A uniform rod is suspended horizontally from its mid-point. A piece of metal whose weight is W is suspended at a distance l from the mid-point. Another weight W_(1) is suspended on the other side at a distance l_(1) from the mid-point to bring the rod to a horizonatl postion. When W is completely immersed in water, W_(1) needs to be kept at a distance l_(2) from the mid-point to get the rod into horizontal postion. The specific gravity of the metal piece is

A uniform rod is suspended horizontally from its mid-point. A piece of

1.	A uniform rod is suspended horizontally from its mid-point. A piece of metal whose weight is W is suspended at a distance l from the mid-point. Another weight W_(1) is suspended on the other side at a distance l_(1) from the mid-point to bring the rod to a horizonatl postion. When W is completely immersed in water, W_(1) needs to be kept at a distance l_(2) from the mid-point to get the rod into horizontal postion. The specific gravity of the metal piece is
Answer» `W/W_(1)` `(Wl_(1))/(Wl-W_(1)l_(2))` `l_(1)/(l_(1)-l_(2))` `l_(1)/l_(2)` Solution :In equilibrium of rod in air, `Wl=W_(1)l_(1)` In equilibrium CONDITION of rod immersed in water, `(W-F_(B))l=W_(1)l_(2)` Where, `F_(B)` = buoyant force on the body of weight `W=W/rho` So, `(W-W/rho)l=W_(1)l_(2)=Wl/l_(1)l_(2)or,1-1/rho=l_(2)/l_(1)` `THEREFORE" "rho=l_(1)/(l_(1)-l_(2))`