The figure above shows a solid of density de floating in a liquid of d

1.	The figure above shows a solid of density de floating in a liquid of density du.(0) What is the relation between de and do in the above case?(II) What is the apparent weight of the floating body?(IM) The same solid floats in water with 3/5th of its volume immersed in it.Calculate the density of the solid. (density of water = 18 cm)
Answer» Answer: i) Relation between the density is GIVEN as $d_o > d_e$ ii) APPARENT weight is the net weight under the condition of flotation $F_{net} = 0$ iii) the density of the OBJECT is given as $\rho_o = 600 kg/m^3$ Explanation: i) As we know that the object is floating in the liquid so we will have $F_{buoyancy} > F_{weight}$ $d_o V g > d_e V g$ $d_o > d_e$ ii) Apparent weight is the net weight under the condition of flotation So we have $F_g - F_b = F_{net}$ $F_{net} = 0$ iii) 3/5 of total VOLUME is immersed into the water so we have $\frac{3}{5}V\rho_w g = \rho_o V g$ $\rho_o = \frac{3}{5}\rho_w$ $\rho_o = 600 kg/m^3$ #Learn Topic : Fluids brainly.in/question/4356825