Question
Find the increase in pressure required to decrease the volume of a water sample by $0.01\%$. Bulk modulus of water $= 2.1 \times 10^9N/m^2.$
✓
Answer
Given:
Bulk modulus of water $B = 2.1 \times 10^9Nm^{-2}$
In order to decrease the volume $(V)$ of a water sample by $0.01\%$, let the increase in pressure be $P.$
$\frac{\text{v}\times0.1}{100}=\triangle\text{V}$
$\Rightarrow\frac{\triangle\text{V}}{\text{V}}=10^{-4}$
From B $=\frac{\text{P}\text{V}}{\triangle\text{V}},$ We have:
$\Rightarrow\text{P}= \text{B}\Big(\frac{\triangle\text{V}}{\text{V}}\Big)$
$=2.1\times10^{9}\times10^{-4}$
$=2.1\times10^{5}\text{N}/\text{m}^2$
Hence, the requred increase in pressure is $2.1 \times 10^5Nm^{-2}.$
Bulk modulus of water $B = 2.1 \times 10^9Nm^{-2}$
In order to decrease the volume $(V)$ of a water sample by $0.01\%$, let the increase in pressure be $P.$
$\frac{\text{v}\times0.1}{100}=\triangle\text{V}$
$\Rightarrow\frac{\triangle\text{V}}{\text{V}}=10^{-4}$
From B $=\frac{\text{P}\text{V}}{\triangle\text{V}},$ We have:
$\Rightarrow\text{P}= \text{B}\Big(\frac{\triangle\text{V}}{\text{V}}\Big)$
$=2.1\times10^{9}\times10^{-4}$
$=2.1\times10^{5}\text{N}/\text{m}^2$
Hence, the requred increase in pressure is $2.1 \times 10^5Nm^{-2}.$
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