Question

Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in figure. The unloaded length of steel wire is 1.5 m and that of brass wire is 1 m.
$\left(Y_{Steel}=2\times10^{11}\,N\,m^{-2}, Y_{Brass}=1\times 10^{11}\,N\,m^{-2}\right)$

The ratio of elongations of the steel wire to that of brass wire is

• A. $\frac{4}{5}$
• B. $\frac{5}{4}$
• C. $\frac{3}{5}$
• D. $\frac{5}{3}$

Answer 1

Answer: (B)

For steel wire,
$L_{S}=1.5\,m, Y_{S}=2\times10^{11}\,N\,m^{-2}$
$r_{S}=\frac{0.25}{2}\,cm=0.125\times10^{-2}\,m$
The stretching force for the steel wire is
$F_{S}=\left(4+6\right)g=10g\,N$
As $Y=\frac{FL}{A\Delta L}=\frac{FL}{\pi r^{2}\,\Delta L}$
$\therefore \Delta L=\frac{FL}{\pi r^{2}\,Y}$
$\therefore \Delta L_{S}=\frac{F_{S}\,L_{S}}{\pi\,r^{2}_{S}\,Y_{S}}$
For brass wire,
$L_B = 1 \,m, Y_B = 1 × 10^{11}\, N\, m^{-2}$
$r_{B}=\frac{0.25}{2}cm=0.125\times10^{-2}\,m$
The stretching force the brass wire is $F_{B}=6g\,N$
$\therefore \Delta L_{B}=\frac{F_{B}\,L_{B}}{\pi r^{2}_{B}\,Y_{B}}$
Their corresponding ratio is
$\frac{\Delta\,L_{S}}{\Delta\,L_{B}}=\frac{F_{S}}{F_{B}} \frac{L_{S}}{L_{B}} \frac{r^{2}_{B}}{r^{2}_{S}} \frac{Y_{B}}{Y_{S}}$
Substituting the given values, we get
$=\frac{10\,g}{6\,g}\times\frac{1.5}{1}\times\frac{0.125\times10^{-2}}{0.125\times10^{-2}}\times\frac{1\times10^{11}}{2\times10^{11}}$
$=\frac{5}{4}$