tạp chí physics for tháng 4 năm 2015

edit Vol XXIII No April 2015 t was a very highly inspiring advice given to the teachers and students by the Hon’ble prime Minister, Shri Narendra Modi ji himself in a recent radio-talk plot 99, Sector 44 institutional area, Gurgaon -122 003 (Hr) Tel : 0124-4951200 e-mail : info@mtg.in website : www.mtg.in Many books have been written on education by very learned authors These are addressed mainly to higher level students and professors But the difficulties Regd Office 406, Taj apartment, Near Safdarjung Hospital, ring road, New Delhi - 110029 : : how to Excel in any competitive Exams, at any Level i Corporate Office : Managing Editor Editor of the students are not spelt out The prime Minister’s advice reaches directly the students of all levels Mahabir Singh anil ahlawat (BE, MBa) The first is the fear of exams which all of us have felt To combat this, the advice is to hold a week-long examination festival, two times a year, with contents Physics Musing (Problem Set-21) satirical poems on exams, cartoon contents and lectures on the psychological effects of exams with debates to pepper the lectures Comparisons with others is wrong Competition should be with yourself Thought Provoking Problems 11 Compete with yourself, BITSAT Full Length 14 Compete for speed, Practice Paper 2015 JEE Advanced Compete to more, 31 Brain Map 46 AIPMT 48 Practice Paper 2015 57 Practice Paper 2015 CBSE Board Class XII Compete to achieve newer heights, Focus on doing better every-time Practice Paper 2015 AIIMS rial 66 The former Ukrainian pole vault champion Sergey Bubka had broken his own record 35 times! perhaps you might have seen the film ‘Bhaag Milkha Bhaag’ Milkha Singh, the famous champion runner was breaking his own record every time Even after winning, he runs round the track at the same speed waving to the crowds! The final advice given by the prime Minister is valid not only to the students but also to everyone working in any field “live in the present, struggle with the present Victory will walk alongside”, Go ahead, students! Victory is yours Solved Paper 2015 Anil Ahlawat Core Concept 80 Physics Musing (Solution Set-20) 83 You Ask We Answer 84 Crossword 85 Owned, Printed and Published by Mahabir Singh from 406, Taj Apartment, New Delhi - 29 and printed by Personal Graphics and Advertisers (P) Ltd., Okhla Industrial Area, Phase-II, New Delhi Readers are adviced to make appropriate thorough enquiries before acting upon any advertisements published in this magazine Focus/Infocus features are marketing incentives MTG does not vouch or subscribe to the claims and representations made by advertisers All disputes are subject to Delhi jurisdiction only Editor : Anil Ahlawat Copyright© MTG Learning Media (P) Ltd All rights reserved Reproduction in any form is prohibited Editor subscribe online at www.mtg.in individual subscription rates yr yrs yrs Mathematics Today 330 600 775 Chemistry Today 330 600 775 Physics For You 330 600 775 Biology Today 330 600 775 combined subscription rates yr yrs yrs PCM 900 1500 1900 PCB 900 1500 1900 PCMB 1000 1800 2400 Send D.D/M.O in favour of MTG Learning Media (P) Ltd Payments should be made directly to : MTG Learning Media (P) Ltd, Plot No 99, Sector 44, Gurgaon - 122003 (Haryana) We have not appointed any subscription agent physics for you | april ‘15 PHYSICS P MUSING hysics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh The aim of Physics Musing is to augment the chances of bright students preparing for JEE (Main and Advanced) / AIIMS / Other PMTs with additional study material In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / various PMTs The detailed solutions of these problems will be published in next issue of Physics For You The readers who have solved five or more problems may send their solutions The names of those who send atleast five correct solutions will be published in the next issue We hope that our readers will enrich their problem solving skills through “Physics Musing” and stand in better stead while facing the competitive exams By : Akhil Tewari 21 A uniform rod of mass m and length l starts rotating with constant angular acceleration a in a horizontal plane about a fixed vertical axis passing through one end The horizontal component of the net force exerted on the rod by the axis when it has rotated by an angle p/2, is l l (a) ma (b) ma + p2 2 m l pa (c) (d) none of these 2 A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R One of the arcs AB of the ring substends an angle q at the centre The value of the magnetic induction at the centre due to the current in the ring is (a) proportional to 2(180° – q) (b) inversely proportional to r (c) zero, only if q = 180° (d) zero, for all values of q A particle is projected with a speed u in air at angle q with the horizontal The particle explodes at the highest point of its path into two equal fragments, one of the fragments moving up straight with a speed u The difference in time in which the two particles fall on the ground is (Assume it is at a height H at the time of explosion.) u (a) 2u (b) u − gH g g (c) u + gH g (d) 2 u + gH g Consider the cube shaped carriage ABCDEFGH of side l and a mass M and it can slide over two frictionless rails PQ and RS A shot of mass m is thrown from corner A such that it lands at corner F The angle of projection as seen from the carriage is 45° While the shot is in the air, the velocity of carriage as seen from the ground is H D A G l C E R F B S Q P (a) m gl 2(M + m) (b) m gl 2(M + m) (c) m gl (M + m) (d) m gl (M + m) A particle of mass m kept at the origin is subjected  ^ to a force F = ( pt − qx) i where t is the time elapsed and x is the x co-ordinate of the position of the particle Particle starts its motion at t = with zero initial velocity If p and q are positive constants, then physics for you | april ‘15 Page (a) the acceleration of the particle will continuously keep on increasing with time (b) particle will execute simple harmonic motion (c) the force on the particle will continuously keep on decreasing with time (d) the acceleration of particle will vary sinusoidally with time Two rods of equal lengths and equal cross-sectional areas are made of materials whose Young’s modulii are in the ratio of 2:3 They are suspended and loaded with the same mass When stretched and released, they will oscillate with time periods in the ratio of (a) : (b) : (c) 3 : 2 (d) : µ0 I (3p + 4)  8pR µ I (c) (3p − 4)  8pR µ0 I (3p + 4) ⊗ 8pR µ I (d) (3p − 4) ⊗ 8pR (a) (b) 10 Two blocks of masses kg and kg are connected through a massless inextensible string The co-efficient of friction between kg block and ground is 0.4 and the coefficient of friction between kg block and ground is 0.6 Two forces F1 = 10 N and F2 = 20 N are applied on the blocks as shown in the figure Calculate the frictional force between kg block and ground (Assume initially the tension in the string was just zero before forces F1 and F2 were applied.) A thin uniform annular disc (see figure) of mass M has outer radius 4R and inner radius 3R The work required to take a unit mass from point P on its axis to infinity is (a) 24 N (c) 18 N (b) N (d) 10 N  solution of march 2015 crossword (a) 2GM (4 − 5) 7R GM (c) 4R (b) − (d) 2GM (4 − 5) 7R 2GM ( − 1) 5R An ideal gas is expanded so that amount of heat given is equal to the decrease in internal energy Find the adiabatic exponent if the gas undergoes the process TV1/5 = constant (a) 7/5 (b) 6/5 (c) 8/5 (d) None of these The magnetic field at the point P is given by F L U I D I T Y G E E R S Y M A 10 U L T I C E L I N E C T 11 12 E HA L L E F F E N X I 15 U C Y C L O N E 16 S C A I E T T 19 H Y P O T H E S I O N O 24 23 M O M E N B 26 R I P P L E 27 A E 29 C P D 31 I K O H 32 VO L C A P B L C O H 34 B L I N D S P O T 35 U R Y G I N R C F 36 L M E A N F R F P E E 38 R H E Y T O D E S I C T A R A S O N I C C T L A R S T N C E F 14 F O L I A T 17 18 F A H R L 20 L U A 22 T W O F I N E R T I T U G 30 R E D S H I F E T L O 33 E L E C T S A N D E S C E N 37 E E T I M E S F L U X S P E C B I O M 6A S S H M E A S L I G 13 O A L M I O N N E N H E I T A 21 R I C A N T W O 25 M A A V P E 28 R E A P S O L S D A Y S O M A G N E T I A M C C E I I A S T I C T S Y D B I T R N L E Winner (March 2015) neha Gupta Solution Senders Sandeep Kumar rana Atriz roy 10 physics for you | april ‘15 Page 10 By : Prof Rajinder Singh Randhawa* I II III –Q Two small identical balls lying on a horizontal plane are connected by a weightless spring One ball (ball 2) is fixed at O and the other (ball 1) is free The balls are charged identically as a resultant of which the spring length increases h = times Determine the change in frequency? O + + + + + A non-conducting ring y of mass m and radius R, the charge per unit length l is shown in – figure It is then placed – – on a rough non– – x conducting horizontal O  ^ plane At time t = 0, a uniform electric field E = E0 i is switched on and the ring starts rolling without sliding Find the frictional force acting on the ring A non-conducting solid cylindrical rod of length L and radius R has uniformly distributed charge Q Find the electric field at point P, a distance L from the centre of the rod R O P L + Q2 A point charge Q1 = –125 mC is fixed at the centre of an insulated h disc of mass kg The disc rests Q1 on a rough horizontal plane Another charge Q2 = 125 m C is fixed vertically above the centre of the disc at a height h = m After the disc is displaced slightly in the horizontal direction, find the time period of oscillation of disc ++ An infinitely long conducting wire of charge density +l and a point charge –Q are at a distance from each other In which of the three regions (I, II or III) are there points that (a) lie on the line passing through point charge perpendicular to the conductor and (b) at which the field is zero? Solution In the region II, the electric field of wire and point charge point in the same direction, +ve x-axis So no point can exist where the field is zero Now, we take a point to the right of the point charge at a distance x from it Resultant field at this point is  ER = Q λ ^ ^ i+ (− i ) 2 πε0 (x + a) πε0 x *Randhawa Institute of Physics, S.C.O 208, First Fl., Sector-36D & S.C.O 38, Second Fl., Sector-20C, Chandigarh, Ph 09814527699 physics for you | april ‘15 11 Page 11 Resultant field is zero if λ Q = ( x + a) x Q Qa =0 or 2lx2 = Qx + Qa or x − x− 2λ 2λ On solving the quadratic equation in x, we get Q Q Qa ± + 4λ 16 λ2 λ Here, there is only one value of x (with +ve sign) because –ve sign would mean that the point is to the left of point charge Now we take a point to the left of wire at a distance x from it The resultant field is  Q λ ^ ^ ER = i (− i ) + 2 πε0 x πε0 (a + x ) The two fields point in the opposite directions, so resultant field can be zero if, Q λ = πε0 x πε0 (a + x )2 x= Q  or x −  − 2a  x + a2 =  2λ  On solving the quadratic equation in x, we get 1 Q 1Q   x =  − 2a  ± − 2a  − a2    λ  2λ If the discriminant of the quadratic equation is real, we have two points where the field is zero Discriminant is +ve for Q ≥ 8al When the balls are uncharged, the frequency of oscillation is k 2π m where k is the force constant of the spring and m = mass of the oscillating ball (ball 1) When the balls are identically charged, q2 (i) = k(ηl − l ) = kl(η − 1) πε0 (ηl )2 where l is the natural length of spring and hl is the new length of spring after its extension q2 (ii) or l = πε0 η2 (η − 1)k υ0 = When the ball is displaced by a small distance x from the equilibrium position to the right, the unbalanced force to the right is given by 12 q2 − k ( ηl + x − l ) πε0 (ηl + x )2 Using Newton’s law, Fr = −2 x q2  m +  − kl(η − 1) − kx =  2 πε0 η l  ηl  dt Expanding binomially and using eqn (i), we get   2q2 d2 x m = − ⋅ + k x dt   πε0 η3l Using eqn (ii), we get   2q2 ⋅ + k  d2 x = −  πε0 m q2 x η dt   πε0 η (η − 1)k    3η −   2(η − 1)  k + k x = −  kx = −  η   η   3η −  k d2 x (iii) ∴ = − x  η  m dt d2 x By definition of simple harmonic motion, d2 x = −ω2 x .(iv) dt From eqns (iii) and (iv), we get  3η −  k  3η −  k ω2 =  ⇒ υ=   η m π  η  m ∴ υ 3η − = υ0 η 3η − times η Here, h = so frequency increases times Thus frequency is increased Let the radius of the Q2 disc be R If the disc F is displaced x, then h  q = x/R The restoring Fsin torque t about the point x of contact of the disc with ground, tP = (Fsinq)R = Ia   MR2 = + MR2  α   Q2 where, F = 4πε0 (h2 + x ) and sin θ = Fcos Q1 P x h + x2 physics for you | april ‘15 Page 12 Hence, or α = R Q x πε0 MR(h + x ) Q x For x < < h, α ≈ − Q θR =− πε0 MRh πε0 MRh3 Negative sign is being introduced because angular acceleration and angular displacement are opposite to each other Thus, α = − Q2θ πε0 Mh3 Hence, ω = or T = π Q2 πε0 Mh3 πε0 Mh3 Q h πε0 Mh Q N + dF – – – + L /2   + L /2 (L − r )dr dr − ∫  ∫ 2 1/2  πR Lε0  − L /2 − L /2 [(L − r ) + R ]  The second integral can be evaluated by substituting (L – r)2 + R2 = t Differentiating both sides, we get, –2(L – r)dr = dt  + L /2 dt  Q ∴ E= [r ]− L /2 + ∫ 1/2  t  πR Lε0  E= + + + + A force of same magnitude but in opposite direction acts on a corresponding element in the region of negative charge \ Equation of motion for pure rolling is π /2 ∫ λRdθE0 R sin θ − fR = (mR )α and f = ma and a = Ra Solving eqns (i), (ii) and (iii), we get f = lRE0 along +ve x-axis Q = dF = lRdqE0  (Q / L)dr  (L − r ) 1−   πR2 (2ε0 )  [(L − r )2 + R2 ]1/2  E= f mg or λR2 E0 − fR = mR2 α L  (Q / L)dr  (L − r ) 1−  2 1/2  − L /2 πR (2ε0 )  [(L − r ) + R ]  Consider a differential element subtending an angle dq at the centre and at angle q as shown in figure – – +L/2 + L /2 T = 0.6 s  r dr E = ∫ dE = ∫ On substituting the given values, we get d P Consider a disc of radius R of thickness dr at a distance r from the centre O of the cylinder Charge on the disc, Q Q dq = × πR2dr = dr L πR L ∵ Electric field due to disc along its axis  x σ  Ex = 1 −  2ε0  (x + R2 )1/2  Hence, dE = = 2π O –L/2 3/2 L   MR2 = + MR2  α πε0 (h2 + x )3/2   Q xR Q πR2 Lε0 [L + t ] + L /2   L + [(L − r )2 + R2 ]1/2  − L /2      πR2 Lε0  Q 1/2 1/2     9L2   L 2  E= L+ +R  − +R    4 πR Lε0          Q .(i) (ii) (iii) physics for you | april ‘15 13 Page 13 EXAM from th th 14 to 29 MAY BITSAT FULL LENGTH PRACTICE PAPER physics 14 Suppose speed of light (c), force (F) and kinetic energy (K) are taken as the fundamental units, then the dimensional formula for mass will be (a) [Kc–2] (b) [KF–2] –2 (c) [cK ] (d) [Fc–2] A sand bag of mass M is suspended by a rope A bullet of mass m is fired at it with speed v and gets embeded in it The loss of kinetic energy of the system is M v2 M mv (a) (b) 2(M + m) 2(M + m) 2 m v (c) (d) (M + m) v 2 (M + m) A steel wire with cross-section cm2 has elastic limit 2.4 × 108 N m–2 The maximum upward acceleration that can be given to a 1200 kg elevator supported by this cable wire if the stress is not to exceed one-third of the elastic limit is (Take g = 10 m s–2) (a) 12 m s–2 (b) 10 m s–2 –2 (c) m s (d) m s–2 A body of density r at rest is dropped from a height h into a lake of density s where s > r Neglecting all dissipative forces, find the maximum depth to which the body sinks before returning to float on the surface hρ h (a) (b) σ σ−ρ hρ hσ (c) (d) σ−ρ σ−ρ Consider two containers A and B containing identical gases at the same pressure, volume and temperature The gas in container A is compressed to half of its original volume isothermally while the gas in container B is compressed to half of its original volume adiabatically The ratio of final pressure of gas in B to that of gas in A is (a) g–1   (c)   − γ  1 (b)   2 γ −1   (d)   γ −  Soap water drips from a capillary When the drop breaks away, the diameter of its neck is mm The mass of the drop is 0.0129 g Find the surface tension of soapy water (Take g = 9.8 m s–2) (a) 12.9 × 10–3 N m–1 (b) 31.2 × 10–3 N m–1 (c) 40.3 × 10–3 N m–1 (d) 58.6 × 10–3 N m–1 An artificial satellite is moving in a circular orbit around the Earth with a speed equal to half the magnitude of escape velocity from the Earth The height of the satellite above the Earth’s surface is (Take radius of Earth = 6400 km) (a) 6000 km (b) 5800 km (c) 7500 km (d) 6400 km A needle placed 45 cm from a lens forms an image on a screen placed 90 cm on the other side of the lens Its focal length and the size of image if the size of the needle is cm are respectively (a) – 30 cm, 10 cm (b) + 30 cm, – 10 cm (c) – 20 cm, 15 cm (d) + 20 cm, – 15 cm In Young’s double slit experiment distance between two sources is 0.1 mm The distance of screen from the source is 20 cm Wavelength of light used is 5460 Å Then, angular position of the first dark fringe is (a) 0.08° (b) 0.16° (c) 0.20° (d) 0.32° 10 When light of wavelength 400 nm is incident on the cathode of a photocell, the stopping potential recorded is V If the wave of the incident light is increased to 600 nm, then the new stopping potential is (a) 1.03 V (b) 2.42 V (c) 4.97 V (d) 3.58 V physics for you | april ‘15 Page 14 11 Two particles A and B describe S.H.M of same amplitude a and frequency u along the same straight line The maximum distance between two particles is a The initial phase difference between the particles is (a) 2p/3 (b) p/6 (c) p/2 (d) p/2 12 A racing car moving towards a cliff sounds its horn The driver observes that the sound reflected from the cliff has a pitch one octave higher than the actual sound of the horn If v be the velocity of sound, the velocity of the car is (a) v/ (b) v/2 (c) v/3 (d) v/4 kinetic energy equals the original kinetic energy of the lighter particle What is the original speed of the heavier particle ? (a) (2 − ) m s–1 (b) 2(1 + ) m s–1 (c) (2 + ) m s–1 (d) 4(1 − ) m s–1 18 An equilateral triangle of side length l is formed from a piece of wire of uniform resistance The current I is fed as shown in the figure The magnitude of the magnetic field at its centre O is Q 13 The rate of cooling at 600 K, if surrounding temperature is 300 K is R Assume that the Stefan’s law holds The rate of cooling at 900 K is 16 (a) (c) 3R (d) R R (b) 2R 3 14 The ratio of specific heat of gas at constant pressure to that at constant volume is g The change in internal energy of one mole of gas when volume changes from V to 2V at constant pressure P is R (a) (b) PV ( γ −1) γ PV PV (c) (d) γ −1 ( γ −1) O P (a) (c) 16 A thin uniform rod AB of mass M and length L is hinged at one end A to the level floor Initially it stands vertically and is allowed to fall freely to the floor in the vertical plane The angular velocity of the rod, when its end B strikes the floor is (g is acceleration due to gravity)  Mg  (a)   L   Mg  (b)   3L  g (c)   L (d)  g   L  1/2 1/2 17 A particle of mass m has half the kinetic energy of another particle of mass m/2 If the speed of the heavier particle is increased by m s–1, its new I µ0 I (b) πl µ0 I 3 µ0 I πl (d) zero πl 19 Two inductors L1 and L2 are connected in parallel and a time varying current flows as shown in figure The ratio of current I1/I2 at any time t is 15 A boy throws a ball upwards with velocity u = 15 m s–1 The wind imparts a horizontal acceleration of m s–2 to the left The angle q with vertical at which the ball must be thrown so that the ball returns to the boy’s hand is (Take g = 10 m s–2) (a) tan–1 (0.4) (b) tan–1 (0.2) (c) tan–1 (0.3) (d) tan–1 (0.15) R I I1 L1 I I I2 (a) (c) L2 L1 L2 (b) L22 (d) L1 L2 L21 (L1 + L2 )2 (L1 + L2 )2 20 An a.c source is connected across an LCR series circuit with L = 100 mH, C = 0.1 mF and R = 50 W The frequency of a.c to make the power factor of the circuit, unity is 10 (a) 2π Hz 103 (b) 2π Hz 10 −4 (c) 2π Hz 10 −3 (d) 2π Hz 21 The electric field (in N C–1) in an electromagnetic wave is given by E = 50 sin w(t – x/c) The energy stored in a cylinder of cross-section 10 cm2 and physics for you | april ‘15 15 Page 15 15 The 16 IC Vin ~ t IB VBB B VBE C Vout RC Vout E IE t VCE VCC CE amplifier with npn transistor Input resistance (ri): This is defined as the ratio of change in base-emitter voltage (DVBE) to the resulting change in base current (DIB) at constant collector-emitter voltage (VCE) 72  ∆V  ri =  BE   ∆I B V CE Current amplification factor (b): This is defined as the ratio of the change in collector current to the change in base current at a constant collector-emitter voltage (VCE) when the transistor is in active state  ∆I  βac =  C   ∆I  B V CE IB (A) VCE = 10.0 V 100 80 60 40 20 IB P VBE 0.2 0.4 0.6 0.8 1.0 VBE(V) To find the input resistance from input characteristic, mark a point P, draw a tangent on this point The reciprocal of the slope of the tangent gives input resistance Collector current (IC) in mA photodiode is special-purpose diode It is fabricated with a transport window to expose its junction to light radiations It works in reverse bias condition below the breakdown voltage When light of frequency u is incident on the junction, such that the energy of its photons is greater than the band gap of the semiconductor (i.e., hu > Eg), additional electron-hole pairs are created due to the conduction band The photogenerated charge carriers increase the conductivity of the semiconductor Larger the intensity of incident light, larger would be the increase in the conductivity of semiconductor Consider the case of an n-type semiconductor Obviously, the majority carrier density (n) is considerably larger than the minority hole density p (i.e., n >>p) On illumination, let the excess electrons and holes generated be Dn and Dp, respectively: n′ = n + Dn p′ = p + Dp Here n′ and p′ are the electron and hole concentrations at any particular illumination and n and p are carrier concentrations when there is no illumination Remember Dn = Dp and n >>p Hence, the fractional change in the majority carriers (i.e., Dn/n) would be much less than that in the minority carriers (i.e., Dp/p) In general, we can state that the fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current Hence, photodiodes are preferably used in the reverse bias condition for measuring light intensity 12.5 10 7.5 2.5 Base current (IB) 60 A 50 A 40 A 30 A 20 A 10 A 1.5 2.5 3.5 0.5 Collector to emitter voltage (VCE) in volts From output characteristic, find the ratio of change in collector current for the corresponding change in the base current It gives the current amplification factor λ λ 17 (a) : Angular width, q = or d = θ d –7 Here, l = 600 nm = × 10 m × π π q = 0.1° = rad = rad 180 1800 d=? × 10−7 × 1800 \ d= = 3.44 × 10–4 m π (b) : Frequency of a light depends on its source only So, the frequencies of reflected and refracted light will be same as that of incident light Reflected light is in the same medium (air) so its wavelength remains same as 500 Å physics for you | april ‘15 Page 72 λ Wavelength of refracted light, lr = µw μw = refractive index of water So, wavelength of refracted wave will be decreased 18 Inductive reactance, XL = wL fo = 15 m   h Eye lens Objective lens Height of object L E ~ Impedance of the circuit, X L2 + R2 = ω2 L2 + R2 (i) When the number of turns in a inductor coil decreases then its inductance L decreases So, the net impedance of the circuit decreases and current through the bulb (circuit) increases Hence brightness (I2R) of bulb increases (ii) When an iron rod is inserted in the inductor, then its inductance L increases So, Z will increase and current through the bulb will decrease Hence, brightness of the bulb will decrease (iii) A capacitor is connected in the series in the circuit, so its impedance, Z= 2 Z = ( X L − XC ) + R Z=R 3.48 × 106 m Rm = 1.74 × 106 m Distance of object = Radius of lunar orbit R0 = 3.8 × 108 cm Distance of image for objective lens is the focal length of objective lens, fo = 15 m Radius of image of moon by objective lens can be calculated R h tan θ = m = R0 fo i.e., Radius of moon Rm = R (Q XL = XC) This is the case of resonance so maximum current will flow through the circuit Hence brightness of the bulb will increase 19 (a) : Microwave are suitable for radar systems used in aircraft navigation These wave are produced by special vacuum tubes, namely klystrons, magnetrons and Gunn diodes (b) Infrared waves are used to treat muscular pain These waves are produced by hot bodies and molecules (c) X-ray are used as a diagnostic tool in medicine These are produced when high energy electrons are stopped suddenly on a metal of high atomic number 20 (i) : Here, fo = 15 m = 1500 cm and fe = 1.0 cm angular magnification by the telescope in normal adjustment 1500 cm f m= o = = 1500 fe 1.0 cm (ii) The image of the moon by the objective lens is formed on its focus only as the moon is nearly infinite distance as compared to focal length h= Rm × fo R0 = 1.74 × 106 × 15 3.8 × 108 h = 6.87 × 10–2 m Diameter of the image of moon, DI = 2h = 13.74 × 10–2 m = 13.74 cm 21 Einstein’s photoelectric equation Kmax = mv2 = hu – f0 = hu – hu0 …(i) Here, Kmax = Maximum Kinetic energy of photoelectron with speed v u = frequency of incident light f0 (= hu0) = work function of the metal Important features of photoelectric equation (i) Kmax depends linearly on u and is independent of intensity of incident light This happen due quantum nature of light (ii) Since Kmax must be positive, equation implies that photoelectric emission is possible only if hu > f0 = hu0 or u > u0 Thus there exists a threshold frequency u0 (= f0/h) for the metal surface, below which no photoelectric emissions possible From eqn (i) hc Kmax = – f0 λ According to question, Kmax = 2Kmax = hc – f0 λ1 hc – f0 λ2 physics for you | april ‘15 …(ii) …(iii) 73 From eqn (ii) and (iii),  hc  hc − φ0  − φ0  =  λ2  λ1 f0 = Also, or 2hc hc  2 − = hc  −  λ1 λ2  λ1 λ2  hc  2 = hc  −  f0 = hc \ λ0  λ1 λ2  λ0 λ1λ2 λ2 − λ1 = ; λ0 = λ2 − λ1 λ0 λ1λ2 22 Trajectory of a particles in coulomb field of target nucleus Nucleus Only a small fraction of  > 90° the number Incident + of incident -particles a-particles (1 in  < 90° 8000) rebound back This shows that the number of a-particles undergoing head-on collision is small This implies that the entire positive charge of the atom is concentrated in a small volume So, this experiment is an important way to determine an upper limit on the size of nucleus mass of nucleus Density of nucleus = volume A × amu ρ= πR where R = R0 A1/3 A × amu amu Density ρ = = 4 πR0 A πR 3 r = 2.97 × 1017 kg m–3 so, nuclear density is constant irrespective of mass number or size OR Nuclear fission : It is the process in which a heavy nucleus (A > 230) when excited gets split up into two smaller nuclei of nearly comparable masses For example, 235 141 92 U + 0n Ba + 36Kr + 30n + Q 92 56 Nuclear fusion : It is the process of fusion of two smaller nuclei into a heavier nucleus with the liberation of large amount of energy For example, 2 H + 1H 1H + 1H 74 He + 24 2He + n MeV Mass deffect, Dm = (2.014102 + 3.016049) – (4.002603 + 1.008665) = 0.018883 Energy released, Q = 0.018883 × 931.5 MeV = 17.589 MeV 23 (i) : Transformer is a device which is used to bring the high voltage down to low voltage of a.c current It works on the principle of mutual induction of two coils in a transformer (ii) Transformer does not work for dc voltage A dc current gives constant magnetic field and constant magnetic flux through the coil of fixed area of cross-section As there is no change in magnetic flux so there is no induced emf in the coil (iii) The value displayed by the students are gaining knowledge and curiosity to learn new things The values displayed by the teacher are providing good education and helpful 24 (a) Waves diffract when they encounter obstacles Applying Huygens principle it becomes clear A wavefront impinging on a barrier with a slit in it, only the points on the wavefront that move into the slit can continue emitting forward moving waves but because a lot of the wavefront has been blocked by the barrier, the points on the edges of the hole emit waves that bend round the edges Before the wavefront strikes the barrier the wavefront generates another forward moving wavefront Once the barrier blocks most of the wavefront the forward moving wavefront bends around the slit because the secondary waves they would need to interfere with to create a straight wavefront have been blocked by the barrier Each point on the wavefront moving through the slit acts like a point source We can think about some of the effect of this if we analyse what happens when two point sources are close together and emit wavefronts with the same wavelength and frequency These two point sources represent the point sources on the two edges of the slit and we can call the source A and source B as shown in the figure physics for you | april ‘15 Page 74 Each point source emits wavefronts from the edge of the slit In the diagram we show a series of wavefronts emitted from each point source The continuous lines show peaks in the waves emitted by the point sources and the dotted lines represent troughs We label the places where constructive interference (peak meets a peak or trough meets a trough) takes place with a solid diamond and places where destructive interference (trough meets a peak) takes place with a hollow diamond When the wavefronts hit a barrier there will be places on the barrier where constructive interference takes place and places where destructive interference happens A B The measurable effect of the constructive or destructive interference at a barrier depends on what type of waves we are dealing with (b) Refer point 6.14 (4 (ii)), (MTG Excel in Physics) (c) On increasing the value of n, the part of slit contributing to the maximum decreases Hence, the maximum becomes weaker OR (a) Refer point 6.5 (5 (ii) and (iii)), (MTG Excel in Physics) (b) Refer point 6.6 (1), (MTG Excel in Physics) 25 (a) Refer point 1.4 (3), (MTG Excel in Physics)  (b) E = xi So flux passes through y faces of cube which are perpendicular to x-axis The magnitude of electric field at the x a left face (x = 0), z EL = The magnitude of electric field at the right face, (x = a), ER = 2a   So, net flux, f = E ⋅ ∆s = EL Ds cos 180° + ERDs cos0° = + 2a × a2 = 2a3 Assume enclosed charge is q q Use Gauss’s law, f = ; q = e0f ε0 q = 2a3e0 OR (a) When a conductor is placed in an external electric field, the free charges present inside the conductor redistribute themselves in such \ a manner that the electric field due to induced charges opposes the external field within the conductor This happens until a static situation is achieved i.e., when the two fields cancel each other and the net electrostatic field in the conductor becomes zero Dielectrics are non-conducting substances i.e., they have no charge carriers Thus, in a dielectric, free movement of charges is not possible It turns out that the external field induces dipole moment by reorienting molecules of the dielectric The collective effect of all the molecular dipole moments is the net charge on the surface of the dielectric which produce a field that opposes the external field, unlike a conductor in an external electric field However, the opposing field so induced does not exactly cancel the external field It only reduces it The extent of the effect depends on the nature of the dielectric The effect of electric field on a conductor and a dielectric is shown below : E0 � + E0 � + � � free E free++ E0 � in + � � E0 + Ein = 0++ E0 � � � � E0 + + + E0 Ein + � E0 + Ein  0+ Conductor Dielectric The dipole moment per unit volume is called polarisation and is denoted by P For linear isotropic dielectrics, P = cE where c is electric susceptibility of the dielectric medium (b) The electric field inside a spherical conducting shell is zero Q So force experienced by the charge at the point C Q F= E=Q×0=0 Q Q C x A 2Q Force experienced by charge 2Q at point A,  Q  FA = 2Q E A = 2Q  4πε x  26 (a) Refer point 3.2 ((1), (4)), (MTG Excel in Physics) OR (a) Refer point 4.2 (2(i)), (MTG Excel in Physics) (b) Refer point 4.2 (2(vi)), (MTG Excel in Physics) (c) Refer point 4.2 (2(ii)), (MTG Excel in Physics)  physics for you | april ‘15 75 …Contd from Page no 30 Since 1.89 g of acid liberates 11946.14 cal of heat, therefore heat liberated by 122 g (mol wt of benzoic acid) of acid = 73 (c) 11946.14 × 122 = 771126.5 cal = 771.12 kcal 1.89 74 (a) : 2C6H5CHO Benzaldehyde alc KCN  C6H5CH(OH)COC6H5 Benzoin The reaction is known as benzoin condensation 75 (a) : The cell is given as Zn|Zn2+(0.1 M)||Pb2+(0.1 M)|Pb \ E°cell = E°right – E°left = – 0.126 – (– 0.763) V = 0.637 V Now, using Nernst equation 2+ Ecell = E°cell – 0.059 V log [Pb + ] Ecell = 0.637 V – \ [Zn ] 0.059 V [0.1] log [0.1] Ecell = 0.637 V 76 (b) : Compound which can stabilise the charge after removal of proton, will be more acidic 77 (a) : 2+ X = Cu KCN quickly decomposes Cu(CN)2 Yellow ppt [Cu(CN)4]3– in excess KCN CuCN + (CN)2 Colourless soluble complex H2S passed Y = Cd Cyanogen No ppt 2+ KCN Cd(CN)2 in excess KCN White ppt H2S passed Yellow ppt CdS 3+ Z = Fe KCN Fe(CN)3 2– [Cd(CN)4] Colourless soluble complex in excess KCN Reddish-brown ppt K3[Fe(CN)6] Yellow solution 78 (a) : 3rd period elements have more DHeg than 2nd period elements and in a period it increases 79 (c) 80 (a) 81 (d) : sin2 x + sin2 y – sin2 z = sin2 x + sin (y + z) sin (y – z) = sin2 x + sin (y + z) sin (π – x) = sin x [sin x + sin (y + z)] = sin x [sin (π + z – y) + sin (y + z)] = sin x [sin (y + z) – sin (z – y)] = sin x cos z sin y 76 82 (c) : Let R = {(x, y) : x + 2y = 8, x, y ∈ N} \ x + 2y = (which must be a natural number) 8−x \ x = {2, 4, 6} \ y = {3, 2, 1} ⇒ y = \ R ={(x, y) : x + 2y = 8} ⇒ R = {(2, 3), (4, 2), (6, 1)}\Range of R = {1, 2, 3} 83 (b) : A = {1, 2, 3, 4}, B = {1, 3, 5} R = {(a, b) : a < b, a ∈A, b ∈A} = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} \ R–1 = {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)} Now (3, 1) ∈ R–1, (1, 3) ∈ R \ (3, 3) ∈ RoR–1 (3, 1) ∈ R–1, (1, 5) ∈ R ⇒ (3, 5) ∈ RoR–1 (5, 1) ∈ R–1, (1, 3) ∈ R ⇒ (5, 3) ∈ RoR–1 (3, 2) ∈ R–1, (2, 3) ∈ R ⇒ (3, 3) ∈ RoR–1 (3, 2) ∈ R–1, (2, 5) ∈ R ⇒ (3, 5) ∈ RoR–1 (5, 2) ∈ R–1, (2, 3) ∈ R ⇒ (5, 3) ∈ RoR–1 (5, 2) ∈ R–1, (2, 5) ∈ R ⇒ (5, 5) ∈ RoR–1 (5, 3) ∈ R–1, (3, 5) ∈ R ⇒ (5, 5) ∈ RoR–1 (5, 4) ∈ R–1, (4, 5) ∈ R ⇒ (5, 5) ∈ RoR–1 \ RoR–1 = {(3, 3), (3, 5), (5, 3), (5, 5)} 84 (c) : Making the equation of the curve homogeneous with the help of the line, we get  hx + ky  x + y − 2(kx + hy)   2hk   hx + ky  + (h2 + k − c )  =  2hk  or 4h2k2x2 + 4h2k2y2 – 4hk2x(hx + ky) – 4h2ky(hx + ky) + (h2 + k2 – c2)(h2x2 + k2y2 + 2hkxy) = This is the equation of the pair of lines joining the origin to the points of intersection of the given line and the curve They will be at right angles if coefficient of x2 + coefficient of y2 = (h2 + k2)(h2 + k2 – c2) = since [h2 + k2 ≠ 0] ⇒ h2 + k2 = c2 85 (d) : The centre of the circle is (1, 1) and radius = 2 From (a, a) must lie outside the circle, so 2a2 – 4a – > ⇒ a < – or a > θ 2 Now, tan = 2a2 − 4a − physics for you | april ‘15 Page 76 π π θ π As ⇒ a2 − 2a − < 3 2a − 4a − ∴ a − 2a − 15 < ⇒ −3 < a < ∴ a ∈ (−3, −1) ∪ (3, 5) sin 9θ cos 9θ sin θ cos θ sin 3θ cos 3θ 86 (b) : β = + + cos 3θ cos θ cos 9θ cos 3θ cos 27θ ⋅ cos 9θ sin 6θ sin 18θ sin 2θ + + ∴ 2β = cos 3θ cos θ cos 9θ cos 3θ cos 27θ ⋅ cos 9θ sin(3θ − θ) sin(9θ − 3θ) sin(27θ − 9θ) = + + cos 3θ cos θ cos 9θ cos 3θ cos 27θ cos 9θ = tan 3q – tanq) + (tan 9q – tan3q) + tan27q – tan9q) = tan27q – tanq = a \ a = 2B 87 (a) : f (x ) = cos x cos(x + 2) − cos (x + 1) = cos(x + – 1)cos(x + + 1) – cos2(x + 1) = cos2(x + 1) – sin21 – cos2(x + 1) by using cos2A – sin2B = cos(A + B) ⋅ cos(A – B) f ( x) = − sin (constant) π ⇒ x = ⇒ f (x) = − sin2 ⇒ f(x) represents a straight line through π   , − sin 1 which is parallel to x-axis as f(x) = – sin21 is a constant 88 (d) : Each coupon can be selected in 15 ways The total number of ways of choosing coupons is 157 If largest number is 9, then the selected numbers have to be from to excluding those consisting of only to 97 − 87     =  =  Desired probability is    15  157 89 (c) : The total numbers of arrangements is 11! 11! = !2 !2 ! The number of arrangements in which C, E, H, I, S 11! 11 ! = appear in that order =     !2 !2 ! ⋅ 5! 11! 11! 1 ∴ Required Probability = ÷ = = ⋅ ! ! 120 90 (c) : 30 ∑ n(Ai ) = × 30 = 150 i =1 Suppose S has m elements ⇒ 150 = 10m ⇒ m = 15 n ∑ n(Bi ) = 3n = 9m ⇒ n = 3m = 45 i =1 91 (d) : 50C6 – 5C1 40C6 + 5C2 30C6 – 5C3 20C6 + 5C4 10C6 = coefficient of x6 in [5C0 (1 + x)50 – 5C1(1 + x)40 + 5C2(1 + x)30 – 5C3(1 + x)20 + 5C4(1 + x)10 – 5C5(1 + x)0] = coefficient x6 in [(1 + x)10 – 1]5 = 5C1 · (10C1)4 (10C2) = 2250000 92 (b) : y = x +1 ⇒ x2 = − y (i) y When x = 0, y = 3; x = ∞, y = x2 = ⇒ x = 1 2(x2 + 1) = ⇒ x = ⇒ x = 2     − 0 ⋅2 Area =  −  ⋅1 +    2 2 ⇒ Definite integral = + = 2 93 (c) 94 (a) : f (x + 4) = f (x + 2) – f (x) ⇒ f (x + 6) = f (x + 4) – f (x + 2) = f (x + 2) – f (x) – f (x + 2) = – f (x) f (x + 12) = f (x) λ +12 ∫ 95 (b) : f (x)dx = λ 12 ∫ f (x)dx sin xdt ∫ sin2 x + (t − cos x)2 −1 sin x  t − cos x  = tan −1   sin x  −1 sin x x x   = tan −1  tan  + tan −1  cot    2 2 x π 0 ⇒ y > dy = [...]... of cooling is proportional to (T 4 – T 40 ), as per Stefan’s law R′ 900 4 − 300 4 \ = R 600 4 − 300 4 gR 2 7 (d) : v 0 = and v e = 2 gR R+h \ 6.6 × 10 − 34 × 3 × 108  1 1  −   −19 −7  4 × 10 1.6 × 10 6 × 10 −7  = 1.03 V (V0)2 = (V0)1 – 1.03 = 6 – 1.03 = 4. 97 V = [∵ h1 = 5 cm] = 9 4 − 34 4 4 = 34 ( 34 − 1) 4 4 = 80 16 = 15 3 6 −3 3 (2 − 1) 16 or R′ = R 3 CP C − CV 14 (c) : As \ P =γ = γ −1 CV CV C... mg 5 F= + mg = mg 2 2 ( 240 )2 Ω 35 (a) : Resistance of 40 W bulb, R1 = 40 ( 240 )2 Resistance of 60 W bulb, R2 = Ω 60 When bulbs are in series, the effective resistance 2 1  ( 240 )2 1 R = R1 + R2 = ( 240 )2  +  = Ω  40 60  24 420 × 24 21 Current I = = A 120 ( 240 )2 Potential difference across 40 W bulb 21 ( 240 )2 = × = 252 V 120 40 Potential difference across 60 W bulb 21 ( 240 )2 = × = 168 V 120 60 Since... [Co(NH3)4SO4]NO2 (b) [Cr(NH3)4SO4]Cl (c) [Cr(NH3)5Cl]SO4 (d) Both (b) and (c) 58 At 25°C, the molar conductivity of 0.001 M hydrofluoric acid is 1 84. 5 W–1 cm2 mol–1 If its L°m is 502 .4 W–1 cm2 mol–1, then equilibrium constant at the given concentration is (a) 3.607 × 10 4 M (b) 5 .40 4 × 10 4 M (c) 2.127 × 10 4 M (d) 6.032 × 10 4 M 59 Zinc on reacting with cold, dil HNO3, gives (a) ZnNO3 (c) NO2 (b) NH4NO3... physics for you | april ‘15 37 to sustain circular motion Bead will reach the point opposite its starting position and then repeatedly retrace its path executing a periodic motion 4 2 (a,b) : Radiant power of body A = eAsTAA 4 Radiant power of body B = eBsTB A The two powers are equal \ eB sTB4 A = e A sTA4 A or e  4  0.01  4 TB4 =  A  TA4 or TB =   × (5802)  0 81  eB  4 4 5802 1 TB4...  R  2 2 T  ⇒ R2 =  2   T1  2 /3 R1 Here, T1 = 1 h, T2 = 8 h, R1 = 1 04 km \ 8 R2 =   1 2 /3 × 10 4 km = 4 × 10 4 km 2πR1 2π × 10 4 = = 2π × 10 4 km h −1 T1 1 2πR2 2π × 4 × 10 4 v2 = = = π × 10 4 km h −1 T2 8 v1 = Relative velocity of S2 with respect to S1 is v = v2 – v1 = (p × 1 04 – 2p × 1 04) km h–1 |v| = p × 1 04 km h–1 33 (c) : Here Px = P and Py = \ 3P Resultant momentum of A and B P =... half-lives n 1 will be N = N 0   2 16 s t = =4 For A : nA = T1/2 4 s 4 1 N A = 10 −2 kg   = 6.25 × 10 4 kg 2 16 s For B : nB = =2 8s 2 1 \ N B = 10 −2 kg   = 2.5 × 10 −3 kg 2 NA 1 = NB 4 30 (c) : The circuit diagram is as shown below : GM mv 2 GMm = ⇒ v2 = 2 R R R 2πR 4 π2R 2 GM Also, v= ⇒ v2 = = 2 T R T 4 2R 3 2 \ T = GM If T1 and T2 are the time periods for satellite S1 and S2 respectively 32... (a) 1.51 eV (b) 3 .4 eV (c) 13.6 eV (d) 12.1 eV 32 Two satellites S1 and S2 revolve around a planet in coplanar circular orbits in the same sense Their periods of revolution are 1 h and 8 h respectively The radius of orbit of S1 is 1 04 km When S2 is closest to S1, the speed of S2 relative to S1 is (a) p × 1 04 km h–1 (b) 2p × 1 04 km h–1 4 –1 (c) 3p × 10 km h (d) 4p × 1 04 km h–1 physics for you | april... g(1) = 5 and ∫ g (t )dt = 2 If f (x) = ∫ ( x − t ) g (t )dt , 2 0 0 then f ′′′(1) − f ′′(1) is (a) 2 (b) 4 (c) 3 (d) 5 97 The least value of the function x  π 3π  F(x) = ∫ (4 sin t + 3 cos t )dt in  ,  4 4  π /6 (a) 4 3− 2 2 (b) 4 3 −3− 2 2 4 3 + 3 + 2 3+ 2 (d) 2 2 2 98 For x – (a + 3)|x| + 4 = 0 to have real solutions, the range of a is (a) (– ∞, –7] ∪ [1, ∞) (b) (–3, ∞) (c) (– ∞, –7] (d) [1,... an elevator can tolerate is 1 T = stress × area of cross-section 3 1 = × (2 .4 × 108) × (3 × 10 4) = 2 .4 × 1 04 N 3 If a is the maximum upward acceleration of elevator then T = m (g + a) or 2 .4 × 1 04 = 1200 (10 + a) On solving, a = 10 m s–2 4 (c) : The speed of the body just before entering the liquid is u = 2 gh The buoyant force FB of the lake (i.e., upward thrust of liquid on the body) is greater... 2.127 × 10 4 M 1− α 0.633 59 (b) : 4Zn + 10HNO3 4Zn(NO3)2 + NH4NO3 + 3H2O 60 (b) : Aromaticity can be predicted by the use of Huckle’s rule which says that (4n + 2) p-electrons are required in delocalisation system to give it aromaticity 61 (d) 62 (c) (dilute) 49 (a) : ρ = Z= Z× M N0 × a 3 or Z = ρ × N0 × a 3 M 23 2.7 × 6.023 × 10 × ( 40 5 × 10 −10 )3 =4 27.0 i.e., number of atoms per unit cell is 4 Hence, ... ( 240 )2 35 (a) : Resistance of 40 W bulb, R1 = 40 ( 240 )2 Resistance of 60 W bulb, R2 = 60 When bulbs are in series, the effective resistance ( 240 )2 R = R1 + R2 = ( 240 )2 + = 40 60 24 420... eBsTB A The two powers are equal eB sTB4 A = e A sTA4 A or e 0.01 TB4 = A TA4 or TB = ì (5802) 81 eB 4 5802 TB4 = ì (5802 )4 or TB = = 19 34 K 3 Option (a) is correct or According... (a) [Co(NH3)4SO4]NO2 (b) [Cr(NH3)4SO4]Cl (c) [Cr(NH3)5Cl]SO4 (d) Both (b) and (c) 58 At 25C, the molar conductivity of 0.001 M hydrofluoric acid is 1 84. 5 W1 cm2 mol1 If its Lm is 502 .4 W1 cm2 mol1,