A motor cycle driver doubles its velocity when he is having a turn. The force exerted outwardly will be

The correct answer is 4 times.

To understand why the outwardly exerted force changes by this factor, we can analyze the physics of circular motion.
When a vehicle (like a motorcycle) takes a turn of radius r at a velocity v, it undergoes centripetal acceleration. The outwardly directed force experienced in the rotating frame of reference is the centrifugal force, which is equal in magnitude to the centripetal force required to keep the motorcycle on its curved path. This force ($F$) is given by the formula:

$F=\frac{m{v}^2}{r}$

where:
- $m$ is the mass of the motorcycle and the driver,
- $v$ is the velocity of the motorcycle, and
- $r$ is the radius of the turn.

From this relation, we can see that the force is directly proportional to the square of the velocity:
$F\propto v^2$

Let the initial velocity be $v_1$ and the initial force be $F_1$. When the driver doubles the velocity, the new velocity becomes:
$v_2=2v_1$

Let us calculate the new force $F_2$:
$F_2=\frac{m{\left(v_2\right)}^2}{r}=\frac{m{\left(2v_1\right)}^2}{r}$

Simplifying the equation gives:
$F_2=\frac{m\cdot 4v_1^2}{r}=4\cdot \left(\frac{m{v}_1^2}{r}\right)=4F_1$

Therefore, when the driver doubles the velocity, the outwardly exerted force becomes 4 times the original force.