Welcome to Lecture 80 of our comprehensive JEE Physics Master Series. In this lecture, we transition into fluid mechanics, exploring the foundational laws governing fluids at rest (hydrostatics): the nature of fluid pressure, hydrostatic variation with depth, Pascal’s principle, hydraulic machinery, accelerated fluid systems, and manometer dynamics.
1. Detailed Concept Card
1.1 Fundamental Nature of Fluid Pressure
A fluid is a substance that cannot sustain a static shear stress; it deforms continuously under the application of tangential forces. Consequently, when a fluid is at rest in static equilibrium, the force exerted by the fluid on any surface in contact with it must be strictly perpendicular (normal) to the surface at every point.
Pressure is defined as the magnitude of normal force per unit contact area:
$$P = \lim_{\Delta A \to 0} \frac{\Delta F_\perp}{\Delta A}$$
Scalar Nature of Pressure: Although the force resulting from pressure has a specific direction (normal to the surface on which it acts), pressure itself is an isotropic scalar quantity. At any given point in a static fluid, pressure acts equally in all directions.
Units and Standard Values:
- SI Unit: $\text{N/m}^2$ or Pascal ($\text{Pa}$), where $1\text{ Pa} = 1\text{ N}\cdot\text{m}^{-2} = 1\text{ kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}$.
- CGS Unit: $\text{dyne/cm}^2$, where $1\text{ Pa} = 10\text{ dyne/cm}^2$.
- Standard Atmosphere ($\text{atm}$): $1\text{ atm} = 1.01325 \times 10^5\text{ Pa} \approx 1.013 \times 10^5\text{ Pa} = 1.013\text{ bar}$.
- Torr and Barometric Height: $1\text{ torr} = 1\text{ mmHg} \approx 133.32\text{ Pa}$. Hence, $1\text{ atm} = 760\text{ torr} = 760\text{ mmHg} = 76\text{ cm of Hg}$.
1.2 Variation of Pressure with Depth in a Static Fluid
Consider an infinitesimal cylindrical fluid element of cross-sectional area $A$ and vertical thickness $dy$ at a vertical height $y$ above a reference datum. Let the pressure at height $y$ be $P$ and at height $y + dy$ be $P + dP$.
The forces acting along the vertical direction are:
- Upward force on the bottom face: $F_{\text{up}} = P A$
- Downward force on the top face: $F_{\text{down}} = (P + dP) A$
- Downward gravitational weight of the fluid cylinder: $dW = (\rho A dy) g$
For static vertical equilibrium: $\sum F_y = 0$
$$P A – (P + dP)A – \rho g A dy = 0 \implies \frac{dP}{dy} = -\rho g$$
For an incompressible fluid (constant density $\rho$), integrating between depth $0$ (surface at atmospheric pressure $P_0$) and depth $h$ below the surface ($y = -h$):
$$P(h) = P_0 + \rho g h$$
Gauge Pressure vs. Absolute Pressure:
- Absolute Pressure ($P_{\text{abs}}$): The total actual pressure at a point, including atmospheric pressure: $P_{\text{abs}} = P_0 + \rho g h$.
- Gauge Pressure ($P_{\text{gauge}}$): The pressure measured relative to ambient atmospheric pressure: $P_{\text{gauge}} = P_{\text{abs}} – P_0 = \rho g h$.
The Hydrostatic Paradox: The pressure at the bottom of a container depends strictly on the vertical depth $h$ and fluid density $\rho$, completely independent of the shape, cross-sectional area, or total volume of fluid in the vessel. However, the total force exerted by the fluid on the base equals $P \cdot A_{\text{base}}$, which may differ from the total weight of the fluid due to vertical force components exerted by slanted container walls.
1.3 Pascal’s Principle and Hydraulic Systems
Statement of Pascal’s Law: Any external pressure applied to an enclosed, incompressible, and static fluid is transmitted undiminished to every point in the fluid as well as to the walls of the containing vessel.
Mathematical Formulation of the Hydraulic Lift:
Consider a hydraulic press with two connected vertical cylinders of cross-sectional areas $A_1$ (input piston) and $A_2$ (output piston), where $A_2 \gg A_1$. Applying a force $F_1$ on piston 1 creates an excess pressure:
$$\Delta P = \frac{F_1}{A_1}$$
By Pascal’s principle, this exact pressure $\Delta P$ is transmitted to the underside of piston 2, producing an upward thrust:
$$F_2 = \Delta P \cdot A_2 = F_1 \left( \frac{A_2}{A_1} \right) = F_1 \left( \frac{d_2}{d_1} \right)^2$$
The Mechanical Advantage (MA) is:
$$\text{MA} = \frac{F_2}{F_1} = \frac{A_2}{A_1} = \left( \frac{d_2}{d_1} \right)^2$$
Conservation of Work: Because the liquid is incompressible, the volume displaced by piston 1 equals the volume accepted beneath piston 2: $V = A_1 d_1 = A_2 d_2$. Hence, the displacement ratio satisfies $d_2 = d_1 (A_1 / A_2)$. The ideal work input equals the work output:
$$W_{\text{in}} = F_1 d_1 = \left( F_2 \frac{A_1}{A_2} \right) \left( d_2 \frac{A_2}{A_1} \right) = F_2 d_2 = W_{\text{out}}$$
Hydraulic machines magnify force at the proportional expense of distance, strictly obeying energy conservation.
1.4 Accelerated Fluid Systems & Effective Gravity
When a container filled with fluid undergoes uniform acceleration $\vec{a}$, we analyze the fluid in the accelerating reference frame by introducing an inertial pseudo-force per unit volume $-\rho \vec{a}$. The effective gravitational field is:
$$\vec{g}_{\text{eff}} = \vec{g} – \vec{a}$$
1. Free Surface Orientation: The free surface of a liquid always orients itself strictly perpendicular to $\vec{g}_{\text{eff}}$.
- Horizontal Acceleration ($a_x \hat{i}$): $\vec{g}_{\text{eff}} = -a_x \hat{i} – g \hat{j}$. The angle $\theta$ between the inclined liquid surface and the horizontal is: $$\tan\theta = \frac{a_x}{g}$$ The liquid climbs up at the trailing wall and drops at the leading wall.
- Pressure Gradient in Accelerated Container: $$\frac{\partial P}{\partial x} = -\rho a_x, \quad \frac{\partial P}{\partial y} = -\rho g$$ Along any horizontal line of length $L$ separated along the direction of acceleration: $$P_{\text{rear}} – P_{\text{front}} = \rho a_x L$$
- Vertical Acceleration ($a_y \hat{j}$): Effective gravity becomes $g_{\text{eff}} = g + a_y$ (accelerating upward) or $g_{\text{eff}} = g – a_y$ (accelerating downward). Pressure at depth $h$ is: $$P(h) = P_0 + \rho (g \pm a_y) h$$ In free fall ($a_y = -g$), $g_{\text{eff}} = 0$, so hydrostatic gauge pressure vanishes everywhere throughout the fluid!
1.5 U-Tube Manometry and Immiscible Liquids
The Fundamental Isbaric Rule: Pressure is strictly equal at any two points lying at the same horizontal height provided that the points can be connected by a continuous path staying entirely within the same homogeneous fluid at rest.
For a U-tube containing two immiscible fluids of densities $\rho_1$ and $\rho_2$ open to the atmosphere on both sides, choosing the reference datum at the interface level where fluid 1 meets fluid 2:
$$P_0 + \rho_1 g h_1 = P_0 + \rho_2 g h_2 \implies \rho_1 h_1 = \rho_2 h_2$$
1.6 Key Insights & Critical Pitfalls for JEE
- Pitfall 1 (Equating Pressures Across Different Liquids): Never equate $P_A = P_B$ at the same horizontal line if point A and point B are situated in different immiscible liquids or separated by an obstruction. The isobaric condition holds only in a continuous single-fluid medium.
- Pitfall 2 (Atmospheric Pressure on Container Walls): When computing net horizontal force on a vertical dam or tank wall, remember that atmospheric pressure $P_0$ acts on both the liquid surface (transmitted through the fluid) and the outside exposed face of the wall. Thus, net force depends only on gauge pressure: $F_{\text{net}} = \frac{1}{2} \rho g H^2 \cdot W$.
- Pitfall 3 (Accelerated Fluid Piling): When a tanker or beaker accelerates to the right ($+x$), the liquid surface tilts such that the liquid height is higher at the back wall ($x = 0$) and lower at the front wall ($x = L$).
2. Four Graded Solved Examples
Example 1: U-Tube Manometer with Immiscible Liquids (JEE Main Level)
Problem Statement: A U-tube of uniform cross-section contains mercury (density $\rho_m = 13.6 \times 10^3\text{ kg/m}^3$). Water (density $\rho_w = 1.0 \times 10^3\text{ kg/m}^3$) is poured into one arm until the column of water is $h_w = 27.2\text{ cm}$ high. Into the other arm, kerosene oil of density $\rho_k = 0.8 \times 10^3\text{ kg/m}^3$ is poured until the upper surfaces of water and oil are at the same horizontal level. Find the height of the oil column $h_k$. Take $g = 9.8\text{ m/s}^2$.
Solution:
Let the two arms be Arm 1 (water) and Arm 2 (oil). Both upper surfaces are open to the atmosphere and lie at the exact same horizontal level. Let this top level be $y = H$.
Water extends from $y = H$ downward to $y = H – h_w$. Oil extends from $y = H$ downward to $y = H – h_k$. Since water is denser than oil ($\rho_w > \rho_k$), to balance hydrostatic pressure, mercury must rise higher in the oil arm than in the water arm. Hence, the mercury-water interface is lower than the mercury-oil interface, which means $h_w > h_k$.
The mercury-water interface is at depth $h_w$ below the top surface. The mercury-oil interface is at depth $h_k$ below the top surface. In the oil arm, below the oil column, there is a column of mercury of height $(h_w – h_k)$ down to the horizontal level of the water-mercury interface.
Equating pressures in both arms at the horizontal level of the water-mercury interface:
$$P_0 + \rho_w g h_w = P_0 + \rho_k g h_k + \rho_m g (h_w – h_k)$$
Subtracting $P_0$ and dividing by $g$:
$$\rho_w h_w = \rho_k h_k + \rho_m h_w – \rho_m h_k$$
$$(\rho_m – \rho_k) h_k = (\rho_m – \rho_w) h_w$$
Substitute numerical values in $\text{g/cm}^3$ ($\rho_m = 13.6$, $\rho_w = 1.0$, $\rho_k = 0.8$, $h_w = 27.2\text{ cm}$):
$$(13.6 – 0.8) h_k = (13.6 – 1.0) \times 27.2$$
$$12.8 \cdot h_k = 12.6 \times 27.2$$
$$h_k = \frac{12.6 \times 27.2}{12.8} = 12.6 \times 2.125 = 26.775\text{ cm}$$
Key Takeaway: In three-liquid manometry, always establish the reference isobar at the lowest liquid-liquid interface across the continuous fluid.
Example 2: Hydraulic Lift with Piston Heights and Friction (JEE Main / Advanced Linkage)
Problem Statement: A hydraulic automobile lift has an input piston of diameter $d_1 = 4.0\text{ cm}$ and an output load piston of diameter $d_2 = 24.0\text{ cm}$. The lift is filled with hydraulic oil of density $\rho = 800\text{ kg/m}^3$. A car of mass $M = 1440\text{ kg}$ rests on the large piston. In equilibrium, the large piston is at an elevation $h = 1.5\text{ m}$ above the small piston. Determine the force $F_1$ that must be applied to the small piston to keep the system in static equilibrium. Take $g = 10\text{ m/s}^2$.
Solution:
Let the reference level ($y = 0$) be the face of the input piston. The pressure at $y = 0$ inside the small cylinder is:
$$P_1 = P_0 + \frac{F_1}{A_1}$$
The large piston is at height $h = 1.5\text{ m}$ above this reference plane. The total downward force acting on the large piston is the car’s weight $Mg$ plus atmospheric thrust $P_0 A_2$. Hence, the fluid pressure immediately below the large piston at height $h$ is:
$$P_2 = P_0 + \frac{M g}{A_2}$$
Since the fluid is static and continuous between $y = 0$ and $y = h$:
$$P_1 = P_2 + \rho g h$$
$$P_0 + \frac{F_1}{A_1} = P_0 + \frac{M g}{A_2} + \rho g h \implies \frac{F_1}{A_1} = \frac{M g}{A_2} + \rho g h$$
Multiply both sides by $A_1$:
$$F_1 = M g \left( \frac{A_1}{A_2} \right) + \rho g h A_1 = M g \left( \frac{d_1}{d_2} \right)^2 + \rho g h \left( \frac{\pi d_1^2}{4} \right)$$
Calculate the terms individually:
- Diameter ratio squared: $\left( \frac{d_1}{d_2} \right)^2 = \left( \frac{4}{24} \right)^2 = \left( \frac{1}{6} \right)^2 = \frac{1}{36}$.
- Direct weight transmission: $M g \left( \frac{1}{36} \right) = \frac{1440 \times 10}{36} = 400\text{ N}$.
- Hydrostatic head contribution: $$\rho g h = 800 \times 10 \times 1.5 = 12000\text{ N/m}^2$$ $$A_1 = \frac{\pi (0.04)^2}{4} = \frac{\pi \times 1.6 \times 10^{-3}}{4} = 4\pi \times 10^{-4}\text{ m}^2 \approx 1.2566 \times 10^{-3}\text{ m}^2$$ $$\rho g h A_1 = 12000 \times (1.2566 \times 10^{-3}) \approx 15.08\text{ N}$$
Total applied force:
$$F_1 = 400 + 15.08 = 415.08\text{ N}$$
Key Takeaway: When hydraulic pistons operate at different elevations, the hydrostatic head term $\rho g h$ cannot be neglected; it adds directly to the required input pressure.
Example 3: Rectangular Tank Under Combined Accelerations (JEE Advanced Level)
Problem Statement: An open rectangular tank of length $L = 2.0\text{ m}$, width $W = 1.0\text{ m}$, and height $H = 1.5\text{ m}$ is filled with water to a depth of $h_0 = 1.0\text{ m}$. The tank is accelerated uniformly on a horizontal track with acceleration $a_x = 3.0\text{ m/s}^2$ and simultaneously given an upward vertical acceleration $a_y = 2.0\text{ m/s}^2$. Taking $g = 10\text{ m/s}^2$ and density of water $\rho = 1000\text{ kg/m}^3$:
(a) Determine the inclination of the free surface.
(b) Verify whether water spills out of the tank.
(c) Calculate the absolute pressure at the bottom-rear corner and bottom-front corner.
Solution:
Part (a): Free surface angle
In the non-inertial frame of the tank, the effective gravity vector is:
$$\vec{g}_{\text{eff}} = -a_x \hat{i} – (g + a_y) \hat{j} = -3.0\hat{i} – (10 + 2)\hat{j} = -3.0\hat{i} – 12.0\hat{j}\text{ m/s}^2$$
The free surface is perpendicular to $\vec{g}_{\text{eff}}$. Its slope with respect to the horizontal is:
$$\tan\theta = \frac{a_x}{g + a_y} = \frac{3.0}{10 + 2} = \frac{3}{12} = 0.25$$
$$\theta = \arctan(0.25) \approx 14.04^\circ$$
Part (b): Spill check
Since no liquid has spilled yet, the volume of water remains constant. The liquid surface pivots about its midpoint along the length ($x = L/2$). The rise in water level at the rear wall ($x = 0$) and the drop at the front wall ($x = L$) is:
$$\Delta h = \frac{L}{2} \tan\theta = \frac{2.0}{2} \times 0.25 = 0.25\text{ m}$$
The maximum water height along the rear wall is:
$$h_{\text{rear}} = h_0 + \Delta h = 1.0 + 0.25 = 1.25\text{ m}$$
Since $h_{\text{rear}} = 1.25\text{ m} < H = 1.5\text{ m}$, the water does not spill over the top edge of the tank.
Part (c): Corner Pressures
Along the rear wall, the liquid column has height $h_{\text{rear}} = 1.25\text{ m}$. The effective vertical gravity is $g_{\text{eff}, y} = g + a_y = 12\text{ m/s}^2$. The absolute pressure at the bottom-rear corner is:
$$P_{\text{rear, bottom}} = P_0 + \rho (g + a_y) h_{\text{rear}} = 1.013 \times 10^5 + 1000 \times 12 \times 1.25 = 1.013 \times 10^5 + 15000 = 116300\text{ Pa} = 116.3\text{ kPa}$$
At the front wall, the water depth is $h_{\text{front}} = h_0 – \Delta h = 1.0 – 0.25 = 0.75\text{ m}$. The absolute pressure at the bottom-front corner is:
$$P_{\text{front, bottom}} = P_0 + \rho (g + a_y) h_{\text{front}} = 1.013 \times 10^5 + 1000 \times 12 \times 0.75 = 1.013 \times 10^5 + 9000 = 110300\text{ Pa} = 110.3\text{ kPa}$$
Check along the bottom surface horizontally: $P_{\text{rear}} – P_{\text{front}} = 15000 – 9000 = 6000\text{ Pa}$. Also, $\rho a_x L = 1000 \times 3.0 \times 2.0 = 6000\text{ Pa}$. The horizontal and vertical pressure formulations match flawlessly.
Example 4: Rotating Fluid Column & Free Surface Paraboloid (JEE Advanced Edge Case)
Problem Statement: A cylindrical vessel of radius $R = 10\text{ cm}$ containing water of density $\rho = 1000\text{ kg/m}^3$ is rotated about its vertical axis of symmetry with a constant angular velocity $\omega = 20\text{ rad/s}$.
(a) Derive the equation of the free liquid surface $z(r)$, where $r$ is the radial distance from the axis of rotation.
(b) Calculate the total difference in height $\Delta z$ between the lowest point of the surface at the center ($r = 0$) and the highest point at the wall ($r = R$). Take $g = 10\text{ m/s}^2$.
Solution:
Part (a): Derivation of the Free Surface
In the rotating reference frame of the cylinder, an element of liquid of mass $dm$ at coordinates $(r, z)$ experiences two forces:
- Downward gravitational force: $dF_g = (dm) g$
- Outward centrifugal pseudo-force: $dF_c = (dm) \omega^2 r$
The free surface is an equipotential surface, which means the tangent to the free surface at any point $(r, z)$ must be perpendicular to the resultant body force vector $\vec{F} = dm(\omega^2 r \hat{r} – g \hat{k})$.
The slope of the surface in the $r$-$z$ plane is therefore:
$$\frac{dz}{dr} = \frac{dF_c}{dF_g} = \frac{(dm)\omega^2 r}{(dm)g} = \frac{\omega^2 r}{g}$$
Integrating with the boundary condition that $z = z_0$ at $r = 0$:
$$\int_{z_0}^{z} dz = \int_0^r \frac{\omega^2 r}{g} dr \implies z(r) = z_0 + \frac{\omega^2 r^2}{2g}$$
Thus, the free liquid surface forms a revolution paraboloid!
Part (b): Calculation of Height Difference
The height difference between the rim ($r = R = 0.1\text{ m}$) and the vertex at the axis ($r = 0$) is:
$$\Delta z = z(R) – z(0) = \frac{\omega^2 R^2}{2g}$$
Substitute $\omega = 20\text{ rad/s}$, $R = 0.1\text{ m}$, and $g = 10\text{ m/s}^2$:
$$\Delta z = \frac{(20)^2 \times (0.1)^2}{2 \times 10} = \frac{400 \times 0.01}{20} = \frac{4}{20} = 0.20\text{ m} = 20.0\text{ cm}$$
Key Takeaway: In rotating hydrostatics, the parabolic depression obeys $\Delta z = \frac{\omega^2 R^2}{2g}$, and the volume of a paraboloid of revolution is exactly half the volume of its circumscribed cylinder: $V_{\text{paraboloid}} = \frac{1}{2} \pi R^2 \Delta z$.
3. Practice Worksheet (10 Questions)
Questions 1 to 5 are Single Correct (JEE Main level). Questions 6 to 8 are One or More than One Correct (JEE Advanced level). Questions 9 and 10 are Numerical / Integer Answer Type (JEE Advanced level).
Question 1 (JEE Main Single Correct)
A simple open U-tube contains mercury. When $11.2\text{ cm}$ of water is poured into the left arm, what height of oil of relative density $0.8$ must be poured into the right arm so that the mercury levels in both arms remain identical?
(A) $14.0\text{ cm}$
(B) $11.2\text{ cm}$
(C) $8.96\text{ cm}$
(D) $17.5\text{ cm}$
Question 2 (JEE Main Single Correct)
A closed cylindrical vessel of base area $A$ and height $H$ is completely filled with a liquid of density $\rho$. The vessel is placed on a horizontal plane and accelerated horizontally with a constant acceleration $a$. The total force exerted by the liquid on the base of the cylinder is:
(A) $\rho g H A$
(B) $\rho (g + a) H A$
(C) $\rho \sqrt{g^2 + a^2} H A$
(D) $\rho g H A + \frac{1}{2} \rho a H A$
Question 3 (JEE Main Single Correct)
In a hydraulic press, the diameters of the small and large pistons are $2\text{ cm}$ and $10\text{ cm}$ respectively. A force of $60\text{ N}$ is applied to the small piston. If the small piston moves down by $25\text{ cm}$, the distance through which the large piston moves upward (assuming incompressible fluid and no leakage) is:
(A) $5.0\text{ cm}$
(B) $1.0\text{ cm}$
(C) $2.5\text{ cm}$
(D) $0.5\text{ cm}$
Question 4 (JEE Main Single Correct)
An open cubical tank of side $a$ is completely filled with an incompressible liquid of density $\rho$. If it is accelerated horizontally along one edge with acceleration $g/2$, the fraction of liquid that spills out of the tank is:
(A) $1/2$
(B) $1/4$
(C) $1/3$
(D) $1/8$
Question 5 (JEE Main Single Correct)
A barometer tube reads $75\text{ cm}$ of mercury. If the tube is tilted by an angle of $60^\circ$ with the vertical, the length of the mercury column along the tilted tube will be:
(A) $75\text{ cm}$
(B) $150\text{ cm}$
(C) $37.5\text{ cm}$
(D) $86.6\text{ cm}$
Question 6 (JEE Advanced One or More Correct)
A U-tube with limbs of equal cross-sectional area contains liquid of density $\rho$. The horizontal base of the tube has length $L$. The tube is accelerated horizontally with a constant acceleration $a$ along the direction of the horizontal base. Which of the following statements is/are correct?
(A) The difference in liquid level between the two vertical limbs is $h = \frac{a L}{g}$.
(B) The pressure at the bottom-trailing corner is greater than the pressure at the bottom-leading corner by $\rho a L$.
(C) The pressure at the midpoint of the horizontal base equals $P_0 + \rho g h_0$, where $h_0$ is the initial equilibrium height in both limbs.
(D) If the tube is subjected to free fall ($g_{\text{eff}} = 0$) while accelerating horizontally, the liquid will not move.
Question 7 (JEE Advanced One or More Correct)
Consider a hydraulic lift containing an incompressible fluid. A force $F_1$ is applied on the smaller piston of area $A_1$, lifting a load $W_2$ on the larger piston of area $A_2$ ($A_2 > A_1$). Neglect friction and assume the pistons are at the same horizontal height. Which of the following statements must be true?
(A) The force multiplication factor equals $A_2 / A_1$.
(B) The work done by $F_1$ during displacement $d_1$ equals the work done on the load $W_2$.
(C) The pressure energy density imparted to the fluid at the small piston is completely transmitted to the large piston.
(D) If the liquid is replaced with a highly compressible gas, the mechanical advantage remains strictly identical under all operating strokes without requiring additional stroke length.
Question 8 (JEE Advanced One or More Correct)
A vertical cylindrical beaker of radius $R$ containing water is placed on an elevator. Under which of the following conditions will the gauge pressure at a depth $h$ below the surface be strictly zero?
(A) The elevator is moving upward with a constant velocity $v$.
(B) The elevator is accelerating downward with acceleration $a = g$.
(C) The elevator cable breaks and the elevator enters free fall.
(D) The elevator is accelerating upward with acceleration $a = g$.
Question 9 (JEE Advanced Numerical Value)
A hydraulic press has a ram (output piston) of diameter $30\text{ cm}$ and a plunger (input piston) of diameter $3\text{ cm}$. The plunger is operated by a light lever with a mechanical advantage of $5:1$ (i.e., effort arm is 5 times the load arm acting on the plunger). To lift a vehicle of mass $1800\text{ kg}$ on the ram, find the minimum force (in Newtons) that must be applied at the end of the lever handle. (Take $g = 10\text{ m/s}^2$).
Question 10 (JEE Advanced Numerical Value)
A U-tube of uniform cross-section contains mercury. The horizontal arm has a length $L = 20\text{ cm}$. The tube is mounted on a horizontal turntable such that one vertical arm lies exactly on the axis of rotation and the other vertical arm is at distance $L = 20\text{ cm}$ from the axis. When the turntable rotates at constant angular speed $\omega = 10\text{ rad/s}$, the difference in heights of the mercury levels in the two vertical limbs is $h$ cm. Find the value of $h$. (Take $g = 10\text{ m/s}^2$).
4. Detailed Solutions & Answer Key
Quick Answer Key
| Question | Type | Correct Answer |
|---|---|---|
| 1 | Single Correct | (A) |
| 2 | Single Correct | (A) |
| 3 | Single Correct | (B) |
| 4 | Single Correct | (B) |
| 5 | Single Correct | (B) |
| 6 | One or More Correct | (A, B, C) |
| 7 | One or More Correct | (A, B, C) |
| 8 | One or More Correct | (B, C) |
| 9 | Numerical | 36 |
| 10 | Numerical | 20 |
Step-by-Step Solutions
Solution 1:
Both arms are open to the atmosphere ($P_0$). For the mercury levels to remain identical in both limbs, the pressure exerted by the liquid columns on the two mercury surfaces must be exactly equal:
$$P_0 + \rho_w g h_w = P_0 + \rho_{\text{oil}} g h_{\text{oil}}$$
$$\rho_w h_w = \rho_{\text{oil}} h_{\text{oil}} \implies h_{\text{oil}} = \frac{\rho_w}{\rho_{\text{oil}}} h_w = \frac{1}{0.8} \times 11.2 = 1.25 \times 11.2 = 14.0\text{ cm}$$
Correct Answer: (A)
Solution 2:
The vessel is closed and completely filled. The horizontal acceleration creates a horizontal pressure gradient $\frac{\partial P}{\partial x} = -\rho a$, which is linear across the base. However, along the vertical direction, the pressure gradient is solely $\frac{\partial P}{\partial y} = -\rho g$.
Integrating vertically over the depth $H$, the pressure on the base at position $x$ is $P(x) = P_{\text{top}}(x) + \rho g H$. Because the top is closed and horizontal, the average pressure on the top is $P_{\text{avg, top}}$, and the average pressure on the bottom is $P_{\text{avg, bottom}} = P_{\text{avg, top}} + \rho g H$. If the gauge pressure at the top center is zero (or referenced to $P_0$), the vertical downward force exerted by fluid weight on the base is $\int \rho g H dA = \rho g H A$. The horizontal acceleration does not alter the total vertical force on a horizontal base.
Correct Answer: (A)
Solution 3:
Because the fluid is incompressible, the volume displaced by the small piston must equal the volume gained under the large piston:
$$\Delta V_1 = \Delta V_2 \implies A_1 d_1 = A_2 d_2$$
$$d_2 = d_1 \left( \frac{A_1}{A_2} \right) = d_1 \left( \frac{d_1}{d_2} \right)^2$$
Given $d_1 = 2\text{ cm}$, $d_2 = 10\text{ cm}$, and displacement of small piston $d_1 = 25\text{ cm}$:
$$d_2 = 25 \times \left( \frac{2}{10} \right)^2 = 25 \times \left( \frac{1}{5} \right)^2 = 25 \times \frac{1}{25} = 1.0\text{ cm}$$
Correct Answer: (B)
Solution 4:
For a cubical tank of side $a$ accelerating horizontally with $a_x = g/2$, the free surface tilts at an angle $\theta$ given by:
$$\tan\theta = \frac{a_x}{g} = \frac{g/2}{g} = \frac{1}{2}$$
The liquid surface will align with the top-rear edge of the cube and tilt downward toward the front wall. At the front wall, the depth of water drops from $a$ by $\Delta h = a \tan\theta = a/2$.
The air wedge formed at the top has a triangular cross-section in the side profile:
$$\text{Base of air wedge} = a, \quad \text{Height of air wedge} = a \tan\theta = \frac{a}{2}$$
The volume of the air wedge (which equals the volume of liquid spilled) is:
$$V_{\text{spill}} = \frac{1}{2} \times \text{Base} \times \text{Height} \times \text{Width} = \frac{1}{2} \times a \times \frac{a}{2} \times a = \frac{1}{4} a^3$$
The initial total volume was $V_0 = a^3$.
Therefore, the fraction of liquid that spills is:
$$\frac{V_{\text{spill}}}{V_0} = \frac{\frac{1}{4} a^3}{a^3} = \frac{1}{4}$$
Correct Answer: (B)
Solution 5:
Atmospheric pressure supports a vertical height of mercury column $h_v = 75\text{ cm}$.
When the tube is tilted at an angle $\theta = 60^\circ$ to the vertical, the vertical height remains determined purely by atmospheric pressure: $h_v = L \cos\theta$.
$$L = \frac{h_v}{\cos 60^\circ} = \frac{75}{0.5} = 150\text{ cm}$$
Correct Answer: (B)
Solution 6:
Let the trailing limb be Arm 1 and leading limb be Arm 2.
(A) In the frame of the tube, effective gravity is $\vec{g}_{\text{eff}} = -a\hat{i} – g\hat{j}$. The free surfaces in both limbs must be on the same equipotential surface, giving $\tan\theta = \frac{h}{L} = \frac{a}{g} \implies h = \frac{aL}{g}$. Statement (A) is correct.
(B) Along the horizontal base of length $L$, the horizontal pressure gradient is $\frac{\partial P}{\partial x} = -\rho a$. Integrating from front to back gives $P_{\text{rear}} – P_{\text{front}} = \rho a L$. Statement (B) is correct.
(C) By conservation of liquid volume, the level rises by $h/2$ in the rear limb and drops by $h/2$ in the front limb. At the midpoint $x = L/2$, the liquid level corresponds to the initial height $h_0$. Hence, $P(L/2) = P_0 + \rho g h_0$. Statement (C) is correct.
(D) Under free fall with horizontal acceleration, the liquid would be driven entirely to the trailing end because there is no vertical gravity to counterbalance the horizontal pseudo-force. Statement (D) is incorrect.
Correct Answer: (A, B, C)
Solution 7:
(A) By Pascal’s principle, $P = F_1 / A_1 = W_2 / A_2 \implies W_2 / F_1 = A_2 / A_1$. Statement (A) is correct.
(B) For an incompressible fluid with no friction, work input equals work output: $F_1 d_1 = W_2 d_2$. Statement (B) is correct.
(C) Energy density $P$ is conserved along streamlines in static fluids. Statement (C) is correct.
(D) If the fluid is compressible, significant input stroke is consumed compressing the gas before transmission of force occurs. Statement (D) is incorrect.
Correct Answer: (A, B, C)
Solution 8:
Gauge pressure at depth $h$ in a vertically moving fluid is $P_{\text{gauge}} = \rho g_{\text{eff}} h = \rho (g – a) h$ for downward acceleration $a$.
When the elevator accelerates downward with $a = g$ (or in free fall when cables break), $g_{\text{eff}} = g – g = 0$, so $P_{\text{gauge}} = 0$ throughout the fluid.
Constant velocity gives $a = 0 \implies P_{\text{gauge}} = \rho g h \neq 0$. Upward acceleration gives $P_{\text{gauge}} = \rho (g + a) h = 2\rho g h$.
Correct Answer: (B, C)
Solution 9:
Load on the ram: $W = Mg = 1800 \times 10 = 18000\text{ N}$.
Ram diameter $D = 30\text{ cm}$, plunger diameter $d = 3\text{ cm}$.
Force required on the plunger $F_p$ from hydraulic transmission:
$$\frac{F_p}{A_p} = \frac{W}{A_r} \implies F_p = W \left( \frac{d}{D} \right)^2 = 18000 \times \left( \frac{3}{30} \right)^2 = 18000 \times \left( \frac{1}{10} \right)^2 = \frac{18000}{100} = 180\text{ N}$$
The lever has a mechanical advantage of $5:1$, so the effort force $F_{\text{effort}}$ applied at the handle satisfies:
$$F_{\text{effort}} \times 5 = F_p \times 1 \implies F_{\text{effort}} = \frac{F_p}{5} = \frac{180}{5} = 36\text{ N}$$
Correct Answer: 36
Solution 10:
In the horizontal arm, the centrifugal pseudo-acceleration at distance $r$ from the axis is $a_c(r) = \omega^2 r$.
The pressure difference between the axis ($r = 0$) and the outer arm ($r = L$) is found by integrating along the horizontal arm:
$$\Delta P = \int_0^L \rho \omega^2 r dr = \frac{1}{2} \rho \omega^2 L^2$$
This pressure difference supports a difference in vertical mercury columns $h$ between the outer arm and inner arm:
$$\Delta P = \rho g h$$
Equating the two expressions:
$$\rho g h = \frac{1}{2} \rho \omega^2 L^2 \implies h = \frac{\omega^2 L^2}{2g}$$
Given $\omega = 10\text{ rad/s}$, $L = 20\text{ cm} = 0.2\text{ m}$, and $g = 10\text{ m/s}^2$:
$$h = \frac{(10)^2 \times (0.2)^2}{2 \times 10} = \frac{100 \times 0.04}{20} = \frac{4}{20} = 0.2\text{ m} = 20\text{ cm}$$
Correct Answer: 20