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A drop of olive oil of 1 mm diameter, when transferred lightly over water dusted lycopodium power, spreads out into a circular thin film of diameter 28.1 cm. Estimate the size of a molecule of the oil, assuming that the film is only one molecule thick.

Here, radius of olive oil drop, r =  mm = 0.5 x 10-3 m
Therefore, volume of the olive oil drop = 
πr3 = π × (0.5 × 10-3)3m3       ----------- (1)
Radius of the olive oil layer on water surface, R =  cm = 14.05
× 10-2 m
Area of the oil layer =
πr2π(14.05 × 10-2)2 m

If d is diameter (in metre) of the olive oil molecule (equal to thickness of the layer), then
Volume of layer = 
π(14.05 × 10-2)2 × d                                              ------------ (2)
From equation (i) and (ii), we have

π(14.05 × 10-2)2 x d = π × (0.5 × 10-3)3
d =  
d = 8.443
× 10-9 m.


A steel wire of length 4.7 m and cross section 3 × 10-5 m2 stretches by the same amount as the copper wire of length 3.5 m and cross section 4 × 10-5 m2 under a given load. What is the ratio of the young's modulus of steel to that of copper?

We know that Y = Stress / Strain

For Steel: 
Length l1 = 4.7m
Area of cross section a1 = 3
× 10-5 m2 
If F Newton is the stretching force and
Δl metre the extension in each case, than
Y1 = =

For copper:
Length l2 = 3.5m
Area of cross section a2 = 4
× 10-5 m2 
                              Y2   =     

Ratio of Young's modulus = Y1/Y2
                                     = 18.8/10.5
                                     = 1.7904 (or) 1.8.


Figure shows the strain-stress curve for a given material. What are
(a) young's modulus and

(b) approximate yield strength for this material?


(a) From the graph fro a stress of 150 × 106 Nm-2 the strain is 0.002
hence Young's modulus of the material

               = 7.5
× 1010 Nm-2

(b) Approximately yield of the material = 3
× 108 Nm-2.


The stress versus strain graphs for two materials A and B are shown below.
(a) Which material has greatest young modulus?

(b) Which material is more ductile?

(c) Which is more Brittle?

(d) Which of the two is the stronger material?


(a) Young’s modulus = Stress/strain = slope of stress-strain graph. Material A has greater slope, so A has greater value of Young’s modulus.

(b) Ductile materials should have high plastic region. So material A is stronger. 

(c) Brittle material has a small plastic region so material B is more brittle than A.

(d) Strength of a material is determined by the amount of stress required to cause fracture. Material A is stronger than material B.


Two different types of rubber are found to have the stress-strain curves shown below.
(a) In which significant ways do these curves differ from the stress-strain curve of a metal wire shown in the question 2?

(b) A heavy machine is to be installed in a factory. To absorb vibrations of the machine, a block of rubber is placed between the machinery and the floor. Which of the two rubbers A and B would you prefer to use for this purpose? Why?

(c) Which of the rubber materials would you choose for a car tyre?


(a) Hooke's law is not obeyed even for small stresses; no permanent strain even for large stresses; large elastic region but with the important difference that the material does not retrace the same curve during unloading. This fracture is known as elastic hysteresis.

(b) One should prefer B to absorb vibrations. The area of the hysteresis loop is proportional to energy dissipated by the material as heat. 

(c) Rubber A to avoid excess heating of the car tyre.


Read each statement below carefully and state, with reasons if it is true or false.
(a) The modulus of elasticity of rubber is greater than that of steel.

(b) The stretching of a coil spring is determined by its shear modulus.

(c) When a material is under tensile stress, the restoring forces are caused by inter atomic attraction, while under compressional stress the restoring forces are due to inter atomic repulsion.

(d) A piece of rubber under an ordinary stress can display 1000% strain; yet when unloaded it returns to its original length. This shows that the elastic restoring forces in a rubber piece are strictly conservative.

(e) Elastic restoring forces are strictly conservative only when Hooke's law is obeyed.

(a) False

(b) True

(c) True

(d) False (Elastic hysteresis signifies non conservative force)

(e) False (Elastic forces are conservative as long as loading and unloading curves are identical even if the curves are not linear).


Two wires of diameter 0.25 cm, one made of steel and other made of brass are loaded as shown in the figure. The uploaded length of steel wire is 1.5m and that of brass wire = 1m. Young's modulus of steel is 2 × 1011 Pa and that of brass is 0.91 × 1011 Pa. Compute the elongation of the steel and brass wire.


For steel:
Mass M = 10kg
Force F = 10 ×9.8 = 98 N
Distance d = 2.5 × 10-3 m
Radius r = 1.5625 × 10-6   m
Area of cross section a = 4.906 × 10-6 m2
Therefore stress = F/a = 98/4.906 × 10-6 m2 = 19974522.29
And strain = l/L = l/1.5
Stress/strain =
Therefore the elongation of the steel bar l = 1.5 × 10-4 m

For brass:

Force F = 58.8 N
Stress = F/a
          = 11984713.38
Strain = l/L
Therefore the elongation of the brass bar l = 1.3 × 10-4 m.


The edges of an aluminium cube are 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to opposite face of the cube. The shear modulus of aluminium is 25 Gpa. What is the vertical deflection of the face? (1 Pa = 1Nm-2).

Length of the aluminium cube, L = 10 cm = 10 × 10-2 m
Force applied, F = 100 kg = 100
× 10 N
Shearing stress = =
Shear modulus of elasticity = 10
× 104 Nm-2
                                      = 25 Gpa = 25
× 109 Nm-2 

Shear strain = =
×109 =      
Change in length
ΔL =
                             = 4
× 10-7 m. 


A silica glass rod has a diameter of 1 cm and is 10 cm long. Estimate the largest mass that can be hung from it without breaking it.

The ultimate strength of glass is 50 × 106 Nm-2. Therefore the largest mass that can be hung from the glass rod is
= [(50
× 106 Nm-2) × (πd2/4)] / g kg
= 392.5 kg


Compute the bulk modulus of water from the following data:
Initial volume = 100 ltrs, Pressure increase = 100 atm., Final volume = 100.5 ltrs (one atm = 1.013
× 105 Pa). Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large. Given: Bulk modulus of air = 0.01 × 107 Pa.

Initial volume V = 100 litres
Final volume = 100.5 litres
ThereforeΔV= Initial volume - Final volume
= (100.5 - 100) = 0.5litres
Pressure increase Δ P = 100 × 1.013 × 105 Pa = 10130000 Pa

Bulk modulus K1 = =
= 2.026 × 109 Pa.
The Bulk modulus of air K2 = 0.01 × 107 Pa
Therefore the ratio of the Bulk modulus =  = 20260.

The large ratio indicates that the bulk modulus of liquids is very large as compared to that of gases. This is because the intermolecular distances in the case of liquids are very small as compared to that of gases. Consequently, there are larger interatomic forces in liquids than in gases.


What is the density of ocean water at a depth where the pressure is 80 atm given that its density at the surface is 1.03 × 103 kg m-3 ? (Compressibility of water = 45.8 × 10-11 Pa-1)

Pressure at the depth of the ocean ‘Δ P’ = 80 atm = 80 × 1.013 × 105 Pa = 8104000 Pa
Density at the surface of the ocean ρ = 1.013 × 103 kg m-3 
Let m be the mass of the water, then
Volume at the surface of the ocean V
Volume at the given depth V =
Therefore decrease in volume, Δ V = =

We know that K = =
45.8 × 10-11 =
Therefore ρ =  
                    = 1.034 × 103 kg m-3.


You will appreciate from question 3 that data on chemical combinations can at best yield ratio of atomic masses of different elements. That is why atomic masses are defined with respect to some 'reference atom'. In early chemistry, hydrogen was chosen to be the reference and assigned a unit mass. In the so-called chemical scale, oxygen was assigned a mass of 16 units. By intermolecular agreement, the unified atomic mass unit (u) is defined to be such that the slope 12C has a mass exactly equal to 12u. The difference between these different scales is slight but important for precision measurements. The mass of 137Cs atom is 136.90707 on the unified atomic scale. The mass of 1H is 1.0078252 on the same scale. What is the the mass of 137Cs atom with respect to the hydrogen reference scale?

Unified atomic mass unit is defined to be such that the isotope C12 has a mass exactly equal to 12. In other words, one a.m.u is the quantity of mass equal to th of the mass of carbon-12 (C12) atom.

Weight of one C12 atoms =  = 1.99 x 10-23 g
... 1 a. m. u = th of the weight of a carbon atom
                 = 1.66 x 10-24g
1.66 x 10-27kg.


Why are atomic masses not exact integers? Why is there a slight difference in the atomic mass of an element on different scales?

The atomic masses are not integers because of the existence of isotopes of an element. The hydrogen has three isotopes namely protium (1H1 or H), deuterium and tritium (1H3 or T) having atomic masses as 1,2 respectively. When atomic mass of hydrogen is calculated by taking their relative abundance into consideration atomic mass of hydrogen comes out to be in fractions. It is also true for other elements.

There is a slight difference in the atomic mass of an element on different scales, because on the unified scale, we have chosen 1/2 mass of C-12 atom as a unit mass, as the mass of C-12 is exactly = 12.0000 amu. On the other hand, we have chosen mass of hydrogen atom as unit, which is not exactly 1.0000 amu. Due to this reason there is a slight difference in atomic mass of an element on different scales.


(a) If data on chemical combination give only relative atomic masses, how does one know the absolute mass of an atom of say oxygen?

(b) Express 1 u in terms of kg, given that the experimental value of Avogadro's hypothesis number is NA = 6.022045
1023 mol-1.

(a) Mass of one mole of oxygen atom = 16 g
Therefore, absolute value of oxygen atom = g = 2.656
× 10-23 g

(b) Atomic mass unit is defined to be such that to isotope has a mass exactly equal to 12 amu. In other words.
one amu is the quantity of mass equal to 1/12th of the mass of the carbon-12 (C12) atom.

Weight of one C12 atom = g = 1.99
× 10-23 g

1 a. m. u = 1/12th of the weight of a carbon atom = g = 1.66 × 10-27 kg.


(a) If interatomic forces are electrical nature, why do they not obey Coulomb's Law i.e. why do they fall by for a large distance (instead of as expected on the basis of Coulomb's law)
(b) What is the physical origin of the attractive and repulsive part of interatomic forces?
(c) If the potential energy is minimum at r = R0 = 0.74Å, is the force attractive or repulsive at r = 0.5 Å and 1.9 Å,


(a) When two atoms come close, then repulsive forces between their nuclei, repulsive forces between their electron clouds and attractive forces between nucleus of one atom and electron clouds become quite appreciable. These forces are electrical in nature but result from interaction of charges distributed over a space and hence variation with distance is not as . The force would vary as , if charges are points in nature as Coulomb's law is valid point charges.

(b) When atoms approach, then initially the average distance between like charges is more than the average distance between unlike charges. Hence, the force is attractive in nature. On the other hand, when atoms approach too close, the average distance between like charges becomes lesser than the average distance between unlike charges. Likewise, the interatomic force becomes repulsive.

(c) It is repulsive at r = 0.5 Å as r < r0, attractive at r =1.9 Å as r > r0 and zero at


How much energy is needed to dissociate 0.05% of hydrogen gas occupying a volume of 5.6 litres under S. T. P? The binding energy of a hydrogen molecule is 4.75 eV.

The binding energy i.e. energy needed to dissociate one molecule of hydrogen gas = 4.75 eV

Total volume of the gas = 5.6 litre      (at S. T. P.)
Volume of hydrogen gas to be dissociated = 5.6 ×  = 2.8 × 10-3 litre

Avogadro's hypothesis, number of hydrogen molecules
= 7.528 × 1019

Therefore, total energy required to dissociate hydrogen molecules
= 7.525 × 1019 × 4.75 = 3.576 × 1020 eV

We know that      1 eV = 1.6 × 10-19 joule
Therefore,   energy required = 3.576 × 1020 × 1.6 × 10-19 = 57.216 J.


The interatomic separation in a neutral H2 molecule is 0.74 Å and the binding energy is 4.75 eV. If an electron of the molecule is removed, the resulting molecular ion has a binding energy of 2.8 eV. Would you expect the separation between two protons in will become greater than 0.74 Å.

When binding energy of a molecule decreases, the separation between the two atoms will increase. Therefore, separation between two protons in will become greater than 0.74 Å. Experimentally, the separation tends to be 1.6 Å.


(a) An energy of about 1.3eV is needed to create free Na+ and Cl- ions. [By this we mean that the Na+ Cl- configuration at large distances has energy 1.3 eV higher than neutral Na and Cl atoms at large.] In spite of this initial investment the sodium chloride molecule prefers ionic binding. Why? 

(b) A related question is why does H2 prefer a covalent boding? Why does it not prefer ionic bonding H+ H- like Na+ Cl-?

(a) Sodium has 11 electrons. According to Pauli's exclusion principle, first K and L shells will be completed be 2 and 8 electrons respectively. The eleventh electron will occupy the M shell and it will be a valence electron. The electronic configuration of sodium is 1s2 2s2 2p6 3s1. The s- sub shell of M shell will be completed if it gains one extra electron or it will be empty by giving one valence electron. On the other hand, chlorine will have 17 electrons in its electronic configuration will be 1s2 2s2 2p6 3s2 3p6. That is the seven electrons will be in M shell. The M shell would require one electron to saturate it. This is done when one valence electron of sodium is transferred to chlorine atom which results in ionic bonding.

(b) Since the K shell is saturated by two electrons therefore two hydrogen atoms share electrons of each other. This results in covalent bonding between them.


The interatomic separation in an O2 molecule is 1.2 Å and its binding energy is about 4.4 eV. The intermolecular potential energy between two oxygen molecules has a minimum at 2.9 Å. The two potential energy curves has roughly similar shapes. Would you expect the minimum of intermolecular potential energy in oxygen to be greater or less than 4.4 eV.

We know that intermolecular forces are about fifty to a hundred times weaker than interatomic forces and hence is case of intermolecular forces; the minimum occurs at much larger distance (As in given, it is 1.2 Å in case of interatomic forces and 2.9 Å in case of intermolecular forces). Hence, the value of intermolecular potential energy in oxygen will be less than 4.4 eV.


In which principle aspects are intermolecular forces different from interatomic forces? In which aspects are they similar.

The two types of forces are similar in following aspects;

(i) The variation of the magnitude of force and the potential energy in the two cases is more or less similar.

(ii) Both are short range forces.

The two types of forces are different in following respects.

(i) The interatomic forces are electrical in nature and are the result of either the sharing of electrons or the transfer of electrons between the two atoms. On the other hand, intermolecular forces are due to small electric dipole current developed by the molecules due to a small shift in the centre of mass of positive and negative charges and are due to different values of separation between them due to different orientations.

(ii) The intermolecular forces are weaker than interatomic forces by a factor of 50 to 100.


Estimate the average intermolecular binding energy per molecule of (i) mercury (ii) water if their latent heats of vaporisation are 2.72 × 106 J kg-1 and 2.26 × 106 J kg-1respectively. Is it correct to think that this average binding energy is the same, as that required pulling two molecules of the substance apart from one another, i.e. the negative of the minimum potential energy between two molecules?

(i) For mercury, the latent heat of vaporization = 2.72 × 105 J kg-1 .
Atomic mass of the mercury = 200. 
Therefore, 200 g or 0.2 kg of mercury contains 6.023
× 1023 molecules
Therefore, number of molecules of mercury in 1 kg = = 3.01

If E is average intermolecular binding energy per molecule of mercury, then 3.01
× 1024
E = 2.72
× 105
E = J = = 0.56 eV

(ii) For water, latent heat of vaporization = 2.26
× 106 J kg-1

Now, 18 g or 0.018 kg of water contains 6.02
× 1023 molecules
Number of molecules in 1 kg of water =
If E is the average intermolecular binding energy of water, then  
E = 2.26
× 106
... E = = 0.422 eV.


Answer the following:
(a) If intermolecular potential energy has a minimum at some separation r = r0, what is it that prevents molecules of the given substance from collapsing to the condensed state, where every pair has separation equal to r0.
(b) Intermolecular attraction is the crucial thing for a gas to liquid transition, while a repulsive potential energy at short distances is crucial for liquid to solid transition. Explain this important qualitative insight into the nature of phase transitions.
(c) Give an example of an unusual phase of matter in nature which arises due to the highly non-spherical shape of its molecules.

(a) As the separation tends to decrease further, the force between molecules becomes strong and repulsive, which prevents the molecules of the given substance from collapsing to the condensed state.

(b) When the separation is greater than r0, force between the molecules is attractive. Due to intermolecular attraction gas to liquid transition takes place. On the other hand, when the separation is less than r0, force between the molecules is repulsive. Due to intermolecular repulsion, liquid to solid transition takes place.

(c) Ammonium oleate (C17 H33 COONH4) molecules exist in liquid crystal phase - an unusual phase of matter. The reason is that ammonium oleate molecules have highly non-spherical shape. These molecules are rod shaped or disc-shaped molecules having a two dimensional symmetry in contrast to ordinary crystals possessing three dimensional symmetry.


Four identical hollow cylinder columns of steel support a big structure of mass 50.000 kg .the inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column. The young's modulus of steel is 2.0 × 1011 Pa [1 Pa = 1 Nm-2].

Young's modulus Y = 2.0 × 1011 Pa (or) N/m2
The inner radius of a column r1 = 30 cm = 0.3 m
The outer radius of a column, r2 = 60 cm =0.6 m
Therefore, cross-section of each column
= =0.85 m2
The whole weight of the structure will be shared by four columns.
Therefore, compressional force on one column
F = N
Y = Pa
Compression strain = = 2.8
× 10-6.


Anvils made of single crystals of diamond, with the shape as shown in Figure are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.5 mm and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil?    

Diameter, d= 0.5 mm
Radius, r = 0.25mm
              = 0.25
× 10-3m
Area =
πr2 =
                = 1964.2
× 10-10

Pressure = = 2.5
× 1011 Pa (or) N/m2.


A composite wire of uniform diameter 3.0 mm consists of a copper wire of length 2.2 m and a steel wire of length 1.6 m stretches under a load by 0.7 mm. Calculate the load, given that the Young's modulus for copper is 1.1 × 1011 Pa and that for steel is 2.0 × 1011 Pa, (1 Pa =1 Nm-2).

Young's modulus for copper Yc= 1.1 × 1011 Pa
Young's modulus for steel Ys = 2.0 × 1011 Pa
Length of the copper wire lc = 2.2m, 
Length of the steel wire ls= 1.6m
We know that

The total extension :

Δlc + Δls = 7.0 × 10-4m                                   -------(2)
From equation (1), we get

Δlc = 2.5 Δls                                                   -------(3)
Substituting equation (3) in (2), we get
Δls + Δls = 7.0 × 10-4m   
Δls = 7.0 × 10-4m
Δls = 2.0 × 10-4m
Δlc = 5.0 × 10-4m   
Therefore W =
                  = N
                  = 1.8 × 102 N.

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