Pressure due to Liquid Column and Barometer Questions in English

(B) Let the total depth of the lake be $h$. The pressure at the bottom of the lake is given by $P_{bottom} = P_0 + h\rho g$.
The pressure at half the depth $(h/2)$ is given by $P_{half} = P_0 + (h/2)\rho g$.
According to the problem,$P_{half} = \frac{2}{3} P_{bottom}$.
Substituting the expressions: $P_0 + \frac{h}{2}\rho g = \frac{2}{3}(P_0 + h\rho g)$.
Multiplying by $3$: $3P_0 + \frac{3}{2}h\rho g = 2P_0 + 2h\rho g$.
Rearranging the terms: $P_0 = 2h\rho g - 1.5h\rho g = 0.5h\rho g$.
Thus,$h = \frac{2P_0}{\rho g}$.
Substituting the values $P_0 = 10^5 \ Pa$,$\rho = 10^3 \ kg/m^3$,and $g = 10 \ m/s^2$:
$h = \frac{2 \times 10^5}{10^3 \times 10} = \frac{2 \times 10^5}{10^4} = 20 \ m$.

(B) The difference in pressure between sea level and the top of the hill is given by the change in the mercury column height:
$\Delta P = (h_1 - h_2) \times \rho_{Hg} \times g = (75 - 50) \times 10^{-2} \ m \times \rho_{Hg} \times g$ ... $(i)$
The pressure difference is also equal to the weight of the air column of height $h$:
$\Delta P = h \times \rho_{air} \times g$ ... $(ii)$
Equating $(i)$ and $(ii)$:
$h \times \rho_{air} \times g = 25 \times 10^{-2} \times \rho_{Hg} \times g$
$h = 25 \times 10^{-2} \times \left( \frac{\rho_{Hg}}{\rho_{air}} \right)$
Given $\frac{\rho_{Hg}}{\rho_{air}} = 10^4$,we get:
$h = 25 \times 10^{-2} \times 10^4 = 2500 \ m$
Converting to kilometers,the height of the hill is $2.5 \ km$.

(C) The pressure $P$ at a depth $h$ in a liquid of density $\rho$ is given by the formula $P = h\rho g$,where $g$ is the acceleration due to gravity.
From this formula,it is clear that the pressure depends on the height of the liquid column $(h)$,the density of the liquid $(\rho)$,and the acceleration due to gravity $(g)$.
It does not depend on the area of the bottom surface of the tank.
Therefore,the correct option is $C$.

(A) The pressure $P$ exerted by the mercury column in a barometer is given by the formula $P = h \rho g$,where $h$ is the height of the mercury column,$\rho$ is the density of mercury,and $g$ is the acceleration due to gravity.
Since the atmospheric pressure $P$ remains constant and the density $\rho$ of mercury is constant,we have $h \propto \frac{1}{g}$.
Taking the logarithmic derivative,we get $\frac{\Delta h}{h} = -\frac{\Delta g}{g}$.
Given that $g$ decreases by $2\%$,we have $\frac{\Delta g}{g} = -0.02$.
Substituting this into the equation,$\frac{\Delta h}{h} = -(-0.02) = 0.02$.
Therefore,the height $h$ increases by $2\%$.

(D) The pressure $P$ exerted by a liquid column of height $h$ is given by the formula $P = \rho g_{eff} h$,where $\rho$ is the density of the liquid and $g_{eff}$ is the effective acceleration due to gravity.
For a barometer,the atmospheric pressure $P_{atm}$ is balanced by the liquid column,so $P_{atm} = \rho g_{eff} h$,which implies $h = \frac{P_{atm}}{\rho g_{eff}}$.
When the elevator is stationary,$g_{eff} = g$.
When the elevator accelerates upwards with an acceleration $a$,the effective acceleration becomes $g_{eff} = g + a$.
Since $g_{eff}$ increases,the height $h$ of the mercury column must decrease to maintain the same atmospheric pressure $(h \propto \frac{1}{g_{eff}})$.
Therefore,the reading will be less than $76 \ cm$.

(A) The vertical height $h$ of the mercury column in a barometer remains constant regardless of the inclination of the tube,as it is determined by the atmospheric pressure.
Given,the vertical height $h = 76 \ cm$.
When the tube is inclined at an angle $\theta = 60^\circ$ with the vertical,the length of the mercury column $l$ along the tube is related to the vertical height $h$ by the formula:
$h = l \cos \theta$
Substituting the given values:
$76 = l \cos 60^\circ$
$76 = l \times \frac{1}{2}$
$l = 76 \times 2 = 152 \ cm$.
Therefore,the length of the mercury column will be $152 \ cm$.

(B) Let the height of the liquid be $h$ and the radius of the cylindrical vessel be $r$. The density of the liquid is $\rho$.
$1$. The pressure at the bottom of the vessel is $P_{bottom} = h\rho g$. The force exerted on the bottom is $F_{bottom} = P_{bottom} \times A_{bottom} = (h\rho g)(\pi r^2)$.
$2$. The pressure varies linearly with depth along the vertical side. The average pressure on the side wall is $P_{avg} = \frac{0 + h\rho g}{2} = \frac{1}{2}h\rho g$. The area of the side wall is $A_{side} = 2\pi rh$. Thus,the force on the side wall is $F_{side} = P_{avg} \times A_{side} = (\frac{1}{2}h\rho g)(2\pi rh) = \pi \rho g r h^2$.
$3$. According to the problem,$F_{bottom} = F_{side}$.
Therefore,$h\rho g \pi r^2 = \pi \rho g r h^2$.
$4$. Simplifying the equation: $h r^2 = r h^2$,which gives $h = r$.

(D) The pressure exerted by a liquid column at a depth $h$ is given by the formula $P = h \rho g$,where $\rho$ is the density of the liquid and $g$ is the acceleration due to gravity.
This formula shows that the pressure at the bottom of a container is independent of the shape of the container or the surface area of the base.
It only depends on the vertical height $h$ of the liquid column.
In the given figure,both tanks $(a)$ and $(b)$ are filled with water to the same vertical height $h$.
Since $h$,$\rho$,and $g$ are identical for both tanks,the pressure at the bottom of both tanks must be the same.

(A) liquid boils when its vapour pressure becomes equal to the external atmospheric pressure.
At this point,the bubbles of vapour can form throughout the bulk of the liquid and rise to the surface,allowing the liquid to transition into the gaseous phase.
Therefore,the correct option is $A$.

(B) Let the depth of the lake be $h$ and the atmospheric pressure be $P_0$. The density of water is $\rho = 10^3 \ kg/m^3$ and $g = 10 \ m/s^2$.
The pressure at depth $h/2$ is $P_1 = P_0 + (h/2)\rho g$.
The pressure at the bottom (depth $h$) is $P_2 = P_0 + h\rho g$.
According to the problem,$P_1 = (2/3)P_2$.
Substituting the values: $P_0 + (1/2)h\rho g = (2/3)(P_0 + h\rho g)$.
Multiplying by $6$ to clear denominators: $6P_0 + 3h\rho g = 4P_0 + 4h\rho g$.
Rearranging the terms: $2P_0 = h\rho g$.
Thus,$h = (2P_0) / (\rho g)$.
Using $P_0 = 10^5 \ Pa$,$\rho = 10^3 \ kg/m^3$,and $g = 10 \ m/s^2$:
$h = (2 \times 10^5) / (10^3 \times 10) = 2 \times 10^5 / 10^4 = 20 \ m$.

(B) The pressure $P$ at the bottom of the container due to a liquid column of height $h$ is given by $P = h \rho g$.
The force $F$ exerted on the bottom area $A$ is given by $F = P \times A$.
Substituting the values: $h = 0.4 \ m$,$\rho = 900 \ kg/m^{3}$,$g = 10 \ m/s^{2}$,and $A = 2 \times 10^{-3} \ m^{2}$.
$F = (h \rho g) \times A$
$F = 0.4 \times 900 \times 10 \times (2 \times 10^{-3})$
$F = 3600 \times 2 \times 10^{-3}$
$F = 7200 \times 10^{-3} = 7.2 \ N$.

(B) The pressure difference between the base and the top of the mountain is equal to the weight of the air column of height $h$ and density $\rho_{air}$.
This pressure difference is also equal to the difference in pressure exerted by the mercury columns.
Let $h$ be the height of the mountain,$\rho_{Hg}$ be the density of mercury,and $\rho_{air}$ be the density of air.
The pressure difference is given by: $\Delta P = (h_1 - h_2) \rho_{Hg} g = h \rho_{air} g$.
Substituting the given values: $(75 \, cm - 50 \, cm) \times \rho_{Hg} = h \times \rho_{air}$.
$25 \times 10^{-2} \, m \times \rho_{Hg} = h \times \rho_{air}$.
$h = 25 \times 10^{-2} \times \left( \frac{\rho_{Hg}}{\rho_{air}} \right)$.
Given $\frac{\rho_{Hg}}{\rho_{air}} = 10^4$.
$h = 25 \times 10^{-2} \times 10^4 = 2500 \, m$.
Therefore,the height of the mountain is $2.5 \, km$.

(D) At depth $h$,the total pressure $P_1$ is given by $P_1 = P + \rho gh$,where $P$ is the atmospheric pressure,$\rho$ is the density of water,and $g$ is the acceleration due to gravity.
At the surface,the pressure $P_2$ is equal to the atmospheric pressure $P$.
Since the temperature is constant,we apply Boyle's Law: $P_1 V_1 = P_2 V_2$.
Here,$V_1 = V_0$ and $V_2 = V$ (the volume at the surface).
Substituting the values: $(P + \rho gh) V_0 = P V$.
Solving for $V$: $V = V_0 \frac{(P + \rho gh)}{P} = V_0 (1 + \frac{\rho gh}{P})$.

(A) The atmospheric pressure is $P_{atm} = 75 \, cm$ of $Hg$.
Since the tube is only $50 \, cm$ above the mercury level,the mercury column inside the tube is $50 \, cm$ high.
The pressure at the top of the tube $(P_{top})$ plus the pressure due to the $50 \, cm$ mercury column must equal the atmospheric pressure.
$P_{top} + 50 \, cm \, Hg = 75 \, cm \, Hg$.
$P_{top} = 75 \, cm - 50 \, cm = 25 \, cm$ of $Hg$.
To convert $25 \, cm$ of $Hg$ to $kPa$,we use the relation $76 \, cm \, Hg \approx 101.3 \, kPa$.
$P_{top} = (25 / 76) \times 101.3 \, kPa \approx 33.3 \, kPa$.

(D) The pressure at the bottom of a container containing a liquid of variable density is given by $P = P_0 + \int_{0}^{H} \rho(h) g \, dh$,where $H = 10\, m$ is the total depth.
Substituting the given values: $P = 10^5 + \int_{0}^{10} (100 + 6h^2) \times 10 \, dh$.
$P = 10^5 + 10 \times [100h + \frac{6h^3}{3}]_{0}^{10}$.
$P = 10^5 + 10 \times [100(10) + 2(10)^3]$.
$P = 10^5 + 10 \times [1000 + 2000]$.
$P = 10^5 + 10 \times 3000 = 10^5 + 30000 = 100000 + 30000 = 130000\, Pa$.
Thus,$P = 1.3 \times 10^5\, Pa$.

(C) The pressure exerted by the mercury column is given by $P = \rho gh$,where $\rho$ is the density of mercury,$g$ is the acceleration due to gravity,and $h$ is the height of the mercury column. This pressure balances the atmospheric pressure $(P_{atm})$.
$1$. Changes in atmospheric pressure $(P_{atm})$ directly cause the height $h$ to change to maintain equilibrium $(P_{atm} = \rho gh)$.
$2$. Changes in the value of $g$ will cause the height $h$ to adjust to keep the pressure $\rho gh$ equal to $P_{atm}$.
$3$. Leakage of air into the tube introduces air pressure above the mercury column,which pushes the mercury down,thus changing the height $h$.
$4$. Evaporation of mercury from the reservoir does not change the atmospheric pressure acting on the surface of the reservoir,nor does it change the density or the gravitational acceleration. Therefore,it does not cause the height of the mercury column to vary.

(A) The lift is moving downward with a deceleration $\alpha$. This means the net acceleration of the lift is directed upward.
The effective acceleration due to gravity $g_{eff}$ in the frame of the lift is given by $g_{eff} = g + a$,where $a$ is the magnitude of the deceleration.
In a barometer,the atmospheric pressure $P_0$ is balanced by the pressure exerted by the mercury column of height $h$,which is given by $P = \rho g_{eff} h$.
Equating the two,we get $P_0 = \rho (g + \alpha) h$.
Solving for $h$,we get $h = \frac{P_0}{\rho (g + \alpha)}$.

(A) $1$. Point $1$ is at the surface of the liquid in the container,which is exposed to the atmosphere. Therefore,the pressure at point $1$ is $P_1 = P_0$.
$2$. Point $3$ is the exit point of the siphon,which is also exposed to the atmosphere. Therefore,the pressure at point $3$ is $P_3 = P_0$.
$3$. $A$ siphon works based on the pressure difference between the two ends. For the liquid to flow out,the exit point must be lower than the liquid level in the container,meaning $h_3 > 0$. Thus,the statement 'Siphon works when $h_3 < 0$' is incorrect.
$4$. Point $2$ is at a height $h_3$ above the exit point $3$. Using the hydrostatic pressure formula $P_3 = P_2 + \rho g h_3$,we get $P_2 = P_3 - \rho g h_3 = P_0 - \rho g h_3$.

(D) The pressure at a depth $h$ below the surface of a liquid is given by the sum of the atmospheric pressure acting on the surface and the hydrostatic pressure exerted by the liquid column.
Mathematically,this is expressed as $P = P_a + h\rho g$,where $P_a$ is the atmospheric pressure,$\rho$ is the density of the liquid,$g$ is the acceleration due to gravity,and $h$ is the depth.

(B) The pressure at any point in a static fluid is given by $P = h'\rho g$,where $h'$ is the depth of the point below the free surface of the liquid.
In the given figure,the total height of the liquid column is $H$ and the height of point $P$ from the bottom is $h$. Therefore,the depth of point $P$ below the free surface is $H - h$.
The normal thrust per unit area is equivalent to the hydrostatic pressure at that point.
Thus,the normal thrust per unit area at point $P$ is $(H - h)\rho g$.

(D) The pressure at a depth $d$ in a fluid is given by $P = P_{atm} + \rho g d.$
For two different depths $d_1$ and $d_2,$ the pressure difference is given by $\Delta P = P_2 - P_1 = \rho g (d_2 - d_1).$
Given $P_1 = 5.05 \times 10^6 \, Pa$ and $P_2 = 8.08 \times 10^6 \, Pa.$
The difference in pressure is $\Delta P = (8.08 - 5.05) \times 10^6 \, Pa = 3.03 \times 10^6 \, Pa.$
Using the formula $\Delta P = \rho g (d_2 - d_1)$ where $\rho = 10^3 \, kg/m^3$ and $g = 10 \, m/s^2$:
$3.03 \times 10^6 = 10^3 \times 10 \times (d_2 - d_1)$
$3.03 \times 10^6 = 10^4 \times (d_2 - d_1)$
$d_2 - d_1 = \frac{3.03 \times 10^6}{10^4} = 3.03 \times 10^2 = 303 \, m.$
Therefore,the value of $d_2 - d_1$ is $303 \, m.$

(A) To allow the blood to enter the vein,the hydrostatic pressure exerted by the column of blood must be equal to the gauge pressure of the vein.
The formula for hydrostatic pressure is $P = h \rho g$,where:
$P = 2000\,Pa$ (gauge pressure),
$\rho = 1.06 \times 10^3\,kg/m^3$ (density of blood),
$g = 9.8\,m/s^2$ (acceleration due to gravity).
Rearranging the formula to solve for height $h$:
$h = \frac{P}{\rho g}$
Substituting the values:
$h = \frac{2000}{1.06 \times 10^3 \times 9.8}$
$h = \frac{2000}{10388}$
$h \approx 0.1925\,m$
Rounding to three decimal places,we get $h = 0.192\,m$.

(C) Atmospheric pressure,$P_{0} = 76\,cm$ of $Hg$.
The difference between the levels of mercury in the two limbs of the manometer gives the gauge pressure.
From the figure,the gauge pressure is $P_{g} = 20\,cm$ of $Hg$.
Absolute pressure is the sum of atmospheric pressure and gauge pressure.
$P_{abs} = P_{0} + P_{g} = 76\,cm + 20\,cm = 96\,cm$ of $Hg$.
Thus,the absolute pressure is $96\,cm$ of $Hg$ and the gauge pressure is $20\,cm$ of $Hg$.

(A) The pressure exerted by a barometer column is given by $P = h \rho g$,where $h$ is the height of the column,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
Since there is no atmosphere on the Moon,the atmospheric pressure on the Moon is $0 \, Pa$.
Therefore,the height of the mercury column $h$ must be $0 \, mm$ because there is no external pressure to support the column.

(A) The pressure exerted by a fluid column is given by the formula $P = \rho h g$,where $P$ is the pressure,$\rho$ is the density,$h$ is the height,and $g$ is the acceleration due to gravity.
Given:
$P = 1.013 \times 10^5 \ Pa$
$\rho = 1.29 \ kg/m^3$
$g = 9.81 \ m/s^2$
Substituting these values into the formula:
$1.013 \times 10^5 = 1.29 \times h \times 9.81$
$h = \frac{1.013 \times 10^5}{1.29 \times 9.81}$
$h \approx \frac{101300}{12.6549}$
$h \approx 8004.8 \ m$
Converting to kilometers,$h \approx 8 \ km$.

(B) Consider an element of the liquid of height $dh$ and area of cross-section $A$ at a depth $h$ below the surface of the liquid. Let $p$ and $p+dp$ be the liquid pressures at the upper and lower surfaces of the element, respectively.
The mass of the liquid in the element is $dm = A \cdot dh \cdot \rho$.
The net upward force on the element is given by $F_{net} = (p+dp)A - pA - dm \cdot g$.
Since the element moves upward with an acceleration $a$, by Newton's second law, $F_{net} = dm \cdot a$.
Therefore, $A \cdot dp - dm \cdot g = dm \cdot a$.
$A \cdot dp = dm(g+a)$.
Substituting $dm = A \cdot dh \cdot \rho$, we get $A \cdot dp = (A \cdot dh \cdot \rho)(g+a)$.
$dp = \rho(g+a) dh$.
Integrating both sides from the surface $(h=0, p=0)$ to depth $h$, we get $p = \int_0^h \rho(g+a) dh = \rho(g+a)h$.

(C) The atmospheric pressure is $P_{atm} = 760\, mm$ of $Hg$.
The pressure in the lungs is $P_{lungs} = 750\, mm$ of $Hg$.
The pressure difference created is $\Delta P = P_{atm} - P_{lungs} = 760\, mm - 750\, mm = 10\, mm$ of $Hg$.
This pressure difference supports a column of water of height $h$ in the straw.
Using the hydrostatic pressure formula $\Delta P = h_w \rho_w g = h_{Hg} \rho_{Hg} g$,we have:
$h_w \rho_w = h_{Hg} \rho_{Hg}$
Given $h_{Hg} = 10\, mm = 1\, cm$,$\rho_{Hg} = 13.6\, g/cm^3$,and $\rho_w = 1\, g/cm^3$:
$h_w \times 1\, g/cm^3 = 1\, cm \times 13.6\, g/cm^3$
$h_w = 13.6\, cm$.

(B) The total pressure $P$ at a depth $h$ is given by the formula $P = P_{a} + \rho gh$,where $P_{a}$ is the atmospheric pressure,$\rho$ is the density of water,and $g$ is the acceleration due to gravity.
Given: $h = 10\; m$,$\rho = 1000\; kg/m^3$,$g = 10\; m/s^2$,and $P_{a} \approx 1.01 \times 10^5\; Pa$.
Substituting the values:
$P = 1.01 \times 10^5\; Pa + (1000\; kg/m^3 \times 10\; m/s^2 \times 10\; m)$
$P = 1.01 \times 10^5\; Pa + 1.00 \times 10^5\; Pa$
$P = 2.01 \times 10^5\; Pa$.
Since $1\; atm = 1.013 \times 10^5\; Pa$,the pressure is approximately $2\; atm$.

(A) The pressure exerted by a liquid column is given by $P = \rho gh$.
For normal atmospheric pressure,the pressure exerted by the mercury column must be equal to the pressure exerted by the French wine column.
Let $\rho_1$ and $h_1$ be the density and height of the mercury column,and $\rho_2$ and $h_2$ be the density and height of the French wine column.
Given: $\rho_1 = 13.6 \times 10^3 \; kg \; m^{-3}$,$h_1 = 0.76 \; m$,and $\rho_2 = 984 \; kg \; m^{-3}$.
Equating the pressures: $\rho_1 g h_1 = \rho_2 g h_2$.
Thus,$h_2 = \frac{\rho_1 h_1}{\rho_2}$.
Substituting the values: $h_2 = \frac{13.6 \times 10^3 \times 0.76}{984} \approx 10.5 \; m$.
Therefore,the height of the French wine column is $10.5 \; m$.

(A) The maximum allowable stress for the structure is $P_{max} = 10^9 \; Pa$.
The depth of the ocean is $d = 3 \; km = 3 \times 10^3 \; m$.
The density of seawater is $\rho \approx 10^3 \; kg/m^3$.
The acceleration due to gravity is $g = 9.8 \; m/s^2$.
The pressure exerted by the seawater at depth $d$ is given by $P = \rho g d$.
Substituting the values: $P = (10^3 \; kg/m^3) \times (9.8 \; m/s^2) \times (3 \times 10^3 \; m) = 2.94 \times 10^7 \; Pa$.
Since the maximum allowable stress of the structure $(10^9 \; Pa)$ is significantly greater than the pressure exerted by the ocean $(2.94 \times 10^7 \; Pa)$,the structure is suitable for use.

(A) For figure $(a)$:
Atmospheric pressure,$P_{0} = 76 \; cm$ of $Hg$.
The difference between the levels of mercury in the two limbs gives the gauge pressure. Since the level in the right limb is higher,the gauge pressure is $20 \; cm$ of $Hg$.
Absolute pressure $= P_{0} + \text{Gauge pressure} = 76 + 20 = 96 \; cm$ of $Hg$.
For figure $(b)$:
The level in the right limb is lower than the left limb,so the gauge pressure is $-18 \; cm$ of $Hg$.
Absolute pressure $= P_{0} + \text{Gauge pressure} = 76 - 18 = 58 \; cm$ of $Hg$.
$(b)$ When $13.6 \; cm$ of water is poured into the right limb,it exerts an additional pressure. Since the relative density of mercury is $13.6$,a column of $13.6 \; cm$ of water is equivalent to $1 \; cm$ of $Hg$.
Let $h$ be the new difference between the levels of mercury in the two limbs.
The pressure in the right limb at the level of the mercury surface in the left limb is $P_{R} = P_{0} + 1 \; cm$ of $Hg = 76 + 1 = 77 \; cm$ of $Hg$.
The pressure in the left limb is $P_{L} = P_{\text{gas}} + h = 58 + h$.
Equating the pressures at the same horizontal level: $58 + h = 77 \implies h = 19 \; cm$.
Thus,the new difference between the levels of mercury in the two limbs will be $19 \; cm$.

(C) Given: Gauge pressure in the vein,$P = 2000 \; Pa$. Density of blood,$\rho = 1.06 \times 10^{3} \; kg \; m^{-3}$. Acceleration due to gravity,$g = 9.8 \; m \; s^{-2}$.
For the blood to enter the vein,the hydrostatic pressure exerted by the blood column in the container must be at least equal to the gauge pressure in the vein.
The pressure exerted by the blood column is given by $P = h \rho g$,where $h$ is the height of the container.
Rearranging for $h$,we get $h = \frac{P}{\rho g}$.
Substituting the values: $h = \frac{2000}{1.06 \times 10^{3} \times 9.8}$.
$h = \frac{2000}{10388} \approx 0.1925 \; m$.
Rounding to the nearest significant value,the height must be approximately $0.2 \; m$.

Consider a fluid at rest in a container. The density of the fluid is $\rho$ and the height of the fluid column is $h$,as shown in the figure.
The weight of the fluid in the cylinder is $W = mg \ldots(1)$
Since the mass of the fluid $m = \text{volume} \times \text{density} = Ah\rho$,
$\therefore W = (Ah\rho g) \ldots(2)$
The pressure produced at the bottom of the container due to this weight is
$P = \frac{\text{Weight } W}{\text{Area } A}$
$\therefore P = \frac{Ah\rho g}{A}$
$\therefore P = h\rho g \ldots(3)$
This equation shows that the pressure produced due to a fluid column depends on the height of the fluid column $h$ and the density of the fluid $\rho$.

(N/A) Suppose a fluid of density $\rho$ is in static equilibrium in a container.
Consider a cylindrical element of fluid having area of base $A$ and height $h$. The pressure at points $1$ and $2$ are $P_{1}$ and $P_{2}$ respectively.
The forces acting on the liquid column are as follows:
$(1)$ The force at point $1$ is $F_{1} = P_{1} A$ (downward).
$(2)$ The force at point $2$ is $F_{2} = P_{2} A$ (upward).
$(3)$ The weight of the fluid column is $W = mg = A h \rho g$ (downward).
Since the fluid column is in equilibrium,the downward forces must equal the upward forces:
$F_{1} + W = F_{2}$
$P_{1} A + A h \rho g = P_{2} A$
Dividing by $A$ on both sides:
$P_{2} = P_{1} + h \rho g$
This equation shows that the pressure difference depends on the vertical distance $h$ between the points,the mass density of the fluid $\rho$,and the acceleration due to gravity $g$,but it does not depend on the area of the cross-section $A$.
If the effect of gravitation is neglected $(g = 0)$,then $P_{2} = P_{1}$,indicating that the pressure at every point in the fluid is the same.

(N/A) The Italian scientist Evangelista Torricelli devised the first method for measuring atmospheric pressure.
This device is shown in the figure.
$A$ long glass tube of $1 \text{ m}$,closed at one end and filled with mercury,is inverted into a trough of mercury. The thumb is kept on the open part of the tube.
If the thumb is removed from the tube,the level of the mercury column decreases slightly.
The space above the mercury column in the tube contains only mercury vapour,whose pressure $P$ is so small that it may be neglected. Thus,$P \approx 0$.
The pressure inside the column at point $A$ is equal to the pressure at point $B$,which is at the same horizontal level. The atmospheric pressure at point $A$ is given by:
$P_{a} = P + h \rho g$
Since $P \approx 0$,we have:
$P_{a} = 0 + h \rho g$
$\therefore P_{a} = h \rho g$
Where $\rho$ is the density of mercury and $h$ is the height of the mercury column.
In this device,the height of the mercury column at sea level is $76 \text{ cm}$,which is equivalent to one atmosphere.

(N/A) The Italian scientist Evangelista Torricelli devised the first method for measuring atmospheric pressure.
This device is shown in the figure.
$A$ long glass tube of $1 \,m$, closed at one end and filled with mercury, is inverted into a trough of mercury. The thumb is kept on the open part of the tube.
If the thumb is removed from the tube, the level of the mercury column decreases slightly.
The space above the mercury column in the tube contains only mercury vapour, whose pressure $P$ is so small that it may be neglected, i.e., $P = 0$.
The pressure inside the column at point $A$ is equal to the pressure at point $B$, which is at the same level. The atmospheric pressure at point $A$ is given by:
$P_{a} = P + h \rho g$
Since $P = 0$, we have:
$P_{a} = 0 + h \rho g$
$\therefore P_{a} = h \rho g$
Where $\rho$ is the density of mercury and $h$ is the height of the mercury column.
In this device, the height of the mercury column at sea level is $76 \,cm$, which is equivalent to one atmosphere.

(N/A) An open-tube manometer is a simple device used to measure the pressure of a gas contained in a closed vessel.
It consists of a $U$-shaped tube containing a suitable liquid (often mercury or water).
One end of the tube is open to the atmosphere,and the other end is connected to the system whose pressure $(P)$ is to be measured.
Consider a point $A$ in the limb connected to the gas container and a point $B$ at the same horizontal level in the open limb.
According to the laws of fluid statics,the pressure at points at the same horizontal level in a continuous static fluid is the same.
Therefore,the pressure at point $A$ is equal to the pressure at point $B$: $P = P_B$.
The pressure at point $B$ is the sum of the atmospheric pressure $(P_a)$ and the pressure due to the liquid column of height $h$: $P_B = P_a + h \rho g$.
Thus,the absolute pressure of the gas is given by: $P = P_a + h \rho g$.
Here,$P_a$ is the atmospheric pressure,$h$ is the height difference of the liquid levels,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
The term $(P - P_a) = h \rho g$ is known as the gauge pressure,which is directly proportional to the height $h$ of the liquid column.

(A) The pressure $P$ at a depth $h$ in a liquid of density $\rho$ is given by the formula $P = P_0 + \rho gh$, where $P_0$ is the atmospheric pressure, $\rho$ is the density of the liquid, $g$ is the acceleration due to gravity, and $h$ is the depth from the surface.
Since $P_0$, $\rho$, and $g$ are constant, the pressure $P$ is directly proportional to the depth $h$.
Therefore, as the depth $h$ increases, the pressure $P$ also increases.

(N/A) The pressure $P$ at a depth $h$ below the surface of a liquid of density $\rho$ is given by the formula:
$P = P_a + \rho gh$
Where:
$P_a$ is the atmospheric pressure at the surface of the liquid.
$\rho$ is the density of the liquid.
$g$ is the acceleration due to gravity.
$h$ is the depth of the point from the liquid surface.

(N/A) The two common instruments used to measure atmospheric pressure are:
$1$. Mercury Barometer: It uses a column of mercury in a glass tube to balance the atmospheric pressure.
$2$. Aneroid Barometer: It uses a flexible metal box (aneroid cell) that expands or contracts with changes in atmospheric pressure,which is then indicated by a mechanical pointer.

(B) The first method to measure atmospheric pressure was devised by the Italian physicist and mathematician Evangelista Torricelli in $1643$. He used a long glass tube filled with mercury,inverted into a dish of mercury,to demonstrate that the height of the mercury column is supported by the pressure of the atmosphere. This apparatus is known as a mercury barometer.

(B) When oil is poured out from a tin with only one hole,the volume of oil inside decreases,creating a partial vacuum or a region of lower pressure inside the tin.
This internal pressure becomes less than the atmospheric pressure outside,which opposes the flow of oil and causes it to come out in spurts or stop flowing entirely.
When two holes are made,air enters through the second hole,which balances the pressure inside the tin with the atmospheric pressure.
This allows the oil to flow out smoothly and continuously under the influence of gravity.

(N/A) The pressure $P$ at any point in a fluid (like the atmosphere) is given by the formula $P = h \rho g$,where:
$(1)$ Height of the fluid column $(h)$: The pressure is directly proportional to the depth or height of the fluid column above the point. As the height increases,the weight of the fluid column increases,leading to higher pressure.
$(2)$ Density of the fluid $(\rho)$: The pressure is directly proportional to the density of the fluid. $A$ denser fluid exerts more pressure because its mass per unit volume is greater.
$(3)$ Acceleration due to gravity $(g)$: The pressure is directly proportional to the acceleration due to gravity. Since weight is the force exerted by gravity on a mass,a stronger gravitational field results in higher pressure.

(N/A) The pressure inside the pressure cooker is significantly higher than the atmospheric pressure. This increased pressure raises the boiling point of water above its normal value of $100^{\circ} C$. As a result,the water inside the cooker reaches a higher temperature,which allows the food to cook much faster.

(N/A) The air on a mountain is thin,which means the atmospheric pressure is lower than at sea level.
According to the principles of fluid mechanics,the boiling point of a liquid decreases as the external pressure decreases.
At high altitudes,water boils at a temperature significantly lower than $100^{\circ} C$.
Because the water boils at this lower temperature,it does not provide enough heat to properly extract the flavor from the tea leaves,resulting in a poor-quality cup of tea.

(N/A) As the altitude increases,the atmospheric pressure decreases. The ink inside the fountain pen's reservoir is filled at ground-level atmospheric pressure. When the airplane flies at a high altitude,the external atmospheric pressure acting on the ink decreases,while the pressure inside the pen remains higher. This pressure difference causes the ink to leak out of the pen and onto the clothing. Therefore,it is not advisable to keep a fountain pen in a pocket while flying at high altitudes.

(N/A) The pressure exerted by a liquid column of height $h$ is given by the formula $P = h \rho g$,where $\rho$ is the density of the liquid and $g$ is the acceleration due to gravity.
As the depth $h$ increases,the pressure $P$ increases linearly.
Therefore,the pressure is significantly higher at the bottom of the dam compared to the top.
To withstand this increased hydrostatic pressure and prevent structural failure,the walls of the dam are constructed to be thicker at the bottom.

(N/A) The pressure exerted by the water column decreases as the bubble rises from the bottom towards the surface because the depth of the water column above the bubble decreases.
According to Boyle's law,for a given mass of gas at a constant temperature,the volume $V$ of the gas is inversely proportional to the pressure $P$ $(V \propto \frac{1}{P})$.
As the pressure $P$ decreases while the bubble rises,its volume $V$ must increase to satisfy the relationship.
Therefore,the air bubble grows in size as it moves towards the surface.

(N/A) The reasons for using mercury in a barometer are as follows:
$(1)$ Mercury has a high density $(13.6 \times 10^3 \ kg/m^3)$,which allows for a reasonably small and manageable height of the barometer tube (approximately $76 \ cm$ at sea level).
$(2)$ Mercury has a very low vapor pressure at room temperature,meaning it does not evaporate significantly into the vacuum space above the column,ensuring accurate pressure readings.
$(3)$ Mercury does not wet or stick to the glass walls of the barometer tube,which allows for a clean meniscus and precise measurement of the column height.

(N/A) The density of water is much lower than that of mercury. Since the atmospheric pressure $P = h \rho g$,the height $h$ of the liquid column required to balance atmospheric pressure is inversely proportional to its density $\rho$. For water,the required height would be approximately $10.3 \ m$ to $11 \ m$. $A$ barometer of this length is impractical to construct,handle,and transport,making mercury a more convenient choice.