Physics · Young's modulus, bulk modulus and modulus of rigidity
A long aluminium wire \) of diameter supports a mass. In order to have the same
A \( 5 \mathrm{m} \) long aluminium wire \( (\boldsymbol{Y}=\mathbf{7} \times \) \( \left.10^{10} N m^{-2}\right) \) of diameter \( 3 \mathrm{mm} \) supports a \( 40 \mathrm{kg} \) mass. In order to have the same elongation in the copper wire \( (\boldsymbol{Y}= \) \( \left.12 \times 10^{10} N m^{-2}\right) \) of the same length under the same weight, the diameter should now be (in \( \mathrm{mm} \) ).
- A. 1.75
- B. 1.5
- C. 2.5
- D. 5.0
Step-by-step solution
For the same elongation, the product of cross-sectional area and Young's modulus must be equal for both wires. Since area ∝ (diameter)^2, we have d_cu^2 * Y_cu = d_al^2 * Y_al. Solving: d_cu = d_al * sqrt(Y_al / Y_cu) = 3 mm * sqrt((7×10^10)/(12×10^10)) ≈ 2.29 mm. The closest option is 2.5 mm.
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