At 20 oC the densities of fresh water and ethyl alcohol are, respectively, 998 and 789 kg/m3. Find the ratio of the adiabatic bulk modulus of fresh water to the adiabatic bulk modulus of ethyl alcohol at 20 oC.

Answer :

Answer:

The ratio is  [tex]\frac{B_1}{B_2} = 1.265[/tex]

Explanation:

From the question we are told that

   The density of fresh water is  [tex]\rho__{f}} = 998 \ kg/m^3[/tex]

     The density of ethanol is  [tex]\rho_{e} = 789 \ kg /m^3[/tex]

Generally speed of a wave in a substance is mathematically represented as

    [tex]v = \sqrt{\frac{B}{\rho} }[/tex]

Here B is the adiabatic bulk modulus of the substance  while [tex]\rho[/tex]  is the density of the substance

So at constant wave speed

     [tex]\sqrt{\frac{B_1}{\rho_1} } = \sqrt{\frac{B_2}{\rho_2} }[/tex]

=>   [tex]\frac{B_1}{\rho_1} = \frac{B_2}{\rho_2}[/tex]

=>   [tex]B_1 \rho_2 = B_2\rho_1[/tex]    

=>  [tex]\frac{B_1}{B_2} = \frac{\rho_1}{\rho_2}[/tex]

Here  [tex]\rho_1 =\rho__{f}} = 998 \ kg/m^3[/tex]   and  [tex]\rho_2 = \rho_{e} = 789 \ kg /m^3[/tex]

So

  =>  [tex]\frac{B_1}{B_2} = \frac{998}{789}[/tex]    

  =>  [tex]\frac{B_1}{B_2} = 1.265[/tex]    

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