One end of a horizontal thick copper wire of length 2 L and radius 2 R is welded to an end of another horizontal thin copper wire of length L and radius R . When the arrangement stretched by applying

Question

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

TestHub · Accepted Answer

Since the wires are in series, the tensile force $F$ acting on both wires is the same. Both wires are made of copper, so their Young's modulus $Y$ is identical.

The elongation $Δ L$ of a wire is given by the formula:

$Δ L = \frac{F L}{Y A}$

where $F$ is the applied force, $L$ is the original length, $Y$ is Young's modulus, and $A$ is the cross-sectional area. The cross-sectional area of a circular wire is $A = π r^{2}$ , where $r$ is the radius.

Let's denote the properties of the thick wire with subscript 1 and the thin wire with subscript 2.

For the thick wire (wire 1):

Length $L_{1} = 2 L$

Radius $r_{1} = 2 R$

Area $A_{1} = π (2 R)^{2} = 4 π R^{2}$

Elongation $Δ L_{1} = \frac{F ( 2 L )}{Y ( 4 π R ^{2} )}$

For the thin wire (wire 2):

Length $L_{2} = L$

Radius $r_{2} = R$

Area $A_{2} = π R^{2}$

Elongation $Δ L_{2} = \frac{F ( L )}{Y ( π R ^{2} )}$

We need to find the ratio of the elongation in the thin wire to that in the thick wire, which is $\frac{Δ L _{2}}{Δ L _{1}}$ .

$\frac{Δ L _{2}}{Δ L _{1}} = \frac{\frac{F L}{Yπ R ^{2}}}{\frac{F ( 2 L )}{Y ( 4 π R ^{2} )}}$

Cancel out common terms ( $F$ , $Y$ , $π$ , $L$ , $R^{2}$ ):

$\frac{Δ L _{2}}{Δ L _{1}} = \frac{\frac{1}{1}}{\frac{2}{4}} = \frac{1}{\frac{1}{2}} = 2$

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

The final answer is $2.00$ .

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