Bulk Modulus Questions in English

(C) The change in pressure is $\Delta P = 250 \text{ kPa} - 200 \text{ kPa} = 50 \text{ kPa} = 50 \times 10^3 \text{ Pa} = 5 \times 10^4 \text{ Pa}$.
The fractional change in volume is $\frac{\Delta V}{V} = 0.25 \% = \frac{0.25}{100} = 2.5 \times 10^{-3}$.
Bulk modulus $B$ is given by $B = -\frac{\Delta P}{\Delta V / V}$.
Compressibility $K$ is the reciprocal of the bulk modulus,$K = \frac{1}{B} = \frac{\Delta V / V}{\Delta P}$.
Substituting the values: $K = \frac{2.5 \times 10^{-3}}{5 \times 10^4} = 0.5 \times 10^{-7} \text{ m}^2 \text{ N}^{-1} = 5 \times 10^{-8} \text{ m}^2 \text{ N}^{-1}$.

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

(A) Given:
Initial volume,$V = 2000 \text{ cm}^3 = 2000 \times 10^{-6} \text{ m}^3 = 2 \times 10^{-3} \text{ m}^3$.
Change in volume,$\Delta V = 5 \times 10^{-2} \text{ cm}^3 = 5 \times 10^{-2} \times 10^{-6} \text{ m}^3 = 5 \times 10^{-8} \text{ m}^3$.
Pressure,$P = 15 \text{ atm} = 15 \times 10^5 \text{ Nm}^{-2}$.
The formula for Bulk modulus $(B)$ is given by:
$B = \frac{P}{\left(\frac{\Delta V}{V}\right)} = \frac{P \times V}{\Delta V}$
Substituting the values:
$B = \frac{15 \times 10^5 \times 2 \times 10^{-3}}{5 \times 10^{-8}}$
$B = \frac{30 \times 10^2}{5 \times 10^{-8}}$
$B = 6 \times 10^{10} \text{ Nm}^{-2}$.

(B) The hydraulic pressure $P$ applied to the glass slab is $14 \,atm$. Converting this to Pascals $(Pa)$:
$P = 14 \times 1.013 \times 10^5 \,Pa \approx 14.182 \times 10^5 \,Pa$.
Given the Bulk modulus $B = 40 \times 10^9 \,N/m^2$.
The formula for Bulk modulus is $B = -\frac{P}{\Delta V/V}$, where $\frac{\Delta V}{V}$ is the fractional change in volume.
Therefore, the fractional change in volume is given by:
$\frac{\Delta V}{V} = \frac{P}{B}$.
Substituting the values:
$\frac{\Delta V}{V} = \frac{14 \times 1.013 \times 10^5}{40 \times 10^9} = \frac{14.182 \times 10^5}{40 \times 10^9} = 0.35455 \times 10^{-4} = 3.5455 \times 10^{-5}$.
Rounding to the nearest option, we get $3.54 \times 10^{-5}$.

(D) Given: Initial volume $V = 4000 \ cc$.
Fractional change in volume $\frac{\Delta V}{V} = 0.05 \% = \frac{0.05}{100} = 0.0005$.
Bulk modulus $B = 2.2 \times 10^9 \ N/m^2$.
The formula for Bulk modulus is $B = -\frac{P}{\Delta V/V}$,where $P$ is the applied pressure.
Taking the magnitude,$P = B \times \frac{\Delta V}{V}$.
Substituting the values: $P = (2.2 \times 10^9) \times (0.0005)$.
$P = 2.2 \times 10^9 \times 5 \times 10^{-4} = 11 \times 10^5 = 1.1 \times 10^6 \ N/m^2$.

(B) The bulk modulus $K$ is defined as $K = -V \frac{dP}{dV}$.
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 $\gamma P + V \frac{dP}{dV} = 0$.
Rearranging the terms,we find $V \frac{dP}{dV} = -\gamma P$.
Substituting this into the definition of bulk modulus,we get $K = -(-\gamma P) = \gamma P$.

(A) Given:
Temperature of hot tea,$t_2 = 57^{\circ} C$
Normal temperature of tooth,$t_1 = 37^{\circ} C$
Coefficient of linear expansion,$\alpha = 1.7 \times 10^{-5} {}^{\circ} C^{-1}$
Bulk modulus,$B = 140 \times 10^9 \ Nm^{-2}$
Change in temperature,$\Delta t = t_2 - t_1 = 57 - 37 = 20^{\circ} C$
Thermal stress is given by the product of Bulk modulus and volumetric strain:
$\text{Stress} = B \times \frac{\Delta V}{V}$
Since $\frac{\Delta V}{V} = \gamma \Delta t$ and $\gamma = 3\alpha$:
$\text{Stress} = B \times (3\alpha) \times \Delta t$
Substituting the values:
$\text{Stress} = 3 \times (140 \times 10^9) \times (1.7 \times 10^{-5}) \times 20$
$\text{Stress} = 3 \times 140 \times 1.7 \times 20 \times 10^4$
$\text{Stress} = 14280 \times 10^4 = 1.428 \times 10^8 \ Nm^{-2} \approx 1.4 \times 10^8 \ Nm^{-2}$.

(C) The pressure applied on the block is $p = 8 \times 10^8 \ N \ m^{-2}$.
Let the initial volume be $V_1$. The final volume $V_2$ is reduced by $20 \%$,so $V_2 = V_1 - 0.20 V_1 = 0.8 V_1 = \frac{4}{5} V_1$.
The change in volume is $\Delta V = V_1 - V_2 = V_1 - 0.8 V_1 = 0.2 V_1 = \frac{V_1}{5}$.
The Bulk modulus $B$ is defined as $B = \frac{p}{\Delta V / V_1}$.
Substituting the values,$B = \frac{8 \times 10^8}{(0.2 V_1) / V_1} = \frac{8 \times 10^8}{0.2} = 40 \times 10^8 = 4 \times 10^9 \ N \ m^{-2}$.

(B) Given that:
Bulk modulus $(K) = 2 \times 10^8 \ N m^{-2}$
Volume strain $\frac{\Delta V}{V} = 1\% = 0.01$
The formula for Bulk modulus is:
$K = \frac{p}{\frac{\Delta V}{V}}$
Rearranging to solve for pressure $(p)$:
$p = K \times \left( \frac{\Delta V}{V} \right)$
Substituting the given values:
$p = (2 \times 10^8) \times (0.01)$
$p = 2 \times 10^6 \ N m^{-2}$
Therefore,the required pressure is $2 \times 10^6 \ N m^{-2}$.

(C) Given,edge of solid copper cube,$l = 7 \,cm$.
Volume of the cube,$V = l^3 = (7 \,cm)^3 = 343 \,cm^3 = 343 \times 10^{-6} \,m^3$.
Hydraulic pressure,$p = 8000 \,kPa = 8000 \times 10^3 \,Pa = 8 \times 10^6 \,Pa$.
Bulk modulus of copper,$\beta = 140 \,GPa = 140 \times 10^9 \,Pa$.
We know that the Bulk modulus is given by $\beta = \frac{p}{\Delta V / V}$,where $\Delta V$ is the change in volume.
Rearranging for $\Delta V$,we get $\Delta V = \frac{p V}{\beta}$.
Substituting the values:
$\Delta V = \frac{(8 \times 10^6 \,Pa) \times (343 \times 10^{-6} \,m^3)}{140 \times 10^9 \,Pa} = \frac{2744}{140 \times 10^9} \,m^3 = 19.6 \times 10^{-9} \,m^3$.
Converting back to $cm^3$:
$1 \,m^3 = 10^6 \,cm^3$,so $\Delta V = 19.6 \times 10^{-9} \times 10^6 \,cm^3 = 19.6 \times 10^{-3} \,cm^3$.

(B) The pressure at depth $h$ is given by $\Delta P = h \rho g$.
Bulk modulus $B$ is defined as $B = \frac{\Delta P}{\Delta V / V}$.
Therefore,the fractional compression is $\frac{\Delta V}{V} = \frac{\Delta P}{B} = \frac{h \rho g}{B}$.
Given: $h = 3000 \,m$,$\rho = 1000 \,kg/m^3$,$g = 9.8 \,m/s^2$,and $B = 2.2 \times 10^9 \,N/m^2$.
Substituting these values:
$\frac{\Delta V}{V} = \frac{3000 \times 1000 \times 9.8}{2.2 \times 10^9} = \frac{2.94 \times 10^7}{2.2 \times 10^9} = 1.336 \times 10^{-2} \approx 1.34 \times 10^{-2}$.

(D) Consider a volume $V_1$ of liquid at the top surface. Due to pressure at depth $H$,the same mass of liquid occupies a volume $V_2$.
From the conservation of mass,we have $\rho_0 V_1 = \rho V_2$,which implies $\frac{V_1}{V_2} = \frac{\rho}{\rho_0}$.
At depth $H$,the gauge pressure is $P = \rho g H$.
From the definition of bulk modulus $B_0$,we have $B_0 = -\frac{\Delta P}{\Delta V / V_1} = -\frac{P}{(V_2 - V_1) / V_1}$.
Rearranging this,we get $\frac{V_2 - V_1}{V_1} = -\frac{P}{B_0} = -\frac{\rho g H}{B_0}$.
Thus,$\frac{V_2}{V_1} = 1 - \frac{\rho g H}{B_0}$.
Taking the reciprocal,$\frac{V_1}{V_2} = \frac{1}{1 - \frac{\rho g H}{B_0}}$.
Substituting this into the mass conservation equation $\rho = \rho_0 \left( \frac{V_1}{V_2} \right)$,we get $\rho = \frac{\rho_0}{1 - \frac{\rho g H}{B_0}}$.
Comparing this with the given expression $\rho = \frac{\rho_0}{1 + \alpha \rho g H}$,we identify $\alpha = -\frac{1}{B_0}$.

(B) The Bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V/V}$.
Given: $B = 60 \ GPa = 60 \times 10^9 \ Pa$,$\Delta P = 10^7 \ Pa$,$r = 36 \ cm = 0.36 \ m$.
The volume of a sphere 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 this into the Bulk modulus formula: $B = -\frac{\Delta P}{3 \Delta r / r}$.
Rearranging for $\Delta r$: $\Delta r = -\frac{\Delta P \cdot r}{3B}$.
Taking the magnitude: $|\Delta r| = \frac{10^7 \times 0.36}{3 \times 60 \times 10^9}$.
$|\Delta r| = \frac{3.6 \times 10^6}{180 \times 10^9} = \frac{3.6}{180} \times 10^{-3} = 0.02 \times 10^{-3} \ m = 2 \times 10^{-5} \ m$.
Converting to cm: $2 \times 10^{-5} \times 10^2 \ cm = 2 \times 10^{-3} \ cm$.

(D) The increase in pressure at the bottom of the pool due to the water column is given by $\Delta P = \rho g h$.
Substituting the given values: $\Delta P = 1000 \times 10 \times 22 = 2.2 \times 10^5 \ N \ m^{-2}$.
The bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V / V}$.
Therefore,the fractional change in volume is $\left| \frac{\Delta V}{V} \right| = \frac{\Delta P}{B}$.
Substituting the values: $\left| \frac{\Delta V}{V} \right| = \frac{2.2 \times 10^5}{2.2 \times 10^9} = 10^{-4}$.

(D) The formula for bulk modulus $B$ is given by:
$B = -\frac{\Delta P}{\Delta V / V}$
Given:
Change in pressure,$\Delta P = 2 \text{ atm} - 1 \text{ atm} = 1 \text{ atm} = 1.01 \times 10^5 \text{ N/m}^2$.
Fractional change in volume,$\frac{\Delta V}{V} = -2 \% = -0.02$ (negative sign indicates reduction in volume).
Substituting these values into the formula:
$B = -\frac{1.01 \times 10^5}{-0.02}$
$B = \frac{1.01 \times 10^5}{0.02}$
$B = 50.5 \times 10^5 \text{ N/m}^2 = 5.05 \times 10^6 \text{ N/m}^2$
Rounding to the nearest option,we get $B \approx 5 \times 10^6 \text{ N/m}^2$.

(A) The bulk modulus $B$ is defined as $B = -\frac{p}{\Delta V / V}$, where $p$ is the pressure and $\frac{\Delta V}{V}$ is the volumetric strain.
Therefore, the fractional compression is given by $-\frac{\Delta V}{V} = \frac{p}{B}$.
The pressure at the bottom of the well is $p = \rho g h$.
Substituting the values: $\rho = 1500 \, kg/m^3$, $g = 10 \, m/s^2$, $h = 2000 \, m$, and $B = 8 \times 10^8 \, N/m^2$.
Fractional compression (in $\%$) $= \frac{\rho g h}{B} \times 100$.
$= \frac{1500 \times 10 \times 2000}{8 \times 10^8} \times 100$.
$= \frac{3 \times 10^7}{8 \times 10^8} \times 100 = \frac{3}{80} \times 100 = 3.75 \%$.

(B) Given: Bulk modulus $B = 2 \times 10^9 \ N/m^2$.
The fractional change in volume is $\frac{\Delta V}{V} = -2 \% = -0.02 = -\frac{1}{50}$.
The formula for bulk modulus is $B = -\frac{\Delta p}{\Delta V/V}$,where $\Delta p$ is the change in pressure.
Rearranging for pressure: $\Delta p = -B \left( \frac{\Delta V}{V} \right)$.
Substituting the values: $\Delta p = -(2 \times 10^9) \times (-0.02)$.
$\Delta p = 2 \times 10^9 \times 0.02 = 4 \times 10^7 \ N/m^2$.

(C) Given,percentage change in volume $\frac{|\Delta V|}{V} = 0.4 \% = 0.004$.
Bulk modulus of water,$B = 2.0 \times 10^9 \text{ Pa}$.
The formula for Bulk modulus is $B = -\frac{p}{\Delta V / V}$,where $p$ is the applied pressure.
Therefore,the pressure required is $p = B \times \left| \frac{\Delta V}{V} \right|$.
Substituting the values:
$p = (2.0 \times 10^9 \text{ Pa}) \times 0.004 = 8.0 \times 10^6 \text{ Pa}$.
To convert pressure from $\text{Pa}$ to $\text{atm}$,we use the conversion factor $1 \text{ atm} \approx 1.01325 \times 10^5 \text{ Pa}$.
$p = \frac{8.0 \times 10^6}{1.01325 \times 10^5} \text{ atm} \approx 78.95 \text{ atm}$.
Rounding to the nearest option,we get $p \approx 80 \text{ atm}$.

(C) The formula for Bulk modulus $(K)$ is given by $K = -\frac{p}{\Delta V / V}$,where $p$ is the change in pressure and $\Delta V / V$ is the volumetric strain.
Given,Bulk modulus $K = 2 \times 10^9 \ N/m^2$.
The fractional change in volume is $\frac{\Delta V}{V} = 0.1 \% = \frac{0.1}{100} = 10^{-3}$.
Since we are increasing the volume,the pressure change $p$ is negative (tensile stress),but we are looking for the magnitude of pressure required.
Using the magnitude: $p = K \times \left( \frac{\Delta V}{V} \right)$.
Substituting the values: $p = (2 \times 10^9) \times (10^{-3})$.
$p = 2 \times 10^6 \ N/m^2$.

(B) The bulk modulus $(B)$ is defined as the ratio of change in pressure to the fractional change in volume.
Solids have a very high bulk modulus compared to gases because they are densely packed and resistant to volume change.
Therefore,solids are the least compressible,while gases are the most compressible.
Statement $(B)$ claims that gases are least compressible,which is incorrect.
The incompressibility of solids arises from the strong interatomic forces (tight coupling) between neighboring atoms.
Compressibility is defined as the reciprocal of the bulk modulus $(K = 1/B)$.

(A) The bulk modulus $k$ is defined as $k = -\frac{p}{\Delta V / V}$.
Since density $\rho = \frac{m}{V}$,we have $\frac{\Delta \rho}{\rho} = -\frac{\Delta V}{V}$.
Given that the density increases by $0.01 \%$,we have $\frac{\Delta \rho}{\rho} = 0.01 \% = \frac{0.01}{100} = 10^{-4}$.
Therefore,$-\frac{\Delta V}{V} = 10^{-4}$.
Substituting this into the bulk modulus formula: $p = k \times \left( -\frac{\Delta V}{V} \right)$.
$p = k \times 10^{-4} = \frac{k}{10000}$.

(B) The bulk modulus $B$ is defined as the ratio of the change in pressure $\Delta P$ to the volumetric strain $-\frac{\Delta V}{V}$.
Mathematically,$B = -\frac{\Delta P}{\Delta V/V}$.
Rearranging the formula to find the fractional change in volume,we get $\frac{\Delta V}{V} = \frac{\Delta P}{B}$.
Given $\Delta P = 6.3 \times 10^7 \text{ N/m}^2$ and $B = 2.1 \times 10^9 \text{ N/m}^2$.
Substituting the values: $\frac{\Delta V}{V} = \frac{6.3 \times 10^7}{2.1 \times 10^9} = 3 \times 10^{-2} = 0.03$.
To find the percentage decrease,multiply by $100$: $0.03 \times 100 = 3\%$.