Answer:
In the air the scale will measure 5.56 N for the weight of the piece of lead.
In the water the piece of lead will have a weight of 0.65 N.
Step-by-step explanation:
We have to use the Archimedes' principle, that is, the upward force felt by a submerge object in a fluid is equal to the wight of the displaced volume of the object. In other words [tex]F_A=\rho\times V \times g[/tex] where [tex]\rho [/tex] is the density of the fluid in [tex]kg/m^3[/tex], V is the volume of the submerged object in [tex]m^3[/tex] and g is the gravitational constant with a value of [tex]9.81\, m/s^2[/tex].
The lead has a volume density of [tex]\rho_{lead}=11340 Kg/m^3[/tex], the volume of the lead is [tex]V_{lead}=50g/cm^3=5\times 10^{-5} m^3[/tex].
The mass of the lead is given by
[tex]m_{lead}=\rho_{lead}\times V_{lead}=0.567Kg[/tex]
Thus its weight in the air is [tex]W_{lead}=0.567\times 9.81=5.56 N[/tex].
On the other hand when submerged the piece of lead will feel an upward force [tex]F_A[/tex]:
[tex]F_A=\rho_{water}\times V_{lead} \times g= 1000 Kg/m^3\times 5\times10^{-5} m^3\times 9.81 m/s^2=4.91 N[/tex].
The perceived weight by the submerged piece of lead will be:
[tex]W_{perceived}=5.56-4.91=0.65N[/tex]
