A spring balance reacts 200gF when carrying a lump of lead in air. If the lead is now immersed with half of its volume in brine solution, what will be the new reading of the spring balance? specific gravity of lead and brine 11.4 and 1.1 respectively.
The weight of lump is given as,
W1=mg
The volume of a lump is given as,
V=mρl
The weight of lump in brine solution is given as,
W2=mg−ρgV2
=mg(1−ρ2ρl)
The new readings of the spring balance is given as,
W′=mg(1−ρ2ρl)
=200×(1−1.12×11.4)
=190.35gF
Thus, the new reading of the spring balance is 190.35gF.