Charge distribution Charge density Types of Gauss’s surfaces Surface area for which Gauss’s theorem Electric field Intensity Infinitely long straight uniformly charged conductor Linear charge density Cylindrical Lateral surface area Infinitely extended plane sheet of Charge Surface charge density Plane Plane surface = uniformly Charged spherical shell Surface charge density Spherical surface Properties of electric field lines:. 1.The imaginary path along which a unit positive charge placed in the electric field tends to follow is the magnetic line of force 2. The electric lines of force emanate from a positive charge and terminate on a negative charge. The tangent to an electric field line at any point gives the direction of the electric field at that point. 3. No two electric lines of force cross each other. If they do, then at the point of intersection, there will be two tangents. It means there are two values of the electric field at that point, which is not possible. 6. Electric lines of force are closer (crowded) where the electric field is stronger and the lines spread out where the electric field is weaker. 4. Electric lines of force contract lengthwise to represent attraction between two unlike charges and Electric lines of force exert lateral (sideways) pressure to represent repulsion between two like charges. Electrostatic Potential: Work done per unit positive Test charge to move it from infinity to that point in an electric field against the field direction . It is a scalar. SI unit: J/C or V V = W / qo Electric potential at any point at a distance r from a point charge q: Graphs: Variation of E & V due to a point charge at any point in the field with r (Graph-1) and Variation of V with 1/r (Graph-2)
get affected? 17 In the given circuit, assuming point A to be at zero potential, use Kirchhoff’s rules to determine the potential at point B. 2011 18 Show that the electric field at the surface of a charged conductor is given by E = 0) n, where is the surface charge density and $ n is a unit vector normal to the surface in the outward direction. 2007 19 The plot of the variation of potential difference across a combination of three identical cells in series, versus current is as shown below. What is the emf of each cell ? 2008 20 A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable resistor ‘R’. Plot a graph showing the variation of terminal potential ‘V’ with resistance R. Predict from the graph the condition under which ‘V’ becomes equal to ‘E‘. 2009 21 Derive an expression for drift velocity of free electrons in a conductor in terms of relaxation time. 2009 22 A wire of 15 Ω resistances is gradually stretched to double its original length. It is then cut into two equal parts. These parts are then connected in parallel across a 3 00 volt battery. Find the current drawn from the battery. 2009 23 (a) You are required to select a carbon resistor of resistance 47 k10% from a large collection. What should be the sequence of colour bands used to code it? (b) Write the characteristics of manganin which make it suitable for making standard resistance. 2012 THREE MARKS QUESTIONS 1 In the two electric circuits shown in the figure, determine the readings of ideal Ammeter (A) and the ideal voltmeter (V). 2015 2 In the circuit shown in the figure, find the current through each resistor. 2015 3 (a) Deduce the relation between current I flowing through a conductor and drift velocity d of the electrons. (b) Figure shows a plot of current ‘I’ flowing through the cross-section of a wire versus the time‘t’. Use the plot to find the charge flowing in 10s through the wire.. 2014 4 A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable load resistor R. Draw the plots of the terminal voltage V versus (i) R and (ii) the current I. It is found that when R = 4 , the current is 1 A and when R is increased to 9 , the 2013
QUESTION BANK OF PROBABLE QUESTIONS FOR AISSCE 2016 CHAPTER – 6 (ELECTROMAGNETIC INDUCTION) Questions that have been asked one time VERY SHORT ANSWER QUESTIONS (1 MARK) How does the mutual inductance of a pair of coils change when distance between the coils is increased and number of turns in the coils is increased? [CBSE (AI) 2013] Ans. (i) Decreases (ii) Increases The motion of copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping? [CBSE (AI) 2013] Ans. As the plate oscillates, the changing magnetic flux through the plate produces a strong eddy current in the direction which opposes the cause. Also, copper being diamagnetic substance, it gets magnetized in the opposite direction, so the plate motion gets damped. The closed loop PQRS is moving into a uniform magnetic field acting at right angles to the plane of the paper as shown. State the direction of the induced current in the loop. [CBSE (AI) 2012] X X X X X P Q X X X X X X X X X X X X X X X X X X X X X X X X X R S
Φ = (nlBA) Substituting the value of B from (i) Φ = nl (µ0nI ) A = µ0n2AlI Self-inductance of air solenoid L = Φ/I = µ0n2Al If N is total number of turns in length l, then n= N/l Self-inductance L = µ0(N/l)2Al = µ0N2A/l Q. 1. What do you mean by mutual inductance of two nearby coils? Find an expression for mutual inductance of two co-axial solenoid. (5) [CBSE (F) 2013, 2010] Ans. When current flowing in one of two nearby coils is changed, the magnetic flux linked with the other coil changes; due to which an emf is induced in it (other coil). This phenomenon of electromagnetic induction is called the mutual induction. The coil, in which current is changed is called primary coil and the coil in which emf is induced is called the secondary coil. The si unit of mutual inductance is henry. Mutual Inductance: Suppose there are two coils C1 and C2. The current I1 flowing in primary coil c1 ; due to which an effective magnetic flux Φ2 is linked with secondary coil C2 . Φ2=M21I1 , M21 is the mutual inductance of coil 2 w.r.t. coil 1
RAY OPTICS AND OPTICAL INSTRUMENTS QUESTIONS HAVE BEEN ASKED ONE TIME SL. NO. QUESTIONS M.M. YEAR 1 An object is held at the principal focus of a concave lens of focal length f. Where is the image formed? 1 2008 2 A converging lens is kept co-axially in contact with a 1 2010
release of energy in both the processes of nuclear fusion and fission. (c) Write the basic nuclear process of neutron undergoing b–decay. Why is the detection of neutrinos found very difficult?(5) (2006,2008,2009,2010,2011,2013) Text Book : PHYSICS PART II ( NCERT) Chapter 14: Semiconductor Electronics: Materials, devices and Simple circuits Frequency : ( Asked three times or more ) Sl.No. Question Marks Year 1 Draw a simple circuit of a CE transistor amplifier. Explain its working. Show that the voltage gain AV, of the amplifier is given by ,where is the current gain, RL is the load resistance and ri is the input resistance of the transistor. What is the significance of the negative sign in the expression for the voltage gain? 5 2012(D) Questions of similar nature asked in 2008, 2009 2 Explain the function of base region of a transistor. Why is this region made thin and lightly doped? Draw a circuit diagram to study the input and output characteristics of n-p-n transistor in a common emitter (CE) configuration. Show these characteristics graphically. Explain how current amplification factor of the transistor is calculated using output characteristics. OR 5 2006, 2013 (i) Draw a circuit diagram to study the input and output characteristics of an n-p-n transistor in its common emitter configuration. Draw the typical input and output characteristics. (ii) Explain, with the help of a circuit diagram, the working of n-p-n transistor as a common emitter amplifier. 2009(D)