Three identical rigid circular cylinder `A` `B` and `C` are arrenged on smooth inclined surfaces as shown in figure. The laest value of theta that prevent the arrangement from collapes is.

Options

A. `tan^(-1)(1//2)`
B. `tan^(-1)(1//2sqrt3)`
C. `tan^(-1)(1//(3sqrt3)`
D. `tan^(-1)(1//4sqrt3)`

Answer

C

Explanation

W=weight of each sphere
`N_(2)` =noemal reaction between `A` and inclined plane
`N_(1)` =normal reaction between `A` and `C` =normal reaction between `B` and `C`
free body diagram of `C` :
Resolving horizontally and vertically `2N_(1)cos30^(@)=W`
or `N_(1)=(W)/(sqrt(3))` (1)
When the arrangement is on the point of collapsing, the reaction between `A` and `B` is zero.
Free body diagram of A:
Resolving horizontal and vertically
`N_(2)sintheta=N_(1)sin30^(@)` or `N_(2)sintheta=(W)/(2sqrt(3))` (2)
`N_(2)costheta=W+N_(1)cos30^(@)=(3W)/(2)` (3)
Dividing (2) by (3) we get
`tantheta=(1)/(3sqrt(3))` or `theta=tan^(-1)((1)/(3sqrt(3)))` .