Question

A boat floating in fresh water displaces water weighing 35.2kg. The change in volume of displaced water if it were floating in sea water of density $1.1\times {{10}^{3}}kg{{m}^{-3}}$ will be
$\begin{align}
  & a)21 \\
 & b)35.2 \\
 & c)3.2 \\
 & d)0.33 \\
\end{align}$

Answer 1

Hint: As we were told the weight, density of the boat while it is floating in freshwater, let us calculate the volume displaced by the boat in freshwater. Next, let's calculate the volume of water displaced by the boat when it is placed in salt water. The weight doesn’t change. Finally, subtract the volumes of the water displaced when placed in two different liquids.
Formula used:
$\rho =\dfrac{m}{V}$

Complete answer:
We know the weight of the boat, which is always constant and doesn’t get affected by the change in the water. The density of freshwater is given,
The volume displaced by the boat in fresh water is calculated as,
$\begin{align}
  & \rho =\dfrac{m}{V} \\
 & \Rightarrow 1000=\dfrac{35.2\times {{10}^{3}}}{V} \\
 & \Rightarrow V=\dfrac{35.2\times {{10}^{3}}}{1000} \\
 & \Rightarrow V=35.2{{m}^{3}} \\
\end{align}$
Now, let us calculate the volume displaced by the boat when it is placed in salt water whose density is different from that of fresh water. The weight of the boat doesn’t change.
$\begin{align}
  & {{V}^{1}}=\dfrac{m}{\rho } \\
 & \Rightarrow {{V}^{1}}=\dfrac{35.2\times {{10}^{3}}}{1.1\times {{10}^{3}}} \\
 & \Rightarrow {{V}^{1}}=32{{m}^{3}} \\
\end{align}$
Now, we know the volume of liquids replaced by the boat when placed in fresh water and salt water respectively,
Therefore, the difference will be,
$\begin{align}
  & V-{{V}^{1}}=35.2-32{{m}^{3}} \\
 & V-{{V}^{1}}=3.2{{m}^{3}} \\
\end{align}$

So, the correct answer is “Option C”.

Additional Information:
Archimedes principle discovered by Greek mathematician Archimedes, states that anybody which is completely or partially submerged in any fluid at rest is acted by an upward force called buoyant force. This buoyant force is equal to the magnitude of the weight of the fluid displaced by the body. The volume of displaced fluid is equivalent to the volume of an object fully immersed in a fluid or to that fraction of the volume below the surface for an object partially submerged in a liquid. The weight of the displaced portion of the fluid is equivalent to the magnitude of the buoyant force. The buoyant force on a body floating in a liquid or gas is also equivalent in magnitude to the weight of the floating object and is opposite in direction; the object neither rises nor sinks.

Note:
While calculating the buoyancy force, one must take the density of the water, volume of the body in the formula. Many of us mistake by taking the density as the density of the body. Weight acting downwards includes the density of the body. As we are calculating the volume of liquid displaced, density will be the density of the liquid but not the body we use.