Question

Find the density of the object, if $\dfrac{2}{3}$ part of it is present outside the water in case of a floating situation.

Answer 1

Hint: In this question, we will use the concept of the Archimedes’ principle. For solving this problem first calculate the volume of the displaced water. Then apply Archimedes' principle and equate the weight of the body with the weight of displaced water.

Complete step by step solution:
In this question, we have given that \[\dfrac{2}{3}\] body part of an object is outside of water and \[\dfrac{1}{3}\] body part of the object is inside of water. So, we can find the volume of displaced water as,
\[ \Rightarrow {V_w} = \dfrac{1}{3}{V_b}\]
where, \[{V_b}\] is the volume of object
Now by applying the Floating condition that is the weight of the body is equal to the weight of water displaced.
Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.
Let us assume,
\[{\rho _b}\] is the density of object, \[{V_b}\]is the Volume of object, \[{\rho _w}\] is the density of water, and \[{V_w}\] is the Volume of water.

Now we will apply the Archimedes’ principle as
\[ \Rightarrow {\rho _b}{V_b}g = {\rho _w}{V_w}g\]
Now we will rearrange the equation and get,
\[ \Rightarrow {\rho _b} = \dfrac{{{\rho _w}{V_w}}}{{{V_b}}}\]
Now, we will substitute the values in the above equation and get,
\[ \Rightarrow {\rho _b} = 1000 \times \dfrac{1}{3}\]
After simplification we will get,
\[\therefore {\rho _b} = 333.33{\text{ }}kg{m^{ - 3}}{\text{ }}\]

Hence, the density of the object is \[333.33{\text{ }}kg{m^{ - 3}}\].

Note: As we know that Archimedes principle is used in designing ships and submarines. A submarine has a large ballast tank, which is used to control its position and depth from the surface of the sea. A submarine submerges by letting water into the ballast tank so that its weight becomes greater than the buoyant force. Conversely, it floats by reducing water in the ballast tank thus its weight is less than the buoyant force.