Initially both blocks are at rest on a horizontal surface and string is just tight. At t=0, two constant horizontal force `F_(1)` and `F_(2)` start acting on blocks as shown. `f_(1)` and `f_(2)` are frication foces acting on `10 kg` and `20 kg` block (co-efficient of frication between and ground are `0.5`). Values of `F_(1)` and `F_(2)` are given column-I. Then match magnitudes of `f_(1), f_(2)` and direction of `overset(vec)(f_(1))` with corresponding values of `F_(1)` and `F_(2)` given in column-I
`[g = 10 m//s^(2)]`.

Options

A. `{:("Coulmn- I",,"Coulmn- II"),((A) F_(2)=120 N","F_(1)=40 N,,(p) F_(2)=100 N","F_(1)=20 N):}`
B. `{:("Coulmn- I",,"Coulmn- II"),((B) F_(2)=160 N","F_(1)=40 N,,(q) F_(2)=20 N","F_(1)=50 N):}`
C. `{:("Coulmn- I",,"Coulmn- II"),((C) F_(2)=60 N","F_(1)=90 N,,(r) F_(2)=70 N","F_(1)=50 N):}`
D. `{:("Coulmn- I",,"Coulmn- II"),((D) F_(2)=20 N","F_(1)=90 N,,underset((t) "unit vector in direction of " bar(f)_(1)" is " -hat(i))((s) "unit vector in direction "bar(f)_(1) "is "hat(i))):}`

Explanation

`|vecF_(1)+vecF_(2)|lt|f_(1"max")|+|f_(2"max")|`
So, both blocks not move in any case
`|f_(1"max")|=50N , |f_(2"max")|=100N`