IIT JEE Advanced & Mains Physics 2018 : Electricity : Rankfile / expected questions IV

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

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Page  : 4

28 ] A  spherically symmetrical charge distribution has density
{\rho}_{v}=\begin{cases}\begin{matrix}{\rho }_{{0}},0\leq r\leq R\\ 0,r> R\end{matrix}\end{cases}
Then the energy stored in region r < R is – 

A ]  \frac {4\pi {\rho }_{{0}}^{{3}}{R}^{{5}}}{45{\epsilon }_{{0}}}J

B ]  \frac {2\pi {\rho }_{{0}}^{{2}}{R}^{{5}}}{45{\epsilon }_{{0}}}J

C ]  \frac {\pi {\rho }_{{0}}^{{3}}{R}^{{5}}}{45{\epsilon }_{{0}}}J

D ]  \frac {2\pi {\rho }_{{0}}^{{2}}{R}^{{5}}}{90{\epsilon }_{{0}}}J

 

29 ] Two point charges of equal masses and charges  m and  Q respectively are suspended at a common point by two threads of negligible mass and length l .If at equilibrium, the angle of inclination of each thread to the vertical is \alpha  , then  {Q}^{2} is proportional to –

A ]  {{\sin}^{{2}}{\alpha {\tan{\alpha }}

B ] {{\sin}^{{3}}{\alpha {\tan{\alpha }}

C ] {{\tan}^{{2}}{\alpha {\cos{\alpha }}

D ] {{\tan}^{{3}}{\alpha {\sin{\alpha }}

 

30 ] In the above arrangement , if  \alpha is very small , then \alpha is given by – 

A ] \alpha=3\sqrt{{\frac {{Q}^{{2}}}{32\pi {\epsilon }_{{0}}mg{l}^{{2}}}}}

B ] \alpha=4\sqrt{{\frac {{Q}^{{2}}}{16\pi {\epsilon }_{{0}}mg{l}^{{3}}}}}

C ] \alpha=2\sqrt{{\frac {{Q}^{{2}}}{18\pi {\epsilon }_{{0}}mg{l}^{{3}}}}}

D ] \alpha=3\sqrt{{\frac {{Q}^{{2}}}{16\pi {\epsilon }_{{0}}mg{l}^{{2}}}}} 

 

31 ] Consider two conducting planes kept at a particular angle to each other in the xy axis with the point of contact being at the origin . Obviously . work must be done to bring a point charge to the region of contact between the planes. Then , finding the work done ,using the method of images [in the above case ] is  possible if the angle is [ in degrees] –

A ] 20

B ] 30

C ] 40

D ] 50

 

32 ] The issue with the angle in the above case is due to the fact that there is – 

A ] Difficulty in finding the location of the image charge.

B ] Difficulty in finding the location of the equipotential.

C ] Difficulty in finding the location of the charge at infinity .

D ] A and B 

E ] A , B and C 

 

33 ] A charge q is kept in a pyramid ,on the center of it’s base ,which is square shaped . Then the flux through any face of the pyramid is – 

A  ] zero

B ]  \frac{q}{{\epsilon}_{0}}

C ]  \frac{q}{{8\epsilon}_{0}}

D ]  \frac{2q}{{\epsilon}_{0}}

 

34 ]  A hollow cylinder of resistivity \rho has length L .It’s  inner radius is a and outer radius b  . Then the resistance produced when a  potential difference is generated between the ends of the cylinder is – 

A]  R=\frac {\rho L}{\pi \left({{b}^{{2}}-{a}^{{2}}}\right)}

B ]  R=\frac {\rho L}{\pi \left({{b}^{{4}}-{a}^{{4}}}\right)}

C ]  R=\frac {\rho L}{A}

D ]  R=\frac {\rho L}{2A}

 

35 ] In the above case – 

A ] No current flows through the cylinder . 

B ] Current flows parallel to the axis of the cylinder.

C ] Current flows radially outward from the axis of the cylinder

D ] Current flows perpendicular to the axis of the cylinder

 

36 ] Continuing with question no .34 , what if  the potential difference is applied between the inner and outer surfaces of the cylinder ? [ Choose the right combination ]

a ) Current flows parallel to the axis of the cylinder.

b ) Current flows radially outward from the axis of the cylinder

c ) Resistance remains the same .

d ) Resistance becomes \frac {\rho }{2\pi L}In\left({\frac {b}{a}}\right)

e ) Resistance becomes \frac {\rho }{\pi L}In\left({\frac {a}{b}}\right)

f )  No current flows through the cylinder

Answer : – 

A ]  c , e

B }  b , d 

C ]  d , f 

D ]  c , f 

 

IIT JEE Advanced & Mains Physics 2018 : Electricity : Rankfile / expected questions III

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

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Page  : 3

19]  A cone [ of resistivity \rho ] is cut off at a height of h from it’s base.If it’s uniformly charged throughout it’s  cross-sections, then the resistance between the two end of the cut – off cone is –

A ]  \frac {\rho h}{\pi ab}

B ]  \frac {\pi h}{\rho ab}

C ]  \frac {\rho }{\pi h}

D]  \frac {\pi ab}{\rho h}

 

20 ] A charge distribution produces an electric field \vec{E}=\frac {{E}_{{0}}}{{r}^{{2}}}{e}^{{-r∕c}}\hat{r}  where c is a constant. Then the total charge in the distribution is – 

A ] 4\pi {E}_{{0}}{e}^{{\frac {R}{c}

B ] 2\pi {E}_{{0}}{e}^{{\frac {R}{c}

C ] \frac {Q}{{\epsilon }_{{0}}}

D ] \frac {4Q}{{\epsilon }_{{0}}}

 

21 ] Figure below shows the path of an electron where A B – when not under the influence of a field and B’ , C’ and D’ ,under the influence of three different electric fields 1 , 2 ,and 3 respectively . . Then the Drift Velocity of the electron under the second field is indicated by –

A ] AB + AB’

B ] ABB’

C ] AD’ –  AC’

D ] BC’

 

22 ] Earnshaw’s law states that a collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges .Give a simple non-mathematical proof for the same . 

 

23 ] Two conductors are embedded in a material of conductivity {{30}}^{{-4}}\omega /m and dielectric constant \epsilon =90{\epsilon }_{{0}}. The resistance between the two conductors is found to be {{30}}^{{5}}\omega. Find the magnitude of the capacitance between the two conductors .

 

24] Figure below shows an hypothetical infinite conducting plane in the ( x,y,z) plane having charges Q1 , Q2 , Q3 , Q4  located in cavities.What can we predict with certainty regarding the nth cavity containing charge Qn ?

A ] V experienced by Q1 will influence the V of Qn

B ] V  of Q1  = V of Qn

C ] The whole conducting surface up to that containing Qn is an equipotential .

D ] V of the conducting plane cannot be zero .

 

25 ] Parallel to ( x ) and at a distance d .above a grounded conducting plane ( xy ) is placed an infinite wire of uniform charge  \lambda Then the  potential in the region above the plane is – 

A ] V   =\frac {\lambda }{4\pi {\epsilon }_{{0}}}In\{{\frac {{y}^{{2}}+{{\left({z+d}\right)}}^{{2}}}{{y}^{{2}}+{{\left({z-d}\right)}}^{{2}}}}\}

B ] V   =\frac {\lambda }{4\pi {\epsilon }_{{0}}}In\{{\frac {{y}^{{2}}-{{\left({z+d}\right)}}^{{2}}}{{y}^{{2}}-{{\left({z-d}\right)}}^{{2}}}}\}

C ] V   = \frac {\lambda }{2\pi {\epsilon }_{{0}}}In\{{\frac {{y}^{{2}}+{{\left({z+d}\right)}}^{{4}}}{{y}^{{4}}+{{\left({z-d}\right)}}^{{2}}}}\}

D ] V   = \frac {\lambda }{2\pi {\epsilon }_{{0}}}In\{{\frac {{y}^{{2}}-{{\left({z+d}\right)}}^{{4}}}{{y}^{{4}}-{{\left({z-d}\right)}}^{{2}}}}\}

 

26 ]  In the above case,the charge density  \sigma induced on the conducting plane is – 

A } -\frac {\lambda d}{\pi \left({{y}^{{2}}+{d}^{{2}}}\right)}

B ]  -\frac {\lambda d}{\pi \left({{y}^{{4}}+{d}^{{4}}}\right)}

C ]  \frac {\lambda d}{\pi \left({{y}^{{2}}+{d}^{{2}}}\right)}

D ] \frac {\lambda d}{\pi \left({{y}^{{4}}+{d}^{{4}}}\right)}

 

27 ] Diagram below represents a hollow conductor on which a test potential of around 90,000 V is applied.Choose the graph which correctly represents the incidence of potential inside the cavity . The x axis denotes the time elapsed in pico -seconds . 

 

IIT JEE Advanced & Mains Physics 2018 : Electricity : Rankfile / expected questions II

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

Primary Users : Students appearing for the Advanced and Mains Exam for Admission to the BTech Courses at the Indian Institutes of Technology , National Institutes of Technology , Indian Institute of Space science and Technology and other National Engineering Institutes, INDIA .

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Tertiary Users : All those who love Physics !

Answers with full explanations and tips & tricks to solve similar problems to be published in our site https://physicsmynd.in ]

Page  : 2

 

10 .  A soap bubble of radius 1 cm , which is at a potential of 90 volts ,  collapses to a drop of radius 1 mm . Find the change in its electrostatic energy?

 

11.From a  sphere of uniform charge density \rho and radius R ,a small spherical hollow region of radius r is cut out at a distance d from the center of it .
(a) What is the  field E at the center of the hollow sphere ?

 

12 .In addition ,also find the potential \phi at above stated point.

 

13 . Suppose ,in the above case , a particle of charge q is moved from infinity to the center of the spherical shell through the cut out hole . Find the work needed ? [take the thickness of the shell to be t .]

 

14 .Find the  the electric field and potential at the center of a ring of radius r whose net  charge +Q is uniformly distributed on it. 

 

15.  Calculate the point(s) on the axis of a charged ring where the electric field is maximum.

 

16.Figure below shows the discharge curve of a Capacitor . From the graph the valve of time [ t ] when \frac {Q}{{Q}_{{0}}}=10 is-

A ] t=2.303\tau

B ] t=2.400\tau

C ] t=1.303\tau

D ] t=3\tau

 

17 ]   Drift Velocity

The primary charge carriers in sea water  are {Na}^{+} and {Cl}^{-} ions. Consider a non conducting pipe of 3 meters connected to a 15 volt battery at it’s both ends by an electrode.This pipe is then filled with seawater..Find the average drift velocity of the ions, in cm/s?

Note : The resistivity of seawater is about 25{\Omega}-m and there are about 3}\times{10}^{20}/{cm}^{3} of each ions .

 

18 ] Consider the following arrangement .-

An air-filled parallel-plate capacitor of area A , with plates a and b are kept on a smooth surface in such a manner ,that it’s one end is fixed and the other is  connected to a spring having a force constant k .The charges on the plates are respectively  Q + and  Q – .Then the expansion of the spring is – 

A ] x=\frac {{Q}^{{2}}}{2kA{\epsilon }_{{0}}}

B ] x=\frac {{Q}^{{2}}}{kA{\epsilon }_{{0}}}

C ] x=\frac {{Q}^{{3}}}{2kA{\epsilon }_{{0}}} 

D ] x=\frac {{Q}^{{2}}}{2A{\epsilon }_{{0}}}

IIT JEE Advanced & Mains Physics 2018 : Electricity : Rankfile / expected questions

Primary Users : Students appearing for the Advanced and Mains Exam for Admission to the BTech Courses at the Indian Institutes of Technology , National Institutes of Technology , Indian Institute of Space science and Technology and other National Engineering Institutes, INDIA .

Secondary Users : GATE Physics  ,JAM Physics , Physics paper of Civil Services Exam , UGC-NET Physics , National and International Physics Olympiads,International College level Physics Exams, Indian State Board Engineering entrance exams etc.

Tertiary Users : All those who love Physics !

[ Answers with full explanations and tips & tricks to solve similar problems to be published in our site https://physicsmynd.in ]

P A G E  : 1

1.THE ATOMIC HOOK

Recently an Ice berg weighing around  {9}\times{10}^{15} kgs broke off in the Weddell ice sea area in the Antarctic circle .Consider the following idea of an invention : A device consisting of two blocks – from one a proton can be extracted whose attractive force is directed towards an electron of the second block . Suppose the device is to be used as a hook  – that’s placing one block on the main ice and the second on the ice section which shows signs of breaking . What are your ideas regarding the following – [ neglect all other possible physical  forces / effects  ]

a } Would the device work as a hook preventing the breaking off of the above mentioned ice section ? If yes ,how many such devices would be needed  ?

b ]  What should be the effective weight of the blocks ?

c ] What is the effective distance needed between the blocks for such a device to work ?

d ] There is a very serious problem with the above idea .Can you point out what it is? 

 

 

2.A negatively charged particle accelerates from East to West in a uniform electric field. What are the direction and the value of the electric field if the particle has charge q=6 μC, mass m=1 mg, and if the value of its acceleration is 6 mm/{s}^{2}? Select the closest answer:

A) West, 0.001 N/C
B) East, 0.001 N/C
C) West, 3 N/C
D) East, 3 N/C
E) North, 300 N/C

 

3. A negatively charged particle with q= -3 μC is placed at the center of the uniformly charged cube with side a=1 mm and total charge Q=9 C spread out over all six faces of the cube. What is the total force acting on the particle in the center? Find the magnitude and the direction.

(A)  {0.6}\times{10}^{6}N, left
(B)  {0.6}\times{10}^{6}N, right
(C)  {2.9}\times{10}^{-9}N, top
(D)  {2.9}\times{10}^{-9}N, bottom
(E) Zero

 

4 . A tiny particle with mass m=3 mg and positive charge Q=3 mC is placed just near the ground. Find the surface charge density on the ground for keeping the particle above it in a stationary position. Consider the ground to be a non-conductor.

(A) {1.75}\times{10}^{-19} C/{m}^{2}
(B) {1.75}\times{10}^{-16} C/{m}^{2}
(C) {1.75}\times{10}^{-13} C/{m}^{2}
(D) {1.75}\times{10}^{-10} C/{m}^{2}
(E) {1.75} C/{m}^{2}

 

5. Three charges , top of 6C and bottom two of charges of the same magnitude but of opposite sign are placed at the corners of an equilateral triangle with it’s side being d = 1 cm. . Find the value and the direction of the field exactly at the center of the triangle?

(A) {0.17}\times{10}^{-16} N/C, top
(B) {2.35}\times{10}^{-6} N/C, top
(C) {8.55}\times{10}^{-1} N/C, bottom
(D) {3.2}\times{10}^{15} N/C, bottom
(E) None of the above

 

6. Some distance away from a positively charged plate with considerable surface charge density is placed an unit charge [ in mC ]  of mass  m=10 mg  on a wire  at an angle of 45 degrees.. Find the surface charge density .

(A) {1.76}\times{10}^{-16} C/{m}^{2}
(B) {1.76}\times{10}^{-12} C/{m}^{2}
(C) {1.76}\times{10}^{-8} C/{m}^{2}
(D) {1.76}\times{10}^{-4} C/{m}^{2}
(E) {1.76} C/{m}^{2}

 

7 . Charges are placed in a geometrical pattern corresponding to the  hour hands of a clock with their magnitudes being equivalent to that of the hours indicated .Calculate the resulting field ?  

 

8. Consider a semicircle of radius r with it’s left quadrant being negatively charged and the rest being positive. .Find the field generated in this negative quadrant ?

 

9. A disc of radius r has an amount of charge q  uniformly spread on it’s surface  Find the potential at any point on it’s axis of symmetry.

 

[ Answers with full explanations and tips & tricks to solve similar problems to be published in https://physicsmynd.in ]

Free JEE Main and Advanced Mathematics Notes & Fully solved problems

infinityphyWelcome dear Students & Viewers , as part of introducing new services and revamping our operations , we are all set t0 launch a separate web service for JEE Mathematics [ Main & Advanced ] …..  Mathsmynd 

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How to get into IITs, NITs ,IIITs and other engineering colleges from 2013.

A single entrance examination will be held from 2013 for admission to the centrally funded engineering institutes. These include the 15 Indian Institutes of Technology (IITs) and all the National Institutes of Technology (NITs). Equal weightage will be given to marks obtained in the Class XII examinations and those of the entrance tests.

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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 ..like 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

                                                            

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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 –

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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

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.