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:
Answer & explanation
Correct answer: option 2
The correct answer is option (3) : 4T
$F_{cp}=ma_{cp}$
$F_{cp}=m\omega^2r$
$T=m\omega^2r$
Now speed becomes $'2\omega'$
$T'=m(2\omega)^2r$
$T'=4m\omega^2r$
$T'= 4T$