A body of mass m = 10 kg is attached to one end of a wire of length 0.3 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 = 4.8×10⁷ Nm^–2 and area of cross-section of the wire = 10^–2 cm²) is $\_\_\_\_$
 

Given,

Mass of the body =10kg

length of wire = 0.3m

Breaking stress of wire = 4.8×10⁷ Nm^–2

area of cross-section of the wire = 10^–2 cm²

we know that,

$\mathrm{Stress}=\frac{\mathrm{force}}{\mathrm{area}}\Rightarrow \mathrm{force}=\mathrm{stress}\times \mathrm{area}=4.8\times {10}^7\times {10}^{-6}=48$

Force is equal to the tension in the wire,

$48={\mathrm{m\omega }}^2\mathrm{r}\Rightarrow \frac{48}{0.3}\times \frac{1}{10}={\mathrm{\omega }}^2\Rightarrow \mathrm{\omega }=4\mathrm{rad}/\mathrm{s}$

Hence the correct answer is 4.