A ball attached to one end of a string is rotated on a horizontal circular path about a vertical axis. Given that the maximum angular velocity of the ball is `42rads^(-1)` a mass (m) of the ball is 0.8 kg, and length (l) of the string is 0.4m. String can bear maximum tension is `Kxx10^(2)` then K is

Answer

5.65

Explanation

`T cos theta` component cancel is mg

`T sin theta` component will provide necessqry centripetal force to the ball towards centreO.
`:. T si theta=mromega^(2)=m(l sin theta)omega^(2)`
`:.T=m//omega^(2)`
`:.omega=sqrt(T/(ml))`
`:.omega_(max)=sqrt((T_(max))/(ml))`
`:.T_(max)=(omega_(max))^(2)xxmxxl=42xx42`
`xx0.8xx0.4`
`:.T_(max)=564.48N=5.65xx10^(2)`