\ \Rightarrow l\pi {\left(\frac{d}{2}\right)}^2=I\text{'}\pi {\left(\frac{d}{4}\right)}^2 \\ \Rightarrow l=\frac{l\text{'}}{4} \\ \Rightarrow l\text{'}=4l \\ Therefore,thechangedresis\tan ceofthewire \\ R\text{'}=\rho \frac{l\text{'}}{A\text{'}}=\rho \left(\frac{4l}{\pi {\left({\displaystyle \frac{d}{4}}\right)}^2}\right)=16\rho \frac{l}{A}=16R \end{gathered}$$