Question

A student wants to determine the density of titanium. The student measures the mass of a solid chunk of titanium, but when he drops it into the water to determine the volume, some water splashes out of the cylinder. Will this cause the calculated density to be higher or lower than it should be? Explain!​

Answer 1

Answer:

Lower than it should be.

Explanation:

Hello.

In this case, since the density is computed by:

$\rho =\frac{m}{V}$

And we obtain the volume of the solid by substracting the mass of the water and the solid minus the mass of water:

$V_{solid}=V_{solid\ with \ water}-V_{water}$

If some water is splashed out of the cylinder, the volume of water will be lower than originally measured, it means that the volume of the solid will be higher than real. In such a way, since the density is in an inversely proportional relationship with volume, as the volume of the solid is wrongly increased, therefore the density of the solid will be lower than in should be.

Regards.