A metal wire of length 0 . 5 m and cross-sectional area 10 - 4 m 2 has breaking stress 5 × 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.

Question

A metal wire of length 0 . 5 m and cross-sectional area 10 - 4 m 2 has breaking stress 5 × 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 .

TestHub · Accepted Answer

Here in horizontal circular motion, the only centripetal force is tension. So, $T = m ω^{2} r = \frac{m v^{2}}{l} = \frac{10 \times v^{2}}{0.5} = 20 v^{2}$
Now, $T_{\max} =$ Breaking stress $\times$ Area
$\Rightarrow 20 v^{2} = 5 \times 10^{8} \times 10^{- 4}$
$\Rightarrow 20 v^{2} = 5 \times 10^{4}$
Thus, the maximum linear velocity of block is
$v = \sqrt{\frac{1}{4} 10^{4}} = 50 m s^{- 1}$

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