Chapter 7 - Circular Motion and Gravitation - Learning Path Questions and Exercises - Exercises - Page 262: 39

The tension is only 29.5 N, and the breakage limit is 100 N, so the string will surely work.

$v=\frac{2pir}{T}=\frac{2pi\times0.28}{0.75}=2.345m/s$ Hence, centripetal force $=m\frac{v^{2}}{r}=\frac{1.5\times 2.345^{2}}{0.28}=29.5N$ Since the tension is only 29.5 N, and the breakage limit is 100 N, so the string will surely work.