91Ó°ÊÓ

When pressure acts on an object from all sides, its volume decreases by an amount directly proportional to the original volume (\({V_0}\)) and the change in the pressure\(\left( {\Delta P} \right)\), i.e., \(\Delta V = - \frac{1}{B}{V_0}\Delta P\).

Here, B is the constant of proportionality, termed the bulk modulus or Young’s modulus, and the minus sign indicates that volume decreases on increasing the pressure.

In this problem, when the container of alcohol is carried to the bottom of the sea, the water exerts pressure on the object in all directions.

The value of the bulk modulus of the water is\(B{\bf{ = 1}}{\bf{.0 \times 1}}{{\bf{0}}^{\bf{9}}}\;{\bf{N/}}{{\bf{m}}^{\bf{2}}}\).

The original volume of alcohol placed in a flexible container is as follows:

\(\begin{array}{c}{V_0} = 1000\;{\rm{c}}{{\rm{m}}^3}\\ = 1000 \times {10^{ - 6}}\;{{\rm{m}}^3}\\ = 1.0 \times {10^{ - 3}}\;{{\rm{m}}^3}\end{array}\)

Original pressure on the surface of the sea is equal to the atmospheric pressure, i.e.,\({P_0} = 1.0 \times {10^5}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\).

Pressure at the bottom of the sea is\(P = 2.6 \times {10^6}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\).

Let V be the volume of alcohol at the bottom of the sea such that the change in the volume is \(\Delta V = V - {V_0}\).

The change in volume of the alcohol when it is carried at the bottom of the sea is as follows:

\(\begin{array}{c}\Delta V = - \frac{1}{B}{V_0}\Delta P\\ = - \frac{1}{B}{V_0}\left( {P - {P_0}} \right)\\ = - \frac{1}{{\left( {1.0 \times {{10}^9}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}} \right)}} \times \left( {1.0 \times {{10}^{ - 3}}\;{{\rm{m}}^3}} \right) \times \left( {2.6 \times {{10}^6} - 1.0 \times {{10}^5}\;} \right){\rm{N/}}{{\rm{m}}^{\rm{2}}}\\ = - 2.5 \times {10^{ - 6}}\;{{\rm{m}}^3}\end{array}\)

Thus, the volume of alcohol will get decreased by \(2.5 \times {10^{ - 6}}\;{{\rm{m}}^3}\).

The change in volume of the alcohol is given by the following:

\(\begin{array}{c}\Delta V = V - {V_0}\\V = {V_0} + \Delta V\\ = 1.0 \times {10^{ - 3}}\;{{\rm{m}}^3} - 2.5 \times {10^{ - 6}}\;{{\rm{m}}^3}\\ = 997.5 \times {10^{ - 6}}\;{{\rm{m}}^3}\\ = 997.5\;{\rm{c}}{{\rm{m}}^3}\end{array}\)

Thus, the volume of the alcohol at the bottom of the sea is \(997.5\;{\rm{c}}{{\rm{m}}^3}\).