A body of mass 1 kg tied to a string is just able to move in a vertical circular path of the radius 1 m. The difference between the kinetic energies at its highest and lowest positions of the body is45√ J20 J 10 J 30 J

The question seems to be incomplete. However, I can explain how to calculate the difference in kinetic energy between the highest and lowest positions of a body moving in a vertical circular path.

1. First, we need to understand that the kinetic energy (KE) of an object is given by the formula KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.

2. The potential energy (PE) at a height h in a gravitational field is given by PE = m * g * h, where g is the acceleration due to gravity.

3. In a vertical circular motion, the total mechanical energy (sum of kinetic and potential energy) is conserved. This means that the sum of KE and PE is the same at the highest and lowest points.

4. At the highest point, the kinetic energy is minimum (let's call it KE_min) and the potential energy is maximum (PE_max = m * g * h, h is the diameter of the circle in this case, 2r = 2m).

5. At the lowest point, the kinetic energy is maximum (KE_max) and the potential energy is minimum (PE_min = 0, because the body is at the lowest point, h = 0).

6. Since the total energy is conserved, we have KE_min + PE_max = KE_max + PE_min.

7. We can rearrange this equation to find the difference in kinetic energy between the highest and lowest points: KE_max - KE_min = PE_max - PE_min.

8. Substituting the known values, we get KE_max - KE_min = m * g * 2r.

9. Plug in the given values (m = 1 kg, r = 1 m, g = 9.8 m/s^2) to find the difference in kinetic energy.

Please provide the complete question to get the exact answer.