A body of mass $m=10 \mathrm{~kg}$ is attached to one end of a wire of length $0.3 \mathrm{~m}$. The maximum angular speed (in $\mathrm{rad} \mathrm{s}^{-1}$ ) 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________
(4) Given : Wire length, $l=0.3 \mathrm{~m}$
Mass of the body, $\mathrm{m}=10 \mathrm{~kg}$
Breaking stress, $\sigma=4.8 \times 10^{7} \mathrm{Nm}^{-2}$
Area of cross-section, $a=10^{-2} \mathrm{~cm}^{2}$
Maximum angular speed $\omega=$ ?
$\mathrm{T}=\mathrm{M} l \omega^{2}$
$\sigma=\frac{T}{A}=\frac{m l \omega^{2}}{A}$
$\frac{m l \omega^{2}}{A} \leq 48 \times 10^{7}$
$\Rightarrow \omega^{2} \leq \frac{\left(48 \times 10^{7}\right) A}{m l}$
$\Rightarrow \omega^{2} \leq \frac{\left(48 \times 10^{7}\right)\left(10^{-6}\right)}{10 \times 3}=16 \Rightarrow \omega_{\max }=4 \mathrm{rad} / \mathrm{s}$
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