 # Two steel wires having same length are suspended from a ceiling under the same load. Question:

Two steel wires having same length are suspended from a ceiling under the same load. If the ratio of their energy stored per unit volume is $1: 4$, the ratio of their diameters is:

1. $\sqrt{2}: 1$

2. $1: 2$

3. $2: 1$

4. $1: \sqrt{2}$

Correct Option: 1

Solution:

(1) If force $F$ acts along the length $\mathrm{L}$ of the wire of crosssection $A$, then energy stored in unit volume of wire is given by

Energy density $=\frac{1}{2}$ stress $\times$ strain

$=\frac{1}{2} \times \frac{F}{A} \times \frac{F}{A Y} \quad\left(\because\right.$ stress $=\frac{F}{A}$ and strain $\left.=\frac{X}{A Y}\right)$

$=\frac{1}{2} \frac{F^{2}}{A^{2} Y}=\frac{1}{2} \frac{F^{2} \times 16}{\left(\pi d^{2}\right)^{2} Y}=\frac{1}{2} \frac{F^{2} \times 16}{\pi d^{4} Y}$

If $u_{1}$ and $u_{2}$ are the densities of two wires, then

$\frac{u_{1}}{u_{2}}=\left(\frac{d_{2}}{d_{1}}\right)^{4} \Rightarrow \frac{d_{1}}{d_{2}}=(4)^{1 / 4} \Rightarrow \frac{d_{1}}{d_{2}}=\sqrt{2}: 1$