Consider an object making uniform motion around a circle of radium 5 m with tangental velocity 2 `ms^(-1)`. The time it takes to complete 2 revolution and the magnitude of acceleration respectively are

Options

A. `0.2pi` s and 0.8 `ms^(-2)`
B. `0.5pi` s and 1 `ms^(-2)`
C. `10pi` s and 0.8 `ms^(-2)`
D. `5pi` s and 5 `ms^(-2)`

Answer

C

Explanation

Given that, circle of radius = 5 m tangenital velocity = 2m/s

Here, 2 revoutions complete by object.
`because` We know that,
`v = omega R rArr (v)/(R) = omega`
`rArr omega = (2)/(5) ` rad / s
Also, `omega = (2pi)/(T) rArr T = (2pi)/(omega)=(2pi)/(2//5)`
`T = 5pi` s for one revolution in time,
`t = 2 xx 5 xx pi = 10 pi s `
Also acceleration , `alpha = v^(2)// R = (2^(2))/(5)= 0.8 " ms" ^(-2)`