Correct Answer - c
When box just slips then Coefficient of static friction
`mu_(s) = tan30^(@) = (1)/sqrt3 ~~0.6`
When box slides down the plane, coefficient of kinetic friction will be involved Let a be the acceleration of the box sliding down the plane then using
`S = ut + (1)/(2)at^(2)` we have
`4 = 0 + (1)/(2) xxa xx 4^(2)` or `a=(2)/(4) = 0.5 ms^(-2)`
Here, `a =g sin theta -mu_(k) g cos theta`
`0.5 =9.8xx sin30^(@) -mu_(k) xx 9.8 cos 30^(@)`
`0.5 =9.8 xx(1)/(2) -mu_(k) xx 9.8 xx sqrt3/2`
`0.5 =4.9 -mu_(k) xx4.9sqrt3`
`mu_(k) =(4.9 -0.5)/(4.9sqrt3) =(4.4)/(4.9 xx1.73) ~~0.5` .