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(B) Given that:Bulk modulus $(K) = 2 	imes 10^8 \ N m^{-2}$Volume strain $\frac{\Delta V}{V} = 1\% = 0.01$The formula for Bulk modulus is:$K = \frac{

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$2 	imes 10^6 \ N m^{-2}$

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(B) Given that:Bulk modulus $(K) = 2 	imes 10^8 \ N m^{-2}$Volume strain $\frac{\Delta V}{V} = 1\% = 0.01$The formula for Bulk modulus is:$K = \frac{p}{\frac{\Delta V}{V}}$Rearranging to solve for pressure $(p)$:$p = K 	imes \left( \frac{\Delta V}{V} ight)$Substituting the given values:$p = (2 	imes 10^8) 	imes (0.01)$$p = 2 	imes 10^6 \ N m^{-2}$Therefore,the required pressure is $2 	imes 10^6 \ N m^{-2}$.

What is the pressure required to reduce the given volume of water by $1 \%$? (Bulk modulus $(K) = 2 \times 10^8 \ N m^{-2}$)

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