Frequency of revolution of stone $v=40\,\text{rev}/\text{min}=\frac{40}{60}\,\text{rev}/\text{s}$ mass of stone, $m = 0.25\,\text{kg}$ radius of circle, $r = 1.5\,\text{m}$ angular speed of the stone $\omega = 2\pi v$ $= 2\pi \times \frac{40}{60} = \frac{4\pi}{3}\,\text{rad}\,\text{s}^{-1}$ T = tension in the string = ?$T_{\max} =$ maximum tension in the string = 200 N$v_{\max} =$ maximum speed of the stone = ?The centripetal force is provided by the tension (T) in the string $T = \frac{m v^{2}}{r} = m r \omega^{2}$ $(\because v = r\omega)$ $= 0.25 \times 1.5 \times \left(\frac{4\pi}{3}\right)^{2}\,\text{N}$ $=6.58\,\text{N}=6.6\,\text{N}$As the string can withstand a maximum tension of 200 N$\therefore T_{\max} = \frac{m v_{\max}^{2}}{r}$ or $v_{\max}$ $= \sqrt{\frac{r T_{\max}}{m}}$ $= \sqrt{\frac{1.5 \times 200}{0.25}} = \sqrt{1200}$$= 34.64\,\text{m}\,\text{s}^{-1} = 35.0\,\text{m}\,\text{s}^{-1}$$\therefore T = 6.6\,\text{N},\; v_{\max} = 35.0\,\text{m}\,\text{s}^{-1}$