`Asqrt(34)` m long ladder weighing 10 kg leans on a frictionless wall. Its feet rest on the floor `3m` away from the wall as shown in the figure. If `F_(f) and F_(w)` are the reaction forces of the floor and the wall, then ratio of `F_(w)//F_(f)` will be :
(Use `g=10m//s^(2)`)

Options

A. `(6)/(sqrt(110))`
B. `(3)/(sqrt(113))`
C. `(3)/(sqrt(109))`
D. `(2)/(sqrt(109))`

Answer

D

Explanation

`N_(1)=f_(2),N_(2)=mg`
`N_(1)xx5=mgxx(3)/(2)impliesN_(1)=(3)/(10)mg`
`R_(1)=N_(1)=(3)/(10)mg,R_(2)=sqrt(N_(2)^(2)+t_(2)^(2))=(sqrt(109))/(10)mg`
`(R_(1))/(R_(2))=(3)/(sqrt(109))=(F_(w))/(F_(f))=(3)/(sqrt(109))`