Problem
Based Learning in Basic Physics - IV
School Science 51(1) Mar2013
A. K .Mody
V. E.S. College of Arts, Science and
Commerce, Sindhi Society
Chembur, Mumbai – 400 071
H. C. Pradhan
HBCSE, TIFR, V. N. PuravMarg,
Mankhurd
Mumbai – 400 088
In this article- fourth in the series of articles we
present problems for a problem based learning course from the area of heat,
thermodynamics and optics. We present the learning objectives in this area of
basic physics and what each problem tries to achieve with its solution.
In this article, fourth in the series of Problem Based Learning in Basic Physics, we present problems on heat, thermodynamics and optics. Methodology and philosophy of selecting these problems are already discussed. (Pradhan 2009, Mody 2011)
To review methodology in brief, we
note here that this PBL (Problem Based Learning) starts after students have
been introduced to formal structure of Physics. Ideally students
would attempt only main problem. If they find it difficult, then depending upon
their area of difficulty, right auxiliary problem have to be introduced by
teacher who is expected to be a constructivist facilitator. Teacher may choose
as per his/her requirement or may construct questions on the spot to guide
student to right idea and method.
H:
Heat and thermodynamics
Learning
objectives:
1.
To
become familiar with heat as a form of energy (that due to random motion).
2.
To
understand effect of heat on physical properties.
3.
Effect
of heat on parameters like pressure, temperature etc.
4.
Transfer
of heat in different processes especially conduction and radiation.
5.
Black
body radiation and temperature.
6.
To
understand mathematical structure dealing with above mentioned points.
Problems:
1. A
narrow necked vessel contains 6 gm of a gas at 7°C and at a certain
pressure. The vessel is heated to 147°C, when some of
the gas escapes. The pressure of the gas in the vessel is constant. What mass
of the gas has escaped? *
Tasks
involved in this problem are:
a. To
apply PV = nRT before and after.
b. Finding
out n in the final state and hence how much gas escapes.
2. An
air bubble starts rising from the bottom of a lake, 2.5 metres deep. Its
diameter increases from 3.6 mm at the bottom to 4 mm at the surface. If the
temperature of water on the surface is 40°C, what is the
temperature at the bottom of the lake? [Atmospheric pressure = 0.76 m of Hg; rHg
= 13600 kg/m3; rwater
= 1000 kg/m3; g = 9.8 m/s2]
[Mistry]
Tasks
involved in this problem are:
a. Use of PV = nRT knowing that ‘n’
remains constant .
b. Using P = Patm + hρg
3. A
closed vessel contains liquid water in equilibrium with its vapour at 100°C
and 1 atm. One gram of water vapour at this temperature and pressure occupies a
volume of 1670 cm3. The heat of vaporization at this temperature is
2250 J/gm. (a) How many molecules are there per cm3 of vapour. (b) How many vapour molecules strike each cm2
of liquid surface per second? (c) If each molecule which strikes the surface
condenses, how many evaporate from each cm2 per second? (d) Compare
the mean kinetic energy of a vapour molecule with the energy required to
transfer one molecule from the liquid to the vapour phase. *
Tasks involved
in this problem are:
a. To visualize what happens at molecular
level.
a. To
use proportionality and mole concept to find number of molecules.
b. To
use statistical average to estimate number of molecules striking unit area of
the liquid surface per second.
c. To
use energy conservation principles to find number of molecules evaporating.
4. According
to Stefan’s Law of radiation, a black body with surface temperature T
radiates energy σT4 from its unit surface area every second,
where σ = 5.67´10 – 8 W/m2K4
is known as Stefan’s constant. If the earth is in radiative equilibrium with
the sun and the average temperature of the earth’s surface is 300 K, estimate
the surface temperature of the sun. Take radius of the sun is 7´108
m and the mean distance between the earth and that of the sun is 1.5´1011
m.
Tasks involved
in this problem are:
1.
To estimate energy radiated by sun and
fraction of it that is received by the earth.
2.
Energy that is radiated by the earth.
3.
If earth is in radiative equilibrium
quantity in (1) and (2) should be equal wich would allow estimation of Sun’s
surface temperature.
5. The
compression ratio (V1/V2) of a four stroke internal
combustion engine is equal to 9.5. The engine takes in air and gas fuel at
temperature 27oC and pressure of 1 atm. The volume is then
compressed adiabatically from state 1 to state 2.
The fuel mixture is ignited causing an
explosion, which doubles the pressure (state 2 to 3), thus moving the piston
into a position in state 3. From state 3 to 4 the gaseous mixture again expands
adiabatically until the volume becomes 9.5V as shown in PV-diagram
and the exhausting valve in the cylinder opens up allowing the pressure in the
cylinder to return to 1 atm. (γ = 
= 1.4)
Determine
= 1.4)
Determine
(i) The pressure and temperature of the
gaseous mixture in states 1, 2, 3, and 4 respectively.
(ii) The thermal efficiency of the cycle. [IPhO 1976]
- An illustrative problem that deals with thermodynamic processes (isochoric and adiabatic).
Tasks involved
in this problem are:
- To start with initial conditions identifying the processes involved and equations governing them.
- To find needed quantities from given information.
- To calculate pressure and temperature at various stages of the cycle using gas equations.
- To calculate energies involved in different processes to calculate efficiency of a cycle.
Surface Tension:
6. Surface
tension is exhibited by liquids due to force of attraction between molecules of
the liquid. The surface tension decreases with increase in temperature and
vanishes at boiling point. Given that the latent heat of vaporization for water
L v = 540 kcal/kg, the mechanical equivalent of heat J= 4.2 J/Cal ,
density of water ρw = 103 kg l-1, Avogadro’s
No NA = 6.0 x 1023 k mol-1 and the molecular
weight of water MA = 18 kg for 1 k mole.
(a) Estimate
the energy required for one molecule of water to evaporate.
(b) Show
that the intermolecular distance for water is
d = 
and find its value.
and find its value.
(c) 1
g of water in the vapour state at 1 atm occupies 1601 cm3. Estimate
the intermolecular distance at boiling point, in the vapour state.
(d) During
vaporization a molecule overcomes a force F, assumed constant, to go fron an
intermolecular distance d to d’. Estimate the value of F.
(e) Calculate
F/d, which is a measure of the surface tension. [NCERT EP XI]
The
tasks in this problem are already listed as (a), (b),… self explainatory. The
aim is to construct conceptual understanding through simple calculation.
I : Optics:
Learning Objectives:
1. To
understand formation of image, its nature and position due to mirrors, due to
surfaces (plane and curved) separating two media.
2. To
understand formation of image, its nature and position due to multiple surfaces
and lenses.
3. To
understand wave propagation and superposition based on wave nature and path (or
phase) difference.
4. To
understand mathematical structure dealing with above mentioned points.
Problems:
7. Find
the height of the shortest plane mirror (held vertically) in which a man six
feet tall could see his entire image. At what height above the ground should
this mirror be placed in order that the man could see his entire image? *
Tasks
involved in this problem are:
a.
Drawing ray diagram to see how image is
formed in the mirror.
b.
Geometrically establishing relationship
between size of the object and minimum size of the mirror needed.
8. Screen
S is illuminated by two point sources A and B. Another source C sends parallel
beam of light towards P on the screen (see fig.) Line AP is normal to the
screen and the lines AP, BP, and CP are in one plane. The distances AP, BP, and
CP are 3 m, 1.5 m and 1.5 m respectively. The radiant powers of the sources A
and B are 90 W and 180 W respectively. The beam C is of intensity 20 W/m2.
Calculate the intensity at P on the screen.
. [JEE 82]
Tasks
involved in this problem are:
a. To
recognize difference between point source and a parallel beam of light.
b. To
learn the variation in intensity as light travels distance from different type
of sources.
c. To
learn how intensity is related to angle of incidence. This had to be discussed
with them.
9.
A person looking through a
telescope T just sees the point A on the rim at the bottom of a cylindrical
vessel when the vessel is empty (see fig.). When the vessel is completely
filled with a liquid (μ = 1.5), he observes a mark at the centre B, of the
bottom without moving the telescope or the vessel. What is the height of the
vessel if the diameter of its cross-section is 10 cm?
Tasks
involved in this problem are:
a.
Identifying incident and
refracted rays.
b.
Working out angles for each of
the rays.
c.
Application of Snell’s law of
refraction.
10.
An object is placed 21 cm in
front of a concave mirror of radius of curvature 10 cm. A glass slab of
thickness 3 cm and refractive index 1.5 is then placed close to the mirror in
the space between the object and the mirror. Find the position of the final
image formed. (You may take the distance of the nearer surface of the slab from
the mirror to be 1.0 cm) [JEE 80]
Tasks
involved in this problem are:
a.
To draw rays as per the rules
of geometrical optics first due to glass slab.
b.
To treat image formed by one as
the object for the next and continue.
c.
To locate image by mathematical
calculation.
11. The
Young’s double slit experiment is done in a medium of refractive index 4/3. A
light of 600 nm wavelength is falling on the slits having 0.45 mm separation.
The lower slit S2 is covered by a thin glass sheet of thickness 10.4
μm and refractive index 1.5. The interference pattern is observed on a screen
placed 1.5 m from the slit as shown in fig. P(5) Find the location of the
central maximum (bright fringe with zero path difference) on the y-axis. (b)
Find the light intensity at point O relative to the maximum fringe intensity.
(c) Now, if 600 nm light is replaced by white light of range 400 to 700 nm,
find the wavelengths of light that form maxima exactly at point O. [All
wavelengths in this problem are for the given medium of refractive index 4/3.
Ignore dispersion] [JEE 99]
Tasks
involved in this problem are:
a.
To find the additional path
difference introduced by the glass plate.
b.
To locate the point on the
screen where path difference will be zero knowing the fringe width.
c.
To estimate the variation in
intensity on the screen from consideration of path difference.
Solutions:
1.
Gas Law:
Gas Law: PV = nRT :
For the gas inside the vessel V and P are constants.
\ nT = contstant or
nMAT = constant Þ mT = constant where m
is mass of the gas and T its absolute temperature.
m1 = 6 gm at T1 = 7oC
= 280 K and m2
= to be found at T2 = 147oC =
420 K
\
m2 = m1 (T1/T2) = 4 gm
\ Dm
= m1 - m2 = 2 gm
: mass of the gas escaped.
2.
Gas Law:
pV = nRT and
constant for a given bubble
constant for a given bubble
Pdepth = Psurface + hrg where r
is
the density of water.
\ 
and P1 = P2 + hrg
and P1 = P2 + hrg
Þ
T1 = 283.4 K = 10.4 oC
3. Kinetic Theory :
(a) Number
of molecules N = n NA
where n is number of moles and NA
is Avogadro Number.
\ 
per cm3
per cm3
(b) N/3 strike wall area l2
in time 2l/c where 
\ 
per cm2
per sec
per cm2
per sec
(c) Energy delivered per cm2 per sec =
= 1860.8 J
= 1860.8 J
U/N = 6.72´10
– 20 J/molecule needed for evaporation.
Thus 
gets evaporated
every sec from 1 cm2.
gets evaporated
every sec from 1 cm2.
(d) Mean K.E. of vapour molecules = (3/2)kT = 7.72´10 –21
J
Each vaporizing molecule needs energy =
6.72´10
–20 J
5.
Radiation :
Sun
radiates energy sTS4
every
second. Earth obstructs effectively
what is flowing through area pRE2
out of 4pr2. If
Earth’s temperature is TE it radiates 4pRE2s
TE4 every second.
\
In radiative equilibrium : 
Þ 
= 6210.6 K
Þ = 6210.6 K
5.
Thermodynamic Cycle
Given:
P1 = 1 atm, T1 = 300 K, and
V1/V2 = 9.5
(i) Path 1 ® 2 : adiabatic : 
Path 2 ® 3 : isochoric (V3 = V2): P3 = 2P2 = 46.74 atm
T3 = 2T2 = 1476.5 K
Path 3 ® 4 : adiabatic : V4/V3 = 9.5 and 
(ii)
Heat intake = Cv (T3 – T2)
= (3/2)R(738.24) = 1107.4 R
Heat Exhaust
= Cv (T4 – T1) = (3/2)R(300) = 450 R
Work done = (1107.4 – 450) R = 657.4 R
\ Efficiency h
= (W/Heat intake) = 0.594 ®
59.4%
6. Surface Tension:
(a) If Lv is the energy
required for unit mass of the substance to evaporate, MA and NA
are molecular weight and Avogadro Number respectively then energy needed for
one molecule to evaporate is 
(b) If density of
liquid is ρw then volume occupied by a molecule is 
where d is inter molecular distance. This gives
where d is inter molecular distance. This gives
(c) in vapour state
this distance increases by a factor of (1601)1/3 = 11.89 i.e. new intrer molecular distance becomes d’
= 36.3×10 -10 m .
(d) Thus if
intermolecular force per unit distance is F then work done in overcoming it for
evaporation is F (d’ – d) = u . This gives F = 0.2048×10 -10 N
(e) Thus 
is the measure
of surface tension of the water.
is the measure
of surface tension of the water.
7.
Reflection on a plane surface:
Refering to Geometry and laws of
reflection: PQ = (1/2) (HE + EF)
= (1/2)
(HF)
= (1/2) of height of the person
8. Photometry :
A
is a point source, and light is incident normally on screen at P :
which
gives IP,A = EA/dPA2 = 10/4p W/m2
B
is a point source, and light is incident at an angle on screen at P:
which
gives IP,B =
(EB/4pdPB2) cos 60 =
40/4p W/m2
C
sends a parallel beam at P but at an angle
which
gives IP,C = IC cos 60 = 10 W/m2
IP
= IP,A + IP,B + IP,c = 12.5/p + 10
»
14 W/m2
9.
Refraction at a plane surface:
, 
and 
Solving which we get 
10. Refraction and reflection at a curved
surface:
As seen from the mirror, if object is at
O, it appears to be at O’. The
displacement can be given by OO’ = t
(1 – 1/m)
= 1 cm [* This is proved as given
at the end]
Þ
PO’ = PO – OO” = 20 cm
For mirror f = R/2 = 5 cm and u =
20 cm and using 
Þ v = PI’ = 20/3 cm
Þ v = PI’ = 20/3 cm
and
II’ = t (1 – 1/m)
= 1
cm as above which gives PI = 23/3 cm =
7.67 cm from the mirror.
For a curved surface separating two media 
here R = ¥ for both the
surfaces
\ For 1st surface v1 = - m
x
\
For 2nd surface
u = - (m
x + t)
\ 
Þ v = - (x + t/m)
Þ v = - (x + t/m)
\ OO’ =
OB – O’B = (x + t) - (x + t/m)
= t(1 – 1/m)
= 1 cm
11. Young’s Double Slit
Interference :
All
wavelengths here corresponds to m
= 4/3
(a) Fringe width Dx
= lD/d
=
2 mm
t(m
- 1) = nl
and m
= mmg
= 9/8
\ n = t(m
- 1/l) = 
: no of fringes
: no of fringes
\ Shift y
= nDx
=´2
= 
´2
=
(b) Dx = l/6 + 2l Þ Phase shift a
= 2p
Dx
/l = p/3 : acute angle
\ Io = 4A2 and
I =A2(1 + cosa)2
= (9/4)A2 = (9/16)Io
(c) n = t(m
- 1/l) : integer :
for l in nm.
for l in nm.
\ n = 2 Þ l
= 650 nm
and n = 3 Þ l
= 433.3 nm in given range.
References:
* Source unknown
- Mistry N. M. Concepts of Physics for class XI , Uttam Prakashan, Mumbai, 1999
- JEE - Joint Entrance Examination for admission to IIT
- IPhO – International Physics Olympiad
- Mody A. K. &Pradhan H. C., ‘Problem Based Learning in Basic Physics – I, School Science 49 (3) Sept 2011
- Pradhan H.C. &Mody A. K., ‘Constructivism applied to physics teaching for capacity building of undergraduate students’, University News, 47 (21) 4-10, (2009)
- Physics Textbook for class XI part-I, NCERT New Delhi (2006)
- NCERT EP XI : Physics Exemplar Problems class XI, NCERT New Delhi (2009)
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