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## 2.101 Buoyant force on a cylinder

Problem 2.101: A cylinder of radius $R$ is kept embedded along the wall of a container of water of density $\rho$. Length of the cylinder is $L$. Determine the vertical force exerted by water on the cylinder.

Solution:

The cylinder is completely immersed in water. The pressure at the lower point $A$ is greater than that at the upper point $B$.

This difference in pressure creates an upward force on the cylinder which is called the buoyant force.

So the vertical force mentioned in the problem is the buoyant force.

Buoyant force on a body immersed in a fluid is equal to the weight of the liquid displaced.

How much volume of water is displaced by the cylinder?

As the cylinder is fully inside water, volume of water displaced will be equal to the volume of the cylinder.

This will be equal to $\pi\:r^2\:L$.

Mass of this water will be $M\:=\:\pi\:r^2\:L\:\rho$.

Weight of this water will be $Mg\:=\:\pi\:r^2\:L\:\rho\:g$.

So  the vertical force exerted by water on the cylinder is

$\pi\:r^2\:L\:\rho\:g$.

## 2.100 Compressibility of a liquid

Problem 2.100: Find the pressure that must be applied to water at atmospheric pressure to reduce its volume by $1\:%$ .Take compressibility of water $=0.5\times 10^{-9} \:Pa^{-1}$.

Solution:

Compressibility of a liquid is the measure of change in volume under external force.

Gases are more compressible than liquids or solids.

Compressibility $K$ of a liquid is the reciprocal of the bulk modulus $B$.

Then, $K\:=\:\frac{1}{B}$.

Bulk modulus is the ratio of change in pressure to the fractional change in volume.

Or bulk modulus is given by $B\:=\:\frac{\triangle\:P}{\frac{\triangle\:V}{V}}$ where $V$ is the initial volume, $\triangle\:P$ is the change in pressure and $\triangle\:V$ is the change in volume.

Given that change in volume is one percent.

Therefore $\frac{\triangle\:V}{V}\times 100\:=\:1$.

Or, $\frac{\triangle\:V}{V}\:=\:\frac{1}{100}\:=\:0.01$.

It is required to calculate the pressure required.

So let us form an equation for change in pressure $\triangle\:P$.

So, change in pressure $\triangle\:P\:=\:B\times \frac{\triangle\:V}{V}$.

As $B\:=\:\frac{1}{K}$, we can write change in pressure $\triangle\:P\:=\:\frac{\triangle\:V}{V}\times \frac{1}{K}$.

From the given data, compressibility of water $K=0.5\times 10^{-9}$.

Therefore, $\triangle\:P\:=\:0.01\times \frac{1}{0.5\times 10^{-9}}\:=\:0.01\times 2\times 10^{9}\:=\:2\times 10^{7}\:Pa$.

Therefore the required change in pressure is $2\times 10^{7}\:Pa$.

$Change\: in\: pressure \: =\: final\: pressure\: -\: initial \:pressure$.

But water was initially at atmospheric pressure which will be $1.013\times 10^5\:Pa$.

Therefore $final\: pressure\:=\:Change\: in\: pressure\:+\:initial \:pressure$.
So, $final\: pressure\:=\:2\times 10^{7}\:Pa\:+1.013\times 10^5\:Pa\:=\:2.01013\times 10^{7}\:Pa$.

For a change of one percent in volume the pressure of water must be increased to $2.01013\times 10^{7}\:Pa$.

## 2.97 Thermal expansion of a sphere

Problem 2.97:The radius of a metal sphere at room temperature $T$ is $R$, and the coefficient of linear expansion of the metal is $\alpha$ . The sphere is heated a little by a temperature $\triangle\:T$  so that its new temperature is $T\:+\:\triangle\:T$ . Find the approximate increase in the volume of the sphere.

Solution:

Firstly let us assume that the sphere is a uniform one in shape and nature of material.

When temperature is increased, volume also increases.

Coefficient of volume expansion of a body is given by the relation, $\alpha_v\:=\:\frac{1}{V}\frac{\triangle\:V}{\triangle\:T}$......$(1)$

Coefficient of volume expansion $\alpha_v$ of a body is three times the  Coefficient of linear  expansion $\alpha_L$.

Or, $\alpha_v\:=\:3\times \alpha_L$.

So from equation $(1)$, we get $3\times \alpha_L\:=\:\frac{1}{V}\frac{\triangle\:V}{\triangle\:T}$.

From this change in volume,$\triangle\:V\:=\:3V\alpha\:\ \triangle\:T$.

Here $V$ is the initial volume which is equal to $\frac{4}{3}\pi\:R^3$.

So change in volume,$\triangle\:V\:=\:3\times \frac{4}{3}\pi\:R^3\alpha\:\ \triangle\:T$.

Or, $\triangle\:V\:=\:4\:\pi\:R^3\alpha\:\ \triangle\:T$.

## 2.96 Pressure in a liquid

Problem 2.96: An open container has water to a depth of $2\:m$ and above this an oil of relative density $0.9$ for a depth of $1\:m$. Find the pressure at the interface of two liquids. Take density of water $=\: 10^3\:kg\:m^3$, acceleration due to gravity $=\:9.8\:m\:s^{-2}$ and atmospheric pressure $=\:1.013\times 10^5\:N\:m^{-2}$.

Solution:

Pressure at any point inside a liquid is due to the liquid above it and the atmosphere above it.

At the interface pressure is due to the liquid above it. Pressure of water is not experienced at this point.

Relative density $\rho_r$ of a substance is the ratio of its density $\rho$ to the density of water $\rho_w$.

Or, $\rho_r\:=\:\frac{\rho}{\rho_w}$.