A body of mass $\mathrm{m}=10\; \mathrm{kg}$ is attached to one end of a wire of length $0.3\; \mathrm{m} .$ The maximum angular speed (in $rad \;s^{-1}$ ) with which it can be rotated about its other end in space station is (Breaking stress of wire $=4.8 \times 10^{7} \;\mathrm{Nm}^{-2}$ and area of cross-section of the wire $=10^{-2}\; \mathrm{cm}^{2}$ ) is
JEE MAIN 2020, Medium
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$\mathrm{T}=\mathrm{m\omega}^{2} \ell$
Breaking stress $=\frac{\mathrm{T}}{\mathrm{A}}=\frac{\mathrm{mo}^{2} \ell}{\mathrm{A}}$
$\Rightarrow \omega^{2}=\frac{4.8 \times 10^{7} \times\left(10^{-2} \times 10^{-4}\right)}{10 \times 0.3}=16$
$\Rightarrow \omega=4$
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