Problem Based Learning in
Basic Physics - VIII
A. K .Mody
C-14 Beverly Hills, Lam Road, Devlali, Nashik 422401, Maharashtra
H. C. Pradhan
HBCSE, TIFR, V. N. Purav Marg,
Mankhurd
Mumbai – 400 088
[In this article- eighth in the series of articles we
present problems for a problem based learning course based on dimensional
analysis. We present the technique of dimensional analysis in the area of basic
physics and what each problem tries to achieve with its solution. ]
The dimensions of physical quantity
are the powers (or exponents) to which the base quantities are raised to
represent that quantity.2
The applications of dimensional analysis are
1. Checking the dimensional consistency
of equations.
2. Deducing relations among physical
quantities.
The limitation of this is we cannot relate number of
quantities that may result in number of equations less than number of unknowns.
Thus number of quantities involved cannot be greater than number fundamental
quantities whose dimensions are involved.
3. To find dimensions of a new
quantity.
Argument of a Function:
Most of the functions can be
expanded into power series as shown below.
In such cases x have to be a dimensionless quantity
otherwise we have right hand side of an equation in which each term has
different dimension.
Now, consider famous equations like
(i)
from radioactive decay
(ii)
for charging of a
capacitor in an RC circuit
(iii)
for growth of current
in an LR circuit
In all of the cases, quantities in
the exponent have to be dimensionless and thus
have dimensions of
time and are called time constants. These quantities in some sense represent
time scale at which changes occur in respective systems.
(iv)
for one-dimensional
simple harmonic progressive wave has wavelength
which has dimensions
of length and time period of oscillation
which has dimensions
of time.
1. Using dimensional
analysis check the consistency of equation
of time period of a
simple pendulum on mass m of its bob, length l of its length and
acceleration due to gravity g.
2.
Excess (above atmospheric value)
pressure inside a liquor id is expected to depend on depth h below its
surface, density ρ, and gravity g. Using dimensional analysis,
deduce relation between pressure P and h, ρ, g.
3.
Using dimensional analysis establish (i)
dependence of potential energy of a particle of mass m, placed in the
earth’s gravitational field having
acceleration due to gravity g at a height h above the earth’s
surface (ii) dependence of kinetic energy of a particle of mass m,
moving with velocity v. (iii) In
this it is not possible to determine constant of proportionality from
dimensional analysis. However using 2 and 3 above with the law of conservation
of energy, and kinematical equations, determine the ratio of proportionality
constants in 2 and 3.
4.
Viscous force between two layers of liquid
in motion is defined by newton’s formula as
, where A is the area of contact between the two
layers, and
is the velocity
gradient in the liquid, i.e., velocity of the flow changes by dv in
distance dz perpendicular to flow direction. What are the dimensions of η,
known as coefficient of viscosity? When a spherical object of radius R
flows through a viscous medium with speed v, it experiences viscous drag (resistive force).
Find an expression of this viscous force using dimensional analysis, assuming
it to depend on η, R and v . In this case it is not possible to
determine constant of proportionality which turns out to be 6π and the
formula thus obtained is known as Stokes’ law.
5.
A great physicist of previous century
(P. A. M. Dirac) loved playing with numerical values of fundamental constants
of nature. By playing dimensionally with mass of electron me
charge on electron e, Planck’s constant h, and
gravitational constant G we may be able to obtain expressions for what
is known as Planck length lP, Planck time tP
and Planck mass mP. Obtain these expressions dimensionally
and estimate value of these quantities. Particle Physicists and Cosmologists
use these values in trying to understand evolution of our universe.
6.
From Coulomb’s formula
and Bio-Savart’s Law
, find dimensions of ε0 and µ0.
Show that
has dimensions of
velocity. From the values of ε0 = 8.85×10-12
SI units and µ0 = 4π×10 – 7 SI units, estimate
value of this velocity .
7.
A dense collection of equal number of
electrons and positive ions is called neutral plasma. Certain solids containing
fixed positive ions surrounded by free electrons can be treated as neutral
plasma. Let ‘N’ be the number density of free electrons, each of mass ‘m’.
When electrons are subjected to electric field, they are displaced relatively
away from the heavy positive ions. If the electric field becomes zero, the
electrons begin to oscillate about the positive ions with a natural angular
frequency ‘ωp’, which is called plasma frequency. To sustain
the oscillations, a time varying electric field needs to be applied that has an
angular frequency ω, where a part of energy is absorbed and a part of it
is reflected. As ω approaches ωp, all the free
electrons are set to resonate together and all the energy is reflected. This is
explanation of highly reflective metals.
(i)
Taking the electronic charge as ‘e’
and the permittivity as ‘ε0’, use dimensional analysis to
determine the correct expression for ωp.
(a)
(b)
(c)
(d)
(ii)
Estimate the wavelength at which plasma
reflection will occur for a metal having the density of electrons N =
4×1027 m – 3. Take ε0 ≈ 10 – 11
and m ≈ 10 – 30, where these quantities are in proper SI
units.
(a)
800 nm (b) 600 nm
(c) 300 nm (d) 200 nm [JEE
2011]
Hints, Solution, Answers:
1.
[l]
= [L]1 [g] = [L]1[T]
-2. Thus
.
The
given equation is dimensionally correct.
2.
Let
. Here [P] = [F/A] = [M1L1T
– 2]/[L2] = [M]1[L] –
1[T] – 2
[h] = [L]1, [ρ] = [M]1[L]
– 3 and [g] = [L]1
[T] – 2
Taking [P]
= [h]x [ρ]
y [g]z, and equating dimensions of m, L, and T, we get
(power of [M]) : y = 1,
(power of [L]) : x – 3y
+ z = -1
(power of [T]) : -2z = - 2
This gives us z = 1 and x = 1 and we have
3.
We
get
and
using conservation of
energy and kinematical equation
If
we choose system of units such that
then
4.
and
Matching the
powers on two sides, this gives us x = 1, y = 1 and z = 1.
Thus ,
, the constant of
proportionality 6π is determined from some other method.
5.
We
get
,
and
6.
Taking [q] = [I1 T1],
[E] = [F]/[q] and
[B] = [F]/[q][v], we get
[E] = [M1
L1 T - 3 I -1] and
[B] = [M1 T - 2 I
- 1 ]
This gives
us [ε0] = [M
- 1 L-3 T 4 I 2]
and [µ0] = [M1 L1
T -2 I -2]
Thus,
and from the given
values
= 3×108
m/s.
7.
[Ans: (i) - (c) ,
(ii) - (b)]
References:
- JEE - Joint Entrance Examination for admission to IIT
- Physics Textbook for class XI part-I, NCERT New Delhi (2006)
