An object is taken 1.0 , km deep in the sea. | vedclass

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(B) The bulk modulus $B$ is defined as $B = -\frac{P}{\Delta V / V}$.Since the mass $M = \rho V$ is constant,differentiating gives $\rho \Delta V + V

Vedclass · Accepted Answer

$0.64$

Vedclass · Answer

(B) The bulk modulus $B$ is defined as $B = -\frac{P}{\Delta V / V}$.Since the mass $M = ho V$ is constant,differentiating gives $ho \Delta V + V \Delta ho = 0$,which implies $\frac{\Delta ho}{ho} = -\frac{\Delta V}{V}$.Substituting this into the bulk modulus formula,we get $\frac{\Delta ho}{ho} = \frac{P}{B}$.The pressure at depth $h$ is $P = ho_{sea} g h$.Given $ho_{sea} = 1.025 	imes 10^3 \, kg/m^3$,$g = 10 \, m/s^2$,$h = 1000 \, m$,and $B = 1.6 	imes 10^9 \, Pa$ $(1.6 	imes 10^6 \, kPa = 1.6 	imes 10^9 \, Pa)$.$\frac{\Delta ho}{ho} = \frac{1.025 	imes 10^3 	imes 10 	imes 1000}{1.6 	imes 10^9} = \frac{1.025 	imes 10^7}{1.6 	imes 10^9} = 0.6406 	imes 10^{-2} \approx 0.0064$.Percentage change $= \frac{\Delta ho}{ho} 	imes 100 = 0.64 \%$.

An object is taken $1.0 \, km$ deep in the sea. The density of sea water is $1.025 \times 10^3 \, kg/m^3$ and the bulk modulus of the object is $1.6 \times 10^6 \, kPa$. Find the percentage change in the density of the object.

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