The increase in pressure required to decrease | vedclass

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(D) The bulk modulus $B$ is defined as the ratio of the change in pressure to the volumetric strain:$B = -\frac{\Delta P}{\Delta V / V}$Given:Bulk mod

Vedclass · Accepted Answer

$43$

Vedclass · Answer

(D) The bulk modulus $B$ is defined as the ratio of the change in pressure to the volumetric strain:$B = -\frac{\Delta P}{\Delta V / V}$Given:Bulk modulus $B = 2.15 	imes 10^9 	ext{ Nm}^{-2}$Volumetric strain $\frac{\Delta V}{V} = -0.2\% = -0.002$Substituting the values into the formula:$2.15 	imes 10^9 = -\frac{\Delta P}{-0.002}$$\Delta P = 2.15 	imes 10^9 	imes 0.002$$\Delta P = 2.15 	imes 10^9 	imes 2 	imes 10^{-3}$$\Delta P = 4.3 	imes 10^6 	ext{ Nm}^{-2}$We need to express this in the form $	ext{P} 	imes 10^5 	ext{ Nm}^{-2}$:$4.3 	imes 10^6 = 43 	imes 10^5 	ext{ Nm}^{-2}$Comparing this with $	ext{P} 	imes 10^5 	ext{ Nm}^{-2}$,we get $	ext{P} = 43$.

The increase in pressure required to decrease the volume of a water sample by $0.2 \%$ is $\text{P} \times 10^5 \text{Nm}^{-2}$. Bulk modulus of water is $2.15 \times 10^9 \text{Nm}^{-2}$. The value of $\text{P}$ is . . . . . . .

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