Question

A body of density d and volume V floats with volume \[{{V}_{1}}\]of it immersed in a liquid of density \[{{d}_{1}}\]and rest of the volume \[{{V}_{2}}\] immersed in another liquid of density \[({{d}_{2}}<{{d}_{1}})\]then \[{{V}_{1}}\]is,
\[\begin{align}
  & A.\,\left( \dfrac{d-{{d}_{2}}}{{{d}_{1}}-{{d}_{2}}} \right)V \\
 & B.\,\left( \dfrac{d+{{d}_{2}}}{{{d}_{1}}+{{d}_{2}}} \right)V \\
 & C.\,\left( \dfrac{{{d}_{1}}-{{d}_{2}}}{d-{{d}_{2}}} \right)V \\
 & D.\,\left( \dfrac{{{d}_{1}}+{{d}_{2}}}{d+{{d}_{2}}} \right)V \\
\end{align}\]

Answer 1

Hint: The question can be solved using the Archimedes principle. Archimedes principle states that the weight of the displaced portion of the fluid is equivalent to the buoyant force. So, we will equate the weight of the floating body to the weight of the water displaced to find the value of the volume.
Formula used:
\[I={{I}_{0}}\left( 1-{{e}^{-\left( \dfrac{Rt}{L} \right)}} \right)\]

Complete step-by-step solution:
From the given information, we have the data as follows.
A body of density d and volume V floats with volume \[{{V}_{1}}\] of it immersed in a liquid of density \[{{d}_{1}}\] and rest of the volume \[{{V}_{2}}\] immersed in another liquid of density \[({{d}_{2}}<{{d}_{1}})\]
Archimedes principle states that the weight of the displaced portion of the fluid is equivalent to the buoyant force. This buoyant force on the body floating in a fluid or in gas is also equivalent to the weight of the floating body in terms of the magnitude. So, we have,
\[Vd={{V}_{1}}{{d}_{1}}+{{V}_{2}}{{d}_{2}}\] ….. (1)
From the given information, we have,
\[\begin{align}
  & {{V}_{1}}+{{V}_{2}}=V \\
 & \therefore {{V}_{2}}=V-{{V}_{1}} \\
\end{align}\]…… (2)
Substitute the equation (2) in (1)
\[\begin{align}
  & Vd={{V}_{1}}{{d}_{1}}+V{{d}_{2}}-{{V}_{1}}d \\
 & \therefore {{V}_{1}}({{d}_{1}}-{{d}_{2}})=Vd-V{{d}_{2}} \\
\end{align}\]
Rearrange the terms of the above equation in terms of volume, so, we get,
\[\begin{align}
  & {{V}_{1}}=\dfrac{V(d-{{d}_{2}})}{{{d}_{1}}-{{d}_{2}}} \\
 & \therefore {{V}_{1}}=\left( \dfrac{d-{{d}_{2}}}{{{d}_{1}}-{{d}_{2}}} \right)V \\
\end{align}\]
\[\therefore \] A body of density d and volume V floats with volume \[{{V}_{1}}\] of it immersed in a liquid of density \[{{d}_{1}}\]and rest of the volume \[{{V}_{2}}\] immersed in another liquid of density \[({{d}_{2}}<{{d}_{1}})\] then \[{{V}_{1}}\] is,
\[\,\left( \dfrac{d-{{d}_{2}}}{{{d}_{1}}-{{d}_{2}}} \right)V\], thus, option (A) is correct.

Note: Archimedes principle states that the weight of the displaced portion of the fluid is equivalent to the buoyant force. This buoyant force on the body floating in a fluid or in gas is also equivalent to the weight of the floating body in terms of the magnitude.