A particle of mass 10 g moves along a circle of radius 64 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to `8xx10^(-4)J` by the end of the second revolution after the beginning of the motion ?

Options

A. `0.1 m//s^(2)`
B. `0.15 m//s^(2)`
C. `0.18 m//s^(2)`
D. `0.2 m//s^(2)`

Answer

A

Explanation

`1/2 mv^(2) = E rArr 1/2 (10/1000)v^(2) = 8 xx 10^(-4)`
`rArr v^(2) = 16 xx 10^(-2) rArr v = 4 xx 10^(-1) = 0.4 m//s`
Now,
`v^(2) = u^(2) + 2a_(t)s, (s = 4piR)`
`rArr 16/100 = 0^(2) + 2a_(t) (4 xx 22/7 xx (6.4)/100)`
`rArr a_(t) = 16/100 xx (7 xx 100)/(8 xx 22 xx 6.4) = 0.1 m//s^(2)`