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(B) According to the work-energy theorem, the kinetic energy $(KE)$ of the stone equals the elastic potential energy $(U)$ of the rubber band.For the

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$50$

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(B) According to the work-energy theorem, the kinetic energy $(KE)$ of the stone equals the elastic potential energy $(U)$ of the rubber band.For the rubber band:Young's modulus $(Y)$ $= 5 	imes 10^8 \, N/m^2$Initial length $(L)$ $= 2 	imes 10^{-2} \, m$Change in length $(\Delta L)$ $= 2 	imes 10^{-2} \, m$Area of cross-section $(A)$ $= 5 	imes 10^{-6} \, m^2$Elastic potential energy $(U)$ $= \frac{1}{2} 	imes Y 	imes (	ext{strain})^2 	imes 	ext{volume}$$U = \frac{1}{2} 	imes Y 	imes \left(\frac{\Delta L}{L}ight)^2 	imes A 	imes L$$U = \frac{1}{2} 	imes 5 	imes 10^8 	imes \left(\frac{2 	imes 10^{-2}}{2 	imes 10^{-2}}ight)^2 	imes 5 	imes 10^{-6} 	imes 2 	imes 10^{-2}$$U = \frac{1}{2} 	imes 5 	imes 10^8 	imes 1^2 	imes 10 	imes 10^{-8} = 25 \, J$This energy is transferred to the stone of mass $m = 20 \, g = 20 	imes 10^{-3} \, kg$:$KE = U$$\frac{1}{2} m v^2 = 25$$\frac{1}{2} 	imes 20 	imes 10^{-3} 	imes v^2 = 25$$10^{-2} 	imes v^2 = 25$$v^2 = 2500$$v = 50 \, m/s$

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