One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another

Q: Why is the tension force $F$ the same in both wires?

Since the two wires are welded end-to-end (in series), any pulling force applied at the ends must be transmitted fully through both sections to maintain equilibrium. Thus, the tension is uniform throughout.

Q: Does the material of the wires affect the ratio?

Both wires are made of copper, meaning they share the same Young's modulus ($Y$). If they were made of different materials, their respective Young's moduli would not cancel out and would need to be included in the ratio.

Q: What if the wires were connected in parallel?

If connected in parallel, both wires would undergo the same elongation ($\Delta l_1 = \Delta l_2$), but they would share the total force unequally depending on their stiffness.

Q: Why did we use radius instead of diameter in the area formula?

The area of a circle is $A = \pi r^2 = \pi (d/2)^2$. Since the ratio of diameters is the same as the ratio of radii, using either radius or diameter yields the exact same dimensionless ratio.

Q: How does Young's modulus change with stretching?

Young's modulus is a material property and does not change with the dimensions of the wire or the applied stretching force, as long as the deformation remains within the elastic limit.

Question

One end of a horizontal thick copper wire of length $2L$ and radius $2R$ is welded to an end of another horizontal thin copper wire of length $L$ and radius $R$. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is

Options:

(A) 0.25

(B) 0.50

(C) 2.00

(D) 4.00

Sigma Percentile · Accepted Answer

### Introduction to the Elasticity Puzzle

Imagine you have two copper wires of different thicknesses welded end-to-end, forming a single continuous line.

When you pull on both ends with a force $F$, how does each section stretch?

This classic problem from the **JEE Advanced** archives tests our fundamental understanding of **Young's Modulus** and series combinations of elastic materials.

### Analyzing the Setup

Let's break down the physical parameters of our two wires:

- **Thick Wire**: Length $2L$, Radius $2R$
- **Thin Wire**: Length $L$, Radius $R$

Since they are welded in series, any tension applied at the ends must be transmitted equally through both wires.

Therefore, the stretching force $F$ is **identical** for both the thick and thin sections.

Furthermore, because both wires are made of copper, they share the exact same material properties.

This means their **Young's Modulus** $Y$ is also identical.

### The Governing Formula

To find the elongation, we recall the definition of Young's Modulus:

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

Rearranging this equation to solve for the elongation $\Delta l$, we get:

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

Since the cross-sectional area of a wire is given by $A = \pi r^2$, we can substitute this into our equation:

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

### Establishing Proportionality

Since the force $F$, Young's Modulus $Y$, and the constant $\pi$ are identical for both wires, we can establish a direct proportionality:

$$\Delta l \propto \frac{L}{r^2}$$

This elegant simplification tells us that the elongation of any wire in this setup depends solely on the ratio of its length to the square of its radius.

### Calculating the Ratio

Now, let's set up the ratio of the elongation of the thin wire (let's call it $\Delta l_1$) to that of the thick wire ($\Delta l_2$):

$$\frac{\Delta l_1}{\Delta l_2} = \frac{\frac{L_1}{r_1^2}}{\frac{L_2}{r_2^2}}$$

Substituting the given dimensions:

- For the thin wire: $L_1 = L$ and $r_1 = R$
- For the thick wire: $L_2 = 2L$ and $r_2 = 2R$

Let's plug these values into our ratio:

$$\frac{\Delta l_1}{\Delta l_2} = \frac{\frac{L}{R^2}}{\frac{2L}{(2R)^2}}$$

### Simplifying the Expression

Let's simplify the denominator first:

$$(2R)^2 = 4R^2$$

Substituting this back into the ratio:

$$\frac{\Delta l_1}{\Delta l_2} = \frac{\frac{L}{R^2}}{\frac{2L}{4R^2}}$$

We can rewrite this division as multiplication by the reciprocal:

$$\frac{\Delta l_1}{\Delta l_2} = \frac{L}{R^2} 	imes \frac{4R^2}{2L}$$

Notice how beautifully the variables $L$ and $R^2$ cancel out:

$$\frac{\Delta l_1}{\Delta l_2} = \frac{4}{2} = 2$$

Thus, the elongation in the thin wire is exactly **twice** the elongation in the thick wire.

This matches **Option (c)** perfectly!

Select Answer:

Visualized Solution

Visualizing the Welded Wire System

Recalling Young's Modulus

Finding the Proportionality Relation

Setting up the Ratio

Substituting the Dimensions

Simplifying to Find the Ratio

Exploring Variations

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

Introduction to the Elasticity Puzzle

Analyzing the Setup

The Governing Formula

Establishing Proportionality

Calculating the Ratio

Simplifying the Expression

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