Physics (Theory) There are 30 questions in total. Question Nos. 1 to 8 are very short answer type questions and carry one mark each.

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1 Physics (Theory) Time allowed: 3 hours] [Maximum marks:70 General Instructions: (i) All questions are compulsory. (ii) (iii) (iii) (iv) (v) There are 30 questions in total. Question Nos. 1 to 8 are very short answer type questions and carry one mark each. Question Nos. 9 to 18 carry two marks each, question 19 to 27 carry three marks each and question 28 to 30 carry five marks each. There is no overall choice. However, an internal choice has been provided in one question of two marks; one question of three marks and all three questions of five marks each. You have to attempt only one of the choice in such questions. Use of calculators is not permitted. You may use the following values of physical constants wherever necessary: c = ms 1 h = Js e = C 0 = Tm A Nm C 40 Mass of electron m e = kg Q.9. Draw a plot showing the variation of (i) electric field (E) and (ii) electric potential (V) with distance r due to a point charge Q. 2 We know, that for a point charge Q Q 1 Electric potential, V or V 4π 0 r r and Q 1 Electric field, E or E 2 2 4π r r 0 Thus, electric potential shows an inverse relationship while electric field shows an inverse square relationship with r. So, their corresponding plots would be

2 Q.10. Derive the expression for the self inductance of a long solenoid of cross sectional area A and length l, having n turns per unit length. 2 Self-Inductance of a Long Solenoid A long solenoid is one whose length is very large as compared to its radius of cross-section. The magnetic field B at any point inside such a solenoid is practically constant and is given by, Where, μ 0 = Absolute magnetic permeability of free space N = Total number of turns in the solenoid Magnetic flux through each turn of the solenoid coil = B Area of each turn Where, A = Area of each turn of the solenoid Total magnetic flux linked with the solenoid

3 If L is coefficient of self-inductance of the solenoid, then Φ = LI (iii) Equating (ii) and (iii), If core is of any other magnetic material, μ 0 is replaced by if n is the number of turns per unit length then L = 0 n 2 Al This is self inductance of a long solenoid. Q.11. A ray of light, incident on an equilateral prism g 3moves parallel to the base line of the prism inside it. Find the angle of incidence for this ray. 2 It is given that the prism is equilateral in shape. So, all the angles are equal to 60. Thus, the angle of prism, A = 60 The angle of refraction in case of a prism, We can now apply Snell s law So, sin i a g sin r Here, a refractive index of air, n 1 = 1 g refractive index of glass, n2 3 i angle of incidence Thus, g g sin i sin r sin 30 a a sin i A r 30 2

4 So, the angle of incidence is i 60 Q.12. The current in the forward bias is known to be more (~ma) than the current in the reverse bias (~A). What is the reason, then, to operate the photodiode in reverse bias? 2 The photodiode always work under reverse biasing conditions although the current produced is less. This is because in reverse bias, the width if the depletion layer increases which reduces the capacitance across the junction, thereby increasing response time. The sensitivity of a photodiode is thus very high, a property that is certainly desired. Q.13. Distinguish between Analog and Digital signals. 2 Mention the function of any two of the following used in communication system: (i) Transducer (ii) Repeater (iii) Transmitter (iv) Bandpass Filter OR Analog Signal:- It is continuous signal, which varies continuously with variable may be time or distance etc. E.g. Sound of human Digital Signal:- It is a type of signal which has only two values high or low. In digital high mean 1 and low means zero. E.g. Temperature of day Maximum 30 C 1 Minimum 15 C 0 OR (1) Transducer: It is an electric device which converts energy from one form to another form. e.g. microphone, which converts sound energy into electric energy and vice versa. (2) Repeater: It is an electronic device used in transmission system to regenerate the signal. It picks up a signal amplifies it and transmits it to receiver. (3) Transmitter: Transmitter is an electronic device which is used to radiate electromagnetic waves. The purpose

5 of the transmitter is to boost up the signal to be radiated to the required power level, so that it can travel long distances. The most familiar transmitters are mobile transmitter antennas, radio and T.V. broadcasting antennas etc. (4) Bandpass filter: It is an electronic filter, which pass the certain band (range) of frequency and reject rest of all. Q.14. Define mutual inductance between two long coaxial solenoids. Find out the expression for the mutual inductance of inner solenoid of length l having the radius r 1 and the number of turns n 1 per unit length due to the second outer solenoid of same length and r 2 number of turns per unit length. 2 Mutual Inductance The ability of production of induced emf in one coil, due to varying current in the neighbouring coil is called mutual inductance. Magnetic flux, = MI Where, M is called coefficient of mutual induction Mutual Inductance of Two Long Solenoids Consider two long solenoids S 1 and S 2 of same length l, such that solenoid S 2 surrounds solenoid S 1 completely. 21 = M 21 I 1 Where, M 21 is the coefficient of mutual induction of the two solenoids Magnetic field produced inside solenoid S 1 on passing current through it, B 1 = 0 n 1 I 1 Magnetic flux linked with each turn of solenoid S 2 will be equal to B 1 times the area of crosssection of solenoid S 1. Magnetic flux linked with each turn of the solenoid S 2 = B 1 A Therefore, total magnetic flux linked with the solenoid S 2, 21 = B 1 A n 2 l = 0 n 1 I 1 A n 2 l 21 = 0 n 1 n 2 lai 1 M 21 = 0 n 1 n 2 Al Similarly, the mutual inductance between the two solenoids, when current is passed through solenoid S 2 and induced emf is produced in solenoid S 1, is given by M 12 = 0 n 1 n 2 Al M 12 = M 21 = M (say)

6 Hence, coefficient of mutual induction between the two long solenoids M n n Al Q.15. A cell of emf E and internal resistance r is connected to two external resistance R 1 and R 2 and a perfect ammeter. The current in the circuit is measured in four different situations: (i) without any external resistance in the circuit (ii) with resistance R 1 only (iii) with R 1 and R 2 in series combination (iv) with R 1 and R 2 in parallel combination The currents measured in the four cases are 0.42 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily i n the order. Identify the currents corresponding to the four cases mentioned above. 2 The current relating to corresponding situations are as follows: (i) Without any external resistance in the circuit E I1 r The current in this case would be maximum, so I 1 = 4.2A (ii) With resistance R 1 only E I2 r R1 The current in this case will be the second smallest value, so I 2 = 1.05 A (iii) With R 1 and R 2 in series combination E I3 r ( R1 R2) The current in this case will be minimum as the resistance will be maximum, so I = 0.42 A. (iv) With R 1 and R 2 in parallel combination- E I4 RR 1 2 r ( ) R1 R1 The current in this case would be the second largest value, so I = 1.4 A. Q 16. Two identical circular loops, P and Q, each of radius r and carrying equal currents are kept in the parallel planes having a common axis passing through O. The direction of current in P is clockwise and in Q is anti-clockwise as seen from O which is equidistant from the loops P

7 and Q. Find the magnitude of the net magnetic field at O. 2 The standard formula for field at an axial point is given as 2 0 ia B a d 2 So, in this case 2 0Ir 0I B r 2 2 r 2 r ( ) 2 Now, as the current flowing in loop P is clockwise by using right hand thumb s rule the direction of the magnetic field will be towards left and as the current in loop Q is clockwise then the direction of magnetic field is towards left. So the net magnetic field at point O will be the sum of the magnetic fields due to loops P and Q. Also, as the fields produced are at an equal distance to O, B P = B Q, So, net field 0I 0I B = B P + B Q = (2) r (2) r Q 17. A metallic rod of L length is rotated with angular frequency of with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring. 2 Consider the infinitesimally small length dx at a distance x. So speed of this part is ωx. Induced small emf = Bωx dx (since emf = vbl)

The emf between the ends of the rotating rod is http://www.meritnation.com/discuss/question/1785161/can-u-plz-answer-this-a- Q 18. When an ideal capacitor is charged by a dc battery, no current flows.

8 The emf between the ends of the rotating rod is Q 18. When an ideal capacitor is charged by a dc battery, no current flows. However, when an ac source is used, the current flows continuously. How does one explain this, based on the concept of displacement current? 2 Ideal capacitor means infinite resistance for dc. When an ac source is used, the current flows continuously, but we know that the capacitor has dielectric (air) between its plates. So, ideally there is no current, and circuit would be incomplete. In real capacitor is charged due to contribution of varying electric field. The current between the capacitor plates is given by displacement current D= E d Id dt

BLUE-PRINT II XII Physics

BLUE-PRINT II XII Physics S.No. UNIT VSA SA I SA II LA Total (1 Mark) (2 Marks) (3Marks) (5 Marks) 1. Electrostatics 1(1) 4(2) 3(1) - 8(4) 2. Current Electricity - 2(1) - 5(1) 7(2) 3. Magnetic Effect of