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Monday, 29 June 2020

Electrostatic (Previous year Questions)

  June 29, 2020 Lakshman Jangid   Pre. Ques   No comments

Q1. If in the electric field, an electron is kept freely. If the electron is replaced by a proton, what will be the relationship between the forces experienced by them?

Ans. When Electron is replaced by proton then the force on proton will remain the same but the direction of the force gets reverse. 

Q2. How does the Coulombian force between two-point charges depend upon the dielectric constant of the intervening medium?

Ans. If medium has some dielectric constant value, then force will be $ \frac{1}{K} $ times the force when medium is air. K is the dielectric constant.

Q3. Draw electrostatic field lines due to a small conducting sphere having a negative charge on it.

Ans.  Electric fields coming out of the charge.

Q4. An electrostatic field line cannot be discontinuous. Why?

Ans. Electric field lines can not be discontinuous because field is continuous in space and exist at all points in space. Click here

Q5. Two electric field lines never cross each other. Why?

Ans. If electric field lines will intersect then there will be two directions of the electric field at a point which is not possible.

Q6. Sketch the pattern of electric field lines due to an electric dipole.

Ans. Click here    

Q7. Which orientation of an electric dipole in a uniform electric field corresponds to stable equilibrium? 

Ans. When the dipole is placed along the direction of the electric field.

Q8. At what points, is the field due to an electric dipole parallel to the line joining the charges?

Ans. At Equitorial point and axial line, Electric field is parallel to dipole.

Q9. If the radius of the Gaussian surface enclosed a surface is halved, how does the electric flux through the Gaussian surface change?

Ans. There will be no change in electric  flux.  

Q10. Electric field inside a conductor is zero, explain.

Ans. Total charge within the conductor is zero so electric field is zero. 

Q11. The electric field E due to a point charge at any point near it is defined as \[E = \lim_{q_0\to 0 } \frac{F}{q_0}\] Where q0 is the test charge and F is the force acting on it. What is the physical significance of \[\lim_{q_0\to 0 }\] in this expression? 

Ans. The charge is taken as small as possible so that it's presence does not affect the electric field.

Q12. An electric dipole is free two move in a uniform electric field. Explain its motion.

Ans. In a uniform electric field, total force acting on the dipole is zero but torque acting is not zero. Because of the torque,  dipole rotate in the uniform electric field.

Q13. How much work is done in moving a 500 mC charge between two points on an equipotential surface? 

Ans. Work done on moving a charge over equipotential surface is zero irrespective of the magnitude of the charge.

Q14. In which position, a dipole placed in a uniform electric field is in (i)  stable (ii) unstable equilibrium?

Ans. (i) Dipole is stable when placed along the direction of the electric field.

(ii) Dipole is unstable when placed opposite to the direction of the electric field.

Q15. Why does the electric field inside a dielectric decrease when it is placed in an external electric field?

Ans. Electric field decrease because due to external electric field there is induced dipole moment in the dielectric which gives rise to internal electric field that oppose the external field. That is why field decrease inside the dielectric.

Q16. Can two equipotential surfaces intersect each other? Give reasons.

Ans. No, if they intersect then there will be two directions of the electric field which is not possible.

Q17. A parallel plat capacitor is charged to a potential different V by a d.c. source. The capacitor is then disconnected from the source. If a distance between the plate is doubled, state with reason how the following will change; 

(i) electric field between the plates,

(ii) capacitance, and

(iii)energy stored in the capacitor.

Ans. When plate separation is doubled then capacitance will be half of initial value.

(i) Since, Electric field is given by $ E = \frac{\sigma}{\epsilon_0}$ which is independent of the saparation so electric field will be same.

(ii) Since, Capacitance is given by $ C_i = \frac{\epsilon_0 A}{d} $. When d is increased to 2d then C will be half.

(iii) Energy is given by $U = \frac{Q^2}{2C} $. When C will be half then energy will be double of the initial value of energy.   

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Wednesday, 24 June 2020

Electrostatics : Potential and Capacitors

  June 24, 2020 Lakshman Jangid   Physics 12   No comments

1. Electric potential V is a scalar.

2. The electric potential difference between two given points in an electric field is equal to the amount of work done against the electric field in order to bring a unit positive test charge (without acceleration) from one point to the other. Mathematically, the potential difference between the points B and A, i.e., VB – VA  is given by \[V_B - V_A= \frac{W_{A \to B}}{q_0}\]  Where $ {W_{A \to B}} $ is the work done in order to carry a test charge q0 from point A to point B. The value of the test charge should be as small as possible.

3. The electric potential at a point in an electric field is equal to the amount of work done by the external force (against the electric field) in order to bring a unit positive test charge (without acceleration) from infinity to that point. Mathematically, electrostatic potential at a point A is given by \[V_A = \frac{W_{\infty \to A}}{q_0}\]  Where  \[{W_{\infty \to A}}\] is the amount of work done in order to carry a test charge q0 from to the point, A. value of the test charge should be as small as possible.

4. Electrostatic potential, as well as potential differences, are scalar quantities and their SI unit is volt.

5. Thus, the electric potential at a point is said to be 1 volt, if 1 J of work is being done in order to move a positive test charge of 1 C from infinity to that point.

6. The electric potential at a point situated at a distance r from a point charge q is given by \[V = \frac{q}{4 \pi \epsilon r}\]. Thus, potential due to a free, independent +ve point charge is positive because work is being done against the repulsive force experienced by + ve test charge. However, potential due to a -ve point charge is negative because here work is done by the attractive force acting on the +ve test charge.

7. It is possible to maintain a positively charged body at a negative or zero potential and a negatively charged body at zero or even positive potential.

8. Electric potential of earth is considered to be zero. Thus, electric potential of any body connected to earth will also be zero.

9. If a number of point charges are present then electric potential at a given point is equal to the algebraic sum of potentials due to different charges. Thus, if a number of charges q1, q2, q3 ….  are present at distance r1, r2, r3 ….etc. respectively from a given point then total electric at that point is \[V = V_1 + V_2 + V_3 + ......... = \frac{1}{4 \pi \epsilon }[{\frac{q_1}{r_1}+\frac{q_2}{r_2}+\frac{q_3}{r_3}+..........}] = \frac{1}{4 \pi \epsilon } \Sigma_{i=1}^{N} \frac{q_i}{r_i}\].

10. Electrostatic force is a conservative force and electric field is an example of a conservative field. It means that amount of work done in carrying a test charge in an electric field depends only upon the positions of initial point and the final point and is independent of the path followed. Moreover, the work done in carrying a charge in an electric field along a complete cyclic path is zero. Mathematically, \[\oint \vec{E}\vec{dl} = 0\].

11. Electric field at a point due to a given point charge is inversely proportional to the square of distance of given point from the point charge i.e., \[E \propto \frac{1}{r^2}\]. However, electric potential due to a given point charge is inversely proportional to the distance i.e., \[V \propto \frac{1}{r}\]. 

12. Electrostatic potential difference between two points in an electric field may also be defined as negative of the line integral of the electric field between the given points i.e., \[V_B-V_A= - \int_A^B \vec{E}.\vec{dl}\]. The electric potential at a the point in an electric field is equal to negative of the line integral of the electric field from infinity to that point i.e. \[V_A = - \int_{\infty}^A \vec{E}\vec{dl} \]

13. Electric field at a point may, thus, be defined as the negative of the rate of change of electric potential with position (i.e., the negative of the potential gradient) at that point i.e., \[\vec{E} = -\frac{dV}{dl}\]  and \[|E| = \frac{dV}{dl}\]. Moreover, the direction of the electric field is the direction in which electric potential is decreasing at a maximum rate i.e., where the decrease of potential is steepest.

14. The electric potential at a point situated at a distance ‘r’ from the mid-point of a short electric the dipole of dipole moment p inclined at an angle θ from the axial line of dipole is given by \[V = \frac{1}{4 \pi \epsilon} \frac{pcos\theta}{r^2}\].

15. At a point situated on the axial line of electric dipole θ = 0o  or π and hence \[V = \pm \frac{1}{4 \pi \epsilon}\frac{p}{r^2}\]. Here +ve sign is taken for θ = 0o and -ve sign is taken for θ = π. Electric potential at any point situated on the equatorial line of an electric dipole ( θ = 900 ) is zero i.e., V = 0.

16. Electric potential due to an electric dipole, in general, is inversely proportional to the square of the distance of the point from mid-point of the dipole, i.e., \[V \propto \frac{1}{r^2}\]. 

17. An equipotential surface is a surface with a constant value of the potential at all points on the surface. For any charge configuration, the equipotential surface through a point is normal to the electric field at the point.

18. For a point charge, equipotential surfaces are concentric spheres with a given charge point as the center. For a uniform electric field, equipotential surfaces are planes perpendicular to the direction of the electric field. 

19. As potential at all points of an equipotential surface is the same, hence, work done in moving a charge from one point to another along the equipotential surface is always zero.

20. The electric potential the energy of a system of point charges is equal to the amount of work done in assembling the given system of charge by bringing them to their respective positions from infinity. A point to be noted is that the potential energy is characteristic of the present state of assembly (or configuration ) and not the way the state is achieved. SI unit of electric potential energy is joule ( J ). For atomic and sub-atomic particals a unit “electron volt” (1 ev) is frequently used, where 1 eV = 1.60 *10-19J.

21. For a system of two point charges q1 and q2 separated by a distance r the potential energy is given by \[U = \frac{q_1 q_2}{4 \pi \epsilon r}\]. If two charges are like charges then the force between them is repulsive. Work is being done against this repulsive force while bringing the charges to their present position and hence electric potential energy of the system will be positive. If two charges are unlike one, the force between them is attractive and work is being done by the attractive force. Consequently, the potential energy of the system will be negative.

22. For a system of n point charges the total electric potential energy of the system is given by \[U = \frac{1}{2}[ \frac{1}{4 \pi \epsilon } \Sigma_{i=1}^{n} \Sigma_{j = 1, i \neq j}\frac{q_i q_j}{r_{ij}}]\]. Here the factor $ \frac{1}{2} $  has been incorporated on account of the that in the summation each term has been counted twice ij and ji in the above expression.

23. The electrostatic potential energy of a single charge q in an external electric field E is given by U(r) = qV(r), Where V(r) is the potential at the given point due to the external electric field.

24. Electrostatic potential energy of a system of two charges q1 and q2 located at $ \vec{r_1} $ and $ \vec{r_2} $ respectively in an external electric field is given by \[U = q_1V(r_1) + q_2V(r_2) + \frac {q_1 q_2}{4\pi \epsilon r_{12}}\].

25. Work done for rotating an electric dipole of dipole moment p in a uniform electric field E from orientation θ1 to orientation θ2 is given by \[W = - pE(cos \theta_2 - cos \theta_1 ) = pE(cos \theta_1 - cos \theta_2 )\]. The electrostatic potetial energy of an electric dipole in a uniform electric field E is given by \[W = - \vec{p}.\vec{E} = - pEcos\theta\] Where $ \theta $ is the angle which axis of given dipole makes with the direction of electric field E.

26. An electric dipole is in a state of stable equilibrium when dipole is placed along the direction of external electric field (i.e., p and E are in same direction) because in that orientation torque on dipole is zero and potential energy of dipole in minimum having a value -pE. On the other hand electric dipole is in a state of unstable equilibrium when p and E are in mutually opposite directions because in that orientation torque on dipole is zero and potential energy of dipole is maximum having a value +pE.

27. Conductors are the materials which possess large number of free electron and, therefore, allow flow of electric charge through them easily. Silver, copper, aluminium and other metals, mercury etc., are example of good conductors of electricity. 

28. When a conductor is placed in an electric field, it exhibits the following properties :               

(i). Net electric field inside the conductor is zero.  

(ii). Electric charge always reside on the outer surface of the conductor only.                                     

(iii). Net electric charge in the interior of the conductor is zero in equilibrium state.

(iv). Just outside the surface of a conductor, the electric field is perpendicular (normal) to the surface at every point.                                                                                                                            

(v). Electric field lines do not pass through the interior of the conductor.

(vi). Electric potential at all points of the conductor, situated inside as well as on its surface, is uniform. Moreover, it has the same value as on its surface.                                                                 

(vii). Electric field at the surface of a charged conductor is given by \[\vec{E} = \frac{\sigma}{\epsilon_0}\widehat{n}\], Where σ is the surface charge density and n is a unit vector normal to the surface. For σ > 0, electric field is normal to the surface outward but for σ < 0, electric field is normal to the surface inward.

29. As electric charge, as well as electric field inside a cavity of any conductor, is zero, the cavity remains shielded from outside electric influence. It is known as electrostatic shielding. Thus, electrostatic shielding (or screening) is the phenomenon of maintaining a certain region in space completely free from external electric fields. Property of electric shielding is made use of in protecting sensitive instruments from outside electric influences.

30. Insulators are those materials that cannot conduct electricity. Insulators possess a negligibly small number of free electrons.

31. Dielectrics are non-conducting substances. When a dielectric is held in an electric field. Small induced charges appear on the surface of the dielectric. However, there is no free movement of charges inside a dielectric. Dielectrics are of two types: polar and non-polar dielectrics. Non-polar dielectrics e.g., O2, N2, H2, CO4, etc., consist of non-polar molecules in which the centre of positive charge exactly coincides with the centre of negative charge and dipole moment of a molecule is zero. Polar dielectrics e.g.  H2O, HCl, NH3, alcohol are made of polar molecules in which centre of positive charge does not coincides with the centre of negative charge and each molecule has some intrinsic electric dipole moment.

32. When a dielectric is placed in an external electric field, the field induces dipole moment by stretching or re-orienting molecules of the dielectric. As a collective effect of these molecular dipole moments, some net charge is developed on the surface of the dielectric which produces a field that opposes the external electric field. The opposing field so induced reduce the external field. The electric dipole moment developed per unit volume in a dielectric when placed in an external electric field E is called “polarisation” or polarisation vector P. For linear isotropic dielectrics, \[\vec{p} = \chi_e \vec{E}\]  where $ \chi_e $ is a constant, known as the electric susceptibility of the given dielectric, whose value depends on the nature of the dielectric and is a characteristic of the dielectric.

33. The potential of a given charged conductor is directly propositional to its charge. The ratio of charge Q of an isolated conductor to its potential V is called a capacitance C of the given conductor. Thus, \[C = \frac{Q}{V}\]. Alternately capacitance of a given conductor is equal to the amount of charge given to the conductor in order to raise its electric potential by unity. SI unit of capacitance is farad (F). 

34. Capacitance of a conductor depends upon its dimensions and shape. Capacitance of an isolated spherical conductor is given by \[C = 4 \pi \epsilon_0 R\] in free space, where R is the radius of conductor. Capacitance of a conductor is independent of the amount of charge given to it. However, capacitance of a conductor depends on the nature of the surroundings.

35. Capacitance of a system of two-conductor, besides their geometrical configuration (shape and size), also depends on (i) the separation between the two conductors, and (ii) the nature of the dielectric separating the two conductor.

36. A parallel plate capacitor is simplest type of capacitor.it consists of two parallel metal plates separated by a thin layer of dielectric. Capacitance of a parallel plate capacitor is \[C = \frac{\epsilon_0 A}{d}\] if free space is the intervening dielectric. Here, A = surface area of either plate and d = separation between the two plates of capacitor.

37. When capacitors is connected in series then the resultant capacitance Cs is given by \[\frac{1}{C_S} = \frac{1}{C_1} +\frac{1}{C_2}+\frac{1}{C_3}+............\] where C1, C2, C3… etc. are the capacitances of individual capacitors.In series combination charge on all the capacitors is same but potential difference between the plates of difference capacitors is inversely proportional to their capacitances. In a series combination of capacitors, resultant capacitance is less then the capacitance of anyone capacitor. However, the combination may withstand a higher potential difference (voltage).

38. In parallel grouping of capacitors resultant capacitance is the sum of individual capacitance of capacitors joined in parallel i.e., \[C_P = C_1+C_2+C_3+........\]. In parallel arrangement potential difference across all the capacitors is same but charges on individual capacitors are directly proportional to their capacitances.

40. On filling a dielectric medium of dielectric constant K between the plates of a parallel plate capacitor, due to polarisation of the dielectric, the net electric field and hence the potential difference between the plates of capacitor is reduced to $ \frac{1}{K} $ times its precious value. Consequently, the capacitance of the capacitor increases K times. Thus, $  K = \frac{Capacitance in presence of dielectric medium Cm­­}{Capacitance when free space is the medium C_0 }$.

41. The capacitance of a parallel plate capacitor with a dielectric medium introduced between the plates is given by \[C = \frac{K\epsilon_0 A}{d} \]

42. If a dielectric medium of dielectric constant K and thickness t (t < d) is filled between the plates of a capacitor then its capacity is given by \[C = \frac{\epsilon_0 A}{d - t - \frac{t}{K}}\].

43. If a conducting sheet of thickness t (t < d) is introduced between the plates of a capacitor, without touching either plate of a capacitor, then the capacitance of the arrangement is given by \[C = \frac{\epsilon_0 A}{d-t} .\]

44. The energy stored in a charged capacitance is \[U = \frac{1}{2}QV = \frac{1}{2}CV^2 = \frac{Q^2}{2C}\] where Q = charge given to capacitance C, and V = potential difference between the plates of capacitor.

46. The energy density (energy per unit volume) of electric field in a capacitor is \[E' = \frac{1}{2} \epsilon_0 E^2\] electric field between the plates of capacitor.

47. When two charged capacitor are joined together, they share their charges till they acquire same “common potential” V, which is given by \[V = \frac{Total Charge}{Total Capacitance} = \frac{C_1V_1+C_2V_2}{C_1+C_2}\]. Total charge remains the same during this process.

48. The sharing of charges between two capacitors is always accompanied by some loss of electrical energy. Loss of electrical energy is given by \[\Delta U = U_2 - U_1 = \frac{C_1C_2(V_1-V_2)^2}{2(C_1+C_2)}\]. 

49. Dielectric strength of a dielectric is the maximum value of electric field (or potential gradient) which it can tolerate without its electric break-down. For reasons of safety a maximum electric field equal to 10% of dielectric strength of the material is actually applied.



Video Lectures:

1. Electric Potential energy, Electric potential definition, Electric potential unit, Electric force conservative nature watch video

2. Electric potential due to group of the charges, Electric potential due to electric dipole watch video

3. Equipotential surfaces and relation between Electric field and electric potential watch video

4. Conductor properties, Capacitor and Capacitance, spherical capacitor watch video

5 Parallel plate capacitor with conducting slab and dielectric slab watch video

6 Grouping of the capacitors watch video

7 Energy stored in capacitor, Common potential and Energy loss in charge sharing watch video



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Monday, 22 June 2020

Unit and Dimensions

  June 22, 2020 Lakshman Jangid   Physics 11   No comments
1. Measurement of any physical quantity involves comparison with a certain basic, arbitrary chosen, widely accepted reference standard called Unit. 
Mathematically, a measure of a quantity Q = nu, where u is the size of the unit, and n is the numerical value of the given measure.  

2. Fundamental quantities: Fundamental quantities are the base quantities. There are 7 fundamental quantities: 
(i) Length 
(ii) Mass
(iii) Time
(iv) Electric Current
(v) Thermodynamic Temperature
(vi) Amount of substance
(vii) Luminous Intensity.

3. Derived quantities: These quantities are formed using fundamental quantities like density, volume, force, etc.

4. Length: Unit is metre (m). Meter is defined as the length of the path traveled by light in vacuum during a time interval of $ \frac{1}{299792458} $ part of a second. 
 1 fermi  = $1f =   10^{-15} m $
 1 angstrom =  $ 1A = 10^{-10} m$
 1 nano-metre = $1nm = 10^{-10}m$
 1 micro-metre = $ 1\mu m = 10^{-6}m$
 1 mili-metre = $1mm = 10^{-3} m$
 1 Astronomical unit = $ 1AU  = 1.496 \times 10^{11}m$
 1 light-year = $ 1ly = 9.46 \times 10^{11} m$
 1 parsec = $ 3.08 \times 10^{16}m $

5. Mass: Unit is Kilogram(kg). The mass of a cylinder made of platinum-iridium alloy kept at the International Bureau of Weights and Measures is defined as 1 kg.

6. Time: Unit is second(s). One second is the time taken by 9 192 631 770 oscillations of the light (of a specified wavelength) emitted by a cesium-133 atom.

7. Electric Current: Unit is Ampere. If equal currents are maintained in the two wires so that the force between them is $ 2 x 10^{-7} $ newton per meter of the wires, the current in any of the wires is called 1 A

8. Thermodynamic Temperature: Unit is Kelvin(K). The fraction $ \frac{1}{273.16} $ of the thermodynamic temperature of the triple point of water is called 1 K.

9. Amount of the Substance: Unit is mole(mole). The amount of a substance that contains as many
elementary entities as there is the number of atoms in 0.012 kg of carbon-12 is called a mole. 

10. Luminous Intensity: Unit is Candela(cd). The SI unit of luminous intensity is 1 cd which is the luminous intensity of a blackbody of surface area $ \frac{1}{600 000} m^{2} $ placed at the temperature of freezing, platinum, and at a pressure of 101,325 $ {N/m^{2}} $, in the direction perpendicular to its surface. 

11. Dimensions: Dimensions are the powers to which fundamental quantities are raised to represent that quantity. It is represented by using a square bracket. 

 Physical Quantities

 Dimensions

 Distance, Length, Displacement

 $[M^0LT^0]$

 Velocity, Speed

 $[M^0LT^{-1}]$

 Acceleration

 $[M^0LT^{-2}]$

Force 

 $[MLT^{-2}]$

 Linear momentum, Impulse

 $[MLT^{-1}]$

 Torque, Work, Kinetic Energy, Potential Energy, Energy, 

 $[ML^2T^2]$

 Power

 $[ML^2T^{-3}]$

Pressure, Stress, Modulus of Elasticity 

 $[ML^{-1}T^{-2}]$


12. Principle of homogeneity of Dimensions: A correct dimensional equation must be homogeneous i.e. dimensions on both sides are the same. 

13. Use of Dimension: To convert a unit from one system to another system, To find the relation between various physical parameters and to check whether the formula is dimensionally correct or not.


Example1:  Find the dimension of the constants a and b in Van Der Wall Equation
i.e. $ (P + \frac{a}{V^2})(V-b) = RT $         
Solution:  Using principle of homogeneity,  
Dimension of b  = Dimension of V (volume) = $ [{ L^3 }] $

Dimension of P (pressure) = Dimension of $ (\frac{a}{V^{2}}) $                               
Dimension of a = dimension of $ PV{^2} $  = $[ML^{-1}T^{-2}] [L^{3}]^{2} $= $ [ML^{5}T^{-2}] $          
 


Example 2: The value of the gravitational constant is G = $ 6.67 * 10^{-11} $ $ Nm{^2}kg^{-2} $. Convert it into a system based on kilometer, tonne and hour as base units.  
Solution: Dimnsional formula of  G is $ [M^{-1}L^{3}T^{-2}] $

$ n_2 = n_1 [\frac{M_1}{M_2}]^{-1}[\frac{L_2}{L_1}]^{3} [\frac{T_2}{T_1}]^{-2} $  

$ n_1 = 6.67 * 10^{-11}, M_1 = 1 kg, M_2 = 1 tonne = 1000kg, $
$T_1 = 1s, T_2 = 3600s, L_1 = 1m and L_2 = 1000m $

$ n_2 = 6.67*10^{-11}[\frac{1}{1000}]^{-1}[\frac{1}{1000}]^{3} [\frac{1}{3600}]^{-2} = 8.64 * 10^{10} $


Example 3: The frequency f of a stretched string depends upon the Tension (T), length (l) and the linear mass density $ /mu $. Find the relation for frequency. 
Solution: Let frequency depends on T, l, and $ \mu $ as follow:
                           $ f = kT^{a}l^{b}{\mu ^{c}} $                      where k is constant.
            
writing dimension formula of both sides,
$ [M^{0}L^{0}T^{-1}] $ = $ [MLT^{-2}]^{a}[L]^{b}[ML^{-1}]^{c} $  =  $ [M^{a+c}L^{a+b-c}T^{-2a}] $

Comparing dimensions on both sides, 
                                     a + c  =  0
                                a + b - c  =  0
                                        -2a  =  -1
solving these we get,  a = $\frac{1}{2} $, b = -1 and c = $ \frac{-1}{2} $

so relation will be,   $f = \frac{k}{l} \sqrt {\frac{T}{\mu}} $ 



Video Lecture:
Fundamental and Derived Quantities, Dimensions, How to find dimension of any physical Quantity, Formula validation by dimensions, Deriving relation between physical quantities Unit conversion Watch video





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