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(B) The Bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V / V}$.For water,the fractional change in volume is $\frac{\Delta V_w}{V_w} = 0.0

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$1$

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(B) The Bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V / V}$.For water,the fractional change in volume is $\frac{\Delta V_w}{V_w} = 0.01 \% = \frac{0.01}{100} = 10^{-4}$.Thus,$B_w = \frac{P}{10^{-4}} = 10^4 P$.For the liquid,the fractional change in volume is $\frac{\Delta V_l}{V_l} = 0.03 \% = \frac{0.03}{100} = 3 	imes 10^{-4}$.Thus,$B_l = \frac{P}{3 	imes 10^{-4}} = \frac{10^4 P}{3}$.The ratio of the Bulk modulus of water to that of the liquid is $\frac{B_w}{B_l} = \frac{10^4 P}{10^4 P / 3} = 3$.Given that the ratio is $\frac{3}{x}$,we have $\frac{3}{x} = 3$,which implies $x = 1$.

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