A massles spool of inner radius r and other radius R is placed against vertical wall and tilted split floor as shown. A light inextensible thread is tightly wound around the spool through which a mass m is hanging. There exists no friction at point A, while the coefficient of friction between spool and point B is `mu`. The angle between two surface is `theta`

Options

A. the magnitude of force on the spool at `B` in order to maintain equilibrium is `mgsqrt((r/R)^(2)+(1-orr/R)^(2)1/(tan^(2)theta)`
B. the magnitude of force on the spool at `B` in order to maintain equilibrium is `mg(1-r/R)1/(tantheta)`
C. the minimum value of `mu` for the system to remain in equilibrium `(cottheta)/((R/r)-1)`
D. the minimum value of `p` for the system to remain equilibrium is `(tantheta)/((R/r)-1)`

Answer

A, D

Explanation

`mgr=fR`……….i
`N_(1) sintheta+f=mg`…………ii
`N-1costheta=N_(2)`…………iii
from i , `f=(mgr)/R`
from eqn ii and iii

`N_(2)=(mg-f)/(tantheta)=(mg)/(tantheta)[1-r/R]`
Net force at `B:f_(B)=sqrt(f^(2)+N_(2)^(2))`
For minimum value of `mu:flefN_(2)`
`implies(mgr)/Rle(mumg)/tantheta[1-r/R]impliesmuge(tantheta)/(R//r-1)`