A mass m is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when:

Options

A. The mass is at the highest point b
B. The wire is horizontal
C. The mass is at lowest point
D. Inclined at an angle of `60^@` from vertical

Answer

C

Explanation

Now, at point P from figure
`T-mg cos theta=(mv^2)/(r)`
`therefore` T=mg cos `theta+(mv^2)/(r)`
From figure,cos`theta=(r-h)/(r)`
So T will be maximum when cos `theta` will be 1
`therefore(r-h)/(r)=1`
`implies h=0`
The tension is maximum when the mass is at the lowest position of the verticle circle, so the chance of breaking is maximum