Question

A certain string breaks under a tension of $45kgwt$. A mass of $100g$ is attached to this string of length $500cm$ and whirled in a horizontal circle. Find the maximum number of revolutions per second without breaking the string.

• A. $25.36$
• B. $4.729$
  
• C. $2.36$
  
• D. $23.6$
  

Answer 1

Answer: (B)

$4.729$

As given that breaking tension of string is $45kgwt=45\times 10N=450N$ so the tension in the string when we rotate the mass should be less than $450N$

$\Rightarrow T\ge m{\omega }^{2}R\Rightarrow \omega \le \sqrt{\frac{T}{mR}}$

given that $m=0.1kg$, $R=5m$

$\Rightarrow \omega \le \sqrt{\frac{450}{0.1\times 5}}$

$\Rightarrow \omega \le \sqrt{900}\Rightarrow \omega \le 30rad/s$

as we know $r.p.s.\left(revolutionpersec\right)\times 2\pi =\omega \Rightarrow r.p.s.=\frac{\omega }{2\pi }=\frac{30}{2\pi }=4.729$