A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume

Q: Why doesn't the mass of the cylinder $M$ appear in the expression for the force on the bottom?

The question asks specifically for the force exerted *by the liquid* on the bottom of the cylinder. This force depends purely on the hydrostatic pressure of the liquid, which is determined by the depth, liquid density, and geometry, completely independent of the material or mass of the suspended cylinder itself.

Q: Is the pressure on the curved bottom surface uniform?

No, the pressure varies with depth along the curved hemispherical surface. However, by using the global force balance ($F_B = F_2 - F_1$), we completely bypass the need to integrate this varying pressure, obtaining the total force $F_2$ directly and elegantly.

Q: Does this formula hold if the cavity was of a different shape?

Yes! As long as the remaining volume of the cylinder is $V$ and the top surface is a flat circle of radius $R$ at depth $h$, the relation $F_2 = \rho g (V + \pi R^2 h)$ remains perfectly valid, regardless of the shape of the bottom cavity.

Q: What would be the tension in the suspending string?

If you analyze the vertical equilibrium of the cylinder, the forces are: upward tension $T$, upward liquid force $F_2$, downward gravity $Mg$, and downward liquid force $F_1$. Thus, $T + F_2 = Mg + F_1$, which simplifies to $T = Mg - F_B = Mg - V\rho g$.

Q: Why did we use gauge pressure instead of absolute pressure?

If we use absolute pressure, we would add atmospheric pressure $P_0$ to both the top and bottom forces. Since $P_0$ acts downwards on the top surface over area $\pi R^2$ and upwards on the bottom surface over the same projected area $\pi R^2$, its net effect cancels out, leaving the liquid force unchanged.

Question

A hemispherical portion of radius $R$ is removed from the bottom of a cylinder of radius $R$. The volume of the remaining cylinder is $V$ and mass $M$. It is suspended by a string in a liquid of density $\rho$, where it stays vertical. The upper surface of the cylinder is at a depth $h$ below the liquid surface. The force on the bottom of the cylinder by the liquid is

Options:

(A) $Mg$

(B) $Mg - V\rho g$

(C) $Mg + \pi R^2 h \rho g$

(D) $\rho g (V + \pi R^2 h)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine holding a beautifully crafted metallic cylinder under water.

Now, imagine that someone has carefully carved out a perfect **hemispherical cavity** of radius $R$ from its bottom.

This leaves us with a solid of remaining volume $V$ and mass $M$.

It is suspended vertically by a thin string, completely submerged in a liquid of density $ho$.

The top surface of this cylinder lies at a depth $h$ below the free surface of the liquid.

Our quest is to find the total upward force exerted by the liquid on the bottom of this cylinder.

At first glance, this might look like a daunting integration problem.

After all, the bottom surface is curved, and pressure varies at every point along its height!

But as Richard Feynman would say, nature always has a simpler, more elegant way of looking at things.

Let's find that elegant path.

### The Master Equation: Archimedes' Principle

Instead of trying to integrate the pressure over the complex curved bottom surface, let's look at the **net effect** of the liquid on the entire cylinder.

According to **Archimedes' Principle**, any object submerged in a fluid experiences an upward buoyant force (or upthrust) equal to the weight of the fluid it displaces.

What is the volume of the liquid displaced by our cylinder?

Since the remaining volume of the cylinder is $V$, it displaces exactly a volume $V$ of the liquid.

Therefore, the net upward buoyant force $F_B$ exerted by the liquid is:

$$F_B = V ho g$$

This buoyant force is not a magic force; it is simply the vector sum of all the hydrostatic pressure forces acting on the outer surfaces of the cylinder.

### Connecting Upthrust to Surface Forces

Let's break down these pressure forces.

Due to horizontal symmetry, the horizontal pressure forces acting on the sides of the cylinder cancel out completely.

Thus, we only need to worry about the vertical forces:

1. The liquid pushes **downward** on the flat top surface with a force $F_1$.

2. The liquid pushes **upward** on the curved bottom surface with a force $F_2$.

Since pressure increases with depth, the upward force $F_2$ must be greater than the downward force $F_1$.

The difference between these two vertical forces is precisely what we call the buoyant force!

$$F_B = F_2 - F_1$$

Rearranging this equation to solve for our target, the upward force on the bottom surface $F_2$, we get:

$$F_2 = F_1 + F_B$$

This is our master key!

To find $F_2$, we only need to calculate the downward force on the flat top surface $F_1$ and add it to the buoyant force $F_B$.

### Calculating the Top Force $F_1$

Calculating the force on the top surface is incredibly straightforward because it is flat and lies at a constant depth $h$.

The gauge pressure $P_1$ at a depth $h$ in a liquid of density $ho$ is:

$$P_1 = ho g h$$

The area of the flat circular top surface of radius $R$ is:

$$A = \pi R^2$$

Therefore, the downward force $F_1$ exerted by the liquid on this top surface is:

$$F_1 = P_1 	imes A = ho g h \pi R^2$$

### Final Calculation and Simplification

Now, let's substitute our expressions for $F_1$ and $F_B$ back into our master key equation:

$$F_2 = F_1 + F_B$$

$$F_2 = (ho g h \pi R^2) + (V ho g)$$

To make this look like the options provided, let's factor out the common term $ho g$:

$$F_2 = ho g (V + \pi R^2 h)$$

This matches **Option (d)** perfectly!

Notice how the complex geometry of the curved bottom surface completely disappeared from our calculations, replaced by the simple and powerful logic of force balance.

This is the beauty of physics—by choosing the right perspective, a seemingly complex problem collapses into a few lines of elegant algebra.

Select Answer:

Visualized Solution

Understanding the Physical Setup

The Force Balance of Submerged Bodies

Upthrust as the Net Pressure Force

Calculating the Downward Force on the Top Surface

Expressing the Upward Force on the Bottom Surface

Simplifying to the Final Form

Alternative Perspective: Equilibrium of the Cylinder

The Sigma Insight: Buoyancy and Archimedes' Principle

Analyzing the Setup

The Master Equation

Connecting Upthrust to Surface Forces

Calculating the Top Force $F_{1}$

Final Calculation and Simplification

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