A uniform metallic wire is elongated by 0.04m when subjected to a linear force F. The elongation, if its length and diameter is doubled and subjected to the same force will be……………….cm.

If $\mu$  is Poisson’s ratio,

Y = 3K (1 - 2 $\mu$ )      ……… (1)

and Y = 2 $n\left(1+\mu \right)$  ……… (2)

With the help of equations (1) and (2), we can write

  $\frac{3}{Y}=\frac{1}{\eta }+\frac{1}{3k}\Rightarrow K=\frac{\eta Y}{9\eta -3Y}$

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 * 2

Energy density (u) = $\frac{1}{2}\times \sigma \times \epsilon =\frac{1}{2}\times Y\epsilon \times \epsilon =\frac{1}{2}{\epsilon }^{2}Y$ $\frac{}{}\frac{}{}\frac{}{}{}^{}$

$\Rightarrow energystored/m^2=\frac{1}{2}{\epsilon }^{2}YA$

Strain $\left(\epsilon \right)=\frac{\Delta \mathcal{l}}{{\mathcal{l}}_0}=\alpha \Delta T=10^{-5}\times 10=10^{-4}$ $\frac{}{{}_{}}{}^{}{}^{}$

$\Rightarrow energystored/m^2=\frac{1}{2}\times 10^{-8}\times 10^{11}\times 10^{-2}=5$