Question

A cubical block of wood weighing `200 g` has a lead piece fastened underneath. Find the mass of the lead pieces which will just allow the block to float in water. Specific gravity of wood is `0.8` and that of lead is `11.3`.
(b) Solve the previous problem if the lead piece is fastened on the top surface of the block and the block is to float with its upper surface just dipping into water.

Answer 1

`(M_omega + M_l)g = rho_omega (V_omega + V_l)g`
`M_omega + M_l = V_omega + V_l = (M_omega)/(sigma_omega) + (M_l)/(sigma_l)`
`200 + M_l = (200)/(0.8) + (M_l)/(11.3)`
`M_l(1 - (1)/(11.3)) = (200)/(0.8) - 200 = 50`
`M_l = (50 xx 11.3)/(10.3) = 54.8 g`
(b) `(M_omega + M_l) g = rho_omega V_omega g`
`200 + M_l = (M_omega)/(sigma_omega) = (200)/(0.8)`
`M_l = 30 g`.
(a)
(b) .