01 Units and Measurements
Physical Quantities
All quantities that can be measured are called physical quantities.
eg. time, length, mass, force, work done, etc. In physics we study about
physical quantities and their inter relationships.
MEASUREMENT
Measurement is the comparison of a quantity with a standard of the
same physical quantity.
UNITS
All physical quantities are measured w.r.t. standard magnitude of the
same physical quantity and these standards are called UNITS. eg. second,
meter, kilogram, etc.
So the four basic properties of units are:-
They must be well defined.
They should be easily available and reproducible.
They should be invariable e.g. step as a unit of length is not
invariable.
They should be accepted to all.
SET OF FUNDAMENTAL QUANTITIES
A set of physical quantities which are completely independent of each
other and all other physical quantities can be expressed in terms of
these physical quantities is called Set of Fundamental Quantities.
| Physical Quantity |
Units(SI) |
Units(CGS) |
Notations |
| Mass |
kg(kilogram) |
g |
M |
| Length |
m(meter) |
cm |
L |
| Time |
s (second) |
s |
T |
| Temperature |
K (kelvin) |
\[\ ^{\circ}C\] |
\[\theta\] |
| Current |
A (ampere) |
A |
I or A |
| Luminous intensity |
cd (candela) |
- |
cd |
| Amount of substance |
mol |
- |
mol |
Physical Quantity (SI Unit)
Length (m)
Mass (kg)
Time(s)
Definition
The distance travelled by light in vacuum in \(\frac{1}{299,792,458}\) second is called 1
metre.
The mass of a cylinder made of platinum-iridium alloy kept at
International Bureau of Weights and Measures is defined as 1
kilogram.
The second is the duration of 9,192,631,770 periods of
the radiation corresponding to the transition between the two hyperfine
levels of the ground state of the cesium133 atom.
If equal currents are maintained in the two parallel infinitely long
wires of negligible cross-section, so that the force between them is
\(2 \times 10^{- 7}\) newton per metre
of the wires, the current in any of the wires is called 1 Ampere.
The fraction \(\frac{1}{273.16}\) of
the thermodynamic temperature of triple point of water is called 1
Kelvin 1 candela is the luminous intensity of a blackbody of surface
area \(\frac{1}{600,000}{\text{
}m}^{2}\) placed at the temperature of freezing platinum and at a
pressure of \(101,325\text{ }N/m^{2}\),
in the direction perpendicular to its surface.
The mole is the amount of a substance that contains as many elementary
entities as there are number of atoms in 0.012 kg of carbon- 12 .
There are two supplementary units too:
Plane angle (radian) angle \(=\)
arc \(/\) radius \(\theta = \mathcal{l}/r\)
Solid Angle (steradian)
DERIVED PHYSICAL QUANTITIES
The physical quantities those can be expressed in terms of
fundamental physical quantities are called derived physical quantities.eg. speed \(=\) distance/time.
DIMENSIONS AND DIMENSIONAL FORMULA
All the physical quantities of interest can be derived from the base
quantities.
DIMENSION
The power (exponent) of base quantity that enters into the expression
of a physical quantity, is called the dimension of the quantity in that
base.
To make it clear, consider the physical quantity "force".
Force \(=\) mass \(\times\) acceleration
\[\begin{matrix}
= \text{~}\text{mass}\text{~} \times
\frac{\text{~}\text{length}\text{~}/\text{~}\text{time}\text{~}}{\text{~}\text{time}\text{~}}
\\
= \text{~}\text{mass}\text{~} \times \text{~}\text{length}\text{~}
\times (\text{~}\text{time}\text{~})^{- 2}
\end{matrix}\]
So the dimensions of force are 1 in mass, 1 in length and -2 in time.
Thus
\[\lbrack\text{~}\text{Force}\text{~}\rbrack =
{\text{~}\text{MLT}\text{~}}^{- 2}\]
Similarly energy has dimensional formula given by
\[\lbrack\text{~}\text{Energy}\text{~}\rbrack =
{ML}^{2}{\text{ }T}^{- 2}\]
i.e. energy has dimensions, 1 in mass, 2 in length and -2 in
time.
Such an expression for a physical quantity in terms of base quantities
is called dimensional formula.
DIMENSIONAL EQUATION
Whenever the dimension of a physical quantity is equated with its
dimensional formula, we get a dimensional equation.
PRINCIPLE OF HOMOGENEITY
According to this principle, we can multiply physical quantities with
same or different dimensional formulae at our convenience, however no
such rule applies to addition and subtraction, where only like physical
quantities can only be added or subtracted. e.g. If \(P + Q \Rightarrow P\) & Q both
represent same physical quantity.
Example :
Calculate the dimensional formula of energy from the equation \(E = \frac{1}{2}mv^{2}\).
Sol. Dimensionally, \(E =\) mass \(\times
(\text{~}\text{velocity}\text{~})^{2}\).
Since \(\frac{1}{2}\) is a number and
has no dimension.
or, \(\lbrack E\rbrack = M \times \left(
\frac{L}{T} \right)^{2} = ML^{2}T^{- 2}\).
Example :
Kinetic energy of a particle moving along elliptical trajectory is
given by \(K = \alpha s^{2}\) where
\(s\) is the distance travelled by the
particle. Determine dimensions of \(\alpha\).
Sol. \(K = \alpha s^{2}\)
\[{\lbrack\alpha\rbrack = \frac{\left(
{ML}^{2}{\text{ }T}^{- 2} \right)}{\left( L^{2} \right)}
}{\lbrack\alpha\rbrack = M^{l}L^{0}T^{- 2}
}{\lbrack\alpha\rbrack = \left( MT^{- 2} \right)}\]
Example :
The position of a particle at time \(t\), is given by the equation, \(x(t) = \frac{v_{0}}{\alpha}\left( I - e^{- at}
\right)\), where \(v_{0}\) is a
constant and \(\alpha > 0\). The
dimensions of \(v_{0}\&\alpha\) are
respectively.
(A) \(M^{0}L^{l}T^{0}\& T^{-
l}\)
(B) \(M^{0}L^{l}T^{- l}\& T\)
\[\left( C^{*} \right)M^{0}L^{l}T^{- l}\&
T^{- l}
\](D) \(M^{l}L^{l}T^{- l}\& LT^{-
2}\)
Sol.
\[{\left\lbrack V_{0} \right\rbrack = \lbrack
x\rbrack\lbrack\alpha\rbrack
}{\&\lbrack\alpha\rbrack\lbrack t\rbrack = M^{0}L^{0}T^{0}
}{= M^{0}L^{l}T^{- l}
}{\lbrack\alpha\rbrack = M^{0}L^{0}T^{- l}}\]
Example :
The distance covered by a particle in time \(t\) is going by \(x = a + bt + ct^{2} + dt^{3}\); find the
dimensions of \(a,b,c\) and \(d\).
Sol. The equation contains five terms. All of them should have the
same dimensions. Since \(\lbrack x\rbrack
=\) length, each of the remaining four must have the dimension of
length.
Thus, \(\lbrack a\rbrack =\) length
\(= L\)
\[\begin{matrix}
\lbrack bt\rbrack = L, & \text{~}\text{or}\text{~} & \lbrack
b\rbrack = LT^{- 1} \\
\left\lbrack ct^{2} \right\rbrack = L, & \text{~}\text{or}\text{~}
& \lbrack c\rbrack = LT^{- 2}
\end{matrix}\]
and \(\ \left\lbrack dt^{3} \right\rbrack =
L\ \) or \(\ \lbrack d\rbrack = LT^{-
3}\)
Uses Of Dimensional Analysis
(I) To Convert Units of a Physical Quantity From One System
Of Units To Another:
It is based on the fact that,
Numerical value \(\times\) unit \(=\) constant
So on changing unit, numerical value will also gets changed. If \(n_{1}\) and \(n_{2}\) are the numerical values of a given
physical quantity and \(u_{1}\) and
\(u_{2}\) be the units respectively in
two different systems of units, then
\[\begin{matrix}
n_{1}u_{1} = n_{2}u_{2} \\
n_{2} = n_{1}\left\lbrack \frac{M_{1}}{M_{2}} \right\rbrack\left\lbrack
\frac{L_{1}}{{\text{ }L}_{2}} \right\rbrack^{b}\left\lbrack
\frac{{\text{ }T}_{1}}{{\text{ }T}_{2}} \right\rbrack^{c}
\end{matrix}\]
Example
Young's modulus of steel is \(19 \times
10^{I0}\text{ }N/m^{2}\). Express it in dyne/cm \(\ ^{2}\). Here dyne isthe CGS unit of
force.
Sol. The unit of Young's modulus is \(N/m^{2}\).
This suggest that it has dimensions of \(\frac{\text{~}\text{Force}\text{~}}{(\text{~}\text{dis}\text{~}tance)^{2}}\).
Thus, \(\lbrack Y\rbrack = \frac{\lbrack
F\rbrack}{L^{2}} = \frac{{MLT}^{- 2}}{{\text{ }L}^{2}} = ML^{- l}T^{-
2}\).
\(N/m^{2}\) is in SI units,
So, \(\ 1\text{ }N/m^{2} = (1\text{
}kg)(1\text{ }m)^{- 1}(1\text{ }s)^{- 2}\)
and 1 dyne \(/{cm}^{2} = (1\text{ }g)(1\text{
}cm)^{- 1}(1\text{ }s)^{- 2}\)
so, \(\frac{1\text{
}N/m^{2}}{1\text{~}\text{dyne}\text{~}/{cm}^{2}} = \left( \frac{1\text{
}kg}{1\text{ }g} \right)\left( \frac{1\text{ }m}{1\text{ }cm} \right)^{-
1}\left( \frac{1\text{ }s}{1\text{ }s} \right)^{- 2} = 1000 \times
\frac{1}{100} \times 1 = 10\)
or, \(\ 1\text{ }N/m^{2} = 10\) dyne
\(/{cm}^{2}\)
or, \(\ 19 \times 10^{10}\text{ }N/m^{2} = 19
\times 10^{11}\) dyne \(/m^{2}\).
Example :
The dimensional formula for viscosity of fluids is,
\[\eta = M^{l}L^{- l}T^{- l}\]
Find how many poise (CGS unit of viscosity) is equal to 1 poiseuille
(SI unit of viscosity).
Sol. \(\ \eta = M^{l}L^{- l}T^{-
l}\)
1 CGS units \(= g{cm}^{- 1}{\text{ }s}^{-
I}\)
\[\begin{matrix}
1\text{~}\text{SI units}\text{~} & \ = kg{\text{ }m}^{- 1}{\text{
}s}^{- 1} \\
& \ = 1000\text{ }g(100\text{ }cm)^{- 1}{\text{ }s}^{- 1} \\
& \ = 10{gcm}^{- 1}{\text{ }s}^{- 1}
\end{matrix}\]
Thus, 1 Poiseuilli = 10 poise
Example :
A calorie is a unit of heat or energy and it equals about 4.2J,
where \(1\text{ }J = 1\text{ }kg\text{
}m/s^{2}\). Suppose we employ a system of units in which the unit
of mass equals \(\alpha kg\), the unit
of length equals \(\beta\) metre, the
unit of time is \(\gamma\) second. Show
that a calorie has a magnitude \(4.2\alpha^{-
1}\beta^{- 2}\gamma^{2}\) in terms of the new units.
Sol. \(\ 1cal = 4.2\text{ }kg{\text{
}m}^{2}{\text{ }s}^{- 2}\)
| \[SI\] |
New system |
| \[n_{l} = 4.2\] |
\[n_{2} = ?\] |
| \[M_{l} = 1\text{ }kg\] |
\[M_{2} = \alpha kg\] |
| \[L_{l} = 1\text{ }m\] |
\(L_{2} = \beta\) metre |
| \[T_{l} = 1\text{ }s\] |
\(T_{2} = \gamma\) second |
Dimensional formula of energy is [ \(ML^{2}T^{- 2}\) ]
Comparing with \(\left\lbrack M^{a}L^{b}T^{c}
\right\rbrack\), we find that \(a = 1,b
= 2,c = - 2\)
Now, \(n_{2} = n_{1}\left\lbrack
\frac{M_{1}}{M_{2}} \right\rbrack^{a}\left\lbrack \frac{L_{1}}{{\text{
}L}_{2}} \right\rbrack^{b}\left\lbrack \frac{T_{1}}{{\text{ }T}_{2}}
\right\rbrack^{c}\)
\[= 4.2\left\lbrack \frac{1\text{
}kg}{\alpha\text{ }kg} \right\rbrack^{1}\left\lbrack \frac{1\text{
}m}{\beta\text{ }m} \right\rbrack^{2}\left\lbrack \frac{1\text{
}s}{\gamma\text{ }s} \right\rbrack^{- 2} = 4.2\alpha^{- 1}\beta^{-
2}\gamma^{2}\]
(ii) To Check The Dimensional Correctness Of A Given Physical
Relation:
It is based on principle of homogeneity, which states that a given
physical relation is dimensionally correct if the dimensions of the
various terms on either side of the relation are the same.
(i) Powers are dimensionless
(ii) \(sin\theta,e^{\theta},cos\theta,log\theta\)
gives dimensionless value and in above expression \(\theta\) is dimensionless
(iii) We can add or subtract quantity having same dimensions.
Example :
Let us check the dimensional correctness of the relation \(v = u + at\).
Here 'u'represents the initial velocity, 'v'represents the final
velocity, 'a'the uniform acceleration and ' \(t\) ' the time.
Dimensional formula of ' \(u\) ' is
\(\left\lbrack M^{0}LT^{- I}
\right\rbrack\)
Dimensional formula of ' \(v\) ' is
\(\left\lbrack M^{0}LT^{- 1}
\right\rbrack\)
Dimensional formula of 'at' is \(\left\lbrack
M^{0}LT^{- 2} \right\rbrack\lbrack T\rbrack = \left\lbrack M^{0}LT^{- 1}
\right\rbrack\)
Here dimensions of every term in the given physical relation are the
same, hence the given physical relation is dimensionally correct.
Example :
Let us check the dimensional correctness of the relation
\[x = ut + \frac{1}{2}at^{2}\]
Here ' \(u\) ' represents the
initial velocity, ' \(a\) ' the uniform
acceleration, ' \(x\) ' the
displacement and ' \(t\) ' the
time.
Sol. \(\ \lbrack x\rbrack = L\)
\[\begin{matrix}
& \lbrack ut\rbrack = \text{~}\text{velocity}\text{~} \times
\text{~}\text{time}\text{~} =
\frac{\text{~}\text{length}\text{~}}{\text{~}\text{time}\text{~}} \times
\text{~}\text{time}\text{~} = L \\
& \left\lbrack \frac{1}{2}at^{2} \right\rbrack = \left\lbrack at^{2}
\right\rbrack = \text{~}\text{accelecration}\text{~} \times
(\text{~}\text{time}\text{~})^{2}
\end{matrix}\]
( \(\therefore\frac{1}{2}\) is a
number hence dimentionless)
\[=
\frac{\text{~}\text{velocity}\text{~}}{\text{~}\text{time}\text{~}}
\times (\text{~}\text{time}\text{~})^{2} =
\frac{\text{~}\text{length/tim}\text{~}e}{\text{~}\text{time}\text{~}}
\times (\text{~}\text{time}\text{~})^{2} = L\]
Thus, the equation is correct as far as the dimensions are
concerned.
(III) To Establish A Relation Between Different Physical
Quantities:
If we know the various factors on which a physical quantity depends,
then we can find a relation among different factors by using principle
of homogeneity.
Example :
Let us find an expression for the time period \(t\) of a simple pendulum. The time period
\(t\) may depend upon (i) mass \(m\) of the bob of the pendulum, (ii) length
\(\mathcal{l}\) of pendulum, (iii)
acceleration due to gravity \(g\) at
the place where the pendulum is suspended.
Sol. Let (i) \(t \propto
m^{a}\)
(ii) \(t \propto
\mathcal{l}^{b}\)
(iii) \(t \propto g^{c}\)
Combining all the three factors, we get
\[t \propto {\text{
}m}^{a}\mathcal{l}^{b}{\text{ }g}^{c}\ \text{~}\text{or}\text{~}\ t =
{Km}^{a}\mathcal{l}^{b}{\text{ }g}^{c}\]
where \(K\) is a dimensionless
constant of proportionality.
Writing down the dimensions on either side of equation (i), we get
\[\lbrack T\rbrack = \left\lbrack M^{a}
\right\rbrack\left\lbrack L^{b} \right\rbrack\left\lbrack LT^{- 2}
\right\rbrack^{c} = \left\lbrack M^{a}L^{b + c}T^{- 2c}
\right\rbrack\]
Comparing dimensions, \(a = 0,b + c = 0, -
2c = 1\)
\[\therefore\ a = 0,c = - 1/2,b = 1/2
\]From equation (i) \(t =
Km^{0}\mathcal{l}^{/2}g^{- 1/2}\ \) or \(\ t = K\left( \frac{\mathcal{l}}{g} \right)^{1/2}
= K\sqrt{\frac{\mathcal{l}}{g}}\)
Example :
When a solid sphere moves through a liquid, the liquid opposes the
motion with a force \(F\). The
magnitude of \(F\) depends on the
coefficient of viscosity \(\eta\) of
the liquid, the radius \(r\) of the
sphere and the speed vof the sphere. Assuming that \(F\) is proportional to different powers of
these quantities, guess a formula for \(F\) using the method of dimensions.
Sol. Suppose the formula is \(F =
k\eta^{a}r^{b}v^{c}\)
Then, \(MLT^{- 2} = \left\lbrack ML^{- 1}T^{-
1} \right\rbrack^{a}L^{b}\left( \frac{\text{ }L}{\text{ }T}
\right)^{c}\)
\[= M^{a}L^{- a + b + c}T^{- a -
c}\]
Equating the exponents of \(M,L\)
and \(T\) from both sides,
\[\begin{matrix}
a = 1 \\
- a + b + c = 1 \\
- a - c = - 2
\end{matrix}\]
Solving these, \(a = 1,b = 1\) and
\(c = 1\)
Thus, the formula for \(F\) is \(F = k\eta rv\).
Example :
If \(P\) is the pressure of a gas
and \(\rho\) is its density, then find
the dimension of velocity in terms of \(P\) and \(\rho\).
(A) \(P^{l/2}\rho^{- l/2}\)
(B) \(P^{l/2}\rho^{l/2}\)
(C) \(P^{- 1/2}\rho^{1/2}\)
(D) \(P^{- 1/2}\rho^{- 1/2}\)
[Sol. \(v \propto P^{a}\rho^{b}\)
\[v = kP^{a}\rho^{b}
\]\(\left\lbrack LT^{- 1} \right\rbrack
= \left\lbrack ML^{- 1}T^{- 2} \right\rbrack^{a}\left\lbrack ML^{- 3}
\right\rbrack^{b}\) (Comparing dimensions)
\[a = \frac{1}{2},b = -
\frac{1}{2}\ \Rightarrow \ \lbrack V\rbrack\ = \left\lbrack
P^{l/2}\rho^{- l/2} \right\rbrack\]
Units And
Dimensions Of Some Physical Quantities
| Quantity |
SI Unit |
Dimensional Formula |
| Density |
\[kg/m^{3}\] |
M/L \(\ ^{3}\) |
| Force |
Newton (N) |
ML/T \(\ ^{2}\) |
| Work |
Joule (J)(=N-m) |
\[{ML}^{2}/T^{2}\] |
| Energy |
Joule(J) |
\[{ML}^{2}/T^{2}\] |
| Power |
Watt (W) (=J/s) |
\[{ML}^{2}/T^{3}\] |
| Momentum |
\[kg - m/s\] |
ML/T |
| Gravitational constant |
\[N - m^{2}/{kg}^{2}\] |
\[L^{3}/{MT}^{2}\] |
| Angular velocity |
radian/s |
\[T^{- 1}\] |
| Angular acceleration |
radian \(/s^{2}\) |
\[T^{- 2}\] |
| Angular momentum |
\[kg - m^{2}/s\] |
\[{ML}^{2}/T\] |
| Moment of inertia |
\[kg - m^{2}\] |
ML \(\ ^{2}\) |
| Torque |
\[N - m\] |
\[{ML}^{2}/T^{2}\] |
| Angular frequency |
radian/s |
\[T^{- 1}\] |
| Frequency |
Hertz(Hz) |
\[T^{- 1}\] |
| Period |
s |
T |
| Surface Tension |
N/m |
\[M/T^{2}\] |
| Coefficient of viscosity |
\[N - s/m^{2}\] |
M/LT |
| Wavelength |
m |
L |
| Intensity of wave |
\[W/m^{2}\] |
M/T \(\ ^{3}\) |
| Temperature |
kelvin(K) |
K |
| Specific heat capacity |
J/(kg-K) |
\[L^{2}/T^{2}\text{ }K\] |
| Stefan's constant |
W/(m2-K4) |
\[M/T^{3}{\text{ }K}^{4}\] |
| Heat |
J |
\[{ML}^{2}/T^{2}\] |
| Thermal conductivity |
W/(m-K) |
ML/T \(\ ^{3}\text{ }K\) |
| Current density |
\[A/m^{2}\] |
\[I/L^{2}\] |
| Electrical conductivity |
\[1/\Omega - m( = mho/m)\] |
\[I^{2}{\text{
}T}^{3}/{ML}^{3}\] |
| Electric dipole moment |
C-m |
LIT |
| Electric field |
V/m(=N/C) |
ML/IT \(\ ^{3}\) |
| Potential (voltage) |
volt (V) (=J/C) |
\[{ML}^{2}/{IT}^{3}\] |
| Electric flux |
V-m |
\[{ML}^{3}/{IT}^{3}\] |
| Capacitance |
farad (F) |
\[I^{2}{\text{
}T}^{4}/{ML}^{2}\] |
| Electromotive force |
volt (V) |
ML \(\ ^{2}/{IT}^{3}\) |
| Resistance |
ohm( \(\Omega\) ) |
\[{ML}^{2}/I^{2}{\text{
}T}^{3}\] |
| Permittivity of space |
\[C^{2}/N - m^{2}( = F/m)\] |
\[I^{2}{\text{
}T}^{4}/{ML}^{3}\] |
| Permeability of space |
N/A \(\ ^{2}\) |
ML/I \(I^{2}{\text{ }T}^{2}\) |
| Magnetic field |
Tesla \((T)\left( = Wb/m^{2}
\right)\) |
\[M/{IT}^{2}\] |
| Magnetic flux |
Weber (Wb) |
\[{ML}^{2}/{IT}^{2}\] |
| Magnetic dipole moment |
\[N - m/T\] |
IL \(\ ^{2}\) |
| Inductance |
Henry (H) |
\[{ML}^{2}/I^{2}{\text{
}T}^{2}\] |
Limitations Of Dimensional
Analysis
(i) Dimension does not depend on the magnitude. Due to this reason
the equation \(x = ut + {at}^{2}\) is
also dimensionally correct. Thus, a dimensionally correct equation need
not be actually correct.
(ii) The numerical constants having no dimensions connot be deduced by
the method of dimensions.
(iii) This method is applicable only if relation is of product type. It
fails in the case of exponential and trigonometric relations.
SI Prefixes: The magnitudes of physical quantites vary over a wide
range. The mass of an electron is \(9.1 \times
10^{- 31}\text{ }kg\) and that of our earth is about \(6 \times 10^{24}\text{ }kg\). Standard
prefixes for certain power of 10 . Table shows these prefixes :
| Power of 10 |
Prefix |
Symbol |
| 12 |
tera |
T |
| 9 |
giga |
G |
| 6 |
mega |
M |
| 3 |
kilo |
k |
| 2 |
hecto |
h |
| 1 |
deka |
da |
| -1 |
deci |
d |
| -2 |
centi |
c |
| -3 |
milli |
m |
| -6 |
micro |
\[\mu\] |
| -9 |
nano |
n |
| -12 |
pico |
p |
| -15 |
femto |
f |
Order-Of Magnitude
Calculations
If value of phycal quantity P satisfy
\[0.5 \times 10^{x} < P \leq 5 \times
10^{x}
\]x is an integer
x is called order of magnitude
Example :
The diameter of the sun is expressed as \(13.9 \times 10^{9}\text{ }m\). Find the
order of magnitude of the diameter?
Sol. \(\ \) Diameter \(= 13.9 \times 10^{9}\text{ }m\)
Diameter \(= 1.39 \times 10^{10}\text{
}m\)
order of magnitude is 10 .
Symbols And There Usual
Meanings
The scientific group in Greece used following symbols.
| \[\theta\] |
Theta |
| \[\alpha\] |
Alpha |
| \[\beta\] |
Beta |
| \[\gamma\] |
Gamma |
| \[\delta\] |
Delta |
| \[\Delta\] |
Delta |
| \[\mu\] |
Mu |
| \[\lambda\] |
Lambda |
| \[\omega,\Omega\] |
Omega |
| \[\pi\] |
Pi |
| \[\phi,\varphi\] |
Phi |
| \[\varepsilon\] |
epsilon |
| \[\psi\] |
Psi |
| \[\rho\] |
Roh |
| \[\nu\] |
Nu |
| \[\eta\] |
Eta |
| \[\sigma\] |
Sigma |
| \[\tau\] |
Tau |
| \[\kappa\] |
Kappa |
| \[\chi\] |
chi |
| \[\cong\] |
Approximately
equal to |
Solved Examples
Q. 1 Find the dimensional formulae of follwoing quantities :
(a) The surface tension S ,
(b) The thermal conductivity k and
(c) The coefficient of vescosity \(\eta\).
Some equation involving these quntities are
\(S = \frac{\rho grh}{2}\ Q = k\frac{A\left(
\theta_{2} - \theta_{1} \right)t}{d}\ \) and \(\ F = - \eta A\frac{v_{2} - v_{l}}{x_{2} -
x_{l}};\)
where the symbols have their usual meanings. ( \(\rho\) - density, g - acceleration due to
gravity, r - radius, h height, A - area, \(\theta_{1}\&\theta_{2}\) -
temperatures, t - time, d - thickness, \(v_{1}\& v_{2}\) - velocities, \(x_{1}\& x_{2}\) - positions.)
Sol. (a) \(S = \frac{\rho
grh}{2}\)
or \(\lbrack S\rbrack =
\lbrack\rho\rbrack\lbrack g\rbrack L^{2} = \frac{M}{L^{2}} \cdot
\frac{L}{T^{2}} \cdot {\text{ }L}^{2} = {MT}^{- 2}\).
(b) \(Q = k\frac{A\left( \theta_{2} -
\theta_{l} \right)t}{d}\)
or \(k = \frac{Qd}{A\left( \theta_{2} -
\theta_{l} \right)t}\).
Here, Q is the heat energy having dimension \({ML}^{2}{\text{ }T}^{- 2},\theta_{2} -
\theta_{1}\) is temperature, A is area, d is thickness and t is
time. Thus,
\(\lbrack k\rbrack = \frac{ML^{2}T^{-
2}}{L^{2}KT} = {MLT}^{- 3}{\text{ }K}^{- 1}\).
(d) \(F = - hA\frac{v_{2} - v_{1}}{x_{2} -
x_{1}}\)
or \({MLT}^{- 2} = \lbrack\eta\rbrack
L^{2}\frac{L/T}{L} = \lbrack\eta\rbrack\frac{L^{2}}{T}\)
or, \(\lbrack\eta\rbrack = {ML}^{- 1}{\text{
}T}^{- 1}\).
Q. 2 Suppose \(A = B^{n}C^{m}\), where
A has dimensions \(LT,B\) has
dimensions \(L^{2}{\text{ }T}^{- 1}\),
and C has dimensions \({LT}^{2}\). Then
the exponents \(n\) and \(m\) have the values:
(A) \(2/3;1/3\)
(B) \(2;3\)
(C) \(4/5; - 1/5\)
(D*) \(1/5;3/5\)
(E) \(1/2;1/2\)
Sol. \(\ LT = \left\lbrack L^{2}{\text{
}T}^{- 1} \right\rbrack^{n}\left\lbrack {LT}^{2}
\right\rbrack^{m}\)
\[LT = L^{2n + m}T^{2\text{ }m -
n}\]
\[\begin{array}{r}
2n + m = 1\#(i) \\
\ - n + 2m = 1\#(ii)
\end{array}\]
On sovling \(\ n = \frac{1}{5},\text{ }m =
\frac{3}{5}\)
Q. 3 If energy(E), velocity(V) and time (T) are chosen as the
fundamental quantities, then the dimensions of surface tension will be.
(Surface tension \(=\) force \(/\) length
(A) \(EV^{- 2}{\text{ }T}^{- 1}\)
(B) \({EV}^{- 1}{\text{ }T}^{-
2}\)
(C) \(E^{- 2}{\text{ }V}^{- 1}{\text{ }T}^{-
3}\)
(D*) \({EV}^{- 2}{\text{ }T}^{-
2}\)
Sol. [surface tension] \(= \lbrack\)
force/length \(\rbrack = M^{1}{\text{
}L}^{0}{\text{ }T}^{- 2}\)
suppose [surface tension] \(=
E^{a}V^{b}T^{c}\)
\[\therefore M^{1}{\text{ }L}^{0}{\text{
}T}^{- 2} = \left\lbrack M^{1}{\text{ }L}^{2}{\text{ }T}^{- 2}
\right\rbrack^{a}\left\lbrack L^{1}{\text{ }T}^{- 1}
\right\rbrack^{b}\lbrack T\rbrack^{c}
\]Matching dimensions of \(M
\Rightarrow a = 1\)
Matching dimensions of \(L \Rightarrow 2a + b
= 0 \Rightarrow \text{ }b = - 2\)
Matching dimensions of \(T \Rightarrow - 2a -
b + c = - 2 \Rightarrow c = - 2\)
\(\therefore\) [surface tension] \(= {EV}^{- 2}{\text{ }T}^{- 2}\)
Q. 4 Given that \(ln(\alpha/p\beta) = \alpha
z/K_{B}\theta\) where \(p\) is
pressure, \(z\) is distance, \(K_{B}\) is Boltzmann constant and \(\theta\) is temperature, the dimensions of
\(\beta\) are (useful formula Energy
\(= K_{B} \times\) temperature)
(A) \(L^{0}M^{0}{\text{ }T}^{0}\)
(B) \(L^{1}M^{- 1}{\text{
}T}^{2}\)
(C*) \(L^{2}M^{0}{\text{
}T}^{0}\)
(D) \(L^{- 1}M^{1}{\text{ }T}^{-
2}\)
Sol. \(\ \mathcal{l}n\left(
\frac{\alpha}{p\beta} \right) = \frac{\alpha
z}{k_{B}\theta}\)
\[\begin{matrix}
\lbrack\alpha z\rbrack = \left\lbrack k_{\beta}\theta \right\rbrack
& \text{~}\text{Also}\text{~}\lbrack\alpha\rbrack = \lbrack
p\beta\rbrack \\
\lbrack p\beta z\rbrack = \left\lbrack k_{\beta}\theta \right\rbrack
& \\
& \lbrack\beta\rbrack = \frac{\left( k_{\beta}\theta \right)}{(pz)}
= \frac{ML^{2}T^{- 2}K^{- I}K}{ML^{- l}T^{- 2}L} = L^{2}
\end{matrix}\]
Q. 5 The SI and CGS units of energy are joule and erg respectively.
How many ergs are equal to one joule ?
Sol. Dimensionally, Energy \(=\)
mass \(\times
(\text{~}\text{velocity}\text{~})^{2}\)
\[= \text{~}\text{mass}\text{~} \times
\left( \frac{\text{~}\text{length}\text{~}}{\text{~}\text{time}\text{~}}
\right)^{2} = {ML}^{2}{\text{ }T}^{- 2}\]
Thus, 1 joule \(= (1\text{ }kg)(1\text{
}m)^{2}(1\text{ }s)^{- 2}\)
and \(1erg = (1\text{ }g)(1\text{
}cm)^{2}(1\text{ }s)^{- 2}\)
\[\frac{1\text{~}\text{joule}\text{~}}{1\text{~}\text{erg}\text{~}}
= \left( \frac{1\text{ }kg}{1\text{ }g} \right)\left( \frac{1\text{
}m}{1\text{ }cm} \right)^{2}\left( \frac{1\text{ }s}{1\text{ }s}
\right)^{- 2}
\]\(= \left( \frac{1000\text{
}g}{1\text{ }g} \right)\left( \frac{1000\text{ }cm}{1\text{ }cm}
\right)^{2} = 1000 \times 10000 = 10^{7}\).
So 1 joule \(= 10^{7}\) erg.
Q. 6 Young's modulus of steel is \(19 \times
10^{10}\text{ }N/m^{2}\). Express it in dyne \(/{cm}^{2}\). Here dyne is the CGS unit of
force.
Sol. The unit of Young's modulus is \(N/m^{2}\).
This suggests that it has dimensions of \(\frac{\text{~}\text{Force}\text{~}}{\text{~}\text{(area)}\text{~}}\).
Thus, \(\lbrack Y\rbrack = \left\lbrack
\frac{F}{L^{2}} \right\rbrack = \frac{MLT^{- 2}}{L^{2}} = {ML}^{-
1}{\text{ }T}^{- 2}\).
\(N/s^{2}\) is in SI units.
So, \(1\text{ }N/m^{2} = (1\text{ }kg)(1\text{
}m)^{- 1}(1\text{ }s)^{- 2}\)
and 1 dyne \(/{cm}^{2} = (1\text{ }g)(1\text{
}cm)^{- 1}(1\text{ }s)^{- 2}\)
So, \(\frac{1\text{
}N/m^{2}}{1\text{~}\text{dyne}\text{~}/{cm}^{2}} = \left( \frac{1\text{
}kg}{1\text{ }g} \right) = \left( \frac{1\text{ }m}{1\text{ }cm}
\right)^{- 1}\left( \frac{1\text{ }s}{1\text{ }s} \right)^{-
2}\)
\[= 1000 \times \frac{1}{100} \times 1 =
10\]
or, \(\ 1\text{ }N/s^{2} = 10\) dyne
\(/{cm}^{2}\)
or, \(19 \times 10^{10}\text{ }N/m^{2} = 19
\times 10^{11}\) dyne \(/{cm}^{2}\)
Q. 7 If velcoity, time and force were chosen as basic quantities, find
the dimensions of mass.
Sol. Dimensionally,Force \(=\) mass
\(\times\) acceleration
\[\text{~}\text{Force}\text{~} =
\text{~}\text{mass}\text{~} \times
\frac{\text{~}\text{velocity}\text{~}}{\text{~}\text{time}\text{~}}\]
or, mass \(=
\frac{\text{~}\text{Force}\text{~} \times
\text{~}\text{time}\text{~}}{\text{~}\text{velocity}\text{~}}\)
or,
\[\lbrack mass\rbrack = {FTV}^{-
1}\]
Q. 8 The dimension of \(\frac{a}{b}\) in the equation \(P = \frac{a - t^{2}}{bx}\) where P is
pressure, x is distance and t is time are
\(\_\_\_\_\) ?
Sol. \(\ P = \frac{a -
t^{2}}{bx}\)
\[{\Rightarrow \ Pbx = a - t^{2}
}{\Rightarrow \ \lbrack Pbx\rbrack = \lbrack a\rbrack = \left\lbrack
T^{2} \right\rbrack
}\]or \(\ \lbrack b\rbrack =
\frac{\left\lbrack T^{2} \right\rbrack}{\lbrack P\rbrack\lbrack
x\rbrack} = \frac{\left\lbrack T^{2} \right\rbrack}{\left\lbrack ML^{-
1}T^{- 2} \right\rbrack\lbrack L\rbrack} = \left\lbrack M^{- 1}{\text{
}T}^{4} \right\rbrack\)
\[\therefore\ \left\lbrack \frac{a}{b}
\right\rbrack = \frac{\left\lbrack T^{2} \right\rbrack}{\left\lbrack
M^{- l}T^{4} \right\rbrack} = \left\lbrack {MT}^{- 2}
\right\rbrack\]