Question

13. Physics students use a spring scale to measure the weight of a piece of lead. The experiment was performed two times: once in the air and once in water. If the volume of lead is 50 cm3, what is the difference between the two readings on the scale?

Answer 1

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.

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
`F_A=\rho* V * g` where
`\rho` is the density of the fluid in
`kg/m^3`, V is the volume of the submerged object in
`m^3` and g is the gravitational constant with a value of
`9.81\, m/s^2`.

The lead has a volume density of
`\rho_(lead)=11340 Kg/m^3`, the volume of the lead is
`V_(lead)=50g/cm^3=5* 10^(-5) m^3`.

The mass of the lead is given by

`m_(lead)=\rho_(lead)* V_(lead)=0.567Kg`

Thus its weight in the air is
`W_(lead)=0.567* 9.81=5.56 N`.

On the other hand when submerged the piece of lead will feel an upward force
`F_A`:

`F_A=\rho_(water)* V_(lead) * g= 1000 Kg/m^3* 5*10^(-5) m^3* 9.81 m/s^2=4.91 N`.

The perceived weight by the submerged piece of lead will be:

`W_(perceived)=5.56-4.91=0.65N`