A particle of mass 10g 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 begining of the motion?

Options

A. `0.15ms^(-2)`
B. `0.18ms^(-2)`
C. `0.2ms^(-2)`
D. `0.1ms^(-2)`

Answer

D

Explanation

Given, mass of particle, m = `0.01` kg.
Radius of circle along which particle is moving, `r=6.4` cm.
`because` Kinetic energy of particle, KE=`8xx10^(-4)J`
`rArr" "(1)/(2)mv^(2)=8xx10^(-4)J`
`rArr" "v^(2)=(16xx10^(-4))/(0.01)=16xx10^(-2)`...(i)
As it is given that KE of particle is equal to `8xx10^(-4)` J by the end of second revolution after the beginning of motion of particle. It means, its initial velocity (u) is `0ms^(-1)` at this moment.
`because` By Newton's third equation of motion,
`rArr" "v^(2)=2a_(t)s" or "v^(2)=2a_(t)(4pir)`
`(because "particle covers 2 revolutions")`
`rArr" "a_(t)=(v^(2))/(8pir)=(16xx10^(-2))/(8xx3.14xx6.4xx10^(-2))`
`[because"from Eq. (i), v"^(2)=16xx10^(-2)]`
`therefore" "a_(t)=0.1ms^(-2)`