Pressure due to Liquid Column and Barometer Questions in English

(B) falling height of the mercury column in a barometer indicates a decrease in atmospheric pressure.
This decrease in pressure is often caused by an increase in the amount of water vapour in the atmosphere,which is less dense than dry air.
Consequently,a falling barometer reading is a common indicator of the possibility of rainfall or an approaching storm.

(B) rising height of a barometer indicates an increase in atmospheric pressure.
This change typically suggests that the weather is becoming fair or dry,as high-pressure systems are generally associated with clear skies and lower humidity.

(C) The effective acceleration due to gravity inside an elevator accelerating upwards with acceleration $a$ is given by $g_{eff} = g + a$.
The pressure exerted by a liquid column of height $h$ and density $\rho$ is given by the formula $P = h \rho g_{eff}$.
Substituting the value of $g_{eff}$,we get the pressure inside the elevator as $P = h \rho (g + a)$.
Since $a > 0$,the pressure measured by the barometer will be greater than the standard atmospheric pressure $h \rho g$.

(B) The effective acceleration due to gravity inside an elevator accelerating downwards with acceleration $a$ is $g_{eff} = g - a$.
The pressure exerted by a liquid column of height $h$ is given by $P = h \rho g_{eff}$.
Since the barometer is calibrated for standard gravity $g$,the reading $h'$ will adjust such that the pressure balance remains consistent with the local effective gravity.
However,the pressure $P$ at the base of the column in the accelerating frame is $P = h \rho (g - a)$.
Since $(g - a) < g$,the effective weight of the mercury column decreases.
Therefore,the height of the mercury column $h$ in the barometer will decrease compared to the standard atmospheric pressure reading at rest,meaning the measured pressure is less than the atmospheric pressure $76 \text{ cm of Hg}$.

(N/A) Since the air is less dense at higher altitudes,the pressure is also lower.
$(a)$ Consider a horizontal portion of air with cross-sectional area $A$ and height $dh$. Let the pressure on the top surface be $P$ and on the bottom surface be $P + dP$. If the portion is in equilibrium,the net upward force must be balanced by the weight.
$(P + dP)A - PA = -mg$ (where mass = volume $\times$ density)
$(dP)A = -\rho(A dh)g$
$dp = -\rho g dh$ ... $(1)$
The negative sign indicates that pressure decreases as height increases.
$(b)$ Let the density of air on the earth's surface be $\rho_0$. Given $P \propto \rho$,we have $\frac{P}{P_0} = \frac{\rho}{\rho_0}$,so $\rho = \left(\frac{P}{P_0}\right)\rho_0$ ... $(2)$
Substituting $(2)$ into $(1)$:
$dP = -\left(\frac{P}{P_0}\right)\rho_0 g dh$
$\frac{dP}{P} = -\frac{\rho_0 g}{P_0} dh$
Integrating both sides from $0$ to $h$:
$\int_{P_0}^{P} \frac{dP}{P} = -\frac{\rho_0 g}{P_0} \int_{0}^{h} dh$
$\ln\left(\frac{P}{P_0}\right) = -\frac{\rho_0 g h}{P_0}$
$P = P_0 e^{-\frac{\rho_0 g h}{P_0}}$
$(c)$ Given $P = \frac{P_0}{10}$,$\ln\left(\frac{1}{10}\right) = -\frac{\rho_0 g h}{P_0}$
$h = \frac{P_0 \ln(10)}{\rho_0 g} = \frac{1.03 \times 10^5 \times 2.303}{1.29 \times 9.8} \approx 18750 \text{ m} \approx 18.75 \text{ km}$.
$(d)$ The assumption is that the density of air is proportional to pressure,which implies an isothermal atmosphere (constant temperature),whereas the actual temperature of the atmosphere varies with height.

(C) The pressure exerted by a liquid column is given by the formula $P = h \rho g$,where $h$ is the height,$\rho$ is the density,and $g$ is the acceleration due to gravity.
For the mercury barometer,the pressure is $P = h_{Hg} \rho_{Hg} g = 0.76 \; m \times 13600 \; kg/m^3 \times g$.
For the liquid barometer,the pressure is $P = h' \rho' g = h' \times 760 \; kg/m^3 \times g$.
Since the atmospheric pressure is the same in both cases,we equate the two expressions:
$h' \times 760 = 0.76 \times 13600$
Solving for $h'$:
$h' = \frac{0.76 \times 13600}{760}$
$h' = \frac{10336}{760} = 13.6 \; m$.

(A) The pressure difference between two points at different depths in a fluid is given by the hydrostatic pressure formula: $\Delta P = \rho g h$.
Here,the density of water $\rho = 1000 \, kg/m^3$,the acceleration due to gravity $g = 10 \, m/s^2$,and the height difference $h = 10 \, cm = 0.1 \, m$.
Substituting these values into the formula:
$\Delta P = 1000 \times 10 \times 0.1$
$\Delta P = 1000 \, Pa$.

(A) The total pressure $P$ at a depth $d$ is given by $P = P_{0} + \rho gd$,where $P_{0}$ is the atmospheric pressure.
Given $P_{1} = 3 \times 10^{5} \; Pa$ and $P_{0} = 1 \times 10^{5} \; Pa$.
So,$\rho gd = P_{1} - P_{0} = 3 \times 10^{5} - 1 \times 10^{5} = 2 \times 10^{5} \; Pa$.
If the depth is doubled $(d' = 2d)$,the new pressure $P_{2}$ is $P_{2} = P_{0} + \rho g(2d) = P_{0} + 2(\rho gd)$.
Substituting the values: $P_{2} = 1 \times 10^{5} + 2(2 \times 10^{5}) = 1 \times 10^{5} + 4 \times 10^{5} = 5 \times 10^{5} \; Pa$.
The percentage increase in pressure is given by $\frac{P_{2} - P_{1}}{P_{1}} \times 100$.
Percentage increase $= \frac{5 \times 10^{5} - 3 \times 10^{5}}{3 \times 10^{5}} \times 100 = \frac{2 \times 10^{5}}{3 \times 10^{5}} \times 100 = \frac{200}{3} \%$.

(D) Consider a strip of width $dy$ at a height $y$ from the bottom. The depth of this strip below the surface is $(h-y)$.
The pressure at this depth is $P = \rho_w g(h-y)$.
The force on the strip of area $dA = L \cdot dy$ is $dF = P \cdot dA = \rho_w g(h-y) L \cdot dy$.
The torque $d\tau$ about the bottom axis (at $y=0$) is $d\tau = dF \cdot y = \rho_w g L (h-y) y \cdot dy$.
Integrating from $y=0$ to $y=h$ for the total torque $\tau_1$:
$\tau_1 = \int_{0}^{h} \rho_w g L (hy - y^2) dy = \rho_w g L \left[ \frac{hy^2}{2} - \frac{y^3}{3} \right]_{0}^{h} = \rho_w g L \left( \frac{h^3}{2} - \frac{h^3}{3} \right) = \frac{\rho_w g L h^3}{6}$.
For the second case,the height is $h' = h/2$ and length is $L' = L/2$. The torque $\tau_2$ is:
$\tau_2 = \int_{0}^{h/2} \rho_w g L' (h' - y) y \cdot dy = \int_{0}^{h/2} \rho_w g \left(\frac{L}{2}\right) \left(\frac{h}{2} - y\right) y \cdot dy$
$= \frac{\rho_w g L}{2} \left[ \frac{h}{2} \frac{y^2}{2} - \frac{y^3}{3} \right]_{0}^{h/2} = \frac{\rho_w g L}{2} \left[ \frac{h}{4} \left(\frac{h^2}{4}\right) - \frac{1}{3} \left(\frac{h^3}{8}\right) \right]$
$= \frac{\rho_w g L}{2} \left[ \frac{h^3}{16} - \frac{h^3}{24} \right] = \frac{\rho_w g L}{2} \left[ \frac{3h^3 - 2h^3}{48} \right] = \frac{\rho_w g L h^3}{2 \cdot 48} = \frac{\rho_w g L h^3}{96}$.
Therefore,$\frac{\tau_1}{\tau_2} = \frac{\rho_w g L h^3 / 6}{\rho_w g L h^3 / 96} = \frac{96}{6} = 16$.

(A) The pressure difference across the window is due to the hydrostatic pressure of the water column above it.
Given:
Area $A = 30 \times 30 \,cm^2 = 900 \times 10^{-4} \,m^2 = 0.09 \,m^2$.
Depth $h = 100 \,m$.
Density $\rho = 1.03 \times 10^3 \,kg/m^3$.
Acceleration due to gravity $g = 10 \,m/s^2$.
The pressure difference $\Delta P$ is given by $\rho gh$.
The force $F$ acting on the window is $F = \Delta P \times A = \rho ghA$.
Substituting the values:
$F = (1.03 \times 10^3) \times 10 \times 100 \times 0.09$
$F = 1.03 \times 10^3 \times 10^3 \times 0.09$
$F = 1.03 \times 10^6 \times 0.09 = 0.0927 \times 10^6 = 9.27 \times 10^4 \,N$.
Rounding to two significant figures,$F \approx 9.3 \times 10^4 \,N = 0.93 \times 10^5 \,N$.

(C) Given:
Blood pressure at heart level,$P_{\text{heart}} = 13.3 \,kPa = 13300 \,Pa$.
Density of blood,$\rho = 10^3 \,kg/m^3$.
Acceleration due to gravity,$g = 10 \,m/s^2$.
Height of heart from foot,$h_1 = 1.3 \,m$.
Height of head from heart,$h_2 = 1.7 \,m - 1.3 \,m = 0.4 \,m$.
Pressure at the foot level is given by $P_{\text{foot}} = P_{\text{heart}} + \rho g h_1$.
$P_{\text{foot}} = 13300 + (10^3 \times 10 \times 1.3) = 13300 + 13000 = 26300 \,Pa = 26.3 \,kPa$.
Pressure at the head level is given by $P_{\text{head}} = P_{\text{heart}} - \rho g h_2$.
$P_{\text{head}} = 13300 - (10^3 \times 10 \times 0.4) = 13300 - 4000 = 9300 \,Pa = 9.3 \,kPa$.
The ratio of $BP$ in the foot region to that in the head region is:
$\text{Ratio} = \frac{P_{\text{foot}}}{P_{\text{head}}} = \frac{26.3}{9.3} \approx 2.828$.
This value is closest to $3$.

(D) The total pressure $P$ at a depth $h$ below the surface of a liquid 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.
Given:
Atmospheric pressure $P_0 \approx 1.013 \times 10^5 \, Pa$ (or $1 \, atm$).
Density of water $\rho = 1000 \, kg/m^3$.
Acceleration due to gravity $g \approx 9.8 \, m/s^2$.
Depth $h = 20 \, m$.
Calculating the gauge pressure: $P_{gauge} = \rho gh = 1000 \times 9.8 \times 20 = 1.96 \times 10^5 \, Pa$.
Since $1 \, atm \approx 1.013 \times 10^5 \, Pa$,the gauge pressure in $atm$ is approximately $1.96 \, atm \approx 2 \, atm$.
The total pressure $P = P_0 + P_{gauge} = 1 \, atm + 2 \, atm = 3 \, atm$.
Therefore,the correct option is $D$.

(B) The pressure at a depth $x$ from the surface of the liquid is given by $P(x) = \rho g x$.
To find the average pressure on the vertical wall of height $H$,we integrate the pressure over the depth and divide by the total height.
The total force per unit width is the integral of pressure with respect to depth: $F = \int_0^H \rho g x \, dx$.
$F = \rho g \left[ \frac{x^2}{2} \right]_0^H = \frac{1}{2} \rho g H^2$.
The average pressure $P_{avg}$ is defined as the total force per unit area. For a wall of height $H$ and unit width,the area is $H$. Thus,$P_{avg} = \frac{F}{H} = \frac{\frac{1}{2} \rho g H^2}{H} = \frac{1}{2} \rho g H$.

(D) The pressure measured by a barometer is given by $P = h \rho g_{eff}$,where $g_{eff}$ is the effective acceleration.
When the elevator accelerates upwards with an acceleration $a$,the effective acceleration becomes $g_{eff} = g + a$.
Since the barometer reads $76 \, cm$ of $Hg$ under this condition,the actual atmospheric pressure $P_0$ (which corresponds to the reading when $a = 0$) would be related to the observed reading $h'$ by $P_0 = h' \rho (g + a)$.
However,the question asks for the pressure reading in the accelerating frame. If the barometer reads $76 \, cm$ while accelerating upwards,it implies that the effective weight of the mercury column has increased.
Since $P = h \rho (g + a)$,for a fixed atmospheric pressure $P_0$,the height $h$ would decrease as acceleration increases. If the reading is $76 \, cm$ during upward acceleration,the actual atmospheric pressure must be higher than $76 \, cm$ of $Hg$ equivalent.
Given the options,if the barometer is calibrated to read $76 \, cm$ at rest,and it reads $76 \, cm$ while accelerating,it suggests the pressure is effectively higher than the standard $76 \, cm$ reading at rest. Thus,the most likely value greater than $76 \, cm$ is $77 \, cm$.

(A) sudden drop in the mercury level of a barometer indicates a rapid decrease in atmospheric pressure.
When atmospheric pressure drops significantly,it creates a pressure gradient that causes air from surrounding high-pressure regions to rush into the low-pressure area at high velocities.
This high-velocity movement of air is characteristic of a storm.
Therefore,a drop of $10 \,mm$ or more is a precursor to a storm.

(A) The pressure at any point in a static liquid depends only on the depth of the point below the free surface of the liquid,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 of the point.
In the given figure,both points $x$ and $y$ are at the same depth $h$ below the free surface of the liquid.
Since the liquid is the same (same $\rho$) and the depth $h$ is the same,the hydrostatic pressure at both points must be equal.
Therefore,$p_x = p_y$.

(C) The correct option is $C$.
When an elevator accelerates upward with an acceleration $a$, an observer inside the elevator experiences a pseudo force acting downwards.
This effectively increases the acceleration due to gravity from $g$ to $g^{\prime} = g + a$.
The pressure exerted by the mercury column in the barometer is given by $P = \rho g h$.
Since the atmospheric pressure $P$ remains constant, we have $\rho g h = \rho (g + a) h^{\prime}$, where $h^{\prime}$ is the new reading.
Since $(g + a) > g$, it follows that $h^{\prime} < h$.
Therefore, the new reading $h^{\prime}$ will be less than $76 \, cm$.

(B) The vertical height of the mercury column in a barometer is determined by the atmospheric pressure and remains constant regardless of the inclination of the tube,provided the open end remains submerged in the mercury reservoir.
Let $h = 75 \,cm$ be the vertical height of the mercury column.
Let $x$ be the length of the mercury column along the tube when it is inclined at an angle $\theta = 30^{\circ}$ with the horizontal.
The relationship between the vertical height $h$ and the length $x$ along the inclined tube is given by $h = x \sin(\theta)$.
Here,$\theta = 30^{\circ}$,so $h = x \sin(30^{\circ})$.
Substituting the values: $75 = x \times (1/2)$.
Therefore,$x = 75 \times 2 = 150 \,cm$.

(A) The atmospheric pressure is given by $P = h \rho g$.
Standard atmospheric pressure is $P = 1.013 \times 10^5 \,Pa \approx 10^5 \,Pa$.
This pressure is equivalent to a mercury column of height $h_{Hg} = 0.76 \,m$ with density $\rho_{Hg} = 13600 \,kg/m^3$.
For the barometric liquid (tribromomethane),the density is $\rho_{TBM} = 2.9 \times 1000 \,kg/m^3 = 2900 \,kg/m^3$.
Using the principle $P = h_{Hg} \rho_{Hg} g = h_{TBM} \rho_{TBM} g$,we have:
$h_{TBM} = \frac{h_{Hg} \rho_{Hg}}{\rho_{TBM}}$
$h_{TBM} = \frac{0.76 \times 13600}{2900} \approx 3.56 \,m$.
Given the approximation $10^5 \,Pa$ and standard values,the calculated height is approximately $3.52 \,m$.

(C) Let the initial height of water in the tank be $h$. The pressure at the bottom is the sum of atmospheric pressure and gauge pressure.
$P_{\text{bottom}} = P + \rho_w g h = 4 P$
Therefore,the gauge pressure due to the water column is $\rho_w g h = 3 P$ ... $(1)$
When water is drawn out such that the level decreases by $\frac{3}{5}$ of its initial height,the remaining height of the water is $h' = h - \frac{3}{5} h = \frac{2}{5} h$.
The new pressure at the bottom is $P' = P + \rho_w g h'$.
Substituting $h' = \frac{2}{5} h$ into the equation:
$P' = P + \rho_w g (\frac{2}{5} h) = P + \frac{2}{5} (\rho_w g h)$.
Using equation $(1)$,substitute $\rho_w g h = 3 P$:
$P' = P + \frac{2}{5} (3 P) = P + \frac{6 P}{5} = \frac{5 P + 6 P}{5} = \frac{11 P}{5}$.

(B) The velocity head of water is given by $\frac{V^2}{2g}$.
The pressure head of mercury is given as $h_{Hg} = 40 \, cm = 0.4 \, m$.
The pressure exerted by this mercury column is $P = \rho_{Hg} g h_{Hg}$.
Equating the pressure head of water to the pressure head of mercury:
$\frac{P}{\rho_w g} = \frac{\rho_{Hg} g h_{Hg}}{\rho_w g} = \frac{\rho_{Hg}}{\rho_w} h_{Hg}$.
Given that the velocity head $\frac{V^2}{2g}$ is equal to this pressure head:
$\frac{V^2}{2g} = \frac{\rho_{Hg}}{\rho_w} h_{Hg}$.
$V^2 = 2g \times \frac{\rho_{Hg}}{\rho_w} \times h_{Hg}$.
Substituting the values: $V^2 = 2 \times 9.8 \times 13.6 \times 0.4$.
$V^2 = 106.5984$.
$V = \sqrt{106.5984} \approx 10.32 \, m/s$.

(C) When a beaker moves upward with an acceleration $a$, a pseudo force acts on the liquid particles in the downward direction.
This pseudo force adds to the gravitational force, effectively increasing the acceleration due to gravity.
The effective acceleration due to gravity is $g' = g + a$.
The pressure $P$ at a depth $h$ is given by the formula $P = \rho g' h$.
Substituting the value of $g'$, we get $P = \rho(g + a)h$.

(C) The correct option is $C$.
The variation of atmospheric pressure $P$ with height $h$ above the Earth's surface is described by the barometric formula:
$P_h = P_0 e^{-mgh / RT}$
where $P_0$ is the pressure at sea level,$m$ is the molar mass of air,$g$ is the acceleration due to gravity,$R$ is the universal gas constant,and $T$ is the absolute temperature.
Since the pressure depends on the height $h$ through an exponential function,the decrease in atmospheric pressure with increasing height is exponential.

(A) The pressure exerted by a liquid column at the base of a vessel is given by the formula $P = h \rho g$,where $h$ is the height of the liquid column,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
Since both vessels $A$ and $B$ are filled to the same height $h$ with the same liquid (water),the pressure at the base depends only on the height $h$,the density $\rho$,and the constant $g$.
Because $h$,$\rho$,and $g$ are identical for both vessels,the pressure at the base of vessel $A$ must be equal to the pressure at the base of vessel $B$.

(A) The absolute pressure $P$ at a depth $h$ is given by the formula: $P = P_{atm} + \rho gh$.
Given values are: $h = 1 \,km = 1000 \,m$,$\rho = 10^{3} \,kg/m^{3}$,$g = 10 \,m/s^{2}$,and $P_{atm} = 1.01 \times 10^{5} \,N/m^{2}$.
Calculating the gauge pressure (pressure due to water column): $P_{gauge} = \rho gh = 10^{3} \times 10 \times 1000 = 10^{7} \,N/m^{2}$.
Now,adding the atmospheric pressure: $P = 1.01 \times 10^{5} + 10^{7} = 0.0101 \times 10^{7} + 10^{7} = 1.0101 \times 10^{7} \,N/m^{2}$.
Rounding to the appropriate significant figures,we get $P \approx 1.011 \times 10^{7} \,N/m^{2}$.

(D) The pressure $P$ at the bottom of a liquid tank is given by the formula $P = h \rho g$,where $h$ is the height of the liquid column,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
From this formula,it is clear that the pressure is directly proportional to the height of the liquid $h$,the density of the liquid $\rho$,and the acceleration due to gravity $g$.
The pressure does not depend on the area of the liquid surface or the shape of the container.
Therefore,the pressure is not proportional to the area of the liquid surface.

(B) The gauge pressure $P_g$ at a depth $h$ is given by the formula $P_g = \rho g h$.
Given:
Density of sea water $\rho = 1025 \,kg \,m^{-3}$
Acceleration due to gravity $g = 10 \,ms^{-2}$
Depth $h = 50 \,m$
Substituting the values into the formula:
$P_g = 1025 \times 10 \times 50$
$P_g = 1025 \times 500$
$P_g = 512500 \,Pa$.

(A) The gauge pressure $P_g$ at the bottom of a fluid column is given by the formula: $P_g = \rho gh$.
Here,the density of oil $\rho = 850 \,kg \,m^{-3}$,the height of the oil column $h = 4 \,m$,and the acceleration due to gravity $g = 10 \,m \,s^{-2}$.
Substituting these values into the formula:
$P_g = 850 \,kg \,m^{-3} \times 10 \,m \,s^{-2} \times 4 \,m$
$P_g = 34000 \,Pa$
Since $1 \,kPa = 1000 \,Pa$,we have $P_g = 34 \,kPa$.

(D) Given:
Depth, $h = 3 \,m$
Density of water, $\rho = 1000 \,kg \,m^{-3} = 10^3 \,kg \,m^{-3}$
Acceleration due to gravity, $g = 10 \,m \,s^{-2}$
The pressure $P$ at the bottom of the pool due to the water column is calculated using the hydrostatic pressure formula:
$P = \rho g h$
Substituting the given values:
$P = 10^3 \,kg \,m^{-3} \times 10 \,m \,s^{-2} \times 3 \,m$
$P = 30 \times 10^3 \,Pa$
Thus, the pressure at the bottom is $30 \times 10^3 \,Pa$.

(C) The pressure exerted by a liquid column is given by the formula $p = \rho g h$.
Here, the height of the mercury column is $h = 1 \,mm = 10^{-3} \,m$.
The density of mercury is $\rho = 13.6 \times 10^3 \,kg \,m^{-3}$.
The acceleration due to gravity is $g = 9.8 \,m \,s^{-2}$.
Substituting these values into the formula:
$p = (13.6 \times 10^3) \times 9.8 \times 10^{-3}$
$p = 13.6 \times 9.8$
$p = 133.28 \,Pa \approx 133.3 \,Pa$.

(B) The pressure at a depth $y$ from the free surface of water is $P = P_{atm} + \rho g y$. The force $dF$ on a small strip of height $dy$ at depth $y$ is $dF = P \cdot dA = (P_{atm} + \rho g y) \cdot (a \cdot dy)$,where $a$ is the width of the dam. The total force $F$ is the integral of $dF$ from $y=0$ to $y=h$. The point of application $y_R$ (center of pressure) is given by $y_R = \frac{\int y dF}{\int dF}$. For a rectangular gate of height $h$ submerged in water,the center of pressure is at a depth of $\frac{2}{3}h$ from the free surface. Since the question asks for the point of application relative to the base $O$ (which is at depth $h$ from the surface),the distance from the base is $h - \frac{2}{3}h = \frac{h}{3}$.

(D) The total pressure $P$ at a depth $h$ below the surface of a liquid is given by the formula: $P = P_0 + \rho gh$.
Here,$P_0$ is the atmospheric pressure,$\rho$ is the density of water $(1000 \ kg \ m^{-3})$,$g$ is the acceleration due to gravity $(10 \ ms^{-2})$,and $h$ is the depth $(10 \ m)$.
Substituting the values:
$P = 1.01 \times 10^5 + (1000 \times 10 \times 10)$
$P = 1.01 \times 10^5 + 10^5$
$P = 1.01 \times 10^5 + 1.00 \times 10^5 = 2.01 \times 10^5 \ Nm^{-2}$.
Rounding to the nearest significant value provided in the options,we get $2 \times 10^5 \ Nm^{-2}$.

(B) Let the density of the liquid be $\rho_l = 1.2 \ g \ cm^{-3}$ and the density of the oil be $\rho_o = 0.9 \ g \ cm^{-3}$.
When the liquid rises by $15 \ cm$ on the right side,it must have fallen by $15 \ cm$ on the left side relative to the initial equilibrium level.
Thus,the total difference in the liquid levels between the two arms is $h_l = 15 \ cm + 15 \ cm = 30 \ cm$.
Let $h_o$ be the height of the oil column on the left side. The pressure at the interface of the oil and liquid on the left side must equal the pressure at the same horizontal level on the right side.
Using the hydrostatic pressure balance: $h_o \rho_o g = h_l \rho_l g$.
Substituting the values: $h_o \times 0.9 = 30 \times 1.2$.
$h_o = \frac{30 \times 1.2}{0.9} = 40 \ cm$.
The oil column stands $40 \ cm$ above the initial liquid level on the left. Since the liquid on the left fell by $15 \ cm$,the top of the oil column is $40 - 15 = 25 \ cm$ above the initial level.
The liquid on the right is $15 \ cm$ above the initial level.
Therefore,the difference in height between the oil level and the right liquid level is $25 \ cm - 15 \ cm = 10 \ cm$.

(B) The formula for absolute pressure is $P = P_{atm} + \rho g h$.
Here, $P = 100 \text{ atm} = 100 \times 10^5 \text{ Pa}$, $P_{atm} = 1 \text{ atm} = 1 \times 10^5 \text{ Pa}$, $\rho = 1000 \text{ kg/m}^3$, and $g = 10 \text{ m/s}^2$.
Substituting the values: $100 \times 10^5 = 1 \times 10^5 + (1000)(10)h$.
$100 \times 10^5 - 1 \times 10^5 = 10^4 h$.
$99 \times 10^5 = 10^4 h$.
$h = \frac{99 \times 10^5}{10^4} = 99 \times 10 = 990 \text{ m}$.
Thus, the submarine can go $990 \text{ m}$ deep.