Ratio of Elongation in Stretched Copper Wires of Different Dimensions

To find the ratio of the elongation produced in the thin copper wire to that in the thick copper wire, we apply Hooke's law for elastic deformation.

The elongation $\Delta L$ of a wire under tensile force $F$ is given by: $\Delta L = \frac{F \cdot l}{A \cdot Y}$

where:

• $F$ is the tension applied to the wire,
• $l$ is the original length of the wire,
• $A = \pi r^2$ is the cross-sectional area of the wire of radius $r$,
• $Y$ is the Young's modulus of the material (copper).

Since both wires are connected in series and subjected to tensile forces at the outer ends, the internal tension force $F$ and the Young's modulus $Y$ are identical for both wires.

1. For the thin copper wire:

   • Length, $l_{\text{thin}} = L$
   • Radius, $r_{\text{thin}} = R$
   • Cross-sectional area, $A_{\text{thin}} = \pi R^2$

   The elongation produced in the thin wire is: $\Delta L_{\text{thin}} = \frac{F \cdot L}{\pi R^2 \cdot Y}$

2. For the thick copper wire:

   • Length, $l_{\text{thick}} = 2L$
   • Radius, $r_{\text{thick}} = 2R$
   • Cross-sectional area, $A_{\text{thick}} = \pi (2R)^2 = 4\pi R^2$

   The elongation produced in the thick wire is: $\Delta L_{\text{thick}} = \frac{F \cdot (2L)}{(4\pi R^2) \cdot Y} = \frac{2FL}{4\pi R^2 Y} = \frac{FL}{2\pi R^2 Y}$

3. Ratio of elongations: $\frac{\Delta L_{\text{thin}}}{\Delta L_{\text{thick}}} = \frac{\frac{FL}{\pi R^2 Y}}{\frac{FL}{2\pi R^2 Y}} = \frac{1}{\frac{1}{2}} = 2.00$

Thus, the ratio of the elongation produced in the thin wire to that in the thick wire is $2.00$.

Correct Option: C