The Young’s modulus for steel is much more than that for rubber. For the same longitudinal strain, which one will have greater tensile stress?

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    Answered by

    Payal Gupta | Contributor-Level 10

    4 months ago

    This is a short answer type question as classified in NCERT Exemplar

    Young’s modulus Y= stress/longitudinal strain

    For same longitudinal strain, Y s t e e l Y r u b b e r = s t r e s s s t e e l s t r e s s r u b b e r

    Ysteel>Yrubber

    so we can say that  stresssteel > stressrubber

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Vishal Baghel

If μ  is Poisson’s ratio,

Y = 3K (1 - 2 μ )      ……… (1)

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Vishal Baghel

dm = (m/L)dx
∴ T = (mω²/2L) (L² - x²)
∴ ΔL = ∫? (mω²/2Lπr²Y) (L² - x²)dx
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Vishal Baghel

Initially S? L = 2m
S? L = √2² + (3/2)²
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? x = S? L - S? L = 0.5 m
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Vishal Baghel

Loss in elastic potential energy = Gain in KE
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0.5 × (0.5×10? × 10? / 0.1) × (0.04)² = 20×10? ³ v²
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0.5 * ( (0.5e9 * 1e-6) / 0.1) * (0.04)^2 = 0.5 * (5e2) * 1.6e-3 = 4.
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As we know that

Y = F L A ? L          

? L = 0 . 0 4 m = F L A Y . . . . . . . . . . . . . . . ( i )           

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? L ' = F . 2 L 4 A . Y = F L 2 Y = 0 . 0 2 m = 2 c m       

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