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. to 8 are very short answer type questions and carry one mark each. Question Nos. 9 to 8 carry two marks each, question 9 to 7 carry three marks each and question 8 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. However, you may use log tables if necessary. You may use the following values of physical constants wherever necessary: c = ms h = Js e = C 0 = Tm A 9 90 Nm C 40 Mass of electron m e = kg Q 8. (a) (b) (a) (b) Describe briefly, with the help of a diagram, the role of the two important processes involved in the formation of a p-n junction. Name the device which is used as a voltage regulator. Draw the necessary circuit diagram and explain its working. Explain briefly the principle on which a transistor-amplifier works as an oscillator. Draw the necessary circuit diagram and explain its working. Identify the equivalent gate for the following circuit and write its truth table.

2 (a) Two important process involved in the formation of a p-n junction are: Diffusion and, Drift In n-type semiconductor electrons are the majority carriers and holes are minority carriers. In the same way, in p-type semiconductor holes are majority and electrons are minority carriers. During the formation of p-n junction, due to concentration gradient across p- and n-side, holes diffuses from p-side to n-side and electrons diffuses from n-side to p-side. This motion gives rise to diffusion current across the junction. When an electron diffuses from n- to p-side, it leaves behind a positive charge. In such a manner a positively charged layer forms on n-side of the junction. Similarly, when a hole diffuses from p- to n-side, it leaves behind a negative charge and a negatively charged layer forms on p-side of the junction. This space-charge region is known as depletion legion. An electric field directed from positive charge towards negative charge develops. Due to this field, electrons on p-side of the junction move to n-side and holes on n-side of the junction move to p-side. This motion of charge carriers due to the electric field is called drift. Drift current is opposite in direction to the diffusion current. Initially, diffusion current is large and drift current is small. Space-charge region on either side increases as the diffusion process continues. This increases the electric field and hence the drift current. This process continues until the diffusion current equals the drift current. Thus a p-n junction is formed. (b) Zener diode is used as a voltage regulator.

3 Voltage regulator converts an unregulated dc voltage into a constant regulated dc voltage using Zener diode. The unregulated voltage is connected to the Zener diode through a series resistance R S such that the Zener diode is reverse biased. If the input voltage increases, the current through R S and Zener diode increases. Thus the voltage drop across R S increases without any change in the voltage drop across Zener diode. This is because in the breakdown region, Zener voltage remain constant even though the current through Zener diode changes. Similarly, if the input voltage decreases, the current through R S and Zener diode decreases. The voltage drop across R S decreases without any change in the voltage across the Zener diode. Thus any change in input voltage results the change in voltage drop across R S without any change in voltage across the Zener Diode. Thus, Zener diode acts as a voltage regulator. (a) Transistor amplifier as an oscillator: In an oscillator, the output at a desired frequency is obtained without applying any external input voltage. The common emitter n-p-n transistor as an oscillator is shown in the following figure. A variable capacitor C of suitable range is connected in parallel to coil T to give the variation in frequency. Oscillator action: As in an amplifier, the base-emitter junction is forward biased while the base collector junction is reverse biased. When the switch S is put on, a surge of collector flows in the coil T. The

4 inductive coupling between coil T and T cause a current to flow in the emitter circuit i.e., feedback from input to output. As a result of positive feedback, the collector current reaches at maximum. When there is no further feedback from T to T, the emitter current begins to fall and collector current decreases. Therefore, the transistor has reverted back to its original state. The whole process now repeats itself. The resonance frequency (f) of the oscillator is given by: f LC The tank of tuned circuit is connected in the oscillator side. Hence, it is known as tuned collector oscillator. (b) Output, Y A + B Here A Output of A from NOT gate, = A B Output of B from NOT gate, = B Thus, Output Y A+ B A B = A B So, the equivalent gate for the given circuit is AND GATE. Truth Table A B Y = A B

5 Q 9. (a) (b) Write the expression for the force, F, acting on a charged particle of charge q, moving with a velocity v in the presence of both electric field F and magnetic field B. Obtain the condition under which the particle moves undeflected through the fields. A rectangular loop of size l b carrying a steady current I is placed in a uniform magnetic field B. Prove that the torque acting on the loop is give by = m B, where m is the magnetic moment of the loop. (a) (b) Explain, giving reasons, the basic difference in converting a galvanometer into (i) a voltmeter and (ii) an ammeter. Two long straight parallel conductors carrying steady currents I and I are separated by a distance 'd'. Explain briefly, with the help of a suitable diagram, how the magnetic field due to one conductor acts on the other. Hence deduce the expression for the force acting between the two conductors. Mention the nature of this force. (a) Force F acting on a charge q moving with velocity in the presence of both electric field E and magnetic field B, F q E q B Consider a region in which magnetic field, electric field and velocity of charge particle are perpendicular to each other. To move charge particle undeflected the net force acting on the particle must be zero i.e. The electric force must be equal and opposite to the magnetic force. qe qvb E v B The direction of electric and magnetic forces are in opposite direction. Their magnitudes are in such a way they cancel out each other to give net force zero so that the charge particle does not deflect. (b) Plane of the loop is at an angle with the direction of the magnetic field, Let the angle between the field and the normal is θ. The forces on BC and DA are equal and opposite and they cancel each other as they are collinear. Force on AB is F and force on CD is F, thus

6 F = F = IbB Magnitude of torque on the loop as in the figure: l l F sin F sin IlbBsin or, IAB sin (Where area,a=lb) If there are n such turns the torque will be niab sinθ Magnetic moment of the current, m = IA mb (a) (i) In converting a galvanometer into a voltmeter, a very high suitable resistance is connected is series to its coil. So, the galvanometer gives full scale deflection. (ii) In converting a galvanometer into an ammeter, a very small suitable resistance is connected in parallel to its coil. The remaining pair of the current i.e. (I I g ) flows through the resistance. Here I = Circuit current I g = Current through galvanometer (b) Assumption: Current flows in the same direction. Using Right hand thumb rule, the direction of the magnetic field at point P due to current I is perpendicular to the plane of paper and inwards. Similarly, at point Q on X Y, the direction of magnetic field due to current I is perpendicularly outward. Using Fleming s left hand rule we can find the direction of forces F and F which are in

7 opposite directions thus, By Ampere s circuited law, we have, 0 I B 4π d Now, F = I LB (Where L length of the conductors) 0 I I L 0 I I L F 4π d π d In similar manner we get, 0 IIL F... π d From above we get the magnitude of forces F and F are equal but in opposite direction. So, F F Therefore, two parallel straight conductors carrying current in the same direction attract each other. Similarly, we can prove if two parallel straight conductors carry currents in opposite direction, they repel each other with the same magnitude as equation (). Q 30. (a) In Young's double slit experiment, derive the condition for (i) constructive interference and (ii) destructive interference at a point on the screen. (b) A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interference fringes in a Young's double slit experiment on a screen placed 4 m away. If the two slits are separated by 0 8 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide. (a) How does an unpolarized light incident on a polaroid get polarized? Describe briefly, with the help of a necessary diagram, the polarization of light by reflection from a transparent medium. (b) Two polaroids A and B are kept in crossed position. How should a third Polaroid C be placed between them so that the intensity of polarized light transmitted by polaroid B reduces to /8 th of the intensity of unpolarized light incident on A? (a) Young s double slit experiment: Consider two narrow rectangular slits S and S placed perpendicular to the plane of paper. Slit S is placed on the perpendicular bisector of S S and is illuminated with monochromatic light. The slits are separated by a small distance d. A screen is placed at a distance D from S, S.

8 Consider a point P on the screen at distance x from O. The path difference between the waves reaching P from S and S is: P = S P S P Draw S N perpendicular to S P. Then, P = S P S P = S P NP = S N SN From right-angled S S N, sin SS P = S N = S S sin = d sin From COP, When is small, x sin tan D xd P D For constructive interference, xd n, n 0,,,3,... D Position of n th nd D D 3D bright fringe, xn 0,,,,... d d d d When n = 0, x n = 0, central bright fringe is formed at O. For destructive interference, xd (n ) D D D 3 D 5 D or xn (n),,,... d d d d Thus, alternate bright and dark fringes are formed on the screen. (b) Given: = 800 nm = m = 600 nm = m D =.4 m d = 0.8 mm = m Let n th maximum corresponds to coincides with n th maximum corresponds to. Then,

9 D D n n d d n or, n The minimum integral value of n is 3 and of n is 4. Therefore, the minimum value of y is, 9 D ymin nl d y mm min (a) Polaroid is made up of a special material which blocks one of the two planes of vibration of an electromagnetic wave. Because of its chemical composition it allows only those vibrations of the electromagnetic wave which are parallel to its crystallographic axis. An ordinary beam of light on reflection from a transparent medium becomes partially polarised. The degree of polarisation increases as the angle of incidence is increased. At a particular value of angle of incidence, the reflected beam becomes completely polarised. This angle of incidence is called the polarising angle (p). (b) Let AC, angle between the transmission axis of Polaroid A and Polaroid C. CB, angle between the transmission axis of Polaroid C and Polaroid B. Then, AC + CB = 80 (As, Polaroid A and B are kept in crossed position.) Or, AC = 80 CB... ()

10 Intensity of unpolarized light = I 0 I, I and I 3 are the intensities of light on passing through the A, B and C polarises respectively. Now, I I 0... I I cos A C I 0 cos A C I cos 80 0 C B from equation I I 0 sin C B... 3 and, I I cos 3 C B I 3 I0 sin C B cos C B or, I 3 I 0 sin C B cos C B A s, given, I 3 I 0 8 T herefore, I I sin 8 4 or, sin CB CB or, C B or, CB 45 Thus, Polaroid C must be placed at angle 45 with Polaroid B.

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