The fractional compression left(frac{Delta V} | vedclass

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(D) The pressure at a depth $h$ is given by $P = \rho gh$.The Bulk modulus $B$ is defined as $B = \frac{P}{\left(\frac{\Delta V}{V}\right)}$.Therefore

Vedclass · Accepted Answer

$1.25$

Vedclass · Answer

(D) The pressure at a depth $h$ is given by $P = ho gh$.The Bulk modulus $B$ is defined as $B = \frac{P}{\left(\frac{\Delta V}{V}ight)}$.Therefore,the fractional compression is $\frac{\Delta V}{V} = \frac{P}{B} = \frac{ho gh}{B}$.Substituting the given values: $ho = 10^3 \ kg m^{-3}$,$g = 10 \ m s^{-2}$,$h = 2.5 \ km = 2.5 	imes 10^3 \ m$,and $B = 2 	imes 10^9 \ N m^{-2}$.$\frac{\Delta V}{V} = \frac{10^3 	imes 10 	imes 2.5 	imes 10^3}{2 	imes 10^9} = \frac{2.5 	imes 10^7}{2 	imes 10^9} = 1.25 	imes 10^{-2}$.To express this as a percentage: $\frac{\Delta V}{V} 	imes 100 = 1.25 	imes 10^{-2} 	imes 100 = 1.25 \%$.

The fractional compression $\left(\frac{\Delta V}{V}\right)$ of water at a depth of $2.5 \ km$ below the sea level is . . . . . . $\%$. Given: the Bulk modulus of water $B = 2 \times 10^9 \ N m^{-2}$,density of water $\rho = 10^3 \ kg m^{-3}$,and acceleration due to gravity $g = 10 \ m s^{-2}$.

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