Question 438: Properties of Matter

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

Accepted Answer

To solve this problem, we need to use Hooke's Law, which states that the elongation (or stretch) of a material is directly proportional to the force applied and the length of the material, and inversely proportional to the cross-sectional area and the Young's modulus of the material. Mathematically, this relation can be expressed as:
$$ \Delta L = \frac{F \cdot L}{A \cdot Y} $$
where:
$$ \Delta L $$ is the elongation
$$ F $$ is the force applied
$$ L $$ is the original length of the material
$$ A $$ is the cross-sectional area of the material
$$ Y $$ is the Young's modulus of the material
In this problem, the Young's modulus ($$ Y $$) and the force ($$ F $$) are the same for both wires because they are made of the same material (copper) and the forces applied are equal. Therefore, the elongation is dependent on the length and cross-sectional area of each wire.
For the thick wire with length $$ 2L $$ and radius $$ 2R $$, the cross-sectional area $$ A_1 $$ is given by:
$$ A_1 = \pi (2R)^2 = 4\pi R^2 $$
For the thin wire with length $$ L $$ and radius $$ R $$, the cross-sectional area $$ A_2 $$ is given by:
$$ A_2 = \pi R^2 $$
Using the formula for elongation for both wires, we get:
Elongation of the thick wire ($$ \Delta L_	ext{thick} $$):
$$ \Delta L_	ext{thick} = \frac{F \cdot 2L}{4\pi R^2 \cdot Y} = \frac{F \cdot 2L}{4\pi R^2 Y} = \frac{FL}{2\pi R^2 Y} $$
Elongation of the thin wire ($$ \Delta L_	ext{thin} $$):
$$ \Delta L_	ext{thin} = \frac{F \cdot L}{\pi R^2 \cdot Y} = \frac{FL}{\pi R^2 Y} $$
To find the ratio of the elongation in the thin wire to that in the thick wire, we divide $$ \Delta L_	ext{thin} $$ by $$ \Delta L_	ext{thick} $$:
$$ 	ext{Ratio} = \frac{\Delta L_	ext{thin}}{\Delta L_	ext{thick}} = \frac{\frac{FL}{\pi R^2 Y}}{\frac{FL}{2\pi R^2 Y}} = \frac{FL}{\pi R^2 Y} \cdot \frac{2\pi R^2 Y}{FL} = 2 $$
Therefore, the ratio of the elongation in the thin wire to that in the thick wire is 2. The correct answer is:
Option C: 2.00

Answer

0.25

Answer

0.50

Answer

2.00

Answer

4.00

Properties of Matter

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

Choose the correct answer: