Answer
The average density of the coin is $19,100~kg/m^3$
Since this density is very close to the density of gold ($\rho = 19,300~kg/m^3$), it is quite possible that the coin is made of gold.
Work Step by Step
We can find the buoyant force exerted on the coin:
$F_b = (0.0497~kg)(9.80~m/s^2)-(0.0471~kg)(9.80~m/s^2)$
$F_b = 0.02548~N$
According to Archimedes' principle, the buoyant force is equal to the weight of the water that is displaced. We can find the volume of the water that is displaced. Note that this volume is also the volume of the coin:
$m_w~g = F_b$
$\rho_w~V~g = F_b$
$V = \frac{F_b}{\rho_w~g}$
$V = \frac{0.02548~N}{(1000~kg/m^3)(9.80~m/s^2)}$
$V = 2.6\times 10^{-6}~m^3$
We can find the density of the coin:
$\rho = \frac{m}{V} = \frac{0.0497~kg}{2.6\times 10^{-6}~m^3} = 19,100~kg/m^3$
The average density of the coin is $19,100~kg/m^3$
Eureka! Since this density is very close to the density of gold ($\rho = 19,300~kg/m^3$), it is quite possible that the coin is made of gold.
