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(A) Given:Water pressure at the bottom,$p = 1.1 	imes 10^{8} \; Pa$Initial volume of the steel ball,$V = 0.32 \; m^{3}$Bulk modulus of steel,$B = 1.6

Vedclass · Accepted Answer

$2.2 	imes 10^{-4} \; m^{3}$

Vedclass · Answer

(A) Given:Water pressure at the bottom,$p = 1.1 	imes 10^{8} \; Pa$Initial volume of the steel ball,$V = 0.32 \; m^{3}$Bulk modulus of steel,$B = 1.6 	imes 10^{11} \; N/m^{2}$The formula for Bulk modulus is given by $B = -\frac{p}{\Delta V / V}$,where $\Delta V$ is the change in volume.Rearranging for $\Delta V$,we get $\Delta V = \frac{p V}{B}$.Substituting the values:$\Delta V = \frac{(1.1 	imes 10^{8} \; Pa) 	imes (0.32 \; m^{3})}{1.6 	imes 10^{11} \; N/m^{2}}$$\Delta V = \frac{0.352 	imes 10^{8}}{1.6 	imes 10^{11}}$$\Delta V = 0.22 	imes 10^{-3} \; m^{3} = 2.2 	imes 10^{-4} \; m^{3}$.Thus,the change in volume of the ball is $2.2 	imes 10^{-4} \; m^{3}$.

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