Pressure and Density (of Mixure) Questions in English

(B) The pressure at a depth $h$ in a static fluid is given by the formula $P = P_0 + \rho gh$,where $P_0$ is the pressure at the surface,$\rho$ is the density of the fluid,and $g$ is the acceleration due to gravity.
In a continuous static fluid,the pressure is the same at all points at the same horizontal level.
Looking at the figure,points $B$,$C$,and $D$ are located at the same horizontal level within the continuous mercury column. Therefore,the pressures at these points are equal: $P_B = P_C = P_D$.
Point $A$ is located at a higher vertical position than points $B$,$C$,and $D$. Since pressure decreases with height in a fluid column,the pressure at $A$ must be less than the pressure at the level of $B$,$C$,and $D$.
Thus,the correct relationship is $P_B = P_C = P_D > P_A$.

(B) Let the density of water be $\rho_w = 1 \,g/cm^3$ and the density of mercury be $\rho_m = 13.6 \,g/cm^3$.
When water of height $h_w = 13.6 \,cm$ is poured into one arm,the mercury level in that arm drops by $y$,and it rises by $y$ in the other arm.
The difference in the mercury levels between the two arms becomes $2y$.
According to the principle of hydrostatic pressure at the same horizontal level:
$P_{\text{water}} = P_{\text{mercury}}$
$\rho_w \cdot g \cdot h_w = \rho_m \cdot g \cdot (2y)$
Substituting the values:
$1 \cdot g \cdot 13.6 = 13.6 \cdot g \cdot (2y)$
$13.6 = 13.6 \cdot 2y$
$2y = 1$
$y = 0.5 \,cm$
Therefore,the mercury rises by $0.5 \,cm$ in the other arm.

(D) The term 'fluid' refers to any substance that has the ability to flow and does not have a fixed shape.
Since both liquids and gases can flow under the influence of an external force,they are collectively classified as fluids.
Therefore,the correct option is $D$.

(D) The correct answer is $D$.
$1$. Pressure is a scalar quantity because it is defined as force per unit area and does not follow vector addition rules.
$2$. Pressure is always compressive in nature,meaning it acts to compress the fluid or object it is applied to.
$3$. According to Pascal's Law,the pressure at a point in a fluid at rest is the same in all directions.
Since all the statements ($A$,$B$,and $C$) are correct,the wrong statement is none of these.

(D) Gauge pressure is defined as the difference between the absolute pressure $(P)$ and the atmospheric pressure $(P_{atm})$,given by the formula: $P_g = P - P_{atm}$.
Since the absolute pressure $(P)$ can be greater than,less than,or equal to the atmospheric pressure $(P_{atm})$,the gauge pressure $(P_g)$ can be positive,negative,or zero respectively.
Therefore,all the given options are correct.

(B) Let the height difference between the two liquid levels be $h$.
According to the principle of hydrostatics,the pressure at the same horizontal level in a continuous static fluid is the same.
Considering the horizontal level at the lower liquid surface,the pressure on the left side is $P + \rho g h$,where $\rho$ is the density of the liquid and $g$ is the acceleration due to gravity.
The pressure on the right side is $p$.
Equating the two,we get $P + \rho g h = p$.
Since $\rho, g, h > 0$,it follows that $P + \rho g h > P$.
Therefore,$p > P$,which means $P < p$.

(C) Points at the same horizontal level in a continuous static liquid have the same pressure. Points at different depths have a pressure difference given by $\Delta p = \rho g \Delta h$.
Let the radius of the circle be $r$ and the depth of point $A$ from the free surface be $h$. Let $p_0$ be the atmospheric pressure.
The absolute pressures at the points are:
$p_A = p_0 + h \rho g$
$p_B = p_D = p_0 + (h + r) \rho g$
$p_C = p_0 + (h + 2r) \rho g$
Now,let us evaluate the expressions in the options:
$1$. $p_D = p_B$ is correct as they are at the same horizontal level.
$2$. Since $h < h+r < h+2r$,it follows that $p_A < p_B = p_D < p_C$,which is correct.
$3$. Calculating the average of $p_A$ and $p_C$:
$\frac{p_C + p_A}{2} = \frac{(p_0 + h \rho g + 2r \rho g) + (p_0 + h \rho g)}{2} = \frac{2p_0 + 2h \rho g + 2r \rho g}{2} = p_0 + (h + r) \rho g = p_B = p_D$.
Thus,the statement $p_D = p_B = \frac{p_C + p_A}{2}$ is correct,and the statement $p_D = p_B = \frac{p_C - p_A}{2}$ is incorrect.
Therefore,option $(c)$ is the incorrect statement.

(A) Let $V$ be the initial volume of the air bubble at the bottom of the lake and $2V$ be its volume at the surface.
Let $P_0$ be the atmospheric pressure and $h$ be the depth of the lake.
According to Boyle's Law,since the temperature is assumed to be constant,$P_1 V_1 = P_2 V_2$.
At the surface,the pressure is $P_2 = P_0$.
At the bottom,the pressure is $P_1 = P_0 + \rho_w g h$,where $\rho_w$ is the density of water.
Given $P_1 V = P_2 (2V) \Rightarrow P_1 = 2 P_0$.
Substituting $P_1$,we get $P_0 + \rho_w g h = 2 P_0 \Rightarrow \rho_w g h = P_0$.
We know $P_0 = h_m \rho_m g$,where $h_m = 0.75 \, m$ and $\rho_m$ is the density of mercury.
So,$\rho_w g h = h_m \rho_m g \Rightarrow h = h_m \left( \frac{\rho_m}{\rho_w} \right)$.
Given $\frac{\rho_m}{\rho_w} = \frac{40}{3}$ and $h_m = 0.75 \, m = \frac{3}{4} \, m$.
$h = \frac{3}{4} \times \frac{40}{3} = 10 \, m$.

(B) Let the height of the glycerine column be $h = 10 \, cm$ and its relative density be $\rho_g = 1.2 \, g/cm^3$. The density of water is $\rho_w = 1.0 \, g/cm^3$.
Let the difference in the levels of the free surfaces of the two liquids be $x$.
At the interface of the glycerine and water in the $U$-tube,the pressure at the same horizontal level must be equal.
Let the pressure at the level of the bottom of the glycerine column be $P$.
Pressure on the left limb at this level = $P_{atm} + \rho_g \cdot g \cdot h$.
Pressure on the right limb at the same level = $P_{atm} + \rho_w \cdot g \cdot (h - x)$.
Equating the pressures:
$\rho_g \cdot g \cdot h = \rho_w \cdot g \cdot (h - x)$
$1.2 \cdot 10 = 1.0 \cdot (10 - x)$
$12 = 10 - x$
Wait,let's re-evaluate the geometry. The glycerine column of height $h$ pushes the water down by some amount,and the water level rises in the other arm. The difference in height between the top of the glycerine and the top of the water is $x$.
From the hydrostatic balance at the level of the glycerine-water interface:
Pressure due to glycerine column = Pressure due to water column of height $(h - x)$.
$1.2 \times 10 = 1.0 \times (10 - x)$ is incorrect based on the diagram. Looking at the diagram,the water level in the right arm is higher than the interface by $(h - x)$.
Actually,the pressure balance is $\rho_g \cdot g \cdot h = \rho_w \cdot g \cdot h_{water}$.
Here,the water column height balancing the glycerine is $h_{water} = h - x$.
$1.2 \times 10 = 1.0 \times (10 - x) \Rightarrow 12 = 10 - x \Rightarrow x = -2$. This implies the water level is higher on the glycerine side,which is wrong.
Correct approach: The pressure at the level of the glycerine-water interface (left side) is equal to the pressure at the same level in the right side.
Pressure (left) = $\rho_g \cdot g \cdot h$.
Pressure (right) = $\rho_w \cdot g \cdot (h - x)$ where $x$ is the difference in free surface levels.
$1.2 \times 10 = 1.0 \times (10 - x) \Rightarrow 12 = 10 - x$. This suggests the water level in the right arm must be higher than the glycerine surface.
Actually,$h_{water} = h \times (\rho_g / \rho_w) = 10 \times 1.2 = 12 \, cm$.
The difference in levels $x = h_{water} - h = 12 - 10 = 2 \, cm$.

(C) Let the mass of the liquid with density $\alpha$ be $M_1$ and the mass of the liquid with density $\beta$ be $M_2$.
Total volume $= V$.
Total density of the mixture $= \sigma$.
Total mass $= M_1 + M_2 = V\sigma$.
Therefore,$M_2 = V\sigma - M_1 \dots (1)$.
The density of the mixture is given by $\sigma = \frac{\text{Total Mass}}{\text{Total Volume}} = \frac{M_1 + M_2}{\frac{M_1}{\alpha} + \frac{M_2}{\beta}} \dots (2)$.
Substituting equation $(1)$ into equation $(2)$:
$\sigma = \frac{V\sigma}{\frac{M_1}{\alpha} + \frac{V\sigma - M_1}{\beta}}$.
$\frac{M_1}{\alpha} + \frac{V\sigma - M_1}{\beta} = \frac{V\sigma}{\sigma} = V$.
$\frac{M_1}{\alpha} - \frac{M_1}{\beta} = V - \frac{V\sigma}{\beta}$.
$M_1 \left( \frac{\beta - \alpha}{\alpha\beta} \right) = V \left( \frac{\beta - \sigma}{\beta} \right)$.
$M_1 = \frac{V(\beta - \sigma)}{\beta} \times \frac{\alpha\beta}{\beta - \alpha} = \frac{\alpha V(\beta - \sigma)}{\beta - \alpha}$.

(D) Let the mass of each liquid be $M$.
The density of a substance is given by $\text{density} = \frac{\text{mass}}{\text{volume}}$,so the volume of each liquid is $V_1 = \frac{M}{d_1}$ and $V_2 = \frac{M}{d_2}$.
The total mass of the mixture is $M_{\text{mix}} = M + M = 2M$.
The total volume of the mixture is $V_{\text{mix}} = V_1 + V_2 = \frac{M}{d_1} + \frac{M}{d_2} = M \left( \frac{d_1 + d_2}{d_1 d_2} \right)$.
The density of the mixture is $d_{\text{mix}} = \frac{M_{\text{mix}}}{V_{\text{mix}}} = \frac{2M}{M \left( \frac{d_1 + d_2}{d_1 d_2} \right)} = \frac{2 d_1 d_2}{d_1 + d_2}$.

(C) Let the gate be of height $H = 1\,m$ and width $W = 1\,m$. The hinge is at the mid-point,i.e.,at a depth of $0.5\,m$ from the top edge of the gate. Let $y$ be the depth from the top of the gate.
The pressure at depth $y$ is $P = \rho g y$.
The force on an elemental strip of height $dy$ at depth $y$ is $dF_p = P \times W \times dy = \rho g y \times 1 \times dy = \rho g y dy$.
The torque $d\tau$ due to this force about the hinge (located at $y = 0.5\,m$) is $d\tau = dF_p \times (0.5 - y) = \rho g y (0.5 - y) dy$.
To keep the gate stationary,the total torque about the hinge must be zero. The force $F$ is applied at the bottom edge $(y = 1\,m)$,so its lever arm relative to the hinge is $0.5\,m$.
$\int_0^1 \rho g y (0.5 - y) dy - F \times 0.5 = 0$
$\rho g \int_0^1 (0.5y - y^2) dy = 0.5 F$
$\rho g [0.5 \frac{y^2}{2} - \frac{y^3}{3}]_0^1 = 0.5 F$
$\rho g [0.25 - 0.333] = 0.5 F$
$\rho g [\frac{1}{4} - \frac{1}{3}] = 0.5 F$
$\rho g [-\frac{1}{12}] = 0.5 F$
Taking the magnitude,$F = \frac{\rho g}{12 \times 0.5} = \frac{\rho g}{6}$.

(B) Let the initial height of water in each arm be $H = 0.29 \ m$. The total volume of water is constant. When kerosene of height $h_k = 0.1 \ m$ is added to the left arm,the water level in the left arm drops by $x$ and rises by $x$ in the right arm.
Total height of liquid in left arm: $h_1 = (H - x) + h_k = 0.29 - x + 0.1 = 0.39 - x$.
Height of liquid in right arm: $h_2 = H + x = 0.29 + x$.
Equating the pressure at the bottom of the $U$-tube:
$P_{left} = P_{right}$
$P_0 + \rho_k g h_k + \rho_w g (h_1 - h_k) = P_0 + \rho_w g h_2$
$\rho_k h_k + \rho_w (h_1 - h_k) = \rho_w h_2$
$800 \times 0.1 + 1000 \times (h_1 - 0.1) = 1000 \times h_2$
$80 + 1000 h_1 - 100 = 1000 h_2$
$1000 (h_1 - h_2) = 20 \implies h_1 - h_2 = 0.02 \ m$.
We have the system of equations:
$1$) $h_1 + h_2 = (0.29 - x + 0.1) + (0.29 + x) = 0.68 \ m$.
$2$) $h_1 - h_2 = 0.02 \ m$.
Adding the equations: $2h_1 = 0.70 \implies h_1 = 0.35 \ m$.
Subtracting the equations: $2h_2 = 0.66 \implies h_2 = 0.33 \ m$.
Therefore,the ratio $\frac{h_1}{h_2} = \frac{0.35}{0.33} = \frac{35}{33}$.

(A) The pressure at depth $h = 3 \ m$ on the water side is $P_w = P_0 + \rho_w gh$,and on the liquid side is $P_{\ell} = P_0 + \rho_{\ell} gh$.
The force exerted by water on the door is $F_w = P_w A = (P_0 + \rho_w gh) A$.
The force exerted by the liquid on the door is $F_{\ell} = P_{\ell} A = (P_0 + \rho_{\ell} gh) A$.
For the door to remain closed,the external force $F_{ext}$ applied on the water side must satisfy $F_{ext} + F_w = F_{\ell}$.
Therefore,$F_{ext} = F_{\ell} - F_w = (P_0 + \rho_{\ell} gh) A - (P_0 + \rho_w gh) A = (\rho_{\ell} - \rho_w) ghA$.
Given: $\rho_{\ell} = 1500 \ kg/m^3$,$\rho_w = 1000 \ kg/m^3$,$g = 10 \ m/s^2$,$h = 3 \ m$,and $A = 100 \ cm^2 = 100 \times 10^{-4} \ m^2 = 0.01 \ m^2$.
Substituting the values: $F_{ext} = (1500 - 1000) \times 10 \times 3 \times 0.01 = 500 \times 30 \times 0.01 = 150 \ N$.

(A) Water vapour in the atmosphere is confined only to its lowest layer,i.e.,the troposphere.
This is the reason why all weather phenomena,such as cloud formation,rain,and storms,occur only in this layer.

(C) Let the mercury level in the $U$-tube be initially at the same height in both arms. When $11.2 \,cm$ of water is poured into the left arm, the mercury level in the left arm drops by $x \,cm$ and rises by $x \,cm$ in the right arm. The total difference in the mercury levels between the two arms becomes $2x \,cm$.
Equating the pressure at the same horizontal level (the interface of water and mercury in the left arm, point $A$, and the corresponding level in the right arm, point $B$):
$p_A = p_B$
$h_{water} \times \rho_{water} \times g = h_{Hg} \times \rho_{Hg} \times g$
Given $h_{water} = 11.2 \,cm = 0.112 \,m$, $\rho_{water} = 1000 \,kg/m^3$, and $\rho_{Hg} = 13600 \,kg/m^3$.
$0.112 \times 1000 = 2x \times 13600$
$112 = 27200x$
$x = \frac{112}{27200} \,m \approx 0.004117 \,m = 0.41 \,cm$.
Thus, the mercury rises by $0.41 \,cm$ in the other arm.

(C) The force exerted by a liquid on the base of a vessel is equal to the weight of the liquid contained in it,provided the vessel has vertical walls (like a cubical vessel).
The force $F$ is given by $F = mg$.
Since the masses of the liquids in all three vessels are equal $(m_{A} = m_{B} = m_{C} = m)$,the force exerted on the base of each vessel is $F_{A} = F_{B} = F_{C} = mg$.
Therefore,the force exerted by the liquid on the base is the same in all the vessels.

(A) The pressure at the bottom of a vessel containing a liquid of density $\rho$ at a height $h$ is given by the formula $P = P_0 + \rho gh$,where $P_0$ is the atmospheric pressure.
Since both vessels are filled with water (same density $\rho$) to the same height $h$,and are kept on the same horizontal plane (same atmospheric pressure $P_0$),the pressure at the bottom of both vessels depends only on the height $h$ and density $\rho$.
Therefore,the pressure at the bottom of vessel $A$ is $P_A = P_0 + \rho gh$ and the pressure at the bottom of vessel $B$ is $P_B = P_0 + \rho gh$.
Thus,$P_A = P_B$.
The ratio of the pressures at the bottom of the vessels $A$ and $B$ is $P_A : P_B = 1 : 1$.

(A) The pressure difference between the head and the feet is given by $\Delta P = \rho gh$.
Here, $\rho = 1.1 \times 10^3 \,kg \,m^{-3}$, $g = 10 \,ms^{-2}$, and $h = 1.65 \,m$.
$\Delta P = 1.1 \times 10^3 \times 10 \times 1.65 = 1.815 \times 10^4 \,Pa$.
The force $F$ required to balance this pressure on the blood vessel is $F = \Delta P \times A$, where $A$ is the surface area of the blood vessel.
The blood vessel is a cylinder with length $L = 1 \,cm = 10^{-2} \,m$ and diameter $d = 1 \,mm = 10^{-3} \,m$ (radius $r = 0.5 \times 10^{-3} \,m$).
The surface area $A$ (lateral surface area) is $A = 2 \pi r L$.
$A = 2 \times 3.14159 \times 0.5 \times 10^{-3} \times 10^{-2} = 3.14159 \times 10^{-5} \,m^2$.
$F = (1.815 \times 10^4) \times (3.14159 \times 10^{-5}) \approx 0.57 \,N$.

(B) In static equilibrium,the pressure at the same horizontal level in a continuous fluid is the same.
Let the pressure at the interface of liquid $A$ and liquid $B$ in the left arm be $P$.
For the left arm,the pressure at this level is $P_0 + \rho_A g h_A$,where $P_0$ is the atmospheric pressure.
For the right arm,the pressure at the same horizontal level is $P_0 + \rho_B g h_B$.
Equating the pressures:
$P_0 + \rho_A g h_A = P_0 + \rho_B g h_B$
$\rho_A h_A = \rho_B h_B$
Given that the density of liquid $A$ is twice the density of liquid $B$,i.e.,$\rho_A = 2\rho_B$.
Substituting this into the equation:
$(2\rho_B) h_A = \rho_B h_B$
$2 h_A = h_B$
$h_A = \frac{h_B}{2}$

(C) When a vessel containing a fluid of density $\rho$ up to height $h$ is accelerated downwards with acceleration $a_0$,the effective gravitational acceleration acting on the fluid is given by:
$g^{\prime} = g - a_0$ ...$(i)$
The pressure $p$ at the bottom of the vessel is the sum of atmospheric pressure $p_0$ and the gauge pressure due to the fluid column under effective gravity:
$p = p_0 + \rho g^{\prime} h$
Substituting the value of $g^{\prime}$ from equation $(i)$:
$p = p_0 + \rho(g - a_0)h$

(C) Let the depth of the lake be $h$. The atmospheric pressure $P_0$ is equivalent to $10 \ m$ of water column,so $P_0 = 10 \rho g$.
At the bottom of the lake,the pressure $P_1$ is the sum of atmospheric pressure and the pressure due to the water column of depth $h$:
$P_1 = P_0 + h \rho g = 10 \rho g + h \rho g = \rho g(10 + h)$.
The volume of the bubble at the bottom is $V_1 = \frac{4}{3} \pi r^3$.
At the surface,the pressure $P_2$ is equal to the atmospheric pressure:
$P_2 = P_0 = 10 \rho g$.
The volume of the bubble at the surface is $V_2 = \frac{4}{3} \pi (\frac{5r}{4})^3$.
Since the temperature is constant,we apply Boyle's Law $(P_1 V_1 = P_2 V_2)$:
$\rho g(10 + h) \cdot \frac{4}{3} \pi r^3 = 10 \rho g \cdot \frac{4}{3} \pi (\frac{5r}{4})^3$.
$(10 + h) = 10 \cdot \frac{125}{64}$.
$10 + h = \frac{1250}{64} = 19.53125$.
$h = 19.53125 - 10 = 9.53125 \ m$.
Thus,the depth of the lake is approximately $9.53 \ m$.

(B) Given: Depth $h = 100 \ m$,Compressibility $k = 0.5 \times 10^{-9} \ N^{-1} \ m^2$,Density of water $\rho = 10^3 \ kg/m^3$,Acceleration due to gravity $g = 10 \ m/s^2$.
Pressure at the bottom of the river is given by $P = \rho gh$.
$P = 10^3 \times 10 \times 100 = 10^6 \ N/m^2$.
Compressibility $k$ is defined as the reciprocal of Bulk Modulus $B$,where $k = \frac{1}{B} = \frac{(\Delta V / V)}{P}$.
Therefore,the fractional compression (fractional change in volume) is $\frac{\Delta V}{V} = k \times P$.
$\frac{\Delta V}{V} = (0.5 \times 10^{-9}) \times 10^6 = 0.5 \times 10^{-3}$.

(D) Let $M$ be the mass of chamber $B$ and $m$ be the mass of each ball. The total downward force exerted by the chamber and the balls on the gas is $F = (M + nm)g$.
If $A$ is the cross-sectional area of the chamber,the pressure $P$ of the gas is given by $P = F/A = (M + nm)g / A$.
When one ball is removed,the new force is $F^{\prime} = (M + (n-1)m)g$.
The new pressure $P^{\prime}$ is $P^{\prime} = F^{\prime} / A = (M + (n-1)m)g / A$.
The difference in pressure is $P - P^{\prime} = (M + nm)g/A - (M + nm - m)g/A = mg/A$.
Thus,$(P - P^{\prime}) / P = (mg/A) / ((M + nm)g/A) = m / (M + nm)$.
Assuming the mass of the chamber $M$ is negligible compared to the total mass of the balls (or that the problem implies the weight is primarily due to the balls),we have $M \approx 0$.
Then,$(P - P^{\prime}) / P = m / (nm) = 1/n$.

(D) The pressure $P$ at the bottom of a vessel filled with a liquid of density $\rho$ to a height $h$ is given by the formula: $P = \rho g h$.
Since the height $h$ is the same for all three vessels and $g$ is constant,the pressure $P$ is directly proportional to the density $\rho$ of the liquid $(P \propto \rho)$.
Given the densities are $\rho_A > \rho_B > \rho_C$,it follows that the pressure at the bottom of the vessels will be $P_A > P_B > P_C$.
Therefore,the pressure is maximum in the vessel containing liquid $A$.

(A) The pressure at the bottom of a vessel due to a liquid column is given by $P = h \rho g$,where $h$ is the height of the liquid column,$\rho$ is the density,and $g$ is the acceleration due to gravity.
Since the vessels are identical,their cross-sectional area $A$ is the same.
The mass $m$ of the liquid is given by $m = \rho V = \rho A h$.
Given that the masses are equal $(m_A = m_B = m_C = M)$,we have:
$M = \rho_A A h_A = \rho_B A h_B = \rho_C A h_C$.
This implies $\rho_A h_A = \rho_B h_B = \rho_C h_C = \frac{M}{A} = \text{constant}$.
The pressure at the bottom is $P = \rho g h = g (\rho h)$.
Since the product $\rho h$ is constant for all three liquids,the pressure $P$ at the bottom of each vessel is the same.

(D) Let the side length of the cubical block be $L = 10 \,cm = 0.1 \,m$.
The block floats at the interface. Let $h_w = 1.5 \,cm = 0.015 \,m$ be the depth of the block in water and $h_o = 10 \,cm - 1.5 \,cm = 8.5 \,cm = 0.085 \,m$ be the height of the block in oil.
The pressure at the lower face (in water) is $P_{lower} = P_{interface} + \rho_w g h_w$.
The pressure at the upper face (in oil) is $P_{upper} = P_{interface} - \rho_o g h_o$.
The difference in pressure between the lower and upper face is $\Delta P = P_{lower} - P_{upper} = (P_{interface} + \rho_w g h_w) - (P_{interface} - \rho_o g h_o) = \rho_w g h_w + \rho_o g h_o$.
Substituting the given values:
$\Delta P = (1000 \,kg/m^3 \times 10 \,m/s^2 \times 0.015 \,m) + (800 \,kg/m^3 \times 10 \,m/s^2 \times 0.085 \,m)$
$\Delta P = 150 \,Pa + 680 \,Pa = 830 \,Pa$.

(D) The pressure variation in a fluid with varying density is given by the hydrostatic law: $dP = -\rho(z) g dz$.
Integrating from the bottom ($z=0$,$P=P_1$) to the top ($z=H$,$P=P_2$):
$\int_{P_1}^{P_2} dP = -\int_{0}^{H} \rho(z) g dz$
$P_2 - P_1 = -g \int_{0}^{H} \rho_0 \left[ 2 - \left( \frac{z}{H} \right)^2 \right] dz$
$P_1 - P_2 = g \rho_0 \int_{0}^{H} \left( 2 - \frac{z^2}{H^2} \right) dz$
$P_1 - P_2 = g \rho_0 \left[ 2z - \frac{z^3}{3H^2} \right]_{0}^{H}$
$P_1 - P_2 = g \rho_0 \left( 2H - \frac{H^3}{3H^2} \right)$
$P_1 - P_2 = g \rho_0 \left( 2H - \frac{H}{3} \right)$
$P_1 - P_2 = g \rho_0 \left( \frac{5H}{3} \right) = \frac{5}{3} \rho_0 g H$.

(C) Consider the horizontal level at the bottom of the liquid with density $\rho_3$ in the left arm. The pressure at this level in both arms must be equal.
In the left arm,the pressure is due to the liquid of density $\rho_1$ (height $h$) and the liquid of density $\rho_3$ (height $h/2$).
$P_{left} = P_{atm} + \rho_1 gh + \rho_3 g(h/2)$
In the right arm,the pressure at the same horizontal level is due to the liquid of density $\rho_2$ (height $h$).
$P_{right} = P_{atm} + \rho_2 gh$
Equating the pressures: $P_{atm} + \rho_1 gh + \rho_3 g(h/2) = P_{atm} + \rho_2 gh$
Dividing by $gh$: $\rho_1 + \frac{\rho_3}{2} = \rho_2$
Rearranging for $\rho_3$: $\frac{\rho_3}{2} = \rho_2 - \rho_1 \implies \rho_3 = 2(\rho_2 - \rho_1)$

(A) Let the vertical line passing through the center be the reference. The interface between the two liquids is at an angle $\theta$ from the vertical.
Since the tube is in a vertical plane,the pressure at the lowest point must be the same from both sides.
Let the density of the liquid on the left be $\delta$ and on the right be $\rho$.
The vertical height of the center of mass of the liquid column of density $\delta$ from the lowest point is $h_1 = R(1 - \cos \theta)$.
The vertical height of the center of mass of the liquid column of density $\rho$ from the lowest point is $h_2 = R(1 - \cos \theta)$.
For equilibrium,the pressure at the lowest point of the tube must be equal from both sides.
The pressure exerted by a liquid column in a circular tube is proportional to the vertical depth of its center of mass.
The condition for equilibrium is given by:
$\delta g R(\cos \theta + \sin \theta) = \rho g R(\cos \theta - \sin \theta)$
Dividing both sides by $gR$:
$\delta(\cos \theta + \sin \theta) = \rho(\cos \theta - \sin \theta)$
$\delta \cos \theta + \delta \sin \theta = \rho \cos \theta - \rho \sin \theta$
$\sin \theta(\rho + \delta) = \cos \theta(\rho - \delta)$
$\tan \theta = \frac{\rho - \delta}{\rho + \delta}$
$\theta = \tan ^{-1}\left(\frac{\rho - \delta}{\rho + \delta}\right)$