The increase in the pressure required to decr | vedclass

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(B) The bulk modulus $B$ is defined as the ratio of the change in pressure $\Delta P$ to the volumetric strain $-\frac{\Delta V}{V}$.
Mathematically,

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$3$

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(B) The bulk modulus $B$ is defined as the ratio of the change in pressure $\Delta P$ to the volumetric strain $-\frac{\Delta V}{V}$.
Mathematically, $B = -\frac{\Delta P}{\Delta V/V}$.
Rearranging the formula to find the fractional change in volume, we get $\frac{\Delta V}{V} = \frac{\Delta P}{B}$.
Given $\Delta P = 6.3 	imes 10^7 	ext{ N/m}^2$ and $B = 2.1 	imes 10^9 	ext{ N/m}^2$.
Substituting the values: $\frac{\Delta V}{V} = \frac{6.3 	imes 10^7}{2.1 	imes 10^9} = 3 	imes 10^{-2} = 0.03$.
To find the percentage decrease, multiply by $100$: $0.03 	imes 100 = 3\%$.

The increase in the pressure required to decrease the volume $(\Delta V)$ of water is $6.3 \times 10^7 \text{ N/m}^2$. The percentage decrease in the volume is . . . . . . . (Bulk modulus of water = $2.1 \times 10^9 \text{ N/m}^2$.) (in $\%$)

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