The Future of Physics – the next 100 years : Michio Kaku

As our dear visitors would be knowing , Physicsmynd is not only about serious learning , it’s also about some mi (y)nd teasers how about what would happen in Physics in the next hundred years ? This is not science fiction , they are the predictions of a top-notch theoretical physicist , a celebrated author as well as a well known science communicator – Dr.Michio Kaku , who had his education from Princeton , Harvard and University of California , Berkeley. This is about his latest best seller – ” Physics of the Future : How Science Will Shape Human Destiny and Our Daily Lives by 2100


See his video chat here

The ‘Relativistic’ growth of Physicsmynd……

Physicsmynd - the antibiotic to all physics ' headaches' .

 We are happy to inform the student community of the decision to revamp the Physicsmynd web services . In this task , we are much encouraged by the  well wishes , support and feedback being received fom the students / parents , without which this endeavor wouldn’t have been possible. The revamp would be done in a two tier mode , with the first one featuring , in addition to all the currently available free resources –

  • IIT JEE Physics Free Preparatory notes / problems
  • AIEE Physics Free resources

In the second tier , we would be looking at quality resources / services relating to the following syllabus levels –

  • NCERT / CBSE  [ +2 ] Physics
  • ICSE  [ +2 ] Physics
  • IGCSE Physics [ University of Cambridge – School Level International Examinations ]

As we prepare ourselves to expand our horizons, our motto will remain the same – i.e, of a physics online service which is not just meant for exam / competitive purpose , but to enable students to elucidate , explore  and enjoy the wonderful science which governs the rules encompassing everything between the atoms to distant galaxies and more called PHYSICS

We welcome your suggestions / feedback 

IIT JEE Main / Advanced Physics 2018 – Magnetism: Last Minute Revisioner [ LMR] / Cheat Sheet

Given below is a last minute revisioner of Magnetism for the IIT JEE Main/ Advanced Physics paper . Note that the points covered are based on the exam point of view , and is not meant to be a comprehensive review of the entire magnetism portion.

  • A moving charge / flowing electric current produces a magnetic feild ( which is a vector feild ) in addition to the electric force / feild . If a charged particle moves parallel to the magnetic feild , the magnetic force on the charge is zero . The magnetic force differs from an electric force in the fact that a) it depends on the particle velocity b) it’s direction is perpendicular to both |v| and |B| and c) it does no work in displacing a charged particle , provided the magnetic feild is constant.
  • Consider a charged particle of mass m moving in an uniform magnetic feild | B | having an initial velocity vector | v | perpendicular to the feild. Then , the path radius = r = mv / qB , angular speed ω = qB/m and time period, T = 2 π m/qB.If the motion in the uniform magnetic feild is at some arbitrary angle θ with respect to B , then the path becomes helical – here there are two velocity components, the perpendicular component moving the charge in a circular path of radius given by r = mv1/qB and the parallel component moving along the feild lines. Pitch p = 2 π m vΦ/qB.
  • Consider a charged particle moving in an electric feild which is perpendicular to the magnetic feild . Here , the particle velocity is perpendicular to both the feilds. The magnetic feild will rotate the particle in a circle in the x-z plane , || to the magnetic feild resulting in a helical path with increasing pitch . vx = vo cos ( Bqt/m ) and vz = vo sin ( Bqt/m ) .
  • A current carrying conductor of any arbitrary shape in an uniform magnetic feild will experience a force |F| = I|L|x|B| where |L| is the length vector joining initial and final points of a conductor. If it’s a closed loop then  |F| = zero. If the magnetic feild is not uniform , various elements of the loop will experience different forces – means that if the loop is free enough , the loop could have a circular motion , ie , |Fr| = zero, but |ζ | may or may not be zero.
  • Considering the force between two infinite , parallel current carrying conductors , the force / unit length = μo I I’/2 π r . Note that the if the direction of current flow is the same , the wires will attract each other.
  • As the magnetic feild pattern produced by a small current loop is like a bar magnet, it acts like a magnetic dipole . Here , if the loop is not lying in a single plane , two equal and opposite currents are to be assumed in a single branch [ note that the net change is zero ] and likewise the required loops are to be considered in different planes . For eg , for a cube of side I carrying a current i  , |M| net = – i I² ( iˆ + kˆ ) .
  • If a magnetic dipole changes it’s orientation in a magnetic feild by an infinitesimal angular displacement d Φ , the feild does work dW given by ζ d Φ  =  dU , the change in potential energy . The energy present in the loop is U = – |M| |B| .
  • A moving point charge with velocity v creates circular magnetic feilds centered on the line of v and which are perpendicular to it. The magnetic feild , |B| = ( μo /4π ) q|v|rˆ/r² .
  • For a current carrying straight wire , the feild is zero for points along the length of the wire , but not on it. Here , as the feild is perpendicular to the plane containing both the wire and the point , the lines of force are concentric circles which encircle the wire. For the infinite long wire , if the point is near the end , B = ( μo /4π ) 2I/d and if the point is near one end , then, B = ( μo /4π ) I/d .
  • The magnetic feild at the centre of a current carrying arc is B = μo I/2 R b)  a point inside a long solenoid B = μo n I c) for a point at one end of a long solenoid B =  μo n I /2 and d) for a point at a distance R from the centre of a flat strip of width a along it’s perpendicular bisector, B = ( μo i /π a ) tan -¹ ( a/2R )  Note – in cases where the points considered are far from the strip , B = ( μo /2π ) ( i/R ) .

IIT JEE Main / Advanced Physics 2018 – Electricity : Last Minute Revisioner [ LMR] / Cheat Sheet

The last minute revisioner given below covers the salient points of the relevant sections from the exam point of view . It’s not meant to be a comprehensive review of the entire chapters.

  • Charge is an intrinsic ( it’s own ) , individualistic and indestrucible property of a body/particle which results in electric forces.  No of charges can be reduced or increased by friction , conduction or induction. Charges are quantized , conserved and additive. A non-accelerated moving charge also causes magnetic forces in addition to the electric forces. If it’s accelerated , the charge also radiates energy.
  • The dielectric constant ( K = εr = ε /εo)  in Couloumb’s law is usually ≥ 1 . When a charge is placed in a medium , the force decreases K times. Coulomb’s law a) applies only to point charges b) obeys Newton’s third law c) is not applicable to distances below 10 -15 [ – fifteen ] m and d) follows the superposition principle. Note that the interaction between two charges is not influenced by a third charge .
  • The period of revolution of a particle of mass m of charge – q1 moving around a fixed charge + q2 in a circular path of radius r is given by T = √  ( 16 π³ εo m r³ / q1 q2 )
  • Consider a thin fixed ring of radius a having a positive charge q uniformely distributed over it . Now , if a particle of mass m and having a negative charge Q is placed on the axis at a distance of x ( x << a ) from the ring’s centre , the negative charge  undergos oscillatory motion nearing simple harmonic as the restoring force is not linear. T = √ 2 π ( 4 π εo m a³ /q Q )
  • To keep the original charge distribution unchanged , test charges are usually of a small value. If a test charge doesn’t experience a force at a point  , the net electric feild at that point should be zero. Note that a charged particle is not affected by it’s own feild.
  • Electric feild lines helps us to visualize the nature of electric feild in a given space. They a) never cross each other b) never form closed loops and c) doesn’t pass through a conductor. Eventhough the tangent to the line of force in an electric feild at a point gives the direction of force / acceleration  acting on a positive charge at that point , it isn’t necessary that the charge will move in the said direction of force.
  • The maximum value of electric feild of a ring shaped conductor having radius r ,carrying a total uniformely distributed charge Q at a point P which lies on the axis of the ring at a distance x from it’s center is 1/4 π εo ( 2Q /[ 3 √ 3 R² ] )
  • The electric feild produced by an infinite plane sheet of charge is independent of the distance from the sheet . This feild is uniform and it’s direction is perpendicular to the sheet and away from it.
  • The net electric feild of a linear charge / charged ring / semicircular charge and a circular charged arc is ∫ d E cos θ . The net electric feild of a charged rod of fixed length having charged density λ along x axis is λ / 4 π εo ro ( sin θ1 + sin θ2 ) and along y axis is λ / 4 π εo ro ( cos θ1 – cos θ2 )  , of a semicircular ring having charge density λ – x axis λ 2 π εo r and along y axis is zero , of a quarter circular ring of charge density λ – x axis and y axis  is λ4 π εo r.
  • Electric feild intensity of a short dipole at some general point is [ p / 4 π εo r³ ] √ (1 + 3 cos² θ ) . Here the resultant feild intensity vectors E2 / E1 = tan Φ
  • Electric flux through a surface describes whether electric feild points into or out of the surface. In an uniform feild , the flux through a closed surface is zero. For a surface of area A tilted at an angle Φ , flux is E A cos Φ and volume flow rate through A is dV/ dt = v A cos Φ .
  • Gauss’s  law is most useful in dealing with charge distributions of spherical / cylindrical symmentry or when charge distribution is uniform over a plane. For point / spherical charge distributions gaussian surface is spherical while that for a line of charge it’s cylindrical. For feild of a line charge it’s 1 / 2 π εo ( λ/r ) , for an infinite plane sheet of charge  σ / 2 εo and at the surface of a conductor it’s σ / εo , inside a uniformely charged sphere ρ r / 3 εo and outside is  ( 1/4π εo ) ( Q / r² ) . Electric feild of a long uniformely charged cylinder is ρ R²2 εo r and that inside a uniform volume charged plane is  ρ r/εo and that of the outside is ( ρ /εo ) d/2
  • The work done by an electric force to move a point charge through a small displacement dl depends only on the change in the distance dr between the charges ( i.e , the end points are important and not the path taken ), which is the radial component of the displacement. Hence the force acting is conservative. The potential energy U = 1/4π εo (q qo/r ) .
  • Electric potential is a scalar quantity where the principle of superposition can be applied . This means that for a discrete distribution of charges where V = V1 + V2 + V3 +… , = ( 1/4π εo ) Σ q/r .In case the potential is caused by a continuous charge distribution , then , V = ∫ dq /4π εo .
  • Equipotential surfaces can be volumes , surfaces or lines . Note – a) they will never cross each other as the potential at a point can’t have two values b) for rest charges , the conducting surfaces are always equipotential c) equipotential surfaces are always perpendicular to the electric lines of force and d) the work done in moving a charge from a point to another point in an equipotential surface is zero.
  • The cartesian components of electric feild w.r.t the electric potential are Ex = – ∂V/∂x , Ey = – ∂V/∂y and Ez = – ∂V/∂z . Note – the equation indicates that the negative of the rate of change of potential with position in any direction is the component of E in that direction. Negative sign means that E is pointing in the direction of decreasing V.
  • For an infinite rod , the potential at any point near the rod is infinite . But the potential difference between any two points a and b is Va -Vb  =  ( λ / 2 π εo  ) In ( b/a ) .

IIT JEE Main / Advanced Physics 2018 – Geometrical Optics : Last Minute Revisioner [ LMR] / Cheat Sheet

IIT JEE Advanced & Mains Physics 2018 : Optics : Rank file / expected questions / Physicsmynd Elite Series       

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Given below is a last minute revision note / cheat sheet / ready reckoner of the Geometrical optics section for the IIT JEE Main / Advanced  Physics paper. Students should note that the material is not meant to be a comprehensive review of the section. Only those points which we feel are more relevant in the exam point of view are included .

  • Light experienced by us is the effect of neural stimulation of certain cell types ( cones & rods ) caused by a paticular section ( visible ) of the electromagnetic spectrum. The point is , we as such doesn’t see light , we experience it…  for eg. , rays of light one see when shining a torch in a dark room is due to the scattering of light by dust particles. The inference here is that only when an object scatters / reflects light we will be able to see it . If the light rays pass through an object without any change in it’s nature , i.e  if it’s perfectly transparent , we will not be able to see the object. Interestingly , if an objects reflects light perfectly [ say a perfect mirror ] , then again one will be seeing the reflected light which will not give any idea of the said object. Hence , to see an object , it has to partially reflect or absorb the incident light.
  • According to Ray optics , light consists of particles / corpuscles travelling in straight lines . This branch of optics can explain reflection , refraction and dispersion.
  • According to Wave optics , light is of an electromagnetic wave like nature. It can explain phenomenon like diffraction,interference and polarization.
  • According to the Quantum theory , light consists of quantized particles called photons . Their momentum and energy are finite and have zero rest mass. Light has the properties of both particles and waves. This theory accounts for phenomena like photoelectric effect and spectral lines.
  • For real objects , the incident rays diverge from a point [ converges in virtual objects ] . In the case of real images , the rays meet a point after their contact with the optical element , hence the image can be seen on a screen [ for virtual images , the rays appear to come from a point , hence such images cannot be captured on a screen ]   .
  • Eventhough virtual images cannot be captured on a screen , we can see ourselves in front of a plane mirrorthis is because of the focusing of the divergent rays by the eye lens. Same is the case with cameras.
  • A light ray , in a homogeneous media , propagates in a straight line and doesn’t interfere with another ray at intersections . The ray also can retrace it path on reflection .
  • When light is incident normally , i = r = 0 and δ = 180° . If the incident ray strikes tangentially,  i = r = 90° and δ = 0°.
  • The angle of deviation of an incident ray getting reflected at an angle i can be calculated by δ = π – 2 i or  δ = π + 2 i .
  • When a mirror is rotated through an angle θ , in an axis which lies in the plane of the mirror and perpendicular to the plane of incidence , a fixed incident ray gets reflected through an angle 2 θ.
  • An extended object in front of a plane mirror forms an image such that  a) size of the image and object are equal b) parallely placed object creates an upright image and c) perpendicularly placed object creates an inverted image .
  • For an observer to see his full image on a plane mirror , the mirror has to be at leat half the size of the observer.
  • For a plane mirror , velocity of an object is equal to the velocity of the related image , when parallel to the mirror surface. If vo m = velocity of object with respect to mirror , vi m = velocity of image w.r.t mirror , vi= velocity of image w.r.t ground and vo = velocity of object w.r.t ground , then , vo m = vo – vm and  vi m = vi – vm .
  • To find the number of images formed by two inclined plane mirrors , inclined at an angle θ , a) if 360°/θ = even number , then the number of images =  360°/θ – 1  , and b)  if 360°/θ = odd number , then the no of images = 360°/θ – 1 , if the object is placed on the angle bisector and it’s  360°/θ  if the object is not placed on the angle bisector .
  • The sign convention used to solve problems in the case of spherical mirrors , for both reflection and refraction are a) distances are to be measured from the pole b) distances measured in the direction of incident rays are taken as positive c) distances measured opposite to the direction of incident rays are taken as negative and d) distances above the principal axis are taken as positive and those below the principal axis are taken as negative.
  • To find the image of an object graphically , any two of the following rays are to be considered – a) the ray passing through the centre of curvature which gets reflected back along itself  b) the ray initially parallel to the principal axis which gets reflected through the focus of the mirror  c) the ray initially passing through the focus which gets reflected parallel to the principal axis and  d) the ray incident at the pole which is reflected symmetrically.
  • The optical power of a mirror is P =  – 1/f where f is the focal length in metres. Note that P is expressed in diopters.
  • If fo – focal length of a mirror and x is the distance of the object from the mirror , then , magnification for a concave mirror is  m = fo/( fo – x ) and that of a convex mirror is m = fo/( fo + x ) . Note – if a virtual object is placed between F and P , then a convex mirror gives a real image . Here , m = – f/( fo – x ).
  • For spherical mirrors , the object and image velocities are related by vi = – m² vo where vi = velocity of the image w.r.t the mirror , vo = velocity of the object w.r.t the mirror and m = lateral / transverse magnification a) if the object is at the centre of curvature , |dv/dt| = |-du/dt| , b) if the object is moving between the focus and the centre of curvature  |dv/dt| > |-du/dt| c) if the object is moving between the focus and the mirror pole , then [dv/dt ] = [ v²/u² ] [du/dt ] here , as |m| > 1 , the speed of the image will be greater than the speed of the object and d) for a small object placed along the principal axis , dv = – m² du . [ Note – in all the above cases , |dv/dt| and  |du/dt| are velocities w.r.t the mirror and not with that of the ground ].
  • Differentiation of the mirror formula gives  dv/du = v²u² . Here dv/du is called longitudinal magnification . It should be noted that in this case , the focus should not lie between the initial and final points of the object.
  • For an object moving perpendicular to the principal axis , h2/h1 = v/u  , hence h2 = ( -v/u) h1. Here , if we are dealing with a point object , then as the x co-ordinates of both the object and image are constant , we get on differentiation , dh2/dt = – [ v/u ] [ dh2/dt]
  • Newton’s formula is useful when distances are asked related to the focus . XY = f ² , where X and Y are distances along the principal axis of the object and the image respectively.
  • When using Snell’s law in cases of refraction , use n1 sin i = n2 sin r . Refraction doesn’t change the frequency of the light. The refractive index of a medium relative to vaccum is √ (μr εr )
  • Apparent shift of an object due to refraction is given by = real depth x [ 1 – μ21] . If μ1 < μ2 , then image distance is greater than object distance and the shift is negative.
  • If the angle of inidence is small ,the apparent shift of an object due to refraction is  d’ = d / n , here d’ = apparent depth , d = real depth and n = ni/nr = refr.index of incident medium / refr.index of refractive medium.
  • Apparent shift of an object due to refraction of a parallel slab is given by s = [ x+ t/μ ] – [ (x+t)] = t [ 1 – ( t/μ ) ] . The object appears to shift along the perpendicular of the slab by a distance t [ 1 – ( t/μ ) ] .Note that the final image is at a distance x + ( 1/μ ) behind the second interface.

IIT JEE Main / Advanced : Physics 2018 – PHYSICSMYND Brainwave(r).6 – The Case of the Distant Murmurs .

In the sixth series in our Brainwaver posts , we are intruding into some ‘ novel ‘ territory. Anyone who has read literature scenarios dealing with vast landscapes and mountain scenes will remember the author musings regarding the unsual nature of sounds and its effects being mentioned with thrilling accroutrements. And which fan of Sherlock Holmes will forget the chilling sound effects reverberating in the environs of the  ‘Hound of the Baskervilles’ ?

The Case of the Distant Murmurs

Well , coming to the point , it is not unusual to hear even faint sounds / murmurs originating from a distant place , in plain landscapes and hilly countryside especially when a wind is blowing towards you from the area of the sound source.

( a ) What is the reason behind this…. is it because the velocity of sound is enhanced by the blowing wind ? (b) Wind blowing across the ground has a vertical velocity gradient given by the formula v = k y² , where y is the height above the ground and k is a constant which depends on the wind speed outside the boundary layer where the parabolic velocity profile is a good approximation. Now , for a given value of k and speed of sound vs , find the distance s , downwind from a sound source , where the maximum enhancement of sound intensity happens ?  Assume that the sound rays follow low , arc like paths represented by y = h sin ( π x / s ) . (c) Even without any wind , one is able to hear more clearly , sounds coming across a water body like a lake – what is happening here ?

Ans . (a) The clarity of the sounds / murmurs heard cannot be simply attributed to the wind blowing in the observer’s direction , as , if this were the case , then any observer in any path of the wind’s direction would hear the sounds with the same clarity. Hence , one has to look at the refractive effects of sound brought about by the variation of the sound velocity, with respect to a stationary observer , at different points of the medium. The key factors to consider here are the velocity gradient and the temperature gradient of the moving wind.  Now , the velocity of compressible waves in a gaseous medium varies with temperature  T as √ T . Variations also occurs if the velocity of the medium itself varies. Now , the refraction of sound waves will change the direction of its wavefront . Considering the surface of the earth in contact with the blowing wind , both gradients may be present and the path of the sound waves could be refracted / bend in multiple ways , causing a distant observer to hear the sounds with remarkable clarity.

(b) The wind velocity near the ground is assumed to be horizontal with a vertical gradient  v = vx = ky².  Here the medium is considered to consist of horizontal layers with different sound velocities. Now , if θ is the angle between the direction of sound wave in the layer and the vertical , and V is the velocity of sound with respect to the ground , then by the law of refraction ,  sin θ / V = k ( a constant ).

θ1  =  θ1        ,         V 2  = vx +  vx sin θ  =  vx+ v sin θ .

θ2 =  θ + d θ1  ,   V2  = vx + ( v + d v ) sin ( θ + d θ )

Now , using the law of refraction –

[  vx + ( v + dv ) sin ( θ + d θ ) ] /  [ vx + v sin θ ]  = [  sin θ + d θ ] / sin θ  .  Now , sin ( θ + d θ )   ≈ sin  θ + cos θ  d θ , and by retaining the lowest order terms , we get  –

dv / vx =   d sin  θ / sin² θ    and so    ∫h0 2 ky ( dy / vx)  =   ∫π/2θ0 ( d sin θ / sin²  θ )

[ Note –  h is the max height of the parabolic wave front from the horizontal ]

=  k h² / vx =  1 / sin θ0 – 1   . Now , the given sound path gives   cot  θ  = dy / dx = ( π h /s ) cos  ( π x / s )   or

1 / sin θ  =   √ ( 1 + cot² θ )  =   √  ( 1 + [ π h /s ]²  cos² [  π x /s ] )

and   1 / sin θ0  =   √  ( 1 + [ π h /s ]² ) .  Using this equation provides the path length s , down wind from the sound source where the maximum enhancement of sound intensity occurs as   –

s   =    π vs /  √  ( k [ 2 vs + kh² ] ).

(c) Now , the speed of sound in a gaseous medium changes with absolute temperature T as  √ T . In the case of a lake surface , it could happen , that a temperature gradient is formed some distance above the surface due to the effect of water ( evaporation ). [ Note that in the problem it’s clearly inferred of a windless situation – this is necessary to maintain the stability of a temperature gradient ] . In such a case , sound waves will get refracted resulting in the clarity of sound heard by an observer on the lake shore .

IIT JEE Main / Advanced Physics 2018 : Free Study Material – Feynman Lectures

In the second part of our series on free study material / videos / podcasts of top notch quality ,  we proudly present the lecture notes of one of the most celebrated physicists – Richard Feynman .

Richard Feynman

He recieved the Nobel Prize in Physics in 1965 for his contributions to the development of quantum electrodynamics and developed a widely used pictorial representation scheme for the mathematical expressions governing the behavior of subatomic particles , which later became known as the Feynman diagrams . He was a member in the panel which investigated the Space Shuttle Challenger disaster. His other major contributions where in the feilds of quantum computing and in conceptualizing nanotechnology.

He was the Richard Chace Tolman Prof. at the California Institute of Technology. For physics students worldwide he is well known by his lecture series – The Feynman Lectures in Physicsconsidered one of the best and fundamental physics study material and one among the essential reading notes for IIT JEE Physics preparation .

PDF Download :

IIT JEE Main / Advanced Physics 2018 : Geometrical Optics – kind of questions to expect /practice tests.

IIT JEE Advanced & Mains Physics 2018 : Optics : Rank file / expected questions / Physicsmynd Elite Series       

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A . Objective questions with only one correct answer  –

1  . Rays of light from the sun falls on a biconvex lens of focal length f and the circular image of the Sun of radius r is formed on the focal plane of the lens . Then ,

  • (a)  area of image is π r² and area is directly proportional of  f  .
  • (b)  area of image is π r² and area is directly proportional to r².
  • (c)  intensity of image increases if  f is increased.
  • (d)  if lower half of the lens is covered with black paper , area will become half .

2  . A ray of light travelling in water is incident on its surface open to air . The angle of incidence is θ , which is less than the critical angle . Then , there will be –

  • (a)  only a reflected ray and no refracted ray .
  • (b)  only a refracted ray and no reflected ray .
  • (c)  a reflected ray and a refracted ray and the angle between them would be less than 180° – 2 θ .
  • (d)  a reflected ray and a refracted ray and the angle between them would be greater than 180° – 2 θ .

3  . A convex lens is in contact with a concave lens . The magnitude of the ratio of their focal lengths is 2/3 . Their equivalent focal length is 3o cm . What are their individual focal lengths ?

  • (a)  – 15 , 10
  • (b)  – 10 , 15
  • (c)  75 , 50
  • (d)  – 75 , 50

4  . A container is filled with water ( μ = 1.33 ) upto a height of 33.25 cm . A convex mirror is placed 15 cm above the water level and the image of an object is placed at the bottom is formed 25 cm below the water level . Focal length of the mirror is –

  • (a)  15 cm
  • (b)  20 cm
  • (c)  – 18.31 cm
  • (d)  10 cm

5  . A short linear object of length b lies along the axis of a concave mirror of focal length f at a distance u from the pole of the mirror . The size of the image is approximately equal to

  • (a)  b ( u – f / f )½
  • (b)  b ( b / b – f )½
  • (c)  b ( u – f / f )
  • (d)  b ( f / u – f ) ²

6  . A thin prism P1 with angle 4° and made from glass of refractive index 1.54 is combined with another thin prism P2 made from glass of refractive index 1.72 to produce dispersion without deviation . The angle of the prism P2 is –

  • (a)  5.33°
  • (b)  4°
  • (c)  3°
  • (d)  2.6°

7  . A ray enters a glass slab at an angle α from air where the refractive index varies along the slab as μ = μ0– k r² , where r is the distance measured along the normal and μ0 , k are positive constants . If the glass slab is sufficiently thick , how far along the normal will the ray go before it is reflected back ?

  • (a)  μ0 – sin θ / k
  • (b)  μ0 + sin θ / k
  • (c)  √ ( μ0 – sin θ / k )
  • (d) √ ( μ0+ sin θ / k )

8 . A bird is flying up at an angle sin-1 ( 3/5 ) with the horizontal. A fish in a pond looks at the bird when the bird is vertically above it . The angle at which the bird appears to fly , to the fish is  –   ( nwater= 4/3 )

  • (a)  sin-1 ( 3/5 )
  • (b)  sin-1 ( 4/5 )
  • (c)  45°
  • (d)  sin-1 ( 9/16 )

9  . A glass sphere of radius 5.0 cm and refractive index 1.6 is used to construct a paperweight by slicing through the sphere on a plane that is 2.0 cm from the centre of the sphere . Moreover , this plane is perpendicular to a radius of the sphere that passes through the centre of the circle formed by the intersection of the plane and the sphere . The paperweight is placed on a table and viewed from directly above by an observer who is 8.0 cm from the tabletop. When viewed through the paperweight , how far away does the table top appear to the observer ?

  • (a)  7.42 cm
  • (b)  6.5 cm
  • (c)  3 cm
  • (d)  4.44 cm

10 . A fish rises up vertically in a pond with a speed of 4 cm/s and notices a bird which is diving downwards towards the fish. The speed of the bird appears to be 16 cm/s , to the fish . If the refractive index of the water is 4/3 , the actual velocity of the bird is –

  • (a)  4 cm/s
  • (b)  9 cm/s
  • (c)  16 cm/s
  • (d)  6.2 cm/s

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.

Radioactive  Decay  Law

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.

url :

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