Unit 6: Gravitation & Properties of Matter

This unit packs three sub-units that JEE treats as one combined mechanics module: gravitation (1–2 questions in JEE Main, often a multi-step JEE Advanced problem), elasticity (1 question, conceptual or numerical), and fluids + viscosity + surface tension (2 questions, often paired with thermodynamics or oscillations later).

Typical question types:

• JEE Main: satellite orbital speed, escape velocity, Bernoulli's tube (Venturi), terminal velocity in a viscous fluid, capillary rise, Young's modulus from stress-strain data.
• JEE Advanced: gravitational PE of a system (rod + point mass), variation of $g$ with depth/latitude, gravitational field of a non-uniform sphere, capillary problem with weight, terminal-velocity-with-buoyancy, multi-cylinder pressure.

Concept Map

• Gravitation
  • Kepler's three laws (proof of 3rd for circular orbit)
  • Newton's universal law; superposition
  • Gravitational field due to ring, shell (state)
  • Variation of $g$: altitude, depth, latitude (rotation), shape
  • Gravitational PE
  • Escape velocity, orbital velocity, satellite total energy
  • Geo-stationary satellites; weightlessness
• Elasticity
  • Stress, strain (types)
  • Hooke's law; Young's, bulk, shear moduli; Poisson's ratio
  • Stress-strain curve; elastic limit, yield point, breaking
  • Elastic PE
• Fluids
  • Hydrostatic pressure; Pascal's law
  • Archimedes (buoyancy); iceberg, mixed liquids
  • Equation of continuity
  • Bernoulli's principle; Venturi, Torricelli, aerofoil
  • Viscosity: Newton's law, Stokes' law, terminal velocity
  • Surface tension: capillary rise, excess pressure (drop & bubble)

Topic 1: Gravitation

Sub-topic A: Kepler's Laws

1. Law of orbits. Planets move in elliptical orbits with the Sun at one focus.
2. Law of areas. The line joining a planet to the Sun sweeps equal areas in equal times (i.e. areal velocity $dA/dt$ = const).
3. Law of periods. $T^2 \propto a^3$, where $a$ is the semi-major axis.

Proof of 3rd law for circular orbit. Centripetal: $G M_\odot m/r^2 = m v^2/r\Rightarrow v = \sqrt{GM_\odot/r}$. Period $T = 2\pi r/v = 2\pi r^{3/2}/\sqrt{GM_\odot}$. So $T^2 = 4\pi^2 r^3/(GM_\odot)\propto r^3$.

Proof of 2nd law. The gravitational force is central (along $\vec r$), so torque about the Sun is zero $\Rightarrow L$ is conserved. $dA/dt = \tfrac12 \vert \vec r\times\vec v\vert = L/(2m)$ = const.

Sub-topic B: Newton's Universal Law

$\vec F = -\frac{Gm_1 m_2}{r^2}\hat r,$

attractive, along the line joining the masses. $G = 6.674\times 10^{-11}$ N·m²/kg².

Sub-topic C: Gravitational Field

The field at a point due to mass $M$ at distance $r$: $\vec E_g = -GM/r^2\,\hat r$ (toward $M$).

Superposition applies — field due to multiple masses = vector sum.

Field due to a uniform ring (radius $R$, mass $M$) at a point on the axis at distance $x$ from centre:

$E_g = \frac{GMx}{(R^2+x^2)^{3/2}}.$

Derivation: by symmetry, only the axial component survives. Each element $dm$ contributes $G dm/(R^2+x^2)$ along its line; multiply by $\cos\theta = x/\sqrt{R^2+x^2}$ and integrate. Maximum at $x = R/\sqrt 2$.

Field due to a uniform thin spherical shell (mass $M$, radius $R$):

• For $r > R$: $E_g = GM/r^2$ (as if all mass at centre).
• For $r < R$: $E_g = 0$ (Newton's shell theorem).

Field due to a uniform solid sphere (mass $M$, radius $R$):

• $r > R$: $E_g = GM/r^2$.
• $r < R$: $E_g = GM r/R^3$ (using only the mass enclosed within radius $r$).

Sub-topic D: Variation of $g$ on Earth

Let $g$ at the surface = $GM_E/R_E^2 \approx 9.81$ m/s².

Altitude $h$ (small $h\ll R_E$):

$g_h = \frac{GM_E}{(R_E+h)^2} \approx g\left(1-\frac{2h}{R_E}\right).$

Depth $d$: treat Earth as a uniform sphere of density $\rho$. At depth $d$, only the mass within radius $R_E-d$ acts:

$g_d = g\left(1-\frac{d}{R_E}\right).$

At the centre ($d=R_E$), $g=0$.

Latitude (Earth's rotation): at latitude $\lambda$, the centrifugal acceleration is $\omega^2 R_E\cos\lambda$, with horizontal component $\omega^2 R_E\cos^2\lambda$. Net apparent $g$:

$g_\lambda = g - \omega^2 R_E\cos^2\lambda.$

At equator: $g_{eq} = g - \omega^2 R_E$. At pole: $g_{pole}=g$.

Shape of Earth: equatorial radius $>$ polar radius by about $21$ km, so $g$ at pole $> g$ at equator.

Sub-topic E: Gravitational Potential Energy

For two point masses $m, M$ separated by $r$:

$U(r) = -\frac{GMm}{r}.$

(Zero at infinity by convention.)

For a system of point masses, $U_{sys} = \sum_{i<j} -Gm_im_j/r_{ij}$.

Near earth's surface, taking the surface as reference, $U = mgh$ (for $h\ll R_E$). This emerges from expansion of the general formula.

Sub-topic F: Escape Velocity

The minimum speed required at the surface of a body of mass $M$, radius $R$, to escape to infinity:

$\tfrac12 m v_e^2 = \frac{GMm}{R}\Rightarrow v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}.$

For Earth: $v_e\approx 11.2$ km/s.

Sub-topic G: Orbital Velocity & Satellite Energetics

For a satellite in circular orbit of radius $r$:

$\frac{GMm}{r^2} = \frac{mv_0^2}{r}\Rightarrow v_0 = \sqrt{\frac{GM}{r}}.$

KE = $\tfrac12 m v_0^2 = GMm/(2r)$.

PE = $-GMm/r$.

Total energy $E = KE + PE = -GMm/(2r)$. Negative → bound.

To escape from orbit, gain KE = $GMm/(2r)$, i.e. $v_{esc-orbit} = \sqrt 2 v_0$.

Sub-topic H: Geostationary Satellites

A geostationary satellite has the same angular speed as Earth ($T = 24$ h). Orbital radius from $T^2 = 4\pi^2 r^3/(GM_E)$:

$r_{geo} = \left(\frac{GM_E T^2}{4\pi^2}\right)^{1/3}\approx 4.22\times 10^7\text{ m}.$

Height above surface $\approx 36000$ km. Plane of orbit = equatorial plane. Direction = west to east.

Sub-topic I: Weightlessness

A body in free-fall feels no apparent weight (normal force = 0). Satellites are essentially in continuous free-fall; astronauts inside feel weightless.

Worked Examples (JEE Main level)

Example 1.1. Find the height where $g = g_0/4$.

$g_0/(1+h/R)^2 = g_0/4\Rightarrow (1+h/R)^2 = 4\Rightarrow h = R$. (Height = Earth's radius.)

Example 1.2. Orbital speed at low Earth orbit ($r\approx R$)?

$v_0 = \sqrt{GM/R} = \sqrt{gR}\approx \sqrt{10\cdot 6.4\times 10^6}=\sqrt{6.4\times 10^7}\approx 8000$ m/s.

Example 1.3. A satellite is at radius $2R_E$. Time period in hours?

$T = 2\pi\sqrt{(2R_E)^3/(GM_E)} = 2\pi\sqrt{8 R_E^3/(GM_E)} = 2\sqrt 2 \cdot T_{LEO}$. $T_{LEO}\approx 84$ min. So $T \approx 4$ h.

Worked Examples (JEE Advanced level)

Example 1.A1. A uniform thin rod of mass $M$ and length $L$ lies along the $x$-axis from $0$ to $L$. A point mass $m$ is at $x = L+a$. Force between them?

$\lambda = M/L$. Force on $m$ due to element $dm = \lambda\,dx$ at position $x$:

$dF = G m\,dm/(L+a-x)^2$, attractive (toward the rod).

$F = Gm\lambda\int_0^L \frac{dx}{(L+a-x)^2} = Gm\lambda\left[\frac{1}{L+a-x}\right]_0^L = Gm\lambda\left(\frac{1}{a}-\frac{1}{L+a}\right) = \frac{GMm}{a(L+a)}.$

Example 1.A2. A particle is projected from the surface of Earth with speed $u$. Maximum height attained (with $u < v_e$)?

Energy conservation: $\tfrac12 mu^2 - GMm/R = -GMm/(R+h)\Rightarrow$

$\frac{1}{R+h} = \frac{1}{R} - \frac{u^2}{2GM} = \frac{1}{R} - \frac{u^2}{2gR^2}\Rightarrow h = \frac{u^2 R}{2gR - u^2}.$

For $u\ll v_e$: $h\approx u^2/(2g)$. For $u = v_e$: $h\to\infty$.

Example 1.A3. Two satellites in same circular orbit; one is slowed slightly. What happens?

It drops to a lower orbit (smaller $r$). By Kepler's 3rd law, smaller $r$ → smaller $T$ → it moves faster in the new orbit (paradoxically, slowing it down actually speeds it up, after the orbit adjusts).

Topic 2: Elasticity

Sub-topic A: Stress and Strain

• Stress = restoring force per unit area = $F/A$. Units: Pa.
• Strain = relative deformation (dimensionless).

Types:

Sub-topic B: Hooke's Law and Moduli

Within elastic limit, stress $\propto$ strain.

• Young's modulus $Y = (F/A)/(\Delta L/L) = FL/(A\Delta L)$.
• Bulk modulus $K = -\Delta P/(\Delta V/V)$. Compressibility $= 1/K$.
• Shear modulus (rigidity) $\eta = F_{\parallel}/(A\phi)$.

For an isotropic solid, $Y, K, \eta$ are related via Poisson's ratio $\sigma$:

$Y = 3K(1-2\sigma) = 2\eta(1+\sigma).$

Poisson's ratio $\sigma = -\text{lateral strain}/\text{longitudinal strain}$. Typical: $0\le\sigma\le 0.5$. Rubber $\approx 0.5$ (incompressible); cork $\approx 0$.

Sub-topic C: Stress-strain curve

A typical metallic curve has:

1. Proportional region (Hooke's law): straight line from origin to proportional limit.
2. Elastic limit: max stress for which material returns to original shape on removal of load.
3. Yield point: deformation becomes plastic (irreversible).
4. Ultimate strength: maximum stress.
5. Fracture point: material breaks.

Ductile materials (copper, aluminum) have long plastic regions; brittle (glass) break shortly after elastic limit.

Sub-topic D: Elastic PE in a stretched wire

Energy stored per unit volume:

$u = \tfrac12\cdot\text{stress}\cdot\text{strain} = \tfrac12 Y\varepsilon^2 = \tfrac{1}{2Y}\sigma^2.$

Total elastic PE in a wire of length $L$ and area $A$ stretched by $\Delta L$:

$U = \tfrac12 F\Delta L = \tfrac12 (YA\Delta L/L)\Delta L = \frac{YA(\Delta L)^2}{2L}.$

Worked Examples (JEE Main level)

Example 2.1. A steel wire ($Y = 2\times 10^{11}$ Pa) of length $1$ m and area $10^{-6}$ m² stretches by $0.5$ mm under load. Force?

$F = YA\Delta L/L = 2\times 10^{11}\cdot 10^{-6}\cdot 5\times 10^{-4}/1 = 100$ N.

Example 2.2. A wire of $Y = 2\times 10^{11}$ Pa, area $1$ mm², length $2$ m elongates by $0.1$ mm. Energy stored?

$U = \tfrac12 F\Delta L = \tfrac12(YA\Delta L/L)\Delta L = \tfrac12\cdot 2\times 10^{11}\cdot 10^{-6}\cdot 10^{-4}\cdot 10^{-4}/2 = 5\times 10^{-4}$ J? Recompute: $F = 2\times 10^{11}\cdot 10^{-6}\cdot 10^{-4}/2 = 10$ N. $U = \tfrac12\cdot 10\cdot 10^{-4} = 5\times 10^{-4}$ J.

Worked Examples (JEE Advanced level)

Example 2.A1. A heavy wire of length $L$ is suspended from the ceiling. Find its elongation due to its own weight (density $\rho$, area $A$, Young's modulus $Y$).

At height $y$ from the bottom, tension = weight of wire below $= \rho g A (L-y)$? No, length below is $y$ (from bottom). Reset: let $y$ = distance from top. Tension at $y$ = weight below = $\rho gA(L-y)$.

Stress at $y$: $\sigma(y) = \rho g(L-y)$. Strain $= \sigma/Y$. Elemental elongation $d(\Delta L) = \sigma(y)/Y\cdot dy$.

$\Delta L = \int_0^L \frac{\rho g(L-y)}{Y}dy = \frac{\rho g}{Y}\cdot\frac{L^2}{2} = \frac{\rho g L^2}{2Y}.$

Note: this is equivalent to the elongation if the total weight $\rho g A L$ were applied at the midpoint, i.e. effective weight $\rho g A L/2$ acting at the end.

Example 2.A2. A steel rod of length $1$ m is heated from $20^\circ$ C to $100^\circ$ C. If ends are rigidly fixed, find the compressive stress ($\alpha = 1.2\times 10^{-5}$ /K, $Y = 2\times 10^{11}$ Pa).

If free, $\Delta L = L\alpha\Delta T = 9.6\times 10^{-4}$ m. Since fixed, this strain is forced as compression. Stress $= Y\alpha\Delta T = 2\times 10^{11}\cdot 9.6\times 10^{-4}/1 = 1.92\times 10^8$ Pa.

Topic 3: Fluid Statics

Sub-topic A: Hydrostatic Pressure

In a fluid at rest in a gravity field, pressure increases with depth:

$dP/dz = -\rho g\Rightarrow P(z) = P_0 + \rho g h\quad (\text{at depth }h).$

Pascal's Law: pressure applied to an enclosed fluid is transmitted undiminished to every part. Basis for hydraulic jack: $P_1=P_2\Rightarrow F_1/A_1 = F_2/A_2$, so a small force on a small piston creates a large force on a large piston.

Sub-topic B: Buoyancy (Archimedes)

An object immersed in a fluid experiences an upward force equal to the weight of fluid displaced:

$F_B = \rho_{fl} V_{displaced}\, g.$

Floating: weight = buoyancy ⇒ $\rho_{obj} V_{obj} = \rho_{fl} V_{submerged}$. Fraction submerged = $\rho_{obj}/\rho_{fl}$.

Iceberg example: $\rho_{ice}/\rho_{water}=0.9$, so $90\%$ of an iceberg is submerged.

Mixed liquids: if a body floats with part in oil, part in water, weight = sum of buoyancies from each.

Worked Examples (JEE Main level)

Example 3.1. A wooden block of density $600$ kg/m³ floats in water. Fraction submerged?

$0.6$ → $60\%$ submerged.

Example 3.2. A hydraulic press has piston areas $A_1 = 0.01$ m², $A_2 = 1$ m². A force of $10$ N on the smaller piston lifts what load on the larger?

$F_2 = F_1\cdot A_2/A_1 = 10\cdot 100 = 1000$ N.

Topic 4: Fluid Dynamics — Equation of Continuity & Bernoulli

Sub-topic A: Continuity (mass conservation)

For an incompressible fluid in steady flow,

$\rho A v = \text{const}\Rightarrow A_1 v_1 = A_2 v_2.$

Smaller cross-section → higher velocity (e.g. narrowing pipe).

Sub-topic B: Bernoulli's Equation

Along a streamline for an inviscid incompressible fluid in steady flow:

$P + \tfrac12\rho v^2 + \rho g h = \text{const}.$

Derivation sketch: apply work-energy theorem to a fluid element. Net work by pressure + gravity = $\Delta KE$. The terms above are work per unit volume (P), KE per unit volume ($\tfrac12\rho v^2$), PE per unit volume ($\rho gh$).

Sub-topic C: Applications

Venturi meter. A horizontal tube with a constriction. Continuity: $A_1 v_1 = A_2 v_2$. Bernoulli: $P_1 + \tfrac12\rho v_1^2 = P_2 + \tfrac12\rho v_2^2$. From these:

$v_1 = A_2\sqrt{\frac{2(P_1-P_2)}{\rho(A_1^2 - A_2^2)}}.$

Pressure drop in the throat allows flow-rate measurement.

Torricelli's Theorem. A small hole at depth $h$ below the free surface of a large tank. Surface velocity is negligible; pressure at the hole = atmospheric. Bernoulli gives

$v = \sqrt{2gh}.$

Same as a freely falling body from height $h$. The horizontal range from a hole at height $h$ above ground (and depth $H$ from surface) is $R = 2\sqrt{h(H-h)}$ — maximum at $h = H/2$ giving $R_{\max}=H$.

Aerofoil / wing lift. Air moves faster over the curved upper surface than below → lower $P$ above → net upward force. (Qualitative; full quantitative needs more.)

Magnus effect. A spinning ball moving through air entrains different speeds on its two sides → pressure difference → sideways force.

Worked Examples (JEE Main level)

Example 4.1. Water flows in a horizontal pipe whose cross-section narrows from $4$ cm² to $1$ cm². If pressure at the wider section is $2\times 10^5$ Pa and velocity is $1$ m/s, find pressure at the narrower section.

$v_2 = A_1 v_1/A_2 = 4$ m/s. $P_2 = P_1 + \tfrac12\rho(v_1^2 - v_2^2) = 2\times 10^5 + 500(1-16) = 2\times 10^5 - 7500 = 1.925\times 10^5$ Pa.

Example 4.2. A tank has a hole at $5$ m below the water surface. Speed of efflux ($g=10$)?

$v = \sqrt{2\cdot 10\cdot 5} = 10$ m/s.

Worked Examples (JEE Advanced level)

Example 4.A1. A tank of height $H$ filled with water has a small hole at depth $h$. Find the depth at which the horizontal range of the issuing stream is maximum (assuming the tank is at rest).

Range $R = v\cdot t = \sqrt{2gh}\cdot \sqrt{2(H-h)/g} = 2\sqrt{h(H-h)}$. Max when $h = H/2$, giving $R_{\max} = H$.

Example 4.A2. Water flows out of a tap at the rate of $4$ cm³/s. Density $1000$ kg/m³. Find the rate of momentum efflux per unit cross-section if cross-section is $0.5$ cm².

Velocity $v = (4\,\text{cm}^3/\text{s})/(0.5\,\text{cm}^2) = 8$ cm/s = $0.08$ m/s. Rate of momentum efflux = $\rho A v^2 = 10^3\cdot 0.5\times 10^{-4}\cdot 0.0064 = 3.2\times 10^{-4}$ N.

Topic 5: Viscosity

Sub-topic A: Newton's Law of Viscous Flow

For laminar flow with velocity gradient $dv/dy$ perpendicular to the flow:

$F = -\eta A \frac{dv}{dy},$

where $\eta$ is the coefficient of viscosity. SI unit: Pa·s (or N·s/m²); CGS: poise = 0.1 Pa·s.

Sub-topic B: Stokes' Law

A small sphere of radius $r$ moving with speed $v$ through a fluid of viscosity $\eta$ experiences a viscous drag

$F_v = 6\pi\eta r v.$

(Valid for low Reynolds number — small, slow spheres.)

Sub-topic C: Terminal Velocity

A sphere of density $\rho$ falling in a fluid of density $\sigma$ (with $\rho>\sigma$):

Net downward force = weight - buoyancy - drag.

$mg - \tfrac{4}{3}\pi r^3\sigma g - 6\pi\eta r v = 0$ at terminal.

$\tfrac{4}{3}\pi r^3(\rho-\sigma)g = 6\pi\eta r v_t\Rightarrow$

$v_t = \frac{2 r^2 (\rho-\sigma)g}{9\eta}.$

Note $v_t\propto r^2$. So bigger spheres fall faster (until Stokes' law breaks down).

If a bubble (less dense than fluid), it rises with terminal velocity.

Sub-topic D: Poiseuille's Law (for completeness)

Volume flow rate through a horizontal cylindrical pipe (laminar):

$Q = \frac{\pi r^4 \Delta P}{8\eta L}.$

JEE Main occasionally tests this directly.

Worked Examples (JEE Main level)

Example 5.1. A steel ball of radius $0.001$ m falls in glycerin ($\rho_{steel} = 8000$, $\rho_{gly} = 1300$ kg/m³, $\eta = 1$ Pa·s, $g=10$). Terminal velocity?

$v_t = 2(10^{-6})(6700)(10)/(9\cdot 1) = 134/(9\cdot 10^6)\cdot 10 = 1.49\times 10^{-2}$ m/s. Recompute: $2\cdot (10^{-3})^2\cdot 6700\cdot 10/(9\cdot 1) = 2\cdot 10^{-6}\cdot 6.7\times 10^4/9 = 1.49\times 10^{-2}$ m/s $\approx 0.015$ m/s.

Worked Examples (JEE Advanced level)

Example 5.A1. Two identical drops of radius $r$ each fall at terminal velocity $v_t$ in air. They coalesce. Find the new terminal velocity.

New radius $r' = (2)^{1/3} r$. Since $v_t\propto r^2$:

$v_t' = v_t \cdot (r'/r)^2 = v_t\cdot 2^{2/3}\approx 1.59 v_t$.

Topic 6: Surface Tension

Sub-topic A: Concept

The surface of a liquid behaves like an elastic membrane in tension. Surface tension $T$ is force per unit length acting tangent to the surface and perpendicular to a line drawn on the surface. Equivalently, surface energy per unit area equals $T$ (numerically).

Cause: molecules at the surface are pulled inward by molecules below (no molecules above), so the surface tends to minimize area.

Sub-topic B: Excess Pressure in Drops and Bubbles

Liquid drop (one surface): $\Delta P = 2T/r$.

Derivation: half-drop in equilibrium under surface tension force $T\cdot 2\pi r$ (around the equatorial circle) balanced by pressure difference times area $\pi r^2$.

$\Delta P\cdot \pi r^2 = T\cdot 2\pi r\Rightarrow \Delta P = 2T/r.$

Soap bubble (two surfaces, inner & outer): $\Delta P = 4T/r$.

Air bubble inside liquid (one surface): $\Delta P = 2T/r$.

Smaller drops/bubbles have higher excess pressure. Two soap bubbles in contact: air flows from the smaller (higher P) to the larger.

Sub-topic C: Capillary Rise

A capillary tube of radius $r$ dipped in a wetting liquid of density $\rho$, contact angle $\theta$, surface tension $T$. The liquid rises to height $h$:

Force balance. Weight of the column = component of surface tension force.

Weight $= \pi r^2 h\rho g$.

Vertical comp. of surface tension = $T\cdot 2\pi r\cos\theta$.

Equating:

$\boxed{h = \frac{2T\cos\theta}{r\rho g}.}$

(For $\theta< 90^\circ$, $h>0$, liquid rises; for $\theta>90^\circ$ (non-wetting, e.g. mercury in glass), $h< 0$, liquid depresses.)

Sub-topic D: Energy released when small drops merge

When $n$ drops of radius $r$ merge into a single drop of radius $R$ (volume conserved): $R = n^{1/3} r$. Surface area decreases from $4\pi n r^2$ to $4\pi R^2 = 4\pi n^{2/3}r^2$. Energy released:

$\Delta E = 4\pi T r^2(n - n^{2/3}).$

Worked Examples (JEE Main level)

Example 6.1. Capillary tube radius $0.1$ mm dipped in water ($T = 0.072$ N/m, $\rho = 1000$, $\theta = 0$, $g=10$). Height of rise?

$h = 2\cdot 0.072/(10^{-4}\cdot 1000\cdot 10) = 0.144/1 = 0.144$ m $= 14.4$ cm.

Example 6.2. Excess pressure inside a soap bubble of radius $1$ cm. $T = 0.04$ N/m.

$\Delta P = 4T/r = 4\cdot 0.04/0.01 = 16$ Pa.

Worked Examples (JEE Advanced level)

Example 6.A1. Two soap bubbles of radii $r_1$ and $r_2$ ($r_1<r_2$) are joined by a tube. Which way does air flow, and what is the radius of the new equilibrium configuration?

$P_1 = P_0 + 4T/r_1$, $P_2 = P_0 + 4T/r_2$. Since $r_1<r_2$, $P_1 > P_2$, so air flows from $1$ to $2$. The smaller bubble shrinks until it's a flat film. The combined bubble has the larger radius $r_2$ (approximately).

Example 6.A2. A capillary tube is dipped in water. The water rises to $h$. If a tube of half the radius is used, what height? If the tube is broken at $h/2$, what happens?

Half radius: $h' = 2h$ (since $h\propto 1/r$).

If the tube length is only $h/2$: water rises to the top, then the meniscus radius adjusts to accommodate. The new meniscus has radius $r' = 2T/(\rho g h/2)/\cos\theta = 2(2T\cos\theta/(\rho g h))\cdot(...)$; simplification: the meniscus flattens (larger radius of curvature). Water does NOT overflow. (Standard JEE Advanced trap.)

Problem-Solving Heuristics

1. Gravitation: always check whether the situation is point-mass (use $-GMm/r$) or extended-body (need integration or shell theorem).
2. For escape velocity, set total energy = 0 (KE just balances PE in magnitude). Doesn't depend on direction (in absence of other forces).
3. For satellite problems, total energy = $-GMm/(2r) = -KE = PE/2$ (an instance of the virial theorem).
4. Elasticity: identify the type of deformation (longitudinal, bulk, shear). For mixed problems, treat each independently if linear.
5. For a wire stretched by its own weight, use the average tension trick: as if half the weight is at the bottom and half at the top, so effective load is $W/2$ at the end.
6. Bernoulli + Continuity: write both equations together for any pipe-flow problem. They give 2 equations in 2 unknowns.
7. For terminal velocity of a sphere, use the net force = 0 principle. Don't forget buoyancy.
8. Capillary rise: $h\propto 1/r$. If tube radius halved, $h$ doubles.
9. Excess pressure: $2T/r$ for one surface (drop), $4T/r$ for two surfaces (soap bubble).
10. For multiple drops merging: volume conservation gives the new radius. Surface area change gives the energy released.

Common Traps & Mistakes

• $g$ at depth $d$: linear $g(1-d/R)$, not quadratic. Different from altitude formula.
• Total energy of bound satellite is negative, not positive.
• Kepler's 3rd law for elliptical orbits uses semi-major axis $a$, not $r$. For circular orbits $r=a$.
• For an iceberg, $90\%$ is submerged, $10\%$ is above water — many students reverse.
• In Bernoulli, the velocity terms refer to fluid velocities along the streamline, not bulk averages (in JEE we usually assume uniform speeds in cross-sections).
• For a bubble inside a liquid (one surface): $\Delta P = 2T/r$. For a soap bubble (two surfaces): $4T/r$.
• Capillary rise formula assumes wetting ($\theta < 90^\circ$). For mercury in glass ($\theta\approx 140^\circ$), $h< 0$: mercury depresses.
• $v_t\propto r^2$ in Stokes' regime: a sphere of double the radius has 4× the terminal velocity (not 8×, which would be the volume ratio).
• Geostationary vs geosynchronous: geo-stationary is in the equatorial plane (always above the same point); geo-synchronous can be inclined but has 24-h period.
• $Y = 3K(1-2\sigma) = 2\eta(1+\sigma)$: this is one identity, not two. Use it to relate moduli.

Quick Revision Card

• $\vec F = -GMm/r^2\,\hat r$; $U = -GMm/r$.
• $v_e = \sqrt{2gR}$, $v_0 = \sqrt{GM/r}$.
• Satellite total energy $E = -GMm/(2r) = -KE$.
• $g_h = g(1-2h/R)$, $g_d = g(1-d/R)$, $g_\lambda = g - \omega^2 R\cos^2\lambda$.
• Kepler: $T^2 = 4\pi^2 a^3/(GM)$.
• $Y = FL/(A\Delta L)$; $u = \tfrac12 Y\varepsilon^2$.
• Elongation under own weight: $\Delta L = \rho g L^2/(2Y)$.
• Pressure: $P = P_0+\rho gh$. Pascal's: $F_1/A_1 = F_2/A_2$.
• Continuity: $Av =$ const. Bernoulli: $P+\tfrac12\rho v^2 + \rho gh =$ const.
• Torricelli: $v=\sqrt{2gh}$.
• Stokes: $F_v=6\pi\eta r v$; terminal velocity $v_t = 2r^2(\rho-\sigma)g/(9\eta)$.
• Capillary: $h = 2T\cos\theta/(r\rho g)$.
• Excess P: drop $2T/r$, soap bubble $4T/r$.

Formula Sheet