A metal wire of length 0 . 5m and cross-sectional area 10 ^ - 4 m ^...

Here in horizontal circular motion, the only centripetal force is tension. So,$T = m{\omega}^{2}r = \dfrac{m{v}^{2}}{l} = \dfrac{10 \times {v}^{2}}{0 . 5} = 20{v}^{2}$

Now, {T}_\max = Breaking stress $\times$ Area

$⇒ 20{v}^{2} = 5 \times {10}^{8} \times {10}^{- 4}$

$⇒ 20{v}^{2} = 5 \times {10}^{4}$

Thus, the maximum linear velocity of block is

$v = \sqrt{\dfrac{1}{4}{10}^{4}} = 50m{s}^{- 1}$