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?
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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1 Answer
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This is a short answer type question as classified in NCERT Exemplar
Young’s modulus Y= stress/longitudinal strain
For same longitudinal strain,
Ysteel>Yrubber
so we can say that stresssteel > stressrubber
Similar Questions for you
If is Poisson’s ratio,
Y = 3K (1 - 2 ) ……… (1)
and Y = 2 ……… (2)
With the help of equations (1) and (2), we can write
dm = (m/L)dx
∴ T = (mω²/2L) (L² - x²)
∴ ΔL = ∫? (mω²/2Lπr²Y) (L² - x²)dx
= ΔL = mω²L²/3πr²Y
Initially S? L = 2m
S? L = √2² + (3/2)²
S? L = 5/2 = 2.5 m
? x = S? L - S? L = 0.5 m
So since λ = 1 m. ∴? x = λ/2
So white listener moves away from S? Then? x (= S? L − S? L) increases and hence, at? x = λ first maxima will appear.? x = λ = S? L − S? L.
1 = d - 2 ⇒ d = 3 m.
Loss in elastic potential energy = Gain in KE
½ (YA/L)x² = ½mv²
0.5 × (0.5×10? × 10? / 0.1) × (0.04)² = 20×10? ³ v²
0.5 × (5×10²) × 1.6×10? ³ = 20×10? ³ v²
0.4 = 20×10? ³ v²
v² = 20 => v = √20 ≈ 4.47 m/s
(Re-checking calculations)
0.5 * ( (0.5e9 * 1e-6) / 0.1) * (0.04)^2 = 0.5 * (5e2) * 1.6e-3 = 4.
0.5 * 20e-3 * v^2 = 10e-3 v^2
4 = 10e-3 v^2
v^2 = 400 => v = 20 m/s
As we know that
If length and diameter both are doubled
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