Question

According to legend, the following challenge led Archimedes to the discovery of his famous principle: Hieron, king of Syracuse, was suspicious that a new crownthat he had received from the royal goldsmith was not pure gold, as claimed. Archimedes was ordered to determine whether the crown was in fact made of pure gold, with the condition that only a nondestructive test would be allowed. Rather than answer the problem in the politically most expedient way (or perhaps extract a bribe from the goldsmith), Archimedes thought about the problem scientifically. The legend relates that when Archimedes stepped into his bath and caused it to overflow, he realized that he could answer the challenge by comparing the volume of water displaced by the crown with the volume of water displaced by an amount of pure gold equal in weight to the crown. If the crown was made of pure gold, the two volumes would be equal. If some other (less dense) metal had been substituted for some of the gold, then the crown would displace more water than the pure gold.

A similar method of answering the challenge, based on the same physical principle, is to compute the ratio of the actual weight of the crown, W_actual , and the apparent weight of the crown when it is submerged in water, W_apparent . See whether you can follow in Archimedes' footsteps. The figure shows what is meant by weighing the crown while it is submerged in water.

Part A
Take the density of the crown to be ρ_c. What is the ratio of the crown's apparent weight (in water) W_apparent to its actual weightW_actual ?
Express your answer in terms of the density of the crown ρ_c and the density of water ρ_w .
W_apparent/W_actual=____________

Answer 1

In water, the crown will experience the Buoyancy force also, caused by the weight of displaced water. Hence its apparent weight will be its actual weight minus the buoyancy force.

Now if "V" is the volume of the crown then its Mass = Density* Volume

\(M_{c}=\rho_{c} \times V=V \rho_{c}\)

and the Mass of displaced water will be \(M_{w}=\rho_{w} \times V=V \rho_{w}\)

Then Actual Weight of Crown is, \(W_{c}=M_{c} g=V \rho_{c} g_{\ldots \ldots . .(1)}\)

and Weight of displaced water, \(W_{w}=M_{w} g=V \rho_{w} g\)

and Apparent Weight of crown in water,

$$ W_{a}=W_{c}-W_{w}=V \rho_{c} g-V \rho_{w} g=V g\left(\rho_{c}-\rho_{w}\right)_{\ldots \ldots . .(2)} $$

Then using equation 1 and 2 ,

\(\frac{W_{a}}{W_{c}}=\frac{V g\left(\rho_{c}-\rho_{w}\right)}{V g \rho_{c}}=\frac{\rho_{c}-\rho_{w}}{\rho_{c}}_{(\mathrm{ANS})}\)