AIR FCE SCHOOLS PRE-BOARD EXAMINATION CLASS XII PHYSICS (Theory) ANSWER KEY A. The physical quantity is Electric field. It s another SI unit is Vm - A2. For accelerating potential V, λ = For accelerating potential 4V, λ = h (2meV ), h (2me 4V), hence λ = λ A3. Nothing is displaced in displacement current. Its SI unit is ampere (A). A4.A common emitter amplifier is not an energy generating device The gain in energy is at the expense of the cell in the circuit. A5. Increasing N means increasing length of wire hence resistance increases. It increases current sensitivity but nut voltage sensitivity. A6. The expression for magnetic component will be B = 0.x0-8 sin [2πx0 8 (t- x )] kt c 0.x0-8 [ mark], T [ mark], k[ mark] 2 A7. Material A is paramagnetic while material B is diamagnetic. S A8. T /2 = 5 years, no of half-lives = 20/5 = 4 In 4 half-lives N = N o (/2) 4 = N o /6, or /6 A9.Magnetic flux φ = Blx, dφ = dblx dt dt = -Blv = ε (-ve sign due to velocity)
Force F = IlxB = εlb, where r is resistance of PQ, F = B2 l 2 v r r Work done in changing current through an inductor is dw dt = ε di, Substitute ε = LI, dt dw dt = LI di dt, W = LI2 /2 Energy in an inductor is stored as magnetic potential energy. A0. Given: G = 200Ω, Ig = 5mA = 5x0-3 A, V = 25V V GIg = RIg, 25-200x5x0-3 = R x5x0-3 25- = Rx5x0-3, or R = 4.8x0 3 = 4800Ω A. According to graph E C, Energy of a capacitor E =Q2 /2C. Hence Q is kept constant. E = CV 2, On joining the capacitors resultant potential is V/2. Total energy = C(V/2) 2 +C(V/2) 2 = CV 2 /8 + CV 2 /8 = CV 2 /4 A2. mark for each refracted or reflected ray correctly drawn. For angle i = 40 0 refracted and reflected ray are for mark each. A3. Zener diode is a heavily doped semiconductor as compared to a normal diode. 2 ` If input voltage V i increases, the current through Rs and Zener diode increases. This increases voltage across Rs without any change across the diode as it is working in the breakdown voltage. Thus, voltage across Zener diode remains unaffected and it acts as a voltage regulator.
A4. (i) Modulation index µ= 5/20 = 0.25 (ii) Sidebands 500 + 5 = 55kHz and 500-5 = 485 khz (iii) Peak potential of sideband = µa c /2 = 0.25x20/2 = 2.5V A5.Resonant frequency ω o means X L = X C, or V L = V C, current is in phase with v For ω <ω o, X C > X L, /ωc>ωl, which means V C > V L, hence current leads potential. For ω >ω o, X C < X L, /ωc<ωl, which means V C < V L, hence current lags potential. A6.Each point on a wavefront is a source of secondary wavelets spreading in all directions with the speed of the wave. A common tangent drawn to these secondary wavelets gives position of new wavefront. Let AB be the incident wavefrontat t=0 which strikes the surface of separation of two media at point A. Time t later point B of the wavefront touches the surface at point C and point A travels in second medium and becomes point D. Therefore CD is the new wavefront at time t. From ABC: sin i = BC AC,from ADC: sin r = AD AC sin i = BC = v t sin r AD v 2 t = µ 2 (a constant) A7.Emf: It is the potential difference across the terminals of a cell when no current is drawn from it. Potential difference: It is the potential difference across the terminals of a cell when current is drawn from it. E = IR + Ir, I = E R+r V = E R+r R,
A8. (a) Principle: A transformer works on the principle of mutual inductance. Laminated core helps in reducing loss of energy due to eddy currents. Soft iron core increases magnetic flux. (b) N N 2 = V V 2, I = 0x000 200 200 = 200, V 2 = 000V 000 V 2 = 50A A9. Line integral of the magnetic field around any closed path in free space is equal to absolute permeability times the net current passing through any surface enclosed by the closed path. B. dl = µ o I Let the toroid be of mean radius R, has N turns and current flowing through it is I (i)for inner radius r =r B. dl = Bdl=B dl = Bx2πr = µ o I, but net current enclosed is zero. Hence B = 0, Magnetic field inside the toroid is zero. (ii)for r = r 2 B. dl = Bdl = B dl = Bx2πr = µ o I, but net current enclosed is zero. Hence B = 0, Magnetic field outside the toroid is zero. For r = R B. dl = Bdl= B dl = Bx2πR = µ o NI, wheretotal current through the loop is NI. B = µ o NI/2πR =µ o ni (a)electric field accelerates the charged particle, while magnetic field rotates the charged particle so that it passes through the electric field multiple times. (b) According to Lorentz force F = q (E + vx B), If E = Ej, B = Bk and v = vi, Then F = qej - qvbj, for no deviation F = 0, which means qej = qvbj, or v = E / B A20. (a)
2 (b) λ = 600nm = 6x0-7 m, a = 80cm = 0.8m Limit of resolution =.22λ/a =.22x6x0 7 0.8 A2. V = Q 4πεoR, V 2 = = 9.5 x0-7 Q 4πεo2R, V 2 = /2V, total charge= 2 Q On connecting the two V = V 2 =V, and charges on each of them are Q and Q 2 Q 4πεoR = Q2 Q or = 2Q Q 4πεo2R R 2R, Q = 2 3 Q and Q 2 = 4 3 Q A22. (a) Intercept on y-axis = Wo, since intercept for metal B is greater hence work function of e B is greater. (b) Slope of both the lines is same as they are parallel and slope = h/e (c) Intercept on frequency axis represents threshold frequency υ o. A23. (a) Ira is a good observer and inquisitive, while her mother has knowledge and likes to share it with her daughter. 2 (b) (all values are in ev)3.6 3.4 = 0.2, 3.6.5 = 2.09, 3.6 0.85 = 2.75, 3.4.5 =.89, 3.4 0.85 = 2.55,.5 0.85 = 0.66 are maximum possible transitions. Out of these 2.09 and 2.75 are not possible due to lack of sufficient energy. Remaining transitions are possible. (x4)=2 A24. (a)drift velocity: It is the average velocity acquired by a free electron in a conductor opposite to the direction of applied electric field. (b) On the application of E, force on electrons F = ee, a = F/m = ee/m. Also v = u + at for N electrons <v> = <u> + <at> which gives v d = eeτ/m. Where v d is drift velocity and τ is average time between two successive collision of electrons. Current is total charge flowing through the wire /time. I = neav d = neaeeτ/m Or I = (ne 2 Aτ/m) E, I E, substituting E = V/l, I = (ne 2 Aτ/ml)V, I V (c ) (i)as temperature increases τ decreases hence current decreases
(ii) As temperature increases τ for semiconductors decreases but n increases. Increase in n is such that it shows an overall increase in current. A25. (a) Consider ΔAOB and ΔA OB, < O are equal and <B A O = < BAO which means ΔAOB ΔA OB, AB AB = AO AO = -v/u..() Consider ΔOMF 2 and ΔA B F 2 the two triangles are similar. Therefore, AB AB AB OA OF2 = OF 2 OM = AF2 OF 2 = v f..(2) f Using equations () and (2) -v/u = v f, -vf = uv uf. Divide the equation by uvf f /f = /v /u (b) Given distance between object and screen, D = 90cm, distance between two positions of the object =20cm. v-u = 20cm or D -2u = 20cm u = D ± (D2 4fD) 2, (2u-D) 2 = D 2-4D, (-20) 2 = 90(90-4f), On solving f = 2.4cm (a) y = a cos ωt, y 2 = a cos (ωt + φ), intensity of each wave is I o = ka 2 y = y +y 2 = a[cos ωt +cos (ωt + φ)] = 2a cos (φ/2)cos(ωt + φ/2) The amplitude of the resultant wave is 2a cos (φ/2) and therefore, intensity of the wave is I = 4I o cos 2 φ/2 If φ = 0,±2π, ±4π, ±6π. corresponds to a maximum intensity or constructive interference whereas φ = ±π, ±3π, ±5π..corresponds to a minimum intensity or destructive interference. If φ remains constant then interference pattern remains constant but if φ keeps on changing with time, then interference pattern will also change with time. If φ is changing with time, then <I > = 4I o <cos 2 φ/2>
= 4I o x/2 = 2I o at all points, which means a simple addition of the intensities of two sources. (b)()an equally spaced fringe pattern is replaced by a central bright and wide fringe followed by equally spaced fringes. (2) All equally bright fringe pattern is replaced by a central very bright fringe followed by fringes of decreasing brightness. (3) At angle λ/a is a maxima replaced by a minima in single slit diffraction pattern. (any two) mark each A26. (a)emitter base junction is forward biased hence its junction resistance decreases while base collector junction is reverse biased hence its junction resistance increases.,, (b)x is a NAND gate while X may be a NAND or N gate (no marks for NOT gate as there are two inputs) (a), By using centre tap and two diodes in such a way that at least one diode is in forward bias when circuit is switched on. For first half cycle D is forward biased and D 2 is reverse biased, so D conducts while D 2 does not. For second half cycle D 2 is forward biased while D is reverse biased, so D 2 conducts but D does not. (b) Gallium Arsenide Phosphide (GaAs -x P x ) provides the whole range of visible spectralcolours, hence it is used extensively for making LED. (c)sun light is absorbed by PbS near the surface but does not reach near depletion region. Therefore, it cannot be used for fabricating solar cells.