ALI RAZA KAMAL EXERCISE NOTES

Transcription

1 ALI RAZA KAMAL EXERCISE NOTES Physics 2 nd Year Ali Raza

2 Electrostatics 2 QUESTIONS 2.. The potential is constant throughout a given region of space. Is the electric field is zero or non-zero in this region? Ans The electric field will be zero in the given region of space. Reason The electric field intensity at a point can also be defined as negative potential gradient of that point i.e. E = - ( V r ) In other words, electric field intensity is the rate of change of potential with distance while negative sign shows the direction of E along the direction of decreasing potential. Since, potential V is constant throughout the region, so we can say; Therefore, V = 0 or E = - ( 0 r ) E = 0 Hence, we can say that if potential is constant throughout the given region of space then electric field will be zero in this region Suppose that you follow an electric field line due to a positive point charge. Do electric field and the potential increases or decreases? Ans Both the electric field and the potential will decrease. Reason If we follow an electric field line due to a positive point charge, then it means that we are moving away from it as shown in the figure. Thus, the distance from the charge increases. Or E = k q r 2 Or V = k q r E r 2 V r As it is clear from above relations that both electric field and potential will decrease due to increase of distance from positive charge.

3 2.3. How can you identify that which plate of a capacitor is positively charged? Ans A device called Gold Leaf Electroscope (GLE) detects the presence of charge on a body. Identification: As we know, the disc of gold leaf electroscope is positively charged. There may be following two cases: Case I: When we touch the disc with any plate of capacitor and the divergence of the gold leaves decreases, then it means the plate of capacitor is negatively charged. Case II: When we touch the disc with any plate of capacitor and the divergence of the gold leaves increases, then it means the plate of capacitor is negatively charged Describe the force or forces on a positive point charge when placed between parallel plates; (a) with similar and equal charges. (b) with opposite and equal charges. Ans We placed the positive charge between the plates and deduced following results; (a) Similar and Equal Charges When a positive charge is placed between parallel plates with similar and equal charges, then the net force on the charge will be zero. Thus, the value of resultant electric intensity E is zero because the electric intensity E due to one plate is equal in magnitude but opposite in direction of electric intensity E 2 due to other plate i.e. E = E E 2 = 0 Hence, the net force on the positive charge is zero. Thus, it will remain at rest (b) Opposite and Equal Charges When a positive charge is placed between parallel plates with opposite and equal charges, then the value of resultant electric intensity E will be; E = E + E 2 It is because the electric intensity E due to one plate is equal in magnitude but in same direction of electric intensity E 2 due to other plate. Hence, the net force on the positive charge will be from positive to negative plate. Thus, it will be accelerated towards negative plate Electric lines of forces never cross each other. Why? Ans Electric lines of forces never cross each other. Electric field lines reveal information about the direction and the strength of an electric field within a region of space. If the lines cross each other at a given location, then there must be two distinctly different values of electric field with their own individual direction at that given location. This could never be the case. Therefore, the lines representing the field cannot cross each other at any given location in space. Concisely, if they did cross, they would represent a location with two different strong electric field vectors, something that cannot exist in nature.

4 2.6. If a point charge q of mass m is released in a non-uniform electric field with field lines pointing in the same direction, will it make a rectilinear motion? Ans Yes, it will make a rectilinear motion. Rectilinear motion is defined as a motion along a straight line. If a point charge q of mass m is placed in a non-uniform electric field, it moves along the field lines. However, in the present case, the field lines are pointing in the same direction i.e. radially outward due to positive point charge +q, it will experience a repulsive force. Hence, the charge will make rectilinear motion along the field lines outward due to force of repulsion Is E necessarily zero inside a charged rubber balloon if balloon is spherical? Assume that charge is distributed uniformly over the surface? Ans Yes, E is necessarily zero inside a charged rubber balloon, if balloon is spherical. Proof: Imagine a Gaussian surface inside the balloon. Then it does not enclose any charge i.e. q = 0. Applying Gauss s law, we have ϕ e = q ε o Therefore, ϕ e = 0 Since, ϕ e = E. A or E. A = 0 Then we can say, here, (i) A 0 (ii) E = 0 Hence, electric field will be zero inside a spherical balloon. Gaussian Surface 2.8. Is it true that Gauss s law states that the total number of lines of forces crossing any closed surface in the outward direction is proportional to the net positive charge enclosed within surface? Ans Yes, it is true that Gauss s law states that the total number of lines of forces crossing any closed surface in the outward direction is proportional to the net positive charge enclosed within surface. Proof: According to the Gauss s law, the flux i.e. the number of electric field lines through a closed surface, is /ε o times the net charge enclosed in it. Flux = x total charge enclosed ε o or Flux = constant x total charge enclosed or Flux α total charge enclosed It follows that the total number of electric field lines through a closed surface is directly proportional to the net positive charge enclosed in it Do electrons tend to go to region of high potential or of low potential? Ans Electrons tend to go to the region of high potential. The electrons are negatively charged particles. When they enter the electric field, they tend to go from a region of low potential i.e. negative end to the region of high potential i.e. positive end. But in case of conventional current, positive charge flows from higher to lower potential regions.

5 PROBLEMS Problem 2. F e = [. q q 2 F g 4πε o r 2 ] [G m m 2 r 2 ] Problem 2.2 F e =. q q 2 4πε o r 2 Problem 2.3 E = Problem 2.4 E = Problem 2.5 4πε o. q r 2 4πε o. q r 2 r E =. q 4πε o r 2 Problem 2.6 E = F q = W q = mg q Problem 2.7 K. E. = q V Problem 2.8 m = ρ 4 3 πr3 q = mgd V Problem 2.9 V = Ed W = q V U = (Work done) K. E. = q V K. E. = 2 mv2 Problem 2.0 V =. q 4πε o r Problem 2. V =. q 4πε o r =. e 4πε o r Total energy = K. E. +P. E. = ( 2 mv2 ) + ( Ve) E i = E E ground Problem 2.2 E = 2 CV2 Problem 2.3 Q = CV n = Q e

6 Current Electricity 3 QUESTIONS 3.. A potential difference is applied across the ends of a copper wire. What is the effect on the drift velocity of free electrons by: (i) increasing the potential difference. (ii) decreasing the length and the temperature of the wire. Ans A small uniform velocity that an electron gains in the presence of an electric field of the battery is called the drift velocity in the direction of E. Case I: The drift velocity of an electron increases with the increase in the potential difference. As we know, the drift velocity of free electrons in conductor of length l and area of cross section A is given by, I v d = nea We also know, I = V/R V v d = near This equation shows that by increasing potential difference, I increases and hence v d increases. Case II: The drift velocity of an electron increases by decreasing the length and the temperature of the wire. As we know, the drift velocity of free electrons in conductor of length l and area of cross section A is given by, I v d = nea We also know, I = V/R V v d = near We also know, A = πr 2 l V v d = neπr 2 lr Here, R α l and R α T Hence, by decreasing length and temperature of wire, R decreases due to which I increase. Therefore, v d increases Do bends in a wire affect its electrical resistance? Explain. Ans No, bends in the wire do not affect its electrical resistance. As we know, the resistance R of wire is given by R = ρ L A As it is obvious from the equation, that resistance of a wire depends upon length, area and nature of the wire. Therefore, the resistance R remains constant.

7 3.3. What are the resistances of the resistors given in the figures A and B? What is the tolerance of each? Explain what is meant by the tolerance? Ans Definition of tolerance: The possible variation in the resistance of a carbon resistor from the marked value is called tolerance. For example, a 000 Ω resistor with the tolerance ± 0% will have an actual value anywhere between 900 Ω and 00 Ω. Resistances of given Resistors Resistor A: Brown Green Red Gold According to colour code, Brown has value, Green has value 5, and Red has value 2. Gold band shows tolerance equal to ± 5% Therefore, R = 500 Ω (± 5%) Resistor B: According to colour code, Yellow White Orange Silver Yellow has value 4, White has value 9, and Orange has value 3. Silver band shows tolerance equal to ± 0% Therefore, R = Ω (± 0%) 3.4. Why does the resistance of a conductor rise with temperature? Ans The resistance of a conductor rise with temperature. As the temperature of the conductor rises; The amplitude of vibration of the atoms in the lattice increases. The collision of the atoms increases. The atoms offer a bigger target. The probability of their collision with the free electrons also increases. Hence, resistance of conductor increases What are the difficulties in testing whether the filament of lighted bulb obeys Ohm s law? Ans The main difficulty is the rise in temperature of filament with increase in current. Ohm s law states that the potential difference is proportional to the current provided, when there is no change in the physical state of the conductor. The physical state means the density, elasticity, temperature etc. of conductor. As the temperature increases, resistance of the filament goes on increasing according to the relation. R t = R o ( + αt) Where α is the temperature co-efficient and t is the rise of temperature. It means that temperature of the lighted bulb does not remain constant, so its filament does not obey Ohm s law Is the filament resistance lower or higher in a 500 W, 220 V light bulb than in 00 W, 220 V bulb? Ans The resistance of filament is lower in a 500 W, 220 V light bulb. Bulb with 500 W, 220 V: As P = 500 W V = 220 V R =? Applying the formula for power, 2 V P = R or R = V 2 = (220)2 P 500 = 96.8 Ω (i)

8 Bulb with 00 W, 220 V: As P 2 = 00 W V 2 = 220 V R 2 =? Applying the formula for power, or R 2 = V 2 2 P 2 = = (220)2 P 2 00 V 2 2 R 2 Result: Comparing equation (i) and (ii), we come to know R 2 > R. = 484 Ω (ii) 3.7. Describe a circuit, which will give a continuously varying potential? Ans A potentiometer can be used to give a continuously varying potential. Explanation: Consider a resistance R in the form of a wire on which a terminal C can slide as shown in figure. The resistance between A and C can be varied from 0 to R, as C slides from A to B. R A r C B If we connect a battery of emf E across a resistance R as shown in figure, the current flowing through it is, I = E R If we represent the resistance between A and C by r, the potential drop between these points will be, V = I r Putting above equation in it, V = E R r Thus, as C slides from A to B, r varies from 0 to R, and the potential drop between A and C changes from zero to E. this arrangement by which potential can be varied continuously from 0 to E is known as a potential divider. E R A r B C G 3.8. Explain why the terminal potential difference of a battery decreases when a current drawn from it is increased? Ans The terminal potential difference of a battery decreases when a current drawn from it is increased. As, we already know a relation for terminal potential difference of a battery, that is I R = E Ir or V t = E Ir Here, E = emf of battery r = internal resistance of battery Ir = potential difference across internal resistance This equation shows that when I is increased, the factor Ir becomes large and V t becomes small. Thus, potential difference of a battery decreases when the current drawn from it is increased.

9 3.9. What is the Wheatstone bridge? How can it be used to determine an unknown resistance? Ans Wheatstone bridge is the electrical circuit, which provides an accurate method for the measurement of an unknown resistance. Explanation: The circuit of Wheatstone bridge is shown in the figure. It consists of four resistances R, R 2, R 3 and R 4 connected in such a way as to form loop or mesh ABCDA as in figure. A battery of emf E is connected between points A and C. A galvanometer of resistance R g is connected between the points B and D. When the bridge is balanced, it satisfied the following relation R = R 3 R 2 R 4 or R 4 = R 2 R R 3 If the values of R, R 2 and R 3 are known then R 4 can be calculated by using equation provide the bridge is balanced. It means that terminals B and D are at same potential so the galvanometer shows no deflection. B R R2, A I I C R3 S G D E R4 Problem 3. I = Q t = ne t Problem 3.2 I = Q t Problem 3.3 ; n = It e PROBLEMS R e = R + R 3 = ( R + R 2 ) + R 3 I = V R e I = V AB R I 2 = V AB R 2 I 3 = I Problem 3.4 R = ρ L A Problem 3.5 = R t R o R o t Problem 3.6 V t = E Ir Problem 3.7 Applying Kirchoff s Second Rule on both loops Problem 3.8 Applying Kirchoff s Second Rule P = I 2 R

10 Electromagnetism 4 QUESTIONS 4.. A plane-conducting loop is located in a uniform magnetic field that is directed along the x-axis. For what orientation of the loop, is the flux a maximum? For what orientation, is the flux a minimum? Ans As we know, flux through any surface is expressed as; ϕ B = B. A = BA cos θ (a) Maximum Flux: When the plane of loop is held perpendicular to direction of B, flux is maximum. In other words, when the loop is in yz-plane, flux is maximum. As, in the present case, θ = 0 So, ϕ B = B. A = BA cos θ = BA cos 0 = BA () = BA cos 0 = Hence, in this case flux is maximum. (b) Minimum Flux: When the plane of loop is held parallel to direction of B, flux is maximum. In other words, when the loop is in xy-plane or xz-plane, flux is minimum. As, in the present case, θ = 90 So, ϕ B = B. A = BA cos θ = BA cos 90 = BA (0) = 0 cos 90 = 0 Hence, in this case flux is minimum A current in a conductor produces a magnetic field, which can be calculated using Ampere s law. Since current is defined as the rate of flow of charge, what can you conclude about the magnetic field due to stationary charges? What about moving charges? Ans As we know, Ampere s law is given as; B. L = μ o I (a) Stationary Charges: Magnetic field due to stationary charges is zero. As, in the case of stationary charges, rate of flow of charges is zero i.e. current I us zero. So, from Ampere s law, B. L = μ o I = μ o (0) = 0 Here, L 0 B = 0 Hence, stationary charges do not produce magnetic field i.e. they produce electric field only. (b) Moving Charges: Moving charges produce magnetic field. In case of moving charges, there is rate of flow of charge. Therefore, moving charges produce current. Flow of current produces magnetic field field in the surrounding space. Hence, moving charges produce magnetic field along their path of motion, which can be calculated by using Ampere s law.

11 4.3. Describe the change in the magnetic field inside a solenoid carrying a steady current I, if (a) the length of the solenoid is doubled but the number of turns remains same and (b) the number of turns is doubled, but the length remains same. Ans As we know, expression for magnetic field due to solenoid is; B = μ o ni As n = N l N B = μ o l I (i) (a) Length of the solenoid is doubled but number of turns remains same: In this case, magnetic field is reduced to half. Proof: When the length of solenoid is doubled and the number of returns same then l = 2l N = N Then, from (i), B ' N = μ o N l I) = 2 (B) 2l I = 2 (μ o In this, strength of magnetic field becomes half. (b) Number of turns is doubled but length of the solenoid remains same: In this case, magnetic field is increased to double. Proof: When the number of returns is doubled and the length of solenoid same then l = l N = 2N Then, from (i), B ' 2N = μ o l I = 2 (μ N I) = 2(B) o l In this, strength of magnetic field is doubled At a given instant, a proton moves in the positive x direction in a region where there is magnetic field in the negative z direction. What is the direction of the magnetic force? Will the proton continue to move in the positive x direction? Explain. Ans (a) Direction of Magnetic Field: Magnetic field is directed along positive y-direction. As, force on proton in magnetic field is, F = +e ( v B ) Given proton is moving along positive x-direction so, v = v î Magnetic field is in the negative z-direction so, B = B(-k ) Then, from (i), F = +e {(v î) B(-k )} F = -evb ( î k ) F = -evb (- ĵ ) F = evb ĵ (i) ( î k ) =- ĵ Hence, magnetic force on proton will be directed in positive y-direction. (b) Movement in positive x-direction: No, the proton will not continue to move in the positive x direction but it moves in a circle. Since, the magnetic force is at right angle to the motion of the proton, therefore it will move along a circular path in xy plane. Thus, the magnetic force (F = qvb) provides the centripetal force. z

12 4.5. Two charged particles are projected into a region where there is a magnetic field perpendicular to their velocities. If the chargers are deflected in opposite directions, what can you say about them? Ans Two charged particles are projected into a region where there is a magnetic field perpendicular to their velocities. If the chargers are deflected in opposite directions, there are two possibilities; Possibility I: Then one charge is proton and other charge is electron and they are projected from same direction perpendicular to magnetic field then they will be deflected in opposite directions. Explanation: Opposite charges projected from same direction. Let B is along y-axis then of B = Bĵ Charges are moving along x-axis i.e. v = vî Then, for proton F = +e ( v B ) F = +e (vî Bĵ) F = evb ( î ĵ) F = evb (k ) ( î ĵ) = k Similarly, for electron F = -e ( v B ) F = -e (vî Bĵ) F = -evb ( î ĵ) F = -evb (k ) ( î ĵ) = k Force on electron is along positive z-axis while force on electron is in negative z-axis. Therefore, electron and proton when projected from same direction perpendicular to magnetic field, they are deflected in opposite directions. Possibility II: When both charges are protons or both charges are electrons and they projected from opposite directions perpendicular to magnetic field then they will be deflected in opposite direction. Explanation: Same charges projected from opposite directions. When both are protons one enters into magnetic field B = Bĵ from positive x-direction so v = v(-î) and other proton enters into magnetic field from negative x-direction so v = vî Then, force on proton from positive x-direction is, F = +e ( v B ) F = +e {v(- î) Bĵ} F = +evb (- î ĵ) F = +evb (- k ) ( - î ĵ) =- k Similarly, force on proton from negative x-direction is, F = +e ( v B ) F = +e (vî Bĵ) F = evb ( î ĵ) F = evb (k ) ( î ĵ) = k Therefore, one proton is deflected along positive z-axis while other along negative z-axis. Similarly same explanation for electron can be provided Suppose that a charge q is moving in a uniform magnetic field with a velocity v. Why is there no work done by the magnetic force that acts on the charge q? Ans If a charge q is moving in a uniform magnetic field with a velocity v, then there is no work done by the magnetic force that acts on the charge q. The magnetic force acting on a charged particle q is given by, F = q ( v B )

13 v F 90 Magnetic force is perpendicular to velocity. Displacement is always directed along the direction of velocity hence we can say that direction of velocity and displacement are same. Hence, the angle between F and d is 90. In this case, work done by magnetic force is zero i.e W = F. d = Fd cos θ = Fd cos 90 = Fd (0) = 0 cos 90 =0 We can say that, circle is a closed path and displacement in a closed path is always zero so W = If a charged particle moves in a straight line through some region of space, can you say that the magnetic field in the region is zero? Ans When magnetic force acting on particle is zero it follows the straight path. Following possibilities may arise; Possibility I: F = q ( v B ) = qvb sin θ = qv(0) sin θ = 0 When in some region of space, magnetic field is zero then charge moves in a straight path. Possibility II: When charged particle moves in the direction of magnetic field then angle between v and B is 0, then, F = q ( v B ) = qvb sin θ = qvb sin 0 = 0 sin 0 =0 Thus, particle moves in a straight path. When charged particle moves in the direction opposite of magnetic field then angle between v and B is 80, then, F = q ( v B ) = qvb sin θ = qvb sin 80 = 0 sin 80 =0 In this case, the charged particle moves in a straight path. Possibility III: When charge particle enters into a region where magnetic field and electric field are simultaneously acting perpendicular to each other, then they cancel the effect of each other and particle moves in straight path. For instance, motion of charge particle in velocity selector Why does the picture on a TV screen become distorted when a magnet is brought near the screen? Ans The picture on a TV screen becomes distorted, when a magnet is brought near the screen. The beam of electrons forms the picture on TV screen. When the magnet is near brought the TV screen, the magnetic field of magnet interacts with the magnetic field around the beam of electrons. So a magnetic force acts on the electrons, which is given by F = q ( v B ) Due to this force, electrons beam is deflected. Hence, the picture is distorted Is it possible to orient a current loop in a uniform magnetic field such that the loop will not tend to rotate? Explain. Ans Yes, it possible to orient a current loop in a uniform magnetic field such that the loop will not tend to rotate. Explanation: When a current carrying loop is placed in magnetic field, it experiences a torque which is given by, τ = BINA cos α

14 where α is the angle between B and plane of the loop. When plane of the loop is placed at right angle i.e. perpendicular to magnetic field B, then α = 90. τ = BINA cos 90 = BINA (0) = 0 cos 90 =0 As the value of torque is zero in uniform magnetic field so loop will not tend to rotate How can a current loop be used to determine the presence of a magnetic field in a given region of space? Ans When a current carrying loop is placed in uniform magnetic field, a torque is produced in a loop due to the current. If the loop is deflected in a given region of space, then it confirms the presence of magnetic field otherwise not. If magnetic field B makes an angle α with the plane of loop, the magnitude of the torque is given by, τ = BINA cos α 4.. How can you use a magnetic field to separate isotopes of chemical element? Ans An apparatus called Aston Mass Spectrograph uses magnetic field to separate the isotopes of chemical element. In such apparatus, when isotopes of certain element are projected perpendicular to magnetic field then they follow circular path due to centripetal force provided by magnetic force. For circular motion, Magnetic Force = Centripetal Force mv 2 qvb sin 90 = r mv 2 qvb = sin 90 = r mv r = qb As isotope of an element, have same charge q but different mass m so, for constant q, v and B. r m Isotope with greater mass will move in a circle of large radius and isotope having less mass will move in circle of small radius. Thus, isotopes in magnetic field can be separated on the basis of their different masses and radii What should be the orientation of a current carrying coil in a magnetic field so that torque acting upon the coil is (a) maximum (b) minimum? Ans Torque acting on current carrying coil in a magnetic field is τ = BINA cos α (i) (a) Maximum When the plane of current carrying loop is held parallel to the field, torque will be maximum. Proof: As we know, when plane of loop is parallel to magnetic field then α = 0. Hence, from (i) τ max = BINA cos 0 = BINA cos 0 = (b) Minimum When the plane of current carrying loop is held perpendicular to the field, torque will be minimum. Proof: As we know, when plane of loop is perpendicular to magnetic field then α = 90. Hence, from (i) τ min = BINA cos 90 = 0 cos 90 = A loop of wire is suspended between the poles of a magnet with its plane parallel to the pole faces. What happened if a direct current is put through the coil? What happens if an alternating current is used instead? Ans Whenever a current carrying loop of wire is placed in a magnetic field it tends to orient itself so that maximum flux is passes through it. When plane of loop is parallel to the poles of the magnet then it will not rotate and τ = 0 because already maximum flux is passing through it. Plane of loop makes an angle α = 90 with B so,

15 τ = BINA cos 90 = 0 cos 90 =0 Therefore, no torque is produced whether DC is passed through loop or AC is passed through it. B N S 4.4. Why the resistance of an ammeter should be very low? Ans The resistance of an ammeter should be very low. Ammeter is a low resistance device. It is always connected in series so that it does not produce any change in current or resistance in the circuit. If its resistance is not small it will decrease the circuit current. Under such condition, current measured will not be accurate Why the voltmeter should have a very high resistance? Ans The voltmeter should have a very high resistance. Voltmeter is high resistance device. It is always connected in parallel in circuit to measure voltage in volts between two points. Its resistance is kept large as compared to the resistance of circuit so that circuit current does not change and accurate potential difference between two points can be measure. If the resistance of voltmeter is not very large then current will flow through it. As a result, circuit current will change. Hence, potential difference to be measured will decrease. Under such condition, potential difference measured will not be accurate. Problem 4. Problem 4.2 Problem 4.3 PROBLEMS F = ILB sin α ; F max = ILB ; B = F max IL F = q (v B ) = qvb sin θ ; v = E = Bv ; Problem 4.4 v = E B τ = BINA cos α ; τ max = BINA Problem 4.5 B = μ oi 2πr Problem 4.6 B = μ o ni Problem 4.7 B = μ o ni = μ oni L Problem 4.8 R s = I gr g I I g Problem 4.9 R h = V I g R g Problem 4.0 R s = I gr g I I g where I = I, I 2, I 3 F qb sin θ = mg qb sin θ

16 Electromagnetic Induction 5 QUESTIONS 5.. Does the induced emf in a circuit depend on the resistance of the circuit? Does the induced current depend on the resistance of the circuit? Ans (a) Dependence of Induced emf: No, the induced emf in a circuit does not depend on the resistance of the circuit. Since by Faraday s law ε = -N φ = -N B.A t t Thus, induced emf depends upon, (i) Number of turns of coil (ii) Rate of change of flux ( B.A) through the coil where increase in flux increases induced emf ε while decrease in flux decreases induced emf ε. (b) Dependence of Induced Current: Yes, the induced current in a circuit depends on the resistance of the circuit. Product of current I and resistance R is always constant called induced emf ε i.e. I R = ε ε I = R Induced current depends upon resistance. If resistance increase, current will decrease while decrease in resistance causes increase in current A square loop of wire is moving through a uniform magnetic field. The normal to the loop is oriented parallel to magnetic field. Is an emf induced in the loop? Give reasons. Ans emf is not induced in the loop. Reason I: Since by Faraday s law ε = -N φ t Constant number of magnetic lines of force are passing through square loop so there is no change in magnetic flux with time. Thus, induced emf depends upon i.e. φ = 0 t Then from Faraday s law, ε = -N(0) ε = 0 Therefore, induced emf in the loop is zero. Reason II: As normal, to loop is zero. So velocity of loop is parallel to B. Angle between B Hence, ε = -vbl sin 0 = 0 sin 0 = 0 A A B B and B is 0.

17 5.3. A light metallic ring is released from above in to a vertical bar magnet (in the fig). Viewed for above, does the current flow clockwise or anticlockwise in the ring? Ans The current flows in clockwise direction in the ring. When metallic ring is released from above into a vertical bar magnet, magnetic flux in the ring is changed. (i) According to Faraday s law, change in flux produces an induced emf and hence an induced current in the ring. Induced emf produces magnetic field. (ii) According to Lenz s law, the direction of induced current is opposite to the cause, which produces it. Therefore, the side of ring, which is toward N-pole of bar magnet, must be N-pole of induced magnetic field. (iii) Right hand rule shows that this type of induced magnetic field is produced when side of ring facing N-pole of bar magnet have current in clockwise direction. When ring is viewed from above, current is in clockwise direction What is the direction of the current through resistor R in the fig? When switch S is; (a) closed (b) opened Ans (a) Switch is Closed: When switch S is closed, current in the circuit increases from zero to maximum, which produced changing magnetic flux. During this interval, magnetic flux linked with secondary coil also changes and a n induced current is produced in it. Magnetic field produced by induced current will be opposite to magnetic filed in first coil, when current flows in anticlockwise direction in R. in this way, by Lenz s law it opposes the cause, which produces it. (b) Switch is Opened: When switch S is opened, the current in first circuit decreases from maximum to zero, flux linked with second coil decreases and an induced current is produced but opposite to first direction i.e. current in R flows in clockwise direction Does the induced emf always act to decrease the magnetic flux through a circuit? Ans No, induced emf does not always act to decrease the magnetic flux through the circuit. Explanation: Since, emf is produced by change of magnetic flux. When magnetic flux in circuit decreases then (i) By Faraday s law, induced emf is produced. (ii) By Lenz s law, induced emf opposes the decrease in magnetic flux. Therefore, induced emf tries to increase magnetic flux in this case When the switch in the circuit is closed a current established in the coil and the metal ring jumps upward (see the fig). Why? Describe what would happen to the ring if battery polarity were reversed? Ans When the switch in the circuit is closed a current established in the coil and the metal ring jumps upward. When polarity of battery is reversed, same effect is observed. Explanation: When switch S is closed, magnetic field is produced around the coil, which changes from zero to maximum. Magnetic flux of coil linking with ring also changes and an induced current is produced in the ring. This induced current produces induced magnetic field, which is opposite to the magnetic field of coil. The end of ring which faces N-pole of coil becomes N-pole. Similar poles produce repulsion. Hence, metal ring jumps upward momentarily. When current in the soil becomes constant, there is no longer change in magnetic flux so induced current lost and therefore induced magnetic field in ring lost and ring comes back to its original position. Hence, when polarity of battery is reversed, same effect is observed.

18 5.7. The figure shows a coil of wire in the xy plane with a magnetic field directed along the y- axis. Around which of the three coordinate axes should the coil be rotated in order to generate an emf and a current in the coil? Ans When the coil is rotated about x-axis, an emf and current is induced in the coil. (i) When the coil of wire is rotated along x-axis then in this case, coil of wire will cut magnetic field lines and change of magnetic flux occur. This change of magnetic flux will produce induced emf and hence induced current in the coil of wire. (ii) If the coil is rotated about y-axis, even then the flux passing through the coil remains zero because plane of the coil is at all times parallel to the lines of magnetic field B. (iii) If coil of wire is rotated along z-axis, it will not cut any magnetic field lines because in this case plane of coil will be parallel to direction of B. Therefore, no change of magnetic flux will produce and hence no induced emf and induced current How would you position a flat loop of wire in a changing magnetic field so that there is no emf induced in the loop? Ans If a flat loop of wire is placed parallel to the changing magnetic field, there will be no emf induced in the loop. When flat loop of wire is placed in such a way that its plane is parallel to B then direction of vector area A makes an angle of 90 with B. Therefore, magnetic flux is ϕ = B. A = B A cos 90 = 0 cos 90 = 0 According to Faraday s law, Here, as ε = -N φ t A A B B φ = 0 Then from Faraday s law, ε = -N 0 t Hence, no emf is induced in loop of wire. ε = In a certain region the earth s magnetic field point, vertically down. When a plane flies due north, which wingtip is positively charged? Ans West wingtip is positively charged. Magnetic force on electron is F = -e ( v B ) Direction of force on electron tells us which wingtip is positively charged. Consider right hand rule, Let forefinger gives the direction of plane toward north, middle finger points in the direction of field B then thumb points in the direction of force on electron present the wing of plane. As thumb point towards west. So, left wingtip towards west is positively charged. Plane is a conductor it contains free electrons. West North x x x x x x x x x x x x x x x x x x x x x x v 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 x x x x x x x x x x x x x x x x x x x South East

19 5.0. Show that ε and Φ have the same units. t Ans As we know, Then, unit of ε is given as, Unit of ε = ε = W q Unit of Work Unit of Charge = J C = Joule Coulomb = Volt (i) Similarly, we know Φ t = B A t Then, unit of Φ is given as, t Unit of Φ = Unit of ( F t IL A t ) = N A.m m2 s Comparing (i) and (ii), = F IL A t = N.m Cs -.s = Nm C = J C = Joule Coulomb = Volt (ii) Unit of ε = Unit of Φ t = J C = Joule Coulomb = Volt 5.. When an electric motor, such as an electric drill, is being used, does it also act as a generator? If so what is the consequences of this? Ans Yes, it acts like a generator. When motor is running then coil of motor called armature rotates in the magnetic field. Due to rotation of armature, change of magnetic flux takes place and on emf is induced which is called back emf. Back emf opposes the rotation of armature. So running motor acts like generator. Consequences: In the absence of load, motor rotates quite fast. Magnitude of back emf increases with the speed of motor and so current flowing through motor becomes small. When motor derives load then back emf decreases due to decrease in speed of motor. This allows motor to draw more current. If motor is overloaded beyond its limit, the current rises to very high value and motor burns out Can a D.C. motor be turned into a D.C. generator? What changes required to be done? Ans Yes, a D.C. motor can be turned into a D.C. generator. Explanation: When electric current is passed through motor. Armature of motor starts rotating in the magnetic field. So magnetic field passing through armature produces change in magnetic flux which produces an induced emf called back emf. Thus electric motor acts like generator because due to induced emf, induced current will flow as output. Two changes are required to convert D.C. motor into D.C. generator. (i) The magnetic field must be provided by a permanent magnet and not by an electromagnet. (ii) As back emf produced opposes the motion of armature. So an arrangement should be made to rotate the armature Is it possible to change both the area of the loop and the magnetic field passing through the loop and still not have an induced emf in the loop? Ans Yes, it is possible to change both the area of the loop and the magnetic field passing through the loop and still not have an induced emf in the loop. Explanation: Here θ = 0 Then the change of magnetic flux through a loop of area A placed parallel to magnetic field B is given by,

20 ϕ = B. A = B A cos θ = B A cos 0 = B A cos 0 = (i) It shows that B and A are inversely proportional to each other. If we change B and A in such a way that the product B A remains constant, thus the change in magnetic flux through the loop will be zero, i.e. φ = 0 As we know that, ε = -N φ t Then from Faraday s law, ε = -N 0 t ε = 0 (ii) Therefore, the induced emf is zero. Hence, no induced emf in the loop will be produced Can an electric motor be used to drive an electric generator with the output from the generator being used to operate the motor? Ans No, an electric motor cannot be used to drive an electric generator with the output from the generator being used to operate the motor. It is not possible because when motor derives generator, output of generator is less than the input, the reason for this is the dissipation of energy due to frictional losses. If the output of generator is feedback to the motor, it will be insufficient to operate the motor. If it becomes possible then combination of motor and generator will be a self operating system, which does not get any energy from outside source. In this case, it also violates law of conservation of energy A suspended magnet is oscillating freely in horizontal plane. Oscillations are strongly damped when a metal plate is placed under the magnet. Explain why this occurs? Ans A suspended magnet is oscillating freely in horizontal plane. Oscillations are strongly damped when a metal plate is placed under the magnet. If suspended magnet oscillates freely in horizontal plane and a metal plate is placed under magnet then due to oscillations of magnet, magnetic flux passing through the metallic plate changes. The change of magnetic flux produces induced emf in the plate and hence, an induced current. This induced current according to Lenz s law opposes the oscillatory motion of bar magnet by producing its own magnetic field. Magnetic field due to induced current decreases the oscillation of bar magnet. In this way, oscillatory motion of bar magnet is damped Four unmarked wires emerge from a transformer. What steps would you take to determine the turns ratio? Ans Four unmarked wires emerge from a transformer. We would take following steps to determine the turns ratio. Step I: First, the continuity test is made to identify the two ends of same wire. Step I: Use ohmmeter to find their electrical resistance. For step down transformer (say) the greater resistance coil is secondary, then other coil will be primary. Step III: Supply known voltage of AC to primary coil. Suppose it is V p. Measure voltage V s across secondary with the help of voltmeter. Step IV: Finally, turns ratio can be found by using following relation, N s N p = V s V p

21 5.7. (a) Can a step-up transformer increase the power level? (b) In a transformer, there is no transfer of charge from the primary to the secondary. How is, then the power transferred? Ans (a) No, it cannot increase the power level. A step-up transformer can only increase the voltage level but it cannot increase the power level. In actual transformer, dissipation of energy in the coil takes place due to Eddy currents so output power is always less than input power i.e. P s < P p Ans (b) In a transformer, there is no transfer of charge from the primary to the secondary. Then the power is transferred magnetically. There is no electrical connection between the primary and secondary coils of transformer, they are magnetically linked with each other i.e. electrical power or energy is transferred from primary to secondary coil with the help of magnetic field. When some input is given to the primary coil, the change in magnetic flux of primary coil occurs. This flux is linked with secondary coil. Therefore, change in magnetic flux of secondary coil produces emf. Thus, change in magnetic flux of secondary coil transfers power from primary to the secondary coil When the primary of a transformer is connected to AC mains the current in it (a) is very small if the secondary circuit is open, but (b) increases when the secondary circuit is closed. Explain these facts. Ans (a) When the primary of a transformer is connected to AC mains the current in it is very small if the secondary circuit is open. If the secondary circuit is open then it means there is no load on the transformer so I s = 0. As we know, in a transformer, Power input = Power output V pi p = V si s Then, according to given condition, V pi p = V s(0) V pi p = 0 Here, V p 0 But I p = 0. In short we can say that, output power P s = V si s is zero. Thus, primary draws very small current from AC mains. Ans (b) When the primary of a transformer is connected to AC mains, the current in it increases when the secondary circuit is closed. If secondary circuit is closed then it means transformer is provided with load, so output power will be increases. As for transformer, Power input = Power output V pi p = V si s Therefore, transformer will draw large current from AC mains to increase the primary power. Hence, greater current is needed in primary to equalize power in secondary coil. Problem 5. ε ε 2 = B B 2 ; B 2 = B ε 2 ε PROBLEMS

22 Problem 5.2 Angle with horizontal = 60 Angle with vertical = θ = = 30 ε = Bvl sin θ Problem 5.3 ε = N φ t Problem 5.4 Problem 5.5 = N B. A t Angle between plane of coil and magnetic field = θ = 40 Angle between vector area and magnetic field = θ = = 50 ε = N φ t = N (φ f φ i ) = N (B f. Acosθ B i. Acosθ ) = N (B f B i )Acosθ t t t ε s = M I p t ; M = ε s I p / t Problem 5.6 ε s = M I p t ε s = N s φ t ; φ = ε s t N s Problem 5.7 L = Nφ I ; φ = LI N ε L = L I t Problem 5.8 L = μ o n 2 la = μ on 2 la l 2 = μ on 2 A l n = N l ε L = L I t Problem 5.9 ε L = L I t U m = 2 L I2 = 2 L(I f 2 I i 2 )

23 Problem 5.0 U m = B 2 (Al) 2 μ o Problem 5. ε o = BNAω ; ω = ε o BNA Problem 5.2 ε o = BNAω = BNω(l b) ; l = ε o BNωb Problem 5.3 ε o = BNAω = BNA(2πf) ; B = Problem 5.4 ε 2 ε = BNAω 2 BNAω = ω 2 ω ; ε 2 = ω 2 ω ε Problem 5.5 ε = V IR Problem 5.6 ε = N φ t = N B. A t B. πr2 = N t ε o NA(2πf) Ohms Law ε = IR ; I = ε R Problem 5.7 ε = N φ t Problem 5.8 = N B. A t P s = V s I s ; I s = P s V s V p I p = V s I s ; I p = V si s V p N s N p = V s V p

24 Alternating Current 6 QUESTIONS 6.. A sinusoidal current has rms value of 0 A. What is the maximum or peak value? Ans The maximum or peak value of current is 4.4 A. Explanation: According to the given data; Root mean square value of sinusoidal current = I rms = 0 A Maximum or peak value of current = I o =? Using the formula; or I rms = I o 2 I o = I rms 2 = (0) 2 = 4.4 A 6.2. Name the device that will (a) permit flow of direct current but oppose the flow of alternating current (b) permit flow of alternating current but not the direct current. Ans Device (a): An inductor is a device, which permit the flow of direct current but oppose the flow of alternating current. As we know, X L = 2πfL For D.C., f = 0 so X L = 0, hence Current I is maximum Device (b): A capacitor is a device, which permit the flow of alternating current but oppose the flow of direct current. As we know, X C = 2πfC For D.C., f = 0 so X C become infinite, hence Current I is zero How many times per second, will an incandescent lamp reach maximum brilliance when connected to a 50 Hz source? Ans It reaches maximum brilliance 00 times per second. The brilliance of lamp will become maximum twice in one cycle because the current also becomes maximum two times in a cycle (i.e. for positive half cycle and negative half cycle). A = Maximum Brilliance As the frequency f of AC cycle = 50 Hz Maximum brilliance = Twice the frequency = 2f = 2 x 50 = 00 Hz Hence, the brilliance will be maximum 00 times in one second. A = Maximum Brilliance

25 6.4. A circuit contains an iron-cored inductor, a switch and a D.C. source arranged in series. The switch is closed and after an interval, reopened. Explain why a spark jumps across the switch contacts? Ans A circuit contains an iron-cored inductor, a switch and a D.C. source arranged in series. The switch is closed, and after an interval, reopened. Then a spark jumps across the switch contacts. When the switch is closed, then the current increases from zero to maximum value. This changing current produces change of magnetic flux and an induced emf is produced. When a switch is reopened after an interval, then again current changes from maximum value to zero. Once again, the emf is produced across the inductor. It is in the form of back emf i.e. alternating voltage. As a result, it is flux change, which causes an induced emf leading to a spark across the switch contacts How does doubling the frequency affect the reactance of (a) an inductor (b) a capacitor? Ans (a) Inductor The capacitance of inductor becomes double. As reactance of an inductor is given by, X L = 2πfL If f = 2f then, X L = 2π(2f)L X L = 2(2πfL) X L = 2 X L (b) Capacitor The capacitance of capacitor becomes half. As reactance of a capacitor is given by, If f = 2f then, X C = 2πfC X C ' = 2π(2f)C X C ' = 2 2πfC X C ' = 2 X C 6.6. In an R-L circuit, will the current lag or lead the voltage? Illustrate your answer by a vector diagram. Ans In R-L circuit, current lags the voltage by π/2 = 90. Illustration: We can also say that voltage leads the current by θ which is given by; θ = tan - V L V R θ = tan - I rmsx L I rms R VL = Irms XL V = Irms Z θ = tan - X L R VR = Irms R 6.7. A choke coil placed in series with an electric lamp in an AC circuit causes the lamp to become dim. Why is it so? A variable capacitor added in series in this circuit may be adjusted until the lamp glows with normal brilliance. Explain, how this is possible? Ans Dimness of lamp: Let an electric lamp connected to a source of alternating voltage V in AC circuit. When there is no inductance or capacitance in the circuit, the impedance is equal to the resistance of the circuit, say R. it means that the current flowing through the lamp is,

26 I = V (i) R If now, a choke coil of inductive reactance X L is placed in series with the electric lamp, the new impedance of the circuit will be, Z = R 2 + X L 2 Therefore, the current flowing through the circuit in this case is V V I = = (ii) Z R X L From the comparison of equation (i) and (ii) it is obvious that I is smaller than I and that is why the electric lamp is dimmed on placing a choke coil in the circuit. Normal brilliance of lamp: When a variable capacitor also is in series with the circuit, its capacitive reactance X C opposes X L and thus the impedance of circuit is, Z 2 = R 2 + (X L - X C ) 2 Therefore, the current flowing through the circuit in this case is V V I 2 = = (iii) Z 2 R 2 + (X L - X C ) 2 If the capacitance of capacitor is also adjusted that X C = X L, then Z 2 = R 2 = R Hence, the current flowing through the circuit is V I 2 = = V (iv) Z 2 R From the comparison of equation (iii) and (iv) it is obvious that I 2 becomes equal to I as if there is no reactance in the circuit and hence the lamp glows with normal brilliance Explain the conditions under which electromagnetic waves are produced from a source? Ans When alternating voltage i.e. V = V o sin 2πft is applied across the ends of a metallic antenna, an oscillating electric field comes into existence which accelerates the electrons again and again as the polarities of the antenna change after half a cycle. If the electric charge on the antenna is +q at the moment, it becomes q at the other moment. The accelerated electrons radiate energy carried by electric field. The changing electric field produces changing magnetic field. Thus, each field produces the other turn by turn and so this process continues all along the direction of propagation. The changing electric and magnetic fields both confined to a plane perpendicular to the direction of propagation of electromagnetic waves How the reception of a particular radio station is selected on your radio set? Ans A particular radio station can be selected on a radio set by tuning it. Explanation: Radio programmes of different stations are present in the form of electromagnetic waves of different frequencies. We can choose the programme of our interest by adjusting the frequency of out receiving set equal to the frequency of the radio station of our choice. This is done by rotating the tuning knob to vary capacitance and thus the frequency of the radio set. It is given by, f = 2π LC By doing so, the phenomenon of resonance takes place. The current of this radio signal becomes maximum because impedance becomes minimum and so, the signal becomes detectable What is meant by A.M. and F.M.? Ans (a) A.M. A.M. stands for Amplitude Modulation.

27 Definition: It is a type of modulation in which amplitude of the carrier wave is increased or decreased as the amplitude of the superposing modulating signal increases or decreases. Range: The A.M. transmission frequencies range from 540 khz to 600 khz. (b) F.M. F.M. stands for Frequency Modulation. Definition: It is a type of modulation in which frequency of the carrier wave is increased or decreased as the amplitude of the superposing modulating signal increases or decreases but the amplitude of the carrier wave remains constant Range: The F.M. transmission frequencies are much higher and range between 88 MHz to 08 MHz. Problem 6. I = I o sin 2πft I rms = 0.7 I o Problem 6.2 I rms = 0.7 I o I = I o sin 2πft Problem 6.3 X L = ωl = 2πfL PROBLEMS I = V X L Problem 6.4 X L = ωl = 2πfL Z = (X L ) 2 + (R) 2 Problem 6.5 Z = (X L ) 2 + (R) 2 = (ωl) 2 + (R) 2 = (2πfL) 2 + (R) 2 I o = V o Z I rms = 0.7 I o θ = tan X L R Problem 6.6 P = I rms V X L = ωl = 2πfL Z = (X L ) 2 + (R) 2 I rms = V rms Z Problem 6.7 Problem 6.8 I = V X C = V(2πfC) = tan ωl R θ = tan X C R = tan 2πfCR Problem 6.9 f r = 2π LC Problem 6.0 f max = 2π LC min f min = 2π LC max 2πfL = tan R

28 Physics of Solids 7 QUESTIONS 7.. Distinguish between crystalline, amorphous and polymeric solids. Ans Solid is a state of matter in which atoms are very close to each other. Atoms are rigidly bounded so they can only vibrate about their mean position. CLASSIFICATION OF SOLIDS Solids are classified into following types: Crystalline Solids Polymeric Solids Amorphous or Glassy Solids A. CRYSTALLINE SOLIDS Definition: Solids in which atoms, ions, or molecules are arranged in regular manner repeatedly and periodically are called crystalline solids. Properties: (i) They have strict long-range ordered structure of atoms, ions or molecules due to cohesive forces, which exist between atoms, ions or molecules of same substance. (ii) Study of crystalline solids using various X-ray techniques tells us that atoms, ions, or molecules in crystalline solids are not static. (iii) In metal crystal, atoms vibrate about their mean positions. Therefore, atoms of crystalline solids possess vibrational motion whose amplitude increases with the increase in temperature. (iv) In crystalline solids, change of state from solid to liquid at higher temperature is abrupt and discontinuous. So structure of solid beaks suddenly and solid melts. (v) Every crystalline solid have definite melting point. (vi) Pattern of crystalline solids is three-dimensional which repeats itself repeatedly. Examples: Metals: Copper, Iron, Zinc etc Ionic Compound: NaCl Ceramics: Zirconia B. AMORPHOUS or GLASSY SOLIDS Definition: Solids in which atoms, ions, or molecules have no regular arrangement are called amorphous or glassy solids. Properties: (i) They have no regular arrangement of molecules. (ii) These solids are more like liquid with frozen disordered structure. (iii) On heating, they become soft and attain paste like state.

29 (iv) At about 800 C, they are converted into viscous liquid. (v) They have no definite melting point. (vi) Molecules have short-range non-repetitive order. Examples: Ordinary glass, rubber, plastic, glue etc. C. POLYMERIC SOLIDS Definition: Material whose structure is intermediate between solid (order) and liquid (disorder) are called polymeric solids. They are called polymers because they are formed by polymerization reaction. Simple Molecules Polymerization Massive Long Chain Molecules / Three Dimensional Structure Properties: (i) Their structure is partially and poorly crystalline. (ii) They have limited arrangement. (iii) They have low specific gravity. (iv) They have good strength to weight ratios, as is observed in polythene bag. (v) These solids are combination of carbon with oxygen, hydrogen, nitrogen and other metallic or non-metallic elements Examples: Cellulose is natural polymer, which provides strength and stiffness to roots, stems and leaves of plants. Natural rubber is a polymer composed of hydrocarbon with formula (C 5H 6) n. Polythene, polystyrene, nylon and synthetic rubber are manmade artificial polymers Define stress and strain. What are their SI units? Differentiate between tensile, compressive and shear modes of stress and strain. Ans STRESS Definition: The force applied on unit area to produce any change in the shape, volume or length of a body is called stress. Mathematically, it is described as: Force (F) Stress (σ) = Area (A) SI Unit: The SI unit of stress is newton per square meter (Nm -2 ), which is given the name pascal (Pa). Types: (i) Tensile Stress or Compressive Stress: A stress that causes the change in length of an object is called tensile stress. (ii) Shear Stress: A stress that causes the change in shape of an object is called shear stress. (iii) Volume Stress: A stress that causes the change in volume of an object is called volume stress. STRAIN Definition: Strain is the measure of deformation of a solid when stress is applied to it. For the case of onedimensional deformation, strain is defined as the fractional change in length. If l is the change in length and l is the original length, then the strain is given by: Change in length ( l) Strain (ε) = Original length (l) SI Unit: Since strain is the ration of two same quantities, it is dimensionless, and therefore, has no units. Types: (i) Tensile Strain and Compressive Strain: If the strain is due to tensile stress, it is called tensile strain. If the strain is due to compressive stress, it is called compressive strain.

30 (ii) Shear Strain: A strain produced in the object when it is subjected to shear stress is called shear strain. When the opposite faces of a rigid body are subjected to shear stress, the shear strain produced is given by Shear Strain (γ) = a a = tan θ (ii) Volumetric Strain: When the applied stress changes the volume, then the change in volume per unit volume is called volumetric strain. Thus, Volumetric Strain = 7.3. Define modulus of elasticity. Show that the units of modulus of elasticity and stress are the same. Also discuss its three kinds. V V o Ans MODULUS OF ELASTICITY Definition: The ratio of stress to strain is a constant for a given material, provided the external applied force is not too great, called modulus of elasticity. Mathematically, it is described as: Stress Modulus of Elasticity = Strain SI Unit: Since the strain is a dimensionless quantity, the units of modulus of elasticity are the same as that of stress, i.e., Nm -2 or Pa. Proof As we know Modulus of Elasticity = Stress Strain = Nm -2 no units = Pa no units Then we can say that units of modulus of elasticity and stress are same. Types: (i) Young s Modulus: For the case of linear deformation, the ration of tensile stress to tensile strain is called Young s Modulus (Y). Tensile Stress F Young's Modulus (Y) = Tensile Strain = A l l (ii) Shear Modulus: When the shear stress τ = (F/A) and shear strain (γ = tan θ) are involved, then their ratio is called shear modulus (G). Shear Stress F Shear Modulus (G) = Shear Strain = A tan θ (iii) Bulk Modulus: For three-dimensional deformations, when volume is involved, then the ratio of applied stress to volumetric strain is called bulk modulus (K). Bulk Modulus (K) = Where V is the change in original volume V Volumetric Stress Volumetric Strain = F A V V 7.4. Draw a stress-strain curve for a ductile material, and then define the terms: Elastic limit, Yield point and Ultimate tensile stress? Ans In tensile test, metal wire is extended at a specified deformation rate. The stresses generated in the wire during deformation are continuously measured by a suitable electronic device fitted in the mechanical testing machine. Stress-strain curve is plotted automatically on XY chart recorder. A typical stress strain curve for a ductile material is shown in the figure. (i) Proportional Limit (σ p): In the initial stages of deformation, stress is increased linearly with strain until point A on stress-strain curve. This is called proportional limit. Hence, proportional limit is defined as the greatest stress that a material can endure without losing straight-line proportionality between stress and strain.

31 (ii) Yield Stress or Elastic Limit (σ e): From A to B, the stress and strain are not proportional, but nevertheless, if the load is removed at any point between O and B, the curve will be retraced and the material will return to its original length. The point B is called yield point and the value of stress at this point is known as yield stress or elastic limit. Hence, elastic limit is the greatest stress that a material can endure without any permanent deformation. (iii) Ultimate Tensile Stress UTS (σ m): If the stress is increased beyond elastic limit, the specimen becomes permanently deformed. This kind of behavior is called plasticity. The region of plasticity is represented by the portion of the curve from B to C. The point C represents ultimate tensile strength (UTS). Therefore, ultimate tensile stress is the maximum stress that a material can withstand. Once point C corresponding to UTS is crossed, the material breaks at point D, responding to fracture stress What is meant by strain energy? How can it be determined from force-extension graph? Ans Strain Energy: The amount of potential energy stored in a material due to displacement of its molecule from its equilibrium position, under the action of stress, is called strain energy. Derivation of the expression of Strain Energy: Consider a wire whose one end is attached to a fixed support, is stretched vertically by connecting a weight at its lower end. The suspended weight acts as a stretching force. The extension in the wire is increased by increasing the stretching force. The graph plotted between extension l for the different value of stretching force is shown in the figure. F2 C H F F F B A D G O x l l2 It is clear from the figure that the force is constant in producing extension l but is changing linearly from 0 to F. In order to calculate the work done for extension l by a certain force F, it is convenient to find the work done by graphical method. The work done for extension l by a certain force F will be equal to the area under force extension curve, which is equal to the area of triangle OAB. Therefore, Work done = Area of triangle OAB Work done = 2 (base)(altitude) Work done = 2 OA AB Work done = 2 l F This work done is appeared as strain energy inside the wire. So, Strain Energy = 2 l F (i) I If A is the cross-sectional area of the wire of length L, then the modulus of elasticity E of the wire can be described as: Stress Modulus of Elasticity = Strain

32 E = F A L l Putting in (i) Strain Energy = F = EAl L 2 l ( EAl L ) = 2 2 (EAl L ) 7.6. Describe the formation of energy bands in solids. Explain difference amongst electrical behaviour of conductors, insulators and semiconductors in terms of energy band theory. Ans ENERGY BAND THEORY The electrical properties of solids determine its ability to conduct electric current. The conventional free electron theory based on Bohr Model failed to explain completely the vast diversity in the electrical behavior of solids. On the other hand, the energy based on wave mechanical model has been found successful in resolving this problem. (i) Energy Band: When the large numbers of atoms are brought together, as in a crystal, they interact with one another to form a solid. As the result, each energy level splits up into several sub-levels due to the action of force exerted by other atoms in the solid. A group of such energy sublevels is called an energy band. (ii) Forbidden Energy Band: There is a large number of energy states between two consecutive permissible energy bands, which cannot be occupied by electrons. These are called forbidden energy states. The energy bands are separated by gaps in which there is no energy level. Such energy gaps are called forbidden bands. TYPES OF ENERGY BANDS (a) Valence Band: The electrons in the outermost shell of an atom are called valance electrons. Therefore, the energy band occupied by valance electrons is called the valance band. The valance band may be either completely filled or partially filled with the electrons but can never be empty. (b) Conduction Band: The energy band next to the valance band is called the conduction band. The valence and conduction bands are separated by forbidden energy gaps. The conduction band may be empty or partially filled. The electrons in the conduction band can drift freely in the materials and are called free or conduction electrons. (c) Completely Filled Energy Band: The bands below the valence energy band are normally completely filled. They do not take part in conduction process. Therefore we consider valence and conductive while discussing the electrical conductivity. APPLICATION OF ENERGY BANDS The width of forbidden energy gap between valance and conduction band decide whether a material is a conductor, insulator or a semiconductor. (i) Insulators: Insulators are those materials in which valance electrons are bound very tightly to their atoms and are not free. In terms of energy bands, it means that an insulator has: (a) An empty conduction band (no free electron) (b) A full valence band (c) A large energy gap (several ev) between them Empty Conduction Band Forbidden Band Conduction Band Overlap Conduction Band Forbidden Band Valence Band Valence Band Valence Band Insulator Conductor Semi-conductor

33 (ii) Conductors: Conductors are those, which have plenty of free electrons for electrical conduction. In terms of energy bands, conductors are those materials in which valence and conduction band largely overlap each other. There is no physical distinction between the two bands, which ensures the availability of free electrons. That is why; the conductions are good conductors of electricity. (iii) Semiconductors: In terms of energy bands, semiconductors are those materials, which at room temperature have (a) A partially filled conduction band (b) A partially filled valence band (c) Very narrow forbidden energy gap (of order of ev) between conduction and valence bands At 0 K, there are no electrons in the conduction band and their valence band is completely filled. It means at 0 K a piece of Ge or Si is a perfect insulator. However, with the increase of temperature, some electrons posses sufficient energy to jump across the small energy gap from the valance band to conduction band. This transfers some free electrons in the conduction band and creates some holes in the valance band. The vacancy of electron in the valence band is known as a hole. It behaves like a positive charge. Thus at room temperature, Ge or Si crystals becomes a semiconductor Distinguish between intrinsic and extrinsic semi-conductors. How would you obtain n-type and p-type material from pure silicon? Illustrate it by schematic diagram. Ans INTRINSIC AND EXTRINSIC SEMI-CONDUCTORS There are two types of semi-conductors. (a) Intrinsic semi-conductors (b) Extrinsic semi-conductors (i) INTRINSIC SEMICONDUCTOR A semiconductor in its extremely pure form is known as intrinsic semiconductors. Pure elements of silicon and germanium are intrinsic semiconductors. Doping The electrical behavior of semiconductors is extremely sensitive to the purity of the material. It is substantially changed on introducing a small impurity into pure semi-conductor. This process is called doping. (ii) EXTRINSIC SEMI-CONDUCTORS The doped semi-conducting materials are called extrinsic semi-conductors. Explanation: Pure elements of silicon and germanium are intrinsic semiconductors. These semi-conductor elements have atoms with four valence electrons. In solid crystalline form, the atoms of these elements arrange themselves in such a pattern that each atom has four equidistant neighbors. Each atom with its four valence electrons, shares an electrons from its neighbors. This effectively allocates eight electrons in the outermost shell of each atom, which is a stable state. The sharing of electrons between two atoms is called covalent bonds. Due to this, electrons are bound in their respective shells.

34 Types: The conductivity of silicon and germanium can be drastically increased by the controlled addition of impurities to the intrinsic (pure) semi conductive material. There are two types of extrinsic semiconductors. (a) n-type Semi-conductors When a silicon crystal is doped with a pentavalent element, e.g., arsenic, antimony or phosphorous etc., four valance electrons of impurity atom form covalent bond with the four neighboring Si atoms, while the fifth valence electron provides a free electron in the crystal. This extra electron becomes a conduction electron because it is not attached to any atom. Such a doped extrinsic semiconductor is n-type semiconductor. (b) p-type Semi-conductors When a silicon crystal is doped with the trivalent element, e.g., aluminum, boron, gallium or indium etc., three valence electrons of the impurity atom form covalent bond with three neighboring Si atoms. While the one missing electron in the covalent bond with the fourth-neighboring Si atom, is called a hole, which in fact is vacancy, where an electron is accommodated. Such a semiconductor is called p-type semiconductor Discuss the mechanism of electrical conduction by holes and electrons in a pure semi-conductor element. Ans ELECTRICAL CONDUCTION BY HOLES AND ELECTRONS IN SOLIDS It has been observed that whenever a covalent bond is broken, an electron hole pair is created. Explanation: In order to explain it, consider a semiconductor crystal lattice e.g. Ge or Si. The circles indicate the positive ions core and dots are valence electrons. These electrons are bound by covalent bond. At room temperature, they have thermal kinetic motion. This is so strong that the electrons break the

35 covalent bond. They get themselves free and create a hole. Thus, whenever a covalent bond is broken, an electron-hole pair is created. Both of them move in the semiconductor crystal lattice. MECHANISM OF ELECTRICAL CONDUCTION: For this Purpose, we consider a row of Si atoms in crystal lattice. Suppose a hole is present in the valence shell of atom A. The core of atom A has a net positive charge as shown in figure because hole is a deficiency of electron. Thus, this attracts an electron from a neighbouring atom `B. Therefore, the hole moves to B as shown in figure and a hole (positive charge) appears at B which attracts en electron from C. This creates a hole at C and a positive charge at C as shown in figure, which in return attracts an electron from D, thus creating a hole and a positive charge at D as shown figure. From the above discussion, we draw the following conclusions. (i) Stay of Hole: If a hole is present in any valence shell, it cannot stay there but it moves from one atom to the other with the electron moving in opposite direction. (ii) Charge of a Hole: It is observed that hole appears with a positive charge. Thus, a moving hole is equivalent to a moving positive charge. (iii) Random Motion of Holes: It is observed that hole has random motion because positively charged core of the atom can attract an electron from any of its neighbouring atoms. (iv) Kinds of Charge Carriers: There are two kinds of charge carriers in semiconductors (a) A free electron i.e. (-e) (b) A hole i.e. (+e) A B C D ELECTRICAL CONDUCTION BY HOLES AND ELECTRONS IN AN ELECTRIC CIRCUIT: The phenomenon of electrical conduction due to holes and electrons in a circuit is proved by considering the circuit as shown in figure (a) (b) (c) (d) + When a battery is connected to a semiconductor, it produces an electric field across it due to which a directed flow of electrons and holes takes place. The electrons flow towards the positive end and the holes flow towards the negative end of the semiconductor as shown in figure. The current flowing through the semiconductor is carried by both electrons and holes. The electrical current and the hole current add up together to give the total current. I = I e + I h

36 7.9. Write a note on superconductors. Ans SUPERCONDUCTORS Definition: Materials whose resistivity becomes zero below a certain temperature are called superconductors. Explanation: A class of elements such as metals, alloys and ceramics show a remarkable change in resistivity with temperature. Thus, superconductors are alloys that at certain temperature conduct electricity with zero resistance. In these, no energy is dissipated and the current at once setup continues to exist indefinitely without the source of an emf. The superconductor offers no resistance to electric current and is called perfect conductor. Critical Temperature: The temperature at which the resistivity of a material falls to zero is called critical temperature. It is denoted by T c. Types of superconductors: There are two types of superconductors (i) Low Temperature Superconductors Kmaerlingh Ornes discovered the first superconductor in 9, when it was observed that resistance of mercury disappears suddenly as the temperature is reduced below 4.2 K. (ii) High Temperature Superconductors Any superconductor having a critical temperature above 77 K (the boiling point of liquid nitrogen) is referred as high temperature superconductor. APPLICATIONS: Superconductors have many technological applications such as in (a) Magnetic Resonance Imaging (MRI) (b) Magnetic Levitation Trains (c) Powerful but small electric motors (d) Fast computer chips Material Critical Temperature T c LOW TEMPERATURE SUPERCONDUCTORS Aluminium.8 K Tin 3.72 K Lead 7.2 K HIGH TEMPERATURE SUPERCONDUCTORS Class of Ceramics 25 K Yttrium Barium Copper Oxide (YBa2Cu3O7) 63 K or 0 C 7.0. What is meant by para, dia and ferromagnetic substances? Give examples for each. Ans (A) Paramagnetic Substances: The solids in which the orbital and spin axes of the electrons in an atom are so oriented that their fields support each other are called paramagnetic substances. In these solids, each atom behaves like a tiny magnet. For example: Aluminium, Manganese, Oxygen gas, Uranium and Plantinum (B) Diamagnetic Substances: In diamagnetic substance, there is no resultant field as the magnetic field produced by both orbital and spin motions of the electron might add up to zero. For example: the atoms of Water, Copper, Bismuth and Antimony are diamagnetic. (C) Ferromagnetic Substances: In ferromagnetic substances, the atoms cooperate with each other in such a way so as to exhibit a strong magnetic effect. For example: Fe, Co, Ni, Chromium dioxide and Alnico In ferromagnetic substance, there exist small regions called domains (contain 0 2 to 0 6 atoms). Within each domain, the magnetic fields of all spinning electrons are parallel to one another, i.e., each domain is magnetized to saturation. Each domain behaves as a small magnet with its own north and south poles. Following are two types of ferromagnetic substances: (i) Soft Ferromagnetic Substances In soft ferromagnetic substances, the domains are easily oriented on applying an external field and readily return to random positions when the field is removed. This is desirable in an electromagnet and in transformers. Iron is a soft magnetic material.

37 (ii) Hard Ferromagnetic Substances In hard ferromagnetic materials, domains are no so easily oriented to order. They required very strong external fields but once oriented retain the alignment. These solids are used to make permanent magnet. Steel and Alnico V is an example of hard ferromagnetic material. 7.. What is meant by hysteresis loss? How is it used in the construction of a transformer? Ans Definition: The area of the loop is the measure of the energy needed to magnetize and demagnetize the specimen during each cycle of the magnetizing current. This is the energy required to do work against internal friction of the domains. This work is dissipated as heat. It is called hysteresis loss. The loss of the energy depends on the hysteresis loop. Explanation: Hard magnetic material like steel cannot be easily magnetized and demagnetized, so they have large loop area as compared to soft magnetic materials such as iron, which can easily be magnetized. The energy dissipated per cycle, thus, for iron is less than for steel. Construction of transformer: The cores of electromagnets used for alternating current where the specimen repeatedly undergoes magnetization and demagnetization should have narrow hysteresis curve of small area to minimize the waste of energy. This is the principle used in the construction of transformer to reduce energy loss. Problem 7. Problem 7.2 Stress = σ = F A = mg πd 2 /4 = 4mg πd 2 change in length Tensile strain = ε = original length = l l Percentage Elongation = ε 00 = l l 00 Problem 7.3 change in length Tensile strain = ε = original length = l l Young s Modulus = Y = Fl A l Problem 7.4 Problem 7.5 Energy = E = 2 F l Young s Modulus = Y = stress strain = σ ε Stress = σ = F A ; Force = F = σa PROBLEMS Work Done = W = Average Force l = 2 F l Young s Modulus = Y = Fl A l Problem 7.6 ; Stress = σ = Yε ; l = Fl YA ε c = l c l ε s = l s l Young s Modulus = Y c = Fl ; F = Y ca l c A l c l Young s Modulus = Y s = Fl ; F = Y sa l s A l s l

38 Electronics 8 QUESTIONS 8.. How does the motion of an electron in a n-type substance differ from the motion of holes in a p-type substance? Ans n-type substance: As we know, electrons are majority charge carriers and holes are minority charge carrier in n-type substances. In such substances, current is produced by the flow of electrons and is called electronic current. When a battery is connected in electronic circuit, the electrons move from lower potential (negative end) to the higher potential (positive end) in an electric field. p-type substance: As we know, holes are majority charge carriers and electrons are minority charge carrier in p-type substances. In such substances, current is produced by the flow of holes and is called conventional current. When a battery is connected in conventional circuit, the holes move from higher potential (positive end) to the lower potential (negative end) in an electric field What is the net charge on a n-type or a p-type substance? Ans The net charge on both, n-type and p-type substance is zero. n-type substance is formed when small amount of pentavalent impurity is added to pure semiconductors. Similarly, p-type substance is formed when a small amount of trivalent impurity is added to pure semi-conductors. Since all the atoms in n-type or p-type substance are electrically neutral so net charge on n-type or p-type substance is zero The anode of a diode is 0.2 V positive with respect to its cathode. Is it forward or reverse biased? Ans The anode of a diode is 0.2 V positive with respect to its cathode. It is forward biased. As the anode (p-type) of the diode is at high potential with respect to cathode (n-type), so it is said to be forward biased. In this case, potential of anode is 0.2 V than that of cathode. Therefore, this diode (pn-junction) is forward biased. Moreover, in the case of forward biased junction, the electrons always flow from lower to higher potential against the electric field, so the flow of electrons takes place from cathode to anode. Hence, the diode is forward biased Why charge carriers are not present in the depletion region? Ans Charge carriers are not present in the depletion region. In the case of pn-junction, n-region contains free electrons as majority charge carriers and p-region contains holes as majority charge carriers. The electrons in the n-region due to their random motion diffuse in the p-region forming positive ion behind. As a result, of their diffusion, the recombination of

39 electrons and holes takes place within the depletion region and negative ion is formed at p-region. Since, the charge carriers are removed, so a charge-less region is formed around the junction in which charge carriers are not present. Thus we, can say that depletion region has no net charge. p-region Depletion region n-region 8.5. What is the effect of forward and reverse biasing of a diode on the width of depletion region? Ans Forward Biasing: When the pn-junction is forward biased, the width of depletion region reduces. It is because, barrier potential is reduced which allows more current to flow across the junction. Reverse Biasing: When the pn-junction is reversed biased, the width of depletion region increases. It is because, majority carriers are pulled away from the junction Why ordinary silicon diodes do not emit light? Ans The ordinary silicon diodes do not emit light. Height of potential barrier is between p and n side of silicon diode. When electron recombines with a hole, 0.7 ev energy is released in the form of a photon. Here, Energy = E = 0.7 ev = 0.7 x.6 x 0-9 J As, E = hf or f = E h λ = c f or λ = ch E Hence, λ = ch E = (3 08 )( ) 0.7 x.6 x 0-9 = m = 776 nm Hence, photons emitted from silicon diodes have wavelength much greater than visible light whose wavelength ranges from nm. Hence, they lie in infrared region and not visible. Moreover, it is opaque to light and when electrons from n-side cross the junction and recombine the holes on p-side, greater percentage of energy is released in form of heat. That is why they do not emit light Why a photodiode is operated in reverse biased state? Ans A photodiode is operated in reverse biased state. Such diode which coverts light energy into electrical energy is called photodiode. In reverse-biased state, the reverse current is almost negligible in the absence of light. However, a reverse-current starts when its pn-junction is exposed to light. When light enters the junction, it creates electron-hole pair, which causes the flow of small current in the reverse direction. As the intensity of light increases, so the number of electrons and holes also Reverse Current Intensity of Light

40 increase. As a result, reverse current increases as shown in figure. Such a photodiode can turn its current ON and OFF in house circuits in nano-seconds Why is the base current in a transistor very small? Ans The base current in a transistor very small. It is because of following reasons; (i) The base is very thin of the order 0-6 m as compared to either emitter or collector. (ii) The base has very small doping level as compared to emitter and collector. (iii) Biasing voltage between base-collector junction V CC is larger than the voltage between baseemitter junction V BE i.e. V CC > V BE. Thus, it has only a small number of electrons and holes to recombine with the charges already present in the base. In other words, majority of electrons coming from emitter do not combine with some holes and escape out of the base. Almost majority of free electrons are attracted towards the collector. As a result, I C I E and hence I B is very small What is the biasing requirement of the junctions of a transistor for its normal operation? Explain how these requirements are met in a common emitter amplifier? Ans For normal operation of a transformer connected in a circuit, emitter-base junction (E-B input) is forward and collector-base junction (C-B output) is reverse-biased. Common-Emitter (CE) Amplifier: In common-emitter amplifier, input is applied between the base and emitter. The output is obtained between collector and emitter. Thus, emitter-base junction (E-B) is forward-biased while collectorbase junction (C-B) is reverse-biased. Voltage applied across E-C junction is always greater than the voltage applied across E-B junction What is the principle of a virtual ground? Apply it to find the gain of an inverting amplifier. Ans Principle of a Virtual Ground: In an inverting amplifier, the input signal voltage V i to be amplified is applied at inverting terminal (negative) through a resistor R and V o is the output voltage. The non-inverting terminal (positive) is grounded and its potential is zero. As open loop gain A OL is very high ( 0 5 ) so by knowing the relation; A OL = output V o = input V + - V For any value of V o, V + - V - 0 or V + V Since, V + is zero at ground potential so V - is virtually at ground potential i.e. V - 0. This is called principle of virtual ground. R2 R I2 I 0V Vi 0V VO Calculation of Voltage Gain: As V - 0, so Ground Current through R = I = V i- V - R I = V i- 0 = V i R R (i)

41 As V - 0, so Current through R 2 = I 2 = V -- V o R 2 0-V o I 2 = = -V o R 2 R 2 Using Kirchhoff s Current Rule I = I 2 Putting values from (i) and (ii) in above equation or Also Hence, V i R = -V o R 2 V o V i = - R 2 R (ii) Gain of Inverting Amplifier = output input Gain of Inverting Amplifier = - R 2 R 8.. The inputs of a gate are and 0. Identify the gate if its output is (a) 0 (b)? Ans (a) Output = 0 The gate may be AND, NOR or XNOR. Inputs Outputs A B AND Gate NOR Gate XNOR Gate 0 X = A. B X =. 0 X = 0 X = A + B X = + 0 X = 0 X = A. B + A.B X = X = X = + 0 X = 0 (b) Output = The gate may be OR, NAND or XOR. Inputs Outputs A B OR Gate NAND Gate XOR Gate 0 X = A + B X = + 0 X = X = A. B X =. 0 X = X = A. B + A.B X = X = X = + 0 X = 8.2. Tick ( ) the correct answer. (i) A diode characteristic curve is a plot between a. current and time b. voltage and time c. voltage and current d. forward voltage and reverse voltage

42 (ii) The colour of light emitted by a LED depends on a. its forward bias b. its reverse bias c. the amount of forward current d. the type of semi-conductor material used (iii) In a half-wave rectifier, the diode conducts during a. both halves of the input cycle b. a portion of the positive half of the input cycle c. a portion of the negative half of the input cycle d. one half of the input cycle (iv) In a bridge rectifier of fig. when V i is positive at point B with respect to point A, which diodes are ON A D4 D Vi ~ D3 D2 R B a. D 2 and D 4 b. D and D 3 c. D 2 and D 3 d. D and D 4 (v) The common emitter current amplification factor β is given by a. I C IE b. I C IB c. I E IB d. I B IE (vi) Truth table of logic function a. summarizes its output values b. tabulates all its input conditions only c. display all its input/output possibilities d. is not based on logic algebra (vii) The output of two inputs OR gate is 0 only when its a. both inputs are 0 b. either input is c. both inputs are d. either input is 0 (viii) A two inputs NAND gate with inputs A and B has an output 0 if a. A is 0 b. B is 0 c. Both A and B are zero d. Both A and B are (ix) The truth table shown below is for A B X 0 0

43 a. XNOR gate b. OR gate c. AND gate d. NAND gate PROBLEMS Problem 8. I C = βi B I E = I C + I B Problem 8.2 Ratio of Collector Current to Emitter Current = I C I E I B = I C β Problem 8.3 V CC = I B R B + V BE I C = V CC V CE R C I B = I C β Problem 8.4 V C = I C R C or V C = V CC V CE I + I 2 = I 3 Problem 8.5 Gain = G = + R 2 R

44 Dawn of Modern Physics 9 QUESTIONS 9.. What are measurements on which two observers in relative motion always agree upon? Ans The measurements on which two observers in uniform relative motion (i.e. moving with uniform motion w.r.t. each other) will agree upon; (i) Speed of light in free space i.e. 3.0 x 0 8 ms - (ii) Magnitudes of force and acceleration of a moving object 9.2. Does the dilation means that time really passes more slowly in moving system or that it only seems to pass more slowly? Ans The dilation means that time only seems to pass more slowly. As we know, the time dilation is given by the formula, t = - v2 c 2 This result shows that a clock moving with respect to the observer appears to tick less rapidly than it does when it is at rest w.r.t. him. It means that moving clock appears to run slow to the observer at rest. Therefore, time dilation is apparent change and it is not the actual change. In other words, it only seems to pass more slowly but not actually If you are moving in a spaceship at a very high speed relative to the Earth, would you notice a difference (a) in your pulse rate (b) in the pulse rate of people on Earth? Ans (a) In our Pulse Rate: We feel no change in our own pulse rate. As we know, the time dilation is given by the formula, t = - v2 c 2 Inside the spaceship, the clock is at rest w.r.t. moving spaceship so speed of clock is zero. When v = 0 then from formula of time dilation t = t o Hence, no change in pulse rate occurs. (b) In the Pulse Rate of People on Earth: The pulse rate of people on Earth will be slower as measured by the people in space ship. Earth and spaceship are two different frame of references. A person will feel a change in the pulse rate of people on the Earth because the clock on which he is measuring the pulse rate is moving at very high-speed w.r.t. to Earth. According to time dilation formula, the clock inside the spaceship will move slowly due to large speed of spaceship, as compared to the clock on the Earth. t = - v2 c 2 Thus, the person in the spaceship will feel slower pulse rate of the people on Earth. t o t o t o

45 9.4. If the speed of light were infinite, what would the equations of special theory of relativity reduce to? Ans If c, then v 2 c 2 = v2 2 = 0 Therefore, the equations of relativity reduce to; (i) Time Dilation There may be no time dilation. (ii) Length Contraction There will be no length contraction. t = t o - v2 c 2 t = t o (iii) Mass Variation There will be no increase in mass. l = l o - v2 c 2 l = l o m = m = m o - v2 c 2 (iv) Mass-Energy Relation As m = 0 so there will be no increase in energy. E = mc 2 E = 0 m o 9.5. Since mass is a form of energy, can we conclude that a compressed spring has more mass than the same spring when it is not compressed? Ans Yes, there would be an increase in the mass of compressed spring. However, the increase in mass is slightly greater than the mass of uncompressed spring due to condensation of energy of compressed string into mass (i.e. work done in compressing the spring has changed into energy, which increases the mass of the spring). However, this increase in mass is negligibly small. According to Einstein s energy-mass relation E = mc 2 This increase in energy increases the mass as; m = E c 2 As, E is very small, therefore m is in too small to be measured As a solid is heated and begins to glow, why does it first appear red? Ans As a solid is heated and begins to glow, it first appears red. According to Wien s displacement law, wavelength is inversely proportional to temperature i.e. λ When a body is heated, it emits radiations. The nature of radiations depends upon temperature. At low temperature, a body emits radiations of longer wavelengths. Since the longest visible wavelength is red, so it first appears red. As the temperature of solid is further increased, the proportion of radiations of short wavelength increases. This is why the solids when heated look red first What happens to total radiation from a black body if its absolute temperature doubled? Ans The total energy radiated per second is increased to 6 times. Proof: According to Stefan-Boltzmann s law, E = σ T 4 T

46 When absolute temperature is increased to double of its value then T = 2T, E' = σ (2T) 4 E' = σ 6T 4 E' = 6(σT) 4 E' = 6 E Thus, the total radiation energy E will increase sixteen times A beam of red light and a beam of blue light have exactly the same energy. Which beam contains the greater number of photons? Ans A beam of red light contains greater number of photons. As we know, the energy of single photon is given by, E = hf Let n be the number of photons, then the energy becomes E = nhf (i) However, c = f λ or f = c / λ Putting the value of f in equation (i) nhc E = λ or Eλ n = (ii) hc As red and blue light have the same energies, so E, h and c have constant values. Thus, n = constant λ n λ The above equation shows that greater wavelength will have greater number of photons. But wavelength of red light is greater than that of blue light i.e. λ red > λ blue, so beam of red light will contain greater number of photons Which photon, red, green, or blue carries the most (a) energy and (b) momentum? Ans (a) Energy Photon of blue light has most energy. As we know, the energy of the photon is given by, E = hf However, c = f λ or f = c / λ E = hc λ However, h and c have constant values. Thus, E = constant λ E λ This relation shows, smaller the wavelength greater the energy. As the wavelength of photon of blue light is shorter among red and green light photons so, photon of blue light has maximum energy. (b) Momentum Photon of blue light has more momentum. As we know, the momentum of the photon is given by, However, h is constant. Thus, p = h λ p = constant λ p λ

47 This relation shows, smaller the wavelength greater the momentum. As the wavelength of photon of blue light is shorter among red and green light photons so, photon of blue light has maximum momentum Which has the lower energy quanta? Radio waves or X-rays. Ans Radio waves have lower energy quanta. Since, E = hf or E f As frequency of X-rays is greater than that of radio waves or wavelength of X-rays is smaller than that of radio waves, so from the above relation it is obvious that X-rays quantum is greater than that of radio waves. In short, the radio waves have lower energy quanta. 9.. Does the brightness of a beam of light primarily depends on the frequency of photons or on the number of photons? Ans The brightness of a beam of light depends on the number of photons. When the light is visible or not, it depends upon frequency. However, the brightness of a beam light depends upon the number of photons not on the frequency of light or photons. Brightness or intensity means the energy delivered per unit time per unit area. Hence, the brightness of beam of light depends upon the number of photons When ultraviolet light falls on certain dyes, visible light is emitted. Why does this not happen, when infrared light falls on these dyes? Ans When infrared light falls on certain dyes, visible light is not emitted. The ultraviolet light has photons of high energy. When they fall on certain dyes, they can excite the atoms of dyes. When the excited atoms return to their ground state, they emit frequencies, which are detectable by normal human eye. However, when infrared light are made to fall on the atoms of dyes, then they may be excited. However, the de-excitation takes place in such a way that photons of frequencies are emitted whose values are below the least value of frequency of visible spectrum. Thus, they cannot be detected by normal human eye. Hence, the ultraviolet light falling on dyes cause the light visible while the infrared light cannot do so Will bright light eject more electrons from a metal surface than dimmer light of the same colour? Ans Yes, bright light will eject more electrons from a metal surface than dimmer light of the same colour. As we have studied in the phenomenon of photoelectric effect, the number of electrons emitted from the metal surface is directly proportional to the intensity of light i.e. Number of electrons Intensity of Light Therefore, bright light being more intense will emit more electrons from a metal surface than a dimmer light of the same colour and less intensity Will higher frequency light eject greater number of electrons than low frequency light? Ans Both higher and lower frequency lights will eject same number of electrons from a metal surface. The number of photoelectrons does not depend upon frequency but depends on intensity of light i.e. Number of electrons Intensity of Light

48 Therefore, high frequency light will not emit more electrons than low frequency light. However, the higher frequency light ejects electrons whose K.E. will be greater than those electrons emitted by low frequency light When light shines on a surface, is momentum transferred to the metal surface? Ans Yes, when light shines on a surface, momentum is transferred to the metal surface According to Einstein, light behaves like particle. Since, E = pc or E p Transfer of momentum depends upon energy of incident light. For instance, we have observed in photoelectric effect that when a metal surface is exposed to a light of suitable frequency, electrons are emitted out of surface. It is possible only when photons of light transfer their energy and momentum to surface electrons Why can red light be used in a photographic dark room when developing films, blue or white light cannot? Ans Red light can be used in a photographic dark room when developing films. As we know, the energy of the photon is given by, E = hf However, c = f λ or f = c / λ E = Since the frequency of red light is smaller than that of blue or white light so red light has less energy as compared to blue and white light. Therefore, red light cannot affect the photographic plate and make no chemical reaction on other concerned material. That is why red light is preferred in photographic dark room Photon A has twice the energy of photon B. What is the ratio of the momentum of A to that of B? Ans As we know, the energy of the photon is given by, E = hf Then, Energy of photon A = E A = hf A Energy of photon B = E B = ½ hf A or we can say E B = ½ E A (i) As we know, momentum of photon is given by, p = Now, Momentum of photon A = p A = E A/c (ii) Momentum of photon B = p B = E B/c But from equation (i) Dividing equation (ii) with this equation p A p B = or p B = E A c h λ E B c = hc λ hf c = = E c ½ E A c c ½ E A = p A p B = 2 ½ = 2 Hence, momentum of photon A is also two times greater than the momentum of photon B.

49 9.8. Why don t we observe a Compton Effect with visible light? Ans We do not observe a Compton Effect with visible light. Compton Effect is observed with X-ray and γ-ray photons. These photons have energy, which is thousand times greater than visible light photons. During Compton Effect, an X-ray photon knocks an electron out of the atom. Even after this interaction, the X-ray photon has sufficient energy so that it retains its identity. On the other hand, visible light photon loses all its energy in single interaction with an electron in an atom. Therefore, we cannot observe a Compton Effect with visible light Can pair production take place in vacuum? Explain. Ans No, pair production cannot take place in vacuum. During the process of pair production, the law of conservation of energy and law of conservation of momentum are observed. In order to conserve the momentum and energy, the presence of heavy nucleus is essential. As, pair production takes place near the nucleus, which take the recoil to conserve the momentum. Therefore, pair production cannot take place in vacuum because vacuum has no particle to stop the photon Is it possible to create a single electron from energy? Explain. Ans No, it is not possible to create a single electron from energy. In the phenomenon of pair production, when a photon of energy E = hf greater than 2m oc 2 strikes a nucleus, an electron-positron pair is created. A positron is an antiparticle of an electron. It is also called positive electron. Hence, creation of electron will be against the law of conservation of charge, which cannot be violated If electrons behaved only like particles, what pattern would you expect on the screen after the electrons passes through the double slit? Ans If electrons behaved only like particles, then they would pass through either of the two slits straight and strike the screen only just in the front of the slits to produce exact images of the double slit on the screen. No interference pattern would be seen If an electron and a proton have the same de Broglie wavelength, which particle has greater speed? Ans Electron will have greater speed. According to de-broglie, we know that wavelength associated with a particle is given by h λ = mv or h v = mλ as, de-broglie wavelength λ for both the particles is same and h is also constant, so we can write v = constant m v m It shows that greater the mass of the particle, smaller will be its velocity. As mass of photon is greater than the mass of the electron, so speed of electron would be greater than the speed of photon We do not notice the de Broglie wavelength for a pitched cricket ball. Explain why? Ans Electron will have greater speed. According to de-broglie relation λ = h mv

50 Due to large mass of cricket ball and small speed, the value of wavelength λ associated with a moving cricket ball is extremely small. Therefore, it cannot show the wave property and no diffraction can be produced. Hence, we cannot notice the de-broglie wavelength for pitched cricket ball. In other words, wavelength of wave associated with ball cannot be detected If the following particles all have the same energy, which has the shortest wavelength? Electron, alpha particle, neutron, proton. Ans Alpha particle will have the shortest wavelength. According to de-broglie relation h λ = mv Also, the energy of moving particle E = v = 2E m v 2 = (i) 2 mv2 2E m (ii) Putting (ii) in (i) λ = h m 2E m h λ = (iii) 2mE where h is Planck s constant for the same energy value for all particles, then from equation (iii) we get to know that wavelength λ varies inversely as the square root of mass of the particle, λ m This relation shows that the particle, which has the largest mass, will have the shortest wavelength. Since the mass of alpha particle is greatest, therefore, alpha particle will have shortest wavelength associated with it When does light behave as a wave? When does it behave as a particle? Ans (a) Wave-like nature of LIGHT: Light behaves as wave in the phenomenon of (i) Interference (ii) Diffraction (iii) Polarization (b) Particle-like nature of LIGHT: Light behaves as particle in the phenomenon of (i) Photoelectric Effect (ii) Compton Effect (iii) Pair Production What advantages an electron microscope has over an optical microscope? Ans Electron microscope has many advantages over optical microscope. ELECTRON MISCROSCOPE (i) Focusing: Focusing of invisible electron is done by electron and magnetic field. (ii) Picture: Picture of internal structure of an object can be obtained with the help of electron microscope. OPTICAL MICROSCOPE (i) Focusing: Focusing of invisible electron is done by optical lens. (ii) Picture: Picture of internal structure of an object cannot be obtained by optical microscope.

51 (iii) Resolving Power: It is thousand times greater than that of optical microscope. (iv) Magnification: It is thousand times higher than that of optical microscope. (iii) Resolving Power: It is thousand times less than that of electron microscope. (iv) Magnification: It is thousand times less than that of electron microscope If measurements show a precise position for an electron, can those measurements show precise momentum also? Explain. Ans No, these measurements cannot show precise momentum also. According to Heisenberg s uncertainty principle, position and momentum of a particle cannot be measured simultaneously with perfect accuracy. x p h h p x So, if x is very small then p will be large. Problem 9. t = Problem 9.2 t o v2 c 2 m o m = v2 c 2 Problem 9.3 E = hf = hc λ Problem 9.4 K. E. max = V o e φ = hf hf o = hf V o e = hc λ V oe Problem 9.5 λ = h ( cos θ) m o c Problem 9.6 E = hc λ λ = h ; λ = hc E PROBLEMS m o c ( cos θ) ; λ λ = h ( cos θ) m o c Problem 9.7 E = hf = hc λ Problem 9.8 (a)λ = Problem 9.9 λ = h mv Problem 9.0 h mv (b)λ b = h (c)λ m b v e = h b m e v e x p h ; x m v h ; Δv = h mδx

52 Atomic Spectra 20 QUESTIONS 20.. Bohr s theory of hydrogen atom is based upon several assumptions. Do any of these assumptions contradict classical physics? Ans Bohr s first postulate contradicts the classical physics. Explanation: According to first postulate of Bohr s theory, an electron in an orbit revolving around the nucleus does not radiate energy by radiation. However, according to the classical physics, an accelerated electron radiates energy due to its circular motion around the nucleus. As the revolving electron is accelerated, so it radiates energy in the form of electromagnetic waves. Due to the emission of radiation energy, the electron would gradually come closer to the nucleus and ultimately should be absorbed in it. Classical Physics (Rutherford) Modern Physics (Bohr) What is meant by a line spectrum? Explain, how line spectrum can be used for the identification of elements? Ans Line Spectrum: When the gas at much low pressure is excited by passing an electric current (discharge) through it, the spectrum of emitted radiation is in the form of discrete sharp lines. This type of spectrum is called line spectrum. Line spectrum of an element is due to excitation of electrons from lower to higher energy level. The excited electrons come back to their ground state by emitting radiation which they have absorbed during excitation. The emitted radiations when passed through prism and fall on the photographic film, a spectrum in form of discrete spectral lines is observed which is called line spectrum. Identification of Elements: Elements can be identified by observing their line spectrum because each element when heated gives off different coloured spectral lines of its own which have definite wavelength and frequency. For instance,