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