Question

A block of ice $0.9gmc{{c}^{-1}}$ and volume of 100cc flows in water. Their volume inside the water is
a) 10 cc
b) 90 cc
c) 50 cc
d) 9 cc

Answer 1

Hint: It is given in the question that the ice cube flows above the water surface such that some portion of it lies in water whereas some lies above the surface. When an object floats on the surface of the liquid we can say that the density of the object is less than the density of water. When an object floats on water its gravitational force is equal to its buoyant force. Therefore by this condition we will obtain an expression, using which we will determine the volume beneath the water.
Formula used:
$\dfrac{{{V}_{L}}}{{{V}_{O}}}=\dfrac{{{\rho }_{O}}}{{{\rho }_{L}}}$

Complete answer:
Let us say an object i.e. ice has volume ${{V}_{O}}$and density ${{\rho }_{O}}$ flows on the surface of the water. Lets the density of water be ${{\rho }_{L}}=1$ and the volume of the displaced water when the ice was immersed in water be ${{V}_{L}}$. The ice flows on the surface of the water. Hence we can say its gravitational force is equal to its buoyant force. Hence we can write,
$\text{Buoyant force=Weight of the solid}$
The buoyant force on an object is given as the product of the density of water times the acceleration due to gravity(g) times the volume of the object inside water which equals the volume of the displaced liquid. Hence the above equation can be written as,
$\begin{align}
  & \text{Buoyant force=Weight of the solid} \\
 & {{\rho }_{L}}g{{V}_{L}}=mg\text{, where m is the mass of the object} \\
\end{align}$
The mass of an object is the product of density times the volume, hence we can write the mass of ice as in the above equation as,
$\begin{align}
  & {{\rho }_{L}}g{{V}_{L}}={{\rho }_{O}}{{V}_{O}}g \\
 & \Rightarrow {{\rho }_{L}}{{V}_{L}}={{\rho }_{O}}{{V}_{O}} \\
 & \Rightarrow \dfrac{{{V}_{L}}}{{{V}_{O}}}=\dfrac{{{\rho }_{O}}}{{{\rho }_{L}}} \\
 & \Rightarrow {{V}_{L}}=\dfrac{{{\rho }_{O}}}{{{\rho }_{L}}}{{V}_{O}} \\
 & \Rightarrow {{V}_{L}}=\dfrac{0.9gc{{c}^{-1}}}{1gc{{c}^{-1}}}100cc \\
 & \Rightarrow {{V}_{L}}=90cc \\
\end{align}$
Hence the volume of the ice inside the surface of the water is 90cc.

Hence the correct answer of the above question is option b.

Note:
The fraction i.e. $\dfrac{{{V}_{L}}}{{{V}_{O}}}$ represents the fraction of ice below the surface of the water. Hence the fraction of ice above the surface of the water will be equal to $1-\dfrac{{{V}_{L}}}{{{V}_{O}}}$. This is also be expressed as $1-\dfrac{{{\rho }_{O}}}{{{\rho }_{L}}}$.From this we can conclude that if we know the density of the object and the density of the liquid in which the solid is immersed or the volume of the displaced liquid and the volume of the solid, we can determine the volume of the solid below and above the surface of the water.