One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end fo another horizontal thin copper wire of lenth 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. (a) 0.25
B. (b) 0.50
C. (c) 2.00
D. (d) 4.00

Explanation

To find the ratio of the elongation in the thin wire to that in the thick wire, we can use the concept of Young's modulus. Let's break down the solution step by step.

### Step 1: Understand the setup
We have two wires:
- A thick wire of length \(2L\) and radius \(2R\).
- A thin wire of length \(L\) and radius \(R\).

Both wires are made of copper and are welded together at one end. When a force \(F\) is applied at both ends, we need to find the ratio of elongation in the thin wire (\(\Delta L_2\)) to that in the thick wire (\(\Delta L_1\)).

### Step 2: Write the formula for Young's modulus
The Young's modulus \(Y\) is defined as:
\[ Y = \frac{F/A}{\Delta L / L} \]
where \(F\) is the force applied, \(A\) is the cross-sectional area, \(\Delta L\) is the change in length (elongation), and \(L\) is the original length of the wire.

### Step 3: Calculate the cross-sectional areas
- For the thick wire:
\[ A_1 = \pi (2R)^2 = 4\pi R^2 \]
- For the thin wire:
\[ A_2 = \pi R^2 \]

### Step 4: Set up the equations for Young's modulus for both wires
For the thick wire:
\[ Y = \frac{F}{A_1} \cdot \frac{L}{\Delta L_1} \Rightarrow Y = \frac{F}{4\pi R^2} \cdot \frac{2L}{\Delta L_1} \]
Rearranging gives:
\[ \Delta L_1 = \frac{F \cdot 2L}{Y \cdot 4\pi R^2} \]

For the thin wire:
\[ Y = \frac{F}{A_2} \cdot \frac{L}{\Delta L_2} \Rightarrow Y = \frac{F}{\pi R^2} \cdot \frac{L}{\Delta L_2} \]
Rearranging gives:
\[ \Delta L_2 = \frac{F \cdot L}{Y \cdot \pi R^2} \]

### Step 5: Find the ratio of elongations
Now, we can find the ratio \(\frac{\Delta L_2}{\Delta L_1}\):
\[ \frac{\Delta L_2}{\Delta L_1} = \frac{\frac{F \cdot L}{Y \cdot \pi R^2}}{\frac{F \cdot 2L}{Y \cdot 4\pi R^2}} \]

### Step 6: Simplify the ratio
Cancelling out common terms:
\[ \frac{\Delta L_2}{\Delta L_1} = \frac{4\pi R^2}{2\pi R^2} = \frac{4}{2} = 2 \]

### Conclusion
Thus, the ratio of the elongation in the thin wire to that in the thick wire is:
\[ \frac{\Delta L_2}{\Delta L_1} = 2 \]

### Final Answer
The ratio of the elongation in the thin wire to that in the thick wire is \(2:1\).

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