by Rajan M V | Nov 10, 2015 | Properties of matter |

**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$.

by Rajan M V | Jun 21, 2015 | Properties of matter |

**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 $2.01013\times 10^{7}\:Pa$.

by Rajan M V | Jun 16, 2015 | Properties of matter |

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

by Rajan M V | Jun 16, 2015 | Properties of matter |

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

From this density of the liquid, $\rho\:=\:$\rho_r \times \rho_w $.

Given that relative density of the liquid $0.9$.

Density of water $ =\: 10^3\:kg\:m^3$.

Therefore density of the liquid, $\rho\:=\:0.9\times \rho_r \times 10^3\:kg\:m^3\:=\:9\times 10^2\:kg\:m^3 $.

Pressure $P$ due to a liquid of density $\rho$ is given by $P_L\:=\:H\:\rho\:g$ where $H$ is the height of the liquid column.

The total pressure at a point inside $P$ is the sum of pressure due to liquid, $P_L$ and atmospheric pressure $P_a$.

Or, $P\:=\:P_L\:+\:P_a$.

Or, $P\:=\: H\:\rho\:g \:+\:P_a$.

Given that atmospheric pressure $=\:1.013\times 10^5\:N\:m^{-2}$.

So, $P\:=\: 1\times \:9\times 10^2\times 9.8 \:+\:1.013\times 10^5$.

Or, $P\:=\: 1.1012\times 10^{+5}\:Nm^{-2}$.

by Rajan M V | Jun 16, 2015 | Properties of matter |

**Problem 2.95: Two narrow tubes of inner diameters $3.0\:mm$ and $6.0\:mm$ are joined together to form U-tube open at both the ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Take density of water$ =\: 10^3\:kg\:m^3$ and surface tension of water $=\:0.0736\: N\:m^{-1}$. Consider angle of contact as zero.**

**Solution:**

In the absence of surface tension, the water level in the two tubes will be the same. This level is indicated by the doted line.

Due to surface tension, water will rise upward in the tubes. The ascent of water is more in the thinner tube of the two.

The height through which water rises is given by, $h\:=\:\frac{4Tcos\theta}{D\rho g}$ where $T$ is the surface tension and $\rho$ is the density of water.

For the first tube, let $h_1$ be the height.

Then, $h_1\:=\:\frac{4Tcos\theta}{D_1\rho g}$.

Using the given values, we get,

** **

$h_1\:=\:\frac{4\times 0.0736\times cos\:0^o }{3\times 10^{-3}\times 10^3\times 9.8}\:=\:10^{-2}\:m\:=\:1\:cm$.

Similarly, for the first tube, let $h_2$ be the height.

Then, $h_2\:=\:\frac{4Tcos\theta}{D_2\rho g}$.

Using the given values, we get,

** **

$h_1\:=\:\frac{4\times 0.0736\times cos\:0^o }{6\times 10^{-3}\times 10^3\times 9.8}\:=\:0.5\times 10^{-2}\:m\:=\:0.5\:cm$.

Difference in height, $h_1\:-\:h_2\:=\:1\:cm\:-\:0.5\:cm\:=\:0.5\:cm$.

**So the water levels in the two tubes will have a difference of five millimeter.**

by Rajan M V | Jun 15, 2015 | Properties of matter |

**Problem 2.94: A liquid bubble $2\:cm$ in radius has an internal pressure of $13\:Pa$. Calculate the surface tension of liquid film.**

**Solution:**

A bubble has two surfaces that come in contact with air. There are molecules on both these surface that produce a tangential force.

Resultant of these tangential forces will act towards the centre of the drop.

Such a net force will increase the inner pressure of the drop. This pressure difference is called excess pressure.

The external pressure in a bubble is given by $P_e\:=\:\frac{4\:T}{R}$ where $T$ is surface tension and $R$ is the radius.

It must be noted that the inner and outer radii are different.

The difference in the two radii is equal to the thickness of the bubble.

For a liquid drop the thickness of the wall is relatively very small.

Hence inner radius = outer radius.

From the above equation, surface tension $T\:=\:\frac{P_e\:R}{4}$.

Radius of the bubble, $R\:=\:2\:cm\:=\:2\times 10^{-2}\:m$.

Excess pressure, $P_e\:=\:13\:P_a\:=\:13\:nm^{-2}$.

Therefore surface tension $T\:=\:\frac{13\times 2\times 10^{-2} }{4}\:=\:0.065\:Nm^{-1}$.