Question

An ornament weighing 36 grams in air weighs only 34 grams in water. Assume that some Copper is mixed with the gold to prepare the ornament, find the amount of copper in it. (The specific gravity of gold is 19.3 and that of copper is 8.9.)

Answer 1

Hint: Apply the Archimedes principle which says that the loss in weight is equal to the buoyancy force which is equal to the weight of the displaced liquid by the ornament. From the given values calculate the buoyancy force and equate to the weight of the volume of the displaced liquid.

Complete step-by-step answer:
The given weight of the ornament in Air is 36 g
Given the weight of the ornament in the water is: 34 g
The decrease in the weight of the ornament is: 2g
We are given the specific gravity of the gold and the specific gravity of the copper and let us consider the volume of the copper as $V_c$ and the volume of the gold as $V_g$ .

Now we know that the weight of the ornament in air is 36 g so we can write the equation as :
Since the weight of the body can be written in terms of volume as $W = V\rho g$
We get the required equation for the given weight in air as : $36 = V_c \rho_c g + V_g \rho_g g$
We get $m_c + m_g = 36$
Now we have obtained an equation in terms of masses of the gold and copper
Let us now calculate the weight of the displaced liquid. We know that the volume of the displaced liquid should be equal to the volume of the ornament in water. So we can write it the weight of the displaced liquid as
$W_w = (\dfrac{m_g}{\rho_g} + \dfrac{m_c}{\rho_c} ) \rho_w g$
We know that the weight of the displaced liquid should be equal to the difference in the weight of the ornament so we can write as:
$W_w = (\dfrac{m_g}{\rho_g} + \dfrac{m_c}{\rho_c} ) \rho_w g =2 g$
$(\dfrac{m_g}{19.3} + \dfrac{m_c}{8.9} ) \rho_w = 2$
Now we have found another equation in terms of masses of the gold and copper.
Solving the about two equations we get the mass of the gold and copper as: $m_g = 33.75 g$ and
$m_c = 2.225g$

Note: One of the major mistakes that we can do in this kind of problem is that we may not be able to find the weight of the displaced liquid as it is not given in the question but we need to consider the fact that the volume of the displaced liquid is equal to the volume of the ornament and use the density of the water to find the weight of the displaced liquid.