Chapter 7 - Circular Motion and Gravitation - Learning Path Questions and Exercises - Exercises - Page 262: 40

a). $2.19mg$ at the top of the loop. b). $4.19mg$ at the bottom of the loop,

a). $F_{N}=mg(-1+\frac{250^{2}}{2000\times9.8})=2.19mg$ at the top of the loop. b). $F_{N}=mg(1+\frac{250^{2}}{2000\times9.8})=4.19mg$ at the bottom of the loop.