Two wires \(A\) and \(B\) are of the same materials. Their lengths are in the ratio \(1: 2\) and diameters are in the ratio \(2: 1 .\) When stretched by force \(F_{A}\) and \(F_{R}\) respectively they get equal increase in their lengths. Then the ratio \(F_{A} / F_{B}\) should be

Force \(F=\frac{Y A \Delta l}{l}\)
Substituting area \(A=\pi\left(\frac{d}{2}\right)^{2}=\frac{\pi d^{2}}{4}\)
So, we have \(\quad F=\frac{Y \pi d^{2} \Delta l}{4 l}\)
\(F \propto \frac{d^{2} \Delta l}{l}\left[\because \frac{Y \pi}{4}=\text { constant }\right]\)
Hence, using Eq. (i), we have
\(\begin{array}{l}F_{A} \propto \frac{d_{A}^{2} \Delta l_{A}}{l_{A}} \\F_{B} \propto \frac{d_{B}^{2} \Delta l_{B}}{l_{B}}\end{array}\)
From Eqs. (i) and (ii), we get
\(\frac{F_{A}}{F_{B}}=\frac{d_{A}^{2} \Delta l_{A}}{l_{A}} \times \frac{l_{B}}{d_{B}^{2} \Delta l_{B}}\)
as
\(\Delta l_{A}=\Delta l_{B}=\Delta l\)
Therefore
\(\frac{F_{A}}{F_{B}}=\frac{d_{A}^{2}}{d_{B}^{2}} \times \frac{l_{B}}{l_{A}}\)
Putting
\(\frac{d_{A}}{d_{B}}=\frac{2}{1}\)
and
\(\frac{l_{B}}{l_{A}}=\frac{2}{1}\)
we get \(\quad F_{A}: F_{B}=8: 1\)