Determine the decrease in the cross-sectional area of a copper wire having an initial cross-sectional area of $0.01 \text{ cm}^2$ that is subjected to a tension of $20 \text{ N}$ . The Young modulus of copper is $1.1 \times 10^{11} \text{ N/m}^2$ and its Poisson ratio is $0.32$ .
[Hint: $\dfrac{\Delta A}{A} = 2 \dfrac{\Delta r}{r}$ ]

Question

Determine the decrease in the cross-sectional area of a copper wire having an initial cross-sectional area of

0.01 \text{ cm}^2

that is subjected to a tension of

20 \text{ N}

. The Young modulus of copper is

1.1 \times 10^{11} \text{ N/m}^2

and its Poisson ratio is

0.32

.
[Hint:

\dfrac{\Delta A}{A} = 2 \dfrac{\Delta r}{r}

]

Quantrex Academy · Accepted Answer

Correct answer: A. 1.164 × 10^ -6 cm^2

The longitudinal strain in the wire is given by

\dfrac{\Delta L}{L} = \dfrac{F}{AY}

The lateral strain is given by

\dfrac{\Delta r}{r} = \sigma \dfrac{\Delta L}{L} = \dfrac{\sigma F}{AY}

The fractional change in the cross-sectional area is

\dfrac{\Delta A}{A} = 2 \dfrac{\Delta r}{r} = \dfrac{2\sigma F}{AY}

The decrease in the cross-sectional area is

\Delta A = A \left( \dfrac{2\sigma F}{AY} \right) = \dfrac{2\sigma F}{Y}

Substituting the given values

F = 20 \text{ N}

,

\sigma = 0.32

, and

Y = 1.1 \times 10^{11} \text{ N/m}^2

:

\Delta A = \dfrac{2 \times 0.32 \times 20}{1.1 \times 10^{11}} \text{ m}^2

\Delta A = \dfrac{12.8}{1.1} \times 10^{-11} \text{ m}^2 = 11.636 \times 10^{-11} \text{ m}^2

Converting to

\text{cm}^2

:

\Delta A = 11.636 \times 10^{-11} \times 10^4 \text{ cm}^2 \approx 1.164 \times 10^{-6} \text{ cm}^2

Determine the decrease in the cross-sectional area of a copper wire...