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

Step 1: Identify the given parameters for both wires.

Since both wires are made of copper and connected in series, they share the same Young's modulus $Y$ and experience the same tension force $F$.

• Thin wire: length $L_{\text{thin}} = L$, radius $R_{\text{thin}} = R$
• Thick wire: length $L_{\text{thick}} = 2L$, radius $R_{\text{thick}} = 2R$

Step 2: Express the formula for elongation.

Young's modulus $Y$ is defined as:

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

Rearranging this formula gives the elongation $\Delta$:

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

where $A = \pi r^2$ is the cross-sectional area of the wire.

Step 3: Calculate the elongation for each wire.

For the thin wire:

$$\Delta_{\text{thin}} = \frac{FL}{\pi R^2 Y}$$

For the thick wire:

$$\Delta_{\text{thick}} = \frac{F(2L)}{\pi (2R)^2 Y}$$ $$\Delta_{\text{thick}} = \frac{2FL}{4\pi R^2 Y}$$ $$\Delta_{\text{thick}} = \frac{FL}{2\pi R^2 Y}$$

Step 4: Calculate the ratio of the elongations.

Divide the elongation of the thin wire by that of the thick wire:

$$\frac{\Delta_{\text{thin}}}{\Delta_{\text{thick}}} = \frac{\frac{FL}{\pi R^2 Y}}{\frac{FL}{2\pi R^2 Y}}$$ $$\frac{\Delta_{\text{thin}}}{\Delta_{\text{thick}}} = 2$$

Answer: 2