Bulk Modulus Questions in English

(A) The bulk modulus $(B)$ is defined as the ratio of volumetric stress to volumetric strain,given by $B = -\Delta P / (\Delta V / V)$.
For a given change in pressure $(\Delta P)$,gases undergo a much larger change in volume $(\Delta V)$ compared to solids and liquids because gas molecules are loosely packed.
Since $B$ is inversely proportional to the fractional change in volume $(\Delta V / V)$,a large $\Delta V$ results in a very small value of $B$.
Therefore,gases have the minimum volume elasticity compared to solids and liquids.

(C) For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = \text{constant}$.
Differentiating both sides with respect to $V$,we get $P(\gamma V^{\gamma-1}) + V^{\gamma} \frac{dP}{dV} = 0$.
Dividing by $V^{\gamma-1}$,we get $P\gamma + V \frac{dP}{dV} = 0$.
Rearranging the terms,we find $-V \frac{dP}{dV} = \gamma P$.
The adiabatic elasticity $K_a$ is defined as the bulk modulus for an adiabatic process,given by $K_a = -V \frac{dP}{dV}$.
Therefore,$K_a = \gamma P$.

(D) Fluids (liquids and gases) cannot sustain shearing stress,which means they have no resistance to change in shape. Therefore,Young's modulus,shear modulus,and modulus of rigidity are not applicable to fluids.
Fluids can only resist a change in volume when subjected to uniform pressure from all sides. This property is described by the Bulk modulus $(B)$,which is defined as the ratio of volumetric stress to volumetric strain.
$B = -\frac{\Delta P}{\Delta V / V}$
Thus,the only elastic modulus applicable to fluids is the Bulk modulus.

(A) The compressibility $C$ is defined as the reciprocal of the Bulk Modulus $K$,given by $C = \frac{1}{K} = \frac{\Delta V / V}{\Delta P}$.
Rearranging the formula to find the change in volume $\Delta V$,we get $\Delta V = C \times \Delta P \times V$.
Given:
Compressibility $C = 4 \times 10^{-5} \ atm^{-1}$,
Initial volume $V = 100 \ cm^3$,
Change in pressure $\Delta P = 100 \ atm$.
Substituting these values into the equation:
$\Delta V = (4 \times 10^{-5}) \times 100 \times 100$
$\Delta V = 4 \times 10^{-5} \times 10^4$
$\Delta V = 4 \times 10^{-1} = 0.4 \ cm^3$.
Therefore,the decrease in volume is $0.4 \ cm^3$.

(D) The bulk modulus $K$ is defined as the ratio of the change in pressure $\Delta P$ to the volumetric strain $\Delta V/V$.
Given:
Depth $h = 200 \ m$
Density of water $\rho = 1 \times 10^3 \ kg/m^3$
Acceleration due to gravity $g = 10 \ m/s^2$
Volumetric strain $\frac{\Delta V}{V} = 0.1 \% = \frac{0.1}{100} = 10^{-3}$
The change in pressure at depth $h$ is given by $\Delta P = h \rho g$.
$\Delta P = 200 \times 10^3 \times 10 = 2 \times 10^6 \ N/m^2$.
Now,calculate the bulk modulus $K$:
$K = \frac{\Delta P}{\Delta V/V} = \frac{2 \times 10^6}{10^{-3}} = 2 \times 10^9 \ N/m^2$.
Therefore,the correct option is $D$.

(C) The Bulk Modulus $(K)$ is defined as the ratio of change in pressure to the volumetric strain: $K = -\frac{\Delta P}{\Delta V/V}$.
Compressibility is defined as the reciprocal of the Bulk Modulus.
Therefore,Compressibility $= \frac{1}{K} = -\frac{\Delta V/V}{\Delta P}$.
This represents the fractional change in volume per unit change in pressure.

(C) The bulk modulus $K$ is defined as $K = -\frac{\Delta P}{\Delta V / V}$.
Given,pressure change $\Delta P = 100 \text{ atm}$.
Since $1 \text{ atm} \approx 1.013 \times 10^6 \text{ dyne/cm}^2$,we can approximate $1 \text{ atm} \approx 10^6 \text{ dyne/cm}^2$.
Thus,$\Delta P = 100 \times 10^6 \text{ dyne/cm}^2 = 10^8 \text{ dyne/cm}^2$.
The fractional change in volume is $\frac{\Delta V}{V} = 0.01\% = \frac{0.01}{100} = 10^{-4}$.
Substituting these values into the formula:
$K = \frac{10^8}{10^{-4}} = 10^{12} \text{ dyne/cm}^2$.
Therefore,the correct option is $C$.

(B) The elastic property applicable to all three states of matter (solid,liquid,and gas) is the Bulk modulus ($B$ or $K$).
Bulk modulus is defined as the ratio of volumetric stress to volumetric strain.
It is given by the formula: $B = -\frac{dp}{dv/v}$,where $dp$ is the change in pressure and $dv/v$ is the volumetric strain.
Since liquids and gases can only resist changes in volume under pressure,they possess a Bulk modulus,whereas Young's modulus and Modulus of rigidity are only applicable to solids.

(C) The concept of Bulk modulus,which describes the resistance of a substance to uniform compression,was first defined and introduced by the physicist James Clerk Maxwell.

(A) The bulk modulus $B$ is given by the formula $B = \frac{\Delta p}{\Delta V/V}$.
Here,the change in pressure $\Delta p$ at depth $h$ is given by $\Delta p = h \rho g$.
Given: $h = 200 \ m$,$\rho = 10^3 \ kg/m^3$ (density of water),$g = 9.8 \ m/s^2$,and $\frac{\Delta V}{V} = 0.1\% = \frac{0.1}{100} = 10^{-3}$.
Substituting the values:
$B = \frac{200 \times 10^3 \times 9.8}{10^{-3}}$
$B = \frac{19.6 \times 10^5}{10^{-3}}$
$B = 19.6 \times 10^8 \ N/m^2$.

(C) The isothermal bulk modulus $(K_i)$ of an ideal gas is defined as the pressure $(P)$ of the gas.
For a gas at atmospheric pressure,the pressure is $P = 1\,atm$.
We know that $1\,atm = 1.013 \times 10^5\,N/m^2$.
Therefore,the isothermal bulk modulus is $K_i = 1.013 \times 10^5\,N/m^2$.

(C) For an ideal gas undergoing an isothermal process, the equation of state is $pV = \text{constant}$.
Differentiating both sides with respect to $V$, we get $p + V(dp/dV) = 0$.
This implies $p = -V(dp/dV)$.
The bulk modulus $B$ is defined as $B = -V(dp/dV)$.
Substituting the expression, we find $B = p$.
Therefore, the isothermal bulk modulus of an ideal gas is equal to its pressure $p$.

(D) The Bulk modulus $(K)$ is defined as the ratio of the change in pressure to the fractional change in volume: $K = -\frac{\Delta p}{\Delta V/V}$.
Given,initial pressure $P_1 = 1.01 \times 10^5 \ Pa$ and final pressure $P_2 = 1.165 \times 10^5 \ Pa$.
The change in pressure is $\Delta p = P_2 - P_1 = (1.165 - 1.01) \times 10^5 \ Pa = 0.155 \times 10^5 \ Pa$.
The fractional change in volume is $\frac{\Delta V}{V} = 10\% = 0.1$.
Substituting these values into the formula:
$K = \frac{0.155 \times 10^5}{0.1} = 1.55 \times 10^5 \ Pa$.

(B) The bulk modulus $B$ is defined as $B = -V_0 \frac{\Delta p}{\Delta V}$,which implies $\Delta V = -V_0 \frac{\Delta p}{B}$.
Since $V = V_0 + \Delta V$,we have $V = V_0 \left[ 1 - \frac{\Delta p}{B} \right]$.
Density $\rho = \frac{m}{V} = \frac{m}{V_0 [1 - \frac{\Delta p}{B}]} = \rho_0 [1 - \frac{\Delta p}{B}]^{-1}$.
Using the binomial approximation $(1 - x)^{-1} \approx 1 + x$ for small $x$,we get $\rho = \rho_0 [1 + \frac{\Delta p}{B}]$.
At depth $y$,the pressure difference is $\Delta p = \rho_0 gy$.
Substituting this,we get $\rho = \rho_0 [1 + \frac{\rho_0 gy}{B}]$.

(A) The bulk modulus $B$ is defined as $B = -V \frac{dP}{dV}$.
For an isothermal process,the equation of state for an ideal gas is $PV = \text{constant}$.
Differentiating both sides with respect to $V$,we get $P + V \frac{dP}{dV} = 0$.
This implies $P = -V \frac{dP}{dV}$.
Therefore,the isothermal bulk modulus $B_T = P$.

(A) For an isothermal process, the equation of state is $PV = \text{constant}$.
Differentiating both sides with respect to $V$, we get $P + V(dP/dV) = 0$, which implies $P = -V(dP/dV)$.
The isothermal bulk modulus $E_{\theta}$ is defined as $E_{\theta} = -V(dP/dV)$.
Therefore, $E_{\theta} = P$.
At normal pressure, $P = 1.013 \times 10^5 \ N/m^2$.
Thus, the isothermal bulk modulus is $1.013 \times 10^5 \ N/m^2$.

(C) Given: Pressure $P = 100 \, atm = 100 \times 1.013 \times 10^5 \, N/m^2 \approx 10^7 \, N/m^2$.
Fractional change in volume $\frac{\Delta V}{V} = 0.01 \% = \frac{0.01}{100} = 10^{-4}$.
The bulk modulus $K$ is defined as $K = \frac{P}{\Delta V/V}$.
Substituting the values: $K = \frac{10^7}{10^{-4}} = 10^{11} \, N/m^2$.
Since $1 \, N/m^2 = 10 \, dyne/cm^2$,we have $K = 10^{11} \times 10 \, dyne/cm^2 = 1 \times 10^{12} \, dyne/cm^2$.

(B) The isothermal bulk modulus is defined as $E_\theta = -V (dP/dV)_\theta = P$.
For an adiabatic process,$PV^\gamma = \text{constant}$.
Differentiating with respect to $V$,we get $P(\gamma V^{\gamma-1}) + V^\gamma (dP/dV) = 0$.
This simplifies to $dP/dV = -\gamma P/V$.
The adiabatic bulk modulus is $E_\phi = -V (dP/dV)_\phi = -V(-\gamma P/V) = \gamma P$.
Since $E_\theta = P$,we substitute to get $E_\phi = \gamma E_\theta$.

(A) The pressure at depth $h$ is given by $P = h \rho g$.
Given: $h = 200 \ m$,$\rho = 10^3 \ kg/m^3$ (density of water),$g = 9.8 \ m/s^2$.
The fractional change in volume is $\frac{\Delta V}{V} = 0.1\% = 0.001$.
The bulk modulus $K$ is defined as $K = \frac{P}{\Delta V/V}$.
Substituting the values: $K = \frac{200 \times 10^3 \times 9.8}{0.001}$.
$K = \frac{19.6 \times 10^5}{10^{-3}} = 19.6 \times 10^8 \ N/m^2$.

(B) The bulk modulus $K$ is defined as $K = -V \frac{dP}{dV}$.
Given the pressure equation $P = P_0 e^{\alpha V}$.
Differentiating with respect to volume $V$:
$\frac{dP}{dV} = P_0 e^{\alpha V} \cdot \alpha = P \cdot \alpha$.
Now,substitute this into the formula for bulk modulus:
$K = -V \frac{dP}{dV} = -V (P \alpha) = -\alpha PV$.
Taking the magnitude for the bulk modulus:
$K = \alpha PV$.

(A) The Bulk modulus $B$ is defined as $B = -V \frac{dp}{dV}$.
For a small change in pressure $p$,the change in volume $\Delta V$ is given by $\Delta V = -\frac{pV}{B}$.
The new volume $V^{\prime}$ is $V^{\prime} = V + \Delta V = V - \frac{pV}{B} = V(1 - \frac{p}{B})$.
Since mass $m$ remains constant,the new density $\rho^{\prime}$ is $\rho^{\prime} = \frac{m}{V^{\prime}} = \frac{m}{V(1 - \frac{p}{B})}$.
Substituting $\rho = \frac{m}{V}$,we get $\rho^{\prime} = \frac{\rho}{1 - \frac{p}{B}}$.
Therefore,the ratio $\frac{\rho^{\prime}}{\rho} = \frac{1}{1 - \frac{p}{B}}$.

(B) Bulk modulus $B$ is defined as $B = -\frac{P}{\Delta V / V}$,where $\Delta V / V$ is the volumetric strain.
The volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Differentiating with respect to $r$,we get $\Delta V = 4 \pi r^2 \Delta r$.
The volumetric strain is $\frac{\Delta V}{V} = \frac{4 \pi r^2 \Delta r}{\frac{4}{3} \pi r^3} = 3 \frac{\Delta r}{r}$.
Substituting this into the bulk modulus formula:
$B = -\frac{P}{3 \Delta r / r}$.
Rearranging for the fractional decrease in radius $(-\frac{\Delta r}{r})$:
$-\frac{\Delta r}{r} = \frac{P}{3B}$.

(C) Given:
Depth of ocean,$d = 2700 \, m$
Density of water,$\rho = 10^3 \, kg/m^3$
Compressibility of water,$K = 45.4 \times 10^{-11} \, Pa^{-1}$
Acceleration due to gravity,$g = 10 \, m/s^2$
The excess pressure at the bottom of the ocean is given by $\Delta P = \rho gd$.
$\Delta P = 10^3 \times 10 \times 2700 = 27 \times 10^6 \, Pa$.
Compressibility $K$ is defined as the reciprocal of the Bulk Modulus $B$,i.e.,$K = \frac{1}{B}$.
Since $B = \frac{\Delta P}{(\Delta V/V)}$,we have $\frac{\Delta V}{V} = K \times \Delta P$.
Substituting the values:
$\frac{\Delta V}{V} = (45.4 \times 10^{-11} \, Pa^{-1}) \times (27 \times 10^6 \, Pa)$
$\frac{\Delta V}{V} = 1225.8 \times 10^{-5} = 1.2258 \times 10^{-2} \approx 1.2 \times 10^{-2}$.

(A) The bulk modulus $B$ is defined as $B = -\frac{P}{\Delta V / V}$.
Since $\Delta V = V' - V$,we have $B = -\frac{P}{(V' - V) / V} = -\frac{PV}{V' - V}$.
Rearranging for the change in volume: $V' - V = -\frac{PV}{B}$,which gives $V' = V(1 - \frac{P}{B})$.
Density $\rho$ is defined as $\rho = \frac{m}{V}$,so $V = \frac{m}{\rho}$ and $V' = \frac{m}{\rho'}$.
Substituting these into the volume equation: $\frac{m}{\rho'} = \frac{m}{\rho}(1 - \frac{P}{B})$.
Therefore,$\frac{1}{\rho'} = \frac{1}{\rho}(1 - \frac{P}{B})$,which implies $\frac{\rho'}{\rho} = \frac{1}{1 - P/B}$.
Since the density increases with pressure,the correct expression is $\frac{\rho'}{\rho} = (1 - P/B)^{-1} \approx 1 + P/B$ for small $P$. However,based on the standard derivation,the exact ratio is $\frac{1}{1 - P/B}$.

(B) Compressibility is defined as the fractional change in volume per unit change in pressure.
Mathematically,compressibility $\sigma = \frac{1}{V} \cdot \frac{\Delta V}{P}$.
Given that $\sigma$ is the compressibility per unit atmospheric pressure,we have the relation $\sigma = \frac{\Delta V}{V \cdot P}$.
Rearranging this formula to solve for the decrease in volume $\Delta V$,we get:
$\Delta V = \sigma \cdot V \cdot P$ or $\sigma PV$.

(B) The volume of the sphere is $V = \frac{4}{3} \pi R^3$.
Taking the logarithmic derivative,we get $\frac{\delta V}{V} = 3 \frac{\delta R}{R}$.
The pressure exerted on the liquid by the piston is $P = \frac{mg}{A}$.
By the definition of bulk modulus,$K = -\frac{P}{\delta V / V}$.
Substituting the expressions for $P$ and $\frac{\delta V}{V}$,we have $K = -\frac{mg/A}{3 \delta R / R}$.
Rearranging for the fractional change in radius,we get $\frac{\delta R}{R} = -\frac{mg}{3AK}$.
The negative sign indicates a decrease in radius due to compression.

(B) The bulk modulus $B$ is defined as $B = \frac{P}{\Delta V / V}$,where $P$ is the hydrostatic pressure.
To prevent expansion,the thermal strain must be balanced by the compressive strain.
The thermal expansion of volume is given by $\Delta V = V \cdot 3\alpha \cdot \Delta T$,where $\alpha$ is the coefficient of linear expansion.
Thus,the volumetric strain is $\frac{\Delta V}{V} = 3\alpha \Delta T$.
Substituting this into the bulk modulus formula: $B = \frac{P}{3\alpha \Delta T}$.
Rearranging for pressure $P$: $P = 3 B \alpha \Delta T$.
Given values: $B = 1.4 \times 10^{11} \ Pa$,$\alpha = 1.7 \times 10^{-5} (^{\circ}C)^{-1}$,and $\Delta T = 30^{\circ}C - 20^{\circ}C = 10^{\circ}C$.
Calculating $P$: $P = 3 \times (1.4 \times 10^{11}) \times (1.7 \times 10^{-5}) \times 10 = 7.14 \times 10^7 \ Pa \approx 7.1 \times 10^7 \ Pa$.

(B) The pressure applied to the liquid by the mass $m$ on the piston is given by $\Delta P = \frac{mg}{a}$.
The bulk modulus $K$ is defined as $K = -\frac{\Delta P}{\Delta V/V}$.
For a sphere,the volume is $V = \frac{4}{3}\pi r^3$,so the fractional change in volume is $\frac{\Delta V}{V} = 3\frac{\Delta r}{r}$.
Substituting these into the bulk modulus formula: $K = -\frac{mg/a}{3(\Delta r/r)}$.
Rearranging for the fractional decrement in radius $\left| \frac{\Delta r}{r} \right|$,we get $\frac{\Delta r}{r} = \frac{mg}{3Ka}$.

(A) The bulk modulus $K$ is defined as the ratio of the change in pressure to the volumetric strain:
$K = \frac{P}{\left( \frac{-\Delta V}{V} \right)} \Rightarrow \frac{\Delta V}{V} = \frac{P}{K}$
where $\Delta V$ is the decrease in volume due to pressure $P$.
When the cube is heated by a temperature $\Delta t$,its volume increases due to thermal expansion:
$\Delta V = V_0 \gamma \Delta t$
where $\gamma$ is the coefficient of volume expansion and $V_0$ is the initial volume.
For a solid,the coefficient of volume expansion $\gamma$ is related to the coefficient of linear expansion $\alpha$ by $\gamma = 3\alpha$.
To bring the cube back to its original size,the increase in volume due to heating must equal the decrease in volume due to pressure:
$\frac{\Delta V}{V_0} = \gamma \Delta t = 3\alpha \Delta t$
Equating the two expressions for volumetric strain:
$\frac{P}{K} = 3\alpha \Delta t$
Solving for the temperature change $\Delta t$:
$\Delta t = \frac{P}{3\alpha K}$

(A) The speed of longitudinal waves in a fluid is given by the formula $v = \sqrt{\frac{B}{\rho}}$,where $B$ is the bulk modulus and $\rho$ is the density.
Given:
Density $\rho = 2.7 \ g/cm^3 = 2.7 \times 10^3 \ kg/m^3$.
Speed $v = 5.4 \ km/s = 5.4 \times 10^3 \ m/s$.
Rearranging the formula for $B$:
$B = \rho v^2$
Substituting the values:
$B = (2.7 \times 10^3) \times (5.4 \times 10^3)^2$
$B = 2.7 \times 10^3 \times 29.16 \times 10^6$
$B = 78.732 \times 10^9 \ Pa$
$B \approx 7.9 \times 10^{10} \ Pa$.

(B) The bulk modulus $\beta$ is defined as $\beta = -\frac{\Delta P}{\Delta V / V}$.
Given $\Delta P = 100 \, kPa = 10^5 \, Pa$ and the fractional change in volume $\frac{\Delta V}{V} = 5 \times 10^{-3} \% = \frac{5 \times 10^{-3}}{100} = 5 \times 10^{-5}$.
Substituting these values,$\beta = \frac{10^5}{5 \times 10^{-5}} = 0.2 \times 10^{10} = 2 \times 10^9 \, Pa$.
The speed of sound $v$ in a fluid is given by $v = \sqrt{\frac{\beta}{\rho}}$,where $\rho = 10^3 \, kg/m^3$.
$v = \sqrt{\frac{2 \times 10^9}{10^3}} = \sqrt{2 \times 10^6} = 1000 \sqrt{2} \approx 1414 \, m/s$.

(A) The pressure at a depth $h$ in a lake is given by the hydrostatic pressure formula: $P = h\rho g$.
This pressure acts as the stress on the water.
The energy density $u$ stored in a material under stress is given by the formula: $u = \frac{1}{2} \times \text{stress} \times \text{strain}$.
Since the bulk modulus $K = \frac{\text{stress}}{\text{strain}}$,we have $\text{strain} = \frac{\text{stress}}{K}$.
Substituting this into the energy density formula: $u = \frac{1}{2} \times \text{stress} \times \frac{\text{stress}}{K} = \frac{1}{2} \frac{(\text{stress})^2}{K}$.
Given that compressibility $\sigma = \frac{1}{K}$,we can rewrite the expression as: $u = \frac{1}{2} \sigma (\text{stress})^2$.
Substituting the stress $P = h\rho g$,we get: $u = \frac{1}{2} \sigma (h\rho g)^2$.

(B) The bulk modulus $B$ is defined as $B = -\frac{P}{\Delta V / V}$.
Since the mass $M = \rho V$ is constant,differentiating gives $\rho \Delta V + V \Delta \rho = 0$,which implies $\frac{\Delta \rho}{\rho} = -\frac{\Delta V}{V}$.
Substituting this into the bulk modulus formula,we get $\frac{\Delta \rho}{\rho} = \frac{P}{B}$.
The pressure at depth $h$ is $P = \rho_{sea} g h$.
Given $\rho_{sea} = 1.025 \times 10^3 \, kg/m^3$,$g = 10 \, m/s^2$,$h = 1000 \, m$,and $B = 1.6 \times 10^9 \, Pa$ $(1.6 \times 10^6 \, kPa = 1.6 \times 10^9 \, Pa)$.
$\frac{\Delta \rho}{\rho} = \frac{1.025 \times 10^3 \times 10 \times 1000}{1.6 \times 10^9} = \frac{1.025 \times 10^7}{1.6 \times 10^9} = 0.6406 \times 10^{-2} \approx 0.0064$.
Percentage change $= \frac{\Delta \rho}{\rho} \times 100 = 0.64 \%$.

(B) The bulk modulus $K$ is defined as $K = \frac{\Delta P}{-\Delta V/V}$.
Here,the change in pressure $\Delta P$ at depth $h$ is given by $\Delta P = h \rho g$.
The fractional change in volume is $\frac{\Delta V}{V} = -0.1\% = -\frac{0.1}{100} = -10^{-3}$.
Substituting the values into the formula:
$9.1 \times 10^8 = \frac{h \times 10^3 \times 10}{10^{-3}}$.
$9.1 \times 10^8 = h \times 10^7$.
$h = \frac{9.1 \times 10^8}{10^7} = 91 \, m$.

(D) The formula for Bulk modulus $(B)$ is given by: $B = \frac{\Delta P}{-\Delta V / V}$.
Rearranging for the change in pressure: $\Delta P = B \times \left( -\frac{\Delta V}{V} \right)$.
Given that the volume decreases by $0.004\%$,the volumetric strain is $\left| \frac{\Delta V}{V} \right| = \frac{0.004}{100} = 4 \times 10^{-5}$.
The Bulk modulus $B = 2100\,MPa = 2100 \times 10^6\,Pa = 2.1 \times 10^9\,Pa$.
Substituting the values: $\Delta P = (2.1 \times 10^9\,Pa) \times (4 \times 10^{-5})$.
$\Delta P = 8.4 \times 10^4\,Pa = 84 \times 10^3\,Pa = 84\,kPa$.

(D) The change in pressure $\Delta P$ at a depth $h$ is given by $\Delta P = h \rho g$.
Given: $h = 100\, m$,$\rho = 10^3\, kg/m^3$ (density of water),$g = 10\, m/s^2$,and fractional volume change $\frac{-\Delta V}{V} = 0.1\% = 0.001 = 10^{-3}$.
The bulk modulus $\beta$ is defined as $\beta = \frac{\Delta P}{(-\Delta V/V)}$.
Substituting the values: $\beta = \frac{100 \times 10^3 \times 10}{10^{-3}} = \frac{10^6}{10^{-3}} = 10^9\, N/m^2$.

(B) The formula for bulk modulus $(B)$ is given by $B = -\frac{\Delta P}{\Delta V / V} = -\frac{V \Delta P}{\Delta V}$.
Given:
Initial volume $V = 1.5\,L = 1.5 \times 10^{-3}\,m^3$.
Change in pressure $\Delta P = 140\,kPa = 140 \times 10^3\,Pa$.
Change in volume $\Delta V = -0.2\,mL = -0.2 \times 10^{-6}\,m^3$.
Substituting the values into the formula:
$B = -\frac{1.5 \times 10^{-3} \times 140 \times 10^3}{-0.2 \times 10^{-6}}$
$B = \frac{210}{0.2 \times 10^{-6}} = 1050 \times 10^6\,Pa = 1.05 \times 10^9\,Pa$.

(C) The bulk modulus $B$ is defined as the ratio of volumetric stress to volumetric strain:
$B = -\frac{\Delta p}{\Delta V / V}$
where $\Delta p$ is the change in pressure and $\Delta V / V$ is the volumetric strain.
For an incompressible liquid,the volume does not change even when pressure is applied,which means $\Delta V = 0$.
Substituting this into the formula:
$B = -\frac{\Delta p}{0 / V} = \infty$
Therefore,the bulk modulus for an incompressible liquid is infinite.

(D) The change in volume of the cork is $\Delta V = V_{initial} - V_{final} = \pi (a + \Delta a)^2 b - \pi a^2 b \approx \pi (a^2 + 2a \Delta a) b - \pi a^2 b = 2\pi a b \Delta a$.
The volumetric strain is $\frac{\Delta V}{V} = \frac{2\pi a b \Delta a}{\pi a^2 b} = \frac{2 \Delta a}{a}$.
The pressure $P$ exerted by the cork on the walls of the bottle is given by $P = B \times \text{volumetric strain} = B \left( \frac{2 \Delta a}{a} \right)$.
The normal force $N$ exerted by the cork on the inner surface of the bottle opening is $N = P \times A_{surface} = \left( \frac{2B \Delta a}{a} \right) \times (2\pi a b) = 4\pi B b \Delta a$.
The frictional force $f$ required to push the cork is $f = \mu N = \mu (4\pi B b \Delta a) = (4\pi \mu B b) \Delta a$.

(A) The bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V/V}$,which implies the fractional compression $\frac{\Delta V}{V} = \frac{\Delta P}{B}$.
For a constant pressure change $\Delta P$,the fractional compression $\frac{\Delta V}{V}$ is inversely proportional to the bulk modulus $B$ (i.e.,$\frac{\Delta V}{V} \propto \frac{1}{B}$).
Given bulk moduli:
$B_{\text{ethanol}} = 0.9 \times 10^9 \, Nm^{-2}$
$B_{\text{water}} = 2.2 \times 10^9 \, Nm^{-2}$
$B_{\text{mercury}} = 25 \times 10^9 \, Nm^{-2}$
Since $B_{\text{ethanol}} < B_{\text{water}} < B_{\text{mercury}}$,the fractional compression follows the order:
$\left( \frac{\Delta V}{V} \right)_{\text{ethanol}} > \left( \frac{\Delta V}{V} \right)_{\text{water}} > \left( \frac{\Delta V}{V} \right)_{\text{mercury}}$
Therefore,the correct order is Ethanol $>$ Water $>$ Mercury.

(D) The Bulk modulus $(B)$ is defined as the ratio of the change in pressure $(P)$ to the fractional change in volume $(\frac{\Delta V}{V})$:
$B = \frac{P}{\Delta V / V}$
Rearranging the formula to find the fractional change in volume:
$\frac{\Delta V}{V} = \frac{P}{B}$
Given values:
$P = 10 \, atm = 10 \times 10^5 \, N \, m^{-2} = 10^6 \, N \, m^{-2}$
$B = 37 \times 10^9 \, N \, m^{-2}$
Substituting the values into the equation:
$\frac{\Delta V}{V} = \frac{10^6}{37 \times 10^9}$
$\frac{\Delta V}{V} = \frac{1}{37} \times 10^{-3}$
$\frac{\Delta V}{V} \approx 0.027027 \times 10^{-3}$
$\frac{\Delta V}{V} = 2.7 \times 10^{-5}$

(D) The formula for Bulk modulus $(B)$ is given by $B = \frac{\Delta P}{-\Delta V / V}$.
Rearranging for the change in pressure,we get $\Delta P = B \times \left( -\frac{\Delta V}{V} \right)$.
Given that the volume decreases by $0.004\%$,the fractional change in volume is $\frac{-\Delta V}{V} = \frac{0.004}{100} = 4 \times 10^{-5}$.
The Bulk modulus $B = 2100 \ MPa = 2100 \times 10^6 \ Pa$.
Substituting these values into the equation:
$\Delta P = (2100 \times 10^6 \ Pa) \times (4 \times 10^{-5})$
$\Delta P = 84000 \ Pa = 84 \ kPa$.

(B) The bulk modulus $K$ is defined as $K = \frac{\Delta P}{\Delta V/V}$.
Here, the change in pressure $\Delta P$ at depth $h$ is given by $\Delta P = \rho gh$, where $\rho$ is the density of water $(10^3 \, kg/m^3)$ and $g$ is the acceleration due to gravity $(10 \, m/s^2)$.
The fractional change in volume is $\frac{\Delta V}{V} = 0.1 \, \% = \frac{0.1}{100} = 10^{-3}$.
Substituting the values into the formula:
$9.1 \times 10^8 = \frac{10^3 \times 10 \times h}{10^{-3}}$
$9.1 \times 10^8 = \frac{10^4 \times h}{10^{-3}}$
$9.1 \times 10^8 = 10^7 \times h$
$h = \frac{9.1 \times 10^8}{10^7} = 9.1 \times 10 = 91 \, m$.

(D) Let the side of the cube be $L$. The volume of the cube is given by $V = L^3$.
Taking the logarithmic derivative,we get $\frac{dV}{V} = 3 \frac{dL}{L}$.
Given that the side is decreased by $1\%$,we have $\frac{dL}{L} = -0.01$.
Therefore,the fractional change in volume (bulk strain) is $\frac{\Delta V}{V} = 3 \times (-0.01) = -0.03$.
The magnitude of the bulk strain is $|\frac{\Delta V}{V}| = 0.03$.

(B) The bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V / V}$,where $\Delta P$ is the change in pressure,$V$ is the initial volume,and $\Delta V$ is the change in volume.
Rearranging for $\Delta V$,we get $\Delta V = -\frac{\Delta P \cdot V}{B}$.
The initial volume $V = (60\, mm)^3 = 216000\, mm^3 = 2.16 \times 10^5\, mm^3$.
Given $\Delta P = 2.5 \times 10^7\, Pa$ and $B = 1.25 \times 10^{11}\, N/m^2$.
Substituting the values: $\Delta V = -\frac{2.5 \times 10^7}{1.25 \times 10^{11}} \times (60\, mm)^3$.
$\Delta V = -2 \times 10^{-4} \times 216000\, mm^3 = -43.2\, mm^3$.

(A) The bulk modulus $B$ is defined as the ratio of volumetric stress to volumetric strain: $B = -\frac{P}{\Delta V / V}$.
The compressibility $\sigma$ is the reciprocal of the bulk modulus: $\sigma = \frac{1}{B} = -\frac{\Delta V / V}{P}$.
Given that there is a decrease in volume $V$ due to pressure $P$,we consider the magnitude of the change in volume $\Delta V$: $\sigma = \frac{\Delta V}{PV}$.
Rearranging the formula to solve for the change in volume $\Delta V$,we get: $\Delta V = \sigma PV$.

(B) The bulk modulus $(B)$ is defined as the ratio of hydraulic stress to the volumetric strain.
$B = \frac{\text{Stress}}{\text{Volumetric Strain}} = \frac{\text{Stress}}{\Delta V / V}$
Rearranging the formula to solve for the fractional change in volume $(\Delta V / V)$:
$\frac{\Delta V}{V} = \frac{\text{Stress}}{B}$
Since the hydraulic stress is given as a constant,the relationship between the fractional change in volume and the bulk modulus is:
$\frac{\Delta V}{V} \propto \frac{1}{B}$

(C) The formula for bulk modulus $K$ is given by $K = \frac{P}{\Delta V/V}$, where $P$ is the change in pressure, $V$ is the initial volume, and $\Delta V$ is the change in volume.
Rearranging for $\Delta V$, we get $\Delta V = \frac{PV}{K}$.
The pressure at depth $h$ is $P = h\rho g = 200 \times 10^3 \times 10 = 2 \times 10^6\, N/m^2$.
The bulk modulus is $K = 22000\, \text{atm} = 22000 \times 10^5 = 2.2 \times 10^9\, N/m^2$.
Given $V = 1\, m^3$, substituting the values:
$\Delta V = \frac{2 \times 10^6 \times 1}{2.2 \times 10^9} = \frac{2}{2200} = \frac{1}{1100} \approx 9.09 \times 10^{-4}\, m^3$.
Rounding to the nearest option, $\Delta V \approx 9.1 \times 10^{-4}\, m^3$.

(A) The pressure $p$ exerted by a $3000\; m$ column of water at the bottom of the ocean is given by the formula $p = h \rho g$.
Given: depth $h = 3000\; m$,density of water $\rho = 1000\; kg\; m^{-3}$,and acceleration due to gravity $g = 10\; m s^{-2}$.
$p = 3000\; m \times 1000\; kg\; m^{-3} \times 10\; m s^{-2} = 3 \times 10^{7}\; N m^{-2}$.
The bulk modulus $B$ is defined as $B = -\frac{p}{\Delta V / V}$,so the fractional compression is $\frac{\Delta V}{V} = \frac{p}{B}$.
Substituting the values: $\frac{\Delta V}{V} = \frac{3 \times 10^{7}\; N m^{-2}}{2.2 \times 10^{9}\; N m^{-2}} \approx 1.36 \times 10^{-2}$.
Thus,the fractional compression is $1.36 \times 10^{-2}$.

(N/A) Initial volume,$V_{1} = 100.0 \; L = 100.0 \times 10^{-3} \; m^{3}$.
Final volume,$V_{2} = 100.5 \; L = 100.5 \times 10^{-3} \; m^{3}$.
Change in volume,$\Delta V = V_{2} - V_{1} = 0.5 \times 10^{-3} \; m^{3}$.
Increase in pressure,$\Delta p = 100.0 \; atm = 100 \times 1.013 \times 10^{5} \; Pa = 1.013 \times 10^{7} \; Pa$.
Bulk modulus $B = -\frac{\Delta p}{\Delta V / V_{1}} = \frac{\Delta p \cdot V_{1}}{\Delta V}$.
$B = \frac{1.013 \times 10^{7} \times 100.0 \times 10^{-3}}{0.5 \times 10^{-3}} = \frac{1.013 \times 10^{6}}{0.5 \times 10^{-3}} = 2.026 \times 10^{9} \; Pa$.
The bulk modulus of air is approximately $1.0 \times 10^{5} \; Pa$.
The ratio of the bulk modulus of water to that of air is $\frac{2.026 \times 10^{9}}{1.0 \times 10^{5}} \approx 2.026 \times 10^{4}$.
This ratio is very large because water is a liquid and is nearly incompressible,whereas air is a gas and is highly compressible due to the large intermolecular spaces.