Question

A steel wire of length 3.6 m and cross-sectional area 2.5×10-6m2 stretches by the same amount as a copper wire of length 2.4 m and cross-sectional area of 3.2×10-5m2 under a given load. What is the ratio of the Youngs modulus of steel to that of copper

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Answer 1

Answer:

96 : 5

Explanation:

For steel wire,

L₁ = 3.6 m

A₁ = 2.5 × 10⁻⁶ m²

For copper wire,

L₂ = 2.4 m

A₂ = 3.2 × 10⁻⁵ m²

As both stretches by same amount under a given load,

F₁ = F₂ = F (say)

∆L₁ = ∆L₂ = ∆L (say)

Therefore we can say,

$\frac{Y1}{Y2} = \frac{ \frac{F \: \times \: L1}{A1 \: \times \: ∆L} }{ \frac{F \: \times \: L2}{A2\: \times \: ∆L} } = \frac{L1}{A1} \times \frac{A2}{L2} = \frac{3.6}{2.5 \times {10}^{ - 6} } \times \frac{3.2 \times {10}^{ - 5} }{2.4} = \frac{480}{25} =19.2$

Hence, required ratio = 96 : 5 or 19.2

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