Question

A raft of mass $600 \mathrm{kg}$ floats in calm water with $7 \mathrm{cm}$ submerged. When a man stands on the raft, $8.4 \mathrm{cm}$ are submerged, the man's mass is :
(A) 30 kg
(B) 60 kg
(C) 90 kg
(D) 120 kg

Answer 1

Hint: We know that a body at rest in a fluid is acted upon by a force pushing upward called the buoyant force, which is equal to the weight of the fluid that the body displaces. If the body is completely submerged, the volume of fluid displaced is equal to the volume of the body. If the body is only partially submerged, the volume of the fluid displaced is equal to the volume of the part of the body that is submerged. Based on this concept we have to answer this question.

Complete step by step answer
It is given that:
Mass or m is 600 kg
Distance of h is 7 cm
The weight of the body is equal to the buoyant force.
This is given as:
$mg=\rho gh$
In the above expression we get that:
m is the mass, $\rho$is the density and h is the height.
Therefore, when we put the values, we get that:
$600\times g=\rho \times g\times 7........(1)$
When a body of mass m is placed on a raft the submerged raft changes 8.4 cm.
Hence, the expression is given as:
$(600+{{m}^{/}})g=\rho \times g\times 8.4........(ii)$
Now we have to divide I by II to get:
$\begin{align}
  & \dfrac{600g}{(600+{{m}^{/}})g}=\dfrac{\rho g\times 7}{\rho \times g\times 8.4} \\
 & \Rightarrow 5040=4200+7{{m}^{/}} \\
 & \Rightarrow m=120kg \\
\end{align}$

So the correct answer is option D.

Note: To answer such a question, it should be known to us that Archimedes' principle is very useful for calculating the volume of an object that does not have a regular shape. The oddly shaped object can be submerged, and the volume of the fluid displaced is equal to the volume of the object. It can also be used in calculating the density or specific gravity of an object.
Let us explain with the help of an example, for an object denser than water, the object can be weighed in air and then weighed when submerged in water. When the object is submerged, it weighs less because of the buoyant force pushing upward. The object's specific gravity is then the object's weight in air divided by how much weight the object loses when placed in water. But most importantly, the principle describes the behaviour of anybody in any fluid, whether it is a ship in water or a balloon in air.