## IIT JEE Main / Advanced Physics 2018 – Tips & Trends – Modern Physics : Radioactivity

Def : Refers to the spontaneous disintegration of certain atomic nuclei which is followed by the emission of alpha particles ( helium nucleus ) , beta particles  ( electrons or positrons ) or gamma radiation ( short wavelength electromagnetic waves ).

Radioactivity is caused by unstable nuclear configurations resulting from nuclear size , limited range of nuclear forces or due to imbalances in small nuclides.

Measurement  –

Ac = –  dN/dt  where N is the number of nuclei present in a radioactive sample at a particular instant. The value is negative as with time the no of elements gets reduced. Unit is Bequeral = 1 disintegration / sec.

Points to note  –

• Radioactivity is a random process . The chance of a particular atom in a radioactive sample undergoing disintegration is predicted by applications of probability.
• Radioactivity is a nuclear process . Extra-nuclear parts of the atom are not involved in the process.
• Radioactivity is not a chemical process because the electronic configuration of the element is not changed. A chemical process involving radioactive elements results in radioactive products.
• Radioactive disintegration is uniform throughout the sample.

States that ‘ the activity of a radioactive element at any instant is directly proportional to the number of undecayed active atoms ( parent atoms ) present at that instant’ .

Ac =  | dN/dt | = λ N  ,  here λ is the decay constant for a decay process .

This equation can be rewritten as dN / dt = –  λ N   , the negative sign means than dN /dt  i.e, the rate at which the active elements are disintegrating decreases with time . Now , integrating with time limits t = 0  to  t = t , we arrive at –

N  =  N0 e -λ t –  this is the radioactive decay equation , which gives the number of active parent atoms N present at time t in the sample.

λ N  = λ N0 e – λ t Ac =  Ac0 e – λ t – this equation can be used to find the activity of a radioactive substance at any instant .

Half – life –  It’s the time required for half the original nuclides to decay , i.e   N = N0/2

therefore , N0/2 =  N0 e – λ t or        In [ 1/2 ] = – λ t      or     In[ 2 ]  = λ t     ,   hence  –

T     =    In [ 2 ] / λ     =  0.693 / λ

Decay equation in terms of Half – life time   –

N  =  N0 (2)  v t

Average  Life

Tavg =    Sum of the ages of all the nuclei / N0 ( ∫0 t  λ N0 e – λ t ) / N0

Points  to  Note  –

• The number of undecayed nuclei left after n mean life  =  N ( 0.37)n No =  [ 1/e ]n N0
• Consider a nuclide decaying simultaneously by two different processes having decay constants  λ1  and  λ2 . In such a case , the effective decay constant of the nuclide  = λ = λ1 +  λ2N  =  N0 e –  ( λ1 + λ2 ) t

On to some problems

1 . The half life of the radioactive radon is 3.8 days . The time , at the end of which 1/20 th of the radon sample will remain undecayed is  ( given log10 e = 0.4343 )  ( IIT JEE 1981 )

• (a)   3.8 days
• (b)   16.5 days
• (c)   33 days
• (d)   76 days

Ans .  N  =  N0 e -λ t and  λ   =  In 2 / t ½ =  In (2) / 3.8   . Therefore , N0 / 20   =  N0 e – In (2) / 3.8

Solving by using the given data gives  t  =  16.5 days .  Hence correct option is (b)

2 . During a negative beta decay  –   ( IIT JEE 1987 )

• (a)  an atomic electron is ejected
• (b)  an electron which is already present within the nucleus is ejected
• (c)  a neutron in the nucleus decays emitting an electron
• (d)  a part of the binding energy of the nucleus is converted into an electron

Ans . The reaction mentioned here is 0n1 →  1H1 + -1e0 + antineutrino   .   Hence correct option is (c)

3 . A freshly prepared radioactive source of half life 2 hr emits radiation of intensity which is 64 times the permissible safety level. The minimum time after which it would be possible to work safely with this source is      (  IIT JEE 1988 )

• (a) 6 h
• (b) 12 h
• (c)  24 h
• (d)  128 h

Ans .   R  = R0 [ 1/2 ]n , therefore  1 = 64 [ 1/2 ] n or   n = 6 no of  half lives ,  hence t = n . t = 6 x 2 = 12 h. So , correct option is (b)

4 . A small quantity of solution containing Na 24 radio nuclide ( half – life = 15 h ) of activity 1.0 microcurie is injected into the blood of a person . A sample of the blood of volume 1 cm³ taken after 5 h shows an activity of 296 disintegrations per minute . Determine the total volume of blood in the body of the person . Assume that the radioactive solution mixes uniformly in the blood of the person. ( 1 curie = 3.7 x 1010 disintegrations per second )  ( IIT JEE 1994 )

Ans .  Disintegration constant = λ  ,  0.693 / t½=  0.693 / 15 h   =  0.0462 / h

Now , the initial activity is 1 micro curie , i.e ,  3.7 x 104 disintegrations / second. Let r denote the activity in 1 cm³ of blood at t = 5h .

=   296 / 60  disintegrations / second  =  4.93  disintegrations / second. Let R be the activity of the whole blood of the person at  t = 5 h.

So, the total vol. of blood =  V  = R/r  =   R0 e – λ t / r    =   [ 3.7 x 104/4.93 ] e – ( 0.0462 ) ( 5 ) cm³  .

V    = 5.95 L

## IIT JEE Main / Advanced Physics 2018 -Free videos,podcasts,study material -Fundamentals of Physics – Prof.Ramamurti Shankar [ Yale ]

The internet , as is well known , is a treasure trove of invaluable learning material . But , not so often , the problem is to find the stuff of required quality and level. For an exam like IIT JEE which requires wide and uptodate theoretical as well as conceptual perspectives , the task becomes more tedious as , even if the required material is found , it won’t be open source . In this series of posts , we will presenting free material which is of top notch quality , and which would help you in your preparations for IIT JEE exam.

Prof.Ramamurty-shankar , Yale University

The first series deals with the Fundamentals of Physics . The subject expert is Prof. Ramamurti Shankar , presently working as the  John Randolph Huffman Professor of Physics at Yale. He received his B. Tech in electrical engineering from the Indian Institute of Technology ( Chennai ) and  his Ph.D. in theoretical particle physics from the University of California, Berkeley.

The course sections are      –

• Newtonian Mechanics
• Vectors in Multiple Dimensions
• Newton’s Laws of Motion
• Newton’s Laws and Inclined Planes
• Work – Energy Theorem and Law of Conservation of Energy
• Kepler’s Laws
• Dynamics of a Multiple-Body System and Law of Conservation of Momentum
• Rotations, Part I: Dynamics of Rigid Bodies
• Rotations, Part II: Parallel Axis Theorem
• Torque
• Introduction to Relativity
• Simple Harmonic Motion
• Simple Harmonic Motion (cont.) and Introduction to Waves
• Waves
• Fluid Dynamics and Statics and Bernoulli’s Equation
• Thermodynamics
• The Boltzmann Constant and First Law of Thermodynamics
• The Second Law of Thermodynamics and Carnot’s Engine
• The Second Law of Thermodynamics (cont.) and Entropy
• Exam material

## IIT JEE Main / Advanced Physics 2018 – PHYSICSMYND Brainwave(r).5

In this 5th post of our Brainwave series , the scenario of action shifts to the space . Well , nothing serious , only the troubles related to transferring cargo between two space ships……. and what’s the great deal ?  Only that they are in space and are travelling at mind boggling speeds .  Of course , don’t be too serious about it , it’s a hypothetical scenario , meant ( as always ) to test the physicist in you… on to the problem ……..

The Case of the  ‘ Space Cargo

As seen from an inertial frame ‘ S ‘ , two space ships are travelling in opposite directions along straight parallel trajectories separated by a distance d . The speed of each ship is c / 2 , where c is the speed of light.

1. Viewed from S , at the instant when the ships are at the points of closest approach , the first ship ejects a  package , which has a speed of 3c/4 as seen from S. From the point of view of an observer in the first ship , at what angle must the package be aimed , so that it can be received by the second ship ?  [ Assume that the observer in the first ship has a co-ordinate system whose axes are parallel to that of  S and the direction of motion is parallel to the y axis ]
2. Calculate the speed of the package as seen by the observer in the first ship ?

Ans  –

1. Considering the inertial frame S , the y component of the velocity of the package must be c/2 , so that the y component of the package will have the same y co-ordinate as the second ship , when the package covers the distance Δ x = d . The velocity of the package in S  is

u = ux ex+ uy ey , here uy= c/2  .      Now , as u = |u| = 3/4 c ,  ux = √ u² – u²y = √5/4 c

Let S’ be the inertial frame of the first ship. In S’ , the package has a velocity u’ = u’xe’x + u’y e’y . Now , S’ has a speed of c/2 relative to S  taken along the  – y direction . Therefore , the velocity of S’  relative to  S is v = – c ey/2 . Considering the velocity transformation –

u’y =  uy – v / [ 1 – v uy / c² ]  =  (c/2 + c/2 )/ ( 1 + 1/4 )  =  4/5 c .

u’x =   [ ux(  √ 1 – v²/c² )  ] /  [ 1 – vux / c² ]  =  ( √5/4 c  ) ( √ 1 – 1/4)  / ( 1 + 1/4  )  = √ 3/ 20 c  .

Let θ’ be the angle between the velocity u’ of the package in S’ and the x’ axis .  Now ,

tan θ’ =  u’y / u’x8 /√15

2. Now , in S’ , the speed of the package  –

u’ =   ( √ u’x + u’y )   = √ 79 / 10 c

## IIT JEE Main / Advanced Physics 2018 Tips & Trends – Motion of the Pendulum

As mentioned in one of our previous posts [ Brainwave series ] simple devices could be used to determine important physical science functions as well as applications. The Pendulum is one  such device .  It consists of a rigid body mounted on a fixed horizontal axis, about which it is free to rotate under the influence of gravity. The period of the motion of a pendulum is quite  independent of its amplitude and depends basically on a ) the geometry of the pendulum and on b) the local value of g, the acceleration of gravity. Pendulums are hence used as the control elements in clocks, or inversely as instruments to measure g.

Pendulum – general representation

In the general representation of the pendulum shown above , O is the axis and C represents the center of mass. The line OC makes an instantaneous angle θ with the vertical. Considering the rotary motion of any rigid body about a fixed axis, the angular acceleration is equal to the torque about the axis divided by the moment of inertia I about the axis. Also , m represents the mass of the pendulum, and the force of gravity can be considered as the weight mg acting at the center of mass C.

Animation : the velocity and acceleration vectors of a simple pendulum.

1. The motion is simple harmonic when the amplitude of motion of the pendulum is small . Period  T, i.e time taken for a complete oscillation is given by   2π √ ( I / mgh )
2. In the case of a simple pendulum the lengths h and L become identical, and therefore , the moment of inertia I equals mL2 . Period  T becomes  2π √ L/g      Note that the period is independent of the mass of the pendulum.
3. If the amplitude is large ,  2π √ [  l/g ( 1 + A²/16 ) ] where A is the amplitude expressed in radians.
4. If the length of the pendulum is large , g becomes directed towards the centre of the earth and T =  2π √ [ 1/g ( 1/ l + 1 / R ) ]
5. If  l << radius of the earth ( R ) , then  T =  2π √ l/g
6. If the length of the pendulum is at infinity , then T = 2π √ R/g
7. If the simple pendulum is in a vehicle accelerating up with acceleration |a| , then T = 2π √ [ l/(g + a) ]
8. If the simple pendulum is in a vehicle accelerating down with acceleration |a| , then T = 2π √ [ l/(g – a) ]
9. If the simple pendulum is in a vehicle accelerating in a horizontal direction with  acceleration |a| , then T = 2π √ [ l/(g² + a² ) ]
10. If the simple pendulum is in a falling lift , then g = 0 and T = ∞  ( no oscillation )

Now on to some questions –

1.The period of oscillation of simple pendulum of length L suspended from the roof of the vehicle which moves without friction , down an inclined plane of inclination α is given by  – [ IIT JEE 2000 ]

• (a) 2π √ L/g cos α
• (b) 2π √ L/g sin α
• (c) 2π √ L/g
• (d) 2π √ L/g tan α

Ans –  This case is a variation of no.8 mentioned above [ geff = g – a ]  , where effective g gets resolved to sin and cos components. Now , if  α is the angle of inclination , g ( acceleration of the bob ) becomes g cos α  and acceleration of the point of suspension is  g sin α .

Hence , geff = g – a   =  √  [ g² +  (g sin α )² + 2 ( g ) ( g  sin α ) cos ( 90° + α ) ]  =   g cos α .  Hence , answer is option (a).

2.A simple pendulum has time period T1. The point of suspension is now moved upward according to the relation y = kt² , ( k = 1m/s²) where y is the verical displacement . The time period now becomes T2. The ratio of T1²/ T2² is ( Take g = 10 m/s² )  [ IIT JEE 2005 ]

• (a) 6/5
• (b) 5/6
• (c) 1
• (d) 4/5

Ans  y = kt² , d²y/dt² = 2k . As k = 1m/s²,  ay = 2 m/s² .

T1 = 2π √ l/g  and   T2  =  2π √ L/g +ay , therefore ,  T1²/ T2² = 10 + 2/10  = 6/5 . So , correct option is (a).

3. The time period of a second’s pendulum at a depth R/2 from the earth’s surface is given by –

• (a) 3 √2  s
• (b) 5 √2  s
• (c) 2 √2  s
• (d) 4 √2  s

Ans  –

At a depth h below the earth’s surface , geff = g (1 – h/R) . Hence  T2/T1 = √ g1/g2  = √ R / (R-h)

T2 =  T1  √  R/R-h , hence  T2  α  1/  √ (R-h) ,  therefore, T R/2 =  2  √ R/( R – R/2 )  = 2√2 s . Hence correct option is (c).

4. The bob of a simple pendulum ( density σ ) is immersed in a liquid of density ρ . The length of the string of the pendulum is l . Then the angular frequency of the small oscillations of the bob is –

• (a) T = 2π √ [ l/( g – Vρg/m) ]
• (b) T = 2π √ [ l/( g + Vρg/m) ]
• (c) T = 2π √ [  g – Vρg/m l ]
• (d) T = 2π √ [  g + Vρg/m l ]

Ans . The key factor here is that the bob experiences buoyant forces due to the liquid in addition to the usual gravitational and tensional forces . Now , the buoyant force = Vρg and so the net force on the bob = F = mg – Vρg . As the bob is undergoing small oscillations , its getting displaced and consequently there is a restoring force. Let the displacement be x ,

F = ( mg – Vρg ) sin θ   =  –  ( mg – Vρg ) x/l  . Now , acceleration =  –  ( g – Vρg /m ) x/l .

The standard form of SHM is  a = – ω² x   and so  ω =  √ [ ( g – Vρg /m )/ l ] .

Hence ,  T =  2π √ [ l/( g – Vρg/m) ] .  option (a) is correct.

## IIT JEE Main / Advanced Physics – PHYSICSMYND CODEX – Intro

In spite of being a breeze in the theory , and a whizkid as far as the mathematical apps and concept interlinking are concerned , do you find yourself floundering somewhere…… unable to reach the excellence you seek despite putting in the required hard work  ? Would you like to reverse the roles…….be in the steps of the examiner rather than the examinee , so that you can exactly come up with the answer the examiner expects from you….

Welcome to PHYSICSMYND CODEX – our treatise on the same , and in many ways , that which could be the game changer as far as our services are concerned. Each codex will deal the excerpts of a specific section / chapter , while the full discourse will be available to subscribers.

Given below is a breif overview of how the codex is created……. Stay tuned for our first codex on Work , Power and Energy.

## IIT JEE Main / Advanced Physics Free Notes 2018 – Alternating Current – All the essentials in one page [ PO’PS.Series.4]

ALTERNATING  CURRENT  [ A.C ]

Refers to the flow of electric charge that reverses periodically( opposite to direct current ) . It starts from zero, grows to a maximum, decreases to zero, reverses, reaches a maximum in the opposite direction, returns again to zero, and repeats the cycle indefinitely. Here , period is the time taken to complete one cycle and Frequency is  the number of cycles per second .The maximum value in either direction is the current’s amplitude. While low frequencies (50 – 60 cycles per second) are used for domestic and commercial power, frequencies of around 100 million cycles per second (100 megahertz) are used in television and of several thousand megahertz in radar and microwave communication. An important advantage of alternating current is that the voltage can be increased and decreased by a transformer for more efficient transmission over long distances.

A.C Current

A current or voltage is called alternating if –

1. its amplitude is constant   and
2. its half cycle alternates between positive and negative

In case the current or voltage changes periodically as the sin or cos function of time , its said to be sinusoidal. The functions are –

v = V sin ω t or   v = V cos ω t . Here , v – instantaneous potential difference , V – maximum potential difference [voltage amplitude ] and ω is the angular frequency [ = 2π x frequency ].

Also,  i = I cos  ω t or   i = I sin ω t , where i – instantaneous current and I – maximum current [ current amplitude ] .

Points to note –

• A rectifier is used to convert AC into a DC . For the reverse process , [ DC to AC ] an inverter is used.
• AC current is not useful for chemical processes like electroplating and electrolysis as large ions cannot follow the frequency of AC current.

PHASOR  DIAGRAMS

a phasor diagram

Rotating vector diagrams used to represent sinusoidally varying voltages / currents are referred to as phasor diagrams . Here , the instantaneous value of a quantity that varies sinusoidally with time is represented by projecting onto a horizontal axis of a vector with a length equal to the amplitude of the quantity.The term ‘ phasor ‘ refers to the vector which rotates counter-clockwise with a constant angular speed ω. As the projection of the phasor into the horizontal axis at time is I cos ω t , it’s the cosine function which is used.

## IIT JEE Main / Advanced Physics 2018 : Previous questions in General Physics / Dimensions

A   . OBJECTIVE  QUESTIONS  [ I ]  Only one correct answer   –

1    . The dimension of ½ ε0 E² ( ε0 – permittivity of free space , E – electric feild ) is  –  [ IIT JEE 2000 ]

• (a)  [ M L T -1]
• (b)  [ M L² T -2 ]
• (c)  [ M L T -2]
• (d)  [ M L² T -1]

2    . A quantity X is given by ε0 L ΔV/Δt , where ε0 is the permittivity of free space , L is a length , ΔV is a potential difference and Δt is a time interval. The dimensional formula for X is the same as that of  –   [ IIT JEE 2001 ]

• (a)  resistance
• (b)  charge
• (c)  voltage
• (d)  current

3    . A cube has a side of length 1.2 x 10 -2 m . Calculate its volume –  [ IIT JEE 2003 ]

• (a)  1.7 x 10 -6
• (b)  1.73 x 10 -6
• (c)  1.70  x 10 -6
• (d)  1.732 x 10 -6

4    . In the relation  p = α/β  e  -αZ/kθ  , p is pressure, k is Boltzmann constant  and θ is the temperature. The dimensional formula of β will be – [ IIT JEE 2004 ]

• (a)  [ M0 L² T0 ]
• (b)  [ M L² T ]
• (c)  [ M L0 T -1 ]
• (d)  [ M0 L² T -1 ]

5    . A wire has a mass ( 0.3 ± 0.003 ) g , radius ( 0.5 ± 0.005 ) mm and length ( 6 ± 0.06 ) cm. The maximum percentage error in the measurement of its density is –  [ IIT JEE 2004 ]

• (a)  1
• (b)  2
• (c)  3
• (d)  4

6    . Which of the following sets have different dimensions ?  [ IIT JEE 2005 ]

• (a)  Pressure , Young’s modulus , Stress
• (b)  Emf , Potential difference , Electric Potential
• (c)  Heat , Work done , Energy
• (d)  Dipole moment , Electric flux , Electric feild

7    . The circular scale of a screw gauge has 50 divisions and pitch pf 0.5 mm. Find the diameter of sphere . Main scale reading is 2 [ IIT JEE 2006 ]

• (a)  1.2 mm
• (b)  1.25 mm
• (c)  2.20 mm
• (d)  2.25 mm

8    . A student performs an experiment to determine the Young’s modulus of a wire , exactly 2 m long , by Searle’s method. In a particular reading , the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0kg.The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of ± 0.01mm. Take g = 9.8 m/s² ( exact ).The young’s modulus obtained from the reading is close to  –   [ IIT JEE 2007 ]

• (a)  ( 2.0 ± 0.3 ) 10¹¹ N/m²
• (b)  ( 2.0 ± 0.2 ) 10¹¹ N/m²
• (c)  ( 2.0 ± 0.1 ) 10¹¹ N/m²
• (d)  ( 2.0 ± 0.05 ) 10¹¹ N/m²

9    . In the experiment to determine the speed of sound using a resonance column  –  [ IIT JEE 2007 ]

• (a)  prongs of the tuning fork are kept in a vertical plane
• (b)  prongs of the tuning fork are kept in a horizontal plane.
• (c)  in one of the two resonances observed , the length of the resonating air column is close to the wavelength of sound in air.
• (d)  in one of the two resonances observed , the length of the resonating air column is close to half of the wavelength of sound in air

10   . Students I , II and III perform an experiment for measuring the acceleration due to gravity ( g ) using a simple pendulum . They use different lengths of pendulum and/or record time for different number of oscillations. The observations are shown in the table [ IIT JEE 2008 ]

Student    length of pendulum (cm)   No of oscillations (n)   Total time for oscillations (s)  Time period (s)

I      64.0      8    128.0   16.0

II     64.0      4      64.0    16.0

III    20.0      4      36.0      9.0

If E1 , E11 and E111 are the percentage errors in g, i.e, ( Δg/g x 100 ) , for students I , II and III respectively ,

• (a)  E1 = 0
• (b)  E1 is minimum
• (c)  E1 = E11
• (d)  E11 is maximum

B   . OBJECTIVE  QUESTIONS  [ II ]  One or more than one correct answer   –

11   . The dimensions of the quantities in one [ or more ] of the following pairs are the same . Identify the pair(s) – [ IIT JEE 1986 ]

• (a)  torque and work
• (b)  angular momentum and work
• (c)  energy and Young’s modulus
• (d)  light year and wave length

12   . The pairs of physical quantities that have the same dimensions is (are) –   [ IIT JEE 1995 ]

• (a)  Reynold’s number and coefficient of friction
• (b)  Curie and frequency of a light wave
• (c)   Latent heat and gravitational potential
• (d)  Planck’s constant and torque

13   . Let [ε0 ] denote the dimensional formula of the permittivity of the vacuum and [ μ0] that of the permeability of the vacuum. If M = mass , L = length , T = time and I = electric current , then –        [ IIT JEE 1998 ]

• (a)   [ε0 ] = [ M -1 L -3 T² I ]
• (b)   [ε0] = [ M – 1 L -3 T4 I² ]
• (c)   [μ0] = [ MLT-2 I -2 ]
• (d)   [μ0] = [ ML² T -1 I ]

14   . The SI unit of the inductance , the henry can be written as  –   [ IIT JEE 1998 ]

• (a)  weber/ampere
• (b)  volt-second/ampere
• (c)   joule/(ampere)²
• (d)  ohm-second

## IIT JEE Main / Advanced Physics 2018 – Thermodynamics,Kinetic theory – All the essentials in one page [ PO’PS.Series.1 ]

All prevailing thermodynamic physics – from the stars to the wave surfer

T H E R M O D Y N A M I C S

The branch of physics which deals with the relationships among heat , work , energy and temperature . Any physical system will spontaneously reach an equilibrium that can be described by specifying its properties, such as pressure , temperature, or chemical composition . Eventhough it’s based on physics , this branch has applications across all scientific streams including life sciences . In fact , physical life itself can be described as a continual thermodynamic cycle  of transformations between heat and energy. But these transformations are never perfectly efficient .

Def  : The branch of physics which deals with the laws that control  –

• the conversion of energy from one form to another
• the direction of the energy / heat flow
• the availability and other parameters required for energy to do work.

It works on the premise there should be a measurable quantity of energy called the internal energy ( U ) in an isolated system , anywhere in the universe . Internal energy here refers to the total kinetic and potential energy of the atoms and molecules of the system of any kind which can be transferred directly in the form of heat – hence , it excludes chemical and nuclear energy .

Internal energy ( U ) changes , when  –

• the system is not isolated   i.e,
• there is transfer of mass to or from the system  or
• there is transfer of heat ( Q ) to or from the system or
• there is work ( W ) being done on or by the system.

KINETIC THEORY OF GASES

Animation : The temperature of an ideal monatomic gas is a measure of the average kinetic energy of its atoms. Shown here is the movement of helium atoms .

Kinetic theory   –  Mind map

Kinetic theory – click to zoom

Kinetic theory of gases was basically devised to develop a model of the molecular behaviour which could account for the observed behaviour of an ideal gas.It deals with both the macroscopic ( like temperature , pressure ) as well as the microscopic ( like kinetic energy of molecules , momentum , speed ) properties of gas molecules .Using the kinetic theory, scientists can relate the independent motion of molecules of gases to their volume, viscosity, and heat conductivity.

Assumptions  –

1. Every gas contains extremely small particles called molecules which are all identical , but different from that of another gas.
2. Molecules of a gas are identical , spherical , rigid and perfectly elastic point masses.
3. Their molecular size is negligible when compared to intermolecular distances.
4. The volume of molecules is negligible when compared to the volume of a gas .
5. In a gas , molecules move in all possible directions with all possible speeds in accordance with Maxwell’s distribution law.
6. The gas molecules undergo perfectly elastic collisions with themselves and with the walls of the vessel.
7. The number of molecules is so large that a statistical treatment can be applied.
8. Collisions are characterized by molecules moving in straight lines with constant speeds.
9. Free path is the distance covered by the molecules between two sucessive collisions . The mean of all free paths is called a mean free path.
10. Except during collisions the interactions among molecules are negligible , i.e, no forces are exerted on one another.
11. The time taken for a molecular collision is negligible when compared to the time between two sucessive collisions.
12. For a gas , the no of collisions / unit volume remains constant.
13. Due to the very small masses and very high speeds of the molecules , gravitational attraction between them is ineffective.
14. Change in momentum caused by the molecules striking the container walls is transferred to the walls.Due to this the gas molecules exerts a pressure on the walls of the container.
15. Density of a gas is constant at all points of the container.
16. The average kinetic energy of the gas molecules depends only on the temperature of the system.
17. Relativistic and quantum – mechanical effects are negligible.
18. The equations of motion of the molecules of an ideal gas are time-reversible.

Ideal gas pressure

Let us consider an ideal gas containing N molecules , each of mass m . It’s confined to a box of sides a, b and c . Now , consider a molecule which hits the wall of the box with velocity v1. After hitting , v1 gets resolved into components vx,vy and vz along x , y and z axis respectively. The change in momentum along the x axis due to the collision is

Δ p = mvx – ( -mvx) = 2mvx .

Now , the time interval between succesive collisions on the same side a is 2a/v . The no of collisions per second by a single molecule on the same side is vx/2a. Hence , the change in momentum / second on the wall due to a single collision of a single molecule is –

vx/2a  x  2mvx =  mvx²/a  . Therefore , the change in momentum for all the collisions on a particular wall will be  ∑ ( mvx²/a )

Now , rate of change of momentum = force [ Newton’s second law ]. Hence , Fx =  ∑ ( mvx²/a )   =   m/a ∑ vx² .  Also , pressure = force/unit area , hence  Px = F/A   =   Fx/bc    =    m/abc ∑  vx²      =     m/V  ∑ vx²  . This means that ,

Px + Py + Pz  =    m/V   ∑ ( vx² +vy² + vz² )  . Now , as P = Px+Py + Pz   and  v²  = vx² +vy² + vz²   , hence   –

3 P = m/V  ∑  v²   . Now , the mean square velocity  =  (|v|)²  =   ( v1²  + v2² + ……. )/N  =   ∑ v² / N

Hence ,  3 P  = (m/V)  x N x  (|v|)²    PV = 1/3 mN  (|v|)² therefore ,

P  =  1/3   mN/V (|v|)²

Points to note  –

• Pressure,volume , mass and temperature are related by –

P  =  1/3 ( mN/V )  v²rms    i.e, [ as v²rms α T ]     P  α ( MN ) T/V

• When V and T are constant ,   P  α m N i.e, pressure α mass of the gas . Hence, when the mass of a gas increases , due to the increase in the no of molecules , the no of collisions / second also increases , leading to increased gas pressure.
• When m and T are constant , P α v²rms α T . This means that as the temperature increases , the mean square speed of the molecules also increases, and hence the rate of collision between themselves and with the walls of the container also increases.

P = 1/3 ( mN/V )  v²rms   . Now , as M = mN = total mass of the gas ,

P = 1/3 ( M/V )  v²rms    =    1/3 ρ v²rms –     eq.1

• For an ideal gas , pressure and kinetic energy are related as  –

K.E = 1/2 M v²rms   =   1/2 [ M/V ] v²rms  ( for unit volume )  =  1/2 ρ v²rms   – eq.2

Combining eqs 1 & 2 ,   P = 2/3 E

• Hence , pressure of an ideal gas is numerically equal to two-thirds of the mean kinetic energy of translation ( kinetic energy due to motion from one location to another ) per unit volume of the gas.

## IIT JEE Main / Advanced Physics 2018 : Simple Harmonic Motion – kind of questions to expect /practice tests.

SECTION  A   OBJECTIVE QUESTIONS   –  Only one correct answer  –

1  . Which of the following doesn’t represent Simple Harmonic Motion  ?

• (a)  y = a sin ω t + b cos ω t
• (b)  y = 1 – 2 sin² ω t
• (c)  y = ( √ a²+b² ) sin ω t cos ω t
• (d)  y = a sin ω t + b cos²  ω t

2  . A particle of mass m is executing oscillations about the origin on the x -axis . It’s potential energy , U (x) = k |x|³ where k is a positive constant . If the amplitude of oscillation is a , then it’s time period, T is

• (a)  proportional to √a
• (b)  proportional to 1 / √a
• (c)  independent of a
• (d)  proportional to a 3/2

3  . A spring of force constant k is cut into two pieces such that one piece is double the length of the other. Then ,the long peice will have a force constant of –

• (a)   2/3 k
• (b)   3/2 k
• (c)   3 k
• (d)   6 k

4  . Two bodies A and B of equal masses are suspended from two separate massless springs of spring constants k1 and k2 respectively . If the two bodies oscillate vertically such that their maximum velocities are equal , the ratio of the one amplitude of vibration of A to that of B is –

• (a)   k2/k1
• (b)   √ k2/k1
• (c)   k1/k2
• (d)   √ k1/k2

5  . One end of a long metallic wire of length L is tied to a ceiling . The other end is tied to a massless spring of spring constant k . A mass m hangs from the free end of the spring . The area of cross-section and the Youngs modulus of the wire are respectively A and Y . If the mass is slightly pulled down and released , it will oscillate with a time period, T equal to –

• (a)   2π [ mYA/kL ]½
• (b)   2π [ m/k ]½
• (c)   2π √ [ m ( YA + kL)/YAk ]
• (d)   2π ( mL/YA )½

6  . A particle which is free to move along the x-axis has potential energy given by U(x) = k [ 1-exp ( -x²)] for – ∞ ≤ x ≤ + ∞ , where k is a positive constant of appropriate dimensions. Then

• (a)   at points away from the origin , the particle is in unstable equilibrium.
• (b)   for any finite non zero value of x , there is a force directed away from the origin.
• (c)   for small displacements from x = 0 , the motion is simple harmonic.
• (d)   if it’s total mechanical energy is k/2 , it has it’s minimum kinetic energy at the origin.

7  . A simple pendulum consists of a small sphere of mass m carrying a positive charge q , and is suspended by a string of length l . The pendulum is placed in an uniform electric feild of strength E directed vertically upwards.Calculate the period with which the pendulum will oscillate , if the electrostatic force acting on the sphere is less than the gravitational force ?

• (a)   2π √ g + qE
• (b)   2π √ g – ( qE²/m )
• (c)   2π √ [1/ (g – qE/m) ]
• (d)   2π √g + ( qE/m)²

8  . A block of mass M attached to a horizontal spring of spring constant k , is undergoing simple harmonic motion . At the instant the block passes through its equilibrium position , a lump of putty of mass m is dropped vertically on to the block from a small height and gets stuck to it. Find the new period with which the block oscillates ?

• (a)   2π √(M+m ) k
• (b)   2π (M+m)²/√k
• (c)   2π √(M+m/k)
• (d)   2π √(M+m)/k²

9  . A person of mass 60 kg stands on a platform executing SHM in the vertical plane . The displacement from the mean position varies as y = 0.5 sin ( 2π f t ) . The value of f for which the person will feel weightlessness at the highest point is ( y is in metres ) –

• (a)    √2g/2π
• (b)    2π √ 2g
• (c)    √g/4π
• (d)    2π √4g

10 . A solid cylinder of density ρ1 having cross-sectional area a and height h is floating in a liquid of density ρ2 ( ρ1 >ρ2 ). The cylinder is pushed down slightly and made to oscillate vertically. The frequency of oscillation is given by –

• (a)   ω = √ g ρ1/h ρ2
• (b)   ω = √ (g ρ1 /ρ2) h
• (c)   ω =  g ( ρ1 +ρ2 )² h
• (d)   ω = √ g ρ1 h/ ρ2

11 . A sphere of mass m and radius r rolls without slipping on a rough concave surface of large radiur R . It undergoes small oscillations about the lowest point . Then , the time period is given by –

• (a)   2π √(R-r)²/4g
• (b)   2π √(R+r)/mg
• (c)   2π √R²/5rg
• (d)   2π √7 (R – r)/5g

12 . A particle of mass m is present in a region where the potential energy of the particle depends on x-co-ordinates according to the expression U = a/x² – b/x , where a and b are positive constants . The particle will perform –

• (a)   simple harmonic motion with time period 2π √ 8a²m/b4 about its mean position for small displacements.
• (b)   oscillatory motion but not simple harmonic motion about its mean position for small displacements.
• (c)   neither simple harmonic motion nor oscillatory motion about its mean position for small displacements.
• (d)   none of the above.

13 . x (t) = 3 sin ( π t + π /4 )  represents the SHM undergone by a body of mass 0.2 kg. Then the total energy of the body if potential energy is zero at the mean position is –

• (a)   1.9 π²
• (b)   3.2 π²
• (c)   7.1 π²
• (d)  0.9 π²

14 . A pendulum hangs from the roof of a car which is moving on a smooth inclined plane of angle Φ . The mass of the bob of the pendulum is m and the length of the pendulum string is l .Then the angular frequency of the small oscillations of the bob is –

• (a)  ω = √ g sin Φ /l
• (b)  ω = √ g cos Φ/l
• (c)  ω = √ g tan Φ/l
• (d)  ω = √ g cos Φ/2l

15 . A particle undergoing SHM having a time period of 3 s is in phase with another particle , also undergoing SHM at t = 0. The time period of the second particle is T ( less than 3 s ) . If they are in the same phase for the third time after 45 s, then the value of T is –

• (a)  2.4 s
• (b)  2.7 s
• (c)  2.5 s
• (d)  3.1 s

16 . The equation x = A sin ωt + B cos ωt represents the motion of a particle . This motion is –

• (a)   simple harmonic with amplitude A + B
• (b)   simple harmonic with amplitude √ A² + B²
• (c)   simple harmonic with amplitude ( A+B)/2
• (d)   not simple harmonic

17 . In simple harmonic motion , the

• (a)   potential energy is never equal to the kinetic energy
• (b)   potential energy is always equal to the kinetic energy
• (c)   the average potential energy in a particular time period is equal to the average kinetic energy in that period
• (d)   the average potential energy in any time interval is equal to the average kinetic energy in that time interval

18 . The equation | r| = (ˆi + 2 ˆj ) A cos ωt represents the motion of a particle in the x-y plane . This motion is –

• (a)   simple harmonic
• (b)   periodic
• (c)   on a straight line
• (d)   on an ellipse

( to continue …… )

## IIT JEE Main / Advanced Physics 2018 : Atomic Physics – mind map

Mind map representation of Atomic physics .

Atomic Physics . click on image to zoom

## IIT JEE Main / Advanced Physics 2018 – Fluid Mechanics – Mind map

Given below is the representation of  IIT JEE Main / Advanced Physics 2013 portion of fluid mechanics as a mind map

Fluid mechanics-mind map.click on image to zoom

## IIT JEE Main / Advanced Physics 2018 Memory Maps – Simple Harmonic Motion [ SHM ]

Memory maps are considered quite useful in recollecting concepts and their connections , more so in the case of a scientific stream like Physics. As far as revising the theoretical portions are concerned , this pedagogical method is considered a valuable time saver. In this first part of our series , we deal with Simple Harmonic Motion. The key terms with the respective equations are provided along with this memory map. Students must note that such methods are best suited for remembering the once learned portions , and hence , they in itself won’t benefit one much , if one hasn’t done the required preparation , especially for an exam like IIT JEE .

Mind map – SHM . Click on image to zoom

Key to the SHM Memory Map  [  from clockwise ]  –

• F = – kx is the standard condition for a  particle to oscillate in a stable equilibrium. Here F is the restoring force and k is a positive constant . For F = -kx , the force is along the negative x axis for x > 0, along the positive x axis for x < 0 and zero for x = 0 .

Kinematics of SHM – it refers to the specifics of the oscillatory motion .  The displacement of a particle undergoing SHM along an axis OX ( O – origin ) , relative to the origin of the co-ordinate system is  given by –

x  = A cos ( ω t + Φ ) , where  ( ω t + Φ )  is called the phase angle and Φ  is the initial phase or phase at t = 0 .

• Period of SHM,  T  =   2 π/ω
• Acceleration,     a  =  – ω²x
• Velocity ,             v  = – ω A sin ( ω t + Φ )
• Frequency ,       v =    ω / 2π

Force in SHM     –

• F  =  – m ω²x

Energy in SHM   –

• Kinetic energy   –  K = ½ k  ( A² –  x² ) [ i.e , K.E  is max at the centre (  x = 0 ) and zero at the extremes (  x ± A ) ]
• Potential energy – U=  ½ m ω² x² [ min at centre and max at the extremes ]
• Total energy –      ½ m ω² A²  =    ½ k  A² [ this is a constant as the force concerned is conservative ]

Fluid column –

• When it’s ‘ V ‘ shaped   T = 2 π √ [  m /ρ g A ( sin θ1 + sin θ2 ) ]
• When it’s ‘ U ‘ shaped   T = 2π √ l / g

Basic differential form of SHM  –  from  d²x/d t² + ω² x = 0  , three forms can be written –

• x  =  A sin ( ω t + Φ )
• x  =  A cos ( ω t + Φ )
• x  =  A sin  ω t + B cos ω t

Finding Time period  –

• Torque [ restoring force ] method   T = 2π √ x / a or   2π √ θ / α
• Energy method    ΔT  =   ½ T α  Δθ

Spring block systems  –

• A single spring to a block ,  T  =   2π √ m /k   =  2π √ l / g
• Two springs in series to a block  T =  2π √ [ m ( k1 + k2 )/ k1 k2 ]
• A spring between two blocks on a horizontal surface ,   T  =   2π √ μ /k where μ = ( m1 m2 / m1 + m2 )

SHM and Uniform circular motion –

• Velocity ,             v  =   – ω A sin ( ω t + Φ )
• Acceleration ,    a  =   – ω² A cos ( ω t + Φ )

Physical Pendulum  –

• T =  2π √ leff /g , where  leff = K² + l² / l = length of an equivalent simple pendulum .