Bulk Modulus Questions in English

$(1.034 \times 10^3 \; kg \; m^{-3})$ Let the pressure at the surface be $p_1 = 1.013 \times 10^5 \; Pa$ and at depth $h$ be $p_2 = 80.0 \; atm = 80.0 \times 1.013 \times 10^5 \; Pa$.
Change in pressure $\Delta p = p_2 - p_1 \approx 80.0 \times 1.013 \times 10^5 \; Pa = 8.104 \times 10^6 \; Pa$.
Bulk modulus $B = -\frac{\Delta p}{\Delta V / V_1}$, so volumetric strain $\frac{\Delta V}{V_1} = \frac{\Delta p}{B}$.
Given compressibility $\frac{1}{B} = 45.8 \times 10^{-11} \; Pa^{-1}$.
$\frac{\Delta V}{V_1} = \Delta p \times \frac{1}{B} = (8.104 \times 10^6) \times (45.8 \times 10^{-11}) \approx 3.71 \times 10^{-3}$.
Since $\rho_2 = \frac{m}{V_2}$ and $\rho_1 = \frac{m}{V_1}$, we have $V_2 = V_1(1 - \frac{\Delta V}{V_1})$.
$\rho_2 = \frac{m}{V_1(1 - \Delta V / V_1)} = \frac{\rho_1}{1 - \Delta V / V_1} \approx \rho_1 (1 + \frac{\Delta V}{V_1})$.
$\rho_2 = 1.03 \times 10^3 \times (1 + 3.71 \times 10^{-3}) \approx 1.0338 \times 10^3 \; kg \; m^{-3} \approx 1.034 \times 10^3 \; kg \; m^{-3}$.

(C) The Bulk modulus $(B)$ of glass is given by $B = 37 \times 10^{9} \; N/m^2$.
The formula for Bulk modulus is $B = \frac{p}{\Delta V / V}$,where $p$ is the hydraulic pressure and $\Delta V / V$ is the fractional change in volume.
Rearranging the formula,we get the fractional change in volume: $\frac{\Delta V}{V} = \frac{p}{B}$.
Given $p = 10 \; atm = 10 \times 1.013 \times 10^{5} \; Pa = 1.013 \times 10^{6} \; Pa$.
Substituting the values: $\frac{\Delta V}{V} = \frac{1.013 \times 10^{6}}{37 \times 10^{9}}$.
$\frac{\Delta V}{V} \approx 0.02737 \times 10^{-3} = 2.737 \times 10^{-5}$.
Rounding to the nearest option,the fractional change in volume is $2.73 \times 10^{-5}$.

(D) The bulk modulus of copper is $B = 140 \times 10^{9} \; Pa$.
The formula for bulk modulus is $B = \frac{p}{\Delta V / V}$,where $p$ is the pressure,$V$ is the original volume,and $\Delta V$ is the change in volume.
Rearranging for volume contraction $\Delta V$,we get $\Delta V = \frac{p V}{B}$.
The original volume of the cube is $V = l^{3} = (10 \; cm)^{3} = (0.1 \; m)^{3} = 10^{-3} \; m^{3}$.
Substituting the values: $\Delta V = \frac{7.0 \times 10^{6} \; Pa \times 10^{-3} \; m^{3}}{140 \times 10^{9} \; Pa}$.
$\Delta V = \frac{7.0 \times 10^{3}}{140 \times 10^{9}} = 0.05 \times 10^{-6} \; m^{3} = 5 \times 10^{-8} \; m^{3}$.
Since $1 \; m^{3} = 10^{6} \; cm^{3}$,we have $\Delta V = 5 \times 10^{-8} \times 10^{6} \; cm^{3} = 5 \times 10^{-2} \; cm^{3}$.
Thus,the volume contraction is $5 \times 10^{-2} \; cm^{3}$.

(A) The fractional change in volume is given by $\frac{\Delta V}{V} = 0.10\% = \frac{0.10}{100} = 10^{-3}$.
The bulk modulus $B$ of water is approximately $2.2 \times 10^{9} \;Nm^{-2}$.
The formula for bulk modulus is $B = \frac{p}{\Delta V / V}$,where $p$ is the change in pressure.
Rearranging for $p$,we get $p = B \times \left( \frac{\Delta V}{V} \right)$.
Substituting the values: $p = (2.2 \times 10^{9} \;Nm^{-2}) \times (10^{-3})$.
$p = 2.2 \times 10^{6} \;Nm^{-2}$.
Thus,the pressure on the water should be changed by $2.2 \times 10^{6} \;Nm^{-2}$.

(A) Given:
Water pressure at the bottom,$p = 1.1 \times 10^{8} \; Pa$
Initial volume of the steel ball,$V = 0.32 \; m^{3}$
Bulk modulus of steel,$B = 1.6 \times 10^{11} \; N/m^{2}$
The formula for Bulk modulus is given by $B = -\frac{p}{\Delta V / V}$,where $\Delta V$ is the change in volume.
Rearranging for $\Delta V$,we get $\Delta V = \frac{p V}{B}$.
Substituting the values:
$\Delta V = \frac{(1.1 \times 10^{8} \; Pa) \times (0.32 \; m^{3})}{1.6 \times 10^{11} \; N/m^{2}}$
$\Delta V = \frac{0.352 \times 10^{8}}{1.6 \times 10^{11}}$
$\Delta V = 0.22 \times 10^{-3} \; m^{3} = 2.2 \times 10^{-4} \; m^{3}$.
Thus,the change in volume of the ball is $2.2 \times 10^{-4} \; m^{3}$.

(N/A) When a solid body is placed in a fluid under high pressure,it is compressed uniformly on all sides.
The force applied by the fluid acts in a perpendicular direction at each point of the surface,and the body is said to be under hydraulic compression. As a result,this leads to a decrease in its volume without any change in its geometrical shape.
The body develops internal restoring forces that are equal and opposite to the forces applied by the fluid.
The internal restoring force per unit area in this case is known as hydraulic stress (or volume stress),and its magnitude is equal to the hydraulic pressure (applied force per unit area).
The strain produced by hydraulic pressure is called volume strain and is defined as the ratio of the change in volume $(\Delta V)$ to the original volume $(V)$.
$\text{Volume strain} = \frac{\Delta V}{V}$
It has no units or dimensional formula.

(N/A) The ratio of shearing stress to the corresponding shearing strain is called the shear modulus of the material and is represented by $G$. It is also called the modulus of rigidity.
$G = \frac{\text{Shearing stress } (\sigma_s)}{\text{Shearing strain } (\theta)} = \frac{F/A}{\Delta x/L} = \frac{FL}{A \Delta x}$
For small angles,$\frac{\Delta x}{L} = \tan \theta \approx \theta$,so $G = \frac{F/A}{\theta} = \frac{F}{A \theta}$.
The shearing stress $\sigma_s$ can be expressed as $\sigma_s = G \times \theta$.
The $SI$ unit of shear modulus is $N m^{-2}$ or $Pa$.
Generally,the shear modulus is less than Young's modulus,and for most materials,$G \approx \frac{Y}{3}$.
When a body is submerged in a fluid,it undergoes hydraulic stress (equal in magnitude to the hydraulic pressure). This leads to a decrease in the volume of the body,producing a volume strain.
The ratio of hydraulic stress to the corresponding hydraulic strain is called the bulk modulus,denoted by $B$.
$B = -\frac{p}{\Delta V / V}$
The negative sign indicates that an increase in pressure leads to a decrease in volume (if $p > 0$,then $\Delta V < 0$). Thus,for a system in equilibrium,$B$ is always positive.
The $SI$ unit of bulk modulus is $N m^{-2}$ or $Pa$,and its dimensional formula is $[M^1 L^{-1} T^{-2}]$.

(A) Compressibility is defined as the reciprocal of the bulk modulus $(B)$. It represents the fractional change in volume per unit increase in pressure.
The formula for compressibility $(k)$ is:
$k = \frac{1}{B} = -\frac{1}{V} \left( \frac{\Delta V}{\Delta p} \right)$
Unit:
The $SI$ unit of compressibility is $Pa^{-1}$ or $m^2/N$.
Dimensional Formula:
Since $B = \frac{\text{Force}}{\text{Area}} \times \frac{\text{Volume}}{\text{Change in Volume}}$,the dimensions of $B$ are $[M L^{-1} T^{-2}]$.
Therefore,the dimensional formula for compressibility is $[M^{-1} L^1 T^2]$.

(A) The bulk modulus $(B)$ is defined as the ratio of volumetric stress to volumetric strain: $B = -\Delta P / (\Delta V / V)$.
Compressibility $(K)$ is defined as the reciprocal of the bulk modulus: $K = 1 / B$.
For solids,the interatomic forces are very strong,making them very difficult to compress. Thus,solids have a very high bulk modulus $(B)$,which results in a very low compressibility $(K)$.
For gases,the interatomic forces are negligible,making them very easy to compress. Thus,gases have a very low bulk modulus $(B)$,which results in a very high compressibility $(K)$.
Therefore,solids are the least compressible and gases are the most compressible.

(C) Compressibility $(K)$ is the reciprocal of the Bulk Modulus $(B)$,given by $K = 1/B$.
For solids,the Bulk Modulus is typically in the order of $10^{11} \ N/m^2$.
For gases,the Bulk Modulus is typically in the order of $10^5 \ N/m^2$.
The ratio of compressibility of gases $(K_g)$ to solids $(K_s)$ is given by:
$K_g / K_s = B_s / B_g = 10^{11} / 10^5 = 10^6$.
Therefore,gases are $10^6$ times more compressible than solids.

(C) The bulk modulus $B$ is defined as the ratio of volumetric stress to volumetric strain: $B = -\frac{P}{\Delta V / V}$.
For a perfectly rigid body,the change in volume $\Delta V$ is always $0$ regardless of the applied pressure $P$.
Substituting $\Delta V = 0$ into the formula,we get $B = -\frac{P}{0} = \infty$.
Therefore,the bulk modulus of a perfectly rigid body is infinite.

(B) Liquids and gases do not have a definite shape and cannot resist shearing forces,so they do not possess Young's modulus or Shear modulus.
They can only resist compressive forces that change their volume.
Therefore,the only elastic modulus applicable to fluids (liquids and gases) is the Volume elastic modulus,commonly known as the Bulk modulus $(B)$.

(N/A) The elasticity of a material is measured by its bulk modulus $(B)$.
The bulk modulus is defined as the reciprocal of compressibility $(K)$,i.e.,$B = 1/K$.
Air is highly compressible,meaning it has a very high compressibility value.
Water is much less compressible than air.
Since the bulk modulus is inversely proportional to compressibility,the material with lower compressibility has a higher bulk modulus.
Therefore,water is more elastic than air.

(C) The bulk modulus $(B)$ is defined as the ratio of the change in pressure to the fractional change in volume.
Since solids,liquids,and gases all experience a change in volume when subjected to uniform pressure,the bulk modulus is defined for all three states of matter.
In contrast,Young's modulus and Shear modulus require a definite shape and resistance to shearing forces,which are properties primarily associated with solids.
Therefore,the bulk modulus is the only modulus of elasticity applicable to solids,liquids,and gases.

The formula for bulk modulus $B$ is given by $B = \frac{P}{\Delta V / V}$,where $P$ is the pressure,$\Delta V$ is the change in volume,and $V$ is the initial volume.
Rearranging for the change in volume: $\Delta V = \frac{PV}{B}$.
The initial volume of the sphere $V = \frac{4}{3} \pi r^3$,where $r = 10\,cm = 0.1\,m$.
Substituting the values:
$V = \frac{4}{3} \times 3.14 \times (0.1)^3 = \frac{4}{3} \times 3.14 \times 0.001\,m^3$.
Now,calculate $\Delta V$:
$\Delta V = \frac{5 \times 10^8 \times (\frac{4}{3} \times 3.14 \times 0.001)}{3.14 \times 10^{11}}$
$\Delta V = \frac{5 \times 10^8 \times 4 \times 3.14 \times 0.001}{3 \times 3.14 \times 10^{11}}$
$\Delta V = \frac{20 \times 10^5}{3 \times 10^{11}} = 6.66 \times 10^{-6}\,m^3$
$\Delta V \approx 6.7 \times 10^{-6}\,m^3$.

(C) The bulk modulus $(B)$ is defined as the ratio of volumetric stress to volumetric strain,given by the formula $B = -\frac{\Delta P}{\Delta V / V}$.
For a perfectly rigid body,the change in volume $(\Delta V)$ is zero for any applied pressure $(\Delta P)$.
Therefore,$B = -\frac{\Delta P}{0} = \infty$.
Thus,the bulk modulus for a rigid body is infinite.

(A) The percentage decrease in volume is given by $\frac{\Delta V}{V} \times 100 = 0.1$.
Thus,the fractional change in volume is $\frac{\Delta V}{V} = \frac{0.1}{100} = 0.001$.
The pressure exerted by the sea water at a depth $h$ is $P = h \rho g$.
The bulk modulus $B$ is defined as $B = \frac{P}{\Delta V / V}$.
Rearranging the formula to solve for depth $h$,we get $h = \frac{B \times (\Delta V / V)}{\rho g}$.
Substituting the given values: $B = 9.8 \times 10^8 \, N/m^2$,$\frac{\Delta V}{V} = 0.001$,$\rho = 10^3 \, kg/m^3$,and $g = 9.8 \, m/s^2$.
$h = \frac{9.8 \times 10^8 \times 0.001}{10^3 \times 9.8} = \frac{9.8 \times 10^5}{9.8 \times 10^3} = 10^2 = 100 \, m$.
Therefore,the required depth is $100 \, m$.

(N/A) Increase in temperature $\Delta T = 57 - 37 = 20\,^{\circ}C$ or $20\,K$.
Coefficient of linear expansion of copper,$\alpha = 1.7 \times 10^{-5}\,^{\circ}C^{-1}$.
Bulk modulus of copper,$K = 140 \times 10^9\,N/m^2$.
Coefficient of volume expansion of copper,$\gamma = 3\alpha = 3 \times 1.7 \times 10^{-5} = 5.1 \times 10^{-5}\,^{\circ}C^{-1}$.
Thermal stress is given by the formula: $\text{Stress} = K \times \text{volumetric strain} = K \times \frac{\Delta V}{V}$.
Since $\frac{\Delta V}{V} = \gamma \Delta T$,the stress is $\text{Stress} = K \gamma \Delta T$.
Substituting the values: $\text{Stress} = (140 \times 10^9) \times (5.1 \times 10^{-5}) \times 20$.
$\text{Stress} = 140 \times 5.1 \times 20 \times 10^4 = 14280 \times 10^4 = 1.428 \times 10^8\,N/m^2$.

(B) The bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V / V}$.
The magnitude of the fractional change in volume is given by $\left| \frac{\Delta V}{V} \right| = \frac{\Delta P}{B}$.
Substituting the given values,$\Delta P = 4 \times 10^9 \; Pa$ and $B = 8 \times 10^{10} \; Pa$:
$\left| \frac{\Delta V}{V} \right| = \frac{4 \times 10^9}{8 \times 10^{10}} = \frac{1}{20} = 0.05$.
For a cube of side length $\ell$,the volume is $V = \ell^3$. Differentiating,we get $\frac{\Delta V}{V} = 3 \frac{\Delta \ell}{\ell}$.
Therefore,the fractional change in length is $\frac{\Delta \ell}{\ell} = \frac{1}{3} \left| \frac{\Delta V}{V} \right| = \frac{1}{3} \times \frac{1}{20} = \frac{1}{60}$.
The percentage change in length is $\frac{\Delta \ell}{\ell} \times 100\% = \frac{1}{60} \times 100\% = \frac{10}{6}\% \approx 1.67\%$.

(B) The compressibility $K$ is defined as the reciprocal of the bulk modulus $\beta$,given by $K = \frac{1}{\beta} = -\frac{1}{V} \frac{\Delta V}{\Delta P}$.
Rearranging the formula to find the change in volume $\Delta V$,we get: $-\Delta V = K \cdot V \cdot P$.
Given values:
Compressibility $K = 6 \times 10^{-10} \ N^{-1} \ m^2$.
Initial volume $V = 1 \ litre = 1000 \ cc = 10^{-3} \ m^3$.
Pressure $P = 4 \times 10^7 \ N \ m^{-2}$.
Substituting these values into the equation:
$-\Delta V = (6 \times 10^{-10}) \times (10^{-3}) \times (4 \times 10^7)$.
$-\Delta V = 24 \times 10^{-6} \ m^3$.
Since $1 \ m^3 = 10^6 \ cc$,we have:
$-\Delta V = 24 \times 10^{-6} \times 10^6 \ cc = 24 \ cc$.

(D) The hydraulic stress is equal to the hydrostatic pressure $P$ at depth $h$,given by $P = h \rho g$.
Here,$h = 2 \, km = 2000 \, m$,$\rho = 1000 \, kg \, m^{-3}$,and $g = 9.8 \, m \, s^{-2}$.
$P = 2000 \times 1000 \times 9.8 = 1.96 \times 10^{7} \, N \, m^{-2}$.
The hydraulic strain is the fractional compression $\frac{\Delta V}{V} = 1.36 \, \% = 1.36 \times 10^{-2}$.
The ratio of hydraulic stress to hydraulic strain is the Bulk Modulus $B = \frac{P}{\Delta V / V}$.
$B = \frac{1.96 \times 10^{7}}{1.36 \times 10^{-2}} \approx 1.44 \times 10^{9} \, N \, m^{-2}$.

(B) The density $\rho$ is given by $\rho = \frac{M}{V}$,where $M$ is the mass and $V$ is the volume.
Taking the logarithmic derivative,we get $\frac{d\rho}{\rho} = -\frac{dV}{V}$.
The bulk modulus $K$ is defined as $K = -\frac{P}{dV/V}$,which implies $-\frac{dV}{V} = \frac{P}{K}$.
Substituting this into the density relation,we get $\frac{d\rho}{\rho} = \frac{P}{K}$.
Therefore,the magnitude of the increase in density is $d\rho = \frac{\rho P}{K}$.

(C) The bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V/V}$.
Here,the change in pressure $\Delta P$ at depth $h$ is given by $\Delta P = \rho g h$.
The fractional change in volume is $\frac{\Delta V}{V} = -0.5\, \% = -0.005$ (the negative sign indicates a decrease in volume).
Substituting the values into the formula: $B = -\frac{\rho g h}{-0.005}$.
Rearranging for $h$: $h = \frac{B \times 0.005}{\rho g}$.
Substituting the given values: $h = \frac{9.8 \times 10^{8} \times 0.005}{10^{3} \times 9.8}$.
$h = \frac{10^{8} \times 0.005}{10^{3}} = 10^{5} \times 0.005 = 500 \, \text{m}$.
Thus,the depth is $500 \, \text{m}$.

(C) The bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V / V}$.
Given,$B = 3 \times 10^{10} \ Nm^{-2}$.
The fractional change in volume is $\frac{\Delta V}{V} = -2\% = -0.02$ (negative sign indicates reduction in volume).
Rearranging the formula for pressure change $\Delta P$: $\Delta P = -B \times \left(\frac{\Delta V}{V}\right)$.
Substituting the values: $\Delta P = -(3 \times 10^{10}) \times (-0.02)$.
$\Delta P = 3 \times 10^{10} \times 0.02 = 0.06 \times 10^{10} = 6 \times 10^{8} \ Nm^{-2}$.
Thus,the required pressure is $6 \times 10^{8} \ Nm^{-2}$.

(C) 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 that the volume contraction is $0.02 \%$,the volumetric strain is $\frac{\Delta V}{V} = -\frac{0.02}{100} = -2 \times 10^{-4}$.
The change in pressure $\Delta P = 50 \, atm = 50 \times 1.01325 \times 10^5 \, Pa \approx 50.66 \times 10^5 \, Pa$.
Using the formula $B = \frac{\Delta P}{-(\Delta V / V)} = \frac{50 \times 1.01325 \times 10^5}{2 \times 10^{-4}} = 25 \times 1.01325 \times 10^9 = 2.53 \times 10^{10} \, Nm^{-2}$.
However,using the standard approximation $1 \, atm = 1.01 \times 10^5 \, Pa$ and the provided options,we calculate $B = \frac{50 \times 1.01 \times 10^5}{0.02 / 100} = \frac{50.5 \times 10^5}{2 \times 10^{-4}} = 25.25 \times 10^9 = 2.525 \times 10^{10} \, Nm^{-2}$.
Re-evaluating the provided solution logic: If $\frac{\Delta V}{V} = 0.02 \% = 2 \times 10^{-4}$ and $P = 50 \, atm = 50.5 \times 10^5 \, Pa$,then $B = \frac{50.5 \times 10^5}{2 \times 10^{-4}} = 2.525 \times 10^{10} \, Nm^{-2}$.
Given the options,there is a discrepancy in the provided solution's calculation. Based on standard physics,the closest order of magnitude is $10^{10}$.

(D) The Bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V / V}$.
Since density $\rho = \frac{m}{V}$,we have $V = \frac{m}{\rho}$.
Taking the derivative,$\Delta V = -\frac{m}{\rho^2} \Delta \rho$.
Thus,$\frac{\Delta V}{V} = -\frac{\Delta \rho}{\rho}$.
Substituting this into the Bulk modulus formula: $B = \frac{\Delta P}{\Delta \rho / \rho}$,which gives $\frac{\Delta \rho}{\rho} = \frac{\Delta P}{B}$.
Given: $B = 8.0 \times 10^9 \, N/m^2$,$\Delta P = 2.0 \times 10^8 \, N/m^2$,and $\rho_1 = 11.4 \, g/cc$.
$\Delta \rho = \rho_1 \times \frac{\Delta P}{B} = 11.4 \times \frac{2.0 \times 10^8}{8.0 \times 10^9} = 11.4 \times 0.025 = 0.285 \, g/cc$.
The final density $\rho_2 = \rho_1 + \Delta \rho = 11.4 + 0.285 = 11.685 \, g/cc \approx 11.7 \, g/cc$.

(A) The Bulk modulus $B$ is defined by the formula $B = -\frac{\Delta P}{\Delta V / V}$.
Given:
Initial pressure $P_1 = 1.01 \times 10^5 \, Pa$
Final pressure $P_2 = 1.165 \times 10^5 \, Pa$
Change in pressure $\Delta P = P_2 - P_1 = (1.165 - 1.01) \times 10^5 \, Pa = 0.155 \times 10^5 \, Pa$.
Since the pressure increases,the volume decreases by $10 \%$,so $\frac{\Delta V}{V} = -0.10$.
Substituting these values into the formula:
$B = -\frac{0.155 \times 10^5}{-0.10}$
$B = \frac{0.155 \times 10^5}{0.10}$
$B = 1.55 \times 10^5 \, Pa$.
Thus,the Bulk modulus is $1.55 \times 10^5 \, Pa$.

(C) The volume of a cube is given by $V = a^3$,where $a$ is the side length.
Taking the logarithmic derivative,we get $\frac{\Delta V}{V} = 3 \frac{\Delta a}{a}$.
Given that the side decreases by $2 \%$,we have $\frac{\Delta a}{a} = -0.02$.
Substituting this value into the expression for volumetric strain:
$\frac{\Delta V}{V} = 3 \times (-0.02) = -0.06$.
The bulk strain is defined as the magnitude of the volumetric strain,which is $|\frac{\Delta V}{V}| = 0.06$.

(C) The pressure at the bottom of the sea is given by $P = \rho g h$.
Given: $\rho = 1 \,g/cc = 1000 \,kg/m^3$,$g = 10 \,m/s^2$,and $h = 1400 \,m$.
$P = 1000 \times 10 \times 1400 = 1.4 \times 10^7 \,N/m^2$.
The fractional change in volume is $\frac{\Delta V}{V} = -2\% = -0.02$.
The Bulk modulus $B$ is defined as $B = -\frac{P}{\Delta V / V}$.
Substituting the values: $B = -\frac{1.4 \times 10^7}{-0.02} = \frac{1.4 \times 10^7}{0.02} = 70 \times 10^7 = 7 \times 10^8 \,N/m^2$.
Thus,the Bulk modulus is $7 \times 10^8 \,N/m^2$.

(D) The formula for Bulk modulus $(B)$ is given by $B = -\frac{P}{\Delta V / V}$.
Given,the fractional change in volume $\frac{\Delta V}{V} = -\frac{0.02}{100} = -2 \times 10^{-4}$.
The applied pressure $P = 50 \, \text{atm} = 50 \times 1.013 \times 10^5 \, \text{Pa} \approx 5.065 \times 10^6 \, \text{Pa}$.
Using $B = \frac{P}{|\Delta V / V|}$,we get $B = \frac{50 \times 1.013 \times 10^5}{2 \times 10^{-4}}$.
$B = 25 \times 1.013 \times 10^9 = 2.5325 \times 10^{10} \, \text{N/m}^2$.
Rounding to the nearest provided option,the correct answer is $2.5 \times 10^{10} \, \text{N/m}^2$.

(C) The pressure at the bottom of the sea is given by $P = \rho g h$.
Given: $\rho = 1000 \, kg/m^3$,$g = 9.8 \, m/s^2$,and $h = 1000 \, m$.
$P = 1000 \times 9.8 \times 1000 = 9.8 \times 10^6 \, N/m^2$.
The fractional change in volume is $\frac{-\Delta V}{V} = 0.01 \% = \frac{0.01}{100} = 10^{-4}$.
The Bulk modulus $B$ is defined as $B = \frac{P}{(-\Delta V/V)}$.
Substituting the values: $B = \frac{9.8 \times 10^6}{10^{-4}} = 9.8 \times 10^{10} \, N/m^2$.

(B) Given:
Edge of the cube $a = 10 \, cm = 0.1 \, m = 10^{-1} \, m$.
Initial volume $V = a^3 = (10^{-1} \, m)^3 = 10^{-3} \, m^3$.
Pressure $P = 7 \times 10^6 \, Pa$.
Bulk modulus $B = 140 \, GPa = 140 \times 10^9 \, Pa$.
We know the formula for Bulk modulus is $B = \frac{P}{(-\Delta V / V)}$,where $-\Delta V$ is the contraction in volume.
Rearranging for contraction in volume: $-\Delta V = \frac{P \times V}{B}$.
Substituting the values:
$-\Delta V = \frac{(7 \times 10^6 \, Pa) \times (10^{-3} \, m^3)}{140 \times 10^9 \, Pa}$.
$-\Delta V = \frac{7 \times 10^3}{140 \times 10^9} = \frac{1}{20} \times 10^{-6} = 0.05 \times 10^{-6} = 5 \times 10^{-8} \, m^3$.
Thus,the contraction in volume is $5 \times 10^{-8} \, m^3$.

(B) The Bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V / V}$.
For water,the fractional change in volume is $\frac{\Delta V_w}{V_w} = 0.01 \% = \frac{0.01}{100} = 10^{-4}$.
Thus,$B_w = \frac{P}{10^{-4}} = 10^4 P$.
For the liquid,the fractional change in volume is $\frac{\Delta V_l}{V_l} = 0.03 \% = \frac{0.03}{100} = 3 \times 10^{-4}$.
Thus,$B_l = \frac{P}{3 \times 10^{-4}} = \frac{10^4 P}{3}$.
The ratio of the Bulk modulus of water to that of the liquid is $\frac{B_w}{B_l} = \frac{10^4 P}{10^4 P / 3} = 3$.
Given that the ratio is $\frac{3}{x}$,we have $\frac{3}{x} = 3$,which implies $x = 1$.

(B) The bulk modulus $B$ is defined as $B = -V \frac{dP}{dV}$.
Given the relation $P = aV^{-3}$.
Differentiating $P$ with respect to $V$:
$\frac{dP}{dV} = a(-3)V^{-4} = -3 \frac{aV^{-3}}{V} = -3 \frac{P}{V}$.
Substituting this into the formula for bulk modulus:
$B = -V \left( -3 \frac{P}{V} \right) = 3P$.
Therefore,the bulk modulus is $3P$.

(B) The bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V / V}$.
The pressure at the bottom of the ocean is given by the hydrostatic pressure formula: $\Delta P = \rho g h$.
Substituting the given values: $\Delta P = 1000 \ kg m^{-3} \times 10 \ m s^{-2} \times 4000 \ m = 4 \times 10^7 \ N m^{-2}$.
The fractional compression is $\frac{\Delta V}{V} = \frac{\Delta P}{B}$.
Substituting the values: $\frac{\Delta V}{V} = \frac{4 \times 10^7}{2 \times 10^9} = 2 \times 10^{-2}$.
Comparing this with $\alpha \times 10^{-2}$,we get $\alpha = 2$.

(D) The bulk modulus $\beta$ is defined as $\beta = -\frac{\Delta P}{\Delta V / V}$.
Here,the change in pressure $\Delta P$ at a depth $h$ is given by $\Delta P = \rho gh$.
The fractional change in volume is $\frac{\Delta V}{V} = -0.02 \% = -\frac{0.02}{100} = -2 \times 10^{-4}$.
Substituting the values into the formula:
$\rho gh = -\beta \left( \frac{\Delta V}{V} \right)$
$10^3 \times 10 \times h = -(9 \times 10^8) \times (-2 \times 10^{-4})$
$10^4 \times h = 18 \times 10^4$
$h = 18 \ m$.

(B) The pressure at depth $h$ is given by $P = \rho g h$.
Given $h = 5 \text{ km} = 5000 \text{ m}$,$\rho = 10^3 \text{ kg m}^{-3}$,and $g = 10 \text{ m s}^{-2}$,the pressure is $P = 10^3 \times 10 \times 5000 = 5 \times 10^7 \text{ Pa}$.
The bulk modulus $B$ is defined as $B = -V \frac{dP}{dV}$. For a cube of side $a$,$V = a^3$,so $dV = 3a^2 da$.
Thus,$\frac{dV}{V} = \frac{3a^2 da}{a^3} = 3 \frac{da}{a}$.
Substituting this into the bulk modulus formula: $B = -\frac{P}{3 \frac{da}{a}} \implies \frac{da}{a} = \frac{P}{3B}$.
Here,$a = 1 \text{ m}$,$P = 5 \times 10^7 \text{ Pa}$,and $B = 70 \times 10^9 \text{ Pa}$.
$da = \frac{a \times P}{3B} = \frac{1 \times 5 \times 10^7}{3 \times 70 \times 10^9} = \frac{5}{210} \times 10^{-2} \text{ m} = \frac{1}{42} \times 10^{-2} \text{ m} \approx 0.0238 \times 10^{-1} \text{ m} = 2.38 \times 10^{-3} \text{ m} = 2.38 \text{ mm}$.

(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 $B = 2.15 \times 10^9 \text{ Nm}^{-2}$
Volumetric strain $\frac{\Delta V}{V} = -0.2\% = -0.002$
Substituting the values into the formula:
$2.15 \times 10^9 = -\frac{\Delta P}{-0.002}$
$\Delta P = 2.15 \times 10^9 \times 0.002$
$\Delta P = 2.15 \times 10^9 \times 2 \times 10^{-3}$
$\Delta P = 4.3 \times 10^6 \text{ Nm}^{-2}$
We need to express this in the form $\text{P} \times 10^5 \text{ Nm}^{-2}$:
$4.3 \times 10^6 = 43 \times 10^5 \text{ Nm}^{-2}$
Comparing this with $\text{P} \times 10^5 \text{ Nm}^{-2}$,we get $\text{P} = 43$.

(B) The formula for bulk modulus $B$ is given by $B = \frac{\Delta P}{\Delta V / V}$,where $\Delta P$ is the change in pressure,$V$ is the initial volume,and $\Delta V$ is the change in volume.
Given: Edge length $L = 10 \ cm = 0.1 \ m$. Initial volume $V = L^3 = (0.1 \ m)^3 = 10^{-3} \ m^3$.
Pressure change $\Delta P = 7 \times 10^6 \ Pa$.
Bulk modulus $B = 1.4 \times 10^{11} \ Nm^{-2}$.
Rearranging the formula for volume contraction $\Delta V$: $\Delta V = \frac{\Delta P \times V}{B}$.
Substituting the values: $\Delta V = \frac{7 \times 10^6 \times 10^{-3}}{1.4 \times 10^{11}} = \frac{7 \times 10^3}{1.4 \times 10^{11}} = 5 \times 10^{-8} \ m^3$.
Since $1 \ m^3 = 10^9 \ mm^3$,we convert the volume: $\Delta V = 5 \times 10^{-8} \times 10^9 \ mm^3 = 50 \ mm^3$.

(D) The pressure at a depth $h$ is given by $P = \rho gh$.
The Bulk modulus $B$ is defined as $B = \frac{P}{\left(\frac{\Delta V}{V}\right)}$.
Therefore,the fractional compression is $\frac{\Delta V}{V} = \frac{P}{B} = \frac{\rho gh}{B}$.
Substituting the given values: $\rho = 10^3 \ kg m^{-3}$,$g = 10 \ m s^{-2}$,$h = 2.5 \ km = 2.5 \times 10^3 \ m$,and $B = 2 \times 10^9 \ N m^{-2}$.
$\frac{\Delta V}{V} = \frac{10^3 \times 10 \times 2.5 \times 10^3}{2 \times 10^9} = \frac{2.5 \times 10^7}{2 \times 10^9} = 1.25 \times 10^{-2}$.
To express this as a percentage: $\frac{\Delta V}{V} \times 100 = 1.25 \times 10^{-2} \times 100 = 1.25 \%$.

(B) Given: Initial pressure $(P_i) = 1 \ atm$,Final pressure $(P_f) = 5 \ atm$.
Change in pressure $(dP) = P_f - P_i = 4 \ atm = 4 \times 10^5 \ Pa$.
Change in volume $(dV) = -0.8 \ cm^3 = -0.8 \times 10^{-6} \ m^3$.
Bulk modulus $(B) = 2 \ GPa = 2 \times 10^9 \ Pa$.
The formula for bulk modulus is $B = -\frac{dP}{dV/V}$,which implies $V = -\frac{B \cdot dV}{dP}$.
Substituting the values: $V = -\frac{2 \times 10^9 \times (-0.8 \times 10^{-6})}{4 \times 10^5}$.
$V = \frac{1.6 \times 10^3}{4 \times 10^5} = 0.4 \times 10^{-2} \ m^3 = 4 \times 10^{-3} \ m^3$.
Since $1 \ m^3 = 1000 \ litres$,$V = 4 \times 10^{-3} \times 1000 = 4 \ litres$.

(C) Compressibility $\beta$ is defined as the reciprocal of the Bulk Modulus $K$,i.e.,$\beta = \frac{1}{K}$.
Given:
Compressibility $\beta = 5 \times 10^{-10} \ m^2/N$
Change in pressure $\Delta P = 15 \times 10^6 \ Pa$
Initial volume $V = 100 \ ml$
The formula for volumetric strain is $\frac{\Delta V}{V} = -\beta \Delta P$.
Substituting the values:
$\frac{\Delta V}{100 \ ml} = -(5 \times 10^{-10} \ m^2/N) \times (15 \times 10^6 \ Pa)$
$\frac{\Delta V}{100 \ ml} = -75 \times 10^{-4} = -0.0075$
$\Delta V = -0.0075 \times 100 \ ml = -0.75 \ ml$.
The negative sign indicates a decrease in volume.
Therefore,the change in volume is a $0.75 \ ml$ decrease.

(A) The density $\rho$ is given by $\rho = M / V$,where $M$ is mass and $V$ is volume.
Differentiating both sides,we get $\frac{d\rho}{\rho} = -\frac{dV}{V}$.
The bulk modulus $K$ is defined as $K = -\frac{P}{dV/V}$,which implies $-\frac{dV}{V} = \frac{P}{K}$.
Substituting this into the density relation,we get $\frac{d\rho}{\rho} = \frac{P}{K}$.
Therefore,the increase in density is $d\rho = \frac{\rho P}{K}$.

(A) The bulk modulus $K$ is defined as $K = -\frac{\Delta p}{\Delta V / V}$.
Given that the volume decreases by $x \%$,we have $\frac{\Delta V}{V} = \frac{x}{100}$.
The change in pressure $\Delta p$ at a depth $h$ in the sea is given by $\Delta p = \rho g h$.
Substituting these into the bulk modulus formula: $K = \frac{\rho g h}{x / 100}$.
Rearranging for depth $h$: $h = \frac{K \cdot x}{100 \cdot \rho \cdot g}$.
Since $100$ is a constant,the depth $h$ is proportional to $\frac{Kx}{\rho g}$.

(D) The bulk modulus $K$ is defined as $K = -V \frac{dp}{dV}$.
For a small change in pressure $p$,we have $K = -V \frac{p}{\Delta V}$,which gives $\Delta V = -\frac{pV}{K}$.
The new volume $V^{\prime}$ is $V^{\prime} = V + \Delta V = V - \frac{pV}{K} = V(1 - \frac{p}{K})$.
Since density $\varrho = \frac{m}{V}$,the new density is $\varrho^{\prime} = \frac{m}{V^{\prime}} = \frac{m}{V(1 - \frac{p}{K})}$.
Substituting $\varrho = \frac{m}{V}$,we get $\varrho^{\prime} = \frac{\varrho}{1 - \frac{p}{K}}$.
Therefore,the ratio $\frac{\varrho^{\prime}}{\varrho} = \frac{1}{1 - \frac{p}{K}}$.

(C) The compressibility $K$ is defined as the reciprocal of the Bulk modulus $B$, given by $K = \frac{1}{B} = -\frac{\Delta V}{P V}$.
Here, the negative sign indicates that an increase in pressure leads to a decrease in volume.
Given:
Compressibility $K = 6 \times 10^{-10} \,m^{2}/N$
Initial volume $V = 1 \,L = 10^{-3} \,m^{3}$
Pressure change $P = 4 \times 10^{7} \,N/m^{2}$
The decrease in volume $\Delta V$ is given by:
$\Delta V = K \cdot P \cdot V$
$\Delta V = (6 \times 10^{-10} \,m^{2}/N) \times (4 \times 10^{7} \,N/m^{2}) \times (10^{-3} \,m^{3})$
$\Delta V = 24 \times 10^{-6} \,m^{3}$
Since $1 \,m^{3} = 10^{3} \,L = 10^{6} \,mL$, we convert the volume change to millilitres:
$\Delta V = 24 \times 10^{-6} \times 10^{6} \,mL = 24 \,mL$.

(D) The formula for Bulk modulus $K$ is given by $K = -\frac{\Delta p}{\Delta V / V}$.
We are given the Bulk modulus $K = 2100 \, MPa = 2100 \times 10^3 \, kPa$.
The fractional change in volume is $\frac{\Delta V}{V} = 0.008 \, \% = \frac{0.008}{100} = 8 \times 10^{-5}$.
To find the increase in pressure $\Delta p$, we rearrange the formula: $\Delta p = K \times \left( \frac{\Delta V}{V} \right)$.
Substituting the values: $\Delta p = (2100 \times 10^3 \, kPa) \times (8 \times 10^{-5})$.
$\Delta p = 2100 \times 8 \times 10^{-2} \, kPa = 21 \times 8 \, kPa = 168 \, kPa$.

(B) The ratio of hydraulic stress to the corresponding volumetric strain is defined as the Bulk modulus $(B)$.
Mathematically,$B = -\frac{\Delta P}{\Delta V / V}$,where $\Delta P$ is the change in pressure and $\frac{\Delta V}{V}$ is the volumetric strain.

(C) The pressure at depth $h$ is given by $P = \rho gh$.
Given: $h = 1 \ km = 10^3 \ m$,$\rho = 10^3 \ kg \ m^{-3}$,$g = 10 \ m \ s^{-2}$.
$P = 10^3 \times 10 \times 10^3 = 10^7 \ N \ m^{-2}$.
The fractional change in volume is $\frac{\Delta V}{V} = 0.01 \% = \frac{0.01}{100} = 10^{-4}$.
The bulk modulus $B$ is defined as $B = \frac{P}{\Delta V / V}$.
$B = \frac{10^7}{10^{-4}} = 10^{11} \ N \ m^{-2}$.
Wait,checking the options: $10 \times 10^{10} = 10^{11}$. Thus,option $C$ is correct.