Download Dimensional formulae of important physical quantities

Fundamental and Derived Units
The units of fundamental physical quantities are called fundamental units. They are length, mass and time. These units
can neither be derived from one another nor can be resolved into any other units. They are independent of one another.
Units of physical quantities can be expressed in terms of fundamental units and such units are called derived units.
Unit of area can be an example for derived unit. If L is the length of square then L x L = L 2 is its area. Similarly, the
volume of a cube is L x L x L = L3 cubic area. Units of any physical quantity can be derived from its defining equation.
The characteristics of several chosen units are:
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they do not vary with place and time
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they are easily reproducible
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they are well-defined and
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they are of proper size i.e., not too small when compared to the quantities to be measured.
Dimensional Formulae and Dimensional Equation
Dimensional equation is obtained when a physical quantity is equated with its
dimensional formula
In general, [X] = [Ma Lb Tc]
RHS represents dimensional formula of physical quantity X, whose dimensions in mass, length and time are a, b and c
respectively.
Dimensional formula
When velocity is defined using the fundamental units of mass, length and time, we have
when there is no mass, M0 = 1 (algebraic theory of indices).
This is the dimensional formula for velocity and we can draw the following inference.
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Unit of velocity depends on the unit of length and time and is independent of mass.
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In the unit of velocity, the power of L and T are 1 and -1 respectively.
E.g., Formula for density is ML-3
Formula for force is MLT-2
Dimensional formulae of important physical quantities
The dimensional formula of a physical quantity can be obtained by defining its relation with other physical quantities and
then expressing these quantities in terms of mass [M], length [L] and time [T].
The dimensional formulae of some of the physical quantities is given below
Advantages of Dimensional Analysis

Dimensional equations are used to validate the correctness of a physical equation.
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Dimensional equations are used to derive correct relationship between different physical quantities.
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Dimensional equations are used to convert one system of units to another.
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Dimensional equations are used to find the dimension of a physical constant.
Limitations of Dimensional Analysis

Dimensional analysis has no information on dimensionless constants.

If a quantity is dependent on trigonometric or exponential functions, this method cannot be used.

In some cases, it is difficult to guess the factors while deriving the relation connecting two or more physical
quantities.
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This method cannot be used in an equation containing two or more variables with same dimensions.

It cannot be used if the physical quantity is dependent on more than three unknown variables.

This method cannot be used if the physical quantity contains more than one term, say sum or difference of two
terms
Solved problems
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A) Convert a velocity of 48 kmh-1 into ms-1
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b) Convert an acceleration of 48 km min-2 into ms-1
a) v = 48kmh-1
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1 km = 1000 m, 1h = 60 x 60 s
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b) a = 48kmmin-2
1 km = 1000 m, 1min = 60 x 60 s
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Absolute Error
Absolute error is the difference between an estimated or measured value and it's actually value. To find out the
absolute error you would have to measurement or estimate of whatever values. At which point you would like to take the
difference between the approximated value and the original value.
Absolute Error formula = Original number - Approximated number
How to find Absolute Error
For example an approximation, we discover that the approximated value may be more or less than the original number.
The difference between two values is called as the absolute error. An approximation of a correct value x, we define the
absolute value of the difference between the two more.
Solved Examples
Question 1: The number of peoples are in a meeting is 730. Find the absolute error
Solution:
Number of peoples= 730
Number of peoples correct to nearest hundred = 700
Formula = Original number - Approximated number
Therefore, Absolute Error = 730 - 700 = 30
The solution is 30.
Question 2: The number of children are in an orphan is 820. Find the absolute error
Solution:
Number of children = 820
Number of children correct to nearest hundred = 800
Formula = Original number - Approximated number
Absolute Error = 820 - 800 = 20
Therefore absolute error is 20
Question 3: 21.571 is the True value, 20.000 is the Recorded Value. Find the absolute error?
Solution:
Formula = (True value) - (Recorded Value)
Absolute error = 21.571 - 20.000 = 1.571
Therefore absolute error = 1.571
Question 4: What is the absolute error for 56 - 0.70?
Solution:
Formula = Original number - Approximated number
Absolute error = 56 - 0.70 = 55.3
Therefore,
Absolute error = 55.3
Question 5: How do you solve this absolute error problem?
Robert measured the weight of a pineapple approximated as 682.325 grams, but the original weight is 684.075
Solution:
Formula = Original number - Approximated number
Absolute error = 684.075-682.325=1.75
So, absolute error = 1.75
Relative Error and Absolute Error
Relative Error:
Relative error is generally defined by dividing absolute error with its true value; it is then represented
in terms of percentage to find out the difference between absolute error and true value.
Relative error which is calculated by dividing the absolute value by the given value, it is calculated as
shown below
Relative error = |V – Vapproximate| / |V| or | (V – Vapproximate ) / V |
Absolute error:
Absolute error is used to identify the exact error that has been occurred in an approximation, This
can be calculated using the formula
Absolute error = Vabsolute = | V – Vapproximate|
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V is some value which we use in the solution.
V – Vapproximate gives us the absolute or exact value (observed - exact value).
Solved Example
Question: Find the relative error when, True value = 200 & Approximate value = 198.5
Solution: Given,
True value = 200
Approximate value = 198.5
Relative error = absolute
= (true
errortrue value
value–approximate value)true value
= (200–198.5)50
= 1.550
Relative error = 0.03
Percentage relative error = 0.03 X 100 = 3 %
Relative Error Formula
The formula for Relative error can be written as,
Relative error = Absoluteerror
Acceptedvalue x 100
As we know absolute error = true value – approximate value
This is also called as relative error equation. Relative error an be calculated with the above given formula.
Percentage Relative Error Problems
Below are some examples based on percentage relative error
Percentage Relative Error
The percent of error of a measurement is the relative error expressed as a percent.
Percent of error can be used to compare different measurements because, being a percent, it
compare each error in terms of 100. Relative error compares the size of the error to the size of the
object being measured. When relative error given as a percent, it referred to as percent error.
Percentage Relative Error = Absolute ErrorTrue Value x 100
Solved Examples
Question 1: True value = 100
Approximate value = 97.5
Solution:
Relative error = absolute
= (true
errortrue value
value–approximate value)true value
= (100–97.5)100
= 2.5100
Relative error = 0.025
Percentage relative error = 0.025 X 100 = 2.5%
Question 2: True value = 30
Approximate value = 29.7
Solution:
Relative error = absolute
= (true
errortrue value
value–approximate value)true value
= (30–29.75)30
= 0.2530
Relative error = 0.0083
Percentage relative error = 0.0083 X 100 = 0.83%
Question 3: True value = 40
Approximate value = 38.65
Solution:
Relative error = absolute
= (true
errortrue value
value–approximate value)true value
= (40–38.65)40
= 1.3540
Relative error = 0.3375
Percentage relative error = 0.3375 X 100 = 33.75%
Propagation of Errors
Final result of an experiment is calculated from a number of observations taken from different instruments, connected
through a formula.
Maximum permissible error in different cases is calculated as follows
Result involving sum of two observed quantities
X is the sum of 2 observed quantities a and b.
X=a+b
Maximum absolute error in X = Maximum absolute error in a + Maximum absolute error in b
Result involving difference of two observed quantities
Suppose X = a - b
Let Da and Db be absolute errors in measurements of quantities a and b, values of a and b and DX be maximum error in
X.
Maximum absolute error in X = Maximum absolute error in a + Maximum absolute error in b
From equations (1) and (2) it is evident that, when result involves sum or difference of 2 observed quantities, absolute
error is the sum of absolute errors in the observed quantities.
Result involving the product of two observed quantities
Suppose X = ab
Let Da and Db be absolute errors in measurements of quantities a and b, values of a and b and DX be the maximum
possible error in X.
Dividing both sides by X = ab, we get
are relative errors of fractional errors in values of a, b and x. Neglecting
product is very small.
The above result is obtained by logarithmic differentiation.
Take log on both sides,
Log X = log a + log b
Differentiating, we get ,
Thus, maximum relative error in X = maximum relative error in a x maximum relative error in b
Maximum absolute error in X = Maximum absolute error in a + Maximum absolute error in b
as its
Result involving quotient of 2 observed quantities
Let Da and Db be absolute errors in measurement of quantities a and b and DX be maximum possible error in X.
Maximum possible relative error in X,
Maximum relative error in X = maximum relative error in a + maximum relative error in b
Maximum percentage error in X,
i.e., Maximum percentage error in X = maximum percentage error in a + maximum percentage in b. From equations 3, 4,
5 and 6, it is seen that when the result involves the multiplication or quotient of 2 observed quantities, the maximum
possible relative error in the result is equal to the sum of the relative errors in the observed quantities.
Result involving product of powers of observed quantities
Relative error in an is n times the relative error a
It can be proved that maximum relative error in X,
Also, maximum percentage error in X,
Maximum percentage error in X = l times maximum percentage error in a + m times maximum percentage error in b + n
times maximum percentage error in c.
Significant Figures
It is the expression of accuracy of a physical quantity. The value in digits, accurately known in a measurement plus one
digit that is not certain, is called significant figures. Significant figures is directly proportional to the accuracy of
measurement. Eg., a reading of 6.34 is accurate to 3 significant figures, while a figure of 6.342 is accurate to 4 significant
figures. Thus, significant figures are all digits about which we are determining significant figures.
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Non-zero digits are all significant.
Eg., 256, 34 has 5 significant figures.
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Between two non-zero digits, all zeros occurring are significant.
Eg., 567.003 has 6 significant figures.
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Zeros to the right of decimal point and to the left of a non-zero digit, are not significant. Eg., 0.002564 has 4
significant digits.
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Zeros to the right of decimal point and to the left of a non-zero digit are significant. .
Eg., In 1995.00 there are 6 significant figures
In 0.019940, there are 5 significant figures.
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Zeros to the right of the last non-zero digit are not significant.
Eg., In 199500, the significant figure is 4.
Rounding off
The omission of the last digit, which is not required for correctness of a physical quantity, with least deviation from its
original value, is called rounding off.
One or two rules has to be followed in rounding off, like
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The rounded off digits should be removed altogether, at once.
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If the digit dropped is less than 5, then the preceding digit remains unchanged.
Eg.,a) When 6.83 m is rounded off to first decimal point then 3 is dropped. As 3<5 and 8 is retained, i.e., 6.83 m when
rounded off, becomes 6.8 m.
b) When 0.009037 m is rounded off to 2 significant figures, we get 0.0090 m.
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If the digit to be dropped is more than 5, then the next digit retained, is increased by one.
Eg., 5.76kg when rounded off to first decimal place, becomes 5.8 kg.
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If the digit to be dropped is exactly 5 and is followed only by zeros, the preceding digit is increased by one if it is
odd and retained as such if it is even.
Eg., When 8.745 kg is rounded off to 3 significant figures, we get 8.74 g.