A cubical block of mass m and surface area 6A is placed on a thick layer of viscous liquid, of thickness d as shown

Initially the block is at rest. A constant horizontal force F0 starts acting on the block at t = 0. In column - 1 a physical quantity regarding the motion of the block is given and in column-2 corresponding variation with time is given. Match the proper entries from column-2 to column-1 using the codes given below the columns.
`{:("Column-I",,"Column-II"),("(P) X (distance travel by the block as function of time.) ",,(1) alphae^(-betat)(alpha,betane0)),("(Q) V (velocity of block as as function of time.)",,(2) alpha+beta t+gammae^(-deltat)(alpha,beta,gamma,delta ne0)),("(R) A (acceleration of block as as function of time.)",,(3) alphae^(-betat)-gammae^(-deltat)(alpha,beta,gamma,deltane0)),("(S) dK/ dt (rate of change in kinetic energy of block as as function of time.)",,(4) alpha+betae^(gammat)(alpha,beta,gammane0)):}`
(here `alpha,beta,gamma,delta` may have different values in each of options)

Options

A. `{:(,P,Q,R,S),((A),2,1,3,4):}`
B. `{:(,P,Q,R,S),((B),2,4,1,3):}`
C. `{:(,P,Q,R,S),((C),2,4,3,1):}`
D. `{:(,P,Q,R,S),((D),1,2,4,3):}`

Explanation

`F=mA=F_(0)-F_(v)=F_(0)-(etaAv)/d`
`A=(F_(0))/m-(etaA)/(md)v=a-bv`
`(dv)/(dt)=a-bv rArr int_(0)^(v) (dv)/(a-bv)=int_(0)^(t) dt`
`rArr (-1/b)ln((a-bv)/a)=trArr V=a/b (1-e^(-bt))`
`(dx)/(dt)=a/b(1-e^(-bt))rArr int_(0)^(x)dx=a/bint_(0)^(t)(1-e^(-bt))dt`
`x=a/b t -a/(b^(2))+a/(b^(2))e^(-bt)`
`A=ae^(-bt)`
`k=1/2 mv^(2)`
`(dk)/(dt)=mv (dv)/(dt)=(ma)/b(1-e^(-bt))(ae^(-bt))=(ma^(2))/b(e^(-bt)-e^(-2bt))`