Question

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an iron rod of length 2m and cross sectional area of 50 mm2 is streached 0.2 mm when a mass of 250 Kg is hung from its lower and Young's modulus of iron..???

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Answer 1

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19.6 × 1010 (N / m2)

L = 2 m

A = 50 mm² = 50 × 10–6 m²

ℓ = 0.5 mm = 0.5 × 10–3 m

m = 250 kg

$stress \: = \frac{f}{a} \\ = \frac{mg}{a} \\ = \frac{250 \times 10}{50 \times {10}^{ - 6} } \\ = 5 \times {10}^{7} N / \: {m}^{2}$

$strain = \frac{ℓ}{l} \\ = \frac{0.5 \times {10}^{ - 3} }{2} \\ = 2.5 \times 10 ^{ - 4}$

Y = [(stress) / (strain)]

= {(5 × 10^7) / (2.5 × 10^–4)}

= 2 × 1011

Y = 20 × 1010 N/m²

Y ≈ 19.6 × 1010 N/m²

Answer 2

Explanation:

19.6 × 1010 (N / m2)

L = 2 m

A = 50 mm² = 50 × 10–6 m²

ℓ = 0.5 mm = 0.5 × 10–3 m

m = 250 kg

Y = [(stress) / (strain)]

= {(5 × 10^7) / (2.5 × 10^–4)}

= 2 × 1011

Y = 20 × 1010 N/m²

Y ≈ 19.6 × 1010 N/m²