A motor cycle driver doubles its velocity when he is having a turn. The force exerted outwardly will be
A motor cycle driver doubles its velocity when he is having a turn. The force exerted outwardly will be
Double
Half
$4$ times
$\frac{1}{4}$ times
A motor cycle driver doubles its velocity when he is having a turn. The force exerted outwardly will be
$F = \frac{{m{v^2}}}{r}$ $⇒$ $F \propto {v^2}$ i.e. force will become $4$ times.