Question

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

Answer:

The reading of the experiment made in air is 50 g more than the reading of the measurement made in water.

Explanation:

Knowing that the density of lead is $11,3 g/cm^{3}$ and the volume, we can calculate the true weight of the piece of lead:

$weight_{lead}=\rho _{lead}*V_{lead}=11,3 g/cm^{3} *50 cm^{3} = 565 g$

When the experiment is done in air, we can discard buoyancy force (due to different densities) made by air because it's negligible and the measured weight is approximately the same as the true weight.

When it is done in water, the effect of buoyancy force (force made by the displaced water) is no longer negligible, so we have to take it into account.

Knowing that the density of water is 1 g per cubic centimeter, and that the volume displaced is equal to the piece of lead (because of its much higher density, the piece of lead sinks), we can know that the buoyancy force made by water is 50 g, opposite to the weight of the lead.

$Weight_{measured}=weight_{lead}-weight_{water}=\frac{(565 g *9.8 m/s^{2} -50 g*9.8 m/s^{2})}{9,8m/s^{2} } = 515 g$

Now that we have the two measurements, we can calculate the difference:

$Difference= |Weight _{in water}- Weight _{in air}|=|515 g-565 g|=50 g$

The reading of the experiment made in air is 50 g more than the reading of the measurement made in water.