Two separate wires A and B are stretched by 2 mm and 4 mm respectively, when they are subjected to a

Q: Why is Young's modulus the same for both wires?

The problem explicitly states that both wires are made of the same material. Young's modulus is a material property, meaning it depends only on the substance the wire is made of, not its dimensions.

Q: Do I need to convert millimeters to meters for the extensions?

While it's generally a good practice to convert all units to SI, in this specific case, we are taking a ratio ($\Delta L_A / \Delta L_B$). The conversion factors ($10^{-3}$) will cancel out, so you can safely use the values in millimeters.

Q: Why does the area ratio become $1/16$ instead of $1/4$?

The cross-sectional area of a wire is proportional to the square of its radius ($A = \pi r^2$). Since wire B has $4$ times the radius of wire A, its area is $4^2 = 16$ times larger.

Q: What if the applied forces were different?

If the forces were different, they wouldn't cancel out in the ratio equation. You would need to plug in their specific values into the $F_B / F_A$ term to find the correct length ratio.

Q: Can I use the diameter instead of the radius to find the area ratio?

Yes, absolutely! The ratio of the diameters is the same as the ratio of the radii. Since area is proportional to the square of the diameter as well ($A = \pi d^2 / 4$), the area ratio will still be $1/16$.

Question

Two separate wires $A$ and $B$ are stretched by $2	ext{ mm}$ and $4	ext{ mm}$ respectively, when they are subjected to a force of $2	ext{ N}$. Assume that both the wires are made up of same material and the radius of wire $B$ is $4$ times that of the radius of wire $A$. The length of the wires $A$ and $B$ are in the ratio of $a : b$. Then, $a / b$ can be expressed as $1 / x$, where $x$ is ......... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are in a physics lab, looking at two wires, A and B, hanging from the ceiling. Both wires are subjected to the exact same pulling force of $2	ext{ N}$. However, they are not identical in shape. Wire B is significantly thicker; its radius is four times that of wire A. We are given their respective elongations and asked to find the ratio of their original lengths.

### The Master Equation

Since both wires are made of the **same material**, their Young's modulus ($Y$) will be identical. This is a crucial piece of information. Let's recall the formula for Young's modulus. It is defined as the ratio of tensile stress to tensile strain.

$$Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F / A}{\Delta L / L}$$

We can rearrange this formula to express the original length ($L$) in terms of the other variables.

$$L = \frac{Y A \Delta L}{F}$$

### Setting Up the Ratio

We need to find the ratio of their original lengths, $L_A / L_B$. Let's write down the expression for this ratio using our rearranged formula.

$$\frac{L_A}{L_B} = \frac{Y_A}{Y_B} 	imes \frac{A_A}{A_B} 	imes \frac{\Delta L_A}{\Delta L_B} 	imes \frac{F_B}{F_A}$$

Notice how the force $F$ and Young's modulus $Y$ will cancel out because they are the same for both wires. This simplifies our calculation immensely!

### Calculating the Area Ratio

Before we substitute the values, let's look at the cross-sectional areas. The area of a wire is given by $\pi r^2$.

$$A = \pi r^2$$

Since the radius of wire B is four times the radius of wire A ($r_B = 4r_A$), the area of wire B will be sixteen times the area of wire A.

$$\frac{A_A}{A_B} = \frac{\pi r_A^2}{\pi (4r_A)^2} = \frac{1}{16}$$

### Final Calculation

Now, let's plug everything back into our length ratio equation. The Young's modulus and force terms become $1$. The area ratio is $1/16$. And the ratio of their extensions is $2	ext{ mm}$ by $4	ext{ mm}$, which simplifies to $1/2$.

$$\frac{L_A}{L_B} = 1 	imes \frac{1}{16} 	imes \frac{2 	imes 10^{-3}}{4 	imes 10^{-3}} 	imes 1$$

$$\frac{L_A}{L_B} = \frac{1}{16} 	imes \frac{1}{2} = \frac{1}{32}$$

Multiplying these together, we get the ratio of lengths as $1:32$. The question states this ratio is $1/x$. Comparing the two, we can clearly see that **$x$ must be exactly $32$**.

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$Y = \frac{Stress}{Strain}$

$\frac{L _{A}}{L _{B}} = \frac{Y _{A}}{Y _{B}} \times \frac{A _{A}}{A _{B}} \times \frac{Δ L _{A}}{Δ L _{B}} \times \frac{F _{B}}{F _{A}}$

$A = π r^{2}$

$\frac{L _{A}}{L _{B}} = 1 \times \frac{1}{16} \times \frac{2 \times 1 0 ^{- 3}}{4 \times 1 0 ^{- 3}} \times 1$

$\frac{a}{b} = \frac{1}{x}$

$Conclusion$

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

Analyzing the Setup

The Master Equation

Setting Up the Ratio

Calculating the Area Ratio

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Y=StrainStress​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LB​LA​​=YB​YA​​×AB​AA​​×ΔLB​ΔLA​​×FA​FB​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A=πr2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LB​LA​​=1×161​×4×10−32×10−3​×1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ba​=x1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion