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

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Accepted Answer

"

The bulk modulus of a material is defined as:

$ B = \frac{-\Delta P}{\left(\frac{\Delta V}{V}\right)} $

Given:

Bulk Modulus, $ B = 2.15 \times 10^9 \, \text{Nm}^{-2} $

Change in volume percentage, $ \frac{\Delta V}{V} = 0.2\% = \frac{0.2}{100} = 0.002 $

To find the required increase in pressure, $ \Delta P $, we rearrange the formula for the bulk modulus:

$ B = \frac{-\Delta P}{- \left(\frac{\Delta V}{V}\right)} $

Plug in the known values:

$ 2.15 \times 10^9 = \frac{\Delta P}{0.002} $

Solving for $ \Delta P $:

$ \Delta P = 2.15 \times 10^9 \times 0.002 $

$ \Delta P = 4.3 \times 10^6 \, \text{Nm}^{-2} $

Since the problem specifies $ \Delta P = P \times 10^5 \, \text{Nm}^{-2} $, we have:

$ 4.3 \times 10^6 = P \times 10^5 $

$ P = 43 $

The value of $ P $ is therefore 43.

"

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

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

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