SR IPE Physics PDF - Past Paper 2024 - Physics Questions

VERY SHORT ANSWER QUESTIONS (2MARKS) SHORT ANSWER QUESTIONS (4MARKS) 1. RAY OPTICS 2M 1. RAY OPTICS 4M 2. MOVING CHARGES AND 2M 2. WAVE OPTICS 4M MAGNETISM 3. MAGNETISM AND MATTER...

VERY SHORT ANSWER QUESTIONS (2MARKS) SHORT ANSWER QUESTIONS (4MARKS) 1. RAY OPTICS 2M 1. RAY OPTICS 4M 2. MOVING CHARGES AND 2M 2. WAVE OPTICS 4M MAGNETISM 3. MAGNETISM AND MATTER (2+2) M 3. ELECTIRIC FIELDS & CHARGES 4M 4. ELECTRIC POTENTIAL & 4. ALTERNATING CURRENT 2M 4M CAPITANCE 5. MOVING CHARGES AND 5. ELECTROMAGNETIC WAVES 2M MAGNETISM 4M 6. DUAL NATURE OF RADIATION & 6. ELECTRO MAGNETIC (2+2)M 4M MATTER INDUCTION 7. SEMI CONDUCTORS 2M 7. ATOMS 4M 8. COMMUNICATION SYSTEM 2M 8. SEMI CONDUCTORS 4M GRAND TOTAL 20 M GRAND TOTAL 32 M LONG ANSWERS QUESTIONS 1. WAVES 8M 2. CURRENT ELECTRICITY 8M 3. NUCLEAR PHYSICS 8M GRAND TOTAL 24 M  IPE – MARCH 2024 (TS) IPE – maY 2024 (TS) I SECTION – A (10X2=20) I SECTION – A (10X2=20) 1. What is sky wave propagation? 1. How do you convert a moving coil galvanometer into a 2. What are intrinsic and extrinsic semiconductors? voltmeter? 3. Write down Einstein’s photoelectric equation. 2. A small angled prism of 40 deviates a ray through 4. Write down de Broglie’s relation and explain the terms 2.480. Find the refractive index of the mirror. therein. 3. Define magnetic declination. 5. What are the applications of microwaves? 4. Classify the following materials with regard to 6. What is the phenomenon involved in the working of a V magnetism: Manganese, Cobalt, Nickel, Bismuth, transformer? E Oxygen, and Copper. 7. What are the laws of reflection through curved mirrors ? N 5. Write the expression for the reactance of i) an inductor 8. What is the principle of a moving coil galvanometer? K and ii) a capacitor. 9. Define magnetic inclination of angle of dip. 6. What is the principle of production of electromagnetic A 10. What is the magnetic moment associated with a solenoid. waves? T 7. What is photoelectric effect? II SECTION – B (6X4=24) E 8. State Heisenberg’s Uncertainty Principle. 11. Define critical angle. Explain total internal reflection S 9. What is a P-type semiconductor? What are the majority using a neat diagram. H and minority charge carriers in it? 12. How do you determine the resolving power of your eye? 10. Define Modulation. Why is it necessary? 13. Derive the equation for the couple acting on a electric dipole in a uniform electric field. II SECTION – B (6X4=24) 14. Explain the behavior of dielectrics in an external field. 11. Define critical angle. Explain total internal reflection 15. Derive an expression for the magnetic dipole moment of using a neat diagram. a revolving electron. 12. Explain Doppler Effect in light. Distinguish between 16. Describe the ways in which Eddy currents are used to red shift and blue shift. advantage. 13. Define intensity of electric field at a point. Derive an 17. Describe Rutherford atom model. What are the draw to expression for the intensity due to a point charge. advantage. 14. Explain series combination of capacitors. Derive the 18. Distinguish between half- wave and full-wave rectifiers. formula for equivalent capacitance in each combination. 15. State and explain Biot-Savart Law. 16. Obtain an expression for the emf induced across a conductor which is moved in a uniform magnetic field which is perpendicular to the plane of motion. 17. What are the limitations of Hohr’s theory of hydrogen III SECTION – C (2X8=16) 19. How are stationary waves formed in closed pipes? Explain atom? the various modes of vibrations and relations for their 18. What is rectification? Explain the working of a full wave rectifier. frequencies A closed organ pipe 70cm long is sounded. If the velocity III SECTION – C (2X8=16) of sound is 331m/s , what is the fundamental frequency of 19. State the working principle of potentiometer explain vibration of the air column ? 9 20. State Kirchhoff’s law for an electrical network. Using these with the help of circuit diagram how the emf of two 4 primary cells are compared by using the potentiometer. laws deduce the condition for balance in a Wheatstone bridge. 9 20. How are stationary waves formed in closed pipes? 21. Explain the principle and working of a nuclear reactor 0 Explain the various modes of vibrations and relations with the help of a labelled diagram? 3 for their frequencies Calculate the energy equivalent of 1 gr of substance. A closed organ pipe 70cm long is sounded. If the 2 velocity of sound is 331m/s , what is the fundamental 6 frequency of vibration of the air column ? 3 21. Explain the principle and working of a nuclear reactor with the help of a labeled diagram? 0 If one microgram of 92U235 is completely destroyed in 6 an atom bomb. How much energy will be released? ~~~~~ IPE SR PHYSICS STUDY MATERIAL (60/60)~~~~~~~~~~~~~~~~~~~~~~~Page No: 1  2. RAY OPTICS 8. MAGNETISM AND MATTER 1. A small angled prism of 40 deviates a ray through 2.480. 15. What happens to compass needles at the Earth’s Find the refractive index of the mirror. pole? A. D m  A  μ  1  , A  4 0 , D m  2.4 8 0 A. A t p o le   9 0 0 , B H  B c o s 9 0 0  0 Dm 2.4 8 0 it may point out in any direction. Dip Needle. μ  1    0.6 2 16. What are the units of magnetic moment, magnetic A 4 μ  1   0.6 2  μ  1  0.6 2  μ  1.6 2 induction and magnetic field? 2. What is myopia? How can it be corrected? A. 1) Magnetic moment : Am2 A. Myopia (or) Nearsightedness: The defect of the eye lens 2) Magnetic induction : Tesla (or) Wb m-2 to form the image in front of the retina. 3) Magnetic field : Tesla V 17. Define magnetic inclination or angle of dip. It is corrected by using concave lens. 3. What is hypermetropia? How can it be corrected? E A. The angle between the direction of earth’s magnetic A. Hypermetropia (or) Farsightedness: The defect of the N field(BE) and direction of Horizontal magnetic field(BH) eye lens to form the image behind the retina. K is called magnetic inclination or angle of dip. It is corrected by using convex lens. 18. Classify the following materials with regard to A magnetism: Manganese, Cobalt, Nickel, Bismuth, 4. What is ‘dispersion’? Which colour gets relatively more T dispersed? Oxygen, and Copper. A. Dispersion: Splitting of white light into its seven colours, E A. Dia magnetic: Bismuth, Copper is called dispersion. Violet colour is more dispersed. S Para magnetic: Manganese, Oxygen 5. Define ‘power’ of a convex lens. What is its units? H Ferro magnetic: Nickel, Cobalt A. Power of a lens: Reciprocal of the focal length is called 19. Magnetic lines form continuous closed loops. Why? power of lens. Units : Dioptre(D) A. Since, isolated magnetic poles do not exists. 6. A concave mirror produces an image of a long vertical 20. Define magnetic declination. A. The angle between magnetic meridian and geographical pin, placed 40cm from the mirror, at the position of the meridian is called angle of declination object. Find the focal length of the mirror. u   40 cm , v   4 0 cm , 21. What is the magnetic moment associated with a A. solenoid? 1 1 1 1 1 1 1 2         f   2 0cm A. M = N i A M = (Total no.of turns)(current)( area) f u v f 40 40 f 40 7. What are the laws of reflection through curved mirrors? 22. What do you understand by the ‘magnetization’ of a 1)  i   r , 2) Incident ray, R eflected ray sample? A. A. The magnetic moment per unit volume is called and N orm al lie in sam e plane. magnetization. ( )= 7. MOVING CHARGES & MAGNETISM 8. What is the importance of Oersted’s experiment? S.I Units magnetization is A/m. A. Every current carrying conductor produces a magnetic 11. ELECTROMAGNETIC WAVES field around it and which is perpendicular to current 23. If the wavelength of electromagnetic radiation is carrying conductor. doubled, what happens to the energy of photon? 9. What is the smallest value of current that can be A. ∝ , = , = , = measured with a moving coil galvanometer? A. It is used to measure very small current upto 10-9A 24. Give any two uses of infrared rays. 10. How do you convert a moving coil galvanometer into A. 1)Solar water heaters.2)To take Photo in fogconditions. an ammeter? 25. What are the applications of microwaves? A. A low resistance is connected in Parallel to it. A. 1) Micro wave ovens 2) Aircraft navigation. 11. How do you convert a moving coil galvanometer into a 26. What is the principle of production of voltmeter? 9 electromagnetic waves? A. A high resistance is connected in Series to it. 4 A. An accelerated charges produces an E.M. wave. 12. Distinguish between ammeter and voltmeter. 27. Microwaves are used in Radars, why? 9 A. Due to short wavelengths, it is easily penetrate the Ammeter Voltmeter It used to measure It is used measure P.D 0 earth’s atmosphere. for space vehicle communication current Between two points. 3 28. What is the average wavelength of X – rays? Resistance of an ideal Resistance of an ideal A. Range 10nm to 10-4nm. 2 ( 1 0 + 1 0 -4 ) Ammeter is zero. Voltmeter is infinity A v erag e  n m  5.0 0 0 0 5 n m. It is connected in series It is connected in 6 2 29. What is the ratio of speed of infrared rays and in the circuits. parallel in the circuit 3 ultraviolet rays in vacuum? 13. A circular coil of radius ‘r’ having N turns carries a 0 A. The ratio of speed is 1:1, current ‘i’. What is its magnetic moment? Speed of EM waves = 3x108m/s A. M a g n e tic m o m e n t  M   N iA  M  N i   r 2  6 30. Define power factor. On which factors does power 14. What is principle of moving coil galvanometer. factor depend? A. Deflection produced in current carrying coil placed in A. Average power lost over cycle is given by uniform magnetic field is directly proportional to current. = ∅ ,. Cos∅ is called power factor. ( Deflection angle)  (Current in the Coil ) It is depends on Voltage (V), current (i), & phase(∅). ~~~~~ IPE SR PHYSICS STUDY MATERIAL (60/60)~~~~~~~~~~~~~~~~~~~~~~~Page No: 2  12. DUAL NATURE OF RADIATION & MATTER 10. ALTERNATING CURRENT 31. What are “cathode rays”? 48. A transformer converts 200V ac into 2000V ac. A. A stream of fast moving electrons are called cathode rays Calculate the number of turns in the secondary if the 32. What is “work function” primary has 10 turns.(IMP) A. The minimum energy required to escape from the metal N S  VS  NS  2 0 0 0  N  100 S surface is called work function. N P VP 10 2 0 0 33. What is “photoelectric effect”? 49. What type of transformer is used in a 6V bed lamp? A. Eemission of electrons from a metal surface when suitable A. Step down transformer is used in 6V bed lamp. V frequency of light falls on it is called photo electric effect. 50. What is the phenomenon involved in the working of a 34. Write down Einstein’s photoelectric equation. E transformer? A. = ∅+ → N A. Transformer works on the principle of mutual induction. ∅ → , → K 51. Write the expression for the reactance of 35. Write down deBroglie’s relation and explain the terms A i) an inductor and ii) a capacitor. therein. A. 1).  X L   ω L , 2 ).  X C   ω1C T A. The debroglie wavelength associated with a material E 52. When does a LCR series circuit have minimum particle is given by = = where ‘h’ is plank’s impedance? S A. At resonance X =X and Z minimum. L C constant. P : momentum, m : mass of electron, v: velocity H 36. State Heisenberg’ Uncertainty Principle. 15. SEMICONDUCTOR ELECTRONICS A. It is impossible to measure the both the position and 53. What is a p-type semi conductor? What is the momentum of a particle simultaneously to any desire majority and minority charged carriers in it? degree of accuracy. ∆ ∆ ≈ℎ A. If a trivalent impurity is added to a tetravalent 37. Give an example of photo sensitive substances. Why semiconductor is called p-type semiconductor. they are called so? majority charge carriers are holes. A. Example :- Li, Na, K, Zn, Cd, Mg etc minority charge carriers are electrons. Their work function is low, They emit the photo electrons 54. What is an n-type semi conductor? What is the 38. What is important fact did Millikan’s experiment majority and minority charged carriers in it? establish? A. If a pentavalent impurity is added to a tetravalent A. Charge Present on a body is equal to integral multiple of semiconductor is called n-type semiconductor. an electron = ± majority charge carriers are electrons. minority charge carriers are holes. 16. COMMUNICATION SYSTEMS 55. What are intrinsic and extrinsic semiconductors? 39. What are the basic blocks of a communication system? A. Pure form of semiconductors is called intrinsic A. 1) Transmitter. 2) Receiver. 3) Channel. semiconductors. 40. What is “World Wide Web” (WWW)? When impure atoms are added to increase their A. It is an encyclopedia of knowledge accessible to everyone conductivity, they are called extrinsic semiconductors. round the clock through inter net. 56. What is a p-n junction diode? Define depletion layer. 41. Mention the Frequency range of speech signals. A. The junction formed with two electrodes when P-type A. Speech signals frequency range is 300Hz to 3100Hz. and N-type semiconductor are joined is called p-n 42. What is sky wave propagation? junction diode. A. The propagation in which the waves of range a few 1MHz Depletion layer: The narrow region on either side of to 30MHz are received due to total internal reflection takes the junction, due to immobile charge carries is called place at ionosphere is called sky wave propagation. depletion layer. 43. Mention various parts of the ionosphere? 57. Which gates are called universal gates? A. D layer (Part of stratosphere),E layer (Part of stratosphere). 9 A. NAND gates and NOR gates are called universal gates. F1layer (Part of mesosphere),F2layer (Part of thermosphere)4 58. What is Zener voltage (VZ) and how will a Zener 44. Define modulation. Why is it necessary? diode be connected in circuits generally? A. Modulation: The process of combining low frequency 9 A. Reverse bias voltage at which resistance becomes zero audio signal with high frequency audio signals is called 0 and current increases suddenly is called Zener Voltage. modulation Zener diode always connected in reverse bias. 3 Necessary of Modulation: 59. Define amplifier and amplification factor. 1) To reduce size of the antenna. 2 A. Amplifier:- The device used for raising the strength of a 2) To avoid mixing signal from different transmitter’s. 6 weak signal is known as amplifier. 45. Mention Basic methods of modulation. Amplification factor: - The ratio of output power to A. 1) Amplitude modulation (AM) 3 input power is called amplification factor. 2) Frequency modulation (FM), 3) Phase modulation (PM).0 60. Draw the circuit symbols for p-n-p and n-p-n 46. Which type of communication is employed in mobile 6 transistors. phones? A. Wireless communication ( Space wave ) 47. What is meant by wattles component of current? A. The component of current which does not useful work in an A.C circuit is called wattles current ~~~~~ IPE SR PHYSICS STUDY MATERIAL (60/60)~~~~~~~~~~~~~~~~~~~~~~~Page No: 3 1. Define focal length of a concave mirror. Prove that the 4. With a neat labeled diagram explain the formation of radius of curvature of a concave mirror is double its image in a simple microscope. focal length. A. Simple Microscope: A simple microscope consists of A. focal length: The distance between the focus (F) and the only one convex lens of small focal length. pole(P) of the mirror is called the focal length. It is also called as magnifying glass. Consider a ray parallel to the In this, the object is principal axis striking the placed between the mirror at M get reflected V principal focus F passing through the focus (F). E and the optic center The line CM is perpendicular of the convex lens to the mirror N The light ray coming Let ‘ θ ’ be the angle of incidence,∠AMC = ∠MCP K from the object parallel to principal axis refracts at Draw MD⊥ CP A lens and passes through second principal focus F|. MD T The right angled  le MCD Tanθ     (1) Another light ray coming from the object passes CD E through the optic centre which is undeviated. These MD S The right angled  le MFD Tan2θ     (2) two rays meet on the back side of the object to form FD H the image. Hence, the image formed is virtual. Erect Dividing eq(2) by eq(1)  M D  and magnified.   T an 2 θ   F D   C D      (3 ) 5. Define critical angle. Explain total internal reflection T an θ  M D  F D using a neat diagram.    C D  A. Critical angle: when light ray travels from denser If is ‘ θ ’very small, Tanθ  θ and Tan2θ  2θ medium to rarer medium, the angle of incidence, for M lies very close to ‘P’ CD = CP and FD = FP which angle of refraction is 900 is called critical angle. 2θ CP R R Total internal reflection: when a light ray travels from From eq n   2  R  2f denser to rarer medium, the angle of incidence is greater θ FP f f than the critical angle, then it reflects into the same Radius of curvature is equal to double of its focal length. medium is called total internal reflection. 2. Why does the setting sun appear red? Explanation: Consider an object in the denser medium. A. Sunlight passes through the atmosphere before it reaches A ray incident on II1 surface bends away from the us. Sun light is composed of seven colors (VIBGYOR) in perpendicular. As the angle of incidence is increased, the the evenings, where the sun is near the horizon. The rays angle of refraction goes on increasing. For certain angle have to travel longer path in the atmosphere to reach us. of incidence, the refracted ray parallel to II1 surface (900) The dust, smoke and water When the angle of vapor present in the atmosphere incidence is further increased, the ray is deviate away these colors not refracted but is differently depending in their totally reflected back wavelengths. As red has longer into the same medium. wavelength, it is less deviated. Thus the red comes to This phenomenon is straight while other colors get deviated into space. That’s called total internal reflection. why setting sun appears red. 6. Explain the formation of rainbow. 3. Explain the formation of a mirage A. Rainbow: The multicolored arc that appears in the sky, opposite to sun on a rainy day is called rainbow. A. Mirage: Mirage is an optical phenomenon in which light 9 Rainbow is the combined effect of Dispersion, rays are bent to produce a displaced image of distant object 4 Refraction and total internal reflection of sunlight by is called mirage. 9 spherical water droplets of rain in atmosphere. The water 1. Mirage are formed due to 0 drops in atmosphere act as small prisms and cause of total internal reflection of light. dispersion and total internal reflection of sunlight to 3 form rainbow. 2. On hot summer days the density of air is less near the 2 The primary rainbow is a 3 step process. 1) The sunlight is first refracted as it enters a rain drop, ground due to heat. 6 which causes different colors. 3. Hotter air is less dense and smaller refractive index than 3 2) These refracted rays strike the inner surface of the drop cooler air. 0 and get internally reflected 4. In the air is still, the optical density at different layers of 3) These internally reflected rays again get refracted at the 6 opposite surface and come out. As red is less deviated, it air increases with height. 5. Hence the light rays coming from a tall body such as comes straight to observer and appears at top. As violet is more deviated, it comes from lower level drops and tree, bends away from normal and under goes total appears at bottom. internal reflection. The secondary rainbow is also formed due to double internal reflection of sunlight in the rain drops. ~~~~~ IPE SR PHYSICS STUDY MATERIAL (60/60)~~~~~~~~~~~~~~~~~~~~~~~Page No: 4 7. Explain Doppler Effect in light. Distinguish between red 11. What is impact parameter and angle of scattering? shift and blue shift. How are they related to each other? A. Doppler Effect in light: the change in the apparent A. Impact parameter (b): The perpendicular distance of frequency of light, due to relative motion between source the initial velocity of the alpha-particle from the centre of light and observer. This phenomenon is called Doppler of the nucleus is called “impact parameter” shift in light. Angle of Scattering (θ): The angle between the Red shift: When source and observer away from each direction of approach and the direction of the scattering other apparent frequency decreases or apparent wave of alpha particle is called angle of scattering. length increases this is called red shift. The relation between impact parameter and Blue shift:, When source and observer approach each scattering angle other apparent frequency increases or apparent wave 1 Ze 2 θ length decreases this is called blue shift. V b x cot 4 πε 0  1 2  2 8. How do you determine the resolving power of your eye? E  mv  2  A. Resolving Power of Eye: The power of instrument to N produce separate images of two objects lying closer to 12. What are the limitations of Bohr’s theory of hydrogen K each other is called resolving power. atom? A A. 1.It could not explain the fine structure of spectral T lines in hydrogen atom. 1) Let us take a pattern of black strips of equal width E 2. It could not explain the wave properties of electron. separated by white stripes of increasing width from the left S 3. It could not explain the elliptical orbits because Bohr to right on a wall at a height of eye. assumed the circular orbits. H 2) By moving away or closer to the wall, find the position 4. It fails in the case of atoms of the elements for where we can just see some two black strips as separate Which Z > 1. strips with one eye. 13. Write a short note on Debroglie’s explanation of 3) All the black strips would merge into one another and Bohr’s second postulate of quantization. would not be distinguishable. A. Bohr proposed his second postulate as the angular 4) If ‘d’ is the width of the white stripe which is separates momentum of electron in a stationary orbit is quantized. to the two regions and ‘D’ is the distance between the eye nh m v r rn  and the wall, the resolving power of the eye is given by 2π d/D. De Broglie argued that the electron in the stationary 9. Does the principle of conservation of energy hold for orbit acts like a particle wave. As a result it forms interference and diffraction phenomena? Explain briefly stationary waves in the orbit. A. 1) Yes, law of conservation of energy is obeyed. The distance travelled by the 2) Pattern of bright and dark fringes are formed in particle wave along interference and diffraction. circumference should be equal to 3) These patterns obey the principle of conservation of integral multiples of wavelength energy to form stationary waves. 4) In interference and diffraction, light energy is Therefore 2πr = nλ for n = 1, 2, 3……. redistributed. Applying De Broglie hypothesis, we have 5) If it decreases in one region producing a dark fringe. It h h h increases in another region producing bright fringe. λ  (P  mv), nλ  n (Here nλ  2πrn ) p mv n mv n 6) Thus there is no gain or loss of energy. 10. Discuss the intensity of transmitted light when a Polaroid h nh 2πrn  n  m v r rn = sheet is rotated between two crossed Polaroid’s. mv n 2π A. Let I0 be the intensity of polarized light after passing 9 14. Describe Rutherford atom model. What are the draw through the first polarizer P1. Then the intensity of light 4 backs of this model ? after passing through second polarizer P2 will be A. Rutherford model of Atom: I = I cos θ 9 1) Atoms is hollow sphere of radius 10-10m. Where is the 0 2) Total +Ve charge and mass of the atom is concentrated Where is the 3 in a nucleus of radius in the order of 1013to 10-15m. angle between 3) Electrons revolves around the nucleus like planets passing axes P1 and P2since P1 and P2 are crossed the angle 2 around the sun. between the pass axes of P2 and P3 will be − 6 4) Total +Ve charge inside the nucleus is equal to total – Ve charge of revolving electrons around it. Hence the intensity of light emerging from P3 will be 3 Drawbacks of Rutherford’s Model of Atom: I = I cos θ. cos − 0 1) As the revolving electron loses energy continuously, it I = I cos θ. sin θ 6 must spiral inwards and finally merge into the nucleus. I So atom has to collapse. According to electromagnetic I = sin 2 theory it is impossible. 4 2) It could not explain the line spectra of atoms. The transmitted intensity will be maximum when = ~~~~~ IPE SR PHYSICS STUDY MATERIAL (60/60)~~~~~~~~~~~~~~~~~~~~~~~Page No: 5 15. State and explain Coulomb’s inverse square law in 18. Derive an expression for the intensity of the electric electricity. field at a point on the axial line of an electric dipole. A. Coulomb’s law : The force of attraction between two A. Electric field at an axial point of an electric dipole. As electric charges is directly proportional to product of their shown in figure. Consider an electric dipole consisting charges and is inversely proportional to the square of distance between them and acts along the line joining the of charge –q and +q separated by distance 2a and placed charges. in vacuum. Let P be a point on the axial line at distance Explanation: The force between two chargesq and q are r from the center O separated by a distance ‘r’ is given by V F ∝ q q ……. (1) E F ∝ …………. (2) N From (1) and (2) we get F ∝ K Electric field due to charge –q at a point P is F= A 1 −q E = Whereε is the permittivity of free space and T 4πε (r + a) = 9X10 N C E Electric field due to charge +q at a point P is F = (9X10 ) S E = ( ) H Hence the resultant electric field at a point P is 16. Define intensity of electric field at a point. Derive an expression for the intensity due to a point charge. E = E +E A. Intensity of Electric field (E): The force acting per unit positive charge is called intensity of electric field. = − ( ) ( ) Derivation: Let ‘P’ be a point at a distance ‘r’ from Point ( ) ( ) charge ‘q’ at a point ‘O’. = = ( ) ( ) 1 qq 0 The force F  4πε 0 r 2 = ( ) Here (P = q x2a) If r >> a, a2 can be neglected compared to r2 The force acting on a unit positive at ‘P’ due to ‘q’ is  1   2p  F 1  1 qq 0  E a x ia l    3  From the definition E      4πε0   r  q0 q 0  4πε 0 r 2  19. Derive an expression for the intensity of the electric  1 q  field at a point on the equatorial line of an electric E  2   4 πε 0 r  dipole. 17. Derive the equation for the couple acting on a electric A. Electric field at an equatorial point of an electric dipole. dipole in a uniform electric field. As shown in figure. Consider an electric dipole A. Dipole: A pair of equal and opposite charges separated consisting of charge –q and +q separated by distance 2a by a distance (2a) is called dipole. and placed in vacuum. Let P be a point on the equatorial It is placed in a uniform electric field E. making an angle line at distance r from the center O with field direction as shown in the figure. Electric field due to charge –q at a point P is 9 E = 4 Electric field due to 9 charge +q at a point P is 0 E = Due to electric field forces on positive charge (+q) is 3 Clearly the magnitude ofE and E perpendicular to F = +qE and force on negative charge F = -qE. These two equal and opposite charges constitute torque 2 the dipole axis will cancel out. The components parallel (or) moment of couple. 6 to the dipole axis will add up. The total electric field is Torque = Force X Perpendicular distance 3 opposite to E = − E cos θ + E cos θ = (qE)(2a sin θ) = −2E cos θ 0 = 2aq. E sin θ = −2.. Here (P = q x 2a) 6 √ Dipole moment P = q X 2a = 2aq = PE sin θ = − ( ) / In vector form = P X E If r >> a, a can be neglected compared to r2 2  1  p  E equa     3  4πε0   r  ~~~~~ IPE SR PHYSICS STUDY MATERIAL (60/60)~~~~~~~~~~~~~~~~~~~~~~~Page No: 6 20. State Gauss’s law in electrostatics and explain its 23. Explain series combination of capacitors. Derive the importance. formula for equivalent capacitance in each Gauss’ law: The total electric flux through any closed combination. surface is equal to times the net charge enclosed by A. Series combination: In series combination first capacitor second plate is connected to second capacitor q first plate and second capacitor second plate is T h e s u rfac e T o ta l e le c tric flux  E.d s  s ε0 connected to third capacitor first plate and so on first Importance of Gauss’ Law: V capacitor first plate and last capacitor second plate is 1) Valid for closed surface any shape and size. E connected to opposite terminals of a battery 2) Gives the relation for between the electric field and the N 1) In series combination the charge on each capacitor charge. K will be same but potential is different. 3) Applicable to any distribution of charge with in the closed A surface. V = V1 + V2 + V3 21. Derive an expression for the electric potential due to a T point charge. E A. Consider a point charge +Q at point on ‘O’ in free space. S Let us find electric potential at point ‘P’ due to charge +q. H But V = , V = , V = , and V = , let ‘r’ be the position vector of ‘P’ from ‘O’ OP = r,. q q q 1 1 1 V= + + = q + + C C C C C C q  1 1 1   q     C  C1 C2 C3  Consider point A having distance ‘x’ with respect to the 1 1 1 1    point ‘O’. the electric field intensity at that point C C1 C2 C3 1 Q E  24. Explain parallel combination of capacitors. Derive the 4πε0 x 2 formula for equivalent capacitance in each Total amount of work done (W) in bringing unit positive combination. charge from infinite ∞ to r. Parallel combination: If the first plates of all the r r r 1 Q capacitors connected to common terminal and second W   dw    E.d x   .d x    4πε0 x 2 plates of all the capacitors connected to common Q r 1 Q  1 r terminal and these terminals are connected to opposite   .d x     x  terminals of the battery. 4πε 0  x2 4πε 0  r 1) In parallel combination the potential on each Q  1  Q 1 1  W      capacitor will be same but charge is different. 4 π ε 0  x   4 π ε 0  r   Q 1 1   1  Q  q= q + q + q    V     4 π ε 0  r    4πε0   r  22. Derive an expression for the capacitance of a parallel plate capacitor. A. P and Q are two parallel plates of a capacitor separated by But = V, q = C V, q = C V, and q = C V a distance of d. = C V+ C V+ C V The area of each plate = V(C + C + C ) is A. The plate P is C V  V C1  C 2  C 3  charged and Q is 9 C  C1  C 2  C 3 earth connected. 4 The charge on P is +q and surface charge density 25. State and explain Ampere’s law. (7. CHAPTER) 9 = ……….(1) A. Ampere’s law: The line integral of the intensity of 0 magnet field around closed path is equal to times the The electric field due to charge +q is E   2εσ 0 3 net current enclosed by the path. ∮ B⃗ d⃗ = μ i The electric field due to charge -q is = E   2εσ0 2 Proof: Consider a long straight conductor carrying E le c tric fie ld b /w tw o p la te s σ  σ   σ  σ 6 current ‘i’. Magnetic field at a distance ‘r’ from the E     2 .......  2  () 2K ε0  2K ε0   2K ε0  K ε0 3 conductor is given by. = By using capacity formula 0 The value 'B' is parallel to 'dl' θ  00  cos00  1 C  q  q  σA  Kε0A 6    V Ed  σ Kε 0 d d ∮B.dl ∮B.dl cos00 ∮B.dl Captaincy of the parallel plate capacitor is given by  C Kε 0 A B dl  B  2πr    dl  2πr  d μ0  i   B.dl  2π x  2πr    B.dl  μ0i r ~~~~~ IPE SR PHYSICS STUDY MATERIAL (60/60)~~~~~~~~~~~~~~~~~~~~~~~Page No: 7 26. State and explain Biot-Savart law. 29. Derive an expression for the magnetic dipole moment A. Consider a very small element of length dl of a conductor of a revolving electron.7. MOVING CHARGES carrying current (i). Magnetic induction due to small A. Consider an electron revolving in a circular orbit of element at a point P distance r form the element. radius ‘r’ with speed ‘v’. Magnetic induction dB is The time taken by the electron to complete one directly proportional to revolution current (i). 2πr e eV T , Electric current  i    ∝ ( )………… (1) V t 2πr Magnetic induction dB V Magnetic dipole moment  M   NiA is directly proportional E to Length of the  eV  , evr N M  1     r  2 M  element (dl).  2 r  2 K ∝ ………… (2) 30. What are the basic components of a cyclotron? A Mention its uses? 7. MOVING CHARGES Magnetic induction dB is directly proportional to sine angle between r and dl and. T A. Cyclotron: Cyclotron is device used to accelerate ∝ …….. (3) E positively charged particles like protons, deuterons etc. Magnetic induction dB is inversely proportional to the S Two hollow D- shaped metallic chambers D1 and D2 square of the distance from small element to point P H High frequency oscillator. ∝ ………… (4) Strong electromagnet. ( )( )( ) From (1), (2), (3) and (4) dB ∝ Vacuum chamber. ( )( )( ) dB = : Permeability in free space. -1 -1 = 10 Wb m A   Uses: In V e c to r fo rm d B  μ 0 i.d l d l  r   For treatment of chronic diseases. 4π r3 In detection and construction of nuclei. 27. Derive an expression for the magnetic field at the center 31. Desscribe the ways in which Eddy currents are of a current carrying circular coil using Biot-Savart law. used.EMI A. Consider a circular coil of radius ‘r’ and carry a A. Eddy current:- The currents produced in large pieces current ‘i’. Consider a small element ‘dl’. Let ‘O’ is the when they are oscillated in magnetic fields are called center of the coil. By using Biot-Savart Law. eddy currents. μ 0 i.dl.sinθ From Biot-savart law dB  ADVANTAGES:- 4π r2 1.Magnetic Brakes in trains: When strong As dl is perpendicular to 'r' θ  90  sin 90 0  1 0 electromagnets are activated, the eddy currents induced μ 0 i.d l in the rails oppose the motion of train. As a result, dB  4π r2 smooth breaking effect comes into play. μ 0 i.dl 2. Electromagnetic damping: In galvanometer Now  dB   4π r2 electromagnetic damping brings the coil to rest quickly. μ0 i This happens due to eddy currents produced in the core. B   dl 4π r2 3. Induction Furnace: A high frequency alternating μ0 i current is passed through a coil which surrounds the B  4π r 2  2π r    d l  2 πr  metals to be melted. Then the eddy currents generated in μ 0 i If the circular coil has ‘N’ turn μ Ni 9 the metals produce high temperatures. B B 0 4. Electric power meter: The shiny metal disc in the 2 r 2 r 4 electric power meter rotates due to eddy currents. 28. Find the magnetic induction due to a long current carrying a conductor. 9 32. Obtain an expression for the emf induced across a Let ‘P’ be a point at a distance ‘r’ from the long straight 0 conductor which is moved in a uniform magnetic field conductor carrying a current ‘i’ magnetic induction is same which is perpendicular to the plane of motion. EMI at all points on the circle of radius ‘r’ passing through a 3 A. Let conductor PQ is moving with velocity ‘v’ towards point ‘P’ 2 left through a distance ‘x’ on a rectangular conductor PQRS placed in uniform magnetic field ‘B”  B.dl  μ i 0 6 perpendicularly  0 3 The magnetic flux enclosed by the loop PQRS is  B.dl cos 0  μ 0i 0  B  Blx  B dl  μ i 0 6 From Faraday’s law d em f  ε     B lx  μ0 i dt B  2πr   μ 0i B em f  ε    d 2π r  B lv  dt  dx  Induced emf  ε    B l v   v  dt  ~~~~~ IPE SR PHYSICS STUDY MATERIAL (60/60)~~~~~~~~~~~~~~~~~~~~~~~Page No: 8 33. What is rectification? Explain the working of a half wave 36. OR, AND, NOT, NAND, and NOR gates rectifier. A. A. Rectification: The process of converting on alternating current into a direct current is called rectification. V 1) A half wave rectifier can be constructed with a single diode. The AC input signal is connected to the primary coil of aE transformer. The output signal is taken across the load N resistance RL. K 2) During positive half cycle, the diode is forward biased and current flows through the diode. A T 3) During negative half cycle, the diode is reverse biased and no current flows through the diode. E 4) Rectifier efficiency is defined as the ratio of output dcS power to the input ac power. o u tp u t D.C p o w e r 0.4 0 6 R L H 37. Define NAND and NOR gates. Give their truth tables. η   A) NAND gate: When the output of an AND gate is in p u t A.C p o w e r rf  R L connected to the in put of a NOT gate, the resultanting 34. What is rectification? Explain the working of a full wave gate is called NAND gate. rectifier A. Rectification: The process of converting on alternating current into a direct current is called rectification. NOR gate: When the output of an OR gate is connected 1) A full wave rectifier can be constructed with the help of two to the input of a NOT gate, the resultanting gate is called diodes D1and D2. NAND gate. 2) The secondary transformer is centre tapped at C and its ends are connected to the P regions of two diodes D1 and D2. 3) During positive half cycles of AC, the diode D1 is forward biased and current flows through the load resistance R. At this time D2 will be reverse biased and will be in switch off position. 4) During Negative half cycles of AC, the diode D2 is forward biased and current flows through the load resistance R. At 38. Explain the different types of spectral series. this time D1 will be reverse biased and will be in switch off A.. position. 5) Rectifier efficiency is defined as the ratio of output dc power to the input ac power. 9 o u tp u t D.C p o w e r 0.8 1 2 R L 4 η   in p u t A.C p o w e r rf  R L 9 35. Distinguish between half – wave and full – wave rectifier Half wave rectifier Full wave rectifier 0 1) Only one diode is 1) Two diodes are used 3 1. Lyman Series: (n1 = 1) & (n2 = 2, 3, 4 …….…) Used. 2 1 1 1 2) The output is 2) The output is = RH − λ 12 n22 Discontinuous continuous 6 2. Balmer Series: (n1 = 2) & (n2 = 3, 4, 5 …….…) 3) Efficiency is 40.6% 3) Efficiency is 81.2% 3 1 1 1 4) Efficiency is Low. 4) Efficiency is High. = RH − λ 22 n22 5) Only one half of the 5) Both half of the AC 0 3. Paschen Series: (n1 = 3) & (n2 = 4, 5, 6 ………) AC input wave is input wave is conv- 6 1 = RH 1 − 1 converted as DC output -erted as DC output. λ 32 n22 4. Brackett Series: (n1 = 4) & (n2 = 5, 6, 7 ………) 1 1 1 = RH − λ 42 n22 5. Pfund Series: (n1 = 5) & (n2 = 6, 7,8 ………..…) 1 1 1 = RH − λ 52 n22 ~~~~~ IPE SR PHYSICS STUDY MATERIAL (60/60)~~~~~~~~~~~~~~~~~~~~~~~Page No: 9 1. Explain the formation of stationary waves in stretched 1. Current in a circuit falls from 5.0A to 0.0A in 0.1sec. strings and hence deduce the laws of transverse waves in if an average emf of 200V is induced, give an stretched string? estimate of the self-inductance of the circuit. A. Consider a string of length l’ and linear density ’ be fixed A. di = 5.0 – 0.0 = 5A, dt = 0.1sec, emf e = 200V L = ? between two supports under a tension T. a stationary e  L di  L  e dt  0.1   L  200  dt di   4H wave is formed in the string due to the superposition of the  5  waves. At the points where the string was fixed rigidly 2. What is the de-Broglie wavelength associated with an nodes are formed. The velocity of transverse vibration in a electron, accelerated through a potential difference of T stretched string is given by v  μ 100volt? 1St loop: It will have two nodes and one antinodes then the V A. Applied potential (V) = 100V, h 1 2.2 7 1 2.2 7 vibrating length l  λ2 W a ve le n g th  λ      1.2 2 7 n m E p 100 10 λ = 2 l 1  N 4. What is the de-Broglie’s wavelength of a ball of mass T K 0.12Kg moving with a speed of 20m/s? 2  V  μ A. Mass (m) = 0.12Kg. Speed (v) = 20m/s, h = 6.63x 10-34J. A h h Relation b/w V, &  , v     V  3 T d e  B rog lie W a ve le n gth  λ    p mv Substitute eqn (1) and (2) values in eqn (3) we get E 6.6 3  1 0  3 4 6.6 3  1 0  3 4 λ     2.7 6 2  1 0  3 4 m e t 1 T 1 T S 0.1 2  2 0 2.4      0 λ μ 2l μ H 5. A current of 10A passes through two very long wires This is known as fundamental frequency held parallel to each other and separated by a distance 2nd loop: It will have three nodes and two antinodes then of 1m. What is the force per unit length between them? the vibrating length l  2λ2 I1  I 2  1 0 A , d  1 m ,  0  4   1 0  7 H / m 2l  0 I1I 2 4  1 0 7  1 0  1 0 λ  1  F   F   2  1 5 5 2 2 d 2 T V  2  6. A long straight wire carries current of 35A. What is μ Relation b/w V, &  , v     V  3 the magnitude of the field B at a point 20cm from the  Substitute eqn (1) and (2) values in eqn (3) we get wire? A. I = 35A and r = 20cm = 0.2m, 1 T 2 T  0 2i 4  1 0 7 2 3 5     1  or   1  2 0 B . .  3.5  1 0  5 T.  22l  μ 2l μ 4 r 4 0.2 This is known as 1st over tone (or) 2nd harmonic 7. The horizontal component of the earth’s magnetic 3rdloop :It will have four nodes and three antinodes then field at a certain place is 2.6x10-5T and the angle of the vibrating length l  3λ2 dip 600. What is the magnetic field of the earth at this location? (1) 2l λ  1  A. HE = 2.6x10-5T, = 60 BE = ? 3 HE T H E  B E co s  , B E  V  2  cos  μ 9 2.6  1 0  5  2  2.6  1 0  5   5.2  1 0  5 T Relation b/w V, &  , v     V   3 BE  1  4   Substitute eqn (1) and (2) values in eqn (3) we get 2 1 T 3 T 9 8. The sequence of bands marked on a carbon resister   2l  2   o r   2  3 0 0 is Red, Red, Red, and Silver. What is its resistance 3 μ 2l μ and tolerance? st rd This is known as 2 over tone (or) 3 harmonic 3 A. Resistance Red, Red, Red = 22x102 Ώ = 2200 Laws of transverse vibrations: 2 Tolerance = ±10% First law: ( when T and μ are constants ) 9. What is the colour code of a carbon resistor of The fundamental frequency of stretched string is inversely 6 resistance 23Kilo Ohm’s proportional to length of the string.  α 1l  1l1  2l2 3 A. 23 X 103 Ώ = Red, Orange, Orange. 0 10. Two wires of equal length, of copper and manganin, Second law: ( when l and are constants ) have the same resistance. Which wire is thicker? The fundamental frequency of stretched string is directly 6 A. R = proportional to square root of the tension.  α T  1  T1 T2 Since ‘R’ and ‘l’ are same ρ ∝ A 2 For manganin wire ‘ρ’ is more so it is Thicker. Third law: ( when l and T are constants ) 11. Why is house hold appliances connected in parallel? The fundamental frequency of stretched string is inversely A. It is because any electrical device can be turned on or proportional to square root of the linear density off without affecting the operation of other devices. 1 1 μ2  α  μ 2 μ1 ~~~~~ IPE SR PHYSICS STUDY MATERIAL (60/60)~~~~~~~~~~~~~~~~~~~~~~~Page No: 10 2. How arestationary waves formed in closed pipes? Explain 3. Explain the formation of stationary waves in an air the various modes of vibrations and relations for their Column enclosed in open pipe. Derive the equation for frequencies the frequencies of the harmonics produced. A. Closed Pipe:-If one end of the organ pipe is closed and A. Open pipe:- If both ends of the organ pipe is open is the other end is open is Known as closed organ pipe known as open organ Pipe Formation of stationary wave:-When a sound wave is Formation of stationary wave:-When a sound wave is sent to a closed pipe. The wave reflects back at the closed sent to a closed pipe. The wave reflects back at the end of the pipe. The incident wave and reflected wave closed end of the pipe. The incident wave and reflected travelling in opposite direction super impose each other. wave travelling in opposite direction super impose each First harmonic (or) fundamental: other is called stationary wave. It will have one node and one antinodes V First harmonic (or) fundamental: λ It will have one node and two antinodes The vibrating length l  1 E 4 N The vibrating length l  λ 1 2 λ1  4l K λ1  2l Relation between V, ϑ, and λ A Relation between V, ϑ, and V T V   λ  0  1  V λ1 V   λ  0  E λ1 V 0  S V 4l 0  H 2l This is known as first harmonic (or) fundamental Third harmonic (or) first overtone: This is known as first harmonic (or) fundamental It will have two node and two antinodes frequency. Second harmonic (or) first overtone: 3λ The vibrating length l  3 It will have two node and three antinodes 4 The vibrating length l  2 λ 2 4l 2 λ3  2l 3 λ2  2 Relation between V, ϑ, and λ Relation between V, ϑ, and λ V V V  λ  1   1  V   λ  1  V  1  V λ3  4l  λ2  2l       3   2  3V  1  3 0 2V 1  4l 1  1  20 2l This is known as 1st over tone (or) 3rd harmonic This is known as 1st over tone (or) 2nd harmonic Fifth harmonic (or) Second overtone: It will have three node and three antinodes Third harmonic (or) second overtone: It will have three node and four antinodes The vibrating length l  5λ 5 3λ3 4 The vibrating length l  4l 2 λ5  5 2l λ3  Relation between V, ϑ, and λ 3 V  λ  2  V  2  V Relation between V, ϑ, and λ λ5  4l  9 V V   V  λ  2   2   5  4 λ3  2l      5V  2  5 0 9  3  2 4l 3V 2  30 rd th This is known as 2 over tone (or) 5 harmonic 0 2  2l The frequency ratio of the closed pipe is given by 3 This is known as 2nd over tone (or) 3rd harmonic  V   0 : 1 :  2  1   V   :3   :5   V    1:3 :5 2 The frequency ratio of the open pipe is given by  4l   4l   4l  6  V  0 :1 :2  1   V   :2   V   :3    1:2 :3 A steel wire 0.72m long has a mass of 5.0 X 10-3 kg. If  2l   2l   2l  the wire is under a tension of 60 N, what is the speed of 3 A closed organ pipe 70cm long is sounded. If the transverse waves on the wire? 0 velocity of sound is 331m/s , what is the fundamental M = 5.0X10-3 kg L = 0.72m T = 60 N 3 6 frequency of vibration of the air column ? M 5.0 X 1 0 μ    6.9 X 1 0  3 kg/m A. l = 70cm = 70 x 10-2 m v = 331m/s l 0.7 2 V 331 The speed of the wave on the wire is given by     118.2 H z T 60 4l 4  70 X 10  2  V    9 3 m /s μ 6.9 X 1 0 3 ~~~~~ IPE SR PHYSICS STUDY MATERIAL (60/60)~~~~~~~~~~~~~~~~~~~~~~~Page No: 11 4. What is Doppler Effect? Derive an expression for the 5. What is Doppler Effect? Derive an expression for the apparent frequency heard when the source is in motion apparent frequency heard when the observer is in and the observer is at rest motion and the source is at rest A. The phenomenon of apparent change in frequency due to A. The phenomenon of apparent change in frequency due relative motion between the source and observer is called to relative motion between the source and observer is Doppler effect called Doppler effect Expression for the apparent frequency heard when the Expression for the apparent frequency heard when Source is in motion and the Observer is at rest the Observer is in motion and the Source is at rest Let us consider a source S producing a sound note the Let us consider a source S producing a sound note the frequency let the velocity of sound in air be’ V’ frequency let the velocity of sound in air be’ V’ V=ϑ V V = ϑλ Let the source S is moving with a velocity towards E Let the observer O is moving with velocity towards observer at rest.Then the distance travelled by the source the source at rest.He will receive more number of in time T is equal to the wave is compressed and N waves each second. The distance travelled by him in hence wavelength of the wave is decrease The apparent K one second is the extra waves received by him due wavelength is λ l  λ -V S T A to his motion are equal to in second. V T The apparent frequency is given by T h e a p p a re n t fre q u en c y = λl E ϑ| = ϑ + V  V 1 l   λ  , T S  λ  VST      H ϑ| = ϑ + l V    V VS  =ϑ+ ϑ       l  V       V  V S  ϑ| = 1 + ϑ | Then the apparent frequency is greater than the actual frequency ϑ| = ϑ Similarly when the source is moving away from the Then the apparent frequency | is greater than the observer The wavelength of the wave increases. actual frequency The apparent wavelength is λ l1  λ  VS T Similarly when the observer is moving away from the V T h e a p p a re n t fre q u en c y = source at rest with velocity he receive number of λl V  V 1  waves less in every second l1    λ  , T  The apparent frequency is given by λ+ VS T      ϑ|| = ϑ − V ϑ|| = ϑ − l1    V    V S   =ϑ− ϑ     l1    V  