Question

A copper wire of cross-sectional area 0.01 cm² is under a tension of 20N. Find the decrease in the cross-sectional area. Young modulus of copper = 1.1 × 10¹¹ N m⁻² and Poisson ratio = 0.32.

`["Hint" : (Delta"A")/"A"=2(Delta"r")/"r"]`

Answer 1

Given:
Cross-sectional area of copper wire A = 0.01 cm² = 10⁻⁶ m²
Applied tension T = 20 N
Young modulus of copper Y = 1.1 × 10¹¹ N/m²
Poisson ratio σ = 0.32
We know that: \[Y = \frac{FL}{A ∆ L}\]

\[\Rightarrow \frac{∆ L}{L} = \frac{F}{AY}\]
\[ = \frac{20}{{10}^{- 6} \times 1 . 1 \times {10}^{11}} = 18 . 18 \times {10}^{- 5} \]

\[\text{ Poisson's ratio }, \sigma = \frac{\frac{∆ d}{d}}{\frac{∆ L}{L}} = 0 . 32\]
\[\text{ Where d is the transverse length }\]
\[\text{ So }, \frac{∆ d}{d} = \left( 0 . 32 \right) \times \frac{∆ L}{L}\]
\[ = 0 . 32 \times \left( 18 . 18 \right) \times {10}^{- 5} = 5 . 81 \times {10}^{- 5} \]
\[\text{ Again }, \frac{∆ A}{A} = \frac{2 ∆ r}{r} = \frac{2 ∆ d}{d}\]
\[ \Rightarrow ∆ A = \frac{2 ∆ d}{d}A\]
\[ \Rightarrow ∆ A = 2 \times \left( 5 . 8 \times {10}^{- 5} \right) \times \left( 0 . 01 \right)\]
\[ = 1 . 164 \times {10}^{- 6} {\text{ cm }}^2\]

Hence, the required decrease in the cross -sectional area is \[1 . 164 \times {10}^{- 6} {\text{ cm } }^2\]