**MEASUREMENT PHYSICS FORM 5**

Measurement is the process of assigning numbers to a given physical quantity.

**Physical Quantity**

In describing the behavior of objects around us we have to consider to matter, space and time. A moving body covers distance with time and for an object to move energy is required. For the motion to take place, force must be applied.

When an object is in the course of motion and changes its speed within a given time interval we said that it is undergoing acceleration. In all this, we have physical quantities which are measurable and whose values can be used in the mathematical expressions to give a numerical description of the object in a question.

The physical quantities are divided into two categories which are fundamental/basic quantities and derived quantities.

**(a)Fundamental quantities**

These are independent physical quantities such as mass, length, and time. These quantities have both dimensions and standard units which can be expressed dimensionally. The dimensions of mass, length, and time are represented as M, L, and T respectively. The term dimension is used to denote the nature of physical quantity.

**(b)Derived quantities**

The physical quantities which are obtained from fundamental quantities are called derived quantities. An example of derived quantities are such as area, volume, density, speed, and momentum. These quantities can be obtained by combining the fundamental quantities in one way or the **other**. The following are the few examples:

(i) Area =Length Length

[A]= L × L =L²

(ii) Volume =Length × Length × Length

[V] = L × L × L

(iii) Density = Mass/Volume

[ρ] = M/V

(iv)Speed = Distance/Time

[V] = L/T =LT^{-1}

**DIMENSION**

Dimension is the way in which the physical quantities are related to fundamental physical quantities.

**DIMENSIONAL ANALYSIS**

Dimensional analysis is the way of showing how physical quantities are related to each other.

The alphabets used to represent particular unit may be called a symbol. There are various systems in use for the same unit , however the symbol M,L and T are dimensionally used for mass, length and time respectively.

The dimensions of a physical quantity refers to a fundamental units contained in it. Any quantity which can be measured in mass unit only, may said to have the dimension of mass.

The derived units are based on the fundamental quantities and in many cases it involves more than one fundamentals in such case the dimension of such quantity is expressed in general as K(M)^{X}(L)^{Y}(T)^{Z} where K is the pure numeral of x, y and z which indicate how many times a particular unit is involved.

The power to which the fundamental units are raised can be obtained and are called the dimension of the derived unit.

For example the area of a square whose sides are in m **each**.

1mx1m=1m^{2}

The dimension of the area of a square 1m^{2} is;

(L) x(L)=L^{2}; then area has the dimension of length.

The dimension of velocity can be obtained from the definition of velocity which is;

Velocity is the rate of change in displacement. Its unit is meter per second.

The dimension of the velocity V=L/T=LT^{-1}

**USES OF DIMENSION**

Dimensions of physical quantities can be used in the derivation of formula, checking of homogeneity of the formula etc

**Derivation of Formula**

Dimensions are sometimes used as a tool in establishing relationship between physical quantities.

For example through observation one would like to establish the connection between mass (m), its velocity (v) and the work done(w) on it.

**The following are steps to follow:**

Form a statement that: Work is proportional to mass and velocity

i.e. W = k

…………………(i)

**Where k is the proportionality constant**

**Dimensions**

Work = Force × Distance

W = F × S

Where F = ma

[W] = [F] [S] = [m] [a] [s]

= MLT^{-2}L

= ML²T^{-2}

[V] = LT^{-1}

**Substitute the dimension in equation (i)**

M¹L²T^{-2} = kM^{x}L^{y}T^{-z}

**Compare and equate the indices of corresponding dimensions**

For M: x = 1

L: y = 2

T: -y = -2 or = 2

**Substitute for x and y in equation (i)**

W = km¹v² or W = kmv²

This is an empirical expression for the work done to move the body **and** the body acquiring kinetic energy.

Through the experiment or mathematical analysis it can be shown that k = ½** **

**Checking of homogeneity of the formula**

Another area which dimensions can be useful is to check consistence of the equation. An equation with several of terms each with number of variables is consistent if every term has dimensions.

In the process of proving consistency or homogeneity, we are supposed to show that the left hand side of the equation is dimensionally equal to the right hand side of that equation. As an example let us consider the third Newton´s equation of motion

v² = u² + 2as

Where v, u,

and s are final velocity, initial velocity, acceleration and distance respectively. Dimensionally

[v²] = [u²] + 2[a] [s]

We get,

[LT¯¹]² = [LT¯¹]² +2[LT¯²][L]

L²T ¯² = L²T¯² + 2L²T¯²

This shows that each term in the equation above represents dimensions of the square of velocity.

**To convert a physical quantity from one system of units to another**

The value of a physical quantity can be obtained in some other system, when its value in one system is given by using the method of dimensional analysis.

Measurement of a physical quantity is given by X = nu,

u – Size of unit,

n – Numerical value of physical quantity for the chosen unit.

Let u_{1} and u_{2} be units for measurement of a physical quantity in two systems and let n_{1} and n_{2} be the numerical values of physical quantity for two units.

n_{1}u_{1} = n_{2}u_{2}

Let a, b and c be the dimensions of physical quantity in mass, length and time

M_{1}, L1, T1 and M2, L2, T2 are units in two systems of mass, length and time.

This equation is used to find the value of a physical quantity in the second or the new system, when its **value** in the first or the given system is known.

**Determination of units**

At time when solving problem using an expression involves number of variables raised to some powers, the final unity of quantity being calculated may not be immediately recalled.

One way of finding what unit is, is by making use of dimension. For example if unit of gravitation constant g after substituting the numerical values in the formula and evaluate the value, from the basic equation we have;

G = Fr²/Mm

**Dimensionally,**

[G] = [F][r²]/[M][m]

Where, [F] = MLT^{-2}

[r²] = L²

[M] = M

[m] = M

**Therefore**, [G] = MLT^{-2}L²/M²

= M^{-1}L³T²

Since M¯¹ = Kg^{-1} , L³ = m³ and T^{-2} = s^{-2} , the units of G are m³s^{-2}kg^{-1} which can also be written as Nm²kg^{-1}

**Limitations of dimensional analysis**

There are shortcomings in the use of dimensions for the cases we have considered above. In the case of deriving empirical expressions, dimensional method cannot be used to derive trigonometric, logarithmic and exponential formula.

As far as checking homogeneity is concerned, dimension method cannot detect the presence of dimensionless constant in the equation, can not be used to find dimensions of physical quantities with more than three fundamental quantities. It is restricted to only mass, length and time.

**Example 1**

After being deformed and then let free, a drop of liquid vibrate with a frequency which appears to depend on the surface tension

of drop, the density ρ of the liquid and radius r of the drop. By means of dimensions, derive an expression for the frequency of vibration.

**Solution**

Frequency f depends on

(i) Surface tension

(ii) Density

(iii) Radius

These are connected together by the **equation**

f = k

Dimensions [f] = T¯¹, [

] = MT¯², [

] =ML¯³ and [r] = L

Substitute these dimensions in the equation above and simplify to get

M^{}L^{}T^{-1} = k

M^{}L^{â°}T^{-1} = k

Equate the indices of corresponding dimensions

For M: x + y = 0…………….. (i)

L: -3y + z =0……………… (ii)

T: -2x = -1…………………. (ii)

Solving simultaneous equations we have,

x =1/2 , y = -1/2 and z = -1/2

f = k

= k

= k(γ/ρr)^{1/2}

### **Exercise 1**

1. (a) Distinguish between a fundamental physical quantity and derived physical quantity?

(b)What is the dimension of physical quantity?

(c) Write the quantities below in dimension form

(i)The coefficient of viscosity

(ii)The surface tension

(iii)The gravitational constant

2. (a) State the uses and limitations of dimensions

(b)Find out whether or not the equation below is dimensionally homogeneous

V/t = k(Prâ´/

Where k, v, t, p, r, η and L are dimensionless constant, volume, time, pressure, radius, viscosity constant and length respectively

(c) A wave is produced in taut wire by plucking it, the speed of the wave is said to be dependent on tension T of the wire, the mass M and length L. Using this information derive an empirical equation for the speed of the wave in wire.

3. A ball bearing of radius r is released from the surface of the viscous liquid of viscosity constant η in a tall tube. The bearing attains maximum velocity v as it falls through the liquid. Given the friction force opposing the motion of the ball as;

F = k

Find numerical value of x, y, and z

4. (a) A student in examination write s = ut + at² as one of the equations of motion he goes on to check dimensionally its homogeneity and he gets satisfied that he has quoted the write equation. What is the problem with the formula?

(b)The volume of the fluid flowing through a narrow tube per unit time depends on pressure gradient

, viscosity constant η of the fluid and radius r of the tube. Obtain the expression for time rate of flow Q.

5. (a) the periodic time T of a simple pendulum is assumed in the form of

T = k

Where k, m, l and g constant, mass of the bob, length of the thread, and acceleration due to gravity in the same order, Find the numerical values of x, y, and z. Give your comment on the expression.

(b) A force experienced by an object moving in a circle depends on mass m of the object, the velocity v at which it moves and radius r of the circle it describes. By using dimension obtain the expression for the force.

An outstanding share! I have just forwarded this onto a co-worker

who had been doing a little research on this. And he actually ordered me lunch because I discovered it for him…

lol. So allow me to reword this…. Thank YOU for the meal!!

But yeah, thanks for spending the time to discuss this issue here on your

blog.