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Step-by-Step Solution
Step 1: Write the expression for energy stored per unit volume
The energy stored per unit volume (denoted by $ \frac{du}{dv} $) in a stretched wire is given by:
$ \displaystyle \frac{du}{dv} = \frac{1}{2} \times \text{(Stress)} \times \text{(Strain)} $
Step 2: Express stress and strain in terms of load and cross-sectional area
• Stress, $ \sigma = \frac{F}{A} $, where $F$ is the load (force) and $A$ is the cross-sectional area of the wire.
• Strain, $ \epsilon = \frac{\Delta L}{L} $, but when expressed in terms of stress and Young's modulus $Y$, we have $ \epsilon = \frac{\sigma}{Y} = \frac{F}{A \, Y} $.
Step 3: Combine the expressions
Substitute stress and strain into the energy expression:
$ \displaystyle \frac{du}{dv} = \frac{1}{2} \cdot \frac{F}{A} \cdot \frac{F}{A \, Y}
= \frac{F^2}{2 \, A^2 \, Y}
$
This shows that $ \frac{du}{dv} \propto \frac{1}{A^2} $. Since for a circular cross-section $A = \frac{\pi d^2}{4}$, we have $A \propto d^2$. Therefore:
$ \displaystyle \frac{du}{dv} \propto \frac{1}{(d^2)^2} = \frac{1}{d^4}.
$
Step 4: Relate the ratio of energy densities to the ratio of diameters
Given that the ratio of energy stored per unit volume for the two wires is 1 : 4, we write:
$ \displaystyle \frac{\left(\frac{du}{dv}\right)_1}{\left(\frac{du}{dv}\right)_2} = \frac{1}{4}.
$
Since $ \frac{du}{dv} \propto \frac{1}{d^4} $, we have:
$ \displaystyle \frac{\frac{1}{d_1^4}}{\frac{1}{d_2^4}} = \frac{1}{4},
$
or equivalently,
$ \displaystyle \left(\frac{d_2}{d_1}\right)^4 = 4
$
. Taking the fourth root on both sides gives:
$ \displaystyle \frac{d_2}{d_1} = \sqrt{2}
\quad \Rightarrow \quad \frac{d_1}{d_2} = \frac{1}{\sqrt{2}}.
$
Thus, the ratio of their diameters
$ \displaystyle d_1 : d_2
$
is
$ \displaystyle \sqrt{2} : 1.
Final Answer
The correct ratio of the diameters of the two wires is
$ \displaystyle \sqrt{2} : 1.
