The fraction of a floating object on volume ${V_0}$ and density ${d_0}$ above the surface of liquid of density d will be:
A) $\dfrac{{{d_0}}}{{d - {d_0}}}$
B) $\dfrac{{d - {d_0}}}{d}$
C) $\dfrac{{{d_0}}}{d}$
D) $\dfrac{{{d_0}d}}{{d + {d_0}}}$
Answer
259.5k+ views
Hint: Recall that the buoyant force is the force exerted by fluid on an object. It points in an upward direction. The ability of the liquids or fluids to exert this force is known as buoyancy. When an object is placed in a liquid, it exerts pressure in all directions.
Complete step by step solution:
When an object floats, then buoyant force acts on the object. The buoyant force of the liquid will have an opposite direction to that of the floating object. But the magnitude of the forces will be the same.
Let the volume of the object is $ = V$
The fraction of the floating object is $ = {V_0}$
If the object is floating, then as per the condition of buoyancy
$ \Rightarrow $Buoyant force by immersed object = Weight of the object
$ \Rightarrow dVg = {d_0}{V_0}g$---(i)
Where $d$ is the density of the object and ${d_0}$is the density above the surface of liquid
The equation (i) can be written as
$ \Rightarrow V = \dfrac{{{d_0}{V_0}}}{d}$
Let $V'$ is the volume of the object outside water. It can be written as
$V' = {V_0} - V$
Substituting the value of V in the above equation,
$ \Rightarrow V' = {V_0} - \dfrac{{{d_0}{V_0}}}{d}$
$V' = {V_0}(1 - \dfrac{{{d_0}}}{d})$
The fraction of the object outside water to the total volume of the object can be calculated as:
$ \Rightarrow \dfrac{{V'}}{{{V_0}}} = \dfrac{{{V_0}(1 - \dfrac{{{d_0}}}{d})}}{{{V_0}}}$
$ \Rightarrow \dfrac{{V'}}{{{V_0}}} = 1 - \dfrac{{{d_0}}}{d}$
$ \Rightarrow \dfrac{{V'}}{{{V_0}}} = \dfrac{{d - {d_0}}}{d}$
Option B is the right answer.
Note: It is important to remember that the density is related to the buoyancy. It decides whether the object will float or sink. Density is defined as the ratio of mass and volume. An object that has more density than water, will sink. But an object that has less density than that of water will float.
Complete step by step solution:
When an object floats, then buoyant force acts on the object. The buoyant force of the liquid will have an opposite direction to that of the floating object. But the magnitude of the forces will be the same.
Let the volume of the object is $ = V$
The fraction of the floating object is $ = {V_0}$
If the object is floating, then as per the condition of buoyancy
$ \Rightarrow $Buoyant force by immersed object = Weight of the object
$ \Rightarrow dVg = {d_0}{V_0}g$---(i)
Where $d$ is the density of the object and ${d_0}$is the density above the surface of liquid
The equation (i) can be written as
$ \Rightarrow V = \dfrac{{{d_0}{V_0}}}{d}$
Let $V'$ is the volume of the object outside water. It can be written as
$V' = {V_0} - V$
Substituting the value of V in the above equation,
$ \Rightarrow V' = {V_0} - \dfrac{{{d_0}{V_0}}}{d}$
$V' = {V_0}(1 - \dfrac{{{d_0}}}{d})$
The fraction of the object outside water to the total volume of the object can be calculated as:
$ \Rightarrow \dfrac{{V'}}{{{V_0}}} = \dfrac{{{V_0}(1 - \dfrac{{{d_0}}}{d})}}{{{V_0}}}$
$ \Rightarrow \dfrac{{V'}}{{{V_0}}} = 1 - \dfrac{{{d_0}}}{d}$
$ \Rightarrow \dfrac{{V'}}{{{V_0}}} = \dfrac{{d - {d_0}}}{d}$
Option B is the right answer.
Note: It is important to remember that the density is related to the buoyancy. It decides whether the object will float or sink. Density is defined as the ratio of mass and volume. An object that has more density than water, will sink. But an object that has less density than that of water will float.
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