A bob is whirled in a horizontal plane by means of a string with an initial speed of $\omega $ rpm . The tension in the string is $T$. If speed becomes $2\omega $ while keeping the same radius, the tension in the string becomes:

$4T$

The correct answer is option (3) : 4T

$F_{cp}=ma_{cp}$

$F_{cp}=m\omega^2r$

$T=m\omega^2r$

Now speed becomes $'2\omega'$

$T'=m(2\omega)^2r$

$T'=4m\omega^2r$

$T'= 4T$