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 \(1:4\), then the ratio of their diameters is found to be \(\sqrt{k} : 1\). Find the value of \(k\).

Hint:

For same load: \(u \propto \frac{1}{d^4}\). Always relate stress to area first.

Answer 1

Correct Answer: 2

Concept: Strain energy per unit volume: \[ u = \frac{\sigma^2}{2Y} \Rightarrow u \propto \sigma^2 \] Since load is same: \[ \sigma = \frac{F}{A} \Rightarrow \sigma \propto \frac{1}{A} \] \[ \Rightarrow u \propto \frac{1}{A^2} \]

Step 1: Use given ratio \[ \frac{u_1}{u_2} = \frac{1}{4} = \frac{A_2^2}{A_1^2} \] \[ \Rightarrow \frac{A_2}{A_1} = \frac{1}{2} \Rightarrow \frac{A_1}{A_2} = 2 \]

Step 2: Convert to diameters \[ A \propto d^2 \Rightarrow \frac{d_1^2}{d_2^2} = 2 \] \[ \Rightarrow \frac{d_1}{d_2} = \sqrt{2} \]

Step 3: Compare with given form \[ \frac{d_1}{d_2} = \sqrt{k} \Rightarrow \sqrt{k} = \sqrt{2} \Rightarrow k = 2 \] Conclusion \[ k = 2 \]