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.
$\left(Y_{Steel}=2\times 10^{11}N{m}^{-2},Y_{Brass}=1\times 10^{11}N{m}^{-2}\right)$

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,
$L_S=1.5m,Y_S=2\times 10^{11}N{m}^{-2}$
$r_S=\frac{0.25}{2}cm=0.125\times 10^{-2}m$
The stretching force for the steel wire is
$F_S=\left(4+6\right)g=10gN$
As $Y=\frac{FL}{A\Delta L}=\frac{FL}{\pi r^2\Delta L}$
$\therefore \Delta L=\frac{FL}{\pi r^{2}Y}$
$\therefore \Delta L_S=\frac{F_S{L}_S}{\pi r_S^2{Y}_S}$
For brass wire,
$L_B=1m,Y_B=1\times 10^{11}N{m}^{-2}$
$r_B=\frac{0.25}{2}cm=0.125\times 10^{-2}m$
The stretching force the brass wire is $F_B=6gN$
$\therefore \Delta L_B=\frac{F_B{L}_B}{\pi r_B^2{Y}_B}$
Their corresponding ratio is
$\frac{\Delta L_S}{\Delta L_B}=\frac{F_S}{F_B}\frac{L_S}{L_B}\frac{r_B^2}{r_S^2}\frac{Y_B}{Y_S}$
Substituting the given values, we get
$=\frac{10g}{6g}\times \frac{1.5}{1}\times \frac{0.125\times 10^{-2}}{0.125\times 10^{-2}}\times \frac{1\times 10^{11}}{2\times 10^{11}}$
$=\frac{5}{4}$