Question

Two wires of diameter $0.25cm$, one made of steel and other made of brass, are loaded as shown in the figure. The unloaded length of the steel wire is $1.5m$ and that of brass is $1.0m$. Young's modulus of steel is $2.0\times {10}^{11}Pa$ and that of brass is $1.0\times {10}^{11}Pa$. If the ratio of elongations of steel and brass wires $\frac{{\Delta l}_{steel}}{{\Delta l}_{brass}}=\frac{x}{4}$, find $x$.

Answer 1

As the elongation is given as $\Delta l=\frac{Fl}{AY}$, where $F$ is the force of elongation, $Y$ is Young's modulus, $l$ is unstretched length, $A$ is the area.

As area, $A\propto d^2$, where $d$ is the diameter, so, $\Delta l\propto \frac{Fl}{d^{2}Y}$.

Taking ratio of elongation for steel and brass wires,

$\frac{\Delta l_{steel}}{\Delta l_{brass}}=\frac{F_{steel}{l}_{steel}}{d_{steel}^2{Y}_{steel}}÷\frac{F_{brass}{l}_{brass}}{d_{brass}^2{Y}_{brass}}$.

Here, $F_{steel}=(4+6)\times g=10\times g$ and $F_{brass}=6\times g$ where $g$ is acceleration due to gravity.

Using values, we get: $\frac{\Delta l_{steel}}{\Delta l_{brass}}=5/4$

So answer is 5