A uniform wooden bar of length l and mass m hinged on a vertical wall of a containing water, at one end.

`3//5th` part of the bar is submerged in water. Find the ratio of densities of the liquid and the bar.

Explanation

Let `rho` and `simga` be the density of the rod PQ and liquid respectively. If A is the area of cross - section of the rod, then the weight of the rod. `W = A l rho g.` It acts downwards at the cente C of the rod. Upward thrust due to liquid on rod, `F_B = Axx(3)/(5)l sigma g.` It acts at the centre of the dipped part of the rof, i.e,. at B. Taking moment of forces about point P, we have `W (l//2) sin theta -F_B (l-3l // 10) sin theta =0`
or `Al rho g xx (l)/(2) sin theta = Axx(3)/(5) lsigma g xx (7l)/(10)sin theta`
or `(rho)/(2) = (3)/(5)sigma xx (7)/(10) or (sigma)/(rho) = (25)/(21)`