A particle of mass 10 g moves along a circle of radius 6.4 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. `0.15 m//s^2`
B. `0.18m//s^2`
C. `0.2m//s^2`
D. `0.1m//s^2`

Answer

D

Explanation

Kinetic energy of the particle, `E=1/2mv^2=8xx10^(-4)J`
or, `1/2 xx10xx10^(-3)xxv^2=8xx10^(-4)`
`or, v^2=(8xx10^(-4)xx2)/(10^(-2)) =0.16`
`therefore v=sqrt(0.16) =0.4m//s`
From `v^2=u^2+2as`,
`(0.4)^2=0+2axx2(2pir)=8pi ra`
[s= distance covered by the particle `=2(2pi r)` r=radius of the circle]
`or, 0.16 =8xx3.14xx6.4xx10^(-2)xxa`
or, `a=(0.16)/(8xx3.14xx6.4xx10^(-2))=0.1m//s^2`