Two steel wires having same length are suspended from a ceiling under the same load. If the ratio of

Q: Why did we convert Strain into Stress in the energy density formula?

Because the problem states that both wires are under the 'same load' (Force $F$). By converting Strain to $\frac{\text{Stress}}{Y}$, we get an equation entirely in terms of Stress, which is directly related to the constant Force ($F/A$). If the problem had said 'same elongation', we would have converted Stress into Strain instead.

Q: What is the difference between total elastic potential energy and energy density?

Total elastic potential energy ($U$) is the total work done in stretching the entire wire, given by $U = \frac{1}{2} F \Delta L$. Energy density ($u$) is the energy stored per unit volume of the wire, given by $u = \frac{U}{\text{Volume}} = \frac{1}{2} \times \text{Stress} \times \text{Strain}$.

Q: Why is the area proportional to $d^2$?

The cross-section of a wire is a circle. The area of a circle is $A = \pi r^2$. Since radius $r = \frac{d}{2}$, the area becomes $A = \pi (\frac{d}{2})^2 = \frac{\pi d^2}{4}$. Here, $\frac{\pi}{4}$ is a constant, so $A$ is directly proportional to $d^2$.

Q: How did we calculate the fourth root of 4?

Taking the fourth root is the same as taking the square root twice. First, the square root of 4 is 2. Then, the square root of 2 is simply $\sqrt{2}$. So, $4^{(1/4)} = (4^{(1/2)})^{(1/2)} = 2^{(1/2)} = \sqrt{2}$.

Q: What if the wires were made of different materials?

If the materials were different, their Young's moduli ($Y_A$ and $Y_B$) would be different. The ratio equation would then be $\frac{u_A}{u_B} = \left(\frac{d_B^4}{d_A^4}\right) \times \left(\frac{Y_B}{Y_A}\right)$. You would need the ratio of their Young's moduli to solve for the diameters.

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

Options:

(A) $\sqrt{2} : 1$

(B) $1 : \sqrt{2}$

(C) $2 : 1$

(D) $1 : 2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are in a physics lab, looking at two steel wires, let's call them Wire A and Wire B, hanging vertically from a rigid ceiling. Both wires are identical in length $L$ and are made of the exact same material, which means their Young's modulus $Y$ is identical.

Now, we hang identical weights from both wires, subjecting them to the exact same load $F$. However, the problem tells us that the energy stored per unit volume (the energy density) in Wire A is one-fourth of that in Wire B. Our mission is to find the ratio of their diameters.

### The Master Equation for Energy Density

To solve this, we need to connect the dots between energy density, the applied force, and the dimensions of the wire. The elastic potential energy stored per unit volume, denoted by $u$, is given by the fundamental relation:

$$u = \frac{1}{2} 	imes 	ext{Stress} 	imes 	ext{Strain}$$

Since we know the load $F$ but not the extension, it's smarter to express Strain in terms of Stress using Hooke's Law ($	ext{Strain} = \frac{	ext{Stress}}{Y}$):

$$u = \frac{1}{2} 	imes 	ext{Stress} 	imes \left(\frac{	ext{Stress}}{Y}ight) = \frac{	ext{Stress}^2}{2Y}$$

We know that Stress is simply the internal restoring force per unit cross-sectional area, $	ext{Stress} = \frac{F}{A}$. Substituting this into our energy density equation gives us our master formula:

$$u = \frac{F^2}{2 A^2 Y}$$

### Finding the Diameter Ratio

Look closely at this master equation. For both wires, the load $F$ and Young's modulus $Y$ are constants. This reveals a beautiful inverse square relationship: the energy density is inversely proportional to the square of the cross-sectional area ($u \propto \frac{1}{A^2}$).

Let's set up a ratio for the two wires:

$$\frac{u_A}{u_B} = \frac{A_B^2}{A_A^2}$$

But we need diameters, not areas. The cross-sectional area of a wire is $A = \frac{\pi d^2}{4}$, meaning $A \propto d^2$. Substituting this into our ratio, the area squared becomes the diameter to the fourth power:

$$\frac{u_A}{u_B} = \frac{(d_B^2)^2}{(d_A^2)^2} = \frac{d_B^4}{d_A^4}$$

### Final Calculation

The problem states that the ratio of their energy densities $\frac{u_A}{u_B}$ is $\frac{1}{4}$. Equating this to our derived expression:

$$\frac{d_B^4}{d_A^4} = \frac{1}{4}$$

To isolate the diameters, we take the fourth root of both sides. The fourth root of $1$ is $1$, and the fourth root of $4$ is $\sqrt{2}$ (since $\sqrt{4} = 2$, and $\sqrt{2}$ is the square root of $2$).

$$\frac{d_B}{d_A} = \frac{1}{\sqrt{2}}$$

Flipping the fraction to find the ratio of Wire A to Wire B:

**$$\frac{d_A}{d_B} = \frac{\sqrt{2}}{1}$$**

And there we have it! The diameter of Wire A must be $\sqrt{2}$ times the diameter of Wire B to store one-fourth the energy density under the same load.

Animated Solution for Physics - Properties of Solids and Liquids: 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

Select Answer:

Visualized Solution

\text{Physical Setup}

u = \frac{1}{2} \times \text{Stress} \times \text{Strain}

u = \frac{\text{Stress}^2}{2Y}

\frac{u_A}{u_B} = \frac{A_B^2}{A_A^2}

A = \frac{\pi d^2}{4}

\frac{d_A}{d_B} = \frac{\sqrt{2}}{1}

\text{Extensions}

The Sigma Insight: Young's Modulus, Bulk Modulus and Modulus of Rigidity

Analyzing the Setup

The Master Equation for Energy Density

Finding the Diameter Ratio

Final Calculation

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