Young's moduli of two wires A and B are in the ratio 7 : 4. Wire A is 2 m long and has radius R. Wir

Q: Why do we use $\pi r^2$ for the area?

Because wires are typically cylindrical, their cross-section is a circle. The area of a circle is $\pi r^2$.

Q: Why did the force $F$ cancel out?

The problem states that the wires stretch by the same length 'for a given load'. This implies the exact same force $F$ is applied to both wires, allowing it to be canceled from both sides of the equation.

Q: What if the wires had different cross-sectional shapes?

If the shapes were different (e.g., square), you would use the appropriate area formula (e.g., side squared) instead of $\pi r^2$. The fundamental relation $\Delta l = \frac{Fl}{AY}$ still holds.

Q: Why is it important to convert millimeters to meters?

To maintain consistency in SI units. Young's modulus is typically in Pascals ($\text{N/m}^2$), and length is in meters. If you don't convert the radius to meters, your final answer will be off by orders of magnitude.

Q: How do I quickly estimate the square root of $3.04$?

You know that $1.7^2 = 2.89$ and $1.8^2 = 3.24$. Since $3.04$ is roughly in the middle, the square root must be around $1.74$. Looking at the options, $1.7$ is the closest fit.

Question

Young's moduli of two wires $A$ and $B$ are in the ratio $7 : 4$. Wire $A$ is $2	ext{ m}$ long and has radius $R$. Wire $B$ is $1.5	ext{ m}$ long and has radius $2	ext{ mm}$. If the two wires stretch by the same length for a given load, then the value of $R$ is close to

Options:

(A) $1.3	ext{ mm}$

(B) $1.5	ext{ mm}$

(C) $1.9	ext{ mm}$

(D) $1.7	ext{ mm}$

Sigma Percentile · Accepted Answer

### Visualizing the Setup
Imagine two wires, $A$ and $B$, hanging from a rigid ceiling. Wire $A$ is $2	ext{ m}$ long, and wire $B$ is $1.5	ext{ m}$ long. They are made of different materials, meaning they have different Young's moduli ($Y_A$ and $Y_B$), and they have different thicknesses.

However, the problem gives us a fascinating constraint: when we hang the exact same weight $F$ from both wires, they stretch by the exact same amount $\Delta l$. This means the longer, stiffer wire $A$ must have a specific radius $R$ to perfectly match the elongation of the shorter wire $B$.

### The Master Equation
To solve this, we need to bring in Hooke's Law for the elasticity of solid materials. The elongation $\Delta l$ of a wire under a stretching force $F$ is given by:

$$ \Delta l = \frac{Fl}{AY} $$

Since the wires are cylindrical, their cross-sectional area $A$ is simply the area of a circle, $\pi r^2$. Substituting this into our formula gives us the master equation for this problem:

$$ \Delta l = \frac{Fl}{\pi r^2 Y} $$

### Equating the Elongations
We are told that both wires stretch by the same length for a given load. This translates mathematically to:

$$ \Delta l_A = \Delta l_B $$

Let's substitute our master equation for both wires:

$$ \frac{F_A l_A}{\pi r_A^2 Y_A} = \frac{F_B l_B}{\pi r_B^2 Y_B} $$

Since the load is the same for both wires, $F_A = F_B = F$. The force $F$ and the constant $\pi$ appear on both sides of the equation, so they cancel out beautifully. We are left with a much simpler relationship:

$$ \frac{l_A}{r_A^2 Y_A} = \frac{l_B}{r_B^2 Y_B} $$

### Isolating the Unknown
Our goal is to find the radius of wire $A$, which is $r_A = R$. Let's rearrange the equation to isolate $r_A^2$ on one side. By cross-multiplying, we get:

$$ r_A^2 = \left(\frac{l_A}{l_B}ight) \cdot \left(\frac{Y_B}{Y_A}ight) \cdot r_B^2 $$

Notice how we grouped the terms into ratios. This is a powerful technique in physics problems because it often makes the calculation much cleaner, especially when ratios are given in the problem statement!

### Crunching the Numbers
Now, let's plug in the values given in the problem. We know the ratio of Young's moduli is $\frac{Y_A}{Y_B} = \frac{7}{4}$, which means the inverted ratio is $\frac{Y_B}{Y_A} = \frac{4}{7}$. The lengths are $l_A = 2	ext{ m}$ and $l_B = 1.5	ext{ m}$. The radius of wire $B$ is $2	ext{ mm}$, which we must convert to standard SI units: $2 	imes 10^{-3}	ext{ m}$.

Substituting these into our rearranged equation:

$$ r_A^2 = \left(\frac{2}{1.5}ight) \cdot \left(\frac{4}{7}ight) \cdot (2 	imes 10^{-3})^2 $$

Let's simplify the fractions. $\frac{2}{1.5}$ is the same as $\frac{4}{3}$. Squaring the radius gives $4 	imes 10^{-6}$.

$$ r_A^2 = \left(\frac{4}{3}ight) \cdot \left(\frac{4}{7}ight) \cdot (4 	imes 10^{-6}) $$

Multiplying the numerators and denominators:

$$ r_A^2 = \frac{64}{21} 	imes 10^{-6}	ext{ m}^2 $$

Evaluating the fraction $\frac{64}{21}$ gives approximately $3.047$.

$$ r_A^2 \approx 3.047 	imes 10^{-6}	ext{ m}^2 $$

### The Final Answer
To find the radius $r_A$, we simply take the square root of both sides:

$$ r_A = \sqrt{3.047 	imes 10^{-6}} $$

The square root of $10^{-6}$ is $10^{-3}$. The square root of $3.047$ is roughly $1.745$.

$$ r_A \approx 1.745 	imes 10^{-3}	ext{ m} $$

Converting back to millimeters, we get $1.745	ext{ mm}$. Looking at our options, the closest value is **$1.7	ext{ mm}$**.

This result makes perfect physical sense. Wire $A$ is longer and stiffer than wire $B$. To stretch the exact same amount under the same load, it must have a specific thickness to balance out its length and stiffness. This delicate interplay of material properties and geometric dimensions is a fundamental concept in structural engineering!

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Hooke’s Law for Elongation$

$Equating Elongations$

$Rearranging the Equation$

$Substituting Given Values$

$Calculating r_{A}^{2}$

$Final Radius R$

$Physical Significance$

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

Visualizing the Setup Imagine two wires, $A$ and $B$ , hanging from a rigid ceiling

The Master Equation To solve this, we need to bring in Hooke's Law for the elasticity of solid materials

Equating the Elongations We are told that both wires stretch by the same length for a given load

Isolating the Unknown Our goal is to find the radius of wire $A$ , which is $r_{A} = R$

Crunching the Numbers Now, let's plug in the values given in the problem

The Final Answer

Similar Questions

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Select Answer:

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; } Hooke’s Law for Elongation

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating rA2​

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

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