A piece of alloy has a mass 240g in air. When immersed in water, it has an apparent weight of 1.83N and in oil has an apparent weight `2.10N`. Calculate the specific gravity of (a) metal and (b) oil.

Explanation

Weight of metal piece in air,
`W-(a)=mg = 240/1000 xx 9.8 = 2.352N`
Weight of metal piece in water, `W_(w) = 1.83 N`
Weight of metal piece in oil, `W_(oil) = 2.10 N`
Weight of water displaced = `W_(a)-W_(w)`
`=2.352 - 1.829 - 0/522 N`
Mass of water displaced `= (0.522)/(9.8) = 0.053 kg`
Volume of water displaced,
`V= (mass)/("density") = (0.053 kg)/(1000 kg//m^(3)) = 5.3 xx 10^(-5)m^(3)`
As volume of water displaced = volume of metal alloy ltbr. `:. ` density of metal alloy
`=(mass)/(volume) =(0.240 kg)/(5.3 xx 10^(-5)m^(3)) = 4.5 xx 10^(3)m^(-3)`
Sp. gravity of metal alloy = `("density of metal")/("density of water")`
`(4.5 xx 10^(3))/(10^3) =4.5`
(b) Weight of oil displaced `= W_(a)-W_(oil)`
`=2.352 -2.10 = 0.252 N`
The volume of oil displaced , V= volume of metal alloy
`=5.3 xx 10^(-5) m^(3)`
if `rho_(oil)` is the density of oil, then
weight of oil displaced = `V=rho_(oil) g = 0.252`
or `rho_(oil) = (0.252)/(Vg) =(0.252)/(5.3 xx 10^(-5) xx 9.8`
`=0.485 xx 10^(3) kg//m^(3)`
Sp. gravity of oil, =`(rho_(oil))/(rho_w) = (0.485 xx 10^(3))/(10^3) = 0.485`