A metal wire of length 0.5 ~m and cross-sectional area 10^-4 ~m^2 has breaking stress 5 \times 10^8 \,Nm^-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 \underline\hspace2cm ms^-1 .

Properties of Matter PYQ — JEE Main 2022 · 29 Jul · Shift 2 · Q64 with Solution

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Correct answer: 50

We need the maximum linear velocity of the block when it rotates in a horizontal circle.

Breaking stress is

\sigma = \frac{F_{\max}}{A}

So,

F_{\max} = \sigma A = (5 \times 10^8)(10^{-4})

F_{\max} = 5 \times 10^4\,\text{N}

Thus the maximum tension the wire can withstand is

T_{\max} = 5 \times 10^4\,\text{N}

Since the block is rotating in a horizontal circle, the tension provides the centripetal force:

T = \frac{mv^2}{r}

Here the radius is the length of the wire:

r = L = 0.5\,\text{m}

At maximum speed, tension reaches breaking value:

\frac{mv^2}{r} = T_{\max}

Substitute values:

\frac{10\,v^2}{0.5} = 5 \times 10^4

20v^2 = 5 \times 10^4

v^2 = \frac{5 \times 10^4}{20} = 2500

v = 50\,\text{m s}^{-1}

\boxed{50}