1.

A proton of mass 1.6xx10^-(27) kg, revolves in a circular path of radius 0.1 m. Calculate the angular velocity of the proton if it is acted upon by a centripetal force of 2.5xx10^(-12) N.

Answer»

Solution :`F=(mv^2)/r=mrw^2`
`therefore w^2=F/(MR)=(4xx10^(-13))/(1.6xx10^(-27)xx10^(-1)`
`w^2 =25xx10^14 therefore w=5xx10^7 (rad)/s`


Discussion

No Comment Found

Related InterviewSolutions