Two light vertical springs with equal natural length and spring constants `k_(1)` and `k_(2)` are separated by a distance `l`. Their upper ends are rigidly connected and the lower ends are connected to `A` and `B` of a light horizontal rod `AB`. A vertically downwards force `F` is applied at point `C` on the rod. `AB` will remain horizontal in equilibrium if the distance `AC` is

Options

A. `(lk_(1))/(k_(2))`
B. `(lk_(1))/(k_(2) + k_(1))`
C. `(lk_(2))/(k_(1))`
D. `(lk_(2))/(k_(1) + k_(2))`

Answer

D

Explanation

`sum ("Moments about" C) =0`
`:. (k_(1)x) AC=(k_(2)x) BC`
`:. (AC)/(BC) =k_(2)/k_(1)`….(i)
`AC + BC =l`….(ii)
Soving these two equation we get,
`AC=((k_(1))/(k_(1)+k_(2)))l`