Time : 3 hrs.

Marks : 70

GENERAL INSTRUCTIONS (i) All questions are compulsory. (ii) There are 30 questions in total. Questions 1 to 8 carry one mark each, questions 9 to 18 carry two marks each, questions 19 to 27 carry three marks each and questions 28 to 30 carry five marks each. (iii) 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 choices in such questions. (iv) Use of calculators is not permitted. 1. Name the physical quantity whose S.I. unit is J C–1. Is it a scalar or a vector quantity? [From Excel in Physics, Page. 43, Q. 71] 2. A beam of a particles projected along + x-axis, experiences a force due to a magnetic field along the + y-axis. What is the direction of the magnetic field? [From Excel in Physics, Page. 187, Q. 45]

3. Define self-inductance of a coil. Write its S.I. units. [From Excel in Physics, Page. 226, point 4.2, 1(iii), (iv)] 4. A converging lens is kept coaxially in contact with a diverging lens, both the lenses being of equal focal lengths. What is the focal length of the combination? [From Excel in Physics, Page. 376, Q. 91] 5. Define ionisation energy. What is its value for a hydrogen atom? [From Excel in Physics, Page. 506, Q. 10] 6. 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

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is twice that in Y, find the ratio of drift velocity of electrons in the two wires. [From Excel in Physics, Page. 115, Q. 13]

7. Name the part of electromagnetic spectrum whose wavelength lies in the range of 10–10 m. Give its one use. [From Excel in Physics, Page. 322, Q. 17] 8. When light travels from a rarer to a denser medium, the speed decreases. Does this decrease in speed imply a decrease in the energy carried by the light wave? Justify your answer. [From Excel in Physics, Page. 418, Q. 1(b) NCERT Solved] 9. Deduce the expression for the magnetic dipole moment of an electron orbiting around the central nucleus. [From Excel in Physics, Page. 160, Point 3.5(4)] 10. A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q. A charge q is placed at the centre of the shell. (a) What is the surface charge density on the (i) inner surface, (ii) outer surface of the shell? (b) Write the expression for the electric field at a point x > r2 from the centre of the shell. [From Excel in Physics, Page. 31, Q. 54 NCERT solved] 11. Draw a sketch of a plane electromagnetic wave propagating along the z-direction. Depict clearly the directions of electric and magnetic fields varying sinusoidally with z.

12. Show that the electric field at the surface of a s ^ n, where s charged conductor is given by E = e0 is the surface charge density and n^ is a unit vector normal to the surface in the outward direction. [From Excel in Physics, Page. 27, & 42 NCERT solved] 13. Two identical loops, one of copper and the other of aluminium, are rotated with the same angular speed in the same magnetic field. Compare (i) the induced emf and (ii) the current produced in the two coils. Justify your answer. [From Excel in Physics, Page. 270, Q. 28] 14. An a-particle and a proton are accelerated from rest by the same potential. Find the ratio of their de Broglie wavelengths. [From Excel in Physics, Page. 459, Q. 47] 15. Write two factors justifying the need of modulating a signal. A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75%? [From Excel in Physics, Page. 577, point 2; Page 585, Q. 6] 16. Write Einstein’s photoelectric equation. State clearly the three salient features observed in photoelectric effect, which can be explained on the basis of the above equation. [From Excel in Physics, Page. 444, point 7.3] 17. Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions which you can draw regarding the nature of nuclear forces. [From Excel in Physics, Page. 483, point 9] OR Draw a plot of the binding energy per nucleon as a function of mass number for a large number of nuclei, 2 ≤ A ≤ 240. How do you explain the constancy of binding energy per nucleon in the range 30 < A < 170 using the property that nuclear force is short-ranged? [From Excel in Physics, Page. 484, point 15] 18. (i) Identify the logic gates marked P and Q in the given logic circuit.

(ii) Write down the output at X for the inputs A = 0, B = 0 and A = 1, B = 1. [From Excel in Physics, Page. 559, Q. 102]

19. Which mode of propagation is used by short wave broadcast services having frequency range from a few MHz upto 30 MHz? Explain diagrammatically how long distance communication can be achieved by this mode. Why is there an upper limit to frequency of waves used in this mode? [From Excel in Physics, Page. 583, point 10.3] 20. Write any two factors on which internal resistance of a cell depends. The reading on a high resistance voltmeter, when a cell is connected across it, is 2.2 V. When the terminals of the cell are also connected to a resistance of 5 W as shown in the circuit, the voltmeter reading drops to 1.8 V. Find the internal resistance of the cell.

[From Excel in Physics, Page. 88, point 2.3]

21. A network of four capacitors each of 12 mF capacitance is connected to a 500 V supply as shown in the figure. Determine (a) equivalent capacitance of the network and (b) charge on each capacitor.

[From Excel in Physics, Page. 39, Q. 86 NCERT solved] 22. (i) Draw a neat labelled ray diagram of an astronomical telescope in normal adjustment. Explain briefly its working. [From Excel in Physics, Page. 347, point 3] (ii) An astronomical telescope uses two lenses of powers 10 D and 1 D. What is its magnifying power in normal adjustment? [From Excel in Physics, Page. 383, Q. 34] OR (i) Draw a neat labelled diagram of a compound microscope. Explain briefly its working. (ii) Why must both the objective and the eyepiece of a compound microscope have short focal lengths? [From Excel in Physics, Page. 346, point (iv)] physics for you | MaY ’10

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23. In Young’s double slit experiment, the two slits 0.15 mm apart are illuminated by monochromatic light of wavelength 450 nm. The screen is 1.0 m away from the slits. (a) Find the distance of the second (i) bright fringe, (ii) dark fringe from the central maximum. (b) How will the fringe pattern change if the screen is moved away from the slits? [From Excel in Physics, Page. 428, Q. 41] 24. State Kirchhoff’s rules. Use these rules to write the expressions for the currents I1, I2 and I3 in the circuit diagram shown.

[From Excel in Physics, Page. 91, point 2.5; Page. 121, Q. 100] 25. (a) Write symbolically the b– decay process of 32 15 P.

series is required whereas in an ammeter a shunt is used? [From Excel in Physics, Page. 159, point 3.4] OR (a) Derive an expression for the force between two long parallel current carrying conductors. (b) Use this expression to define S.I. unit of current. [From Excel in Physics, Page. 158, point 9] (c) A long straight wire AB carries a current I. A proton P travels with a speed v, parallel to the wire, at a distance d from it in a direction opposite to the current as shown in the figure. What is the force experienced by the proton and what is its direction?

[From Excel in Physics, Page. 498, Q. 29(iii)] (b) Derive an expression for the average life of a radionuclide. Give its relationship with the half-life. [From Excel in Physics, Page. 488, point 15(b)] 26. How does an unpolarised light get polarised when passed through a polaroid? [From Excel in Physics, Page. 414, point 7] Two polaroids are set in crossed positions. A third polaroid is placed between the two making an angle q with the pass axis of the first polaroid. Write the expression for the intensity of light transmitted from the second polaroid. In what orientations will the transmitted intensity be (i) minimum and (ii) maximum? [From Excel in Physics, Page. 423, NCERT Q. 25; Page. 432, Q. 22] 27. An illuminated object and a screen are placed 90 cm apart. Determine the focal length and nature of the lens required to produce a clear image on the screen, twice the size of the object. [From Excel in Physics, Page. 375, Q. 81; Page. 376, Q. 94] 28. (a) With the help of a diagram, explain the principle and working of a moving coil galvanometer. (b) What is the importance of a radial magnetic field and how is it produced? (c) Why is it that while using a moving coil galvanometer as a voltmeter a high resistance in

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[From Excel in Physics, Page. 192, Q. 43]

29. State Faraday’s law of electromagnetic induction. Figure shows a rectangular conductor PQRS in which the conductor PQ is free to move in a uniform magnetic field B perpendicular to the plane of the paper. The field extends from x = 0 to x = b and is zero for x > b. Assume that only the arm PQ possesses resistance r. When the arm PQ is pulled outward from x = 0 to x = 2b and is then moved backward to x = 0 with constant speed v, obtain the expressions for the flux and the induced emf. Sketch the variations of these quantities with distance 0 ≤ x ≤ 2b.

[From Excel in Physics, Page. 239, NCERT Q. 21] OR Draw a schematic diagram of a step-up transformer. Explain its working principle. Deduce the expression for the secondary to primary voltage in terms of the number of turns in the two coils. In an ideal transformer, how is this ratio related to the currents in the two coils?

How is the transformer used in large scale transmission and distribution of electrical energy over long distances? [From Excel in Physics, Page. 252]

30. (a) Draw the circuit diagrams of a p-n junction diode in (i) forward bias, (ii) reverse bias. How are these circuits used to study the V–I characteristics of a silicon diode? Draw the typical V–I characteristics. [From Excel in Physics, Page. 533, points 4, 5] (b) What is a light emitting diode (LED)? Mention two important advantages of LEDs over conventional lamps. [From Excel in Physics, Page. 556, Q. 57] OR (a) Draw the circuit arrangement for studying the input and output characteristics of an n-p-n transistor in CE configuration. With the help of these characteristics define (i) input resistance, (ii) current amplification factor. [From Excel in Physics, Page. 539, point 7] (b) Describe briefly with the help of a circuit diagram how an n-p-n transistor is used to produce self-sustained oscillations. [From Excel in Physics, Page. 541, point 10] SOLUTIONS 1. J C–1 is the S.I. unit of electrostatic potential. It is a scalar quantity. 2. From Lorentz force, F = q(v × B) ⇒ F j^ = q(v i^ × B) ^ It is clear that B is along –z axis, as i^ × (− k) = j^ 3. Self inductance of a coil is defined as the emf induced in the coil when the rate of change of current through the coil is unity. Its S.I. unit is henry (H) or V s A–1. 4. f1 = f (for converging lens) f2 = – f (for diverging lens) \ Focal length of their combination is 1 1 1 1 1 1 = + = − =0 ⇒ F= =∞ 0 F f f f f 1 2 5. Ionisation energy for an atom is defined as the energy required to remove an electron completely from the outermost shell of the atom. The ionisation energy required to remove an electron from ground state of hydrogen atom is E = E∞ – E1 = 0 – (–13.6) eV = 13.6 eV 6. Since the wires are connected in series, current I through both is same. Therefore ratio of drift velocities vX I / nX eAX = vY I / nY eAY

where, nX, nY = respective electron densities AX, AY = cross-sectional areas v n 1 ⇒ X = Y = (Given AX = AY , nX = 2nY ) vY nX 2 ⇒ vX : vY = 1 : 2 7. The given range corresponds to X-rays. X-rays are used for detection of fractures, formations of stones etc. in human bodies. They are also used to study crystal structure of solids. 8. According to relation E = hu, the energy of light wave depends upon frequency u. As the frequency does not change during refraction, energy also remains the same. 9. A revolving electron in an orbit of radius r moving with velocity v behaves as a current loop of effective current I = ue (u is frequency of revolution) ve = 2πr Hence it acts like a magnetic dipole moment ve M = IA = × πr 2 2πr evr = 2 10.

(a) (i) Surface charge density on the inner surface of shell is −q sin = 4πr12 (ii) Surface charge density on the outer surface of shell is Q+q sout = 4πr22 (b) Using, Gauss’s law, 1 Q+q E(x) = 4πe0 x 2 11. An e.m. wave propagating along z-axis is shown below.

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12. Consider an elementary area dS on the surface of the charged conductor. Enclose this area element with a cylindrical gaussian surface as shown in figure.

Now electric field inside a charged conductor is zero. Therefore, direction of field, just out side dS will be normally outward i.e. in direction of n^ . According to Gauss’s theorem, total electric flux coming out is sdS E ⋅ dS = [E is electric field at the surface] e0 sdS ⇒ E dS cos 0° = e0 s ⇒ E= e0

13. (i) Induced emf in both the loops will be same, as the two loops are of same area A and are rotated with same angular speed w in same magnetic field B given by e = BAwsinwt e (ii) As I = , so in copper loop with less resistance, R induced current will be more. h 14. de Broglie wavelength l = 2 mE h = kinetic energy E = qV ] [ 2mqV 2mpqpV la h \ = × lp 2maqaV h mpqp mpqp 1 1 la = = = ⇒ = (4mp ) ⋅ (2qp ) 8 2 2 lp ma qa ⇒ l a : l p = 1 : 2 2 15. Need of modulation : (i) audio/video signals do not have sufficiently high energy to travel upto long distances, because of their lower frequency. (ii) For effective transmission, the size of the l antenna should be at least of the size , where l is wavelength of signal to be sent. 4

Am Ac So peak voltage of modulating signal, Am = mAc ⇒ Am = 0.75 × 12 =9V 16. Einstein’s photoelectric equation is given below. 1 2 hu = mvmax + W0 2 where u = frequency of incident radiation 1 2 mvmax = maximum kinetic energy of an electron 2 emitted W0 = Work function of the target metal Three salient features observed are (i) Below threshold frequency u0 corresponding to W0, no emission of photoelectrons takes place. (ii) As energy of a photon depends on the frequency of light, so the maximum kinetic energy with which photoelectron is emitted depends only on the energy of photon or on the frequency of incident radiation. (ii) As the number of photons in light depend on its intensity, and one photon liberates one photoelectron, so number of photoelectrons emitted depend only on the intensity of incident light. 17. Modulation index, m =

From the above plot, following conclusions can be drawn. (i) Nuclear forces are short range forces (ii) For a separation greater than r0, the nuclear forces are attractive and for separation less than r0, the nuclear forces are strongly repulsive. OR

For an e.m. wave of the frequency of the order of audio signal, we need an antenna of size 3.75 km, which is practically impossible. Hence these low frequency base band signals are first converted into high frequencies, through modulation.

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18. (i)

Gate P is a NAND gate, and gate Q is an OR gate. (ii) Boolean expression for the above logic circuit is X = A⋅B + B = A + B + B

= A+1 X=1 [using boolean identities] Thus output at X is going to be 1 for all the possible inputs at A and B. 19. Sky wave propagation is used by short wave broadcast services having frequency range from a few MHz upto 30 MHz. Sky waves are used for long distance radio communication. The successive reflection of these radiowaves at the earth’s surface and the ionosphere make it possible to transmit these waves from one part to another part of the earth.

These waves are reflected by ionosphere by means of total internal reflection, which arises due to change in refractive indices of different layers of ionosphere. Critical frequency of reflection for a particular layer is given by fc = 9 N max where Nmax = maximum electron density in the given layer. For any frequency greater than fc ionosphere is not able to reflect the radiowave. Hence there exists an upper limits to frequency of waves used in this mode. 20. Internal resistance of a cell depends upon (i) surface area of each electrode. (ii) distance between the two electrodes. (iii) nature, temperature and concentration of electrolyte. Let internal resistance of cell be r.

The circuit given in question can be redrawn as

Initially when K is open, voltmeter reads 2.2 V. i.e. emf of the cell, e = 2.2 V Later when K is closed, voltmeter reads 1.8 V which is actually the terminal potential difference, V. i.e. if I is the current flowing, then e = I(R + r) ⇒ 2.2 = I(5 + r) ...(i) and V = e – Ir 1.8 = 2.2 – Ir ...(ii) Solving (i) and (ii), I = 0.36 A Substituting in (ii) 0.4 r= 0.36 10 r= W 9 21. (a)

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Here C 1, C 2 and C 3 are in series, hence their equivalent capacitance is C′ given by 1 1 1 1 = + + C′ C1 C2 C3 C′ =

12 mF 3

C′ = 4 mF The circuit can be redrawn as shown, above. Since C′ and C4 are in parallel \ Cnet = C′ + C4 = 4 mF + 12 mF = 16 mF (b) Since C′ and C 4 are in parallel, potential

difference across both of them is 500 V. \ Charge across C4 is Q4 = C4 × 500 C = 12 × 10–6 × 500 C = 6 mC Charge across C′, Q′ = C′ × 500 C = 4 × 10–6 × 500 C = 2 mC C1, C2, C3 are in series, charge across them is same, which is Q′ = 2 mC 22. (i) An astronomical telescope in normal adjustment. Ray diagram is shown below.

It is used to see distant objects. It consists of two lenses: n Objective of large aperture and large focal length fo n Eyepiece of small aperture and short focal length fe Working : A parallel beam of light from an astronomical object at infinity is made to fall on objective lens. It forms a real, inverted and diminished image AB of the object. In normal adjustment, AB lies at focus of the eye piece. So a highly magnified, erect image (w.r.t. AB) is formed at infinity. (ii) Here, power of objective lens = 1 D Power of eye piece = 10 D In normal adjustment

f P Magnifying power M = − o = − e fe Po ⇒ M = – 10 OR

(i) Compound microscope is used to see extremely small objects. It consists of two lenses.

n Objective lens of short aperture and short focal length fo

n Eye lens of large aperture and short focal length fe

Ray diagram of a compound microscope is shown below.

Working : A real, inverted and enlarged image A′B′ of a tiny object AB, is formed by objective. Eye lens is so adjusted that A′B′ lies between its optical centre and principle focus Fe. A virtual and magnified image A″B″ (erect w.r.t. A′B′ ) is formed by the eye lens. (ii) Both, the objective and the eye piece of a compound microscope should have short focal lengths to have greater magnifying power as magnifying power of a compound microscope is given by −L D M= 1 + f o fe where L = length of microscope tube D = least distance of distinct vision. 23. Given that distance between the two slits, d = 0.15 mm Wavelength of monochromatic light, l = 450 nm Distance between the screen and slits, D = 1 m (a) (i) Distance of nth bright fringe from central nlD maximum = d 450 × 10 −9 × 1 =2× [ n = 2 here ] 0.15 × 10 −3 = 6 × 10–3 m = 6 mm (ii) Distance of n th dark fringe from central maximum lD = (2n − 1) 2d 450 × 10 −9 × 1 = (2 × 2 − 1) × [ n = 2 here] 2 × 0.15 × 10 −3 3 = × 3 × 10 −3 2 = 4.5 mm (b) Since width of bright or dark fringes is given by lD b= , d physics for you | MaY ’10

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Thus when screen is moved away, D increases and hence fringe width increases. 24. Kirchhoff’s first law of electrical network or junction rule states that at any junction of electrical network, sum of incoming currents is equal to the sum of outgoing currents i.e.,

I1 + I2 + I4 = I3 + I5 Kirchhoff’s second law of electrical network or loop rule states that in any closed loop, the algebraic sum of the applied emf’s is equal to the algebraic sum of potential drops across the resistors of the loop i.e., ∑ e = ∑ IR To find I1, I2, I3 in following diagram.

For loop ABCFA E 1 + I 1r 1 – I 2r 2 – E 2 = 0 ⇒ 2 + 4I1 – 3I2 – 1 = 0 ⇒ 4I1 – 3I2 + 1 = 0 Using loop FCDEF E 2 + I 2r 2 + I 3r 3 – E 3 = 0 ⇒ 1 + 3I2 + 2I3 – 4 = 0 ⇒ 3I2 + 2I3 – 3 = 0 Also using junction rule I3 = I1 + I2 Using (ii) and (iii) 3I2 + 2I1 + 2I2 – 3 = 0 ⇒ 2I1 + 5I2 – 3 = 0 Solving (i) and (iv) 4I1 − 3I 2 + 1 = 0 − 2 × (2I1 + 5I 2 − 3) = 0 0 − 13I 2 + 7 = 0

7 ⇒ I2 = A 13 Substituting in (i) 7 4I 1 − 3 × +1=0 13 8 4I 1 = 13 2 I1 = A 13

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...(i)

...(ii) ...(iii)

⇒ I3 = I1 + I2 9 = A 13 32 P 25. (a) b– decay process of 15 32 32 0 P → S + e + u 15 16 −1 (b) The average or mean life of a radioactive substance is defined as the time for which the active nuclei of the atoms of the radioactive substance exit. In mean life Ta, both number of nuclei, N and rate of disintegration, R reduce to 1/e of their initial values. N i.e., when t = Ta then N = 0 e Using it in equation –lt N = N0e , we get N0 or = N0e −lTa e

or e −1 = e −lTa or 1 = lTa

or Ta =

Let two polaroids P1 and P3 are placed in crossed positions. Let P2 be the polaroid sheet placed between P1 and P3 making an angle q with pass axis of P1. If I1 = intensity of polarised light after passing through P1, then intensity of light after passing through P2 will be I2 = I1cos2q ...(i) π Now angle between P2 and P3 = − q 2

1 l so, finally mean life is reciprocal of decay constant l. 0.693 Also half life T1 / 2 = = 0.693 Ta l 26. When unpolarised light is made to pass through a polaroid, only those electric field vectors, parallel to its crystallographic axis emerge out of it. Thus the emerging light is plane polarised. Such a crystal is called polariser. If the emergent plane polarised light is passed through another crystal called analyser with its plane of transmission normal to that of polariser then no light emerges from it, as it is completely absorbed by the analyser.

...(iv)

[ P1 and P3 are in crossed position] \ Outcoming intensity after P3 is π I 3 = I 2 cos 2 − q 2 π I 3 = I1 cos 2 q ⋅ cos 2 − q 2 = I1cos2qsin2q

Working :

When a rectangular loop PQRS (suspended through a torison head) of sides ‘a’ and ‘b’ carrying current I is placed in uniform magnetic field B, such that area vector A makes an angle q with direction of magnetic field, then forces on the arms QR and SP of loop are equal, opposite and collinear, thereby perfectly cancel each other, whereas forces on the arms PQ and RS of loop are equal and opposite but not collinear, so they give rise to torque on the loop. or t = IABsin q [where A = ab] and if loop has N turns, then t = NIABsinq Due to this torque, the coil is deflected by an angle a, where it is balanced by restoring torque Ca, developed in suspension strip. C is restoring torque per unit deflection or torsional constant of the strip. NIAB = Ca C I= a NAB So, by measuring a, we can measure current I in the coil. (b) In order to make torque on the coil independent of angle q, the plane of coil should always remain parallel to the field. For this purpose a radial magnetic field is applied. (c) A galvanometer can be converted into a voltmeter by connecting high resistance in series with it, so that most of the voltage applied drops across it, enabling the galvanometer to measure much larger voltages.

[Using (i)]

2 1 = I1 sin 2q 2 If I0 = intensity of unpolarised light, then I I1 = 0 2 2 I0 1 ⇒ I 3 = sin 2q 2 2

(i) Maximum outcoming intensity is received π when q = 4 I0 1 2 I0 ⇒ I3 = = 2 2 8 π (ii) Minimum intensity, when q = 2 I3 = 0 27. As the image of the object is formed by the lens on the screen, therefore the image is real. Let the object is placed at a distance x from the lens. As the distance between the object and the screen is 90 cm. Therefore the distance of the image from the lens is (90 – x) According to new cartesian sign conventions, u = – x, v = + (90 – x) v Magnification m = u (90 − x) \ − 2 = −x x = 30 cm \ u = – 30 cm, v = 60 cm Let be f focal length of the lens. According to thin lens formula 1 1 1 − = v u f 1 1 1 − = 60 − (30) f 1 1 1 + = 60 30 f f = + 20 cm A convex lens of focal length 20 cm is required. 28. Principle : Galvanomter works on the principle that when an electric current is passed through a coil placed in a magnetic field, it experiences a torque, whose magnitude is proportional to the strength of electric current passed through it.

V − Rg Ig A galvanometer can be converted into an ammeter by connecting a low shunt resistance in parallel to or

R=

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(c)

it, so that most of the current by passes through the shunt resistance, enabling the galvanometer to measure much larger currents.

or

S=

I g Rg

I − Ig OR (a) When two parallel infinite straight wires carrying currents I1 and I2 are placed at distance r from each other, then current I1 produces magnetic field around it, which at any point on current I2 carrying wire is

m 0 I1 directed inwards perpendicular to plane 2πr of wires. So, current I2 carrying wire then experiences force due to this magnetic field which on its length l is given by m I m II F = I 2lB1 sin 90 = I 2 l × 0 1 or F = 0 1 2 l 2πr 2πr So, force per unit length that each wire exerts on the other is F m II f= = 012 l 2πr n If the direction of electric currents flowing in two parallel straight conductors is same, then they attract each other. n If the direction of electric currents flowing in two parallel straight conductors are opposite to each other, then they repel each other. (b) If I1 = I2 = 1 A and r = 1 m and l = 1 m m then f = 0 = 2 × 10 −7 N/m 2π Thus, electric current through each of two parallel long wires placed at distance of 1 m from each other is said to be 1 ampere, if they exert a force of 2 × 10–7 N/m on each other. This is definition of S.I. unit of current. B1 =

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VS = Vg or (I – Ig)S = IgRg

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Magnetic field due to I at P m I B = 0 into the plane of the paper. 2πd Expression for Lorentz magnetic force F = q(v × B) m I = e v × 0 n^ 2πd m 0 Iev F= away from the wire 2πd 29. Faraday’s law of electromagnetic induction states that whenever there is a change in the magnetic flux linked with a circuit an induced emf is set up in it, which lasts as long as the magnetic flux linked with it is changing and the magnitude of induced emf ‘e’ is directly proportional to the rate of change of magnetic flux linked with it i.e., df | e |∝ dt According to the data given in question, during the motion from x = 0 to x = b Initial flux = 0 Final flux = Blb Motional emf in the arm PQ, e = –Bvl e = − df dt During the motion from x = b to x = 2b, Flux remains constant which is f = Blb \ Motional emf, e = 0 Above values remain same, for the motion of arm PQ, from x = 2b to b. During motion from x = b to x = 0, initial flux f = Blb, final flux = 0 df \ Motional emf e = − = + Blv (direction dt reversed)

Variation of these quantities is shown in following graph.

OR A schematic diagram of step-up transformer is shown below.

Working Principle It works on the principle of mutual induction. It consists of two coils primary P and secondary S wound on a laminated soft iron core. The input voltage is applied across the primary coil and output voltage is obtained across the secondary coil. Magnetic flux fS and fP linked with secondary and primary coils at any instant are proportional to the number of turns NS and NP in secondary and primary coils i.e., fS NS df N −dfP = or − S = S fP N P dt N P dt NS eS NS or eS = e P or = NP eP NP NS where is called transformation ratio of NP transformer. In step up transformer eS > eP and NS > NP and transformation ratio > 1 In an ideal transformer, there is no loss of energy or it is 100% efficient. Then Power input = Power output e I or e P I P = eS IS ⇒ S = P e P IS where IS and IP are currents in secondary and primary coils of transformer. Transformer is mainly used in long distance transmission of electrical energy. At the electric power producing station, a step-up transformer is used which increases the alternating voltage upto several kilo volts, thereby decreasing the electric current flowing through transmission wires, As Joul’s heating is proportional to square of current, so this decreases the loss of electrical energy across transmission wires. Further a step-down transformer is used to decrease the alternating voltage at sub-station before distributing electrical energy for domestic use.

30. (a) Biasing of a pn junction diode: Forward Biasing: When an external voltage (V) is applied across the diode such that p-side is connected with positive or at a higher potential and n-side is connected with negative or at a lower potential, the diode is called “Forward Biased”

Reverse Biasing: When the external voltage (V) is applied across the diode such that p side is connected with negative or at a lower potential and n side is connected with positive or at a higher potential, the diode is called “Reverse Biased”.

By changing the biasing voltage in both circuits above, corresponding readings of voltmeter and ammeter are observed, and graphs are plotted between them for both the circuits. These graphs are known as VI characteristics.

(b) Light emitting diode (LED) is a junction diode made of gallium arsenide or indium phosphide in which when hole-electron pairs recombine at forward biased pn-junction, energy is released in the form of light. physics for you | MaY ’10

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Two advantages of LED over conventional incandescent lamps are: (i) In LED energy is produced in the form of light only, whereas in incandescent lamp energy is produced in the form of heat and light. Thus, there is no energy loss in LED. (ii) To operate LED, very small voltage (≈ 1 V) is required, whereas for the incandescent lamp higher voltages are required. OR (a) Circuit arrangement for studying the input and output characteristics of an npn transistor in CE configuration is shown below. Symbols have their usual meaning..

Output characteristics : A graph showing the variation of collector current IC with collectoremitter voltage V CE at constant base-current I B is called the output characteristic of the transistor.

Current amplification factor (b): This is defined as the ratio of the change in collector current to the change in base current at a constant collectoremitter voltage (VCE) when the transistor is in active state. DI b ac = C DI B VCE (b) Circuit diagram of an npn transistor connected across a tank circuit used to produce self-sustained oscillations is shown below.

Input characteristics : A graph showing the variation of base current IB with base-emitter voltage VBE at constant collector-emitter voltage VCE is called the input characteristic of the transistor.

Input resistance (ri): This is defined as the ratio of change in base-emitter voltage (DVBE) to the resulting change in base current (DIB) at constant collector-emitter voltage (VCE). This is dynamic (ac resistance) and as its value varies with the operating current in the transistor: DV 1 ri = BE = D I Slope of input characteristic B V CE

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physics for you | MAY ’10

When key K is closed, IC starts increasing in the circuit. The changing current through inductance L1 induces an emf across the inductor L, which further increase the base current IB and hence IE. The upper plate of the capacitor gets positively charged. As the current through inductor L 1 becomes steady, the mutual induction stops and induced emf across inductor L becomes zero and capacitor starts getting discharged through inductor L, thereby decreasing the base current IB and hence emitter current IE and collector current IC till IC becomes zero. Thus collector current oscillates between maximum and zero values. mm

Solved Paper - CBSE Class XII Physics 2010

Published on May 13, 2010

Solved Paper - CBSE Class XII Physics 2010

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