Answer:

Now when it will reach at point B then its normal force is just equal to ZERO


Step-by-step explanation:
Since we need to cross both the loops so least speed at the bottom must be

also by energy conservation this is gained by initial potential energy


so we will have

now we have

here we have
R = 7.5 m
so we have


Now when it will reach at point B then its normal force is just equal to ZERO

now when it reach point C then the speed will be
![mgh - mg(2R_c) = \frac{1}{2]mv_c^2]()


now normal force at point C is given as


