A sting breaks under a load of 4.8 kg A mass of 0.5 kg is attached to one end tothe string 2malong and is ratated in a horizontal circle caculate the greatest number of revolutions that the mass can make without breaking the string

A sting breaks under a load of 4.8 kg A mass of 0.5 kg is attached to

1.	A sting breaks under a load of 4.8 kg A mass of 0.5 kg is attached to one end tothe string 2malong and is ratated in a horizontal circle caculate the greatest number of revolutions that the mass can make without breaking the string
Answer» <html><body><p></p>Solution :Given`m=0.5 kg` <br/> `r=2 m` <br/> `g=9.8ms^(-2)` <br/> Themaximumtensionthat the stringcanwithstand <br/> `F=4.8 kg wt=4.8xx 9.8 <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>` <br/><a href="https://interviewquestions.tuteehub.com/tag/let-11597" style="font-weight:bold;" target="_blank" title="Click to know more about LET">LET</a> themaximum <a href="https://interviewquestions.tuteehub.com/tag/number-582134" style="font-weight:bold;" target="_blank" title="Click to know more about NUMBER">NUMBER</a> of revolutions persecond=v<br/> Now F = `mr epsilon^(2)= mr (2 pi v)^(2) = <a href="https://interviewquestions.tuteehub.com/tag/4pi-1882352" style="font-weight:bold;" target="_blank" title="Click to know more about 4PI">4PI</a>^(2) mrv^(2)` <br/> or `v^(2)=(F )/(4pi^(2)mr)` <br/> `=(4.8 xx 9.8)/( 4xx 9.8 xx 0.5 xx 2) =1.215` <br/> or `v = <a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(1.215 )= 1.102 rps` <br/> `=1.102 xx 60 =66 .13 rpm`</body></html>