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

Concept & Intuition

A stress-strain curve tells the story of how a material deforms under load. The ultimate strength is the maximum stress the material can withstand; the fracture point is where it actually breaks. If these two points are very close, the material snaps soon after reaching its peak strength, with almost no necking or stretching. That’s exactly what happens in brittle materials (like glass or cast iron) — they fail suddenly with little warning. In contrast, ductile materials (like copper or mild steel) stretch significantly between ultimate strength and fracture, giving a long “tail” on the curve.

Step-by-step reasoning

1. Recall the shape of stress-strain curves

   • For a ductile material, after the ultimate strength, the material continues to deform plastically (necking) over a large strain range before finally fracturing. So the fracture point is far to the right of the ultimate strength point.
   • For a brittle material, plastic deformation is minimal; the material fractures almost immediately after reaching ultimate strength. Hence the two points are very close on the strain axis.

2. Interpret the given condition

   The problem states: “ultimate strength and fracture points are close.” This means the horizontal distance (strain) between them is small. That implies negligible plastic deformation after the peak stress.

3. Match to material types

   • (A) Ductile → large separation → not correct.
   • (B) Elastomer → these are rubber-like; they have a very different curve (large elastic strain, no clear yield/ultimate point in the same sense) — not characterized by closeness of ultimate and fracture.
   • (C) Brittle → small separation → correct.
   • (D) Plastic → “plastic” is not a material class here; it refers to a deformation regime. Thermoplastics can be ductile or brittle depending on conditions, but the term alone doesn’t match.

4. Confirm with a classic example

   Think of a glass rod: it bends very little, then snaps at its maximum stress. The ultimate strength and fracture are essentially the same point. That’s brittle behavior.

With ρ=1000 kg m⁻³, v=0.10 m s⁻¹, D=0.018 m, η=10⁻³ Pa·s: Re = ρvD/η = (1000×0.10×0.018)/10⁻³ = 1800. Flow is laminar for Re≲1000, turbulent for Re≳2000, and unsteady/transitional in between. Re=1800 lies in that middle band, so the flow is unsteady.

ANSWER: C

Excess pressure inside a soap bubble is P = 4T/r, so the smaller bubble is at higher pressure. When two bubbles are connected by a tube, air flows from high to low pressure, i.e. from the small bubble to the large one. As the small bubble shrinks its pressure rises further, so it collapses completely (r -> 0).

For the combined air, using the standard idealisation (neglecting atmospheric pressure relative to the Laplace pressure, isothermal process):

So the final radii are 0 cm and 17 cm.

ANSWER: C

Concept and Intuition

A venturimeter measures flow speed by converting kinetic energy into pressure change. For an incompressible fluid, Bernoulli’s equation tells us that an increase in speed (at the throat) causes a drop in pressure. That pressure drop is then read as a height difference in two vertical tubes (a manometer). Crucially, the height difference $\Delta h$ is not the pressure difference itself — it’s the pressure difference divided by the fluid’s own weight density ($\rho g$). So if two different fluids flow at the same speed through identical venturimeters, the one with lower density will show a larger height difference for the same pressure drop. That’s the heart of this problem.

Step-by-step solution

1. Apply Bernoulli’s equation to each venturimeter For a horizontal venturimeter (no height change), Bernoulli’s equation between the wide section (area $A_1$, velocity $v_1$) and the throat (area $A_2$, velocity $v_2$) gives:

So the pressure drop is:

2. Relate the velocities By continuity ($A_1 v_1 = A_2 v_2$), we have $v_2 = \frac{A_1}{A_2} v_1$. Since both venturimeters are identical, $A_1/A_2$ is the same constant $k > 1$. Given $v_1 = 10~\mathrm{m/s}$ for both, the pressure drop becomes:

So $\Delta P \propto \rho$ — the pressure drop is directly proportional to the fluid’s density.

3. Interpret the manometer reading Each vertical tube is open to the fluid at that point, so the height difference $\Delta h$ in the two tubes is given by:

because the fluid in the tubes is the same as the flowing fluid. Therefore:

4. Substitute the pressure drop Using $\Delta P \propto \rho$ from step 2:

The density cancels! That would suggest $\Delta h$ is the same for both — but wait, that’s only true if the manometer fluid is the same as the flowing fluid. Here it is: petrol in venturimeter A, diesel in B. So indeed, for identical venturimeters and same inlet velocity, the height difference is independent of density? Let’s check carefully.

Actually, the constant in step 2 is $\frac{1}{2} \cdot 100 (k^2 - 1)$, which does not depend on $\rho$. So:

This is exactly the same for both fluids. So $\Delta h_1 = \Delta h_2$, giving ratio $1:1$.

But the problem gives densities $750$ and $850$ — why would they include them if they cancel? This is the classic pitfall: many students think the height difference depends on density, but here the manometer fluid is the flowing fluid, so the density cancels. The answer should be $1:1$.

5. Confirm with the given numbers For petrol: $\Delta h_1 = \frac{50(k^2-1)}{g}$ For diesel: $\Delta h_2 = \frac{50(k^2-1)}{g}$ Identical. So $\Delta h_1 : \Delta h_2 = 1:1$.

Concept & Intuition

A soap bubble has two surfaces (inner and outer), so the excess pressure inside it is given by $\Delta P = \frac{4T}{r}$, where $T$ is the surface tension and $r$ is the radius. This is double the pressure for a liquid droplet (which has only one surface). The key insight: pressure and radius are inversely related — a smaller bubble has higher internal pressure. Once we know the radii ratio from the pressure ratio, the volume ratio follows directly because volume $V = \frac{4}{3}\pi r^3$ scales with the cube of the radius.

Step-by-step solution

1. Write the excess pressure formula for a soap bubble For a soap bubble of radius $r$, the excess pressure is $\Delta P = \frac{4T}{r}$. Given that the excess pressures are in the ratio $2:3$, we have:

2. Relate pressure ratio to radius ratio Since $\Delta P \propto \frac{1}{r}$, we get:

Therefore:

So the radii are in the ratio $3:2$.

3. Find the volume ratio Volume of a sphere: $V = \frac{4}{3}\pi r^3$. Hence:

So the volumes are in the ratio $27:8$.

The key idea is Hooke’s law for a wire under tension:

where $F$ is the applied force (tension), $L$ the original length, $A$ the cross‑sectional area, and $Y$ Young’s modulus (same for all wires since they are made of the same material).

Because $F$ and $Y$ are identical for all four wires, the elongation $\Delta L$ is simply proportional to $\frac{L}{A}$. Since the wires are circular, $A = \pi r^2$, so

We only need to compare the ratio $L / r^2$ for each wire — the largest ratio gives the greatest elongation.

1. Wire A: $L = 1\ \text{m}$, $r = 0.2\ \text{mm} = 0.2 \times 10^{-3}\ \text{m}$.

2. Wire B: $L = 2\ \text{m}$, $r = 0.4\ \text{mm} = 0.4 \times 10^{-3}\ \text{m}$.

3. Wire C: $L = 3\ \text{m}$, $r = 0.6\ \text{mm} = 0.6 \times 10^{-3}\ \text{m}$.

4. Wire D: $L = 4\ \text{m}$, $r = 0.8\ \text{mm} = 0.8 \times 10^{-3}\ \text{m}$.

Comparing the values:

• A: $2.5 \times 10^{7}$
• B: $1.25 \times 10^{7}$
• C: $0.833 \times 10^{7}$
• D: $0.625 \times 10^{7}$

The largest ratio is for wire A.

The concept first: force lives on the contact line, not on an area

Surface tension $T$ is a force per unit length of the line along which the liquid surface meets the solid. So to get a force you must ask: how long is that line, and in which direction does it pull?

The pull acts along the liquid surface, i.e. at the contact angle $\theta$ to the wall. Resolving it along the straw (vertical) direction brings in the factor $\cos\theta$.

The subtlety that this question is built on: a straw is a tube of negligible thickness, not a solid rod. Liquid climbs on its inner wall and clings to its outer wall. Each contributes a circle of circumference $2\pi R$ (the thickness is negligible, so inner radius $\approx$ outer radius $\approx R$).

Step-by-step

Step 1 — Contact length.

Step 2 — Force per unit length and its direction. Each element of the contact line is pulled with magnitude $T\,dl$ along the liquid surface, which makes angle $\theta$ with the straw's wall. The component along the straw is $T\,dl\cos\theta$; the horizontal components cancel by symmetry around the circle.

Step 3 — Total force.

Step 4 — Put in the number. $\cos 53^\circ = 0.6 = \dfrac{3}{5}$:

Common trap: treating the straw as a solid rod gives $L = 2\pi R$ and $F = \tfrac{6\pi RT}{5}$ — that is exactly the distractor (B). The word straw (hollow) is doing real physics work here.

ANSWER: A

The key concept here is capillary rise — the phenomenon where a liquid climbs up a narrow tube due to adhesive forces between the liquid and the tube wall, balanced by gravity. The classic formula for the height $h$ of the liquid column is:

where:

• $\gamma$ = surface tension of the liquid,
• $\theta$ = contact angle,
• $\rho$ = density,
• $g$ = gravity,
• $r$ = radius of the tube.

Notice: height is inversely proportional to the radius, not to the cross-sectional area. This is the crucial insight. Many students mistakenly think height scales with area, but area is $\pi r^2$, so a change in area changes the radius by the square root.

Let’s work through it step by step.

1. Relate area to radius. The cross-sectional area of a circular capillary is $A = \pi r^2$. If the area becomes $4A$, then:

So the radius doubles.

2. Apply the capillary rise formula. For the original tube:

For the new tube (radius $2r$):

3. Interpret the result. Since the radius is twice as large, the liquid column is only half as high. The surface tension, contact angle, density, and gravity are all unchanged — only the geometry matters.

The key idea here is that a soap bubble is not a single surface — it’s a thin film of liquid with an inside surface and an outside surface. Each surface contributes to the total surface area, and surface energy is simply surface tension times total area.

Many students instinctively treat a bubble like a solid sphere, using $4\pi r^2$ for the area. That’s the trap. A bubble’s film has two sides, so the relevant area is $2 \times 4\pi r^2 = 8\pi r^2$. Surface energy is $E = T \times \text{total surface area}$, and the work needed to blow the bubble is the change in this energy.

Let’s walk through it step by step.

1. Surface area of a bubble For a sphere of radius $r$, the surface area of one side is $4\pi r^2$. Since a soap film has two surfaces (inner and outer), the total surface area of the bubble is

2. Surface energy Surface energy is given by $E = T \times A$, where $T$ is the surface tension. So for a bubble of radius $r$,

3. Initial and final radii

   The initial radius is $R$. Doubling the diameter means the new radius is $2R$.

4. Change in surface energy

   The work required (the surface energy needed) is the difference:

Compute: $(2R)^2 = 4R^2$, so $4R^2 - R^2 = 3R^2$. Thus

Concept. For a wire under tension, Young's modulus gives

because $F$, $Y$ (same material) and the constant $\pi/4$ are identical for both wires.

Calculation.

• Wire A: $L = 10\,\text{cm}, d = 1\,\text{mm} \Rightarrow \dfrac{L}{d^2} = \dfrac{10}{1^2} = 10$.
• Wire B: $L = 70\,\text{cm}, d = 2\,\text{mm} \Rightarrow \dfrac{L}{d^2} = \dfrac{70}{2^2} = 17.5$.

Since $10 < 17.5$, wire A extends less than wire B.

ANSWER: B

Force from surface tension

For a thin square sheet of side $4~\text{cm}$ and negligible thickness, the contact line runs along both faces, so the effective perimeter is

The vertical surface-tension force is $F = T\,L\cos\theta$, added to the sheet's weight $W$ in the balance reading $R = W + T L\cos\theta$.

Two readings

• Contact angle $\theta = 0^\circ$ (fluid wets, pulls the sheet down):

• Contact angle $\theta = 180^\circ$ (after oil, force reverses upward):

Solve for $T$

Subtracting eliminates the weight:

The core idea behind this problem is to correctly associate key principles in fluid mechanics with their defining characteristics or primary applications. Each principle in Column I describes a specific aspect of fluid behavior, and we need to find the most direct and accurate link in Column II.

Let's break down each concept:

1. Stoke's Law: This law quantifies the drag force experienced by a small spherical object moving through a viscous fluid at low Reynolds numbers (laminar flow). The force is directly proportional to the viscosity of the fluid, the radius of the sphere, and its velocity. This force is a type of viscous drag.

   The viscous drag force $F_d$ on a sphere of radius $r$ moving with velocity $v$ in a fluid of viscosity $\eta$ is given by:

   $F_d = 6 \pi \eta r v$

2. Turbulence: This describes a fluid flow regime characterized by chaotic, unpredictable changes in pressure and flow velocity. Unlike smooth, orderly laminar flow, turbulent flow involves eddies and vortices. The transition from laminar to turbulent flow is often predicted using a dimensionless quantity called the Reynolds number.

3. Bernoulli's Principle: This principle is a statement of the conservation of energy for an ideal fluid (incompressible, non-viscous, steady flow) along a streamline. It relates the pressure, fluid velocity, and height at different points in the flow. Essentially, it states that the sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume remains constant.

   For an ideal fluid, Bernoulli's equation states:

   $P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}$

   where $P$ is pressure, $\rho$ is fluid density, $v$ is fluid velocity, $g$ is acceleration due to gravity, and $h$ is height. This equation directly links pressure and energy forms.

4. Pascal's Law: This law states that a pressure change applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of its container. This principle is fundamental to the operation of hydraulic systems, which use fluids to transmit forces. A classic application is the hydraulic lift.

Now, let's make the specific matches:

1. A) Stoke’s law is directly associated with III) Viscous drag. Stoke's law specifically calculates this drag force for a sphere.

2. B) Turbulence is characterized and predicted by the IV) Reynold’s number. A high Reynolds number indicates turbulent flow, while a low one indicates laminar flow.

3. C) Bernoulli’s Principle is a statement of energy conservation for fluids, relating I) Pressure and energy (kinetic, potential, and pressure energy).

4. D) Pascal’s law is the underlying principle for the operation of a II) Hydraulic lift, where pressure is transmitted through a fluid to multiply force.

Combining these matches, we get the sequence:

A $\rightarrow$ III

B $\rightarrow$ IV

C $\rightarrow$ I

D $\rightarrow$ II

This corresponds to the option (A).