A metal wire of length 0.5 m and cross-sectional area $$10^{-4}$$ m$$^2$$ has breaking stress $$5 \times 10^8$$ N m$$^{-2}$$. A block of 10 kg is attached at one end of the string and is rotating in a horizontal circle. The maximum linear velocity of block will be _____ m s$$^{-1}$$

We have a metal wire of length $$L = 0.5$$ m, cross-sectional area $$A = 10^{-4}$$ m$$^2$$, and breaking stress $$\sigma = 5 \times 10^8$$ N/m$$^2$$. A block of mass $$m = 10$$ kg is attached at one end and is rotating in a horizontal circle.

The maximum tension the wire can withstand before breaking is $$T_{\max} = \sigma \times A = 5 \times 10^8 \times 10^{-4} = 5 \times 10^4$$ N.

For circular motion, the tension provides the centripetal force: $$T = \frac{mv^2}{L}$$. At the maximum velocity, $$T_{\max} = \frac{mv_{\max}^2}{L}$$.

Solving for $$v_{\max}$$: $$v_{\max}^2 = \frac{T_{\max} \cdot L}{m} = \frac{5 \times 10^4 \times 0.5}{10} = \frac{2.5 \times 10^4}{10} = 2500$$

Hence $$v_{\max} = \sqrt{2500} = 50$$ m/s.

Hence, the correct answer is 50.