A stone of mass 20g is projected from a rubber catapult of length 0.1m and area of cross section 10?? m² stretched by an amount 0.04 m. The velocity of the projected stone is _______ m/s. (Young's modulus of rubber = 0.5 × 10? N/m²)

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.

As we know that

$Y=\frac{FL}{A?L}$          

$?L=0.04m=\frac{FL}{AY}...............\left(i\right)$           

If length and diameter both are doubled

$?L\text{'}=\frac{F.2L}{4A.Y}=\frac{FL}{2Y}=0.02m=2cm$       

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

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