Welcome to Lecture 81 of our comprehensive JEE Physics Master Series. In this lecture, we investigate the profound consequences of gravity and non-inertial reference frames on fluids: the general hydrostatic pressure field equation, fluid containers undergoing linear and rotational acceleration, Archimedes’ principle, center of buoyancy, floatation dynamics, and the atmospheric barometric lapse rate.
1. Detailed Concept Card
1.1 General Hydrostatic Equation & Pressure Gradient
Consider an infinitesimal fluid element of dimensions $dx, dy, dz$ and mass $dm = \rho\, dx\, dy\, dz$ at equilibrium within an arbitrary body-force field per unit mass $\vec{f}$. The net surface force acting along the $x$-direction is:
$$dF_x = P(x)\, dy\, dz – P(x + dx)\\, dy\, dz = -\frac{\partial P}{\partial x}\, dx\, dy\, dz$$
In vector notation, the net surface force per unit volume is $-\nabla P$. Applying Newton’s second law for a fluid element with acceleration $\vec{a}$:
$$-\nabla P + \rho\, \vec{g} = \rho\, \vec{a} \implies \nabla P = \rho\, (\vec{g} – \vec{a}) = \rho\, \vec{g}_{\text{eff}}$$
This is the Eulerian Hydrostatic Equation. It states that:
- The pressure gradient vector $\nabla P$ points strictly along the direction of the effective gravity $\vec{g}_{\text{eff}} = \vec{g} – \vec{a}$.
- Equipotential (constant-pressure) surfaces are surfaces perpendicular to $\vec{g}_{\text{eff}}$.
- The free surface of a liquid is an isobaric surface ($P = P_0$) and is therefore everywhere orthogonal to $\vec{g}_{\text{eff}}$.
1.2 Fluids in Non-Inertial (Accelerated) Frames
A. Linear Horizontal Acceleration ($\vec{a} = a_x\hat{i}$)
Taking upward as $+y$ and rightward as $+x$, the effective gravity is:
$$\vec{g}_{\text{eff}} = -a_x\hat{i} – g\hat{j}$$
The inclination angle $\theta$ of the free liquid surface relative to the horizontal is:
$$\tan\theta = \frac{a_x}{g}$$
Liquid piles up at the trailing wall and lowers at the leading wall. For any two points at the same depth separated horizontally by distance $L$ along the acceleration vector:
$$P_{\text{rear}} – P_{\text{front}} = \rho\, a_x\, L$$
B. Inclined Plane Acceleration
If a tank slides frictionlessly down an incline of angle $\alpha$, its acceleration is $a = g\sin\alpha$ parallel to the slope. Resolving components along the incline ($x’$) and normal to the incline ($y’$):
$$g_{\text{eff}, x’} = g\sin\alpha – a = 0, \quad g_{\text{eff}, y’} = g\cos\alpha$$
Because the tangential effective gravity vanishes ($g_{\text{eff}, x’} = 0$), the free surface aligns parallel to the inclined plane!
C. Rotating Fluid (Centrifugal Field)
In a container rotating with constant angular velocity $\omega$ about a vertical $z$-axis, the pseudo-acceleration is centrifugal: $a_r = -\omega^2 r$. The pressure gradients in cylindrical coordinates are:
$$\frac{\partial P}{\partial r} = \rho\, \omega^2 r, \quad \frac{\partial P}{\partial z} = -\rho\, g$$
Integrating gives the complete pressure distribution throughout the fluid:
$$P(r, z) = P(0, 0) + \frac{1}{2}\rho\, \omega^2 r^2 – \rho\, g\, z$$
The free surface ($P = P_0$) traces a paraboloid of revolution: $z(r) = z_0 + \frac{\omega^2 r^2}{2g}$.
1.3 Archimedes’ Principle & Buoyancy in General Fields
Physical Origin of Buoyancy: Buoyancy arises from the net imbalance of hydrostatic pressure forces integrated over the closed boundary of an immersed body:
$$\vec{F}_B = -\oint_{\partial V} P\, d\vec{A} = -\int_V \nabla P\, dV$$
Substituting the Eulerian equilibrium condition $\nabla P = \rho_f \vec{g}_{\text{eff}}$:
$$\vec{F}_B = -\rho_f V_{\text{disp}}\, \vec{g}_{\text{eff}}$$
Archimedes’ Generalized Law: The buoyant force acting on a body immersed in a fluid is equal in magnitude and opposite in direction to the weight of the displaced fluid calculated under the prevailing effective gravitational field $\vec{g}_{\text{eff}}$.
Key Consequences:
- Centroid of Action (Center of Buoyancy $B$): The buoyant force acts through the center of mass of the displaced fluid, not the center of mass of the body ($G$).
- Equilibrium of Floating Bodies:
- Translational Equilibrium: $W = F_B \implies M g = \rho_f V_{\text{in}} g \implies \frac{V_{\text{in}}}{V} = \frac{\rho_s}{\rho_f}$.
- Rotational Stability: Determined by the relative position of the metacenter $M$ and the center of gravity $G$. The metacenter must lie above $G$ ($MG > 0$) for stable floating equilibrium.
- Accelerating Lift: In an elevator accelerating upward with $a_y$, $F_B = \rho_f V_{\text{disp}}(g + a_y)$. In free fall ($a_y = g$), $F_B = 0$!
1.4 Barometric Formula for Compressible Atmospheres
For an ideal gas atmosphere ($P = \rho R T / M$), density depends on pressure: $\rho = \frac{P M}{R T}$. The hydrostatic differential equation becomes:
$$\frac{dP}{dz} = -\rho g = -\frac{P M g}{R T} \implies \frac{dP}{P} = -\frac{M g}{R T}\, dz$$
Assuming an isothermal atmosphere ($T = \text{constant}$):
$$P(z) = P_0\, \exp\left(-\frac{M g z}{R T}\right)$$
The pressure drops exponentially with altitude with a scale height $H_s = \frac{R T}{M g} \approx 8.5\text{ km}$ for Earth’s air.
1.5 Important Insights & JEE Pitfalls
- Pitfall 1 (Direction of Bubble Motion): When an aquarium or beaker accelerates rightward ($+x$), the liquid surface tilts up at the rear. An air bubble experiences a buoyant force along $-\vec{g}_{\text{eff}} = a_x\hat{i} + g\hat{j}$. Thus, the bubble moves forward and upward (in the direction of acceleration)!
- Pitfall 2 (Zero Buoyancy at Flat Contact): If a flat-bottomed cube sits squarely on the flat bottom of a tank and no film of liquid can penetrate underneath, the downward fluid thrust on the top face presses the block harder onto the bottom. Buoyancy is zero!
- Pitfall 3 (Melting Floating Ice): When pure ice melts in pure water, the water level remains identical. But if the ice contains an embedded denser material (like a pebble or lead pellet), the total water level drops upon melting because the pebble originally displaced its weight equivalent of water, whereas after melting it only displaces its volume.
2. Four Graded Solved Examples
Example 1: Floating Ice Cube with Embedded Cavity and Pellet (JEE Main Level)
Problem Statement: A block of ice of mass $M_{\text{ice}} = 90\text{ g}$ containing an embedded lead shot of mass $m_{\text{lead}} = 11.3\text{ g}$ floats completely submerged in water such that its top is just level with the water surface in a cylindrical vessel of cross-sectional area $A = 100\text{ cm}^2$. Densities are: $\rho_{\text{water}} = 1.0\text{ g/cm}^3$, $\rho_{\text{ice}} = 0.9\text{ g/cm}^3$, and $\rho_{\text{lead}} = 11.3\text{ g/cm}^3$.
When the ice completely melts, determine the change in water level in the vessel. Take $g = 10\text{ m/s}^2$.
Solution:
Step 1: Initial State
The combined system (ice + lead) is floating in equilibrium in water. By the principle of floatation, the total buoyant force equals the total weight:
$$F_B = (M_{\text{ice}} + m_{\text{lead}})\, g = \rho_{\text{water}}\, V_{\text{disp, initial}}\, g$$
$$V_{\text{disp, initial}} = \frac{M_{\text{ice}} + m_{\text{lead}}}{\rho_{\text{water}}} = \frac{90 + 11.3}{1.0} = 101.3\text{ cm}^3$$
Step 2: Final State after Melting
When the ice melts completely, it converts into liquid water of identical mass $M_{\text{ice}} = 90\text{ g}$. The volume added to the water is:
$$V_{\text{melted water}} = \frac{M_{\text{ice}}}{\rho_{\text{water}}} = \frac{90}{1.0} = 90.0\text{ cm}^3$$
The lead shot sinks to the bottom because $\rho_{\text{lead}} > \rho_{\text{water}}$. Once at the bottom, it displaces only its own physical volume:
$$V_{\text{lead}} = \frac{m_{\text{lead}}}{\rho_{\text{lead}}} = \frac{11.3}{11.3} = 1.0\text{ cm}^3$$
The total volume occupied by the melted water and the submerged lead is:
$$V_{\text{final}} = V_{\text{melted water}} + V_{\text{lead}} = 90.0 + 1.0 = 91.0\text{ cm}^3$$
Step 3: Change in Water Level
The decrease in effective submerged volume is:
$$\Delta V = V_{\text{disp, initial}} – V_{\text{final}} = 101.3 – 91.0 = 10.3\text{ cm}^3$$
The change in water level $\Delta h$ in the cylinder of area $A = 100\text{ cm}^2$ is:
$$\Delta h = -\frac{\Delta V}{A} = -\frac{10.3}{100} = -0.103\text{ cm} = -1.03\text{ mm}$$
The water level drops by $1.03\text{ mm}$.
Key Takeaway: Dense inclusions initially displace their mass-equivalent volume of water. When freed and sunken, they displace only their own true volume, inevitably lowering the liquid level.
Example 2: Tank Sliding Down an Incline (JEE Main / Advanced Linkage)
Problem Statement: A rectangular open tank of length $L = 1.2\text{ m}$ containing water is placed on an inclined plane of inclination $\alpha = 30^\circ$. The coefficient of kinetic friction between the tank and the incline is $\mu_k = \frac{1}{2\sqrt{3}}$. As the tank slides down the incline, find the angle $\phi$ that the free surface of the water makes with the inclined surface. Take $g = 10\text{ m/s}^2$.
Solution:
Let coordinate axes be aligned with the incline:
Axis $x’$ points down along the incline, and axis $y’$ is perpendicular to the incline pointing upward away from the plane.
Step 1: Calculate Tank Acceleration
The forces acting on the tank of mass $M$ along the plane are component of gravity $M g \sin\alpha$ down the plane, and kinetic friction $f_k = \mu_k N = \mu_k M g \cos\alpha$ up the plane:
$$a = g(\sin\alpha – \mu_k \cos\alpha)$$
Substitute $\alpha = 30^\circ$ and $\mu_k = \frac{1}{2\sqrt{3}}$:
$$a = 10 \left( \sin 30^\circ – \frac{1}{2\sqrt{3}} \cos 30^\circ \right) = 10 \left( \frac{1}{2} – \frac{1}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{2} \right) = 10 \left( \frac{1}{2} – \frac{1}{4} \right) = 10 \times \frac{1}{4} = 2.5\text{ m/s}^2$$
Step 2: Effective Gravity Components
In the non-inertial frame of the tank, pseudo-acceleration is $-a\hat{i}’$.
Component of $\vec{g}_{\text{eff}}$ along the incline ($x’$):
$$g_{\text{eff}, x’} = g\sin\alpha – a = 10 \left(\frac{1}{2}\right) – 2.5 = 5.0 – 2.5 = 2.5\text{ m/s}^2$$
Component of $\vec{g}_{\text{eff}}$ normal to the incline ($y’$):
$$g_{\text{eff}, y’} = -g\cos\alpha = -10\left(\frac{\sqrt{3}}{2}\right) = -5\sqrt{3}\text{ m/s}^2$$
Step 3: Free Surface Angle relative to Incline
The free surface is perpendicular to $\vec{g}_{\text{eff}}$. The slope of the free surface with respect to the $x’$-axis (the inclined plane) is:
$$\tan\phi = \frac{g_{\text{eff}, x’}}{|g_{\text{eff}, y’}|} = \frac{2.5}{5\sqrt{3}} = \frac{1}{2\sqrt{3}}$$
Notice that $\tan\phi = \mu_k$! This is an elegant general result for rough inclines: $\tan\phi = \mu_k$.
$$\phi = \arctan\left(\frac{1}{2\sqrt{3}}\right) \approx \arctan(0.2887) \approx 16.1^\circ$$
Key Takeaway: When a tank slides down a rough incline, the free surface tilts with respect to the incline at an angle whose tangent is precisely the coefficient of friction: $\tan\phi = \mu_k$.
Example 3: Oscillating Floating Cylinder in an Accelerating Elevator (JEE Advanced Level)
Problem Statement: A uniform solid cylinder of cross-sectional area $A$, height $H$, and density $\rho_s$ floats vertically in a liquid of density $\rho_f$ (where $\rho_s < \rho_f$) contained inside an elevator. The elevator accelerates upward with a constant acceleration $a_0$.
(a) Find the equilibrium submerged depth $h_e$ of the cylinder.
(b) If the cylinder is slightly pushed down by a small vertical displacement $y$ and released, prove that it executes Simple Harmonic Motion (SHM) and find its angular frequency $\omega$.
Solution:
Part (a): Equilibrium Submerged Depth
In the frame of the elevator, effective gravity is $g_{\text{eff}} = g + a_0$.
The downward effective weight of the cylinder is:
$$W_{\text{eff}} = M g_{\text{eff}} = (\rho_s A H)(g + a_0)$$
The upward buoyant force for submerged depth $h_e$ is:
$$F_{B, \text{eff}} = \rho_f (A h_e)(g + a_0)$$
Equating $W_{\text{eff}} = F_{B, \text{eff}}$:
$$(\rho_s A H)(g + a_0) = (\rho_f A h_e)(g + a_0) \implies h_e = \left( \frac{\rho_s}{\rho_f} \right) H$$
Notice that $(g + a_0)$ cancels out completely! The equilibrium submerged depth $h_e$ is completely independent of the elevator acceleration.
Part (b): Restoring Force and Oscillation Frequency
When the cylinder is depressed further downward by a displacement $y$ from equilibrium, the submerged depth becomes $h = h_e + y$. The net restoring force acting on the cylinder (in the elevator frame) is:
$$F_{\text{net}} = W_{\text{eff}} – F_B’ = (\rho_s A H)(g + a_0) – \rho_f A (h_e + y)(g + a_0)$$
$$F_{\text{net}} = -\rho_f A (g + a_0) y$$
According to Newton’s second law in the elevator frame:
$$M \frac{d^2 y}{dt^2} = -\rho_f A (g + a_0) y \implies (\rho_s A H) \frac{d^2 y}{dt^2} + \rho_f A (g + a_0) y = 0$$
$$\frac{d^2 y}{dt^2} + \left( \frac{\rho_f (g + a_0)}{\rho_s H} \right) y = 0$$
This is the standard equation of SHM: $\frac{d^2 y}{dt^2} + \omega^2 y = 0$, with angular frequency:
$$\omega = \sqrt{\frac{\rho_f (g + a_0)}{\rho_s H}} = \sqrt{\frac{g + a_0}{h_e}}$$
The time period of oscillation is:
$$T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{h_e}{g + a_0}}$$
Key Takeaway: The equilibrium floating position does not depend on acceleration, but the oscillation frequency scales with $\sqrt{g_{\text{eff}}}$.
Example 4: Atmospheric Pressure Lapse Rate with Linear Temperature Gradient (JEE Advanced Edge Case)
Problem Statement: The temperature in the troposphere is not constant but decreases linearly with altitude according to $T(z) = T_0 – \alpha z$, where $T_0$ is sea-level temperature and $\alpha$ is the positive temperature lapse rate ($-\frac{dT}{dz}$). Assuming air behaves as an ideal gas of molar mass $M$ under uniform gravity $g$, derive an expression for the atmospheric pressure $P(z)$ as a function of altitude $z$, given sea-level pressure $P_0$.
Solution:
The hydrostatic differential equation is:
$$\frac{dP}{dz} = -\rho g$$
From the ideal gas equation, $\rho = \frac{P M}{R T(z)} = \frac{P M}{R (T_0 – \alpha z)}$. Substituting this into the hydrostatic equation:
$$\frac{dP}{dz} = -\frac{M g}{R} \frac{P}{T_0 – \alpha z}$$
Separating variables:
$$\frac{dP}{P} = -\frac{M g}{R} \frac{dz}{T_0 – \alpha z}$$
Integrate from $z = 0$ ($P = P_0$) to altitude $z$ ($P = P(z)$):
$$\int_{P_0}^{P} \frac{dP’}{P’} = -\frac{M g}{R} \int_0^z \frac{dz’}{T_0 – \alpha z’}$$
Let $u = T_0 – \alpha z’$, so $du = -\alpha\, dz’$ $\implies dz’ = -\frac{du}{\alpha}$:
$$\ln\left(\frac{P}{P_0}\right) = -\frac{M g}{R} \left[ -\frac{1}{\alpha} \ln(T_0 – \alpha z’) \right]_0^z = \frac{M g}{\alpha R} \ln\left( \frac{T_0 – \alpha z}{T_0} \right)$$
$$\ln\left(\frac{P}{P_0}\right) = \ln\left[ \left( 1 – \frac{\alpha z}{T_0} \right)^{\frac{M g}{\alpha R}} \right]$$
Exponentiating both sides yields the celebrated Polytropic Barometric Formula:
$$P(z) = P_0 \left( 1 – \frac{\alpha z}{T_0} \right)^{\frac{M g}{\alpha R}}$$
Limiting Case Check: As $\alpha \to 0$, using the limit $\lim_{x \to 0} (1 – kx)^{1/x} = e^{-k}$, we recover the isothermal formula: $P(z) = P_0 e^{-M g z / R T_0}$.
3. Practice Worksheet (10 Questions)
Questions 1 to 5: Single Correct (JEE Main level). Questions 6 to 8: One or More than One Correct (JEE Advanced level). Questions 9 and 10: Numerical Value (JEE Advanced level).
Question 1 (JEE Main Single Correct)
A beaker containing liquid of density $\rho$ is placed on the floor of an elevator. When the elevator moves downward with an acceleration $a = g/3$, the pressure at depth $h$ below the surface is:
(A) $P_0 + \frac{4}{3}\rho g h$
(B) $P_0 + \frac{2}{3}\rho g h$
(C) $P_0 + \frac{1}{3}\rho g h$
(D) $P_0 – \frac{2}{3}\rho g h$
Question 2 (JEE Main Single Correct)
A piece of alloy of mass $105\text{ g}$ weighs $90\text{ g}$ in water and $93\text{ g}$ in an unknown liquid. The density of the unknown liquid is:
(A) $0.80\text{ g/cm}^3$
(B) $0.85\text{ g/cm}^3$
(C) $0.90\text{ g/cm}^3$
(D) $0.75\text{ g/cm}^3$
Question 3 (JEE Main Single Correct)
A block of wood floats in water with $4/5$ of its volume submerged. When the same block floats in oil, $9/10$ of its volume is submerged. The relative density of the oil is:
(A) $8/9$
(B) $9/8$
(C) $4/9$
(D) $7/10$
Question 4 (JEE Main Single Correct)
An open cylindrical can of height $H = 40\text{ cm}$ and radius $R = 10\text{ cm}$ is full of liquid. It is rotated about its vertical axis of symmetry. What is the maximum angular velocity $\omega$ (in $\text{rad/s}$) such that no liquid spills over the edge? (Take $g = 10\text{ m/s}^2$).
(A) $0\text{ rad/s}$ (Any rotation causes spilling when initially full)
(B) $10\text{ rad/s}$
(C) $20\text{ rad/s}$
(D) $28.3\text{ rad/s}$
Question 5 (JEE Main Single Correct)
A closed rectangular box filled completely with water of density $\rho$ is accelerated horizontally along its length $L$ with acceleration $a$. The difference in pressure between two points located on the same horizontal line at the rear and front ends of the box is:
(A) $\rho a L$
(B) $\frac{1}{2}\rho a L$
(C) $\rho(g + a)L$
(D) Zero
Question 6 (JEE Advanced One or More Correct)
A solid sphere of density $\rho_s$ and radius $R$ is attached to the bottom of a container filled with liquid of density $\rho_f$ ($\rho_f > \rho_s$) by a light taut string. The container is accelerated horizontally with a constant acceleration $a$. In steady state relative to the container:
(A) The string makes an angle $\theta = \arctan(a/g)$ with the vertical tilted backward.
(B) The string makes an angle $\theta = \arctan(a/g)$ with the vertical tilted forward.
(C) The tension in the string is $T = V(\rho_f – \rho_s)\sqrt{g^2 + a^2}$.
(D) The net buoyant force on the sphere has magnitude $F_B = V \rho_f \sqrt{g^2 + a^2}$.
Question 7 (JEE Advanced One or More Correct)
A body floats in a liquid contained inside a beaker. Which of the following operations will alter the fraction of the body’s volume that is submerged?
(A) The beaker is placed in an elevator accelerating upward with $a = 2\text{ m/s}^2$.
(B) The temperature of the system is increased (assuming non-zero thermal expansions of body and liquid).
(C) The ambient atmospheric pressure is doubled by compressing air above the surface.
(D) A liquid of lesser density that is immiscible with the existing liquid is gently poured on top.
Question 8 (JEE Advanced One or More Correct)
A U-tube containing liquid is rotated about one of its vertical arms with constant angular speed $\omega$. The horizontal distance between the arms is $L$. Which of the following statements is/are correct?
(A) The height difference between the liquid levels in the two arms is $h = \frac{\omega^2 L^2}{2g}$.
(B) The liquid rises in the outer arm and drops in the inner arm by equal amounts $h/2$.
(C) The pressure at the bottom corner of the outer arm is greater than that at the bottom corner of the inner arm by $\frac{1}{2}\rho \omega^2 L^2$.
(D) If the liquid is mercury and both arms are closed with vacuum above, the height difference remains $\frac{\omega^2 L^2}{2g}$.
Question 9 (JEE Advanced Numerical Value)
An open rectangular tank of length $L = 3\text{ m}$, width $W = 1\text{ m}$, and height $H = 2\text{ m}$ contains water up to a depth of $h_0 = 1.2\text{ m}$. Find the maximum horizontal acceleration $a_{\text{max}}$ (in $\text{m/s}^2$) with which the tank can be accelerated along its length without any water spilling over the edges. (Take $g = 10\text{ m/s}^2$).
Question 10 (JEE Advanced Numerical Value)
A body of mass $m = 4\text{ kg}$ and density $\rho_s = 800\text{ kg/m}^3$ is completely submerged in water ($\rho_w = 1000\text{ kg/m}^3$) by holding it down with a vertical downward force $F$. If the container is accelerated upward with acceleration $a = 2.5\text{ m/s}^2$, find the magnitude of the required holding force $F$ (in Newtons). (Take $g = 10\text{ m/s}^2$).
4. Detailed Solutions & Answer Key
Quick Answer Key
| Question | Type | Correct Answer |
|---|---|---|
| 1 | Single Correct | (B) |
| 2 | Single Correct | (A) |
| 3 | Single Correct | (A) |
| 4 | Single Correct | (A) |
| 5 | Single Correct | (A) |
| 6 | One or More Correct | (B, C, D) |
| 7 | One or More Correct | (B, D) |
| 8 | One or More Correct | (A, B, C, D) |
| 9 | Numerical | 5.33 (or $16/3$) |
| 10 | Numerical | 12.5 |
Step-by-Step Solutions
Solution 1:
When the elevator accelerates downward with $a = g/3$, the effective gravitational field in the elevator frame is:
$$g_{\text{eff}} = g – a = g – \frac{g}{3} = \frac{2}{3}g$$
The hydrostatic pressure at depth $h$ below the free surface is:
$$P = P_0 + \rho\, g_{\text{eff}}\, h = P_0 + \frac{2}{3}\rho g h$$
Correct Answer: (B)
Solution 2:
Let the true volume of the alloy be $V$.
Loss of weight in water $= 105 – 90 = 15\text{ g-wt} = V \rho_w g$. Since $\rho_w = 1.0\text{ g/cm}^3$, we have $V = 15\text{ cm}^3$.
Loss of weight in unknown liquid $= 105 – 93 = 12\text{ g-wt} = V \rho_L g$.
$$\rho_L = \frac{12}{V} = \frac{12}{15} = 0.80\text{ g/cm}^3$$
Correct Answer: (A)
Solution 3:
Let $V$ be the total volume of the wood block and $\rho$ its density.
Floating in water: $V \rho g = \left(\frac{4}{5}V\right)\rho_w g \implies \rho = \frac{4}{5}\rho_w$.
Floating in oil: $V \rho g = \left(\frac{9}{10}V\right)\rho_{\text{oil}} g \implies \rho = \frac{9}{10}\rho_{\text{oil}}$.
Equating the two expressions for $\rho$:
$$\frac{4}{5}\rho_w = \frac{9}{10}\rho_{\text{oil}} \implies \frac{\rho_{\text{oil}}}{\rho_w} = \frac{4/5}{9/10} = \frac{4}{5} \times \frac{10}{9} = \frac{8}{9}$$
Correct Answer: (A)
Solution 4:
When a cylinder full of liquid is rotated, any non-zero angular speed $\omega > 0$ creates a parabolic free surface where the liquid rises at the outer boundary ($r = R$) and depresses at the center ($r = 0$). Because the vessel is initially filled to the very brim, any rise at the wall immediately spills liquid over the lip. Thus, the maximum angular velocity without spilling is zero.
Correct Answer: (A)
Solution 5:
In a closed vessel, the horizontal pressure gradient under horizontal acceleration $a$ along $+x$ is $\frac{\partial P}{\partial x} = -\rho a$.
Integrating between $x = 0$ (rear) and $x = L$ (front):
$$P_{\text{rear}} – P_{\text{front}} = \rho a L$$
Correct Answer: (A)
Solution 6:
Effective gravity is $\vec{g}_{\text{eff}} = -a\hat{i} – g\hat{j}$. The buoyant force acts opposite to $\vec{g}_{\text{eff}}$, so $\vec{F}_B = V \rho_f (a\hat{i} + g\hat{j})$, pointing upward and forward.
Gravity on the sphere is $-V \rho_s g \hat{j}$, and pseudo-force is $-V \rho_s a \hat{i}$.
The net force without the string is $(V \rho_f – V \rho_s)(a\hat{i} + g\hat{j})$.
For equilibrium, the string tension must point in the opposite direction (downward and backward), meaning the string tilts forward at angle $\tan\theta = a/g$. Magnitude of tension is $T = V(\rho_f – \rho_s)\sqrt{g^2 + a^2}$, and magnitude of buoyant force is $F_B = V \rho_f \sqrt{g^2 + a^2}$.
Correct Answer: (B, C, D)
Solution 7:
Submerged fraction is $\frac{V_{\text{in}}}{V} = \frac{\rho_s}{\rho_f}$.
(A) In an accelerating elevator, $\vec{g}_{\text{eff}}$ cancels from both sides, leaving $V_{\text{in}}/V$ unchanged.
(B) Thermal expansion modifies both $\rho_s$ and $\rho_f$ differently (unless thermal expansion coefficients match perfectly), altering the submerged fraction.
(C) For an incompressible liquid and solid, pressure does not change densities, so the submerged fraction is unchanged.
(D) Adding a layer of lighter liquid creates an additional buoyant force on the protruding part, reducing the depth immersed in the bottom liquid.
Correct Answer: (B, D)
Solution 8:
(A) The centrifugal pressure difference across distance $L$ is $\Delta P = \frac{1}{2}\rho \omega^2 L^2$, which balances the hydrostatic column $\rho g h \implies h = \frac{\omega^2 L^2}{2g}$.
(B) Conservation of volume in arms of equal cross-section requires the rise in one arm to equal the drop in the other: $\Delta h = h/2$.
(C) $\Delta P = \frac{1}{2}\rho \omega^2 L^2$ represents the exact pressure increase along the bottom horizontal arm.
(D) With vacuum in both limbs, atmospheric pressure is absent on both sides, so the balance between centrifugal force and gravity is identically preserved.
Correct Answer: (A, B, C, D)
Solution 9:
The maximum height the liquid can rise at the rear wall without spilling is $\Delta h_{\text{max}} = H – h_0 = 2.0 – 1.2 = 0.8\text{ m}$.
Since the surface pivots about the center ($x = L/2 = 1.5\text{ m}$):
$$\Delta h = \frac{L}{2}\tan\theta = \frac{L}{2}\left(\frac{a}{g}\right)$$
$$0.8 = \frac{3}{2}\left(\frac{a}{10}\right) \implies 0.8 = 0.15\, a \implies a_{\text{max}} = \frac{0.8}{0.15} = \frac{80}{15} = \frac{16}{3} \approx 5.33\text{ m/s}^2$$
Correct Answer: 5.33
Solution 10:
Volume of the body: $V = \frac{m}{\rho_s} = \frac{4}{800} = \frac{1}{200}\text{ m}^3 = 0.005\text{ m}^3$.
In the upward accelerating elevator with $a = 2.5\text{ m/s}^2$, $g_{\text{eff}} = g + a = 10 + 2.5 = 12.5\text{ m/s}^2$.
The effective downward gravity on the body is $W_{\text{eff}} = m g_{\text{eff}} = 4 \times 12.5 = 50\text{ N}$.
The upward buoyant force is $F_B = \rho_w V g_{\text{eff}} = 1000 \times 0.005 \times 12.5 = 62.5\text{ N}$.
For static equilibrium in the elevator frame:
$$F_B = W_{\text{eff}} + F \implies F = F_B – W_{\text{eff}} = 62.5 – 50.0 = 12.5\text{ N}$$
Correct Answer: 12.5