Q.Fill in the blanks using the word(s) from the list appended with each statement:

Concept and Intuition

Each of these is a classic textbook pairing:

• Pascal's Law: pressure applied to an enclosed fluid is transmitted undiminished in all directions — the working principle of a hydraulic lift/press.
• Torricelli's Law: the speed of efflux of a liquid from an orifice under gravity, $v=\sqrt{2gh}$ — a specific named result (itself derivable from Bernoulli's equation, but conventionally cited by its own name for 'speed of efflux').
• Bernoulli's Principle: relates pressure and velocity along a streamline; explains dynamic lift (faster flow over a curved/airfoil surface → lower pressure → net lift).
• Stoke's Law: gives the viscous drag force on a small sphere moving through a fluid, $F=6\pi\eta rv$.

Step-by-Step Solution

1. a) Hydraulic Lift → Pascal's Law → (iv).
2. b) Speed of Efflux → Torricelli's Law → (ii).
3. c) Dynamic lift → Bernoulli's Principle → (i).
4. d) Viscous drag force → Stoke's Law → (iii).
5. Combined: a–iv, b–ii, c–i, d–iii.

Common Mistakes

• Confusing 'speed of efflux' (Torricelli) with 'dynamic lift' (Bernoulli) since both ultimately stem from Bernoulli's equation — but the question asks for the specifically named law/principle conventionally associated with each application.

ANSWER: D

Concept and Intuition

A soap bubble has two liquid surfaces (inner and outer film surfaces), so its excess internal pressure is $\dfrac{4T}{r}$ (twice the single-surface value $2T/r$ for a droplet). Since the bubbles are in vacuum, there is no external atmospheric pressure to add — the entire internal pressure IS this surface-tension term. Using $PV=nRT$ (constant temperature, so mass $\propto PV$ since $n\propto$ mass) lets us find how mass scales with radius.

Step-by-Step Solution

1. Excess pressure inside a soap bubble (two surfaces): $P = \dfrac{4T}{r}$. In vacuum, this is the total pressure of the trapped air (no atmosphere adds to it).
2. Volume of the bubble: $V = \dfrac{4}{3}\pi r^3$.
3. At constant temperature, mass of enclosed air $m \propto PV$ (from $PV = \dfrac{m}{M}RT$, with $M$, $R$, $T$ same for both):

4. Therefore $\dfrac{m_A}{m_B} = \dfrac{r_1^2}{r_2^2}$, i.e. ratio $= r_1^2 : r_2^2$.

Common Mistakes

• Adding an atmospheric pressure term (irrelevant here since the bubbles are explicitly in vacuum).
• Using the single-surface droplet formula $2T/r$ instead of the two-surface bubble formula $4T/r$ (doesn't change the final ratio's exponent here, but is conceptually wrong).
• Assuming mass simply scales as volume ($r^3$), ignoring that pressure itself depends on $r$.

ANSWER: B

Concept and Intuition

Viscosity is the internal friction between adjacent layers of a fluid moving at different velocities. In a liquid, molecules sit close together, and viscosity is dominated by the cohesive (attractive) intermolecular forces holding neighbouring layers together. Heating a liquid increases the average kinetic energy of its molecules, which overcomes these cohesive bonds more easily — so the resistance to flow (viscosity) falls.

In a gas, molecules are far apart and rarely interact via cohesive forces; instead, viscosity comes from the transfer of momentum as molecules randomly cross between adjacent layers moving at different bulk speeds — faster molecules diffusing into a slower layer speed it up, and vice versa. Kinetic theory gives gas viscosity $\eta \propto \sqrt{T}$ (through the mean molecular speed), so heating a gas makes molecules move faster, transfer momentum more effectively, and the gas viscosity rises.

Step-by-Step Solution

1. Identify the mechanism of viscosity in liquids: cohesive intermolecular forces.
2. Heating a liquid $\Rightarrow$ weaker effective cohesion $\Rightarrow$ liquid viscosity decreases.
3. Identify the mechanism of viscosity in gases: momentum transfer via molecular collisions.
4. Heating a gas $\Rightarrow$ faster molecules, more frequent/vigorous momentum exchange $\Rightarrow$ gas viscosity increases.
5. Combine: gases increase, liquids decrease $\Rightarrow$ option (B).

Common Mistakes

• Assuming viscosity always decreases with heating (true only for liquids, not gases).
• Confusing the microscopic origin of viscosity between gases (momentum transfer) and liquids (cohesive forces).

ANSWER: B

Concept and Intuition

The Reynolds number $Re=\dfrac{\rho v d}{\eta}$ tells us whether flow is smooth (laminar, low $Re$) or chaotic (turbulent, high $Re$, generally $Re\gtrsim2000$–$4000$ for pipe flow). Here we're given volumetric flow rate $Q$ rather than velocity directly, so we substitute $v=Q/A$.

Step-by-Step Solution

1. $A=\dfrac{\pi d^2}{4}=\dfrac{\pi(0.015)^2}{4}\approx1.767\times10^{-4}\ \mathrm{m^2}$.
2. $v=\dfrac{Q}{A}=\dfrac{7.5\times10^{-5}}{1.767\times10^{-4}}\approx0.424\ \mathrm{m/s}$.
3. $Re=\dfrac{\rho v d}{\eta}=\dfrac{1000\times0.424\times0.015}{10^{-3}}\approx6366$.
4. Since $Re\approx6366>6000\gg2000$, the flow is clearly turbulent.

Common Mistakes

• Using the diameter directly in $Re=\rho v r/\eta$ instead of the correct $\rho v d/\eta$ (or vice versa) — mixing up radius and diameter.
• Forgetting to convert diameter to metres.

ANSWER: C

Concept and Intuition

• Gas viscosity arises from momentum transfer via molecular collisions, which increases with temperature (faster molecules, more frequent transfer) — so gas viscosity increases with $T$.
• Liquid viscosity arises from intermolecular cohesive forces, which weaken as thermal agitation increases — so liquid viscosity decreases with $T$.
• Wetting of a solid by a liquid is governed by the angle of contact $\theta$: if $\theta<90^\circ$ (acute), the liquid wets/spreads on the surface; if $\theta>90^\circ$ (obtuse), it does not (beads up).
• Water fails to wet an oily/greasy glass surface because the adhesive force between water and the oily surface is weak compared to water's own cohesive force — not simply because "oil's cohesive force is less than water's" (a different, non-standard comparison).

Step-by-Step Solution

1. Statement A: temperature ↑ ⇒ gas viscosity ↑, liquid viscosity ↓. This is the standard, correct physics — TRUE.
2. Statement C: claims wetting happens when contact angle $>90^\circ$. The correct rule is the opposite (wetting needs $\theta<90^\circ$) — FALSE.
3. Statement B: the reasoning given (cohesive force of oil vs. water) is not the accepted explanation for non-wetting; the correct explanation involves the adhesive force between water and the greasy surface being weaker than water's cohesive force — so B, as an explanation, is FALSE.
4. So A is true; B and C are false — matching option (C).

Common Mistakes

• Assuming the contact-angle rule for wetting is $>90^\circ$ instead of $<90^\circ$.
• Accepting a plausible-sounding but non-standard causal explanation (as in B) at face value.

ANSWER: C

Concept and Intuition

A liquid's surface behaves like a stretched membrane with energy stored per unit area — that's what

surface tension $T$ measures (energy per unit area, equivalently force per unit length). Splitting

one big drop into many small ones necessarily increases the total surface area (you're creating new

surface where none existed), and that new surface costs energy equal to $T\times(\text{area created})$. Volume is conserved in the split, but area is not — smaller drops always have more

total surface area for the same total volume.

Step-by-Step Solution

1. Surface area of the original single drop: $A_{initial}=4\pi R^2$.
2. Surface area of all $n$ smaller drops combined: $A_{final}=n\times4\pi r^2$.
3. Since surface energy $=T\times$ area, the extra energy needed to create the additional surface is $\Delta E = T(A_{final}-A_{initial}) = T(4\pi r^2 n - 4\pi R^2)$.

Common Mistakes

• Confusing surface energy (area $\times T$) with volume-based quantities (option B mistakenly uses volume terms $\frac43\pi r^3$, $\frac43\pi R^3$).
• Getting the order of subtraction backwards (should be final $-$ initial, since breaking up always increases area and always costs energy, i.e. the bracket must be positive).

ANSWER: A

Concept and Intuition

The angle of contact $\theta$ at a liquid–solid boundary is set by the balance of the three relevant surface tensions (solid–air, solid–liquid, liquid–air) at the line of contact. Since this balance is purely local — it depends on what materials are touching and the ambient conditions (impurities, temperature) — tilting the solid surface as a whole does not change that local force balance, so $\theta$ stays the same regardless of inclination. This is a specifically counter-intuitive but standard textbook fact.

Step-by-Step Solution

1. (a) claims $\theta$ depends on the inclination of the solid surface — this is FALSE; $\theta$ is set by the surface-tension balance at the contact line and is independent of how the surface is tilted.
2. (b) is TRUE: if $\theta<90^\circ$ the liquid "wets" the solid and tends to spread over it (adhesive forces dominate).
3. (c) is TRUE: the angle of contact for most liquid–solid pairs increases as the liquid's temperature rises (surface tension and hence the force balance shifts).
4. (d) is TRUE: for pure water on clean glass the angle of contact is very close to $0^\circ$ (water wets clean glass almost completely).
5. Since only (a) is false, the incorrect statement is (a) only.

Common Mistakes

• Assuming a tilted surface must change how a liquid "sits" on it, and therefore wrongly picking (a) as true.
• Confusing angle of contact with the shape of the meniscus, which does visually look different on a tilted surface even though $\theta$ itself is unchanged.

ANSWER: A

Concept and Intuition

Reynolds number characterizes whether fluid flow is dominated by viscous forces (laminar) or inertial forces (turbulent). The standard textbook classification has three regimes: streamline/laminar at low $R_e$, an unstable transitional zone in the middle, and turbulent at high $R_e$. The middle zone is specifically called unstable/unsteady, not steady.

Step-by-Step Solution

1. Recall the standard classification: for $R_e<1000$, flow is streamline/laminar — statement (A) is correct.
2. For $1000<R_e<2000$, the flow becomes unstable/unsteady as it transitions from laminar to turbulent — NOT steady. Statement (B) incorrectly calls this regime "steady".
3. For $R_e>2000$, flow becomes turbulent — statement (C) is correct.
4. Reynolds number is defined as $R_e = \dfrac{\rho v D}{\eta}$, a ratio of quantities with the same net dimensions, hence dimensionless — statement (D) is correct.
5. Therefore the incorrect statement is (B).

Common Mistakes

• Assuming "steady" and "unstable/unsteady" mean the same thing in this context — they are opposites, and the transitional regime is specifically the unstable one.
• Not recognizing this transitional band as a distinct third regime and instead lumping it with either laminar or turbulent.

ANSWER: B

Concept and Intuition

Molecules at the surface of a liquid experience a net inward pull from the molecules below them (since there are no liquid molecules above to balance the attraction), creating a tendency for the surface to contract — this net effect is called surface tension. A surface under tension always tries to minimize its potential energy, and since surface energy is proportional to surface area, minimizing area minimizes energy. This is why free liquid surfaces (like a small drop, unaffected by gravity) tend to be spherical, and why a soap film pulls tight.

Step-by-Step Solution

1. Surface tension gives the liquid surface a form of elastic potential energy proportional to its area: $E = \sigma \times A$, where $\sigma$ is the surface tension coefficient.
2. A system in equilibrium tends toward minimum potential energy.
3. So the free surface adopts the shape of minimum area, consistent with any constraints (volume, external forces).
4. None of viscosity (resists flow), diffusion (molecular mixing), or pressure (a scalar force per area, not an area-minimizing tendency) produce this area-minimizing behaviour.

Common Mistakes

• Confusing surface tension with viscosity — viscosity is about resistance to flow (shear), not surface shape.
• Thinking pressure alone shapes a free liquid surface — pressure differences are actually a CONSEQUENCE of surface tension (Laplace's law), not the cause of area minimization.

ANSWER: B

Concept and Intuition

A liquid wets a surface when the adhesive force between the liquid and the surface exceeds the liquid's own cohesive force (molecules attracting each other) — the liquid then prefers to spread out and stick to the surface. If cohesion dominates over adhesion, the liquid instead minimizes contact with the surface, forming beads (non-wetting), exactly as mercury does on glass, and as water does on an oily/waxy surface.

Step-by-Step Solution

1. Wetting criterion: liquid wets surface if (adhesive force between liquid & surface) $>$ (cohesive force within the liquid).
2. Non-wetting occurs when cohesive force within the liquid $>$ adhesive force between liquid and surface.
3. Here, the surface is oily glass, and the liquid is water. The relevant adhesive force is between water molecules and the oil molecules coating the glass (oil, being non-polar, doesn't attract water strongly).
4. Since this adhesive force (water–oil) is weak, water's own cohesive force dominates, so it doesn't wet the oily surface.

Common Mistakes

• Comparing oil's cohesion to something else, rather than comparing water's own cohesion to the water–oil adhesion (the correct comparison for whether water wets the surface).
• Vague answers like "oil repels water", which isn't the correct physical mechanism (it's a cohesion vs adhesion balance, not repulsion).

ANSWER: D

Concept and Intuition

Surface tension exists because surface molecules of a liquid are pulled inward more strongly than they are pulled by the much sparser vapour above — a net inward cohesive pull. As temperature rises toward the critical temperature $T_c$, the vapour becomes denser and the liquid less dense, until at $T_c$ the two phases merge into one indistinguishable fluid. At that point there is no longer any difference in intermolecular force environment across the "surface" (because there is no surface), so the net inward pull — and therefore the surface tension — is exactly zero.

Step-by-Step Solution

1. Assertion: Surface tension $\to 0$ as $T\to T_c$ — this is an experimentally and thermodynamically established fact (used to derive Eötvös' and van der Waals-type relations for surface tension vs temperature).
2. Reason: At $T_c$, liquid and gas coexistence ends; their densities and average intermolecular forces become equal.
3. Because the forces become equal, there's no preferential inward pull at any interface, so the liquid can expand into the vapour without any restoring (tension) force — directly causing zero surface tension.
4. Hence R correctly explains why A is true.

Common Mistakes

• Thinking surface tension depends only on temperature via a simple linear formula without linking it to the disappearance of the phase boundary at $T_c$.
• Confusing critical temperature with boiling point — surface tension is already small near boiling but only strictly zero at $T_c$.

ANSWER: A

Concept and Intuition

Surface tension is the tendency of a liquid surface to minimize its area, behaving like a stretched elastic membrane. When a shaving brush (or any bunch of fine hairs/bristles) is pulled out of water, a thin film of water clings between adjacent hairs. This film's surface tension pulls the hairs as close together as possible, since a bunched configuration has less exposed liquid surface area than hairs spread apart with the same amount of water between them.

Step-by-Step Solution

1. On removal from water, a film of water is trapped between neighbouring hairs of the brush.
2. This film has two free surfaces exposed to air, each possessing surface tension.
3. The system minimizes its surface energy by minimizing surface area — clustering the hairs together reduces the total area of the water film's surface compared to spread-out hairs.
4. The inward pull caused by this area-minimizing tendency is what visibly draws and holds the hairs together.

Common Mistakes

• Attributing this to viscosity of water — viscosity resists relative motion/flow, it does not explain the clumping/clinging effect.
• Thinking it's some special "attraction between hairs" independent of the water film — the effect vanishes if the hairs are dried, showing it is water's surface tension, not an intrinsic hair property.

ANSWER: B