A rubber cord has a cross-sectional area 10 ^ - 6 m^ 2 and total...

Force constant $K$ of rubber is given by

$K = \dfrac{YA}{l} = \dfrac{5 \times {10}^{8} \times {10}^{- 6}}{0. 1} = 5 \times {10}^{3}Nm^{- 1}$

Now, from conservation of energy,

elastic potential energy of cord = kinetic energy of particle

i.e. $\dfrac{1}{2}K\left(∆ l\right)^{2} = \dfrac{1}{2}mv^{2}$

$⇒ v = \sqrt{\dfrac{K}{m}} \cdot ∆ l$

$= \sqrt{\dfrac{5 \times {10}^{3}}{5 \times {10}^{- 3}}} \left(0. 125 - 0. 1\right)$

$= {10}^{3} \times0. 025 = 25ms^{- 1}</p>$