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

Step 1: Understanding the Concept:
Elastic potential energy per unit volume (energy density) depends on the stress applied to the material and its Young's modulus.
Step 2: Key Formula or Approach:
Energy stored per unit volume (\(u\)) is given by:
\[ u = \frac{1}{2} \times \text{stress} \times \text{strain} = \frac{(\text{stress})^{2}}{2Y} \]
Stress \(\sigma = \frac{\text{Force}}{\text{Area}} = \frac{F}{\pi (d/2)^{2}} = \frac{4F}{\pi d^{2}}\).
Step 3: Detailed Explanation:
Substituting stress into the energy density formula:
\[ u = \frac{1}{2Y} \left( \frac{4F}{\pi d^{2}} \right)^{2} \]
\[ u = \frac{8F^{2}}{Y \pi^{2} d^{4}} \]
Since \(F\), \(Y\), and the length are constant for both wires:
\[ u \propto \frac{1}{d^{4}} \implies d \propto \frac{1}{u^{1/4}} \]
The ratio of energy densities is \(u_{1} : u_{2} = 1 : 4\).
The ratio of diameters is:
\[ \frac{d_{1}}{d_{2}} = \left( \frac{u_{2}}{u_{1}} \right)^{1/4} = \left( \frac{4}{1} \right)^{1/4} = (4^{1/2})^{1/2} = 2^{1/2} = \sqrt{2} \]
Given ratio is \(\sqrt{k} : 1\).
By comparison, \(\sqrt{k} = \sqrt{2}\), which implies \(k = 2\).
Step 4: Final Answer:
The value of \(k\) is 2.