A string of length `l=1m` is fixed at one end carries a mass of 100gm at other end. The string makes `sqrt(5)//(pi)` revolutions per second about a verticle axis passing through its second end. What is the angle of inclination of the string with the vertical?

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `75^(@)`

Answer

C

Explanation

The different force are shown in figure. From figure
`Tsintheta=(mv^(2))/(r)=momega^(2)r=momega^(2)lsintheta` (i)
and `Tcostheta=mg`
From equation (i)
`T=momega^(2)l=momega^(2) ( :. l=1m)`
From equation (ii),
`momega^(2)costheta=mg`
:. `costheta=(g)/(omega^(2))=(g)/(4pi^(2)n^(2))=(10)/(4pi^(2)(5//r^(2)))`
or `costheta=0.5` i.e., `theta=60^(@)`