A block of Ag of mass x kg hanging from a string is immersed in a liquid of relative density 0.72. If the relative density of Ag is 10 and tension in the string is 37.12 N then compute the mass of Ag block.

Explanation

Let the mass of Ag block be x kg,
Tension in the string `T=37.12N`.
Relative density of liquid `R_(liq)=0.72`
Relative density of silver `R_(Ag)=10`
According to the principle of flotation
`Vrhog=mg`
`thereforeVrho=m" "...(1)`
Weight of Ag blok `=mg=Vrhog=R_(Ag)V_(g)`
Weightof Ag block `W=10Vrhog[becausem=Vrho]`
Force of buoyancy `F_(B)=R_(liq)(Vrho)g=0.72Vrhog`
Apparent weight `W_(app)=10Vrhog-0.72Vrhog`
`=9.78Vrhog`
`=9.78mg`
Tension in the string = Apparent weight
`37.12=9.78mg`
Mass `m=(37.12)/(9.28)=4kg`