Two vertical walls are separated by a distance of 2 metres. Wall 'A' is smooth while wall B is rough with a coefficient of friction `mu=0.5`. A uniform rod is propped between them. The length of the longest rod thhat can be probed between te walls is equal to

Options

A. 2 metres
B. `2sqrt(2)` metres
C. `sqrt(5)` metres
D. `(sqrt(17))/(2)` metres

Answer

D

Explanation

Drawing F.B.D. of rod PQ
`sumFx=0`
`sumFy=0impliesmuN=mg` …(1)
`sumtau` about centre of mass=0
`[Nxx(l)/(2)sintheta][2]=muNxx(l)/(2)costhetaimpliestantheta=(mu)/(2)`
Hence `sectheta=(sqrt(mu^(2)+4))//2`
thus, `((l//2)/(1))=(sqrt(mu^(2)+4))/(2)impliesl=sqrt(mu^(2)+4)`
`impliesl=sqrt(0.25+4)=(sqrt(17))/(2)` metres.