Question

The girlfriend of a PHS1282 student proposes to him and offers him a "golden" ring in the process. However, his girlfriend is a student in the Department of Accountancy and Finance, and would struggle to afford a gold ring. Additionally she is known to have a slightly dodgy character. After initially being flattered by the marriage proposal he therefore gets suspicious about whether the ring is actually made of gold. He therefore bring the ring to the PHS「282 lab and measure its mass to be 2.80± 0.02 g and its volume to be 0.16 ± 0.03 cm3 . Gold has a mass density of 19.3 g/cm3. Could the ring be made of gold? (Explain your answer) . Brass has a mass density of 8.4 g/cm3 to 8.7 g/cm3 (depending on the composition of the alloy). Could the ring be made of brass?

Answer 1

The density of the ring is:

`\rho=17.5\pm 3 \, g/cm^3`

This means the ring could very well be made of gold, but it is very unlikely that it is made of brass.

Step-by-step explanation:

For a quantity f(x,y) that depends on other quantities (in this case two) x and y, the error is given by:

`\sigma_f=\sqrt{\left((\partial f)/(\partial x)\right)^2\sigma_x^2+\left((\partial f)/(\partial y)\right)^2\sigma_y^2 }`

where
`\sigma_x` and
`\sigma_y` are the standard deviations on errors of the variables
`x` and
`y`.

In our case
`\rho=f(m,V)=(m)/(V)` where
`m` is the mass and
`V` is the volume.

Knowing that
`\sigma_m=0.02` and
`\sigma_V=0.03` we can estimate the error on the density

`\sigma_(\rho)=\sqrt{\left((1)/(V)\left)^2\sigma_m^2+\left((m)/(V^2)\right)^2\sigma_V^2}`
`\approx 3` (values were directly plugged)

The density is by using the given values

`\rho=(m)/(V)=(2.80)/(0.16)=17.5 \, g/cm^3`

The density with error is given by

`\rho=(m)/(V)\pm \sigma_(\rho)=17.5\pm 3 \, g/cm^3`

Which means it could go as high as 20.5 or as low as 14.5, Meaning that the ring could very well be made of gold, but it is very unlikely that it is made of brass.