Q.At a depth of $1000\ \text{m}$ in an ocean

Concept & Intuition

In a U-tube, the pressure at any horizontal level in a connected liquid is the same. When we pour a lighter liquid (density 800 kg/m³) into one limb, it sits on top of the water. The water level in that limb is pushed down, and the water in the other limb rises. The pressure at the bottom of the lighter liquid column (where it meets water) must equal the pressure at the same depth in the other limb — but that other limb contains only water. So we equate the hydrostatic pressures.

Step-by-step solution

1. Identify the densities and heights

   Density of water, $\rho_w = 1000~\mathrm{kg/m^3}$

   Density of the poured liquid, $\rho_l = 800~\mathrm{kg/m^3}$

   Height of the poured liquid column, $h_l = 25~\mathrm{cm} = 0.25~\mathrm{m}$

2. Understand the water level changes

   When the lighter liquid is poured into the left limb, the water level in that limb falls by some amount $x$, and the water level in the right limb rises by the same $x$ (since the total volume of water is unchanged). So the difference in water levels between the two limbs is $2x$. The poured liquid sits above the water in the left limb.

3. Choose a reference level for pressure equality

   Consider the horizontal line that passes through the interface between the lighter liquid and water in the left limb. At that depth in the right limb, there is only water. The pressure on both sides must be equal.

4. Write the pressure equation

   On the left side: pressure = pressure due to the lighter liquid column of height $h_l$

   On the right side: pressure = pressure due to a water column of height equal to the vertical distance from the reference level to the water surface in the right limb.

   That vertical distance is: the rise in the right limb ($x$) plus the fall in the left limb ($x$) = $2x$.

   So:

   $$$$

The flow of water from a tap is an excellent example of steady, incompressible fluid flow. To solve this problem, we rely on two fundamental principles of fluid dynamics:

1. Conservation of Mass (Continuity Equation): For an incompressible fluid flowing steadily through a pipe or stream, the mass flow rate must be constant. This means the product of the cross-sectional area and the fluid velocity remains constant along a streamline. As the water falls, its velocity increases due to gravity, and consequently, its cross-sectional area must decrease to maintain a constant volume flow rate.
2. Conservation of Energy (Bernoulli's Principle): This principle states that for an ideal fluid in steady flow, the sum of the pressure energy, kinetic energy per unit volume, and potential energy per unit volume is constant along a streamline. It's essentially an application of the work-energy theorem to fluid flow.

We will use these two principles together to find the initial velocity of the water, then calculate the mass flow rate, and finally the total mass.

Step-by-step Derivations

1. Identify Given Parameters and Convert Units:

   First, let's list the given values and convert them to SI units for consistency in calculations.

   • Initial area of cross-section, $A_1 = 1.8\,\text{cm}^2 = 1.8 \times 10^{-4}\,\text{m}^2$.
   • Area of cross-section at $25\,\text{cm}$ below, $A_2 = 1.2\,\text{cm}^2 = 1.2 \times 10^{-4}\,\text{m}^2$.
   • Vertical distance, $h = 25\,\text{cm} = 0.25\,\text{m}$.
   • Time duration, $t = 10\,\text{s}$.
   • Acceleration due to gravity, $g = 10\,\text{ms}^{-2}$.
   • Density of water, $\rho = 1000\,\text{kgm}^{-3}$ (standard value for water).

2. Apply the Continuity Equation:

   Let $v_1$ be the velocity of water at the tap (point 1) and $v_2$ be the velocity at $25\,\text{cm}$ below (point 2). According to the continuity equation, the volume flow rate ($Q = Av$) is constant:

   $A_1 v_1 = A_2 v_2$

   Substituting the given areas:

   $1.8 \times 10^{-4}\,\text{m}^2 \times v_1 = 1.2 \times 10^{-4}\,\text{m}^2 \times v_2$

   $1.8 v_1 = 1.2 v_2$

   $v_2 = \frac{1.8}{1.2} v_1 = 1.5 v_1$

   This equation relates the two velocities. As expected, since the area decreases, the velocity must increase.

3. Apply Bernoulli's Principle:

   We apply Bernoulli's principle between point 1 (at the tap) and point 2 (at $25\,\text{cm}$ below).

   $P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2$

   • Both points are exposed to the atmosphere, so the pressure at both points is atmospheric pressure. Thus, $P_1 = P_2 = P_{\text{atm}}$. These terms cancel out.
   • Let's set the reference level for potential energy ($h=0$) at point 2 (the lower cross-section). Then $h_2 = 0$ and $h_1 = h = 0.25\,\text{m}$.

   The Bernoulli equation simplifies to:

   $\frac{1}{2}\rho v_1^2 + \rho g h = \frac{1}{2}\rho v_2^2 + \rho g (0)$

   Divide by $\rho$:

   $\frac{1}{2} v_1^2 + g h = \frac{1}{2} v_2^2$

   Rearranging to solve for $v_2^2 - v_1^2$:

   $v_2^2 - v_1^2 = 2gh$ …

In an open-tube manometer the pressure difference read by the instrument is

where $h$ is the difference in the liquid levels. Here the manometer reads a difference equal to atmospheric pressure ($\Delta P = P_{atm} = 100\ \mathrm{kPa} = 1\times10^{5}\ \mathrm{Pa}$), with

Solving for the density of the manometer liquid: …

Concept and Intuition

When a capillary tube is open at both ends and held vertically, water can be held inside it only if the surface tension forces at the two menisci (top and bottom) can support the weight of the water column. At the top meniscus, surface tension pulls upward; at the bottom meniscus, it also pulls upward (since the water is hanging, the bottom meniscus is concave upward). So both ends contribute to supporting the column. The maximum height occurs when the weight of the water exactly equals the total upward force from surface tension at both ends. If the column is any longer, it will fall.

Step-by-Step Solution

1. Identify the forces The water column is cylindrical, of height $h$ and radius $r = 0.5~\text{mm} = 5 \times 10^{-4}~\text{m}$. Weight of the column:

where $\rho = 1000~\text{kg/m}^3$ (density of water), $g = 10~\text{m/s}^2$.

2. Surface tension force at each meniscus For a water-air interface, surface tension acts along the circumference. The vertical component of the force at a meniscus is:

where $T = 0.07~\text{N/m}$ and $\theta$ is the contact angle. For water in a clean glass tube, $\theta \approx 0^\circ$, so $\cos\theta = 1$.

Thus each meniscus contributes:

3. Both ends pull upward Since the tube is open at both ends and held vertically, the top meniscus pulls the water up, and the bottom meniscus also pulls the water up (it is concave upward because the water is hanging). So total upward force:

4. Set upward force equal to weight …

Concept & Intuition

A soap bubble has two liquid-air interfaces (inner and outer), so the excess pressure inside is twice that of a simple liquid droplet: $\Delta P = \frac{4T}{R}$. This pressure pushes outward and is balanced here by the hydrostatic pressure of an oil column. Hydrostatic pressure depends only on height, density, and gravity: $P = \rho g h$. Setting them equal lets us solve for surface tension $T$.

Step-by-step solution

1. Write the excess pressure inside a soap bubble For a soap bubble of radius $R$, surface tension $T$ produces an excess pressure

Here $R = 0.5\ \text{cm} = 0.5 \times 10^{-2}\ \text{m} = 5 \times 10^{-3}\ \text{m}$.

2. Write the hydrostatic pressure of the oil column The pressure due to a column of liquid of height $h$, density $\rho$, under gravity $g$ is

Given $h = 4\ \text{mm} = 4 \times 10^{-3}\ \text{m}$, $\rho = 900\ \text{kg m}^{-3}$, $g = 10\ \text{m s}^{-2}$.

3. Equate the two pressures The problem states the bubble’s excess pressure is balanced by the oil column’s pressure, so

4. Solve for $T$

Substitute the values:

5. Simplify step by step First, numerator: $900 \times 10 = 9000$. …

Concept & Intuition

When a cylindrical vessel containing liquid is rotated steadily about its vertical axis, the liquid is forced outward by centrifugal acceleration. The free surface is no longer flat — it becomes a paraboloid of revolution. At the centre, the liquid is depressed; at the walls, it rises. The key is that every point on the free surface has the same pressure (atmospheric), so the sum of gravitational potential energy and centrifugal potential energy is constant along the surface. This leads directly to a simple formula for the height difference.

1. Set up the rotating frame

   In a frame rotating with the cylinder, the liquid is at rest relative to the container. A fluid particle at radius $r$ experiences a centrifugal acceleration $\omega^2 r$ outward. The effective gravity in this frame has components: vertically downward $g$, and radially outward $\omega^2 r$.

2. Condition for the free surface

   The free surface is an equipotential surface of the effective potential. The gravitational potential is $gz$ (taking $z=0$ at the bottom of the vessel, but we only care about differences). The centrifugal potential is $-\frac{1}{2}\omega^2 r^2$ (since its gradient gives $-\omega^2 r$ outward). So the total effective potential is

On the free surface, $\Phi$ is constant. Let that constant be $C$.

3. Find the shape of the surface At the centre ($r=0$), let the height be $z_0$. Then

At a general radius $r$,

Hence

This is a parabola — the surface rises quadratically with radius.

4. Height difference The height at the wall ($r = R$) is $z(R) = z_0 + \frac{\omega^2 R^2}{2g}.$ …

When a body floats, the net vertical force is zero: the downward pull of gravity (weight) is exactly balanced by the upward push of the fluid (buoyant force). Archimedes’ principle tells us that the buoyant force equals the weight of the fluid displaced. For a floating object that is only partly submerged, the displaced volume is just the volume of the part that sits below the surface.

Here the sphere is hollow, so its total weight depends on the volume of the shell material only — the empty cavity inside contributes nothing to the weight. The buoyant force, however, depends on the external volume of the sphere (the whole outer shape), because that’s what pushes fluid aside. And we are told the sphere is exactly half submerged, so the displaced volume is half of that outer volume.

Let’s work through the numbers.

1. Find the volume of the shell material. Outer radius $R = 4\,\text{cm}$, inner radius $r = 2\,\text{cm}$. Volume of the material = volume of outer sphere minus volume of inner cavity:

So $V_{\text{material}} = \frac{224}{3}\pi \,\text{cm}^3$.

2. Find the volume of the displaced liquid. The sphere floats half submerged, so the submerged volume is half the outer volume:

3. Apply the floating condition. Weight of sphere = buoyant force: $$$$

Why the level changes

When ice floats, it displaces a volume of liquid whose weight equals the weight of the ice (Archimedes' principle). When the ice melts, it becomes water with its own density. The key insight: if the surrounding liquid is denser than water, the meltwater will occupy less volume than the ice originally displaced, causing the level to fall.

Let's track the volumes carefully.

Step-by-step solution

1. Find the volume of liquid displaced by the floating ice

   The ice has mass $m = 9\ \mathrm{g}$. For it to float in equilibrium:

So the volume displaced is:

2. Find the volume of water after the ice melts

   Ice melts into water. The mass is conserved ($9\ \mathrm{g}$), but now it's liquid water with density $\rho_{\text{water}} = 1\ \mathrm{g/cm}^3$:

3. Calculate the net change in liquid volume

   Initially, the ice displaced $7.5\ \mathrm{cm}^3$ of the liquid (this volume was "pushed up" into the column). After melting, the ice contributes $9\ \mathrm{cm}^3$ of water to the liquid.

   The net change in total liquid volume is:

   $\Delta V = V_{\text{water}} - V_{\text{displaced}} = 9 - 7.5 = 1.5\ \mathrm{cm}^3$ …

The problem is about a U-tube manometer with two immiscible fluids. When oil is poured on one side, it pushes the liquid down on that side and up on the other. The liquid on the right rises by 15 cm, meaning the interface on the left has dropped by the same amount (since the liquid is incompressible and the tube has uniform cross-section). The oil column sits above the left-side liquid, and we need to find how much higher the top of the oil column is compared to the top of the liquid column on the right.

The central concept is pressure equality at the same horizontal level in a static fluid. Pick a horizontal line through the interface between oil and liquid on the left side. At that depth, the pressure due to the oil column above must equal the pressure due to the liquid column above on the right side (at the same level). That gives a direct relation between heights and densities.

Let’s work it through step by step.

1. Set up the geometry.

   Let the original liquid level be the reference. When oil is poured on the left, the liquid on the right rises by $h = 15\ \text{cm}$. Because the liquid is incompressible and the tube has constant cross-section, the liquid level on the left falls by the same $h = 15\ \text{cm}$ from its original position. So the vertical distance between the left and right liquid surfaces is $2h = 30\ \text{cm}$.

2. Identify the key horizontal level for pressure balance.

   Choose the horizontal plane that passes through the interface between oil and liquid on the left side. On the left, above this plane there is only oil (of height $H$, say). On the right, at the same horizontal level, there is only the liquid (of height $2h = 30\ \text{cm}$ above that plane, because the right-side liquid surface is 30 cm above the interface level).

3. Apply pressure equality.

   Pressure at the interface on the left = pressure at the same depth on the right.

Cancel $g$ and substitute densities: $\rho_{\text{oil}} = 0.9\ \text{g/cm}^3$, $\rho_{\text{liquid}} = 1.2\ \text{g/cm}^3$, $h = 15\ \text{cm}$.