Question

An ornament weighing 36g in air weighs only 34g in water. Assuming that some copper is mixed with the gold to prepare the ornament, find the amount of copper in it. Specific gravity of gold is 19.3 and that of copper is 8.9
2.2 g
4.4 g
1.1 g
3.6 g

Answer 1

Hint: When the ornament is immersed in water, the change in its weight should be equal to the weight of the water displaced. Additionally, the real mass of the ornament will be the addition of mass of gold present in it and copper. Using all this prior knowledge, we can form the following equations and solve for the required answer.
Formula Used:
Loss in weight $ = {\text{ }}{W_{air}} - {W_{water}}$
Weight in water, $W = V \times {\rho _w} \times g = 2 \times g$
Real weight, ${m_{real}} = {m_{Au}} + {m_{Cu}}$

Complete step by step answer:
Given:
The weight of gold ornament in air, ${W_{air}} = 36g$
The weight of gold ornament in water, ${W_{air}} = 34g$
The specific gravity of gold, ${\rho _{Au}} = 19.3g/cc$
The specific gravity of copper, ${\rho _{Cu}} = 8.9g/cc$
The loss of weight of gold when it gets immersed in water is because of the force of buoyancy acting on it in a vertically upward direction due to its weight. Mathematically, this can be written as:
Loss of weight = Weight of displaced water = Weight in air – weight in water
$\eqalign{
  & {\text{Loss of weight }} = {\text{ }}{W_{air}} - {W_{water}} \cr
  & {\text{Loss of weight }} = 36 - 34 = 2g \cr} $
Now, let V represent the volume of the ornament in water. So the weight of the ornament in water is equal to the weight of the displaced water, as these are two forces which must be equal in magnitude and opposite in direction
$ \Rightarrow W = V \times {\rho _w} \times g = 2 \times g \cdots \cdots \cdots \left( 1 \right)$
But we know that,
$\eqalign{
  & {\text{Density, }}\rho = \dfrac{{{\text{Mass, }}m}}{{{\text{Volume, }}V}} \cr
  & \Rightarrow V = \dfrac{m}{\rho } \cr} $
Using the above relation in equation (1), we get:
$\eqalign{
  & \Rightarrow \left( {\dfrac{{{m_{Au}}}}{{{\rho _{Au}}}} + \dfrac{{{m_{Cu}}}}{{{\rho _{Cu}}}}} \right){\rho _w} \times g = 2 \times g \cr
  & \Rightarrow {m_{Au}}{\rho _{Cu}} + {m_{Cu}}{\rho _{Au}} = 2{\rho _{Au}}{\rho _{Cu}}{\text{ }}\left[ {\because {\rho _w} = 1} \right] \cr
  & \Rightarrow 8.9{m_{Au}} + 19.3{m_{Cu}} = 2 \times 8.9 \times 119.3 \cr
  & \Rightarrow 8.9{m_{Au}} + 19.3{m_{Cu}} = 343.54 \cdots \cdots \cdots \left( 2 \right) \cr} $
But the mass of gold in this equation is actually the real mass of the ornament which is a mixture of gold and copper.
$\eqalign{
  & {m_{real}} = {m_{Au}} + {m_{Cu}} = 36g \cr
  & \Rightarrow {m_{Au}} = {m_{real}} - {m_{Cu}} \cr} $
Substituting this value back in equation (2), we get:
$\eqalign{
  & 8.9\left( {{m_{Au}} - {m_{Cu}}} \right) + 19.3{m_{Cu}} = 343.54 \cr
  & \Rightarrow 8.9 \times 36 - 8.9{m_{Cu}} + 19.3{m_{Cu}} = 343.54 \cr
  & \Rightarrow 320.4 + 10.4{m_{Cu}} = 343.54 \cr
  & \Rightarrow 10.4{m_{Cu}} = 343.54 - 320.4 \cr
  & \Rightarrow 10.4{m_{Cu}} = 23.14 \cr
  & \Rightarrow {m_{Cu}} = \dfrac{{23.14}}{{10.3}} \cr
  & \therefore {m_{Cu}} = 2.2g \cr} $
Therefore, the correct answer is A i.e., 2.2 g

Note: When a body is partially or fully dipped into a fluid at rest, the fluid exerts an upward force of buoyancy equal to the weight of the displaced fluid, this is Archimedes principle. This upwards force is also known as buoyancy or buoyant force.