Question

An iron casting has a number of cavities in it. It weighs `6000 N` in air and `4000 N` in water. Determine the total volume of all the cavities in the casting. The density of iron (without cavities) is `8.0 g//cm^(3), rho_(w)= I g//cm^(3)`.

Answer 1

Volume of the cavities `V_("cav")` can be determined by taking difference between the volume `V_("cast")` of the casting as a whole and the volume of the iron in the casting
`V_("cav")=V_("cast")-V_("iron")`
`V_("iron")=W/(rho_(iron)g)`
`W` is the weight of casting `W_("eff")=W-rho_(w)gV_("cast")`
`V_("cast")=(W-W_("eff"))/(rho_(w)g)impliesV_("cav")=(W-W_("eff"))/(rho_(w)g)-W/(8rho_("iron"))`
`(6000-4000)/((10^(3))(10))-6000/((10)(8xx10^(3)))=0.125m^(3)`