XII Physics Study Material (High Achievers) - PDF - 2022-2023

KENDRIYA VIDYALAYA SANGATHAN CHENNAI REGION CLASS – XII PHYSICS SESSION 2022-23 PREPARATION PLANNER & PACKAGE FOR HIGH ACHIEVERS 1  CHIEF PATRON Ms. T.Rukmani Offg.Deputy Commissioner KVS Chennai Region PATRON...

KENDRIYA VIDYALAYA SANGATHAN CHENNAI REGION CLASS – XII PHYSICS SESSION 2022-23 PREPARATION PLANNER & PACKAGE FOR HIGH ACHIEVERS 1  CHIEF PATRON Ms. T.Rukmani Offg.Deputy Commissioner KVS Chennai Region PATRON Mr.P I Thanga Raja Assistant Commissioner KVS Chennai Region COURSE DIRECTOR Mr. R.N.Senthil Kumar Principal Kendriya Vidyalaya, Nagercoil. 2  CONTENTDEVELOPMENT TEAM S.No Name of the Chapter Name of theTeacher Name of the KV Unit - 1 Electrostatics (Chapters 1 MRS. R SUGUNA CHENNAI AFS AVADI 1 & 2) Unit - II Current Electricity (Chapter-3) & Preparation of MRS. GEETHA RAMESH CHENNAI ANNANAGAR 2 Questions Paper for 35 marks from Unit 1 & 2 Unit - III Magnetic Effects of 3 Current and Magnetism (Chapters MR.T.MURALI CHENNAI DGQA 4&5) Unit - IV Electromagnetic Induction and Alternating MRS.RADHA MUKUNDAN CHENNAI MINAMBAKKAM 4 (Chapters 6 & 7) Unit - V Electromagnetic Waves (Chapter-8) & Preparation of MR.SANKARRAMAN CHENNAI MINAMBAKKAM 5 Questions Paper for 35 marks from Unit 3 & 4 6 Unit - VI Optics (Chapters 9&10) MR.V SIVARAMAKRISHNAN COIMBATORE Unit - VII Dual Nature of Radiation and Matter (Chapter- 11) & MR. K. RENGANATHAN TRICHY NO.1 7 Preparation of Questions Paper for 35 marks from Unit 5 & 6 Unit - VIII Atoms and Nuclei MR. C. MURUGAVEL PONDICHERRY NO.1 Shift 1 8 (Chapter 12 & 13) Unit - IX Electronic Devices (Chapter 14) & Preparation of MR SATYA SURYANARAYANA PORT BLAIR NO.2 9 Questions Paper for 35 marks SURMPUDI from Unit 7, 8 & 9 EDITING AND COMPILATION BY RESOURCE PERSONS MR.P.SEENIVASAN, K.V No.1, NARIMEDU, MADURAI MR.K V SRINIVASAN, K.V No.1, TAMBARAM, CHENNAI MR.S DHAMODHARAN, KV GILL NAGAR, CHENNAI MR.S KUMAR, KV CRPF AVADI MR.S CHANDRA KUMAR, KV NAGERCOIL MRS.A. BEULAH JASMINE, KV NAGERCOIL 3  CONTENT& PLAN DAY NAME OF TASKS PAGE Completed THE UNIT NO. or Not Unit - 1 Electrostatics Preparation of Lessons DAY -1 3–4 (Chapters 1 & 2) Revision –Formula chart & Quick revision notes Numericals for \revisions 5 -9 Numericals for practice 9 – 14 DAY -2 Solving-Case study questions 14 – 17 Solving-Assignment-1 17 – 21 Solving- Assignment-2 21 - 23 Unit - II Current Preparation of Lessons DAY -3 Electricity (Chapter-3) Revision –Formula chart & 24 Quick revision notes 25 – 29 Numericals for revisions 30 – 36 Numericals for practice 37 – 42 Solving-Case study questions 43 – 51 Solving-Assignment-1 51 - 60 Solving- Assignment-2 DAY -4 Full Revisions Revision Test - Unit 1 &2 (35 marks) 60 -69 Unit - III Magnetic Preparation of Lessons DAY -5 Effects of Current and Revision –Formula chart & 70 - 72 Magnetism (Chapters Quick revision notes 4&5) Numericals for revisions 73 – 75 DAY -6 Numericals for practice 76 - 77 Solving-Case study questions 78 – 80 Solving-Assignment-1 81 - 83 Solving- Assignment-2 Unit - IV Preparation of Lessons DAY -7 Electromagnetic Revision –Formula chart & 84 – 87 Induction and Quick revision notes Alternating (Chapters 6 Numericals for revisions 87 – 92 & 7) Numericals for practice 93 – 95 Solving-Case study questions 96 – 97 Solving-Assignment-1 98 – 106 Solving- Assignment-2 106 - 111 DAY -8 Full Revisions Revision Test - Unit 3 &4 (35 marks) 112 - 120 4  Unit - V Electromagnetic Preparation of Lessons DAY -9 Waves (Chapter-8) Revision –Formula chart & 121 Unit - VI Optics Quick revision notes (Chapters 9) Numericals for revisions 122 - 124 Numericals for practice 124 –125 Solving-Case study questions 126 – 129 Solving-Assignment-1 130 - 138 Solving- Assignment-2 DAY -10 Unit - VI Optics Preparation of Lessons Continued to Revision –Formula chart & 139 – 140 (Chapters10) Quick revision notes Numericals for revisions 141 – 144 Numericals for practice 145 – 146 Solving-Case study questions 146 – 147 Solving-Assignment-1 148 – 151 Solving- Assignment-2 152 - 154 DAY -11 Full Revisions Revision Test - Unit 5 & 6 (35 marks) 174 - 179 DAY -12 Unit - VII Dual Nature of Preparation of Lessons Radiation and Matter Revision –Formula chart & 155 (Chapter- 11) Quick revision notes 156 – 158 Numericals for revisions 159 – 161 Numericals for practice 162 – 164 Solving-Case study questions 165 – 167 Solving-Assignment-1 168 - 173 Solving- Assignment-2 DAY -13 Unit - VIII Atoms and Preparation of Lessons Nuclei (Chapter 12 & Revision –Formula chart & 180 – 181 13) Quick revision notes 182 – 185 Numericals for revisions 186 – 187 Numericals for practice 188 – 189 Solving-Case study questions 190 – 193 Solving-Assignment-1 194 - 195 Solving- Assignment-2 DAY -14 Unit - IX Electronic Preparation of Lessons Devices (Chapter 14) Revision –Formula chart & 196 Quick revision notes 197 – 200 Numericals for revisions 201 – 203 Numericals for practice 204 – 205 Solving-Case study questions 206 – 208 Solving-Assignment-1 209 - 214 Solving- Assignment-2 DAY -15 Full Revisions Revision Test - Unit 7,8 & 9 (35marks) 215 - 221 5 Unit - 1 Electrostatics (Chapters 1&2) FORMULA CHART 6 7 NUMERICALS FOR PRACTICE WITH SOLUTIONS S.no QUESTION HINT 1 What is the net force and its direction that the charges at the vertices A and C of the right triangle ABC exert on the charge in vertex B? |F| = √(|FAB|2 + |FCB|2) = k × 10-8 √( 0.875 2 + 12 ) = 9.00 × 109 × 10-8 √( 0.875 2 + 12 ) = 1.20 × 102 N θ = arctan(|FCB|/ |FAB| ) = arctan( k × 10-8 / 0.875 k × 10-8) = 48.8° 2 Three charges are located at the vertices of a right isosceles triangle as shown below. What is the magnitude and direction of the resultant electric field at the midpoint M of AC? E = EBM = 1.44 × 107 N/C 3 The distance AB between charges Q1 and Q2 shown If W is the work to be done to below is 5.0 m. How much work must be done to move move Q2 from a position where 8  charge Q2 to a new location at point C so that the distance its potential energy is Ep1 and BC = 2.5 m? kinetic energy 0 (from rest) to another position where its potential energy is Ep2 and kinetic energy 0 (to rest), then by the conservation of energy, we have. Ep1 + W = Ep2 which gives W = Ep2 - Ep1 W = k Q1 Q2 (1/AB - 1/AC) = 9×10-3J 4 A parallel plate capacitor having capacitance 12 pF is Initial Energy of the capacitor, charged by a battery to a potential difference of 10 V Ui = (1/2) CV2 between its plates. The charging battery is now = (1/2) x 12pF x 10 x 10 disconnected and a porcelain slab of dielectric constant = 600 pJ 6.5 is slipped between the plates. Find the work done by the capacitor on the slab. After the slab, the energy of the slab, Uf = (1/2) Q2/C’ Q = CV = (12 pF)(10 V) = 120 p C C’ = kC = 6.5 x 120 x 10-12 F Therefore, Uf = [(1/2) (120 x 10- 12 2 ) ]/[6.5 x 120 x 10-12] Uf = 92 pJ W= Ui - Uf 5 There is a uniform electrostatic field in a region. The ΔV = E.d potential at various points on a small sphere centred at ΔV = Edcosθ = 0.8 x cos 600 P, in the region, is found to vary between the limits 589.0 V to 589.8 V. What is the potential at a point on ΔV = 0.4 the sphere whose radius vector makes an angle of 60° Hence the new potential at the with the direction of the field? point on the sphere is 589.0 + 0.4 = 589.4 9 6 A test charge ‘q’ is moved without acceleration from A to P.D does not depend upon the C along the path from A to B and then from B to C in path along which the test charge electric field E as shown in the figure. q moves (ii) At point C, electric potential will be more as potential decreases in the direction of (i) Calculate the potential difference between A and C. electric field. (ii) At which point (of the two) is the electric potential more and why? 7 Two small identical electrical dipoles AB and CD, each of Answer: dipole moment ‘p’ are kept at an angle of 120° as shown in Resultant dipole moment of both the figure. What X’ is the resultant dipole moment of this dipoles is combination? If this system is subjected to electric field (E→) directed along + X direction, what will be the magnitude and direction of the torque acting on this? Resultant dipole moment (p) makes an angle of 60° with each dipole and 30° with x-axis as shown in the figure. 8 Two uniformly large parallel thin plates having charge The equipotential surface is at a densities + σ and – σ are kept in the X-Z plane at a distance d/2 from either plate in distance ‘d’ apart. Sketch an equipotential surface due to X-Z plane. For a particle of electric field between the plates. If a particle of mass m charge (- q) at rest between the and charge q’ remains stationary between the plates, what plates, then is the magnitude and direction of this field? 10  (i) weight mg acts, vertically downward (ii) electric force qE acts vertically upward 9 Draw 3 equipotential surfaces corresponding to a field that uniformly increases in magnitude but remains constant d2 < d1 for increasing field along Z-direction. How are these surfaces different from and d2 = d1 for uniform field. that of a constant electric field along Z-direction? 10 A charge ‘q’ is moved from a point A above a dipole of potential remains constant dipole movement ‘p’ to a point B below the dipole in equitorial plane without acceleration. Find the work done in the process. (All India 2016) 11 11 Calculate the work done to dissociate the system of three Find charges placed on the vertices of a triangle as shown. initial P.E= Ui Final P.E = Ui =0 W.D = Ui- Ui 12 Figure shows two identical capacitors C1 and C2, each of 2 (i) When switch S is open and µF capacitance, connected to a battery of 5 V. Initially dielectric is introduced, charge switch ‘S’ is left open and dielectric slabs of dielectric on each capacitor will be q1 = constant K = 5 are inserted to fill completely the space C1 V, q2 = C2V between the plates of the two capacitors. How will the q1 = 5CV charge and = 5 × 2 × 5 = 50 µC, q2 = 50 µC Charge on each capacitor will become 5 times (ii) P.d. across C1 is still 5V and across C2, q = (5C) V (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted? 13 A network of four capacitors each of 12μF capacitance is Equivalent capacitance of the connected to a 500 V supply as shown in the figure. network, Determine (a) equivalent capacitance of the network and (b) charge on each capacitor. 14 Two parallel plate capacitors of capacitances C1 and Net capadtance before filling C2 such that C1 = 3C2 are connected across a battery of V the gap with dielectric slab volts as shown in the figure. Initially the key (k) is kept dosed to fully charge the capacitors. The key is now 12  thrown open and a dielectric slab of dielectric constant ‘K’ is inserted in the two capacitors to completely fill the gap between the plates, Energy stored in the combination before introduction of dielectric slab, Find the ratio of (i) the net capacitance and (ii) the energies stored in the combination, before and after the introduction of the dielectric slab. 15 Concentric metallic hollow spheres of radii R and 4R hold charges Q1 and Q2 respectively. Given that the surface charge density of the concentric spheres are equal, Find potential difference V(R) – V(4R). VA=kQ1/R+kQ2/R Potential on the surface of the outer sphere (at B) VB=kQ1/4R+kQ2/4R Potential difference, Δ=VA−VB=3/4(kQ1/R)= [3/16πε0] (Q1/R) NUMERICALS FOR PRACTICE S.No Question Answer 1 What is the direction and magnitude of the electric field at the Direction against midpoint of an electric dipole made of length 2a? 13  EI =EIII = 0, EII= σ/ε0 2 I +σ II - II σ I Write the expression for the electric field in the regions I, II, III shown in the above figure. 3 Show diagrammatically the stable and unstable equilibrium of an electric dipole placed in a uniform electric field Unstable 4 Two electric charges 3μC, -4μC are placed at the two corners of an isosceles right angled triangle of side 1 m as shown in the figure. What is the direction and magnitude of electric field at A due to the two charges? A -4μC 3μC 5 A charge +Q fixed on the Y axis at a distance of 1m from the Force due to both the changes origin and another charge +2Q is fixed on the X axis at a are equal = KQ2& ⊥ r to each distance of m from the origin. A third charge – Q is placed other so the resultant force at the origin. What is the angle at which it moves? will make 45o with X-axis. 6 Draw the graph showing the variation of electric potential with V distance from the centre of a uniformly charged shell. r dist 14 7 Sketch the electric field lines, when a positive charge is kept in + the vicinity of an uncharged conducting plate q - - - 8 A uniformly charged rod with linear charge density λ of length L is inserted into a hollow cubical structure of side ’L’ with ø constant velocity and moves out from the opposite face. Draw the graph between flux and time. time 9 Draw a graph showing the variation of potential with distance from the positive charge to negative charge of a dipole, by V choosing the mid-point of the dipole as the origin. d 10 Three charges +Q, q, +Q are placed respectively, at distance QQ/d2 + Qq/(d/2)2 =0 0, d/2 and d from the origin, on the x-axis. If the net force Q + 4q = 0 experienced by +Q placed at x = 0 is zero, then value of q is or q = -Q/4 11 An electric field of 1000 V/m is applied to an electric dipole E = 1000 V/m , p = 10-29 cm, at an angle of 45°. The value of the electric dipole moment is θ = 450 10–29 Cm. What is the potential energy of the electric dipole? Potential energy stored in the dipole, U = -p.Ecos θ = – 1029- x 1000 x cos450 U=−12×10−26 15  U = – 0.707 x 10-26 J= -7 x 10 -27 J 12 The bob of a simple pendulum has a mass of 2 g and a charge At equilibrium, of 5.0 C. It is at rest in a uniform horizontal electric field of Tcosθ = mg ——-(1) intensity 2000 V m–1. At equilibrium, the angle that the pendulum makes with the vertical is (take g = 10 m s–2) Tsinθ = qE ——–(2) Dividing (2) by (1) tanθ = qE/mg θ = tan-1((5 x 10-6 x 2 x 103) / (2 x 10-3 x 10)) = tan-1(0.5) 13 A capacitor with a capacitance 5 µF is charged to 5 µC. If the Work done = Uf – Ui = plates are pulled apart to reduce the capacitance to 2 µF, how (½)q /Cf – (½)q /Ci 2 2 much work is done? Work done = q2/2[1/Cf – 1/Ci] Work done = [(5 x 10-6)2/2][(1/(2 x 10-6)) – (1/(5 x 10-6))] Work done = 3.75 x 10-6 J 14 A parallel plate capacitor of capacitance 90 pF is connected Induced charge on dielectric, to a battery of emf 20 V. If a dielectric material of dielectric Q = Q(1 – 1/K) ind constant K = 5/3 is inserted between the plates, find the magnitude of the induced charge? Final charge on capacitor, Q = K C0 V Q = (5/3) x 90 x 10-12 x 20 = 3 x 10-9 C = 3nC Qind = 3(1 – ⅗) = 3 x ⅖ = 1.2 nC 15 How will connect seven capacitors of 2µf to obtain an effective 5 in parallel and 2 in series capacitance of 10/11 µf. 16 Two identical metal plates are given positive charges Q1 and V=Q/C Q2,where Q1> Q2. Find the potential difference between them, Total charge= 16  if they are now brought together to form a parallel plate Q1-Q2 capacitor with capacitance C. V = Q1-Q2/C 17 Why is electrostatic potential constant throughout the volume Electric field inside the of the conductor and has the same value (as inside) on its conductor = 0 surface? 18 Two point charges 4Q, Q are separated by lm in air. At what point on the line joining the charges is the electric field intensity zero? Also calculate the electrostatic potential energy of the system of charges, taking the value of charge, Q = 2 × 10-7C (ii) Electrostatic potential energy of the system is 19 The given graph shows variation of charge ‘q’ versus potential Line B corresponds to C1 difference ‘V’ for two capacitors C1 and C2. Both the Reason: Since slope (qv) of capacitors have same plate seperation but plate area of C2 is ‘B’ is less than that of ‘A’ greater than that of C1. Which line (A or B) corresponds to C1 and why? 20 Net capacitance of three identical capacitors in series is 1 pF. Let C be the capacitance of a What will be their net capacitance if connected in parallel? capacitor Find the ratio of energy stored in the two configurations if they Given : C1 = C2 = C3 = C are both connected to the same source. 17  When connected in series: CASE STUDY QUESTIONS (I) Concept of field lines was introduced by Michael Faraday as an aid in visualizing electric and magnetic fields. Electric line of force is an imaginary straight or curved path along which a unit positive charge tends to move in an electric field. Properties of lines of forces observed by the scientist such as: Lines of force start from positive charge and terminate at negative charge, Lines of force never intersect, the tangent to a line of force at any point gives the direction of the electric field E at that point, the number of lines per unit area, through a plane at right angles to the lines, is proportional to the magnitude of E. This means that, where the lines of force are close together, E is large and where they are far apart, E is small. Each unit positive charge gives rise to 1/ ε0 lines of force in free space. Hence number of lines of force originating from a point charge q is N = q/ε0 in free space. 1. Choose correct statement regarding electric lines of force: (a) Emerges from (-ve) charge and meet at (+ve) charge. (b) Electric field in a region is strong when the electric lines of force at that region is closelyspaced. (c) Just as it is shown for a point system in the same way it represents for a solid sphere. (d) has a physical nature. 2. Two electric field lines due to a point charge: (a) Never intersect (b) May intersect near the charge (c) Always intersect at 2 points (d) None of these 3. The tangent at any point on the electric field line gives: (a) The direction of magnetic field at that point (b) The direction of electric field at that point (c) The direction of acceleration due to gravity (d) All of the above 18 4. A metallic sphere is placed in a uniform electric field. The lines of force follow the paths as shown in figure. Identify the correct path of lines of force. (a) I (b) ii (c) iii (d) iv 5. If the direction of the electric field line due to two unlike point charges is from left to right then: (a) Positive charge is at left and negative charge is at right (b) Negative charge is at left and positive charge is at right (c) Both charges are at left (d) Both charges are at right (II) The parallel plate capacitor consists of two parallel metal plates X and Y each of area A, separated by a distance d, having a surface charge density σ as shown in figure. The medium between the plates is air. A charge +q is given to the plate X. It induces a charge –q on the upper surface of earthed plate Y. When the plates are very close to each other, the field is confined to the region between them. The electric lines of force starting from plate X and ending at the plate Y are parallel to each other and perpendicular to the plates. The capacitance is directly proportional to the area (A) of the plates and inversely proportional to their distance of separation (d). The capacitance (C) of the parallel plate capacitor is given by C= ϵ0A/d. if the region between the two plates is filled with dielectric like mica or oil. Its capacitance increased by ϵr times of the medium. 1. The potential difference between the two plates of a parallel plate capacitor, if Q is magnitude of charge on each plate of area A separated by a distance d is (a) Qd/(εoA) (b) dεo/AQ (c) Ad/(εoQ) (d) QA/dεo 2. A capacitor is charged by a battery and the charging battery is disconnected and a dielectric slab is inserted in it. Then for the capacitor 19 (a) Charge remains constant (b) Charge increases (c) Potential difference remains constant (d) Potential difference increases 3. A parallel plate capacitor has a capacitance of 10 μF. If the distance between two plates is doubled then the new capacitance will be (a) 20 μF (b) 15 μF (c) 10 μF (d) 5 μF 4. Capacitance of a parallel plate capacitor does not depend on: (a) Area of the plates (b) Type of metal used for plates (c) Separating distance between the plates (d) Dielectric constant of the medium between the plates 5. A parallel plate air capacitor with no dielectric between the plates is connected to a constant voltage source. What happens to the capacitance if a dielectric of dielectric constant k = 2 is inserted between the plates? (a) Capacitance decreases (b) Capacitance increases by two times (c) Capacitance remains unchanged (d) Insufficient data III Electric field strength is proportional to the density of lines of force i.e., electric field strength at a point is proportional to the number of lines of force cutting a unit area element placed normal to the field at that point. As illustrated in given figure, the electric field at P is stronger than at Q. 1) Electric lines of force about a positive point charge are (a) radially outwards (b) circular clockwise (c) radially inwards (d) parallel straight lines 2) Which of the following is false for electric lines of force? (a) They always start from positive charge and terminate on negative charges. 20 (b) They are always perpendicular to the surface of a charged conductor. (c) They always form closed loops. (d) They are parallel and equally spaced in a region of uniform electric field ASSIGNMENT -I (MCQ ) Q.NO. QUESTIONS ANS WER 1 Three charges +Q, q, +Q are placed respectively, at distance 0, d/2 and d from the (d) origin, on the x-axis. If the net force experienced by +Q placed at x = 0 is zero, then value of q is (a) +Q/4 (b) –Q/2 (c) +Q/2 (d) –Q/4 2 An electric field of 1000 V/m is applied to an electric dipole at an angle of 45°. The (b) value of the electric dipole moment is 10–29 Cm. What is the potential energy of the electric dipole? (a) –10 × 10–29 J (b) –7 × 10–27 J (c) –20 × 10–18 J (d) –9 × 10–20 J 3 Voltage rating of a parallel plate capacitor is 500 V. Its dielectric can withstand a (b) maximum electric field of 106 V m–1. The plate area is 10–4 m2. What is the dielectric constant if the capacitance is 15 pF? (given ε0 = 8.86 × 10–12 C2 N–1 m–2) (a) 3.8 (b) 8.5 (c) 6.2 (d) 4.5 4 The bob of a simple pendulum has a mass of 2 g and a charge of 5.0 C. It is at rest in a (b) uniform horizontal electric field of intensity 2000 V m–1. At equilibrium, the angle that the pendulum makes with the vertical is (take g = 10 m s–2) (a) tan–1 (0.2) 21  (b) tan–1 (0.5) (c) tan–1 (2.0) (d) tan–1 (5.0) 5 A parallel plate capacitor has 1 μF capacitance. One of its two plates is given + 2 μC (d) charge and the other plate, +4 μC charge. The potential difference developed across the capacitor is (a) 3 V (b) 2 V (c) 5 V (d) 1 V 6 Two identical conducting spheres A and B, carry equal charge. They are separated by (a) a distance much larger than their diameters, and the force between them is F. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and then removed. As a result, the force between A and B would be equal to (a) 3F/8 (b) F/2 (c) 3F/4 (d) F 7 Two capacitors C1 and C2 are charged to 120 V and 200 V, respectively. It is found (c) that by connecting them together the potential on each one can be made zero. Then (a) 9C1 = 4C2 (b) 5C1 = 3C2 (c) 3C1 = 5C2 (d) 3C1 + 5C2 = 0 8 An electric dipole is placed at an angle of 30º to a non-uniform electric field. The (d) dipole will experience (a) a torque only (b) a translational force only in the direction of the field (c) a translational force only in a direction normal to the direction of the field (d) a torque as well as a translational force 9 The magnitude of electric field intensity E is such that, an electron placedinitwould (b) experienceanelectricalforce equaltoitsweightisgivenby (a) mge 22  (b) mg/e (c) e/mg (d) e2g/m2 10 Four point charges -Q , -q , 2q and 2Q are placed, one at each corner of the square. The (a) relation between Q and q for which the potential at the centre of square is zero is: (a) Q = -q (b) Q = - 1/q (c) Q = q (d) Q = 1/q 11 What is the flux through the cube of side ‘a’ if a point charge of q is at one corner? (b) (a) 2q/ԑ0 (b) q/8 ԑ0 (c) q/ ԑ0 (d) q 6a2/ ԑ0 12 The electric potential V at any point (x,y,z), all in metres in space is given by V = 4x2 volt. (d) The electric field at the point (1,0,2) in volt/metre, is (a) 8 along positive X -axis (b) 16 along negative X -axis (c) 16 along positive X - axis (d) 8 along negative X - axis 13 The presence of an uncharged conductor near a charged one increases the (b) (a) the potential of the charged conductor (b) the capacity of the charged conductor (c) charge of the charged conductor (d) No effect 14 What is the value of capacitance that must be connected in parallel with 50 pF condenser to (b) make an equivalent capacitance of 150 pF? (a) 200pF (b) 100pF (c) 50pF (d) 150pF 15 The relation between electric polarization and susceptibility indicates that electric (d) polarization is (a) proportional to square root of susceptibility. (b) proportional to susceptibility. (c) inversely proportional to susceptibility. (d) independent of susceptibility. 16 If a third equal and similar charge is placed between two equal and similar charges, then © this third charge will (a) move out of the field of influence of the two charges 23  (b) not be in equilibrium (c) Will be in stable equilibrium (d) be in unstable equilibrium 17 The electric potential at a point in free space due to a charge Q coulomb is Qx1011 V. The (a) electric field at that point is (a) 4πԑ0Qx1022 V/m (b) 12πԑ0Qx1020 V/m (c) 4πԑ0Qx1020 V/m (d) 12πԑ0Qx1022 V/m Directions:Thesequestionsconsistoftwostatements,eachprintedasAassertionandReason. Whileansweringthesequestions,youarerequiredtochooseanyone ofthefollowingfive responses. a) Ifbothassertionandreasonaretrueandthereasonisthecorrectexplanationofhe assertion. b) Ifbothassertionandreasonaretruebutreasonisnotthecorrectexplanationofthe assertion. c) Ifassertionistruebutreasonisfalse. d) Iftheassertionandreasonbotharefalse. 18 Assertion: If a point charge q is placed in front of an infinite grounded conducting plane (a) surface, the point charge will experience a force. Reason : This force is due to the induced charge on the conducting surface which is at zero potential. 19 Assertion : A metallic shield in the form of a hollow shell, can be built to block an ( c) electric field. Reason : In a hollow spherical shell, the electric field inside is not zero at every point. 20 Assertion : Work done in moving a charge between any two points in an electric field is (d) dependent of the path followed by the charge, between them Reason : Electrostatic forces are non conservative in nature. 21 Assertion : Force between two charges decreases when air separating the charges is (a) replaced by water. Reason : Medium intervening between the charges has dielectric constant K >1. 22 Assertion : The whole charge of a body can be transferred to another body (c) Reason :Charge cannot be transferred partially 23 Assertion : The number of electric lines of force emanating from 1µC charge in vacuum (a) is 1.13 x 106 Reason : This follows from Gauss’s theorem in electrostatics 24 Assertion : In a series combination of capacitors, charge on each capacitor is same. (c) Reason : In such a combination, charge cannot move only along one route, 25 Assertion : When the battery across the plates of the charged condenser is off and (a) dielectric slab is introduced between its plates, the energy stored in the capacitor 24  decreases. Reason : The charge stored in capacitor Q remains constant and its capacity increases. ASSIGNMENT- 2 (Descriptive questions) Q.No Question Mark 1 How does the electric flux, electric field enclosing a given charge vary when the 1 area enclosed by the charge is doubled? 2 Name the physical quantities whose SI units are Vm, Vm-1. Which of these are 1 vectors? 3 How much work is done in moving a 500 µC charge between two points 1 separated by a distance of 2cm on an equiotential surface? 4 Two capacitors of 0.1 µF and 0.2 µF are raised to the same potential of 1 50V.Calculate the ratio of the energy stored in each. 5 Consider three charged bodies P, Q and R. If P and Q repel each other and P 1 attracts R, what will be the nature of the force between Q and R? 6 What is the work done in moving a test charge q through a distance of 1 cm along 1 the equatorial axis of an electric dipole? 7 Two capacitors of capacitances 6μF and 12 μF are connected in series with a 1 battery. The voltage across 6μF capacitor is 2 V. Compute the total battery voltage. 8 A hollow metal sphere of radius 5 cm is charged such that the potential on its 1 surface is 10 V. What is the potential at the centre of the sphere? 9 Two equal balls having equal positive charge ‘q’ coulumbs are suspended by two 1 insulating strings of equal length. What would be the effect on the force when a plastic sheet is inserted between the two ? 10 The given graph shows variation of charge ‘q’ versus potential difference ‘V’ for - two capacitors C1 and C2. Both the capacitors have same plate seperation but plate 2 area of C2 is greater than that of C1. Which line (A or B) corresponds to C1 and why? 11 two identical metallic spheres A and B of exactly equal masses are taken. Sphere 2 A is given positive charge of Q coulomb and B is given an equal negative charge. So initially before the charge is given MA=MB=M 25  12 Electric field inside a dielectric decreases when it is placed in an external field. Give reason to support this statement. 2 13 An electric dipole of moment p is aligned parallel to the external electric field. How much work has to be done in rotating the dipole through (a) 900 (b) 1800 2 14 Derive the expression for the electric field intensity dur to an infinitely long straight charged wire. 2 15 Derive the expression for the electric field intensity due to a thin infinite plane 3 sheet of charge, 16 When two charged capacitors having different capacities and different potentials 3 are joined together, show that there is always some loss of energy. MARKING KEY Q.no Marks 1 ∅=constant 1/2 E is halved ½ (1) mark 2 Electric flux ∅-scalar ½ Electric field intensity E-Vector ½ (1) mark 3 Zero (1) mark 4 U1/U2 = 1/2C1V2/1/2C2V2 – ½ = 2----1/2 (1) mark 5 Q attracts R---(1) mark 6 Since potential for equatorial axis V = 0 1/2 ∴ W = qV = 0 -- ½ (1) mark 7 Charge on both capacitors are same 6x2 =12x V2 ---1/2 V2= 1V, battery voltage=3V ½ (1) mark 8 Inside the sphere E =0----1/2 V= constant = 10V—1/2 (1) mark 26 9 force would be reduced by a factor ‘K’ (equal to the value of dielectric constant of plastic sheet)—1/2 --1/2 (1) mark 10 line B corresponds to C1--1 slope (q/v) of ‘B’ is less than that of ‘A’ ---1 (2) marks 11 The process of giving positive charge involves removal of electrons and that of negative charge involves addition of electrons.---1 Hence the mass of the positively charged sphere will be less than that of negatively charged sphereMAR1) and a perfect ammeter are given. The current in the circuit is measured in five different situations: (i) Without any external resistance in the circuit, (ii) With resistance R1 only, (iii) With resistance R2 only, (iv) With both R1 and R2 used in series combination and (v) With R1 and R2 used in parallel combination. The current obtained in the five cases are 0.42A, 0.6A, 1.05A, 1.4A, and 4.2A, but not necessarily in that order. Identify the currents in the five cases listed above and calculate E, r,, R1 and R2. 29 10. A set of n-identical resistors, each of resistance R ohm when connected in series have an effective resistance of Xohm and when the resistors are connected in parallel the effective resistance is Y ohm. Find the relation between R, XandY? ANSWERS TO THE NUMERICALS Q NO SOLUTIONS 1 I =dq/dtor dq=Idt dq=(3t2+2t-1)dt 2 𝑞 = ∫ (3𝑡 2 + 2𝑡 − 𝑖) 𝑑𝑡 = 10 𝐶 0 2 (a) The. The drift speed V is given by V = (I / neA),e = 1.6 × 10–19 C, A = 1.0 × 10–7m2 , I = 1.5 A. The density of conduction electrons, n is equal to the number of atoms per cubic metre (conduction electron per Cu atom) A cubic assuming onemetre of copper has a mass of 9.0 × 103 kg. Since 6.023 × 1023 copper atoms have a mass of 63.5 =8.5×1028m–3 Vd =1.1×10–3ms–1 3 (i) Whencurrentflowsalongthesidel R=𝜌 l/ ah (ii) Whencurrentflowsalongthesideh R=𝜌 h / al (iii) Whencurrentflowsalongtheside a (iv) R = 𝜌 a / hl 30 4 5 6 31  E2 =1.02V, PQ=1m. When switch S open, null position is obtained at a distance of 51 cm from P. problem 7 (i) Potential gradient k = = 0.02V/cm (ii) emf of the cell E1 = k = 2V 32  (iii)When switch S is closed, null point is not affected because no current drawn from cell E1 at the null point. 8 (i)The emf E of a cell is independent of externalresistance (R). E E V = IR = R= r+R r 1+ The terminal p.d. R On increasing R, V increases. E When R = 0, V → 0. When R = r, V = 2 When R →  , V = E. 9 E E E E (i) I 1 = , (ii) I 2 = , (iii) I 3 = , (iv) I 4 = , r r + R1 r + R2 r + R1 + R2 (v) I = E 5 R1 R2 r+ R1 + R2 This is clear that I1  I 5  I 2  I 3  I 4. Hence I1 = 4.2 A, I 5 = 1.4 A, I 2 = 1.05 A, I 3 = 0.6 A, I 4 = 0.42 A. Putting these values in (i) to (v) and on solving, E = 4.2V , R1 = 3, R2 = 6, r = 1 10 n – resistors connected in seriesX= nR—–1) n– Resistorsconnectedinparallel Y= ——2) 33  Multiplyeg.(1)&(2) XY = NUMERICALS FOR PRACTICE : Q.1 Two conducting wires X and Y of same diameter but different materials are joined in series across a battery. If the number density of electrons in X is twice that in Y, find the ratio of drift velocity of electrons in the two wires. Q 2 A wire of resistance 8R is bent in the form of a circle. What is the effective resistance between the ends of the diameter? Q 3A battery of emf 10 V and internal resistance 3Ω is connected to a resistor. If the current in the circuit is 0.5 A, find (i) the resistance of the resistor; (ii) the terminal voltage of the battery. Q 4 Find thevaluesofI1,I2& I3inthegivennetwork. Q 5Two cells of emfs 1.5 V and 2.0 V having internal resistance 0.2 Ω and 0.3 Ω respectively are connected in parallel. Calculate the emf and internal resistance of the equivalent cell. Q 6Find the Potential Difference ( P.D.) across each cell and the rate of energy dissipation in resistor R in the given network. 34 Q 7. A uniform wire of resistance 12 Ω is cut into three pieces so that the ratio of the resistances R1 : R2 : R3 = 1: 2 : 3 and the three pieces are connected to form a triangle across which a cell of emf 8V and internal resistance1Ωis connectedasshown. Calculatethecurrentthrougheachpartofthecircuit Q 8 Two wires, one of copper and the other of manganin, have same resistance and equal thickness. Which wire is longer? Justify your answer Q.9 Nichrome and copper wires of same length and same radius are connected in series. Current I is passed through them. Which wire gets heated up more? Justify your answer Q 10 Plot a graph showing variation of voltage Vs the current drawn from the cell. How can one get information from this plot about the emf of the cell and its internal resistance? Q 11 A 16 Ω resistance wire is bent to form a square. A source of emf 9 V is connected across s one of its sides as shown. Calculate the current drawn from the source. Find the potential difference between the ends C and D. If now the wire is stretched uniformly to double the length and once again the same cell is connected in the same way, across one side of the square formed, what will now be the potential difference across one of its diagonals. 35 Q 12 : Consider two cells which is connected in series. The positive terminal of one cell is connected to negative terminal of the next cell. Here one terminal of two cells are free and the other terminal of two cells are joined together. ε1 and ε2 are the emfs of the cells and r1 and r2 are the internal resistance of the cells respectively. Let I be the current flowing through the cells Q 13 you given three constantan wires P, Q and R of length and area of cross-section respectively. Which has highest resistance? Q 14 Calculate the equivalent resistance between points A and B in the figure given below.Q 15 In a meter bridge, the balance point is found to be 39.5 cm from end A. The known resistance Y is. Determine unknown resistance HINT FOR NUMERICALS PROBLE SOLUTION M NO 1. 2. 3. Answer: Resistance of the resistor is 17Ω and the terminal voltage is 8.5V 36 4. 5. E1 = 1.5 V, E2 = 2.0 V, 37 6. 7. Current through arms AC = current through ABC. (Since the resistance of arm AC = resistance of arm ABC) = 1A 8. Copper Reason: Let l1 and l2 be lengths of copper and manganin wires having same resistance R and thickness i.e., area of cross-section (A). 38 9. 10. 11. 12. Cells connected in series Consider the points A, B and C and let V (A), V (B) and V (C) be the potentials of these points respectively. V (A) - V (B) will be the potential difference between the positive and negative terminals for the first cell. So VAB = V (A) - V (B) = ε1- Ir1. VBC = V (B) - V (C) = ε2 – Ir2. 39  Now the potential difference between the terminals A and C is VAC = V (A) – V(C) = [V (A) - V (B)] + V (B) - V (C)] = ε1- Ir1 + ε2 – Ir2 = ( ε1 + ε2) – I(r1+r2). In case if we replace this combination of cells by a single cell between the points A and C with emf εeqand internal resistance req, VAC = εeq- req. and thus we found out that εeq= ε1 + ε2 and req = r1+r2 from the previous equation. It is clear that the equivalent emf of n number of cells in series combination is the sum of their individual emfs. The equivalent internal resistance of n cells in series combination is the sum of their individual internal resistance. In series combination if the current leaves the cell from the negative electrode, the emf of the cell will be for example VBC = - ε2 – Ir2 and finally the equation for εeq = ε1 - ε2, (ε1 > ε2) 13. Q has the highest resistance. 14. 15. 8.16 OHM 40 CASE BASED QUESTIONS Q 1 Read the source given below and answer any four out of the following questions: The Wheatstone bridge works on the principle of null deflection, i.e. the ratio of their resistances are equal and no current flows through the circuit. And The principle of working of Meter bridge is wheat stone bridge principle and it is used to find the resistance of an unknown conductor or to compare two unknown resistance. i. When a metal conductor connected to the left gap of a meter bridge is heated, the balancing point shifts a. right b. left c. unchanged d. none of these ii. Wheatstone bridge is a/an: a. a.c. bridge b. d.c. bridge c. high voltage bridge d. none of these iii. Wheatstone bridge is used to measure the d.c. the resistance of various types of wires for: a. determining their effective resistance b. computing the power dissipation c. quality control of wire d. none of these iv. By using the variations on a Wheatstone bridge we can: a. measure quantities such as voltage, current, and power b. measure high resistance values c. measure quantities such as complex power d. measure quantities such as capacitance, inductance and impedance v. For a Wheatstone bridge arrangement of four resistances – R1, R2, R3, R4 (Junction of R1 and R2 is connected to anode and Junction of R3 and R4 to the cathode of the cell). The null-point condition is given by a. R1R3=R2R4R1R3=R2R4 b. (R1×R3)=(R2×R4)(R1×R3)=(R2×R4) c. (R1 + R3) = (R2 + R4) d. (R1 - R3) = (R2 - R4) 2 Read the following source and answer any four out of the following questions: Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms. 41 Also Resistivity, the electrical resistance of a conductor of unit cross- sectional area, and unit length. A characteristic property of each material, resistivity is useful in comparing various materials on the basis of their ability to conduct electric currents. I. Resistivity is independent of: A nature of material B temperature C dimensions of material D none of the above IIAs compare to short wires, long wires have resistance. A more B less C same D zero IIIAs compare to thin wires, thick wires have resistance. A more B less C same D zero IV The resistance of a wire depends upon: A cross-sectional area B length of wire C wire's nature D all of the above V A copper wire having the same size as steel wire have: A more resistance B less resistance C same resistance D none of the above 3Read the source given below and answer any four out of the following questions: The Wheatstone bridge works on the principle of null deflection, i.e. the ratio of theirresistancesareequalandnocurrentflowsthroughthecircuit.Undernormalconditions,the bridge is in the unbalanced condition where current flows through the galvanometer. Thebridgeissaidtobeinabalancedconditionwhennocurrentflowsthroughthegalvanometer. Thisconditioncanbeachievedbyadjustingtheknownresistanceandvariableresistance. 42  I The Wheatstone bridge is an arrangement of four resistances – R1, R2, R3, R4. The null- point condition is given by: R1 R3 a. = R2 R4 b. R1+R2=R3+R4 R2 R3 c. R2= 𝑅4+𝑅1 𝑅2 𝑅3 d. = 𝑅4 𝑅1 II. TheWheatstonebridgeisusedfortheprecisemeasurementof. A highresistance B lowresistance C lowcurrent D highcurrent III Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips? A Minimize the resistance B Maximize the resistance C Minimize current D None of these IV. What happens if the galvanometer and cell are interchanged at the balance point of the bridge? a. Current flow b. Show deflection c. Nodeflection d. Lowresistance V In a metre bridge [Fig. below], the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 Ω. Determine the resistance of X. 43  8.2 Ω 8.4 Ω 7.2 Ω 8.6 Ω 4. Read the source given below and answer any four out of the following questions: The rate of flow of charge through any cross-section of a wire is called electric current flowingthrough it. Electric current (I) =q/t. Its SI unit is ampere (A). The conventional direction of t electric current is the direction of motion of positive charge. The current is the same for all cross-sections of a conductor of the non-uniform cross-section. Resistance is a measure of the opposition to current flow in an electrical circuit I An example of non-ohmic resistance is: A tungsten wire B carbon resistance C diode D copper wire II Current is: A scalar quantity B vector quantity C both scalar and vector quantity D none of the above III In a current-carrying conductor, the net charge is: A 1.6 × 10–19 coulomb B. 6.25 × 10–18 coulomb C. zero D. infinite IV The current which is assumed to be flowing in a circuit from the positive terminal to negative is called: 44 A direct current B pulsating current C conventional current D none of these V A current passes through a wire of non-uniform cross-section. Which of the following quantities are independent of the cross-section? The charge crossing drift velocity current density free electron density 5. When a potential difference V is applied across the two ends of a conductor, the free electrons in the conductor experience a force and are accelerated towards the positive end of conductor. On their way, they suffer frequent collisions with the ions/atoms of the conductor and lose their gained kinetic energy and again get accelerated due to electric field and lose the gained kinetic energy in the next collision and so on. The average velocity with which the free electrons get drifted towards the positive end of the conductor under the effect of applied electric field is called drift velocity. i) The motion of electrons between two successive collisions (with the atoms/ions) in the presence of electric field follows: (a) Straight line path (b) Circular path (c) Elliptical path (d) Curved path ii) The drift velocity of the electrons depends on (a) Dimensions of the conductor (b) Number density of free electrons in the conductor (c) Both a and b (d) None of these. iii) When potential difference across a given copper wire is increased, drift velocity of free electrons (a) Decreases b) Increases (c) Remain same (d) Get reduced to zero 45  iv) Two wires of same material having radii in the ratio 1:2, carry currents in the ratio 4:1. The ratio of drift velocities of electrons in them is (a) 2:1 (b) 1:1 (c) 1:4 (d) 16:1 v) If the temperature of a conductor increases, the drift velocity of free electrons (a) Remains same (b) Increases (c) Decreases (d) May increase or decrease. 6.. Electromotive Force You can think of many different types of voltage sources. Batteries themselves come in many varieties. There are many types of mechanical/electrical generators, driven by many different energy sources, ranging from nuclear to wind. Solar cells create voltages directly from light, while thermoelectric devices create voltage from temperature differences. All such devices create a potential difference and can supply current if connected to a resistance. On the small scale, the potential difference creates an electric field that exerts force on charges, causing current. We thus use the name electromotive force, abbreviated emf. Emf is not a force at all; it is a special type of potential difference. Electromotive force is directly related to the source of potential difference, such as the particular combination of chemicals in a battery. 1) Emf of a cell is (a) the maximum potential difference between the terminals of a cell when no current is drawn from the cell. (b) the force required to push the electrons in the circuit. (c) the potential difference between the positive and negative terminal of a cell in a closed circuit. (d) less than terminal potential difference of the cell. 2) A cell of internal resistance r is connected to an external resistance R. The current will be maximum in R, if (a) R = r (b) R < r (c) R > r (d) R = r/2 3) A cell of internal resistance r is connected across an external resistance nr. Then the ratio of the terminal voltage to the emf of the cell is (a)1/(n ) (b) 1/(n + 1) (c) n/(n + 1) (d) n/(n - 1) 4) To draw a maximum current from a combination of cells, how should the cells be grouped? 46  (a) Parallel (b) Series (c) Mixed grouping (d) Depends upon the relative values of internal and external resistances. 5) A battery of 6V and internal resistance 2 Ω is connected to a silver voltameter. If the current of 1.5A flows through the circuit, the resistance of the voltameter is (a)4Ω (b) 2Ω (c) 6Ω (d) 1Ω Solutions I II III IV V 1 a b c d a 2 c a b d A 3 a b a C A 4 c a c c D 5 d d b d C 6 a a c d b Assignment -1 on MCQ questions with answers 1. Consider a current carrying wire current I in the shape of a circle. Note that as the current progresses along the wire, the direction of j (current density) changes in an exact manner, while die current/remain unaffected. The agent that is essentially responsible for is] (a) source of emf. (b) electric field produced by charges accumulated on the surface of wire. (c) the charges just behind a given segment of wire which push them just the right way by repulsion. (d) the charges ahead. 2. Which of the following is wrong? Resistivity of a conductor is (a) independent of temperature. (b) inversely proportional to temperature. (c) independent of dimensions of conductor. (d) less than resistivity of a semiconductor. 3.Drift velocity vd varies with the intensity of electric field as per the relation 47 (a) vd∝ E (b) vd∝ 1E (c) vd = constant (d) vd∝ E² 4.When there is an electric current through a conducting wire along its length, then an electric field must exist (a) outside the wire but normal to it. (b) outside the wire but parallel to it. (c) inside the wire but parallel to it. (d) inside the wire but normal to it. 5.From the graph between current I and voltage V shown below, identify the portion corresponding to negative resistance (a)AB (b)BC (c)CD (d) DE 6. A battery consists of a variable number V of identical cells having internal resistances connected in series. The terminals of battery are short circuited and the current i is measured. Which of the graph below shows the relationship between i and n? 7.Acharge is moving across a junction, then (a) momentum will be conserved. (b) momentum will not be conserved. (c) at some places momentum will be conserved and at some other places momentum will not be conserved. (d) none of these. 8.The I-V characteristics shown in figure represents 48 (a)ohmic conductors (b) non-ohmic conductors (c) insulators (d) superconductors 9. The resistivity of alloy manganin is (a) Nearly independent of temperature (b) Increases rapidly with increase in temperature (c) Decreases with increase in temperature (d) Increases rapidly with decrease in temperature 10.In the series combination of two or more than two resistances (a) the current through each resistance is same. (b) the voltage through each resistance is same. (c) neither current nor voltage through each re-sistance is same. (d) both current and voltage through each resis¬tance are same. 11.Combine three resistors 5 Q, 4.5 Q and 3 Q in such a way that the total resistance of this combination is maximum (a) 12.5 Q (b) 13.5 Q (c) 14.5 Q (d) 16.5 Q 12.If n cells each of emf e and internal resistance r are connected in parallel, then the total emf and internal resistance will be 13.In a Wheatstone bridge if the battery and galvanometer are interchanged then the deflection in galvanometer will (a) change in previous direction (b) not change (c) change in opposite direction (d) none of these. 14. When a metal conductor connected to left gap of a meter bridge is heated, the balancing point (a) shifts towards right 49  (b) shifts towards left (c) remains unchanged (d) remains at zero 15. Consider a current carrying wire (current I) in the shape of a circle. Note that as the current progresses along the wire, the direction of j (current density) changes in an exact manner, while the current I remain unaffected. The agent that is essentially responsible for is (a) source of emf. (b) electric field produced by charges accumulated on the surface of wire. (c) the charges just behind a given segment of wire which push them just the right way by repulsion. (d) the charges ahead. 16. The drift velocity of the free electrons in a conducting wire carrying a current i is v. If in a wire of the same metal, but of double the radius, the current be 2I, then the drift velocity of the electrons will be (a) v/4 (b) v/2 (c) v (d) 4 v 17. Temperature dependence of resistivity ρ(T) of semiconductors insulators and metals is significantly based on the following factors. (a) Number of charge carriers can change with temperature T. (b) Time interval between two successive collisions does not depend on T. (c) Length of material can be a function of T. (d) Mass of carriers is a function of T. 18. Kirchhoff’s junction rule is a reflection of (a) Conservation of current density vector. (b) Conservation of energy. (c) The fact that the momentum with which a charged particle approaches a junction is unchanged (as a vector) as the charged particle leaves the junction. (d) The fact that there is no accumulation of charged at a junction. 19. Figure represents a part of a closed circuit. The potential difference between points A and B (VA – VB) is a. +9v b. -9v c. +3v d. +6v 20.Two resistors of resistance R1 and R2 having R1 > R2 are connected in parallel. For equivalent resistance R, the correct statement is: (a) R > R1 + R2 (b) R1 < R1 < R2 (c) R2 < R1 < (R1 + R2) (d) R < R2 < R1 21.The resistivity of iron is 1 ×10–7 ohm-meter. The resistance of the given wire of a particular thickness and length is 1 ohm. If the diameter and length of the wire both are doubled the resistivity will be (in ohm-meter) (a) 1 ×10–7 (b) 2 ×10–7(c) 4 ×10–7(d) 8 ×10–7 50 22. If charges move without collisions through the conductor , their kinetic energy would also change so that the total energy is a. changed b. unchanged c. doubled d. halved 23.Consider a current carrying wire (current I ) in the shape of a circle. Note that as the current progresses along the wire, the direction of j (current density) changes in an exact manner, while the current I remain unaffected. The agent that is essentially responsible for is a. source of emf. b. electric field produced by charges accumulated on the surface of wire. c. the charges just behind a given segment of wire which push them just the right way by repulsion. d. the charges ahead. 24 The current in a wire varies with time according to the equation I= 4 +2t where I is in ampere and t is in seconds. The quantity of charge which passes through a cross section of the wire a. 40 C b. 48 C c. 38 C d. 43 C 25. Which of the following characteristics of electrons determines the current in a conductor? a. drift velocity alone b. thermal velocity alone c. both drift velocity and thermal velocity d. Neither drift velocity nor thermal velocity MCQ No Answers 1. Explaination: (b) Current density j changes due to electric field produced by charges accumulated on the surface of wire. 2. (a) Resistivity is property of material and inversely proportional to temperature for conductor, ρ=mne2τ. 3. Answer: a Explaination: 51  (a) Drift velocity vd = eEmτ, i.e. vd∝ E 4. Explaination: (c) Electric field parallel to wire inside creates potential difference and electrostic force on electrons 5. Answer: c Explaination: (c) For portion CD slope of the curve is negative i.e. resistance be negative. 6. Answer: d Explaination: (d) I = nEnr=Er. current is independent of n. 7. d none of these. 8. b non-ohmic conductors 9. a Nearly independent of temperature 10. a the current through each resistance is same. 11. a 12.5 Q 12. A Er r/n 13. b not change 14. a shifts towards right 15. (b) electric field produced by charges accumulated on the surface of wire. 16. (b) v/2 17. (a) Number of charge carriers can change with temperature T. 18. (d) the fact that there is no accumulation of charged at a junction. 19. (a)+9V 20. Answer d R < R2 < R1 21. Answer a1 ×10–7 52  22. Answer b unchanged 23. b. electric field produced by charges accumulated on the surface of wire. 24. Answer b. 48 C 25. Answer: a drift velocity alone Solution for Assignment - 1 Assertion and reasoning Read the assertion and reason carefully to mark the correct option out of the options given below: (a)If both assertion and reason are true and the reason is the correct explanation of the assertion. (b)If both assertion and reason are true but reason is not the correct explanation of the assertion. (c)If assertion is true but reason is false. (d)If the assertion and reason both are false. (e)If assertion is false but reason is true. Q 1Assertion: A current flow in a conductor only when there is a electric field within the conductor Reason: The drift velocity of electrons in the presence of electric field decreases Q 2Assertion:In a simple battery circuit the point of lowest potential is positive terminal of the battery Reason: The current flows towards the point of the higher potential as it flows in such circuit from the negative to the positive terminal. Q 3Assertion: The electric bulb glows immediately when the switch is on. Reason: The drift velocity of electrons in a metallic wire is very high Q 4 Assertion: Resistivity of a conductor increases with increase in temperature. Reason: When temperature increases the random motion of free electrons increases and vibration of ions increases which decreases the relaxation time. Q 5 Assertion: There is no current in the metals in the absence of electric field. Reason: Motion of free electron are randomly. Q 6 Assertion: Electric appliances with metallic body have three connections, whereas an electric bulb has a two-pin connection. Reason: Three pin connections reduce heating of connecting wires. 53 Q 7 Assertion: In meter bridge experiment, a high resistance is always connected in series with a galvanometer. Reason: As resistance increases current through the circuit increases. Q 8 Assertion: A person touching a high-power line gets stuck with the line. Reason: The current carrying wires attract the man towards it. Q 9 Assertion: Electric field outside the conducting wire which carries a constant current is zero. Reason: Net charge on conducting wire is zero. Q 10 Assertion: Potential difference across the terminals of a cell is always less than its emf. Reason: Emf of a cell is the maximum potential difference across the terminals of a cell in an open circuit. Q 11 Assertion: A potentiometer of longer length is used for accurate measurement. Reason: The potential gradient for a potentiometer of longer length with a given source of e.m.f. becomes small Q 12 Assertion: The e.m.f. of the driver cell in the potentiometer experiment should be greater than the e.m.f. of the cell to be determined. Reason: The fall of potential across the potentiometer wire should not be less than the e.m.f. of the cell to be determined. Q 13 Assertion: In meter bridge experiment, a high resistance is always connected in series with a galvanometer. Reason: As resistance increases current through the circuit increases. Q 14 Assertion: In a meter bridge experiment, null point for an unknown resistance is measured. Now, the unknown resistance is put inside an enclosure maintained at a higher temperature. The null point can be obtained at the same point as before by decreasing the value of the standard resistance. Reason: Resistance of a metal increases with increase in temperature. Q 15 Assertion: The e.m.f of the driver cell in the potentiometer experiment should be greater that the e.m.f of the cell to be determined. Reason: The fall of potential across the potentiometer wire should not be less than the e.m.f of the cell to be determined. 54 Q 16 Assertion: In a simple battery circuit the point of lowest potential is positive terminal of the battery Reason: The current flows towards the point of the higher potential as it flows in such a circuit from the negative to the positive terminal. Q 17 Assertion: Electric field outside the conducting wire which carries a constant current is zero. Reason: Net charge on conducting wire is zero. Q 18 Assertion: The connecting wires are made of copper. Reason: The electrical conductivity of copper is high. Q 19 Assertion: A potentiometer of longer length is used for accurate measurement. Reason: The potential gradient for a potentiometer of longer length with a given source of e.m.f. becomes small. Q 20Assertion: The e.m.f. of the driver cell in the potentiometer experiment should be greater than the e.m.f. of the cell to be determined. Reason: The fall of potential across the potentiometer wire should not be less than the e.m.f. of the cell to be determined. Answers for assertion and reasoning Q no answers 1. (c) drift velocity is directly proportional to electric field 2. (d) It is quite clear that in a battery circuit, the point of lowest potential is the negative terminal of the battery and the current flows from higher potential to lower potential. 3. Ans. (c) 4. Ans.(a) 5. (a) It is clear that electrons move in all directions haphazardly in metals. When an electric field is applied, each free electron acquires a drift velocity. There is a net flow of charge, which constitute current. In the absence of electric field this is impossible and hence, there is no current. 55 6. (c) The metallic body of the electrical appliances is connected to the third pin which is connected to the earth. This is a safety precaution and avoids eventual electric shock. By doing this the extra charge flowing through the metallic body is passed to earth and avoid shocks. There is nothing such as reducing of the heating of connecting wires by three pin connections. 7. (c) The resistance of the galvanometer is fixed. In meter bridge experiments, to protect the galvanometer from a high current, high resistance is connected to the galvanometer in order to protect it from damage. 8. d) Because there is no special attractive force that keeps a person stuck with a high-power line. The actual reason is that a current of the order of 0.05 A or even less is enough to bring disorder in our nervous system. As a result of it, the affected person may lose temporarily his ability to exercise his nervous control to get himself free from the high-power line 9. (a) When current flows through a conductor it always remains unchanged, hence no electric field is produced outside it. 10. (a) 11. Sensitivity is inversely proportional to potential gradient which directly proportional to length of wire 12. (a) If either the e.m.f. of the driver cell or potential difference across the whole potentiometer wire is lesser than the e.m.f. of the experimental cell, then balance point will not be obtained. 13. (c) The resistance of the galvanometer is fixed. In meter bride experiments, to protect the galvanometer from a high current, high resistance is connected to the galvanometer. 14. (d) With increase in temperature, resistance of metal wire increases, but balance conduction will not change. 15. (a) If either e.m.f. of the driver cell or potential difference across the whole potentiometer wire is lesser than the e.m.f. of then experimental cell, then balance point will not be obtained. 16. Ans. (d) It is quite clear that in a battery circuit, the point of lowest potential is the negative terminal of the battery and the current flows from higher potential to lower potential. 56  17. Ans. (a) When current flows through a conductor it always remains unchanged, hence no electric field is produced outside it. 18. (a) Due to high electrical conductivity of copper, it conducts the current without offering much resistance. The copper being diamagnetic material does not get magnetised due to current through it and hence does not disturb the current in the circuit. 19. (a) Sensitivity∝1/(Potential gradient)∝Length of wire 20. (a) If either the e.m.f. of the driver cell or potential difference across the whole potentiometer wire is lesser than the e.m.f. of the experimental cell, then balance point will not be obtained. Assignment -2 ( 2 marks questions) 1. Using data given in graph determine (i) emf (ii) internal resistance of the cell. (iii) For what current, does maximum power dissipation occur in the circuit? 2. Draw V-I graph for ohmic and non- ohmic materials. Give one example for each. 3.Two primary cells of emf E1 and E2 (E1 > E2) are connected to the potentiometer wire as shown in the figure. If the balancing lengths for the cells are 250 cm and 400 cm. Find the ratio of E1 and E2 4. How does the resistivity of (i) a conductor and (ii) a semiconductor vary with temperature? Give reasons. 5. Explain How does the resistivity of a conductor depend upon (i) number density ‘n’ of free e’s. (ii) Relaxation time ‘ ’. 6. Out of the two bulbs marked 25W and 100W, which one has higher resistance. 7. Explain how electron mobility changes for a good conductor when (i) the temperature of the conductor is decreased at constant potential difference and (ii) applied potential difference is doubled at constant temperature. 8.A cylindrical metallic wire is stretched to increase its length by 5%. Calculate the percentage change in its resistance. 9.Explain how the average velocity of free electrons in a metal at constant temperature, in an electric field, remain constant even though the electrons are being constantly accelerated by this electric field? 57 10.Two metallic wires of the same material have the same length but cross-sectional area is in the ratio of 1:2. They are connected (i) in series and (ii) in parallel. Compare the drift velocities of electrons in the two wires in both the cases. Q NO ANSWERS 1. (i) Emf = 1.4V E −V (ii) Internal resistance of the cell r = = 5 I E (iii)For maximum power dissipation I = r+R =.14A 2. : Ohmic materials –metallic conductors for small current. 3. : E1 - E2 = 250 ф E1 + E2 = 400 ф E1 : E2 = 13 :3 4. Ans: ρ = ρα and ρ α (i) For a conductor, the density of free e’s is almost independent of temperature but the frequency of collision of e’s increases with increase in temperature. Therefore the relaxation time decreases. Hence the resistivity of a conductor increases with increase in temperature (conductivity decreases). 58  (ii) On increasing the temperature of a semiconductor, the density of free e’s increases and the relaxation time decreases. But the increase in ‘n’ is large than the decrease in ‘τ’. Hence the resistivity of a semiconductor decreases with increase in temperature (conductivity increases). 5. ρ= ρα and ρ α 6. Ans: R = Rα. The bulb marked 25W has higher resistance than the bulb marked 100W. 7. Mobility μ = = τ (i) When the temperature of the conductor is decreased at constant potential difference, the relaxation time ‘τ’ of free e’s decreases. Hence ‘μ’ decreases. (ii) Mobility is independent of applied potential difference. 8. 105 ' Al = A ' l '  A = A 100 R1 lA ' = '  R2 = (1.05) R1 l R=  2 A R2 l A R2 − R1 % Change = 100 = 10.25% R1 9. The electron keeps colliding with the positive metal ions and other electrons. The velocity gained by it between two successive collisions due to electric field is lost in next collision. 10. I vd = (i) In series, current in both wires is same. Drift velocity neA , =eV= 1 2 v𝑣𝑑1 d = 𝐴2 (ii) In parallel, p.d. across the both wires is same. Drift velocity ml 𝑣𝑑2 𝐴1 vd 1 l1 1 = = vd 2 l 2 1. Assignment -2 ( 3 marks questions) 1. Define the term resistivity and write its SI unit. Derive the expression for the resistivity of a conductor in terms of number density of free electrons and relaxation time. 59  2.Define the terms resistivity and conductivity and state their S.I. units. Draw a graph showing the variation of resistivity with temperature for a typical semiconductor. 3.Define the temperature coefficient of resistivity. Write its S.I. unit. Plot a graph showing the variation of resistivity of nichrome / copper with temperature. 4.What is meant by drift velocity of free e’ s. Derive ohm’s law on the basis of the theory of electron drift. 5.Draw Circuit diagram for a meter bridge to determine the unknown resistance of a resistor. Obtain the balance condition for a meter bridge. Find the shift in the balance point for a meter bridge when two resistors in its two gaps, are interchanged. 6.Are the paths of e’s straight lines between successive collisions in the (i) absence of electric field (ii) presence of electric field.Establish a relation between drift velocity and current. Hence obtain the relation between current density and drift velocity. 7..In a meter bridge the balance point is found to be 39.5 cm from one end A, when the resistor Y is of 12.5 . Determine the resistance of X. Why are the connections between resistors in a meter bridge made of thick copper strips? What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current? 8. E2 =1.02V, PQ=1m. When switch S open, null position is obtained at a distance of 51 cm from P. Calculate (i) potential gradient (ii) emf of the cell E1 (iii) when switch S is closed, will null point move towards P or Q. Give reason for your answer 9..In the given network find the values of currents I1 , I2 and I3. 10.(i) Find the p.d. between the ends A and B. (ii) Would the method work, if the battery E1 is replaced by a cell of emf of 1V. 60 11. AB=100 cm, RAB=10. Find the balancing length AC. 12. The given figure shows the experimental set up of a meter bridge. The null point is found to be 60cm away from the end A with X and Y in position as shown. When a resistance of 15Ω is connected in series with ‘Y’, the null point is found to shift by 10cm towards the end A of the wire. Find the position of null point if a resistance of 30Ω were connected in parallel with ‘Y’. 13 A cell of emf I and internal resistance I is connected across a variable external resistance I. Plot graphs to show variation of (i) E with R, (ii) Terminal p.d. of the cell (V) with R. Q Answers No 1 The resistivity of a material is equal to the resistance of the unit cube of thatmaterial.SI unit – ohm metre I = neAvd ---------- (1) Vd = eE τ = eV τ ---------(2) m ml ne A 2 V m l Putting the value of vdin (1), I = V R= = 2 -------- (3) ml I ne  A l m Resistance R = ρ ------ (4), comparing (3) & (4), ρ = 2 A ne  2 The reciprocal of resistivity of a material is called its conductivity.S.I. unit – mho/ metre 61 3 The temperature coefficient of resistivity α of a material is defined as the fractional increase in its resistivity per unit increase in temperature. S.I. unit – per kelvin. 4 Drift velocity – The average velocity with which free electrons of a conductor get drifted in a eE eV direction opposite to the direction of applied electric field. v d = τ = τ m ml ne 2 A Current I = neAvd = V m l V = 2 ml I ne  A At a constant temperature, RHS of above equation is constant for a given conductor. V m l  = 2 = Constant I  V  I I ne  A 5 : S is an unknown resistance whose value to be determine. It is connected across one of the gaps. Across the other gap, a resistance box is connected. The four arms AB, BC, AD and DC [with resistances R, S, P and Q] form a Wheatstone bridge with AC as the battery arm and BD the galvanometer arm. After taking out a suitable resistance R from resistance box, the jockey is moved along the wire AC till there is no deflection in galvanometer. This is the balanced condition of Wheatstone bridge. For balanced condition of Wheatstone bridge, P = R -----(1) Q S Let l be the balancing length. Then AD = l and DC = 100 –l P = resistance of arm AD = r l and Q = resistance of arm DC = r (100- l) R l From eq. (1), = S 100 − l Let ‘l’ be the balancing length for a meter bridge from end ‘A’. When two resistors in its two gaps, are interchanged, the balancing length becomes (100-l ). 62 6 (i) In the absence of electric field, the paths of e’s are straight lines between successive collisions. (ii) In the presence of electric field, every electron experiences a force in a direction opposite to the direction of electric field. Therefore, the paths of e’s are curved. 7 lY X= = 8.16 Ω 100 − l Thick copper stripes offer minimum resistance. Therefore the connections between resistors in a meter bridge are made of thick copper strips to minimize the resistances of connections which are not accounted in bridge formula. When the galvanometer and cell are interchanged at the balance point of the bridge, the condition for balanced bridge remains satisfied. The galvanometer will not show any current. 8 E2 Potential gradient k = = 0.02V/cm l2 (ii) emf of the cell E1 = k l PQ = 2V (iii)When switch S is closed, null point is not affected because no current drawn from cell E1 at the null point. 9 : At junction D, I1 = I 2 + I 3 ------------ (1) In loop DCBD, − 3 + 3I 3 + I 3 + 1 − 3I 2 = 0  4I 3 − 3I 2 = 2 ----------- (2) In loop ADBA, I1 + 1 + 3I 2 − 2 + 2I1 = 0  3I1 + 3I 2 = 1 ------------- (3) On solving (1), (2) & (3), I1 = 13/33 A, I2 = -2/33 A & I3 = 5/11 A 10 E (i) ф = 2 ; V AB = ф l AB = 2.5V (ii) No, because E1< E2. Therefore the null point can l AC not be obtained through the potentiometer wire. 11 E1 IR : I = = 0.2A ; ф = AB = 2 x 10 −2 V/cm ; E2 = ф l AC  l AC = 60cm. R AB + R l AB 12 X l X 60 Formula = , =  2 X = 3Y ……………..(1) Y 100 − l Y 40 When a resistance of 15Ω is connected in series with ‘Y’ 63  X 50 =  X = Y + 15 ………(2)On solving (1) & (2), X = 45 Ω , Y + 15 50 Y = 30Ω When a resistance of 30Ω is connected in series with ‘Y’ X l =  l = 75cm from end A. Y + 30 100 − l 13. (i)The emf E of a cell is independent of external resistance I. E E (v) The terminal p.d. V = IR = R= r+R r 1+ R On increasing R, V increases. E When R = 0, V → 0. When R = r, V = 2 When R →  , V = E. Assignment -2 ( 5 marks questions) 1. A cell of unknown emf E and internal resistance r, two unknown resistances R1 and R2 (R2>R1) and a perfect ammeter are given. The current in the circuit is measured in five different situations: (i) Without any external resistance in the circuit, (ii) With resistance R1 only, (iii) With resistance R2 only, (iv) With both R1 and R2 used in series combination and (v) With R1 and R2 used in parallel combination. The current obtained in the five cases are 0.42A, 0.6A, 1.05A, 1.4A, and 4.2A, but not necessarily in that order. Identify the currents in the five cases listed above and calculate E, r,, R1 and R2. 2. (a) Describe the formula for the equivalent EMF and internal resistance for the parallel combination of two cells with EMF  1 and  2 and internal resistances r1 and r2 respectively. What is the corresponding formula for the series combination? (b) Two cells of EMF 1V, 2V and internal resistances 2Ω and 1Ω respectively are connected in (i) series, (ii) parallel. What should be the external resistance in the circuit so that the current through the resistance be the same in the two cases?In which case more heat is generated in cells? 3.. Deduce the condition for balance in a Wheatstone bridge. Using the principle of Wheatstone bridge, describe the method to determine the specific resistance of a wire in the laboratory