In: Physics

Imagine that the apparent weight of the crown in water is \(W_{\text {apparent }}=4.50 \mathrm{~N},\) and the actua weight is \(W_{\text {actual }}=5.00 \mathrm{~N}\). Is the crown made of pure \((100 \%)\) gold? The density of water is \(\rho_{\mathrm{w}}=1.00\) grams per cubic centimeter. The density of gold is \(\rho_{\mathrm{g}}=19.32\) grams per cubic centimeter.

- Yes
- No

$$ \begin{aligned} &\frac{W_{a p p a r e n t}}{W_{a c t u a l}}=\frac{4.5}{5}=1-\frac{\rho_{w}}{\rho_{c}}\\ &\frac{\rho_{w}}{\rho_{c}}=1-\frac{4.5}{5}=0.1\\ &\rho_{c}=\frac{\rho_{w}}{0.1}=\frac{1 / c c g}{0.1}=10 g / c c\\ &\text { As } \rho_{c}<\rho_{\text {gold}} \text { so crown is not made of pure gold. } \end{aligned} $$

The crown is not made of pure gold.

It is said that Archimedes discovered the buoyancy laws when asked by King Hiero of Syracuse to determine whether his new crown was pure gold (SG = 19.3). Archimedes measured the weight of the crown in air to be 11.8 N and its weight in water to be 10.9 N. Was it pure gold?

Take the density of the crown to be \(\rho_{\mathrm{c}} .\) What is the ratio of the crown's apparent weight (in water) \(W_{\text {apparent }}\) to its actual weight \(W_{\text {actual }} ?\)
Express your answer in terms of the density of the crown \(\rho_{\mathrm{c}}\) and the density of water \(\rho_{\mathrm{w}}\)
\(\frac{W_{\text {apparent }}}{W_{\text {actual }}}=1-\frac{\rho_{w}}{\rho_{c}}\)

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