JEEACCELERATED LEARNING SERIES www.mtg.in I March 2016 I Pages 96 I ? 30 Practice Papers PHYSICS PHỴSICS Musing Main& Advanced Trust of morethan 1 Crore Readers Since 1982 ^ppmpppp|pwỉ ỊÉM g Si iằ?Ịr Ồ 02135 m [11] [ị: F3 4Tj r òraue PHỴSICS for Managing Editor MahabirSingh Editor Anil Ahlavvat (BE, MBA) ưì o u Volume 24 No. 3 March 2016 Corporate Office: Plot 99, Sector 44 Institutional area, Gurgaon -122 003 (HR). Tel: 0124-4951200 e-mail:
[email protected] vvebsite : www.mtg.in Regd. Office: 406, Taj Apartment, Near Safdarjung Hospital, New Delhi -110029. Physics Musing Problem Set 32 AIPMT Practice Paper JEE Accelerated Learning Series Exam Prep 2016 JEE Advanced Practice Paper Physics Musing Solution Set 31 Live Physics You Ask We Ansvver Crossvvord 8 12 20 24 31 46 60 67 72 85 87 88 89 Subscribe Online at www.mtg.in Individual Subscription Rates ị Combined Subscription Rates lyr. 2 yrs. 3 yrs. 1 1 yr. 2 yrs. 3 yrs. Mathematics Today yỀ 600 ; 775 PCM 900™” • 300 1900 chrrr stry "ocay r.ov; ■ 775 : 3 7 3 900 ' r>o:.i : 1900 P iysks For You 33.: 60.: 775 ■'CMG 1000 •800 ■ 2300 Biology Today 330 600 775 Send D.D/M.O in íavour 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 subscrìption agent. Owned, Printed and Published by Mahabir Singh from 406, Taj Apartment, New Delhi - 29 and printed by Personal Graphics and Advertisers (P) Ltd., Okhla Industria l Area, Phase-II, New DelhL Readers are adviced to make appropriate thorough enq uiries beíòre acting upon any advertisements published in this magazine. Focus/Infoc us íeatures are marketing incentives. MTG does not vouch or subscribe to the claim s and representations made by advertisers. All disputes are subject to Delhi jur isdiction only. Editor: Anil Ahlawat Copyright© MTG Learning Media (P) Ltd. All rights reserved. Reproduction in any form is prohibited. PHYSICS FOR YOU MARCH 16 © PHYSICS Physics Musing was started in August 2013 issue of Physics For You with the sugg estion of Shri Mahabir Singh. The aim of Physics Musing is to augment the chance s 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 vario us topics of JEE (Main and Advanced) / various PMTs. The detailed solutions of t hese problems will be published in next issue of Physics For You. The readers who have solved five or more problems may send their detailed Soluti ons with their names and complete address. The names of those who send atleast f ive 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. SiNGLE OPTiON CORRECT TYPE Find the inductance of a unit length of two long parallel wires, each of radius a, whose centers are a distance d apart and carry equal currents in opposite dir ections. Neglect the flux within the wire. d - a (a) —ln ( d - a 1 (b) 2n V a ) n (c) ^Olnl( d - a 1 (d) n V a ) 3n d - a 2. From a cylinder of radius R, a cylinder of radius R/2 is removed, as shown in the figure. Current flowing in the remaining cylinder is I. Then, magnetic fiel d strength is (a) zero at point A (b) zero at point B 3o1 (c) 2nR at point A (d) -^oI at point B. 3nR 3. A beam of the light is incident vertically on a glass hemisphere of radius R and refractive index V2, lying with its plane side on a table. The axis of beam coincides with the vertical axis passing through the centre of base of the hemis phere and cross sectional R radius of beam is . The luminous spot formed , - is 2 "p on the table is of radius (a) R (c) ■ẼR (3+1) (b) R 2 (d) (3 + l)R gravitationally. Find their velocity of approach when they are separated by a di stance s. (a) G (l + M2) (b) 2s (c) 2G(M1 + M2 ) (d) GM1M2 Gs 1M1M2 5. A system s receives heat continuously from an electrical heater of power lo W. The temperature of s becomes constant at 50°C when the surrounding temperat ure is 20°C. After the heater is switched off, s cools from 35.1°C to 34.9°C in l minu te. The heat capacity of s is (a) loO J°C-1 (b) 300 J°C-1 (c) 750 J°C-1 (d) l500 J°C-1 6. A flywheel rotating about an axis experiences an angular retardation pro portional to the angle through which it rotates. If its rotational kinetic energ y gets reduced by AE while it rotates through an angle 0, then (a) AE c (c) AE c (b) (d) AE AE Vẽ ẽ ,3/2 COMPREHENSiON TYPE 4. Two masses M1 and M2 at an infinite distance apart are initially at rest. The y start interacting 9 PHYSICS FOR You | MARCH 16 For questions 7 and 8 The value of potential energy at the reference point itself can be set equal to zero because we are always concerned only with differences of potential energy b etween two points and the associated change of kinetic energy. A particle A is f ixed at origin of a fixed coordinate system. Another particle B which is free to move experiences a - ( 2a 3 1 force F = + r due to particle A where r is the V r 3 r2 ) position vector of the particle B relative to A. It is given that the force is c onservative in nature and potential energy at infinity is zero. If it has to be removed from the influence of A, energy has to be supplied for such a 1 s s a a Mad about rehearsing? ĩ 400 Tune. Fine tune. Reach the peak of your readmess for JEE with MTG’s 38+14 Years Ch apterwise Soluticns. It is undoubtedly the most comprehensive 'real'question ban k, complete with detailed Solutions by experts. Studies have shovvn that successíul JEE aspirants begin by íamiliarising themselves with the problems that have appeared in past JEEs as early as 2 years in advance . Making it one of the key ingredients for their success. How about you then? Ge t 38+14Years Chaptervvise Solutions to start your rehearsals early. Visit www.mt g.in to order Online. MsbGr •mom lr«arninn M(tU« these books are . The ionization energy Eo is the work that has to be done by an external agent to move the particle from a distance r0 to infinity slowly. exhaustive practice and actual MCQs. 7. Ifthe capacitor i s charged. What is the potential energy function of particle as a function of r? a B (a) -^2 r2 r (c) -4-Ê r2 r (b) -Ọ. experience force in opposite direction of the field. these boo ks íocus on all the basic needs of a student to efficiently prepare for the exam. There are two capacitors of capacitances C and 2C. then the maximum potential difference that can be applied across the combination for their safely use will be 3 (a) Vo (b) 3 Vo (c) 2 Vo (d) 3 Vo 10. The breakdown potential of ea ch capacitor is Vo. electric field is created inside the dielectric. which capacitor will undergo breakdown first? (a) C (b) 2C (c) Both at same moment (d) None of these MTG Presents the BEST BOOK to Excel in IVIHT-CET STRICTLY BASED ON MAHARASHTRA 12™ HSC BOARD 22000+ CHAPTERVVISE-TOPICVVISE MCQs The wait is over!!! The most reliable and updated source for MHT-CET entrance ex am is now available. Due to this field. Find the ionization energy E0 of the particle B. If a very high field is applied in the dielectric. If they are joined in series. This phenomenon is called dielectric breakdown. Ế2 2a p_2 4a (d) Ễ_ a For questions 9 and 10 A parallel plate capacitor is filled with dielectric material. Ifthe voltage across the parallel combination of these two capacitors is increased. Here r0 is th e equilibrium position of the particle. For more iníormation o r for help in placing your order: Call 0124-6601200 or email:info<5>mtg. 9.Available at all leading book shops throughout the country.+ Ễ r2 r (d) á + Ễ r 2 r (a) (c) (b) 2P2 8. The minimum field at which the breakdown occurs is called the dielectric strength of the material and correspon ding potential is called breakdown potential. With comprehensive theory. An excellent blend of accuracy with authenticity.in WHATIS A GRAVITATIONAL WAVE? It's a ripple in the tabric of space and time Courtesy: The Times of India process. t he electrons (which are not free). the outer electrons m ay get detached from the atoms and then the dielectric behaves like a conductor. circumscribing the triangle while preserving the equilateral triangle. 4. Clwmntry and MatlxMTVitKi only Available at ail leading book shops throughout the country. (a) shifts depending on height of breaking (b) does not shift (c) shifts towards body C (d) shifts towards body B 5. For more iníormation o r for help in placing your orden Call 0124-6601200 or e-mail:info@mtg. So no mo e waiting. A 20 cm long capillary tube is dipped in water.the best selí scorer guỉdes for every MHT-CET aspirant.II Iquestions of higher Standard) from entire chapters to make th e students better equipped • 22000+ Chapter\vise-Topicwise MCQs íraned írom each line o( the Maharashtra syllabus with detailed explanations • Previous years’ (2015 2011) MCQs of MH CET. Three identical bodies of mass M are located at the vertices of an equilateral t riangle of side L. go grab your copy today and ensure your Seat in MHT<ET. the length of water column in the capillary tube will be (a) 8 cm (b) 10 cm (c) 4 cm (d) 20 cm * 6. 3 3 The centre of mass of bodies B and C taken together as compared to centre of mas s of body A. The water rises upto 8 c m. They revolve under the effect of mutual gravitational force i n a circular orbit. 3. a body B of mass 12 — M and a body C of mass — M. Their orbital velocity is gM 3GM 3GM 2 GM (a) L (b) u (c) L (d) u A stone of mass m tied to a string of length L is rotating along a circular path with constant speed v. If the entire arrangement is put in a freely falling elevator. HIGHL1GHTS • Strictly based on Maharashtra 12'” HSC Board syllđbus • Synopsis of chapters complemented withillustrations & concept maps • Simple and lucid language • MCQs divided in three levels: Level-1 (Tapicvvise easy and mediuni level questio ns) and Level. The excess pressure inside a spherical drop of radius r of a liquid of s urface tension T is (a) directly proportional to r and inversely proportional to T (b) directly proportional to T and inversely proportional to r (c) directly proportional to the product of T and r . The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring is (a) -v/3 : 5 (b) 12 : -s/3 (c) 1: V3 (d) V5 : vỏ 2.in YA±GIrdnUun M<i!U *K P Singh SET 1 ROTATiONAL MOTiON | GRAViTATiON | MECHANiCAL PROPERTiES OF SOLiDS AND FLUiDS 1. The torque on the stone is 2 mv mv (a) mLv (b) —— (c) (d) zero LL A body A of mass M while falling vertically downwards under gravity breaks into two parts. AIPMT and JEE Main* (Level-lll) • Three Model Test Papers for the final practice •In Mtyviii. The angular velocity of the system immediately after the collision is 6v 6v 6v 6v (a) Tha (b) (c) (d) 4732 a 33 a 40 a 41 a *A renowned physics expert. the guidebooks have been updated to match the syllabus and the exam pattern of all major medical entrance exams. 7. Four wires of the same material are stretched by the same load.(d) inversely proportional to the product of T and r. then the level of water (a) rises (b) falls (c) remains unchanged (d) either rises or falls. A V The fractional compression. both at national and State levels.24 % 9. No vvonder these guidebooks emerged as bes tsellers in a short period of time. 1 PMT Guides ĩ 650 ? 700 ? 550 MTG's Complete AIPMT Guides are lndia's best selling PMT books!! Rich in theoret ical knovvledge with a vast question bank comprising a wide variety of problems and exercises. 8. of water at the bottom of the ocean (given that the bulk modulus of the water = 2. then the p osition of the centre of gravity of the rod is 7 „ 12 10 9 (a) 3m (b) ym (c) ym (d) ^m 10. Which on e of them will elongate most if their dimensions are as follows ? (a) L = 100 cm.91 % (c) 1. A piece of ice is floating in a jar containing water. these guidebooks ensure siudents are ready to compete in the toug hest of medical entrance tests. KP Institute of Physics. The average depth of Indian ocean is about 3000 m. For more iníormation o r for help in placing your order: Call 0124-6601200 or e-mail:info<a)mtg. r = 1 mm (b) L = 200 cm. 100% NCERT ba sed. When the ice melts . 09872662552 E • P HYSICS FOR you | MARCH 16 Presenting lndia's No.in “Application to read QR codes required 11. Chandigarh. If linear density of a rod of length 3 m varies as l = 2 + X. HIGHLIGHTS: . A uniform rod of length 8a and mass 6m lies on a smooth horizontal surface. 100% NCERT based • Comprehensive unitwise theory complemented with concept maps. Two point masses m and 2m moving in the same plane with speed 2v and v respectiv ely strike the rod perpendicularly at distances a and 2a from the mid point of t he rod in the opposite directions and stick to the rod. flowcharts and eas y-to-understand illustrations • Last 10 years' questions (2005-2014) oí AIPMT • Unitwise MCQs with detailed explanations and Solutions • AIPMT Solved Paper 2015 Included S<annow WI rhyour ỉmcrlphone or tabtet • • Over 50% of questions that appeared In AIPMT 2015 were from MTG's Complptp AIPMT Guidps MsbGIruralnR Mnlli Available at all leading book shops throughout the country.2 X 109 N m-2 and g = 10 m s-2) is (a) 0. r = 3 mm .82 % (b) 0.36 % (d) 1. 5 m s-1 13. The maximum and minimum distances of the planet from the sun are r1 and r2 res pectively. 21. in general .r2)3/2 (d) (r1 + r2)3/2 19. the moment of inertia of the disc about an axis parallel to its diameter and touching the edge of the rim is 35 (a) I (b) 2 I (c) 21 (d) 21 22. L/2 and L/3 are connected in series.(c) L = 300 cm. The ir radii are r.15 m min-1. The work done (in J) in blowing to form a soap bubble of surface area 40 cm2. The surface tension of soap solution is 0. Three capillaries of lengths L. If the liquid flows inside the tube with a speed of 0. A planet of mass m moves around the sun of mass M in an elliptical orbit . Then. The cylindrical tube of a spray pump has a cross-section of 8 cm2. one end of which has 40 fine holes each of area 10-8 m2. In an elliptical orbit under gravitational force. Two drops of equal radius coalesce to form a bigger drop. First particle has an acceleration a-1 = (5j + 5 j)ms-2.03 N m-1. then the (a) pressure difference across the ends of second capillary is 8 p (b) pressure difference across the third capillary is 43 p (c) pressure difference across the ends of the second capillary is 16 p (d) pressure difference across the third capillary is 56 p. Then the angular acceleration of the body is (a) 91 (b) 92 (c) 2 91 (d) 2 92 18. r = 3 mm (d) L = 400 cm.05 m s-1 (d) 0. Then.2 X 10-4 (b) 2. Two capillaries of lengths L and 2L and of radii R and 2R are connected in series. The time period of the planet is proportional to E • PHYSICS FOR you | MARCH 16 (a) r + r2) (b) (rj + r2)1/2 (c) (r1 . if stream-line flow is to be mai ntained and the pressure across the first capillary is p. The angle turned by a body undergoing circular motion depends on time as 9 = 90 + 91 t + 92í2. the speed wit h which the liquid is ejected through the holes is (a) 50 m s-1 (b) 5 m s-1 (c) 0. while the acceleration of the other partic le is zero. X = ) 8 nL 8 9 5 7 (a) 9 X (b) 8X (c) 7X (d) 5X 17. The centre of mass of the two particles moves in a path of (a) straight line (b) parabola (c) circle (d) ellipse 14. r/2 and r/3 respectively. What is ratio of surface energy of bigger drop to smaller one? (a) 21/2 : 1 (b) 1 : 1 (c) 22/3 : 1 (d) None of these 16.4 X 10-4 (c) 12 X 10-4 (d) 24 X 10-4 20. The net rate of flow of fluid through them will be (given rate of the flow through single capillary. Two particles of equal mass have velocities V1 = 4i m s-1 and V2 = 4 j m s-1. r = 4 mm 12. The change in potential energy when a body of mass m is raised to a heig ht nR from earths surface is (R = radius of the earth) n (a) mgR (b) mgR (n -1) 2 n n2 (c) mg^^—~ (d) mgR^n---(n +1) (n2 +1) 15. is (a) 1. The moment of inertia of a thin circular disc about an axis passing thro ugh its centre and perpendicular to its plane is I. A door 1. A thin uniform square lamina of side a is placed in the Xy-plane with it s sides parallel to X and y-axis and with its centre coinciding with origin. A layer of glycerine of thickness 1 mm is present between a large surface area and a surface area of 0.0 X 105 N m-2. then velocity of flow of water is (a) 100 m s-1 (b) 10 m s-1 (c) 1 m s-1 (d) 10\/ĨÕ ms-1 31. When the temperature increases. how high does the mercury rise in the other arm from its initial unit? (a) 0. The average mass density of the planet is 2/3 times th at of the earth. the viscosity of (a) gas decreases and liquid increases (b) gas increases and liquid decreases (c) gas and liquid increase (d) gas and liquid decrease. with the same effo rt.1 m2. A rope 1 cm in diameter breaks. Gravitational acceleration on the surface of a planet 6 _______________________________________'. Two rain drops reach the earth with different terminal velocities having ratio 9 : 4. so that it can move with a velocity of 1 m s-1 ? ( Given that coefficient of viscosity = 0.8 m (c) 0.07 kg m-1 s-1) (a) 70 N (b) 7 N (c) 700 N (d) 0. the strain produced in the two wires will be in the ratio (a) 2 : 1 (b) 1 : 1 (c) 1 : 2 (d) 1 : 4 26. If an 5 astronaut can jump to a maximum height of 1..5 X 105 N m-2. the maximum height he can jump on the planet is (a) 1 m (b) 0. An open U-tube contains mercury.6 m wide requires a force of 1 N to be applied at the free end to open or close it. Then value of d is 7 47 9 51 (a) 3a (b) 12a (c) 5a (d) 12 a 28. 23.41 cm (d) 2. Two wires of same material and radius have their lengths in ratio 1 : 2. the reading of manometer falls to 3.2 cm of water is poured into on e of the arms of the tube. Its moment of inertia about an axis passing through a point on the y-axis at a dist ance y = 2a and parallel to X-axis is equal to its moment of inertia about an ax is passing through a point on the X-axis at a distance X = d and perpendicular t o xy-plane.70 N 2 24...25 m 25.5 m on the earth..4 m distance from t he hinges for opening or closing the door is . When 11. fe“/ nz p n s. The ratio of radii of earth to another planet is — .32 cm 30. The maximum tension that may be given to similar rope of diameter 3 cm is (a) 500 N (b) 3000 N (c) 4500 N (d) 2000 N 32. When the va lve is opened. If these wires are stretched by the same force. 27. if the tension in it exceeds 500 N. is —g .35 cm (c) 0. the escape speed on the surface of the planet in km s-1 will be (a) 5 (b) 7 (c) 3 (d) 11 29. With what force the s mall surface is to be pulled. If the escape speed on the surface of the earth is taken to be 11 km s-1.5 m (d) 1. A manometer connected to a closed tap reads 3. Then the ratio of their volumes is (a) 3 : 2 (b) 4 : 9 (c) 9 : 4 (d) 27 : 8 33. where g is the gravitational acceleration on the surface of the earth. / and the ratio of their mean densities is — . The force that is required at a point 0.56 cm (b) 1.(a) tangential velocity is constant (b) angular velocity is constant (c) radial velocity is constant (d) areal velocity is constant. Two spherical soap bubbles of radii r1 and r2 in vacuum combine under is othermal conditions.e. The ratio of the forces experienced by the tw o particles situated on the inner and outer parts of the ring. A body is released from a point. i. An annular ring with inner and outer radii R1 and R2 is rolling without slip ping with a uniform angular speed. R2 y (d) R2 R1 41. distant r from the centre of earth. then the ter minal velocity of this big drop is (a) 80 cm s-1 (b) 30 cm s-1 (c) 10 cm s-1 (d) 40 cm s-1 40. R. If the drops combine to form a single drop big in size. The potential energy of 4 particles each of mass 1 kg placed at the four vertices of a square of side length 1 m is (in SI units) (a) + 4. A thin circular ring of mass M and radius R rotates about an axis throug h its centre and perpendicular to its plane. Youngs modulus Y and area of cross-section A is extended by X.6 N (c) 2. Eight equal drops of water are falling through air with a steady velocity of 10 cm s-1.0 G (b) . The new angular velocity of the ring will be M (a) 4 w (b) 4mw 4m ( M + 4 m (c) Im )w (d) M ) |w M + 4m ) 38. If p is the density of the planet.5 G . If R is the radius of the earth and r > R.4 N (d) 4 N 34.(a) 1. Then the energy stored in the wire is given by 1 YA 2 (a) X2 2L 1 YL 2 (c) “ — X2 (c) 2 A 1 YA 2 (b) 3 LX (d) 2 f x2 36. the time period of nearby satellite i s given by (a) (b) (c) (d) PHYSICS FORyou | MARCH 16 <15 39. with a constant angular velocity w. The resulting bubble has a radius equal to (a) r1 + r2 2 r1 r2 (b) r1 + r2 (c) 4r1 r2 (d) r12 + r22 37. A wire of natural length L.7. Four small spheres each of mass m (negligible radius) are kept gently to the op posite ends of two mutually perpendicular diameters of the ring. (a) 1 (b) 1 R2 ị is (c) (R ì2 V. then the velocity of the body at the tim e of striking the earth will be (a) yịgR (b) ^2 gR (c) WR (d) 35.2 N (b) 3.. u1 is the frequency of the series limit of Lyman series. The angle of deviation is (a) 25° (b) 30° (c) 45° (d) 35° 4. A body weighs 50 g in air and 40 g in water. Consider the nuclear reaction X200 ^ A110 + B80. t he angle of incidence is found to be twice the angle of refraction. When a number of small droplets combine to form a large drop.u2 = u3 (b) u1 = u2 .5.5? (a) 30 g (b) 35 g (c) 65 g (d) 45 g 45. How much would it weigh in a liquid of specific gravity 1. If the binding energy p er nucleon for X.1 MeV respectively. 5. A ray of light passes through an equilateral prism such that the angle o f incidence is equal to the angle of emergence and the latter is equal to 3/4 th e angle of prism. then (a) total volume decreases (b) thermal energy increases (c) thermal energy decreases (d) surface energy increases. then th e energy released in the reaction is (a) 70 MeV (b) 200 MeV (c) 190 MeV (d) 10 MeV 3. If the energy of the photon is increased by a factor of 4.4 G (d) + 6. (a) KE (b) 2 KE KE (c) — 2 (d) 3KE 2 Three particles each of mass m are kept at vertices of an equilateral triangle o f side L. Then. A and B are 7. A uniform electric field and a uniform magnetic field exist in a region in the same direction. A body is orbiting very close to the earth’s surface with kinetic energy KE. Two light sources are said to be coherent (a) when they have same frequency and a varying phase difference (b) when they have same frequency and a constant phase difference (c) when they have constant phase difference and different frequencies (d) when they have varying phase difference and different frequencies. Then angle o f incidence is (a)cos-1 (mj (b)2cos-1 (m] (c) 2 sin-1 (m) (d) 2sin 1 j 2. An electron is projected with a velocity pointed in the s ame direction. (a) u1 . T he energy required to completely escape from it is 43. The gravitational field at centre due to these particles is _ 3GM 9GM 12 GM (a) zero (b) —— (c) —Y~ (d) Ụ3 ~L} 44. u2 is the frequ ency of the first line of Lyman series and u3 is the frequency of the series lim it of the Balmer series.3 G 42.u3 111 111 (c) —-—+ — (d) —-—+ — u2 u1 u3 u1 u2 u3 7. SET 2 OPTiCS | MODERN PHYSiCS | SEMiCONDUCTOR ELECTRONiCS 1. A ray of light passes from vacuum into a medium of refractive index m. then its mome ntum (a) does not change (b) decreases by a factor of 4 (c) increases by a factor of 4 (d) decreases by a factor of 2 6.(c) . Then the electron will .2 MeV and 8.4 MeV. 8. (d) Emitter is heavily doped. collector is heavily doped and base is moderat ely doped.14 X 10-14 J is (Boltzmann constant = 1. collector is moderately doped and base is ligh tly doped. about doping in a transistor? (a) Emitter is lightly doped. The temperature at which protons in proton gas would have enough energy to overcome Coulomb barrier of 4.414 and refracting angle 30° has one of the refracting surfaces silvered. (b) Emitter is lightly doped.1 (b) 10 (c) 100 (d) ^ .33 Ả (c) 1.2 V -[>1—m3 V (c) 200 A (d) 2 X 10-4 A Wavelengths of light used in an optical instrument are 11 = 4000 Ả and 12 = 5000 Ả.67 Ả (b) 3. A prism having refractive index 1. The ratio of intenSitieS at P and Q will be 8 (a) 3 : 2 (b) 2 : 1 (c) V2 :1 (d) 4 : 1 16.53 Ả) (a) 1. Which of the following is correct. Then the distance of closest approach for the alpha nucleus will be proportional to 1 1 2 (a) v (b) 2 (c) 4 (d) Ze2 mv 13. The focal lengths of the objective and the eye piece of telescope are 100 cm and 10 cm respectively. The radioactivity of a certain material drops to 1/16 of the initial val ue in 2 h. At two points P and Q on screen in Young’s double slit experiment. then ratio of their respective resolving powers (corresponding to 11 and 12) is (a) 16 : 25 (b) 9 : 1 (c) 4 : 5 (d) 5 : 4 10.(a) be deflected to the left without increase in speed (b) be deflected to the right without increase in speed E • PHYSICS FOR You | MARCH 16 9. 11. The de Broglie wavelength of the electron in the ground State of the hyd rogen atom is (Radius of the first orbit of hydrogen atom = 0.06 Ả (d) 0. The half life of this radio nuclide is (a) 10 min (b) 20 min (c) 30 min (d) 40 min 14. An alpha nucleus of energy —mv bombards a heavy nuclear target of charge Ze.38 X 10-23 J K-1) (a) 2 X 109 K (c) 6 X 109 K (b) 109K (d) 3 X 109 K 1 2 12. The current in the circuit shown in the figure considering ideal diode is (a) 20 A (b) 2 X 10-3 A 3. collector is moderately doped and base is heav ily doped. waves f rom slits S1 and s2 have a l path difference of 0 and — respectively. A beam of light incident on the other refracting surface will retrace its path. if the angle of incidence i s (a) 45° (b) 60° (c) 30° (d) 0° 17. (c) Emitter is heavily doped. The magnification of the teleSco pe when final image iS formed at infinity iS (a) 0. collector is lightly doped and base is moderat ely doped. (c) not be deflected but its speed will decrease (d) not be deflected but its speed will increase.53 Ả 15. the angle of refraction i nside the prism will be (a) greater for red colour (b) equal but not 30° for both the colours (c) greater for violet colour (d) 30° for both the colours. |He + 4Be ^ 1n +? The miSSing ion in the given nuclear reaction iS (a) proton (b) oxygen-12 (c) carbon-12 (d) nitrogen-12 22. their equivalent power becomes individual powers (in dioptre) are (a) 4.6 % (d) 81. its de Broglie wavelen gth changeS by the factor 11 (a) yỈ2 (b) 2 (c) 2 (d) 2 19.2 % (c) 40. The distance between th e plates is 0. The nu mber of electrons in the oil drop is (a) 10 (b) 5 (c) 50 (d) 20 28. the one which gives a n emission line of the highest frequency is (a) n = 1 to n = 2 (b) n = 2 to n = 1 (c) n = 3 to n = 10 (d) n = 10 to n = 3 27. Two beams of red and violet colours are made to pass separately through a prism of angle 60°. 26. 7 + ^~ D. When they are separated b y a distance of 20 cm. The velocity of the particle a t the end of 2 X 10-2 m path when it starts from rest is (a) 2n/2 X 105 m s-1 (b) 8 X 105 m s-1 (c) 16 X 105 m s-1 (d) 4V2 X 105 m s-1 24.2 % 20. the input characteristics of a transistor is the variation o f (a) Ib against VBE at constant VCE (b) Ic against Vce at constant VBE (c) Ib against Ic (d) Ie against Iq. In Millikans oil drop experiment. An electron is moving in an orbit of another hydrogen atom from which there can be a maximum of three transitions. The ratio of the velocities of the electron in these two orbits is 1 „ 2 5 3 (a) 2 (b) 2 (c) 4 (d) 3 23. When an unpolarized light of intensity I0 is incident on a polarizing sh eet. A particle of mass M at rest decays into two masses m1 and m2 with non-z ero velocities.I2 t1 (b) I1 . Two thin lenses have a combined power of +9 D. 5 (b) 3. then the number of nuclei that have disinteg rated in the time (t2 . Their 5 (d) 1. If the kinetic energy of a free electron doubles. Of the following transitions in the hydrogen atom.9 cm and potential difference between the plates is 2000 V.6 X 105 V m-1. An a-particle of mass 6.I2 (c) (d) (I1 . 29. the intensity of the light which doeS not get tranSmitted iS I0 I0 (a) y (b) Ỵ (c) zero (d) Iữ 21.4 X 10-27 kg and charge 3.8 X 10-14 kg is stationary between the plates.tị) is proportional to (a) I112 . The radioactivity of a sample is Iị at a time t1 and I2 at a time t2.18. In CE mode.2 X 10-19 C is situate d in a uniform electric field of 1. If t he half life of the sample is t1/2. An electron is moving in an orbit of a hydrogen atom from which there ca n be a maximum of Six transitions. The ratio of 11 de Broglie wavelengths of the particles is m m1 . The maximum efficiency of full wave rectifier is (a) 100 % (b) 25. 8 PHYSICS FORyou | MARCH 16 <17 25.I2) t1/2 T1/2 30. In the minimum deviation position. a charged drop of mass 1. 6 (c) 2. The wavelength of the light used in Young’s double slit experiment is l. where the path difference is . Two media having speeds of light 2 X 108 m s-1 and 2. are separated by a plane surface.5 (c) 0.866 (b) 0. To what height the image of the fish is raise d? (Refractive index of water = 4/3) (a) 9 cm (b) 12 cm (c) 3. Then Anode potential SOLUTiON OF FEBRUARY 2016 CROSSWORD ‘g 2P A T E R A 3D E w 4H 0 5F E M T o 6L 0 u L 7R u SF I N D E R 9m A R E ‘°E N R I c H E M E 11 K 12 A D i3L I D A R E B s A 14 D c G R I5G 16s D 17I 0 N I z A T I 0 N R u E I8P c N s p A D A N R 0 s .75 34.(a) 2 (b) m mm (c) l (d) 1 31.8 cm (d) 3 cm 32. b and c respectively. A fish at a depth of 12 cm in water is viewed by an observer on the bank of a lake. If I0 denoteS the maximum intenSity. Ib and Ic be the intensities and fa .707 (d) 0. then the ratio of I and I0 is (a) 0. Let Ia.4 X 108 m s-1.9L I N A c V s 2°c A L 0 R I E 0 T 21b I p 22 B A R Y 0 N N L T 0 R \ N T I G R A V I T Y 25z B 0 s 0 N N z N I I c 26I N F L A T 0 N . The transition from the State n = 4 to n = 1 in a hydrogen like atom res ults in ultraviolet radiation. What is the angle for a ray going from medium I to medium II? (a) sin-1 (6) (b) sin 1 ' 5 ì V12 ) (c) ”1 ) (d) Sin-1 ' 1 ì V 2.1 The figure shows Ỷ Photocurrent variation of photocurrent with anode potential for a photo-sensitive surface for three different radiations. fb an d fc be the frequencies for the curves a. Infrared radiation will be obtained in the tranSi tion from (a) 2 ^ 1 (b) 3 ^ 2 (c) 4 ^ 2 (d) 5 ^ 3 33. T he intensity at a point on the l screen is I. 5 has power 1 D.6 D (b) 0. When same surface is illuminated by ligh t of wavelength 2l. An object is placed at dist ance 1. then stopping potential becomes vs.__________________________\ WINNERS (Pebruary 2016) • Amey Gupta (UP) • rohit Garg (Haryana) solution senders (January 2016) • Harsh Verma (UP) • Mayank Kumar (Bihar) • Lovedeep singh (Punjab) 58 • PHYSICS FOR You | MARCH 16 (a) fa = fb and Ia * Ib (b) fa = fc and Ia = Ic (c) fa = fb and Ia = Ib (d) fb = fc and Ib = Ic 36. then stopping potential is 3 vs. one will find the electronic charge on the moon ratio —-------------——--------------— to be electronic charge on the earth (a) 1 (b) 0 (c) — (d) gM gM gE 337. 43.6 eV . The collector resistan ce and input resistance are 5 kW and 500 W respectively. If the input voltage is 0.6 eV.6 D 39. fe = 2. then the saturation current and the stopping potential respectively are (a) 4 mA and 1 V (b) 12 mA and 1 V (c) 3 mA and 1 V (d) 3 mA and 0. When a piece of metal is illuminated by a monochromatic light of wavelen gth l. If gE and gM are the acceleration due to gravity on the surfaces of the eart h and the moon respectively and if Millikarís oil drop experiment could be perform ed on the two surfaces. After this. An a-particle and a proton are accelerated from rest by a potential diff erence of 100 V. the energy required to remove the electron from the first excited State of Li2+ is (a) 30.0 mA and 0.01 V.2 V (c) 62 V (d) 620 V 38.4 m away from the photoelectr ic cell.3 cm (c) 6.5 V 40. l p The ratio —.5 cm (d) 6. the fringe width (a) can be changed only by changing the wavelength of incident light (b) can be changed only by changing the separation between the two slits (c) can be changed either by changing the wavelength or by changing the sepa ration between two sources (d) is a universal constant and hence cannot be changed.2 m from a photoelectric cell. their de Broglie wavelengths are 1a and 1p respecti vely. A thin convex lens of crown glass having refractive index 1. For compound microscope. When a monochromatic point source of light is at a distance 0. The value of threshold w avelength for photoelectric emission will be 4 (a) 4 l (b) 8 l (c) l (d) 6 l 41. the saturation current and cut-off voltage are 12.5 cm.3 cm 42.5 cm (b) 8.D 27L I T R E R {.8 D (c) 1. What will be the power of similar convex lens but refra ctive index 1.6? (a) 0.6 eV (b) 13.2 D (d) 1. f0 = 1 cm. In common emitter amplifier.62 V (b) 6. What should be length of microscope for normal adj ustment? (a) 8. In Youngs double slit interference pattern.5 V respectively. is (a) 3 (b) 4 (c) 6 (d) 5 44. the current gain is 62.2 cm from object lens. If the binding energy of the electron in a hydrogen atom is 13.to the nearest integer. the output voltage is (a) 0. If the same source is placed 0. (b) 27. (a) 17. (d) 40.RE CONCEPTon Torque Have you ever given it a thought that-o Why are door knobs always attached towar ds the extreme end. (b) 16. (a) 44. (b) 28. (d) 2. (d) 34. (a) 29. (a) 36. (b) 2. (b) 27. (d) 33.4 eV 45. (a) 43. (a) 35. (d) 13. (a) 21. (d) 6. (c) 26. (d) 33. It is also observed that if we fix F. (c) 9. a force F being applied on a point object whose position vector is r as given here. (a) 3.. (d) 10.. (d) 38. (b) 24. (b) 15. (d) 4. (c) 32. (b) 9. increasing r increases rotation capability and decreasing r decreases the rotati on capability. (a) 44. Let us see for example. then what is the wavelength for doubly ionised lithium ion for same transiti on? (a) 3 (b) 3 l (c) 9 (d) 9 l ANSWER KEYS SET 1 1. (c) 29. (d) 18. (d) 19. the obvious answer that you come across is t hat it becomes easier to rotate them. Is this F capable of changing the orientation of r . and only change the point of application. (b) 25. (d) 40. (b) 20. (d) 34. (d) 20. (d) 32. (b) SET 2 1.(c) 3. (c) 39. (a) 45. But why it is so? The answer is TORQUE! Torque of an applied force represents the rotational capability of the applied f orce to rotate the line joining the point of application of the applied force an d the axis of rotation (AOR). i. (b) 13. (b) 10. (a) 36. (d) 24. whenever we talk of rotation capability of F. (c) 22. i. (a) 7.4 eV (d) 122. (c) 43. (b) 7. is it capable of chan ging f ? To understand this we break two components of the applied force F 1. (d) 31. (d) 11. (a) 37. (a) 41. (c) 42. This component clearly can change the orientation. F± :The component of the applied force which is perpendicular to the position v ector. it changes the distance r. F| | : The component of the applied force which is parallel to the position vector. (a) 41.e. Hence the rotation capability of F. (d) 23. If l is the wavelength of hydrogen atom from the transition n = 3 to n = 1. (b) 31. the torque of F with respect to origin is . (c) 30. we think of Fỵ_. (c) 15. (d) 23. (b) 5. (a) 17. (b) 38. Therefore keep in mind. (b) 42. (a) 14. (b) 25. (b) 18. (b) 4. hence we should try to maximi se its value. (a) 12. (c) ■ ■ PHYSICS FORyou | MARCH 16 09 CQ. (a) 21. (b) 19. (c) 39. This component clearly cannot change the orientation. (b) 28. (a) 3. (d) 30. (b) 5.e. (d) 22. 2. (d) 37. (c) 6. (d) 35. (d) 11. (a) 12. (c) 8. (c) 16. (c) 8. far away from the hinged end? WalỊ Knob o Why are the handle bars of wrenches made large? Now that you have started thinking. (c) 14. (b) 45. (d) 26. Kolkata 50 • PHYSICS FOR you | MARCH 16 2. and the direction in which thumb points gives us the direction of the ax is of rotation (AOR). Hence we use following conventions to re present clockwise/ anticlockwise rotation.t = rF± = rF sin9 = (r sin9)F = r± F where r± can be seen as the component of position vector which is perpendicular to the applied force. Anticlockwise: or o or (k) In vector form. Angular momentum (L) It is a measure of the amount of rotational motion of an object with respect to an observation point in the same way as linear momentum is seen as the amount of translational motion in the object. Applied force is parallel to AOR. when can the torque of a force be zero? Two cases 1. Correlating it with linear momentum p = mv. Pure rotation : The velocity of an elemental mass dm. \ With respect to AOR. For example in the above example the torque is anticlockwise. either clockwise or anticlockwise with respect to an observer. The rotation will also have a direction. Clockwise: -AOR % :> or ® or (-k) Contributed By: Bishwajit Barnwal. Let the plane of paper be the xy plane. Aakash Institute. r X F = 0 2. . L = mvr sin9 = m(v sin9)r or mv(r sin9) = mv±r or mvr± 2. it is clear that MOI has the same ro le to play in rotational mechanics which is played by mass in translational mech anics. Applied force passes through the point of observation/AOR. \ With respect to origin. 1. dLAOR = (dm)(wr)(r) \ dLAOR = (dmr2)w \ laor = ídLAOR =(dmr2 )w • 7 = r w AOR •• Laor = IAOR w where Iaor = í dmr2 represents the moment of inertia (MOI) about the chosen axis o f rotation. Pure translation: L = r X p = r X (mv) = m(r X v) Door ịmg AOR: r±-where r is the position vector of the COM with respect to point of observation. In vector representation. v = wr. So. It is measured in 3 different ways depending upon the type of motion. torque of the applied force is t = r X F Remember that torque of a force is axis/point of observation dependent. Hence mg will not create any torque h ere. I am assuming. Hence changing it means changing t. somewhat as given here. we can find it by using right hand curl rule where we curl the fingers of the right hand in the sense of rotation keeping the thumb st raight. 1. The door is hinged at one of its sides which behaves as AOR and the weight mg te nds to turn its AOR itself which is fixed. In this article. that you know the Standard MOI results of differ ent types of objects. since r is the position vector of the point with respect to origin which was our point o f observation. Hence the equation says that the torque of all the forces acting on the object w ith reSpect to AOR iS equal to the product of MOI about the choSen axiS multipli ed with angular acceleration. \ V = wR is not the condition for pure rolling in all cases. PHYSICS FORyou | MARCH '16 \ For pure rolling. Let us consider an object rolling on ground with instantaneous values of linear velocity of COM VCOM> angular velocity w. To find the acceleration. Note: To differentiate between velocity and acceleration. Remember that with respect to COM each point on sphere has tangential as well as radial acceleration. We have chosen two points A and B. dt 1 This relation is similar to F = ma of translational mechanicS.} dt For an object rotating about an axis. •J^vCOM vCOM + 2vCOM vCOM vCOM = ^VCQM _ ị dt ydt) dt u flCOM. The Sphere of radiuS R iS rolling on a plank which iS alSo moving.3. one on the rolling object and the other on th e plank.aR . it is only when the surface is fixed. Vp = 0 \ V = wR. Pure Rolling It iS a Special caSe of tranSlation and rotation of an object where the point of contact of the object doeS not slip over the surface on which it is kept which is possible only if the velocity of the object at the point of contact is same a s the surface. Translation + Rotation : \ With respect to origin L = rCOM X mvCOM COM + Icom w . L AOR = 1AOR w dLAOR d w — ^ cừ = IAOR dt ^ T AOR = Iaor a where a = = angular acceleration. Let uS find a condition for pure rolling. Relation between torque and angular momentum L = r X p dL 1 dp dr 1 — . Both these points are in contact.1 • = r X + X p = r X F + (v X p) dt dt dt = r X F v is parallel to p} 7 dL d \ t = — {Newton's 2nd law in rotational mechanics. I have shown velocity with straight tail and acceleration with zig-zag tail. we fi rs t fi nd th e acceleration of each point with respect to the COM and then we add the acceleration of COM vectorially to each o f these points. linear acceleration of COM «COM and angu lar acceleration a as shown. Let us find out the velocity and acceleration of any arbitrary point on the sphe re. Let us see an example to understand better. VA = VB coR V vẢ = v-(ữR v-wR = Vp This is the required condition! What if the sphere was rolling on a fixed surface? In such case. t seconds later are.mgt = V0 .(iv) 2R 0 When pure rolling starts.T = maCOM . The advantage is that the torque of such forces will be zero.. . which is on the strai ght string.mgt = 2mgt ^ v0 = 2mgt ^ t = 7 mg \ v = v0 .. The acceleration of point P is zero...(i) 52 • PHYSICS FOR you | MARCH 16 With respect to COM. Now before we start solving questions... Whenever in confusion about the selectio n of the point. About the point P on ground.. arbitrarily chosen. the torque of tension would be zero.. to rque is axis specific. LP initially = LP finally ^ mv0r = mvr + Icom w 2 2._ 5 .. Now.(ii) mg Now. tcom = ^COM a ^ (mmg)R = (2mR2ja ^ aR = ^ ..2 V0 = 5 V0 Alternatively. he nce angular momentum remains conserved. But b e careful in the selection since... Find the acceleration of the COM. v = wR 5 7 2 v0 ^ v0 . Find the velocity of COM after pure rolling starts. aCOM fk = mg fk = maCOM ^ mmg = maCOM ^ aCOM = mg . but we prefer those points through which maximum number of unknown forces p ass through.(i) For rotational mechanics. = mvr + — mR w = mvr + — mvr ^ v = — vn 5 5 ... ap = 0 ^ aCOM . since even if it is accelerated. prefer COM.. v = u + aCOM t = v0 .. Q1 : A string is wrapped around a disc and one end of string tied to a ceiling a nd released. pseudo force will p ass through it and hence would not have any torque.mgt .... Hence.(ii) \ Linear velocity (v) and angular velocity (w). remember that for torque calculations. //////// Soln. hence we get a in one eq uation directly \ = L a p p 3 2 2 ^ mgR = — mR a^aR = — g . and the relation between them is the constraint relat ion.. there is no external torque. let me show you a smart way to solve this question. Hence you can choose any arbitrary point for applying t = Ia.(iii) w = w0+ at = 5mgt (. .. I prefer the point of contact P. Let us solve some questions.. it should not be an accelerated point else we will have to consider a pseudo force on the object which would pass through the COM and might have a torque of its own.. since. «COM and a are related. mg .: For translational mechanics.aR = 0 2 ^ aCOM =aR = 3 g aCOM Q2 A : olid sphere is projected by giving it a translational velocity V0 on a ro ugh horizontal surface with friction coefficient m as shown. the point of contact is tangentially unaccelerated but is radial ly accelerated.... Soln. 2.: Since slipping starts .with respect to COM with respect to ground Note here that. w0 = 0) . kinetic friction would act in backward direction due to which COM decelerates and sphere attains angular acceleration in anticlockwise direction as shown. . 1/4 Soln. COM .(iii) Along the length of rod.m = • 5 35 4 35 16 3 18 3 ^ m=_ = 35 5 35 35 ^ m= 3 16 PHYSICS FORyou | MARCH '16 PRACTICEPẠPẸR ©_ JẺẺ MẦĨN .N = mãcoM = m — 4 „ m 48g 12mg ^ mg . 2l mg sin37°.: The only forces acting on the rod are normal force exerted at the hinge a nd weight.(i) For rotational motion. al mg cos 37° . ^ 1/2 \ Củ cô/ m. friction cannot perform any net work.fk = mw — 3 16mg m 72g ^ mg . Find the speed of the free end when the rod become horiz ontal. The point of application of normal force is always at instantaneous r est hence it cannot perform any net non zero work..: Again till slipping does not start.. so 223 only rotational KE about AOR} ^ wl = -\Ịĩgl = velocity of free end..■ it is a case of pure rotation..N = • = 5 4 35 35 ( 4 12 ì____16 ^ N =1------1 mg =— mg ..... Soln. about the point of contact t = Ia \2 4 l 7ml l ml2 (l ) = + m 1 1 4 l 12 14 ) -a ^ al = 48g 35 a 4 ^ mg-=-----5 48 Along the normal. Hence only gravity is perform ing work.Q3 : The rod is hinged at one end and released by disturbing it from its unstabl e equilibrium position.(ii) . Find the friction coefficient between table and rod. it is found that rod rotated about the extreme edge of the table and started slipping after turning through an angle of 37°. l \ PEloss = KEgain l 1 mí1 2 ■ r ^ mg = w {■.. Q4 : On releasing the rod of mass m and length l from the position shown. 20 . What is the change in the diameter of the hole when the sheet is heate d to 227°C? Coefficient of linear expansion of copper is 1. On the back face of slab is a reflecting plan e mirror. Two particles A and B having charges 8 X 10-6 C and -2 X 10-6 C respecti vely.6 m from A (d)0. Two blocks M1 and M2 having equal mass are to move on a horizontal frictionle ss surface. t hough final momentum is equal to initial momentum. then collision c annot be elastic. The reading on the main scale is 2. An observer sees the image of object in mirror [figure]. (b) While spring is fully compressed the system momentum is not conserved.44 X 10-2 cm (b) 2. A block of mass m = 2 kg is resting on a rough inclined plane of inclina tion 30° as shown in figure.0 X 10-6 m2 and it is carrying a current of 3. A spaceship is launched into a circular orbit close to earth's surface. the relative percentage error in the density is (a) 0. without sliding to a radius R/2.8 m s-2) (a) 11.1 m from B 6. (d) If the surface on which blocks are moving has friction. made of an extensible material is rolled along its l ength in the form of a cylinder of radius R and kept on a rough floor. (c) If spring is massless. the final state of the M2 is state of rest. A hole is drilled in a copper sheet.24 cm at 27. The number density of free electrons in a copper conductor estimated is 8.0 m long to its other end? The area of cross-section of the wire is 2. so that block does not slip on the plane? (g = 10 m s-2) E • PHYSICS FORyou | MARCH 16 8. How long does an electron take to drift from one end of a wire 3 . whose pitch is 0.2 km s-1 (b) 8 km s-1 (c) 3.5 mm and the re are 50 divisions on the circular scale. are held fixed with a separation 20 cm. What minimum force F should be applied perpendicular to the plane on the block. (a) 1.5. The additional velocity that should be imparted to the spaceship in the orbit to overcome the gravitational pull is (Radius of earth = 6400 km and g = 9.2 km s-1 (d) 1. If the measured mass of the b all has a relative error of 2%. Distance of i mage from front face as seen by observer will be (a) h + (b) 2h + 2d — d (c) h + d (d) h + — — 5.44 X 10-2 mm (d) 2. 7.5 km s-1 2. Where should a third charged parti cle be placed so that it does not experience a net electric force? (a) 0.2 m from B (b) 0.0 A. A point luminous object (O) is at a distance h from front face of a glass slab o f width d and of refractive index —.4% (c) 3. The diameter of the hole is 4. Initially M2 is at rest and M1 is moving toward M2 with speed v and collides head-on with M2 . M2 is attached to a massless spring as shown in figure.2% 9.1% (d) 4. (a) 6 h 23 min (b) 7 h 33 min (c) 7 h 43 min (d) 6 h 53 min 4.44 X 10-3 cm (c) 1. The coefficient of friction between the block and the plane — = 0.44 X 10-3 mm 3.5 m from A (c) 0. If the ca rpet is unrolled.9% (b) 2.70 X 10-5°C-1. all the kinetic energy of M1 is stored as potential energy of spring. the decrease in potential ene rgy is 1 7 .Exam Dates OFFLINE : 3rd April ONLINE : 9th & 10th April 1.0°C. Mị = m M2 = m (a) While spring is fully compressed. The di ameter of the ball is measured with a screw gauge.5 X 1028 m-3. The density of a solid ball is to be determined in an experiment. A carpet of mass M.5 mm and that on the circular scale is 20 divisions. if one deci des to balance a resistance of 4X against Y? (a) 50 cm (b) 80 cm (c) 40 cm (d) 70 cm 13. A cyclic process ABCA shown in V-T diagram. then where will be new position of the null point from the same end. If the velocity of light in air is c. Two identical capacitors 1 and 2 are connected in series to a battery as sho wn in figure. In a metre bridge experiment null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y. A solid rod o f length L. Now the dielectric s lab is removed and the corresponding charges are Q' and Q 2. If X < Y. cross-section A and density p is suspended freely in the tank. Then the angle of inclination 9 of the rod with the horizontal in equilibrium positi on is (a) sin -1 (b) sin-1 í V n (c) sin -1 í 0 (d) sin -1 11. the stopping potential is _4". Q1 and Q2 are the charges stored in the capacitors.(a) 7-MgR (b) ^MgR 2 8 53 (c) 5MgR (d) -MgR 84 10. is performed with a constant mas s of an ideal gas. The l ower end of the rod touches the base of the tank and h = L/n (where n > 1). the threshold frequency of photoelectric emission is c c 2c 4c (a)ả (b)ả (c)3: (d) 31 14. When a metal surface is illuminated with light of wavelength l. the stopping potential is V0. When the same surface is illuminated with light of wavelength 2l. A liquid of density p0 is filled in a wide tank to a height h. Which of the following graphs in figure represents the corres ponding process on a P-V diagram? 1 n n (a) A' B' ơ (b) Vr c p (c) I B' A' ơ (d) A' B' 'ơ 12. Capacitor 2 contains a dielectric slab of dielectric constant K as shown. Then Q1 = K + 1 1 1 . A body of mass m thrown horizontally with velocity v. A boy of mass 30 kg starts running from rest along a circular path of ra dius 6 m with constant tangential acceleration of magnitude 2 m s-2. A body of mass 2 m thrown horizontally with velocity v. the resi stance of the coil is (a) 30 W (b) 45 W (c) 105 W (d) 75 W 17. 27 V and 12 V respectively. If the like poles are in the same direction. An inductance coil is connected to an ac source through a 60 W resistanc e in series. Therefore. After 2 s f rom start he feels that his shoes started slipping on ground. To convert this galvanometer into a volt meter having a range of 25 V. from the top of to wer of height h touches the level ground at distance of 250 m from the foot of t he tower. If a force is applied to a wi . Two short bar magnets of magnetic moment M each are arranged at the oppo site corners of a square of side d such that their centres coincide with the cor ners and their axes are parallel. it should be connected with a resistance of (a) 2500 W as a shunt (b) 2450 W as a shunt (c) 2550 W in series (d) 2450 W in s eries 18. The lengths (in cm) of the pendulums are (a) 72 and 50 (b) 60 and 38 (c) 50 and 28 (d) 80 and 58 PHYSICS FOR you | MARCH 16 © 16. They oscillate at the same place s o that one of them makes 30 oscillations and the other makes 36 oscillations dur ing the same time. The friction betwe en his shoes and ground is (Take g = 10 m s 2) 11 (a) T (b) (c) 11 (d) (a) 4p d3 (c) M (d) 4p d3 m0 M3 2p (c)i (d) dĩ 2p 2 3 4 5 19. The value of X is (a) 250 m (b) 500 m (c) 125 m (d) 250V2 m 23. from the top of tower of height 4 h 2 will touch the level ground at a distance X from the foot of tower. voltage across the coil and voltage across the resistance are found to be 33 V. A current of 4 X 104 A gives a deflection of one division. Two pendulums differ in lengths by 22 cm.(a) (b) Q1 OÍL Q2 K K +1 : 2 h V Q2 K +1 Q2 (d) Qj_ = K Q1 2 15. The Poission's ratio of a material is 0. A galvanometer of 50 W resistance has 25 divisions. The source voltage.4. t he magnitude of the magnetic induction at any of the other corners of the square is m0 M m0 2M (b) 22. By (b) Ez. An electromagnetic wave travels along z-axis.2km s-1 \ The additional velocity to be imparted to the orbiting satellite for escaping is 11. Then its time period in seconds is (a) (b) 2nS (a) R = 1 kW. find the relationship between the t wo elements. Which of the following pai rs of space and time varying fields would generate such a wave? (a) Ex.8 = 3.5% 24. L = 10 H 25. C = 5 mF (b) R = 1 kW.5 kW 27. If the circuit 4 consist possibly only of RC or RL in series. Bz (d) Ey. Strong resonances are observed at two successive lengths 0.25 (c) 2. di . If velocity of sound is 340 m s-1. In Youngs double slit experiment.5% (c) 1% (d) 0.30 E • PHYSICS FORyou | MARCH 16 26.=>l2gR ^ ve =yfĩ X vo = 1. the resultant intensity when they inte rfere at a phase diff erence — is given by (a) (c) Im I 1 + 2cos2 (b) Im 1 1 + 4cos2 — 3 2 ) 5 V 2 Im 1 1 + 8cos2 -) (d) Im í8 + cos2 —) 9 2) 9 V 2) For escaping from closed to the surface of earth. GMm 1 2 = — mve R 2 e Í2GM ve = -R. 4p 2d3 4p 2d3 20. How much time (in ms) would it take to t raverse the straight fibre of length 1 km? Air (a) 3. In the circuit as shown in figure.2 km s-1 2. the phase difference between emf (e) and current (i) in the circuit is p observed to be .41 X 8 = 11. The perc entage increase in its length is (a) 3% (b) 2. (a) A (b) — A (c) — A (d) A 32 w 32 32 15 29. the magnitude of its velocity is equal to that of its acceleration. (a) : Given. A tuning fork is struck a nd held over the tube. C = 10 mF (c) R = 1 kW. Bx (c) Ey.50 m and 0. then the frequency of the tuning fork is (a) 128 Hz (b) 256 Hz (c) 384 Hz (d) 500 Hz 21 A particle executes SHM with an amplitude of 2 cm.85 (b) 4.84 m above the surface of water. C = 1 H (d) R = 1 k W. then the maximum resista nce one can use without spoiling zener action is (a) 20 kW (b) 15 kW (c) 10 kW (d) 1.2 . there is a decrease of cross-sectional area by 2%. A long glass tube is held vertically in water. Bx 28. so that it just undergoes a total internal reflection.90 (d) 7. the current in ammeter is 2 Q «r 6 V 42 27 15 32 . When the particle is at 1 cm from the mean position. as shown in figure. A light ray from air is incident as shown in figure at one end of the gl ass fibre making an incidence angle of 60° on the lateral surface. one of the slits is wider than the other. If a zener diode (Vz = 5 V and Iz = 10 mA) is connected in series with a resistance and 20 V is applied across the combination. When an AC source of emf e = e0 sin(100 t) is connected across a circuit . so that the amplitude of the light from one slit is double than that from the ot her slit.re of this material. If Im be the maximum intensity. 525p = pdn 4 4 or d22 = 4.70 X 10-5°C-1 Coefficient of superficial expansion.ameter of the hole. the distance of image I from mm' will be . A = 2 X 10-6 m2 Current I = 3 A Charge on electron.) Drift velocity vd Using the relation. d"] h +— I and as the distance of Virtual mirror from m) d .44 X 10-2 cm 30. a = 1. l = 3 m Area of cross-section of wire.494p[1 + 3.(4. (a) 100 nm (b) 300 nm (c) 50 nm (d) 200 nm SOLUTiONS 1. Ad = d2 . vo = Y = gR \ vo =yj (9.4. 4.40 X 10-5 X (227 .8 X 6..0068 2 = 4.6 X 10-19 C Time taken by electron to drift from one end to another of the wire. I = neAvd I or v“=nA "'(ii) PHYSICS FOR you | MARCH 16 Putting the value from eq.24)2 = 4. the time taken by an electron to drift fr om one end to another end is 7 h 33 min.24 cm Initial temperature.72 X 104 s = 7 h 33 min Thus.27)] = 4.4 X106) = 8 km s-1 3.5 X1028 X 1.2544 cm \ Change in diameter. (ii) in eq. (b) : Given. wavelength of the light incident on the film = 600 nm. Number density of electrons.5 X 1028 m-3 Length of wire.24 = 0. (i).6X10-19 X2 X10-6 t = = I 3 or t = 2. A2 = A1(1 + b • AT) = 4. What is the minimum thickness of a thin film required for constructive inter ference in the reflected light from it ? Given.5. Tị = 27 + 273 = 300 K Final temperature. (c) : For spaceship orbiting close to earth's surface mvị GMm R ~ R2 GM i-e-. mirr or MM' at a depth ídV _ 1 1 from its front face. IneA 3 X8. then A2 = ndi 4.0144 cm = 1. n = 8. the refractive index of the film = 1.e.2544 .525p cm If diameter of hole becomes d2 at 227°C.494P cm2 4 Area of hole at 227°C.525 X 4 or d2 = 4. e = 1.494P X 1. A.40 X 10-5 X 200] = 4.494p[1 + 3.d1 = 4. d1 = 4. Length of the wire l (. (a) : As shown in figure glass slab will form the image of bottom i. b = 2a = 3. So the distance of object O from Virtual mirror Ĩ^Observer mm' will be I h + Now as a plane mirror forms image behind the mirror at the same distance as the object is in front of it. T2 = 227 + 273 = 500 K Coefficient of linear expansion. = pr2 = —— ~ 1 4 = .40 X 10-5°C-1 Area of hole at 27°C. If surface on which blocks are moving has friction. X = 0. M1 comes to rest. also M1 = M2.2 + x)2 But | Fc | = 0 1 Then 4pen (8 X10 6)Q (2 X10 6)Q (0. (8 X10 6)Q (2 X10 6)Q (0. of division on circular scale 0. A and B have charges of opposite nature. instead of M2. the block tends to slip down the plane.5 X 2 X 10 cos 30° = 10 X 0.5 2 F = 20 . Fca = 1 (8X10 6)Q' 4pe0 (0.66 N As mg sin 9 > f. (a) : mg sin 9 = 2 X 10 sin 30° = 10 N and f = m R = m mg cos 9 ^ p = 0.2 m 2 X 6. 2d + = h + 5. 7.7 mm mass M . C should be placed closed to B than A. Choice (d) is correct. (c) : Least count of screw gauge pitch no. the particle should be placed on the line AB. los s of energy is involved. the entire kinetic energy of M1 is not stored as potential energy of spring as M2 may move.2 + x)2 X2 =0 which gives.5 mm + 20 X 0. also A has larger m agnitude of charge than B. Then. Hence.the front face of slab is ’ the distance of image I from front face as seen by obs erver will be d h + d .2 + X)2 _3 y -1 (2X10-6Q and Fcb =—1------T---i 4pe X2 Ẹ ĨCB c FJ 20 cm -+MCA Fc = Fca + Fcb 1 4pe. (d) : While spring is fully compressed.5 mm = = 0.866 = 8. D = MSR + CSR X LC = 2.2 X 0. mg sin 30° = m R = m(F + mg cos 30°) \ F = 2 X10 X H .01 mm = 2. If spring is massles s. From fi gure BC = X (say) and charge on C is Q.32 = 2.01 mm. Hence. R = F + mg cos 30° To avoid slipping. the force due to A and B must be opposite in direction.68 N 8. Collision cannot be elastic. 50 Diameter of ball. velocities of M1 and M2 are interchanged on collision. On applying F perpendicular to plane. (a) : The net electric force on C should be equal to zero.17. As. 11% = 3. V T or — = constant „ PV n T As = R = constant T \ P is constant (A' B' is a straight line || to volume axis). V Hence. then net torque on the rod about point A is zero.7 = 2% + 1. sin 9 = — = l or 9 = sin -1 p0 L / n 1 L l n l 1 p0 np 11. Ll (AL p g) — cos 9. From B to C. Fb = Alp0g acting vertically upwards at the centre of buoyancy D. (i) In the second case. the centre of mass is at a height R/2. \ PV = constant (Boyle's law) So.e.20) 80 4 or Y = 4X . When the carpet unrolls itself and has a radius R/2.l 4X l or Y 100 -1 (Using (i)) 4 X 100 -1 or l = 50 cm 13. The upward thrust on rod. CA' is a curve such that P -1.As density. (a) : From A to B. (b) : According to Einsteins photoelectric equation . which is the mid point of length of rod inside the liquid. let l be the length of rod immersed in liquid. = = — = — Y (100 . 9 be the angle of inclination of rod with horizontal in equilibrium position. From C to A. p = = volume 4p| D j3 Relative error in the density. X 20 20 1 12.(Al p0 g) —cos 9 = 0 or L 2 p0 p L or = l h Now. (c) : Refer to figure. i. temperature T is constant. correct representation is in figure (a). The mass left over unrolled is M p(R/2)2 = M pR2 _ 4 \ The decrease in potential energy = MgR-('7] g ('2 )= 7 MR 10. As the rod is in equilibrium position. (a) : In the first case. The weight of rod = mg = ALpg acting vertically downwards at the centre of gravi ty C of the rod.1% 9.. (b) : The centre of mass ofthe whole carpet is originally at a height R above the floor.. volum e V is constant (B' C is a straight line || to pressure-axis). Ap = AM 3AD p = M D Relative percentage error in the density is E • PHYSICS FORyou | MARCH 16 Ap (am 3AD ì p l M D ) = — X100 + — X100 = 2 + 3 X — X100 M D 2.. — = . .. we get .. Ỵ = eVs +f0 where.(i) 1 hc eV0 = 0 + fn 21 4 Subtract (ii) from (i).l2 = 22 cm Solving the eqs..1 " T l2 where N1 = 30 and N2 = 36 ÍỊ[ .Ni_ .= eV0 + f0 . (c) : Let C be the capacitance of capacitor without slab.N2 T2 N1 \ T.hu = Kmax + f 0 hc . 1 = wavelength of incident light f 0 = work function Vs = stopping potent ial According to given problem hc . (i) and (ii).30 Also.Q2 . (a) : Time period of a pendulum is T ."í Also. we get l1 = 72 cm and l2 = 50 cm 16.(ii) hc " 12 ( 12 1 . (i)..N .r . (b) : L r R = 60Q —'UIMP—W\ÁM--------wm-----27 V------12 V ------------ữ-------------33 V Let r be resistance of the coil. .q2 _ 2K 15.Q2 KCV K +1 After the slab is removed. Before the slab is removed C1 = C and C2 = KC (C)(KC) ( K Cnet C1C2 C1 + C2 C + KC K+1 C 2 PHYSICS FOR you | MARCH 16 © Ql .C Cnet C + C 2 Q1 . l1 .2 2=eV0| 1 -1 1 1 V 4) hc 3 = — eV0 or 40 II 21 2 hc 3 1 Substituting the value of eV0 in eq. C1 = C and C2 = C (C)(C) . ĨL . we get hc 2 hc hc — =-------oròn = — 1 3 1 0 0 31 \ Threshold frequency f0 hc c u° = h = ĩĩh = 31 14.Ci Q1 — Q2 — K +1 Hence> qI .36 l2 . we get 332 . T ro 3 . of div isions = 4 X 10-4 X 25 = 10-2 A Given V = 25 V Hence.v .G . 18. m2 -1 . (ii) 17.50) .s. 2 p 2 p 2p As ro .x2 or.. wx . (ii) . F = mmg 1 or 100 = m X 30 X 10 or m19..122 . A = 2 cm Magnitude of velocity from mean position = <wA2 -~x'2 and acceleration = m2x. ..roV4 -1 or> /3.. Magnetic induction at point E due to magnet at D .272 .332 .. I . (c) : Here.2(l2 -\) .2(0. 12 V Current in the circuit.272 -122 + 2 X12 X Vr V .B1... V 25 R . Resultant magnetic induction (magnitude) at point E is B . . speed of boy will be v = 2 X 2 = 4 m s-1 At this moment.—— 4p d3 It acts along EF.vị + (VR + vr )2 Subtract (i) from (ii). required resistance.272 . (i) ..84-0.B2 4nd 3 20..w/A2 . (b) : After 2 s.9V (2 X12) Vr 9 V Vr = Ir ^ r = ~ĩ = 0. (d) : From v = 2u(l2 .2 A -45 W .—— or T ..2X0. centripetal force on the boy is Tangential force on the boy is Ft = ma = 30 X 2 N = 60 N Total force acting on b oy is F -yjf2 + f2 -^J(80)2 + (60)2 -100 N At the time of slipping. (d) : G = 50 W Ig = Current for full scale deflection = Current per division X total no. Now.vR + v2 + 2VRVr . (i) ..2A 60 W According to voltage formula 272 . . (a) : s i 0 N d è2 E Bi íỉ j N N-------------H Magnetic induction at point E due to magnet at F .vỊ + v2 332 .V2 332 .340 340 u..— ..l1) . . m0 2M (axial point) is B1 . un M (equatorial point) is B24p d3 It acts along FE.50 = 2500 . .0.50 = 2450 W Ig 10-2 This resistance of 2450 W should be connected in series to convert the galvanome ter into a voltmeter.34 = 500 Hz 21. ELECTRONiC DEViCES Classification of Solids Solids can be classified on the basis of conductivity and energy bands. • Insulators : The conduction band is empty and valence band is completely filled. the two energy bands are distinctly separate without any overlapping. and high rank in JEE (Main and Advanced) / BITS AT by reading this column. Thei r electric conductivity lies between 102 S m-1 and 108 S m-1. Their conductivity lies between 10-11 S m-1 and 10-19 S m-1. Forbidden energy gap (Eg) is th e energy gap between the top of the highest valence band and bottom of the lowes t conduction band. solids are classified as follows : • Metals : In metals or good conductors. A solid crystal contains about 1023 atoms/c m3. So each atom is in the electrostatic field of neighbouring atoms. Their conductivity lies between 105 S m -1 and 10-6 S m-1. the energy bands which consist of closely spa ced completely filled energy states at 0 K are called valence bands.E • PHYSICS FORyou | MARCH 16 Contd. on page no. Types of energy bands Energy bands are of two types : • Valence band • Conduction band For semiconductors and insulators. Classification on the basis of energy bands Each electron in an atom has definite energy value. • Insulators : Solids having very low electric conduc-tivity are known as insulato rs. So in soli ds where atoms are closely spaced.This specially designed column is updated year after y ear by a panel of highly qualiíied teaching experts well-tuned to the requirements of these EntranceTests.g -H 5Ị> 85 3 <u ! >3 . 71 ACCELERATED LEARNING SERIES Unit(9) Electronic Devices I Communication Systems Maximize your chance of success. solids can be classified in three categories. • Semiconductors : Solids whose conductivity is intermediate to that of metals and insulators are known as semiconductors. The bands w ith higher energies are called conduction bands. atoms are closely packed. the atomic energy levels of the electrons bro aden and give rise to energy bands. Classification on the basis of conductivity On the basis of conductivity. In solids. • Metals : Solids having very high electric conductivity are known as metals. • Semiconductors : In semiconductors. There is no forbidden energy gap. Overlapping Conduction band Conduction band Valence band <v°> Valence band (a) (b) Difference between the energy bands of a metal PHYSICS FOR you | MARCH 16 o. conduction band is either partially fille d (figure (a)) or overlaps (figure (b)) the valence band. These definite energy values are called energy levels. On the basis of band theory. Extrinsic Semiconductors These are obtained by doping the pure semiconductor with small amount of certain impurities of either trivalent or pentavalent atoms. In p-type semiconductor. i. called holes. Sb or Bi is added to a pure semicond uctor. The trivalent impurity atoms are called acceptor ato ms. number of holes become more than number of electrons. As. where n is called the int rinsic carrier concentration. the fermi level shifts towards the conduction band. It has electrons as majority carriers and holes as minority carriers. When electric field is applied across an intrinsic semiconductor. semiconductor becomes deficient in electrons.Empty conduction band V3eV Valence band Diíĩerence between the energy bands of an insulator ữ tì §3 0J Í-H c w 8 T <3 I Eg<3eV DiAèrence between the energy bands of a semiconductor intrinsic Semiconductor A semiconductor in pure form is called intrinsic semiconductor.05 eV n -type Semiconductor When a pentavalent impurity. Extrinsic semiconductors are of two types: • p-type semiconductor • n-type semiconductor p -type Semiconductor When a trivalent impurity like B. They have covalent bonding. ỉ = ỉe + ỉỵ. Electrons moving to the conduction band leave behind the va cancy of electron. A pure semiconductor has negative temperature co-efficient of resistance. Al. In the p-type semiconductors. holes are majority charge carriers. In n-type semiconductor. where ỉe = free electron current and ỉỵ = hole current. whereas electr ons are minority charge carriers. the number of electrons become more than the holes in the semiconductor a nd such a semiconductor is called n-type semiconductor.. The number density (ne) of electrons in conduction band is equal to the number d ensity (nh) of holes in the valence band ne = nh = ni. such as P. in the valence band. the fermi level shift s towards the valence band. At ord inary temperature some electrons absorb energy from lattices and move to the con duction band.e. electrons and holes move in opposite directions so that conventional current. Mobility of hole is smaller than that of electron. Acceptor Majority ions carriers Conduction band Acceptor level s o o Valence band Eg«leV e: = 0. They are tet ravalent elements. Hole has positive charge e qual to that of electron. This happens due to breaking of covalent bond because of the effec t of thermal energy.I . Ga or In is added to a pure semiconductor. The deliberate addition of a d esirable impurity to intrinsic semiconductor in controlled quantities to promote conductivity is called doping. Germanium (Ge) a nd Silicon (Si) are important examples of intrinsic semiconductors. Such a semiconductor is called p-type semiconductor. The pentavale . 02 mA (d) 0.05 mA (b) 0. The conductivity of an intrinsic semiconductor is si = nie(me + mh) The conductivity of n-type semiconductor is sn = eNdme The conductivity of p-type semiconductor is sp = eNamh p-n Junction When donor impurities are introduced into one side and acceptors into the other side of a single crystal of an intrinsic semiconductor. higher is the probability of Ẽ2 • PHYSiCS FOR you | MARCH 16 their recombination. E)onor Majority ions camers Conductíon band Donor leve] í. It is symbolically represented by o-o The most important characteristic of a p-n junction is its ability to conduct cu rrent in one direction only. Since this region has immobile ions . In forward biasing. a p-n junction is formed . in which free electron further recombine with hole. The current voltage relation of diode is given by I = (e1000 V/T . the region containing the uncompensated acceptor an d donor ions is known as depletion region. what will be the error in the value of current in mA? (a) 0. The recombination occurs due to electron colliding w ith a hole.2 mA (c) 0. If a student makes an error measuring ± 0. At equilibrium. V is negative and high. Ir = . It is also known as junction diode. Apart from the process of generation. The current in the junction diode is given by I = I0 (eeV/kT -1) where k = Boltzmann constant. then forward current. generation of free electron and hole ta kes place. Independent of the amount of d oping by acceptor and donor impurities.5 mA _________________________________(JEE Main 2014) Depletion Region In the vicinity of junction. eeV/kT < < 1.nt impurity atoms are called donor atoms. Hence for a given semiconductor. ne = nh = ni so rate of recombination = Rn2 Rne X nh = Rn2 ^ n2 = ne X nh Under thermal equilibrium. where the applied voltage V is in volts and the temperature T is in degree Kel vin. rate of generation of charge carriers is equal to rate of recomb ination of charge carriers. where R = recombination coefficient For int rinsic semiconductor. then reverse current. Mass action law . There is a depletion of mobile charge s (holes and free electrons) in this region.1) mA . I0 = reverse saturation current.Io C5selfchec^____________________________________ 1. rate of recombination ne X nỵ so rate of recombination = Rne X nh. If = Io (eeV/kT-1) In reverse biasing. the product of the concentration ne of free electrons and the concentration nh of holes is a constant. In the other (reverse) direction it offers very hig h resistance.01 V while measuring the current of 5 mA at 300 K. larger value of ne or nỵ. ne X nh = ni Conductivity of semiconductor The conductivity of semiconductor is given by s = e(neme + nhmh) where me and mh are the electron and hole mobilities and e is the electronic cha rge. V is positive and low. 7" ĩ Valence band Eg = leV Mass Action Law In semiconductors due to thermal effect. a process of recombination also occurs simultaneously. The I-V character istics of a p-n junction are as shown in the figure. if the revers e bias voltage is continuously increased.7 V. It conducts well in the forward direction and poorly in the reverse direction. the width of the depletion region incre ases and barrier height increases. In reverse biasing. The difference of potential from one side of the barrier t o the other side is known as the height of the barrier. whereas for a g ermanium p-n junction it is approximately 0. Knee Voltage In forward biasing. The electric field between the acceptor and the donor ions is known as a barrier. There can be two different causes for the br eak down. This reverse bias voltage is thus known as breakdown voltage.7 V. then the p-n junction is said t o be reverse biased. the width of the depletion region decreases and barrier he ight reduces. However. Breakdown Voltage A very small current flows through p-n junction. when it is reverse biased. In forward biasing. The resistance of the p-n junction becomes low in forward biasing. The reve rse current is almost independent of the applied voltage. • The thickness of the depletion region is of the order of one tenth of a micromet re. It would have been ideal if a dio de acts as a perfect conductor (with zero voltage across s • PHYsics FoR you | MARCH 16 it) when it is forward biased. In forward bias. The flow of the current is due to the movement of minority charge carriers. • The width of the depletion layer and magnitude of potential barrier depends upon the nature of the material of semiconductor and the concentration of impurity a toms. for a certain reverse voltage.which are electrically charged it is also known as the space charge region. F or a Silicon p-n junction. Dynamic Resistance It is defined as the ratio of a small change in voltage AV applied across the pn junction to a small change in current AI through the junction. the voltage at which the current starts to increase rapidly is known as cut-in or knee voltage. PHYsics FoR you | MARCH 16 €3 The resistance of the p-n junction becomes high in reverse biasing. and as a perfect insulator (with no current flow through it) when it is reverse biased. the barrier potential is about 0. reverse Biasing of a p -n Junction When the positive terminal of the external battery is connected to n-side and th e negative terminal to p-side of a p-n junction. it acts as a closed switch whereas in reverse bias it acts as a . • The physical distance from one side of the barrier to the other is known as the width of the barrier. the cur rent through the p-n junction will increase abruptly.3 V. I-V Characteristics of a p -n Junction The I-V characteristics of a p-n junction do not obey Ohm’s law. _ AV r _ AI Ideal diode A diode permits only unidirectional conduction.3 V while for silicon it is 0. The I-V characteristics of an ideal diode is shown in figure below. Forward Biasing of a p -n Junction When the positive terminal of external battery is connected to p-side and negati ve to n-side of p-n junction. One is known as Zener breakdown and the other is known as avalanche br eakdown. then the p-n junction is said to be forward biased . t ỉ o Forward Reverse An ideal diode acts like an automatic switch. For germanium it is 0. input and output waveforms for a full wave rectifier are as shown in the figure.I.1 A and 0. Rectifiers are of two types • Half wave rectifier • Full wave rectifier Half Wave Rectifier The circuit diagram. Peak value of current is I = dc value of current is Id = n rms value of current is Irms = —f= ' 2 Peak inverse voltage is P. Rectifier is based on th e fact that.1 A (b) 0.2 A (d) 0. V. rms value of current is I _ 1 m ~ 2 dc value of current is Idc _ I m n Peak inverse voltage is P. input and output voltage waveforms for a half wave rectifie r are as shown in the following figure.2 A and 0. In an unbiased n-p junction electrons diffuse from n-region to p-region because (a) holes in p-region attract them (b) electrons travel across the junction due to potential difference (c) electron concentration in n-region is more as compared to that in p-regi on (d) only electrons move from n to p region and not the vice-versa (JEE Main 2015) Rectifier It is a device which converts ac voltage to dc voltage.4 A and 0. The forward biased diode connection is (a) (b) (c) (d) (JEE Main 2014) 3. 0 Output voltage Vn I = m m~rf + Rl where Tf is the forward diode resistance. 5—1> 5-[X opSệ Closed • ---•----o Open • •---o 2.2 A (c) 0.2 A and 0. A 2 V battery is connected across AB as shown in the figure. = Vm dc value of voltage is Vdc _Idc Rl = ~mRL Full Wave Rectifier The circuit diagram. The value o f the current supplied by the battery when in one case battery’s positive terminal is connected to A and in other case when positive terminal of battery is connec ted to B will respectively be (a) 0.4 A _(JEE Main 2015) 4. = 2 Vm dc value of voltage is Vd = Id Rl =-nm rl Ripple Frequency ur = u = 50 Hz (half wave rectifier) ur = 2Vị = 100 Hz (full wave rectifier) . a forward bias p-n junction conducts and a reverse bias p-n junctio n does not conduct.V.I.n open switch as shown in the figure below. Rl is the load resistance and Vm is the peak value of the alternating voltage. . I I m m_ rms dc n r = I -1= 1.483 2 Rectifìcation Efficiency The rectification efficiency tells us what percentage of total input ac power is converted into useful dc output power. Thus.Ripple Factor The ripple factor is a measure of purity of the dc output of a rectifier. I 2I m m rms dc 2 n 2 I r_ 'ỊmỊ 2 2I I n -1 = 0. • For a full wave rectifier. rectification efficiency is define d as PHYSICS FOR you | MARCH 16 C5 h = dc power delivered to load ac input power from transformer secondary For a half wave rectifier. dc power delivered to the load is Pdc=I dA=(■ ^) Rl Input ac power is Pac = Iỉms(rf + Rl) = (■ ^) r + RL) Rectification efficiency h = P P„.6%. average or dc value I dc For half wave rectifier.6 1 + rf / Rl % If rf << Rl maximum rectification efficiency.21 . dc (Im (n) Rl X100% = (Im/2)2 (rf + Rl) 40. Im I n For full wave rectifier. h = 40. dc power delivered to the load is Pdc = I lRL = (■ ) RL Input ac power is Pac = Ir2ms(rf + Rl ) = ( I2) (rf + Rl ) Rectification efficiency P (2Im / n)2 Rr . and is defined as rms value of the components of wave r = r =. GaAs is used f or making infrared LED. The symbol of a photodiode is shown in the figure below. Form Factor Form factor = 1 dc • For half wave rectifier. It works on the same principle (photovoltaic effect) as the photodiode. It is designed to operate under reverse bias in the breakdown region and is used as a voltage regulator. The semiconductor used for fabrication of visible LEDs mus t at least have a band gap of 1./g Rs Vị 'QV-r Vo -ả K^key point The Zener diode is always reverse biased and this reverse-bias voltage should be greater than the breakdown voltage. h = 81. The compound semiconductor gallium arsenid e phosphide (GaAsP) is used for making LEDs of different colours. equals (a) (Vt . The I-V characteristic of an LED is ểể7/ (JEE Main 2013) Photodiode A photodiode is a special type p-n junction diode fabricated with a transparent window to allow light to fall on the diode. But the threshold voltages a re much higher and slightly different for each colour.h = PP-=— m 2—L— X100% Pac (Im /V2) )rf + Rl ) 81. The reverse breakdown voltages of LEDs are very low. ( 1self checkỊỊ 6.2 = 1 + rf / Rl % If rf<< Rl maximum rectification efficiency.VL)/n IL . o--------p>|------o nSELF CHECK^ 5. The symbol of a LED is shown in the figure. needed in the dc voltage regulator Circuit sho wn here. A solar cell is basically a p-n junction which generates emf when solar radiation falls on the p-n junction.8 eV. It is a heavily doped p-n junct ion which under forward bias emits spontaneous radiation. The value of the resistor. „ I / V2 n Formfactor = m . It is operated under reverse bias. Form factor = \m /2 = — = 1.57 ỉm / n 2 • For full wave rectifier. W hen it is illuminated with light of photon energy greater than the energy gap of the semiconductor. Zener Diode as a Voltage Regulator The circuit diagram for Zener diode as a voltage regulator is shown in the figure below. except t hat no external bias is applied and the junction area is kept large. RS. electron-hole pairs are generated in near depletion region. The I-V characteristic s of a LED is similar to that of Si junction diode. s • PHYSICS FOR you | MARCH 16 y Zener diode It was invented by C. Zener.2%. _ = = 1. The symbol for Zener diode is shown in the figure.11 2Im / n 2 2 Light Emitting Diode (LED) It converts electrical energy into light energy. typically around 5 V. u o-------1>------o Solar Cell It converts solar energy into electrical energy. . The base is very lightly doped. Condition Emitter junction Collector junction Region of operat ion I. Although the two outer regions are of the same type (n-type or p-type). The two regionS have differ ent physical and electrical properties. A transistor can be operated in any one of the following three configurations : • Common emitter (CE) • Common base (CB) • Common collector (CC) input Characteristics of a Transistor The variation of the input current with the input voltage for a given output vol tage is known as input characteristics of a transistor. RF Reverse biased Forward biased Inverted Transistor as a Switch .. A transistor has two p-n junctions. their functionS cannot be interchanged. Transistor A transistor is basically a Silicon or germanium crystal containing three separa te regions. IB is base current. The schematic representations of a n-p-n and p-n-p transistors are shown in the figure. It has three regions. One junction is between the emitter and the base. Action of a Transistor A transistor has two junctions-emitter junction and a collector junction. RR Reverse biased Reverse biased Cut off IV. FR Forward biased Reverse biased Active II. In most transistors. or simply collector junction. The doping of the collector is between the h eavy doping of the emitter and the light doping of the base. FF Forward biased Forward biased Saturation III. The base passes most of these electrons (holes in case of p-n-p) on to the collector. Emitter Base Collector \ 7 * E°Emitter Base Collector \ ^ -oc E°-°c The symbols for n-p-n and p-n-p transistors are shown in the figure below. The emitter is heavily doped. This condition is often described as forward reverse (FR). Ie = Ib + Ic where IE is emitter current. The function of the emitter is to emit or inject electrons (holes in case of a p-n-p transistor) in to the base. IC is the collector current. and is called the emitter-base junction. In condition I.. The middle region iS called the baSe and the two outer regionS are called the emitte r and the collector. the collector regio n is made physically larger than the emitter region Since it iS required to diSS ipate more heat. where emitter junction is forward biased and collector junction is reverse biased. and is very In operation of a transistor.. The collector has the job of collecting or gathering these ele ctrons (holes in case of a p-n-p) from the base. T he other junction is between the base and the collector. and is called collector -base junction.(b) (V. Output Characteristics of a Transistor The variation of the output current with the output PHYSiCS FOR you | MARCH 16 voltage for a given input current is known as output characteristics of a transi stor. or simply the emitter junction. It can either be n-p-n-type orp-n-p-type.+ vL)/n Il (c) (V. There are four possible ways of biasing these two junctions as shown in the table.vL)/(n + 1) Il (d) (V + vL)/(n + 1) Il (JEE Main 2015) thin. R AV = V =«ac X R Power Gain It is defined as the ratio of output power to the input power. Power gain (in dB) = 10 log P° Common base Amplifier In common base transistor amplifier. “ac I AIj AI Voltage Gain It is defined as the ratio of output voltage to the input voltage. Negative sign represents tha t output voltage is opposite in phase with the input voltage. the CE configurations of . A = Yọ. 3 =AAC ac AI B Voltage Gain It is defined as the ratio of output voltage to the input voltage. A = Output power (Po) _ 3 v A Input power (P.) 3ac x A Note : Voltage gain (in dB) = 20 log10 V1 _ 20 log10 AV P .a 1 + 3 aac ^ AV 3 1 " »■■■■ • Due to large value of the current amplification factor. = V V 3ac x Ro Ri where Ro and Rị are the output and input resistances. Power Gain It is defined as the ratio of the output power to the input power. 3 = — Pdc = T B ac Current Gain It is defined as ratio of change in collector current (AIc) to the change in bas e current (AIb). dc Current Gain It is defined as the ratio of the collector current (Ic) to the base current (Ib ). Transistor as an Amplifier When the transistor is used in the active region. the input signal voltage and the out put collector voltage are 180° out of phase. it acts as an amplifier. Common Emitter Amplifier In the common emitter transistor amplifier. dc Current Gain It is defined as the ratio of collector current (Ic) to the emitter current (IE) . output power (Po) p input power (Pị) Relationship Between a and 3= a 3 • a = 1 -a. = Ic adc = f í L ac Current Gain It is defined as the ratio of change in collector current (AIc) to the change in emitter current (AIE). the input signal voltage and the output col lector voltage are in the same phase.When the transistor is used in the cut off region or saturation region. it acts as a switch. Each logic gate follows a certain logical relationship between input an d output voltage. The block diagram of an oscillator is shown in the figure. OR Gate An OR gate has two or more inputs but only one output. The logic symbol of OR gate is Y The truth table for OR gate is Input Output A B Y 0 0 0 0 1 1 1 0 1 1 1 1 The Boolean expression for OR gate is Y = A + B AND Gate An AND gate has two or more inputs but only one output. There are three basic logic gates : • OR gate • AND gate • NOT gate Truth Table It is a table that shows all possible input combinations and the corresponding o utput combinations for a logic gate. s • PHYSICS FOR you | MARCH 16 Transistor as an Oscillator A transistor can be used as an oscillator. The circuit diagram of the tuned collector oscillator is shown in the figure bel ow. An oscilla tor is a self sustained amplifier in which a part of output is fed back to the i nput in the same phase (called positive feedback). It has only one input and one o utput. NOT gate is also called inverter because it inverts the input. The frequency of the oscillation is given by u = 1 2nyfĩc LOGiC GATES A digital circuit with one or more input signals but only one output signal is k nown as logical gate. The logic symbol of AND gate is The truth table for AND gate is Input Output A B Y 0 0 0 0 1 0 1 0 0 1 1 1 The Boolean expression for AND gate is Y = A-B NOT Gate The NOT gate is the simplest of all logic gates. It is called AND gate be cause output is high only when all the inputs are high. It is called OR gate beca use the output is high if any or all the inputs are high.transistor is preferred over the CB and CC configurations. An oscillator generates ac output signal without any input ac signal. The logic symbol of NOT gate is The truth table for NOT gate is Input Output A Y 0 1 . The logic gates are the basic building blocks of a digital system. ATh = Ã + B NAND is equivalent to bubbled OR gate. The logic symbol of NOR gate is The truth table for NOR gate is Input Output A B Y 0 0 1 0 1 0 1 0 0 1 1 0 The Boolean expression for NOR gate is Y = A + B NAND as a Universal Gate NAND gate is called as universal gate because with the repeated use of NAND gate we can construct any basic gate NOT gate from NAND gate Y = A AND gate from NAND gate Y = A ■ B = A ■ B OR gate from NAND gate Y = A ■ B = A + B = A + B NOR Gate as a Universal Gate NOR gate is called as universal gate because with the repeated use of NOR gate w e can construct any basic gate. 7. The logic symbol for NAND gate is A Ai B' Y The truth table for NAND gate is Input Output A B Y 0 0 1 0 1 1 1 0 1 1 1 0 The Boolean expression for NAND gate is Y = Ã7Ẽ NOR Gate It is an OR gate followed by a NOT gate.1 0 The Boolean expression for NOT gate is Y = A PHYSICS FOR you | MARCH 16 © NAND Gate It is an AND gate followed by a NOT gate. PHYSICS FOR you | MARCH 16 Boolean identities A + B = B + A A-B = B-A A + (B + C) = (A + B) + C A-(B-C) = (A-B)-C A-(B + C) = A-B + A-C A + B-C = (A + B)(A + C) A + 0 = A A-0 = 0 A + 1 = 1 A1 = A A + A = A A-A = A A + A = 1 A ■ A = 0 II II ^ II II ^ A + B = A ■ B A ■ B = A + B A + A-B = A A'(A + B) = A A + A ■ B = A + B A ■ (A + B) = A ■ B ( jSELFCHECK^. NOT gate from NOR gate Y = A AND gate from NOR gate Y = A + B = A ■ B = A ■ B OR gate from NOR gate Y = A + B = A + B De Morgan's Theorems A + B = Ã ■ B NOR gate is equivalent to bubbled AND gate. Truth table for system of four NAND gates as shown in figure is . In an analog s ignal.g.Bandwidth is 20 kHz. current or voltage value varies continuously with time. which pick up the information s ent out by the transmitter is called receiver. . The process of changing some characteristics e. Signal A processed information converted into electrical pulse for transmission is called a signal. • Speech signals .Bandwidth is about 4. The sound waves cannot be transmitted from a radio transmitter by c onverting them into electrical waves (audio signal) directly for the following r easons. There are two forms of signal : • Analog signal : A signal. frequency or phase of carrier wave in accordance with instantaneous value of modulating signal is k nown as modulation. Bandwidth of Transmission Medium For co-axial cable.Bandwidth is 2800 Hz. Bandwidth of Signals The frequency range of a signals is called its bandwidth. • Transmission channel : The medium or the link. • Receiver : The part of the communication system. which has only two levels (either low or high) is called a digital signal. Need of Modulation The sound waves within the range of human hearing have frequency range from 20 H z to 20 kHz. • Digital signal : A signal.425 GHz for Uplink) (3. a high frequency carrier wave is used to carry the informatio n signal over a long distance.925-6. Different types of signals require different ranges of frequencies for proper communication. bandwidth offered is 750 MHz. The combination of gates shown below yields (a) NAND gate (b) OR gate (c) NOT gate (d) XOR gate (AIEEE 2010) COMMUNiCATiON SYSTEMS The set up used for exchanging information between a sender and receiver is call ed communication system.2 MHz. These ranges of frequencies are called bandwidths of the corresponding signals. • Standard AM broadcast (~ 540-1600 kHz) • FM broadcast (~ 88-108 MHz) • Television (~ 54-890 MHz) • Satellite communication (5. which is a discontinuous function of time. Decoding Encoding demodulatìon Block diagram of communication System A communication system is composed of three basic units : • Transmitter : The part of communication system. In modulation. • Video signals . PHYSIcS FoR you | MARCH 16 <D (Normally operated below 18 GHz) For free space communication.7-4. • Transducer : A device which converts energy in one form to another is called a t ransducer. (from 300 Hz to 3100 Hz) • For music .2 GHz for Downlink) Modulation Modulation process is used for transmission of signal from transmitter to receiv er. which is a continuous function of time. which transfer message signal fro m the transmitter to the receiver of communication channel. which sends out the information is called transmitter. amplitude.___________________________________(AIEEE 2012) 8. (ii) where wc = angular frequency of carrier wave.. when the frequency of t he signal to be transmitted is below 15 kHz..e..wm) = 2wm S2 • PHYSICS FOR you | MARCH 16 • For detection of AM wave. This signal is made to pass through a band pas s filter which rejects the d. For a frequ ency of 15 kHz of audio signal. m= m = max ^ Ac Amax + Amin Ac = Peak amplitude of high frequency carrier wave. It also makes the direct transmissi on of audio signal as impractical. (b) Carrier wave (c) Amplitude modulated wave The amplitude modulated signal given by A cm ơ) = Ac sin wct + ^2 cos(Wc-w m)t mA . • Required bandwidth = (wc + wm) . essential condition is 1 . an audio signal cannot be transmitted directly. Am = Peak amplitude of low f requency modulating signal. • The energy radiated from an antenna is practically zero. the audio signal from different transmitting stations will get hopelessly and inseparably mixed up.• For efficient transmission and reception. The following are three types of modulation. It allows a band of frequencies wc . Amplitude Modulation In amplitude modulation the amplitude of carrier wave is varied in accordance with the amplitude of the audio frequency modulatin g signal. wc .. 20 Hz to 20 kHz. m < 1 or else Ac distortion occurs. To set up a vertical antenna of this size is practically impos sible.. the length of the antenna comes out to be of the order of 5000 m. If modulating signal m(t) = Am sinwmt . • Due to the fact that all audio signals from different sources possess frequencie s in the same range i. Jc The amplitude modulated signal contains d. Am. The aforementioned difficulties faced during the transmission of audio signal. It is bec ause.c cos(w +w )t 2 c m This modulated wave contains frequencies (wc . wc and (wc + wm). In amplitude modulation (AM): A A — A • Index of AM. components and also low and high frequencies. The frequency of amplitude modulated wave remains unchanged as that of the carrier wave.wm and wc + wm are called the lower side band and upper side band frequenci es respectively. i f transmitted directly.<< RC where RC = time constant of the Circuit. (i) and carrier wave c(t) = Ac sinmct . wm = angular frequency of modulati ng signal. the transmitting and receiving antenna must have a length equal to quarter wavelength of the audio signal. are overcome by the process of modulation. ỉ 0) TỈ 1 cL . Ac = Peak amplitudes of modulating and carrier waves respectively.wm. = —7m (amplitude modulation index). A ụ.wm).c.(wc .c component and also some different fr equencies which are not required. wc + wm to pass through. wc. I The frequency spectrum of amplitude modulated wave is shown in above figure. The two side band frequencies have equal amplitude (= mAc/2) which never exceeds ha lf the carrier amplitude. The amplitude modulation index (m) define the quality of the transmitted signal. When modulation index (m) is small, variation in carrier amplitude will not be large, therefore, audio signal being transmitted will not be strong. As the modu lation index (m) increases but less than 1, the audio signal on reception become s more clear. Frequency Modulation In frequency modulation, the frequency of carrier wave is varied in accordance w ith the audio frequency signal. The instantaneous values of the voltage of carri er waves and modulating signal can be represented as ec = Ec coswct ... (i) and em = Emcoswmt ... (ii) where ec, Ec, wc are the instantaneous value, peak value, angular frequency of t he carrier and em, Em and wm are the instantaneous value, peak value and the ang ular frequency of the modulating signal. The frequency of modulated signal varies between kfE \ ( kfE \ f . \ =f min \ c 2n and ỉm = fc + 2n V / where kf is the proportionality constant and determines the maximum variation in frequency of modulated wave for a given modulating signal. The maximum swing of frequency of modulated wave from the carrier frequency is c alled frequency deviation (8). 8 = fmax - fc = fc - ./min The modulation index (mf) of the frequency modulating wave is 8 mỉ = L. = m = J_c min fm Bandwidth of FM : In frequency modulated signal, the audio signal is contained i n the sidebands. Since the sidebands are separated from each other by the freque ncy of modulating signal (fm), bandwidth = 2n X frequency of modulating signal where n is the number of signifi cant sidebands pairs. Advantages of frequency modulation : The following are a f ew advantages of frequency modulation over amplitude modulation. • FM reception is quite immune to noise as compared to AM reception. Noise is a fo rm of amplitude variations in the transmitted signal due to atmosphere, industri es, etc. In FM receivers, the noise can be reduced by increasing the frequency d eviation or by making use of amplitude limiters. • FM transmission is highly efficient as compared to AM transmission. • In FM transmission, all the transmitted power is useful; whereas in AM transmiss ion, most of the power goes waste in the transmitted carrier, which contains no useful information. • Due to a large number of sidebands, FM transmission can be used for the stereo s ound transmission. Disadvantages : The following are a few disadvantages of FM transmission: • The bandwidth in FM transmission is about 10 times as large as that needed in AM transmission. As a PHYSICS FOR you | MARCH 16 03 result, much wider frequency channel is required in FM transmission. • FM reception is limited to line-of-sight. Due to this, area of reception for FM is much smaller than that for AM. • FM transmitting and receiving equipments are very complex as compared to those e mployed in AM transmission. c£SfghÈcV____________________________________ 9. A signal of 5 kHz frequency is amplitude modulated on a carrier wave off requency 2 MHz. The frequencies of the resultant signal is/are (a) 2005 kHz, 2000 kHz and 1995 kHz (b) 2000 kHz and 1995 kHz (c) 2 MHz only (d) 2005 kHz and 1995 kHz (JEE Main 2015) Demodulation The process of detection or demodulation is the inverse process of modulation. I n the process of demodulation, audio signal is separated from the modulated sign al. Modem and Fax Modem is contraction of the term modulator and demodulator. A modem acts as a mo dulator in transmitting mode and as a demodulator in receiving mode. The electro nic transmission of a document at a Radio waves The electromagnetic waves of frequency ranging from a few kilohertz to few hundr ed megahertz are called radio waves.. Frequency range and wavelength range of radiowaves are given in the table. distant place via telephone line is known as facsimilie or FAX. Earth's Atmosphere The gaseous envelope surrounding the earth is called earth’s atmosphere. Earth’s atm osphere mainly consists of nitrogen 78%, oxygen 21% along with a littile portion of argon, carbon dioxide, water vapour, hydrocarbons, sulphur compounds and dus t particles. Earth’s atmosphere helps in the propagation of electromagnetic waves from one plac e to another place. The various regions of earth’s atmosphere are as follows: • Troposphere : It extends upto a height of 10 km, over earth’s surface. • Stratosphere : It extends from 10 km to 50 km. There is an ozone layer in strato sphere which mostly absorbs high energy radiations like ultraviolet radiations e tc. coming from outer space. • Mesophere : It extends from 50 km to 65 km. • Ionosphere : It extends from 65 km to 400 km. • In this region, the temperature rises to some extent with height, hence it is ca lled thermosphere. The ionosphere which is composed of ionised matter (i.e., ele ctrons and positive ions) plays an important role in space communication. The io nosphere is subdivided into four main layers as D, E, Fị and F2 as shown in the ta ble. Attempt free Online test Log on to http://test.pcmbtoday.com Name of the layer Approximate height over earth’s surface Exists during Frequencies most affected D 65-75 km Day only Reflects LF, absorbs MF and HF to some d egree E 100 km Day only Helps surface waves, reflects HF F1 170-190 km Day time, merges with F2 at night Partially absorb s HF waves yet allowing them to reach F2. F2 300 km at night, 250-400 km during day time Day and night Efficien tly reflects HF waves, particularly at night 54 • PHYsics FoR you | MARCH 16 S. No. Frequency band Frequency range Wavelength range Main use 1. Very-low frequency (VLF) 3 kHz to 30 kHz 10 km to 100 km Long dis tance point to point communication 2. Low frequency (LF) 30 kHz to 300 kHz 1 km to 10 km Marine a nd navigational purpose 3. Medium frequency (MF) 300 kHz to 3 MHz 100 m to 1 km Marine a nd broadcasting purposes 4. High frequency (HF) 3 MHz to 30 MHz 10 m to 100 m Communications o f all types 5. Very high frequency (VHF) 30 MHz to 300 MHz 1 m to 10 m T.V., Radar and air navigation 6. Ultra-high frequency (UHF) 300 MHz to 3000 MHz 10 cm to 1 m Radar and microwave communication 7. Super-high frequency (SHF) 3 GHz to 30 GHz 1 cm to 10 cm Radar, r adio relays and navigation purposes 8. Extremely high frequency (EHF) 30 GHz to 300 GHz 1 mm to 1 cm Optical fibre communication Space Communication The term space communication refers to sending receiving and Processing of infor mation through space. There are three types of space communication. Ground wave propagation This mode of propagation can exist when the transmitting and receiving antenna a re close to the surface of the earth. For radiating high efficiency signals, the size of the antenna should be of the order of 1/4. 1 = wavelength of the signal . The field component of such a launched wave soon becomes vertically polarised as it glides over the surface of the earth. The electrical fields due to the wave induce charges on the earth’s surface. The ground wave is weakened as a result of energy absorbed by the earth during its propagation. These losses make ground wa ves unsuitable for very long range communication. Ground wave propagation can be sustained only at low frequencies (~ 500 kHz to 1 500 kHz) or for radio broadcast at long wavelengths. • The ground wave propagation is generally used for local band broadcasting and is commonly called medium wave. The maximum range of ground or surface wave propag ation depends on : ► frequency of the radio waves and ► power of the transmitter. Sky wave propagation (ionospheric propagation) A transmitted wave going up is reflected back by the ionosphere which forms an i onised layer of electrons and ions around the earth. The ionosphere (including mesosphere and part of stratosphere) extends from about 65 km to 400 km above th e earths surface. Constituent gases are ionised in it by solar radiation. Throug h out the ionosphere, there are several layers in which the ionisation density e ither reaches a maximum or remains roughly constant. These regions are designate d as D(65-75 km), £(100 km) and £(130 km-400 km) in order of approximate heights abo ve earth’s surface. During day time the £ layer splits into separate layers called £ ( 170-190 km over earth’s surface) and £2 (250-400 km over earths surface). During nig ht £ layer usually disappears. Hc = 8 X 10» nr3 F2layer ne = 3 X 1011 nr3 Fì layer ne=> 2 X 10» rrr3 E layer ne = 109 rrr3 Dlayer Ẻarth Region of high conductivity is confined to a relatively thin layer at the lower edge of £-region and to the upper part of D-region. Therefore high frequencies are attenuated when they penetrate this region. D and £ layer disappears during night and low and medium frequency communication also becomes possible. Ionisation de nsity increases as we go up from D layer to 300 km |)170km I ▲ 100 km 65 km 0 km PHYsics FoR you | MARCH 16 © BRAIN > ReverseBiasCharacteristics Widthofdepletionlayerincreases Effective barrier potential increases High resistanceofferedatthejunctionand lowcurrentflowoftheorderof pA.MAP ► SEMICONDUCTOR ELECTRONICS SEMICONDUCTOR DIODE c TYPES OF SEMICONDUCTORS Reverse Break down Voltage -V ■ ReverseỊ^ Bias • P-N Junctỉon Dỉode: In layman language. ịiỵ = electron and hole mobility JUNCTION TRANSISTOR í It is a semiconductor device containing three separate region having a íundamental action of transter resistor. =^>Here ne >>n^ • Electrical conductivity of a semiconductor is ơ = e[nepe + n/7p/7] ne/n^ = electron and holenumber density ịie. • Forward and Reverse biased characteristic ofp-/i junction diode. Here/Ie = n^ = riị where ne = Electron density /T/7 = Holedensity /i. > p-n-p transistor: A thin layer of /1-type semiconductor is sandwiched be tweentwop-type.7V Reverse Current INTRINSIC SEMICONDUCTORS • The pure semiconductors having thermally generated current carriers. Bias . p-type semiconductor =^> obtained by doping a trivalent impurity atom. On the basis of doping there are two type of extrinsic semiconductors. APPLICATIONS OF DIODE • Dỉode as a rectỉfỉer: A device vvhich rectiíies ac input to unidirectional pulsating output > Halfwaverectifier Averageoutputpercycle +/ Zener Breakdovvn Avalanche Region e.5V Constant Voltage +y Cut offi Forward region. when a p-type semiconductor is brought into contact with an /1-type semiconductor such that structure remainscontinuousatbou ndary/p-/ijunctiondiodeisformed. Junctỉon transỉstors are of two types > n-p-n transistor: A thin layer of p-type semiconductor is sand vviched b etvveen two /1-type. Reverse break down occurs at a high reverse biased voltage where the ord inary diodes get damaged. . =^> Here nh » ne /1-type semiconductor =^> Obtained by doping a pentavalent impurity atom. > Forward BiasCharacteristics VVidth of depletion layer decreases Effective barrier potential decreases Low resistance offered atjunction and high currentflowoftheorderof mA.=Densityof intrinsiccarriers. EXTRINSIC SEMICONDUCTORS • The semiconductor whose conductivity is mainly due to impurity.g7. 0. Transỉstor characterỉstỉcs > Input resistance (jị)5= BE \ \Ị V B ^17^-constant / l/ Ripplefrequency = Frequencyofacinput fdc~~ : Vdc=~ Full waverectifier T Averageoutputpercycle 2/ 2V Ripplefrequency = 2xFrequencyofacinput hc .For CE (n-p-n) /r -constant . LED: Ap-njunction devicethatemitsoptical radiation underforward biasconditi ons.Relationbetweenaand6: 6= — and a=— 1-cc 1 + P DIGITAL ELECTRONICS AND LOGIC GATES Digital electronic Circuit usesdiscretesignals.Adigital Circuit operates in abin arymanneronly in two States designatedasO(off) and 1 (on) usingdifferentlogicgat es. 230 V— 115 V— 0 V — ov -12 V I A DC=0. ị ANDGate VARIOUSTYPES OF LOGIC GATES NANDGate > Saturationregion APPLICATIONS OF TRANSISTORS ._ : ^dc=~~ K lí Zener dỉode: It is a highly doped p-n junction diode used as a voltage regulator. Solar cell: Generates emf of its own due to the effect of sun radiations. Photo dỉode: Ap-n junction diode use to detect light signals operated in reverse b ias.There are three confỉguratỉons of transỉstors > CB(Common Base) > CE(Common Emitter) > CC(CommonCollector). Bỉasỉngruleforajunctỉontransỉstor > Emitter base EB junction must be forward biased and collector base junct ion must be reverse biased.For CB (p-n-p) a= AỊc vA/£ l/(3£-í:onstan1: l/rfì_constart .637 Vpk Load /777 p Outputresistance (rỡ) = CE A/r > Currentamplificationfactor . > Atanypoint in the Circuit /£ = /£ + /cmusthold. fc is called critical frequen cy and it represents the highest frequency that will be reflected back from a pa rticular layer at vertical incidence. (a) 10. N is the electron density in the ionosphere. (c) 5. there Hole current are many electrons in the conduction band of the n-type material and many holes . (d) 8. m is the mass of the el ectron. The maximum line-of-sight distance dM between the two antennas having heights hT and hR above the earth is given by dM = yỊ 2RhT +^2RhR 10. either repeater transmitting stations are necessary or height of t he transmitter is increased by locating it in a satellite. Ap=Avx PứC Rin Transỉstor as a Switch > Atransistor can be used as a svvitch if it is operated in its cutoff and saturation States only. Space wave propagation (Tropospherical propagation) This mode is also known as line of sight communication.There are three regỉons of transỉstor operatỉon: > Cutoffregion > Activeregion • TransỉstorasaVoltageAmplỉfỉer > To operate it as an amplitier we need to fix its operating voltage some where in active region where it increase the strength of input ac signal and pro duces an amplitied outputsignal. e = electronic charge. A radar has a power of 1 kW and is operating at a frequency of 10 GHz. (b) 2. (b) 6. where m' = m0 1 í Ne2 ^ y£0moy J ■11/2 m' = refractive index of particular region of ionospere.= -(] Povưeĩgã\n. f ~ 9 (N )1/2 c max/ Nmax = maximum electron density of the ionosphere. Transỉstor as an Oscỉllator > Anoscillatorisageneratorofanacsignal using positiveíeedback. (b) 9. then dT =^ J2RhT . (d) SCIENCE BEHIND THE JUNCTION LASER I n the arrangement of figure. (b) 3. Therefore the transmitted wave is reflected back in a way similar to the phen omenon of total internal refraction. The maximum distance upto which it can detect object located on the surface of the earth (Radius of earth = 6. > Frequencyofoscillation \sf = ^ / 271 -Jĩc F layer and then decreases. (b) 4. If hf is the height o f transmitting antenna and df is the distance to the horizon from it. > Voltagegain. • Skip zone is a silent zone where no signal can be picked up. Maximum usable frequency (MUF) : It is the highest frequency that can be reflect ed from a particular layer of ionosphere for a given angle of incidence (i) is g iven as MUF = fc sec i. (c) 7. due to which their respective refractive indices var y.4 X 106 m) is (a) 16 km (b) 40 km (c) 64 km (d) 80 km (AIEEE 2012) ANSWER KEYS (SELF CHECK) 1. dT is called radio horizon of transm itting antenna. where R is the radius of the earth. e0 and n0 are permittivity and refractive index respectively for free space. I t is located on a mountain top of height 500 m. To send signals at far a way stations. * • • The distance between the transmitting antenna and point where the sky wave is fi rst received after returning to earth is called the skip distance. Ionosph ere behaves as a rarer medium by which carrier wave is reflected back if its fre quency (f < fc) where fc = critical frequency.Au = .Au = Voltagegain. that is. Suppose the current amplification factor of the trans istor is 100. What would be the percentage power saving if the carrier and one of the side bands were suppressed in AM wave modulated to a depth of 75% before transmi ssion took place? (a) 89. if the ou tput resistance is 500 kW. For the transistor amplifier circuit shown below. where. opposite faces of the p-n junction crystal must be flat and p arallel. Junction lasers are usually designed to operate in the infrared region of the electromagnetic spectrum because optical fibers have two windows i n that region (at l = 1.1 V and 14 mA (c) 0.55 mm) for which the energy absorption per unit length of the fiber is a minimum. The circuit shown in the figure contains three diodes each with forward resistance of 50 W and with infinite backward resistance. The input resistance of a common emitter transistor amplifier. producing a second photon by stimulated emission. by detecting reflection s from the rotating disc. there are more electrons in higher energy levels t han in lower energy levels. The maximum electron density of the iono sphere on that day was near to (a) 1. and power gain is 6.001 V and 100 mA (d) 2. Junction lasers are built into compact disc (cD) players. audio signal voltage across the collector resistance of 2 kW is 2V. Frequencies higher than 10 MHz were found not being reflected by the ion osphere on a particular day at a place. if the base resist ance is 1 kW.98.0 V and 10 mA 5.02% (b) 9.81% (d) 100% 6.in the valence band of the p-type material. a chain react ion of stimulated emission events can occur and laser light can be generated. Thus. Thus.condition for laser action.01 V and 10 mA (b) 0.1% (c) 7. so that light can be reflected back and forth within the crystal.0625 X 106. value of collector bia s voltage Vc should be approximately (Take VBE = 0. the current through the 100 W resistance is (a) 0 (b) 36 mA (c) 43 mA (d) 50 mA 4. there is a population inversio n for the electrons. its light output being highly coher ent and much more sharply defined in wavelength than light from an LED. For a CE-transistor amplifier. the current gain. a = 0. To bring this about. When a single electron moves from the conduction band to the valence band.7 V) (a) 6 V (b) 8 V (c) 10 V (d) 11 V 3. The simplified output Y of the given logic circuit is A B y-1 Y .but not a sufficient . They are also much used in optical communication systems based on optical fibres.5 X 1010 m-3 (b) 1. This photon can stimulate a second electron to fall into the valence band. if the current through the junction is great enough. a p-n junction can act as a junction laser. it can release its energy as a photon. is (a) 200 W (b) 300 W (c) 100 W (d) 400 W 2.24 X 1012 m-3 (c) 3 X 1012 m-3 (d) None of these 7.31 and 1. This is normally a necessary . In this way. (a) 0. they are used to translate microscopic pits in the dis c into sound. If the battery voltage is 6 V. Ẽ • physics for you | march 16 Exam Café QUESTIONS FOR PRACTICE 1. Find the input signal voltage and base current. 4% of the electrons are lost in the base. The angle of refraction is (electron density for D-region is 400 electron cc-1) (a) 60° (b) 15° (c) 45° (d) 30° 9. In bridge rectifier circuit.97 (c) 0. 1010 electrons enter the emitter in 10-6 s.5 W (b) 10 W (c) 12. the input signal should be connected between (a) A and D A (b) B and C (c) A and C (d) B and D 13.64 W or 8. In an npn transistor. What should be the value of the resistance R connected in series with diode for obtaining m aximum current? PHYSICS FOR you | MARCH 16 €3 (a) 2.5 W (d) 15 W 11. The value of the resistance R is R 10. A 24 V. then the ratio of 4 concentration of electrons and holes will be (a)5 25 (c) 49 (b) (d) 7 5 49 25 12.98 (b) 0. 4Q 80 WVWv VWWV<J d2 (a) 6.96 (d) 0.94 15. 5 . across AB i s given by (Assume the diodes to be ideal.24 W 14.(a) A ■ B + A ■ B (b) A ■ B + A ■ B (c) A ■ B + A ■ B (d) A ■ B + A ■ B 8. The current transfer ratio will be (a) 0. The logic circuit shown in figure (i) yields the truth table figure (ii).24 W (b) 5. The following configuration of gates is equivalent to (a) NAND (b) XOR (c) OR (d) AND 16.64 W (c) 8.24 W (d) 5. as shown in figure. as shown in the figure. of electrons and holes is . The ratio of electron and hole currents in a semi-7 conductor is — and the ratio of drift velocities 4 . A sky wave with a frequency of 55 MHz is incident on the D-region of ear th’s atmosphere at 30°. The equivalent resistance of the circuit as shown in figure.) Pi 2ÍÌ 40 HMA/VVA vwwv Am— l 50. Wh at is the gate X in the diagram? (a) OR gate (b) AND gate .5 V at all currents and a maximum power rating of 100 mW. 600 mW zener diode is used to provide a 24 V stabilized supply t o a variable load RL. The diode used in the circuit shown in the figure has a constant voltage drop of 0. (c) NAND gate (d) NOR gate 17. When a p-n junction diode made from germanium or silicon is forward-bias . 1) (b) (1. 1. Nmax of ionosphere = 1012 m-3) (a) ground wave propagation (b) space wave propagation (c) sky wave propagation (d) satellite communication 20. B = 1 and A = 0. A = 1. (a) ABCDE (b) BADEC (c) BDACE (d) BEADC 22. can this beam propagate along the fibre? (a) No (b) Yes (c) Refractive indices have nothing to do with beam propagation (d) The given data is insufficient 23.5 and 1. can the wave be demodulated? (a) Yes (b) No (c) Can not be predicted (d) Data insufficient 24. What is the base resistance Rb in the circuit as shown in figure. 0. 0.01 mF. 1) (c) (1.04 ms to reach a receiver via re-transmission from a satellite 600 km above earths surface. A light beam entering an optical fibre makes an angle of 10° with the fibr e core-fibre clad boundary surface. (Radius of earth = 6. 1. B = 0. Select the output Y of the combination of gates as shown in figure for i nputs A = 1. (d) The relative magnitude of the reverse currents cannot be determined from the given data only. 1) (d) (1. 21. Id entiíy the mode of propagation. (a) (0. A basic communication system consists of (A) transmitter (B) information source (C) user of information (D) channel (E) receiver Choose the correct sequence in which these are arranged in a basic communication system. An audio signal is modulated by a carrier wave of 20 MHz such that the b andwidth required for modulation is 3 kHz.4 X 106 m. (c) The reverse current is identical in the two diodes. A 50 MHz sky wave takes 4. The band gap in Si is larger than that in Ge. A ground receiver station is receiving a signal at 5 MHz. (b) The reverse current in Si is larger than that in Ge. If the fibre core and clad refractive indice s are 1. B = 0 respectively. 18. 0) 25. transmitted fr om a ground transmitter at a height of 300 m located at a distance of 100 km.49 respectively. If R = 10 kW and C = 0. find the distance between source and receiver. (a) 606 km (b) 170 km (c) 340 km (d) 280 km 26. A Si and a Ge diode has identical physical dimensions. An identical reverse bias is applied across the diodes. The output of the given circuit in figure o v„ sin E • PHYSICS FOR you | MARCH 16 (a) would be zero at all times (b) would be like a half-wave rectifier with positive cycle in output (c) would be like a half-wave rectifier with negative cycle in output (d) would be like that of a full-wave rectifier. Assuming re-transmission time by s atellite negligible. if hFE = 90? (a) 29 kW (b) 82 kW (c) 108 kW (d) 55 kW 19. (a) The reverse current in Ge is larger than that in Si. A block of pure silicon at 300 K has a length of 10 cm and an area of 1. then the maximum value of R so that the voltage is above knee point will be R 0. The mobility of holes is 0.98 1 .2 X 10-7A (c) 2.38 X 10-23 J K-1).0.0 kW (c) 4.04 V 29.7 kW (d) 6. Match the column I with column II and select the correct option. It converts light signals into electrical signal s.52 V (d) 1. P Q R S 1 (a) 1 2 3 4 R' (b) 2 3 4 1 (c) 3 4 1 2 (d) 3 1 4 2 1= SOLUTIONS R (a) : Voltage gain.0 X 10-7A (b) 2.6 V (c) 0.98 = 49 Av = 49 3 500X103 R~ í Power gain = bAv = 492 500 X10 R .26 V (b) 2. Q. S. PHYSICS FOR you | MARCH 16 €5 R.6 X 10-7A 28. The saturation current of a p-n junction diode made from germanium at 27°C is 10-5 A.a 1 . b = a 0.3 kW (b) 4. The junction diode in the following circuit requires a minimum current o f 1 mA to be above the knee point (0. It is the loss of strength of signal during its propagation through communicatio n channel. It converts speech signal into electrical signal s. 0 cm2. Attenuation 4.7 V -----vww--------1>-----I——I ý. This energy is in (a) Visible region (b) Infrared region (c) UV region (d) X-ray region 27.4 X 10-7A (d) 2. Column I P. What will be required potential in order to obtain a current of 250 mA in forward bias? (Boltzmann constant. k = 1. AV = p— Ri Also current gain. if VB = 4 V. The hole current is (a) 2. A battery of emf 2 V is connected across it. Piezo-electric 2. 05 m2 V-1 s-1. sensor Column II It converts pressure variation into electrical signals. The vo ltage across the junction diode is independent of current above the knee point.14 m2 V-1 s-1 and their density is 1.ed. (a) 3.5 X 1016 m-3. Microphone 1. energy is released at the junction due to the recombination of electrons and holes.7 V) ofits I-V characteristic curve.6 kW 30. The mobility of electron is 0. (a) 0. Photo-detector 3. IC = b IB = 100 X 17.0625 X 106 Ri= 49 X 5 X105 X 49 6. As diode Dj and 150 W resistor are in series and their combination is in paralle l with the series combination of diode D2 and 50 W resistor. Ib = . Rinput = 1 kW = 1000 W VR output output Q Voltage gain. the correct option is (c). 6V I = V = R 500 = 3.0.7) 106 = 17.1 V vC= VCC -VRc = (18 .3 mA = 1. (a) : Given.7 kW = 8.3 4 and power gain = 6. (c) : IB = VCC VBE RB (18 .. collector resistance Routput = 2 kW = 2000 W Current amplification factor of the transistor bAC=100 Audio signal voltage.3 mA Also. (b) : In the given circuit.73 mA VRc = ICRC = 1..(i) .1) = 9.9 V .0625 X10 R.73 mA X 4.(ii) 49 X 49 X 500 X10J R~ = 6. The equivalent circuit is as shown in the figure. we get 3 ..i = 198 W = 200 W 6 2.6 X10 2A = 36 X 10-3 A = 36 mA 4..0625 X 106 \ From (i) and (ii). the upper two diodes Dj and D2 are forward biased an d the lower diode D3 is reverse biased. Voutput = 2 V Input (base) resistance. R = 100 200 W = 500 W 3 3 Current in the circuit. Their equivalent re sistance is Tý + 150 W Tý + 50 W 1 1 50 W +150 W 50 W + 50 W 1 1 200 —-------1----— or R = W R' 200 W 100 W 3 Total resistance of the circuit. AV = = bAC V Input signal voltage input R input V V output input bAC (Routput / Rinput) 2 100(2000/1000) = 0.10 V So. no current flows through the lowe r diode D3.8. Hence.01V Base (input) current. For the wave not being r eflected from ionosphere . i = 30° N = 400 X 106 m-3 . modulation index. Arn B* Or y • Y Y '_ A ■ B _ A + B Y"_ A ■ (A + B) = A + (A + B) = A + (A ■ B) = A + (A ■ B) Y" '_ B ■ (A + B) _ B + (A + B) _ B + (A ■ B) _ B + (A ■ B) Y _ [ A + (A ■ B)] ■ [B + (A ■ B)] = [A + (A ■ B)] + [B + (A ■ B)] = A ■ (ÃTh) + B ■ AB) = ÃTh (A + B) = (A + B)(A + B) _ A ■ A + A ■ B + A ■ B + B ■ B = 0 + A ■ B + A ■ B + 0 _ _ (v A ■ A = 0 and B ■ B = 0) = A ■ B + A ■ B 8. 73 64 P -X100 Pc 41 32 X100 = Pc 73 X 32 64 X 41 X100 = 89.6 80.6 7.24 X1012 m 3 80. f 80. (a) : Here.Vinput = 0. m = = 10 X 10-6 A = 10 mA 75 = 3 100 = 4 Power produced by the AM transmitter 4^14)=í]+32 ]=X 32 1 3 56 • PHYSICS FOR you | MARCH 16 On suppressing.6 100 12 12 —3 = X1012 = 1. f = 55 X 106 Hz . power saved is Percentage saving = 73 64 P.02% 6.6 N cos2 0° or „T f2 (10 X106)2 N _ _ 80.01 Rinput = 1000 -. (d): For D-region. i = 0. (b) : f 2 = cos2 i where i is angle ofincidence and N is electron density.-6 5.6 N _ 80. 9 X 8 The effective resistance is 4 + -—— = 8.5V Total resistance of the circuit = = 12. (a) : Given. Id = ~ = = 02A Rd 2.2 A Resistance R in circuit = 12. a = . It will be so when input is con nected between B and D.5 W 2.m = í7 m = 1 - 81.45 N f 2 81.= = 25 mA Z 24 V Voltage drop across R = 32 V .5 W = 10 W 11.96 X1010 c 100 ___t ne X e Emitter current.5 W d P 0. PHYSICS FOR you | MARCH 16 2 X 9 The effective resistance is ——.24 W. = 24 V. 2 + 9 When negative terminal of external battery is connected to A and positive terminal of battery to B. will offer maximum resistance.64 W. R > = =---------= 2. P = 10 0 mW = 100 X 10-3 W = 0. The effective circuit will be as shown in figure (ii). will offer least resist ance. Ie = nc X e Collector current.— nh Ih v e 45 5 12. (b) : Voltage drop across diode. Ie _ neve Ih nhvh ne Ie X Vh 74 7 or _ X X _ . (b) : I = nAeVd or I nVd .24 V = 8 V 8V R = = 320 W 25 mA 10. The effective circuit will be as shown in figure (i). the reverse should take place.5 V)2 Diode resistance.2.5 W 0. v2 (0. (d): When positive terminal of external battery is connected to A and ne gative terminal of battery to B then Dị is forward biased.+ 4 = 5. (d): The input signal should be connected between two points of bridge r ectifier such that in positive half wave of input signal.45 X 400 X106 (55 X 106)2 = sin r 0 o CO II or sin i or i = r =1 9. I = c c t Current transfer ratio. (c) : Number of electrons reaching the collector.1 W = Vd = 0. 13. 9 + 8 14.5 V Maximum power rating. then D1 is reverse biased and D2 is forward biased. D2 is reverse biased.5 W . Vd = 0. nc = — X1010 = 0. Pz = 600 mW Current through zener diode P 600X10-3 W I7 = .1 W .5V = Current in the diode. one p-n junction shoul d be forward biased and other should be reverse biased and in negative half wave of input signal. During the negative half-cycle of input ac voltage. Y = A + X = A + B. the value of critical frequency of signal should be 1 1 fc = 9(N max )2 = 9(1012)2 = 9 X 106 Hz = 9 MHz.(A + B) = A.5 .e. The resistance of p-n junction becomes high which will be more than resi stance in series. 17.9933 = sin83°21' m1 1. rb = 3-07 = = 82 X103 W Ib 2. gate X is AND gate. so the communication is through sky waves . (b) : Output of Gj = (A + B) Output of G2 = A . (b) : From the truth table we note that. the p-n junction is reverse biased. B. For optical fibre.. Due to it. 21. Thus.. there will be voltage across p-n junction with nega tive cycle in output.Ic = nc_ Ie ne 0. B + A. B So X = A. A + A. A = A . The resistance of p-n junction is low. VCE = 4 V.-------------= — sin(90°_ 9i) m0 ^8 • PHYSICS FOR you | MARCH 16 or sin 9a = Mi cos 9.. A + B. a maximum potential difference will appear ac ross resistance connected in series of circuit. A + B. = —sin9a = .e. B It is the Boolean of XOR gate. B Output of G3 is Y = (A + B).78 X10 5 = 82 kW 19.4 = IcRc ^ Ic = .96 15. the given configuration of gates is equivalent to XOR gate. B = A. VBE = 0. 22. Since receiver-transmitter distance i s 100 km. there is no output vo ltage across p-n junction. (b) : The block diagram of a communication system is shown in the figure . (b) : Here. B + B.62 X 103 m.7 V . therefore. signal cannot go either by ground wave or space wave. which is AND gate with input as A and B. (c) : During positive half cycle of input ac voltage. b = 90. 20. (a) : Critical angle of core-cladding surface is 1. The current in the circ uit is maximum.(A + A) (•■• A + A = 1) = A + B. Hence. it is OR gate. hFE = forward current ratio i. Y = A + B i. the p-n junction i s forward biased.96 X 1010 1010 0. or A + X = A + B = A + B . m0 or cos9.3 2 X103 or Ic = 2. In this situation. then 9 . (c) 18.(1 + B) + A. Therefore. (c) : Maximum distance covered by space wave communication = yỊ2RhT = V2 X 6. B = A + A. m1 a 1.49 sin c = 2 = = 0.(TB) = (A + B). As the signal frequency 5 MHz < 9 MHz. Due to it.78 X 10 5 A B b 90 Since the transistor operates in active region. Rc = 2 k W If Ic is the collector current. 16.5 mA Th 2. From gate in figure (i).50 or c = 83°21' sin Qa m.4 X106 X 300 . For sky wave.sin10° .5 mA _r IB = Th = = 2. ne = nh = 1. d2 = x2 .= 2 X 0.5 Diode detector can demodulate when — << RC << — fc fm V RC = 10 kW X 1 X 10-8 F = 10-4 s It can be demodulated.05 X 10-6 = 0. Hence.04 X 10-3 s Let x be the distance of satellite from the surface of earth.38 X10 23 X 300 . Total time taken (t) = Total distance travelled (2x) Speed of electromagnetic waves (c) ct (3 X 108)(4. (c) : In a pure semiconductor.0 m s-1 Ih = nh Aevh .126 or V = 10.126 X kT kT e 10. (b) : Ifp-n junction diode is made from germanium or Silicon.5 X 10-7s fc 20 X X ' Bandwidth = 2fm = 3 kHz = 3 X 103 Hz fm = 1.06 . (b) : fc = 20 MHz = 20 X 106 Hz -L = Ả.6 X 10-19) X 1. Total time taken. fm 1. t = 4. (a) : Here. 23.4 X 10-7 A 28.1] eV/kT I 250 X10 3 or eeV ỉkỉ = — +1 =-------—T----+1 = 25001 Tr Iq 10-5 _ _ or — = ln(25001) = 10. transmitter).5 X 1016 m-3 £ = Ị = = 20 Vm -1 l 10X10 2 Vh = mh X £ = 0.e. T = 27 + 273 = 300 K. the bea m cannot propagate along the fibre.0 = 2.06 X 105 m = 606 km ' Let T be the source of electromagnetic waves X ' (i. so the ray will not suffer total internal reflection.1736 = 0.04 X10 3) \ X = — = 22 = 6.06 km Distance between source and receiver = 2d = 2 X 85.h2 = (606)2 . (b) : Here..1157 3 9. (d) : 25. the energy released due to recombination of free electrons and holes is in the infrared re gion. However. if the p-n junction diode is made from gallium arsenide or indium phosphide. = 83°20' As the angle of incidence at the core and cladding surface is just equal to crit ical angle.126 X 1.04 ms = 4. 27.0 X 10-4) X (1. = (1.5 X 1016) X (1. then the energy released due to recombination of free electrons and holes is in the visible region. I = 250 X 10-3 A I = I0 [eeV/kT . 24.170 km 26. R be receiver and S be satellite ____ at locations as shown in figure. X10-6 = 0.05 X 20 = 1. I0 = 10-5 A.5 X 103 Hz — = — X10-3 = 0.7 X 10-3s.(600)2 = 7236 s ị \ \ h \ \ d R d = 85. 2 mm s-1 What is the equivalent resistance across the points A and B in the given circuit ? (a) 8 W (b) 12 W (c) 16 W (d) 32 W >10 Q >10 Í2 >16 Q . (Given. Then the number of electrons striking the target per second i s 16 (b) 5 X 106 17 (d) 4 X 10 2.1 mm s-1 (d) 0. Electrical resistance of these wires will be in the ratio of (a) 1 : 1 : 1 (b) 1 : 2 : 3 (c) 9 : 4 : 1 (d) 27 : 6 : 1 4.3 mm s-1 (b) 0.2 X 106 m s-1. (d) Common sense is nothing more than a deposit of prejudices laid down by the mind before you reach eighteen.5 mm s-1 (c) 0. an electron moves in an orbit of radius 5.2 A 15 (c) 12 A (d) 120 A 1. if the current flowing through a copper wire of 1 mm diameter is 1.5 X 10-3 A (c) 3. What is the drift velocity of electrons. If th e radius of the earth is 6400 km.5 X 10-3 K-1.2 mA. 2 29.15 protons cm-2 s-1 are entering the earth’s atmosphere.1. The potential difference applied to an X-ray tube is 5 kV and the current thr ough it is 3. In cosmic rays 0.1 A? Assume that. the current received by the earth in the form of cosmic rays is nearly (a) 0.3 W.Alb&rt Einsleln PHYSICS FOR you | MARCH 16 €9 ẸXẠ ỊVỊ PRÈP 2016 CHAPTERWiSE MCQs FOR PRACTiCE CURRENT ELECTRICITY (a) 2 X 10 (c) 1 X 10 Useful for All National and State Level Medical/Engg. Find the equivalent current. then the temperature at which the resistance becomes 0. density of Cu = 9 g cm-3 and atomic weight of Cu = 6 3) (a) 0. (a) : As.0 X 10-11 m wit h a speed of 2.3k W 10 3 30. In a hydrogen atom.2 X 10-6 A (d) 1. each atom of copper contri butes one electron.6 X10 -19 = 0.12 X 10-3 A (b) 1.^^ .6 X 10-19 coulomb) (a) 1.26 V. If the temperatu re coefficient of resistance of wire is 1. VB = Vknee + IR or 4 = 0.3 X103 W = 3. The resistance of a wire at 300 K is found to be 0. Entrance Exams I.12 A (b) 1.7 + 10-3R 3 3 or R = —7 = 3. (Given. Masses of the three wires of same material are in the ratio of 1 : 2 : 3 and their lengths in the ratio 3 : 2 : 1.12 X 10-6 A 3. electronic charge = 1.6 W. is (a) 720 K (b) 345 K (c) 993 K (d) 690 K 5. In the figure. 4 W (b) 6 V. 4 V. DB and DC are in the ratio of 3 : 2 : 1 11. then the potential difference across the 2 W resistor is (a) 8 V (b) 10 V (c) 2 V (d) 4 V 252 14 W (d) W 85 3 R 1A JWW—«— D±^10V -ẠrA 9Q.5 m if Ax = 1 cm? (a) 4 X 10-8 T (b) 8 X 10-8 T (c) 4 X 10-5 T (d) 8 X 10-5 T . The equivalent resistance betwe en points A and B is (a) 21 W (b) 7 W (c) 13. the potential at points B and C. at which a current carrying conductor should be connecte d so that the resistance R of the sub-circuit between these points is equal to 8 /3 W. 4 V. 6 W (d) 6 V. DB and DC are in the ratio of 1 : 2 : 3 (d) currents in the paths AD. Three identical bulbs are connected in series and these together dissipa te a power P. 4 W 14. A bridge circuit is shown in the figure. 7 8 PHYSICS FOR You | MARCH 16 9. are connected across a vari able external resistance R. Find the points A and B as shown in the figure.5 Q Dl A ring is made of a wire having a resistance R0 = 12 W. If the power dissipated in the 9 W resistor in the circuit shown is 36 W . Then. 6 W (c) 10 V. In the circuit given here. The end-corrections are 1 cm and 2 cm respectively for the ends A and B. then the power dissipated will be (a) P/3 (b) 3P (c) 9P (d) P/9 MOViNG CHARGES AND MAGNETiSM 16. zero and 1 0 V respectively. A cell having an emf E and internal resistance r. to determine an unknown resistance X using a standard 10 W resistor. 6Q HI||y 2 Q ■AV/H 15. 4 V. and the value of resista nce R are (a) 10 V. (a) the point D will be at a potential of 60 V (b) the point D will be at a potential of 20 V (c) currents in the paths AD. the plot of potent ial difference V across R is given by 10. the points A. B and C are at 70 V. The galvanometer shows null point when tapping-k ey is at 52 cm mark.►12 £2 ►10 Í2>2. A meter bridge is set-up as shown. The determined value of X is 12. An element Al =Axi is placed at the origin and carries a large current I = 10 A. 4 V. As the resistance R is increased. What is the magentic field on the y-axis at a distance of 0. If the bulbs are connected in parallel. 2 Wb m(c) 12 Wb m2 (d) 120 Wb m 18. The ratio of the magnetic a field at — and 2a is 2 (a) 1 : 4 (b) 4 : 1 (c) 1 : 1 (d) 1 : 2 23. The magnetic field produ ced at the centre of the orbit is nearly (a) 0. Two parallel wires P and Q placed at a separation of r = 6 cm carry electric currents Iị = 5 A and I2 = 2 A in opposite directions as shown in figure.5 m in the same plane with the sam .63 X 10-2 T (b) 1. the magnitude o f B is E • PHYSICS FOR you | MARCH 16 m>Lcot f9 4p r V 2 m0 I (1 + cos 9 /2) 2p r (sin 9 /2) (a) 103 Wb m-2 (b) 105 Wb m-2 (c) 1016 Wb m-2 (d) 10-2 Wb m-2 24. where a magnetic field of induction B is along the 7-axis and an electric field of magnitude 104 V m-1 is along the negative Zaxis. A long straight wire of radius a carries a steady current I.51 X 10-2 T (d) 6. A solenoid of length 50 cm and radius of crosssection 1 cm has 1000 turns of wire wound over it. A particle of charge -16 X 10-18 coulomb moving with velocity 10 m s-1 a long the X-axis enters a region.6 X 1015 rev s-1 around the nucleus of radius 53 Ả.26 X 10-2 T (c) 2. the electron is making 6. This is smaller than the magnetic field at the centre by the fraction -2 (a) (2/3)r2/h2 (c) (2/3)h2/r2 (b) (3/2)r2/h2 (d) (3/2)h2/r2 2 2 PHYSiCS FOR you | MARCH 16 19. The magnitude of the magnetic field at O on the bisector angle of these two wire s at a distance r from point Q is (a) —1 sin f9ì (b) 4p r V2) (c) m> 1 tan (A) (d) 4 p r V 2 ) 21. carry equal currents I as shown in figure. One end of both the wires extends to infinity and ZPQR = 9. The kinetic energy of a pro ton that describes circular orbit of radius 0. The current is uniformly distributed across its cross-section.17.3 T 22. A deuteron of kinetic energy 50 keV is describing a circular orbit ofradius 0. Find th e point on the line PQ where the resultant magnetic field is zero. near the centre of the solenoid is approximate ly (permeability of free space m0 = 4p X 10-7 T m A-1) (a) 0. Ij h —<8>-----------<•> p 9 M----r------N (a) 4 cm to the right of Q (b) 9 cm to the left of P (c) 2 cm to the right of P (d) 3 cm to the left of Q 20. The magnetic field normal to the plane of a wire of n turns and radius r which carries a current I is measured on the axis of the coil at a small distan ce h from the centre of the coil.12 Wb m (b) 1. in a plane perpendicular to magnetic field B. If the charged particle continues moving along the X-axis. Two wires PQ and QR. In hydrogen atom. If the current carried is 5 A .5 m. the magnetic field on its axis. the number of electrons Avogadro’s number n = Volume of 63 g of copper 1. when its resistance is increased by a factor of 3. A galvanometer of resistance 100 W gives a full scale deflection for a curre nt of 10-5 A. (a) : Here r = 5. The magnitude of the force per unit length exerted by one wire on the other. What should be the frequency of applied electric field? The mass and charge of proton are 1. a magnetic induction of 1. a ful l scale deflection of 30 units is obtained in the galvanometer.5x 10 3 5.= 2 x 10 16 2. I = neAvd where.67 X 10-27 kg and 1. is (a) m0I (b) m01 2nb2 (c) 2nb (d) m0I b2 '2nb~ 2nb 2nb 29.0 X 10 11 ỉ : Here r = 5. we sh ould connect a resistance of (a) 1 W in parallel (b) 10-3 W in parallel (c) 105 W in series (d) 100 W in series 30.0 X 10 m. the resistance in series will be (a) 4513 W (b) 5413 W (c) 2000 W (d) 6000 W 26.5 X 107 Hz (b) 2. A coil in the shape of an equilateral triangle of side 0.6 X 10-19 C respectively.14 X 107 Hz (c) 3. The voltage sensitivity of g alvanometer changes by a factor (a) 35% (b) 45% (c) 55% (d) None of these 27.1 A is passed through the coil. A galvanometer of resistance 25 W is connected to a battery of 2 V along with a resistance in series.5 X 10-3t ^ t = 1-081_ = 720° C = 993K 1.84 X 107 Hz SOLUTIONS ^ 2 + 81 X 10-3 = 1 + 1.5 X 107 Hz (d) 3. (a) 2. v = 2. In a cyclotron. When the value of this resistance is 3000 W. what is the couple acting? (a) 5^3 X10-7 N m (b) X10-10 N m -v/3 —7 (c) X10 7 N m (d) None of these 28. (a) : As I = q = ne It n = — = e 3.6 X 10-19 C Period of revoluti .02 m is suspended f rom its vertex such that it is hanging in a vertical plane between the pole piec es of permanent magnet producing a uniform field of 5 X 10-2 T.5 X 10-3t ^ 2 + 0. To convert it into an ammeter capable of measuring upto 1 A.2 X 106 m s-1.081 = 1 + 1.2 x 10 3 x 1 1.e magnetic field is (a) 200 keV (b) 50 keV (c) 100 keV (d) 25 keV 25. e = 1. In order to redu ce this deflection to 20 units. (c) : As. If a current of 0. The current sensitivity of a moving coil galvanometer increases by 35 %.6 x 10 _19 .4 T is used to accelerate protons. Two thin long parallel wires separated by a distance b are carrying a curren t I ampere each. R3 and 10 W are in series R4 = 10 + 6 = 16 W Now. (a) : Surface area of earth.2 x 7 x 10 2 x 22 x 5 2px 5. u= = 2.1x 7 6.= 7 x 1015s 1 Current. a = 1.12 X 10-3 A (d): Mass. M = volume X density = Al X d or A = M/ld pl2d Resistance. R ~ l2/M M 32 22 12 = : : =27:6:1 1 2 3 (c) : Given.12 A 1 _ 1 1 _ 5 _ 1 (a) : -— = — + — = — = --.1 mm s-1 6.02 x 10 23 63 x 10 -3 6. R1 10 2.5x 10 3)2 = 0.5 X 10-3 K-1 6.5 10 2 > R1 = 2 W l2 l1 l2 : l2 l2 : l3 Now 2 W and 10 W are in series M1 M2 M3 R2 = 10 + 2 = 12 W R2 and 12 W are in parallel — =-----ị--^ R = 6 W R 12 12 Now. Vd = I neA 1.1 X 10-3 m s-1 = 0.2 x 106 T 2.6 W and T = 300 K = 27°C Temperature coefficient of resistance.02 x 10 29 n= 7 m -3 Also.0 x 10 17 _11 .15 X 1.02 x 10 7 23 cm -3 9 x 10 -3 6. I = J A = J 4pr2 = 0. T = 2pr 2px 5.on of electron.0 x 10 -11 2.02x 1029 x 1.3 W. R = pl /A = pl /(M/ld) = So.2 x 10 6 1 Frequency.6 X 10 19 X 7 X 1015 = 1.15 X 1.6 X 10-19 X 104 X 4 X 3.14 X (6. R300 = 0.15 X 1.6 X 10-19 X 104 m-2 s-1 \ Current. I = eu = 1.4 X 106)2 = 0.6 X 10-19 cm-2s1 = 0.6x 10 19 xpx(0. R4 and 16 W are in parall el . A = 4pr2 Charge entering the earth per second per unit area J = 0. Rt = 0. x R2 8 and 1 2 = W R1 + R2 3 . Rt = R0(1 + at) ^ 0.52 + 2)cm _ 50 53 x 10 or X _ _ 10....(ii) After solving eq. RlR2 = 32 W .(i) Again.3 1 +1.2 l2 r2 2 (c) : Here E = Ỉ(R + r) ^ Ỉ = E/(R + r) Also.0 _3 A Ỉ2 _--------_ 2 A Ỉ3 _■ 20 40 -10 30 _1 A 11.Vb + Vd ..R .5 X 10-3 X 27) .. R = p — \ R ^ l A Here. we get s 3 7 4 PHYSICS FOR you | MARCH 16 (63 9.3 = R0 (1 + 1..5 x 10 " x 27 ^ 2(1 + 1.Vd _ Vd . (d) : Applying Kirchhoff’s law at junction point D.VD _ D + D D 2 3 VD = 40 V Ỉ1 _ 70 . we get Rị = 4 W and R2 = 8 W l1 _ R1 _ 1 Hence> 1 . graph must be as given in option(c).(ii) Dividing eq.. (ii) and eq.40 10 40 . we get 111 — =-ị-^ R = 8 W R 16 16 8.5 X 10-3 X 27) = 1 + 1. 10.R300 = R0(1 + a x 27) ^ 0. Rị and R2 are in parallel) Putting Rj + R2 = 12 W in above equation. (i). 0.6 0.5 x 10 31 -\_3 (d): Resistance of wire. (b) : Using the condition for balanced Wheatstone bridge.. E = ỈR + Ir or E = V + Ir Er E \ E _ V + ——^ V _ E— xr _ R + r R + r ER R + r V_ E 1 + r / R So... we get Ỉ1 = Ỉ2 + Ỉ3 and Va .6 W 50 .5 X 10-3 X t) .5 X 10-3t 2 R.Vc 10 20 30 VD VD -10 70 . we get X _ (52 + 1)cm _ 53 10 W_ (100 . R0 = Rị + R2 = 12 W 1 +1.(i) (. (i) and (ii).6 = R0(1 + 1. then R' = R + R + R = 3 R V2 Power dissipated. (b) : The power dissipated in resistor of 9 W is V12 V12 P _ ^^ 36 _.. Potential difference between the points B and A Vb . (i) and (ii). un Ỉdlsin9 10-7 x 10x 10-2 xsin90° dB _^° " _ 4p r2 (0.12. Vc and Vd respect ively.VD = 1 X 4 = 4 V \ Vc .5 m -ES-+x According to Biot-Savart law. we have P' = 9 P 16 (a) : Here dl = Dx = 1 cm = 10-2 m.0 = 4 or Vc = 4 V Potential difference between the point B and C = 10 .5)2 . so Va = 0 E • PHYSICS FOR you | MARCH 16 \ Vb .1 • V2 _ 36R RR ^ V1 _yj36 x 9 ^ V1 _ 6 x 3 _ 18 V \ Total current through circuit is V _ ỈRa 18 _ Ỉ 9 x 6 18 _ Ỉ 18 5 Ỉ _ 5 A eq b 9 + 6. Potential across 2 W is V2 = ỈR ^ V2 = 5(2) = 10 V 15 (c) : By Joule’s law. (c) : Let the potential at points A. r = y = 0.Va = 10 V As. earth potential is taken zero.4 = 6 V R _ 6 _ 6 W 1 14. Vb. Ỉ = 10 A. \ RAB _ 7x14 7 +14 13.(ii) R/3 From eq. (d) : The Wheatstone bridge is balanced. the power dissipated through a resistor R. having a poten tial difference V is P_ V! R When bulbs are connected in series.. we now have (3 W + 4 W) and (6 W + 8 W) resista nce in parallel. P _ 3R When they are connected in parallel. m0/4p = 107 T m A-1 Ấ 0. B.(i) Power dissipated. because 3 W 6 W 4 W 8 W \ 7 W resistance is ineffective.0 = 10 V or Vb = 10 V Potential difference between the points c and D VC .5 m. then 11 1 1 _ 3 R R" R R R R 3 . P' _ V 2 . C and D be VA. 9 = 90°. _ 2px But B1 = B2 • I1 = I2 r + x x I2r 2 A X 6 cm or x = _ _ = = 4 cm I1 -12 5A . (a) : At the required point. as shown in figure. But the wire Q has smaller current.9 /2) + sin90°] X 2 4p a = (cos9/2 +1) = m°-(1 + cos9/2). Iị h —®--------------0-----------p Q R M-----r------nV---*----N Field due to current I1 at point R.= 4 X 10-8 T The direction of the field dB will be the direction of vector dl x r. 2p r sin 9 /2 2p r sin 9 /2 21. 2 r2 19. m0 I1 B1 = 2p(r + x) Field due to current I2 at point R. (b) : The magnetic field is given by B = m0nI where. (a) : B = m1 ^ X (eu) 2r 2r 4PX10-7 X 1. Let this po int be R at distance x from Q.B2 í h2 ^ -3/2" í 3 h2 1 1 1 + 2 V r 2 = 1 . (d): B1 =m°2pn1 and B = 2nnlr 2 B2 So. dl X r = Dxi X y j = Dxy(i X j) = Dxyk Hence field dB is in the +z-direction.6X1015 2 X 53 X10 = 0. V r 4p (r2 + h2)3/2 Fractional decrease in the magnetic field will be B1 B2 B .6X10-19 X6. 17. (c) : Current density.1 . (d): Perpendicular to O from PQ or QR.3 1 V 2 r2 .26 X 10-2 T 22. the resultant magnetic field will be zero when the fields due to the two wires have equal magnitude and opposite directions. Su ch point should lie either to the left of P or to the right of Q. 2 = B 4 p r í h2 Th3/2 1 + 2 . a = r sin 9/2 Magnetic field inductio n at O due to current through PQ and QR is B = 1 [sin(90° . m012 B 2 = . the point should lie closer to the right of Q.2 A 20.12 Wb m-2 -10 18. m0 = 4p X 10-7 T m A-1 -7 1000 B = 4 PX10 7 X-_ X 5 50 X10-2 B = 1. J = — pa2 . B.e. (c) : In this case. r ^ B = pa At r = a/2..X 50 keV [. (a) : Given q = .. B X 2pr = m0 X J X pr2 m01.16 X 10-18 C. B X 2pr = m0I ^ B = At r = 2a.) or E1 = = . r = = Bq qB 2mE 2m1E1 r = = Bq Bq mE (2m. (a) : Current through galvanometer... The current corresponding to 20 unit s deflection of galvanometer. V = 10j m s' B = B j.From Ampere’s circuital law (Ị)B •dl m0 ■Ienclosed For r < a. mv2 qvB = mv V2mẼ Radius of path. 23. b2 = 2pr 4pa So. Bl =.. (c) : Given Is = Is + Is = Is s s 100 s 100 s Is Initial voltage sensitivity. m = 2m1] = 100 keV 25. magnetic force provides necessary centripetal force i. = 4pa m01 For r > a. PHYSICS FOR you | MARCH 16 (65 m e : 16 X 10-17 B = 16 X 10-14 or B = 103 Wb m-2 24..(i) .. we get R' = 4512. V/ = li R' ^Is .5 W = 4513 W / 35 135 T 26. then I= 22 or X R' + G 3 3025 R' + 25 On solving.(i) g R + G 3000 + 25 where k is figure of merit of galvanometer.(ii) Fe = qE = -16 X10-18 X (-104 k) = 16 X10-14 k N Fm = q (V X B) = -16 X 10-18(10j X Bj) = -16 X10-17 B k N As the particle continues to move along the same direction. E = -104 k Vm-1 . V 2 Ig = = = k X 30 .. B2 . Vs = — New voltage sensitivity. Ig 2 2 2 I = X 20 = Ig = X A 30 3 g 3 3025 If R' is the resistance to be used in series of galvanometer. 14 X107 Hz. = ỂX°Ỉ2L = s X 10-4m2 44 t = 0.100 s) 3R % decrease in voltage sensitivity ( V.s) Vs. (b) : Here.4 T. we should connect a resistance S in parallel to it.00001 1 -10 .14 X 1. Ig X G = (I . (b) : G = 100 W.67 X 10 27 kg.Ig) X s ( Th ^ s = I .1xV3 X10 4 X 5X10 2 sin90° = s/3 X10-7 N m 28. B = 1.V. 9 = 90° A =1X base X height or 1 aVã A = —aX—-— 22 = ỂỊ. t = IAB sin 9. 56 • PHYSICS FOR you | MARCH 16 Advanced PRACTICE PAPER 2016 PAPER-I .20 Vs 1-^ .= 10 3 W 30.34 X 10-8 s.6 X10 19 X 1. therefore F = mọ l 2p b 29.67 X10-27 -8 t = = = 2. e = 1. I = 0. Ig = 10-5 A.1 A.34 X10-8 = 2. m = 1. 1. The frequency of the applied electric field should be fc 1 = 1 2t 2 X 2.34 X10 8 s.0. (a) : Torque.IX100% = 20— X100% = 55% l Vs ) Vs 27.6 X 10-19 C The time required by a charged particle to complete a semicircle in a dee is pm 3. I = 1 A To convert the galvanometer into an ammeter. the direction of electric field should reverse after every 2. (d): Let two long parallel thin wires X and Y carry current I and separa ted by a distance b apart.Ig V g X G = 10 -5 -5 X100 or S = 10 -3 1 . The magnitude of magnetic field B at any point on Y due to current I1 in X is gi ven by B =m0 h_ 2p b The magnitude of force acting on length l of Y is F = I2 Bl = I2 ') l 2 2l2p b) Force per unit length is F = m0 I1I2 l 2p b Given I1 = I2 = I.4 Thus. The sphere has a mass 0. (Molar mass of air is 28 g mol-1) 2. a2 = a1er. 6. In two systems of units. find the air pressure (in atm) in a mine at the depth 5. There are two poSitionS of the lenS at which the Sharp image of the object is formed on the screen. and T F2 = —. 3. In second case the secondary cell is changed as es = 5 V and r = 2 W. A 1. (b). Find the magnitude of the vel ocity of particle P1 after the collision. then zero deflection is o bserved at length PJ = 10 cm.5 kg and radius 5 cm. One of the circuits for the measurement of resistance by potentiometer i s shown in the figure. Let us assume that air is under Standard conditions close to the Earth’s s urface. What is the value of p ? (Neglect the effect of gravity. The distance betwee n the axes of the wire is 20 times greater than the cross-sectional radius of ea ch wire. (Molar mass of KCl is 74.0 g sample of pure KCl from the chemistry stockroom is found to be radioacti ve and to decay at an absolute rate of 1600 counts s-1.SECTiON 1 (MAXiMUM MARKS : 32) • This section contains EIGHT questions • The answer to each question is a single digit integer ranging from 0 to 9. The distance between the centre of the ring and the source is 1 m. the relation between velocity. Find the rate of m ean energy flow (in pW) across the area enclosed by ring if the intensity of sou rce at the centre of ring is equal to I0 = 3 pW m-2. The resultant force of interaction between the wires turn into zero for R = 3. The galvanometer is connected at point A and zero deflect ion is observed at length PJ = 30 cm when es = 10 V and r = 1 W. What is the maximum horizontal force F (in N) that may be applied to the plank o f mass 3/7 kg for which the solid sphere does not slip as it begins to roll on t he plank. 7. the radius of the ring is 0. M F PHYSICS FOR you | MARCH 16 8.18% of normal potassium.9 g mol-1. A thin convergent lens is placed between an object and a screen whose po sitions are fixed. The half-life for this decay is 13 X 10N years. where e and t are constants.24 respec tively. 5.0 km below t he surface. which constitutes 1 .) A point isotropic source is located on a line perpendicular to the plane of a ri ng drawn through the centre of the ring. ONE Or MORE than oNe of th ese four option(s) is(are) correct 9. Find N . Particle P1 moving with velocity 10 m s-1 experienced a head-on-collisio n with a stationary particle P2 of the same mass.25 and 0. The decays are traced to the element potassium and in particular to the isotope 40K. Find the transverse dimension (in mm) of the object if at one poSition the tranSverSe dimenSion of the image iS h1 = 8 mm and at the other h2 = 2 mm. length and time in the two systems is correctly given by . The coefficient of stati c and kinetic friction between the sphere and the plank are 0. (c) and (d). 2 Vie acceleration and force is v2 = 1 .6 X 10p W. Two long parallel wires of negligible resistance are connected at one en d to a resistance R and at other end to a dc voltage source.50 m. Relation et between mass. What is the current (in A) through resistance R wh en PJ = 30 cm ? 4. both inclusive 1.) SECTiON 2 (MAXiMUM MARKs : 40) • This section contains TEN questions • Each question has FOUR options (a). As a result of the collision. Presuming that the temperature and the molar mass of air are independent of height. the kinetic energy of the System decreased by 50%. during which its velocity decreases h times is v0(h . Which of the foll owing statements is/are correct? (a) Threshold frequency for the metal is 1. the maximum kinetic energy of the ejected photoelectrons is measured for various wavelengths of the incident light . shown in figure. 13. which is not possib le (b) mechanical energy is completely converted into heat.968 eV. P. 3qB (c) . 11. 2r (c) The velocity of the motorboat as a function of time is v0e~(r/m)t. In a photoelectric effect experiment. Consider a cycle followed by an engine. (b) Time for which the particle was in magnetic nm field is . which is not possib le (c) curves representing two adiabatic processes do not intersect (d) curves representing an adiabatic process and an isothermal process do no t intersect. Select t he correct alternative(s). A charge particle of charge q and mass m is moving with velocity v as sh own in figure in a uniform magnetic field B along negative z-direction. (b) Work function of the metal is 4. (a) The motorboat moved a distance mv0/r with the shutdown engine. KXXXXXXX /30° XXXXXXXX) ^^XXXXXXXXXXXX) ^xxxxxxxxxxxxx? xxxxx large distance xxxxxix & 58 • PHYSICS FORyou | MARCH '16 (a) Velocity of the particle when it comes out from the magnetic field is v = v cos60° i + v sin60° j . (c) Maximum kinetic energy of photoelectrons corresponding to light of wavel ength 100 nm is nearly 7. xX x Xxx xxxx->Extend upto a .4 eV. (b) The motorboat moved a distance mv° with the shutdown engine. (d) Photoelectric effect takes place with red light. Assume the resistance of water to be proportional to the velocity of the boat F = -rv. A motorboat of mass m moves along a lake with velocity v0.M1 Le2 (a) M = -2^ (b) L = e212 T2 (c) T2 = T (d) L2 = ^3t2 T3 10. 12. (d) The mean velocity of the motorboat over the time interval (beginning wit h the moment t = 0).2 X 1015 Hz. At the moment t = 0 the engine of the boat is shutdown. Figure shows a graph of this maximum kinetic energy Kmax as a function of the wavelength l of the light falling on the surface of the metal.1)/h lnh. 1 3 V 1 to 2 is isothermal 2 to 3 is adiabatic 3 to 1 is adiabatic Such a process does not exist because (a) heat is completely converted into mechanical energy. Q (a) current through r is zero (b) current through r is 2Bwa 5r (c) direction of current in external resistance r is from centre to circumfe rence (d) direction of current in external resistance r is from circumference to c enter. 2 3 4 (a) 25 cm left of s4.2 kPa. Column I and Column II • Match the entries in Column I with the entries in Column II • One or more entries in Column I may match with one or more entries in Column II 19. A narrow beam of parallel light is dire cted into the sphere. One of these is isothermal and the other is adiabatic. 1. Find the final image and its nature.0 m s-1 while at another point B. 15. There is a uniform magnetic field of strength B pointing inward and r is a stationary resistance. A hemisphere of radius R is charged uniformly with surface density s. the speed is 4. Then the pressure at B when water (a) is flowing is 6. PQ is a uniform rod of resistance r. A glass sphere. Th en. Then. has a spherical c avity of radius 5 cm concentric with it.2 kPa.7 kPa. The pressure at A is 20 kPa when the water is flowing and 18 kPa when the water flow stops. then which of the following graphs are correct? SECTIQN 3 (MAXIMUM MARKS : 16) • This section contains TWO questions • Each question contains two columns. R is a fixed conducting ring of negligible resistance and rad ius a.0 m s-1. (d) stops flowing is 8. virtual (c) 15 cm left of s4. virtual 17. and area enclose d by the nth orbit is An. in a hydrogen atom. (c) stops flowing is 10.5 and radius 10 cm. the speed of the water is 3. An ideal gas undergoes two processes A and B. Water is flowing smoothly through a closed pipe system. 2e0 18. (b) is flowing is 8. (a) electric potential at the centre is 2e0 (b) electric potential at the centre is -SR4e0 (c) electric field at the centre is — 4en s (d) electric field at the centre is . radius of nth Bohr orbit is rn. At one point A. nmv 2qB 14. virtual (b) 25 cm right of s4. virtual (d) 20 cm right of s4. refractive index 1.0 m higher.(d) Distance travelled in magnetic field is None of these. If. In figure. PHYSICS FOR you | MARCH 16 © Column-I Column-II (A) P —V (P) Heat supplied during curve A is positive (B) P % V (Q) Work done by gas in both processes positive (C) P Á V (R) Internal energy increase . 16.2 kPa. It is hinged at the centre of the ri ng and rotated about this point in clockwise direction with a uniform angular ve locity w. com E • PHYSICS FOR You | MARCH '16 JỀẺ r/ỈẨỈN Contd. (a) 5. Q. S). D ^ (R. d) 14. (a) : Time of free fall of a body from a height h. ring or a disc]. C ^ (P. A ^ (P. c.s in adiabatic process (D) P ^Ế> V (S) Temperature of gas in process B is constant 20. d) 15. (5) 8. S) Online Test Series Practice Part Syllabus/ Full Syllabus Mock Test Papers for JEE Main Log on to http://test. (5) 4. c. c) 18. D ^ (P. column-I describes some of the physical quantities associat ed with their motions while column-II gives you the relation between a physical quantity for two different bodies [bodies can be cylinder (solid or hollow). (2) 2. c) 19. (a. C ^ (P. B ^ (P. Q). from page no. A ^ (P. Q. Q). Q. S). d) 11. (2) 9. b. S). same diameter but are of different materials (B) Time taken by the two bodies to travel a distance s on the incline would be (Q) Different if A and B are of same diameter and same mass but havi ng different radius of gyration about centre of mass (C) Velocity acquired by A and B after travelling through s would be (R) Same if bodies have same diameter. Q. d) 17. (b. (a. The uniform rigid bodies A and B are allowed to roll without slipping on a r ough inclined plane.pcmbtoday. (5) 10. (a. S) 20. distance X _ __ 2 g . (a. c) 13. (8) 3. (a. (a. S). c) 12. Column-I Column-II (A) Acceleration of centre of mass of A and B (P) Same if A and B have the same radius of gyration about centre of mass. 30 22. (a. sph ere (solid or hollow). _ Í2h g Distance from the foot of the tower 2h d _ vt _ v _ 250 m g v When velocity = — and height of tower = 4h v 2(4h) Then. B ^ (R. b. Match the entries of column-I with the e ntries of column II. (a. (4) 6. (2) 7. b) 16. different radius of gyration about centre of mass and different masses (D) Minimum coefficient of friction required for pure rolling (S) Different if bodies have same diameter but different radius of gyration about ce ntre of mass and different masses (T) Same if they have same mass and diameter but different radius of gyration about centre of mass ANSWER KEYS 1. (a) : EX and Bywould generatea plane electromagnetic wave travelling in z-direction. e.85 ps c 3 X108 26. E.5kQ 10X10 27.2h X _ v _250 m g 23. we get 2d Ad _ — ■ AA So. 1 „ 1 V3 _ 2 ^ sinC _ — ^ sin60°_ — _—— ^ p_ —ị= p p 2 3 The time taken by light to travel some distance in a medium. ơ _ _ 0. Ậ X103 ụx _ 3_ t _ ■ _ 3.4 Al/l Ad X100 _ 1% d n 1 ^ tan _ ^ CmR = 1. ỉ X j_k) \ E is along x-axis and B is along y-axis. r eq . AA n 2Ad 4 A 2 Ad „ _ 2 A nd2/4 d AA Given X100 _ 2% A Ad/d Also.5X 103_1. (b) : Poisson's ratio. B and k form a right handed system where k is along z-axis.5 = 15 V 20 VArR 15 -3 _1. AA _ _ 4 ■AA n d 2ndAd .r7 + E9r. 3 X1 + 6 X 2 __12 21 _ _5V p_ _ eq r1 + r2 2 +1 The effective internal resistance of two cells in parallel circuit. A _ nr _ 2 Ad/d Al /l 2 _ 4A or d _ — 4 n Differentiating both sides. (d) : Voltage available across load resistance R = 20 . (c) : The equivalent emf of the two cells in parallel circuit.4 _ 2 nd Area. 28. (As. a_ 0. The motion of the particle take s place in a plane. If the amplitude is doubled and the time period of oscilla tion decreased to 1/3 of its original value. i = C i. (c) 30. the maximum velocity becomes (a) 18v (b) 12v (c) 6v (d) 3v 3.4 d = 2. 3. only (b) is correct. A particle is acted upon by a force of constant magnitude which is alway s perpendicular to the velocity of the particle. A particle moves in x-y plane according to the equations x = 4Í2 + 5t + 16 and y = 5t where x. 5Ỵ (2/3) Q Ad Al Al 1 Ad or —-_0. (a) : Given. n = 0. (b) : As the current (i) leads the voltage by —. The maximum velocity of a particle executing simple harmonic motion is v. the voltage .e.5 PHYSICS FOR you | MARCH 16 Exam on 1st May r AIPMT 2016 MODELTEST PAPER 1. . tan ò_ —— R Current in ammeter.5 X 1% = 2. I _ 5V 10 ^ ^ 3 _ —A 32 4 CwR As w = 100 rad s-1 ^ CR _ — _-------s w 100 Thus. from all the given options. It follows that (a) its velocity is constant (b) its acceleration is constant (c) its kinetic energy is constant (d) it moves in a straight line. y are in metre and t is in second. 4 XC Hence. 25. n = 0 X X 600 X10-9 2ut_ ^ t_ _ _100nm 2 4|X 4 X 1. If the capacitance is short circuited. the voltage across the resistance.. Ai = 60° a nd x = 1 km = 103 m When the total internal reflection just takes place from lateral surface..5% n 24. (a) : Condition for constructive interference is X 2ụ. In a series LCR circuit.t _ (2n +1) where. capacitance and inductance is 10 V each. The acceleration of the particle is (a) 8 m s-2 (b) 12 m s-2 (c) 14 m s-2 (d) 16 m s-2 4..12 2 X1 2 _ ■_-Q r + r2 2 +1 3 The equivalent circuit will be as shown in figure. C = 60° 29. it is R-C circuit. 2. 1.4 — ^ —X100 _ —---------—X100 d l l 0. For minimum thickness. 2. v is velocity. h is height and g is accelerat ion due to gravity.5 X 104 N (b) 0. the current ob served in the circuit is 0. The arrangement of NAND gates shown below effectively works as (a) AND gate (b) OR gate (c) NAND gate (d) NOR gate 9. Two stars of masses m1 and m2 distance r apart revolve about their centre of mas s.75 X 105 Pa. A body executes simple harmonic motion. At a displacement y.8 X 105 Pa and the pressure on the bottom surface is 0. According to Bernoulli’s equation 1 2 P + pv + pgh = constant where P is pressure.9 A.1 W (b) 0.5 W 7. 2G(m1 + m2) ỉ 2r3 G(m1 + m2) (b) 2p (d) 2p Ír3 (m1 + m2) 2G(m-ím2) G(mx + m2) 3 r 3 r 8. The current in a circuit containing a battery connected to 2 W resistance is 0. Two projectiles A and B thrown with speeds in the ratio 1: -v/2 acquired the same heights. (c) 2n. the acceleration of 1 kg block is g (a) downwards 2 g (b) upwards g (c) downwards 4 (d) upwards 12. its potential energy is U2. The dimensions of the constant are (a) [ML2T-2] (b) [MLT-2] (c) [mL-1T-2] (d) [mL-2T-1] 13.across the inductance will be 10 (a) 10 V (b) V 2 (c) 10V2V (d) 20 V 5. The period of revolution is (a) 2n. its potenti al energy is Uị. the angl e of projection of B will be (a) 0° (b) 60° (c) 30° (d) 45° 10. p is density. The angle made by the vector-y/31 + j with x-axis is (a) 30° (b) 45° (c) 60° (d) 90° 6. If the area of each surfac e is 50 m2.3 A.5 W (c) 1 W (d) 1. When a resistance of 7 W is connected to the same battery. If A is thrown at an angle of 45° with the horizontal. What is the pote ntial energy of the body at a displacement (x + y)? (a) U + U2 (b) (VŨT + 4Ũ2)2 (c) + U22 (d) 4W2 52 • PHYSICS FOR you | MARCH 16 11. The pressure on the top surface of an aeroplane wing is 0. In the system shown in the figure. At a displacement x. Then the internal resistance of the battery is ( a) 0. the dynamic lift on the wing is (a) 0.25 X 104 N (c) 5 X 104 N (d) 25 X 104 N . (b) They have a moderate forbidden energy gap.1 m.025 m s-2 (b) 0. The magnification.5 m s-2 (d) 500 m s-2 20. then r is equal to (a) R (b) 0. The potential energy per unit volume of the two wires will be in the ratio (a) 1 : 1 (b) 4 : 1 (c) 2 : 1 (d) 16 : 1 22. A machine gun is mounted on a 2000 kg car on a horizontal frictionless surface. The acceleration of the car i s (a) 0. Which one of the following statements is not correct regarding a semiconducting material? (a) They have negative temperature coefficient of resistance. If r is the distance between the axis and the centre of th e sphere. In the circuit shown in figure. what distance will it travel in the last second of upward journey? (Take g = 10 m s-2) (a) 5 m (b) 10 m (c) 15 m (d) 20 m 18. At some instant. The radius of gyration of a solid sphere of radius R about a certain axi s is also equal to R.25 m s-2 (c) 0. In a sample of radioactive substance. A light wave and a sound wave have same frequency u and their wavelength s are lị and 1s respectively. each of mass 10 g with a velocity of 500 m s-1. A coaxial cable consists of two thin cylindrical conducting shells of ra dii a and b (a < b). (d) Every semiconducting material is a tetravalent element. (c) Current is carried by electrons and holes both. Two wires of the same material and same length but diameters in the rati o 1 : 2 are stretched by the same force. what fraction of the initial nucle . If the same body is thrown upwards with velocity 80 m s-1.14.5R (c) VÕ6R (d) ^|Õ3R PHYSICS FORyou | MARCH '16 • 73 24. then (a) lị = 1s (b) lị > 1s (c) lị < 1s (d) lị = 2ls 21.b 4b2 container is (a) 11g (b) 22 g (c) 33 g (d) 44 g 15. The magnification produced by an astronomical telescope for normal adjus tment is 10 and the length of the telescope is 1. 17. the current through 2 W resistance is 4Q 8Q (a) zero (c)lìĩ 1A (d) |ỊỊ|A 23. A body is thrown upwards with velocity 40 m s-1 and it covers 5 m in the last second of its upward journey. The inductance per unit length of the cable is (a) m0 (a + b) 2p a (c) m»ln ( b 1 4p | a ) (b) (d) ■^ln ( a 1 4 p | b ) m°ln (b-1 2p | a ) 16. If pressure of CO2 (real gas) in a container is given by P = RT-------—. then mass of the gas in 2V . when th e image is formed at least distance of distinct vision is (a) 6 (b) 14 (c) 16 (d) 18 19. the gun fires 10 bullets/secon d. 5p ms (c) 5p ms (d) 10p ms 26. contains 15 litres of water.i will remain undecayed after half of a half-life of the sample? 111 1 (a) i (b) à (c)ĩ (d) 25. The electric and magne tic fields in the room are ma0 3ma0 (a) ---east. When it is projected towards north with a speed v 0 it moves with an initial acceleration 3a0 towards west. 5 litres of water comes out in ti me tị. If a constant force (-6? . The time taken by the proton to traverse 90° arc is (Mass of proton = 1. the next 5 litres in further time t2 and the last 5 litres in further time t3. When an unpolarized light of intensity I0 is incident on a polarizing sh eet.5 being initially at temperature 29 0 K is adiabatically compressed to increase its pressure 8 times. A linearly polarized electromagnetic wave given as E = E0 cos(kz . 0up e ev0 ma0 2ma0 (c) west.wt) i 30. The rms value of potential difference V shown in the figure is VẠ ^0 o1— TI2 T —►í (a) V1 3 (b) V0 (c) ả V0 (d) 2 33. The temperatur e of the gas after compression will be (a) 580 K (b) 870 K (c) 290-s/ĨK (d) 1160 K 32. A proton is projected with a velocity 107 m s-1 at right angle to a unif orm magnetic field of 100 mT. the reflected wave will be given as (a) Er =-E0 cos(kz . Three identical charges of magnitude 2 mC are placed at the corners of a right angled triangle ABC whose base BC and height BA are respectively 4 cm and 3 cm.wt) i is incident normally on a perfectly reflecting infinite wall at z = a. open at the top.6 X 10-19 C) (a) 0. Then (a) t1 < t2 < t3 (b) t1 > t2 > t3 (c) = t-2 = t3 (d) > t-2 = t3 27. One mole of gas of specific heat ratio 1. the intensity of the light which does not get transmitted is (a) I0 (b) zero (c) 4Iq (d) 110 28. A body of mass 5 kg starts from the origin with an initial velocity U = (30? + 40 ỹ) ms-1.------down e ev0 ma0 2ma0 (b) 0west. The angle between their resultant force and F2 is (a) tan-1 (76 Ị (b) tan-1 (ti6 ị (c) sin_1 ^16j (d) cos-1 ^16j 29.6 X 10-27 kg and charge of proton = 1. the time in wh ich the y-component of the velocity becomes zero is (a) 5 s (b) 20 s (c) 40 s (d) 80 s 31. down e ev0 ma0 3ma0 (d) — east.——°up e ev0 . it starts with an initial acceleration a0 towards west. Assuming that the material of the wall is optically inactive. Forces on the charge at the right angled corner B due to the charges at A and C are respectively Fị and F2.wt) i (b) Er = E0 cos(kz + wt) í (c) Er =-E0 cos(kz + wt) i (d) Er = E0 sin(kz .5|)Nacts on the body. It dra ins out through a small opening at the bottom. A cylindrical drum.05p ms (b) 0. When a proton is released from rest in a room. . The region surrounding a stationary electric dipole has (a) electric field only (b) magnetic field only (c) both electric and magnetic fields (d) neither electric nor magnetic field E • PHYSICS FOR you | MARCH 16 36. A man goes at the top of a smooth inclined plane.34.4°C.(c) hc hc (d) c hu 43. He releases a bag to f all freely and himself slides down on inclined plane to reach the bottom. Two identical flutes produce fundamental notes of frequency 300 Hz at 27°C . (a) —1 (b) 9 1 (c) y1 (d) —1 40. 25 * .. . If the field strength is increased to 4 times of the e arlier field strength. The momentum of photon whose frequency is u is hu hc u (a) (b) 5. The angle of reflection and angle of refraction are r and r respectively. If wavelength of photon emitted due to transition of an electron from th e third orbit to the first orbit in a hydrogen atom is 1. the num ber of beats heard per second will be (a) 3 (b) 2 (c) 1 (d) 4 44. A bullet is fired normally towards an immovable wooden block. The c ritical angle is (a) sin-1(tanr) (b) cos-1(tanr) (c) sin-1(tanr') (d) sec-1(tanr) 37. A ray of light from a denser medium strikes a rarer medium at an angle o f incidence i. The reddish appearance of rising and setting sun is due to (a) reflection of light (b) diffraction of light (c) scattering of light (d) int erference of light .75 s 42.4 kJ (b) 30 kJ (c) 336 kJ (d) 11. The work required by the refrigerator for every 1 kg of water conver ted into ice is (Latent heat of ice = 336 kJ kg-1) (a) 24.5 s (d) 0. then (a) u1 > u2 (b) u1 < u2 (c) u1 = u2 (d) u1 and u2 cannot be compared 38. 128.. If the temperature of the air in one of the flutes is increased to 31°C. The total thickness penetrated by the bullet into the block is (a) 4x (b) 6x (c) 8x (d) 2x 41. It loses 2 5% of its kinetic energy in penetrating through a thickness X of the plank. The reflected and refracted rays make an angle 90° with each other. The ratio of the de Broglie wavelengths associated with proton and alpha particle respectively is (a) 1:2^2 (b) 2 : 1 (c) ^/2 : 1 (d) 4 : 1 35. A proton and an alpha particle are accelerated through the same potentia l difference. 125. then the wavelength of photon emitted due to transition of electron from the fourth orbit to the secon d orbit will be . the time period will be (a) 12 s (b) 6 s (c) 1.2 kJ 39. If u1 and u2 are the respective velocities of the man and bag at the bottom of incline d plane. A Carnot refrigerator extracts heat from water at 0°C and rejects it to ro om at 24. s 36 * . A bar magnet suspended freely in a uniform magnetic field is vibrating w ith a time period of 3 s. the impedance of the circuit is z ' = yỊ R2 + xị = \í-R2 + R2 = V2r and the current in the circuit is = V = 10 V 1 = 2'= VĨR \ The voltage across the inductance is (10 V 'ì 10 R = V VL = IXL = V2r ) 2 ĩ ĩ 5.e. As the magnitude of velo city (i.45. The equivalent capacitance between the points X and Y in the circuit wit h C = 1 mF is c X Lc Y (a) 0. 2 VX = = (4t2 + 5t +16) = 8t + 5 x dt dt dy d y dt dt dvX d . 2. (c) : If A is the amplitude of simple harmonic motion. a =-X = — (8t + 5) = 8 x dt dt dvy d and ay = = (5) = 0 y dt dt c u PHYSICS FORyou | MARCH '16 • 75 The acceleration of the particle is a = axi + ay j = 8ĩ + oj = 8ĩ or a = >/8? = 8 ms-2 (b) : As VR = VL = V = 10 V R = XL = XC and z = R and V = IR = 10~V When the capacitor is short circuited. If ĩ is the unit vector a long x-axis.5 mF (b) 1 mF (c) 2 mF (d) 3 mF SOLUTIONS 1. (a) : As X = 4Í2 + 5t + 16 and y = 5t dx d . (a) : Let 9 be the angle made by 3 i + j with x-axis. so its kinetic energy is constant. speed) remains constant.where T is the time period of oscillation. then 9 (V3ĩ+i •ĩ s V3 V3 cos 9 =----------= = = |V3ĩ + iỊịĩị 32 +12^ 4 2 or 9 = cos -1 . then v = Aw 2p But ro = —. the maximum velocity becomes v' = 2A -^ì = 6( A2p| = 6v T/3 ì 1 T (using (i)) 3. (i) 2p \ v = A-2 T When the amplitude is doubled and the time period T decreased to . (c) : It is a case of uniform circular motion in which the velocity and acceleration vectors change due to change in direction. = mlrl(ữ = m2r2 w r 2 m1 or m1r1 = m2r2 or r2 = 1 rj m2 Substituting this value of r2 in eqn. Then r = rx + r2 .. / 1 Let the distances of the stars with masses mj and m2 from their centre of mass b e rx and r2 respectively.9 A. (c) : Ba> _A .(i) As the necessary centripetal force for their circular motion is provided by the gravitational force between them..(ii) 0.e 7 W + r Dividing eqn.5 W 7. (i) by eqn.(£ ^ V 2 y = 30° (b) : Let e and r be the emf and internal resistance of the battery respectively . m2 2 2 ----.3 A.3 A 2 W + r 2 W + r or 6 W + 3r = 7 W + r or 2r = 7 W . (i)...6 W = 1 W 1W or r = = 0. R = 7 W \ 0.9 A 7 W + r 7 W + r or 3 = . (d): The situation is shown in figure. we get m ( m. Then The current in the circuit is I = -R + r In the first case.9 A = 2 W + r In the second case.3 or w = G(m1 + m2) The period of revolution is T=2p=2P r! w G(m1 + m2) 8. I = 0. we get 0. ^ rj(m2 + mj) r = r + m " r1 = r1 m1 1 m2 ) m Gm1m2 m1m2rw2 r 2 (m1 + m2) 2 G(m1 + m2) or w = . (ii). I = 0... so Gm.(i) . R = 2 W \ 0.3 A = . Potential energy. U = — kỉ? 1 2 U— =1 kĩ? \ U = kx 1 2 and U2 = — ký2 .1 = 30° 2 V 2.0A-B ___B The truth table of the given circuit is A B A A • B A • B (a • B)• B Y = (a • B )• B 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 0 0 0 1 0 1 1 1 0 0 1 1 0 which is the truth table of NAND gate. Thus the given arrangement of NAND gates works as NAND gate. (b) : In simple harmonic motion.(ii) At a displacement (x + y). 10.. Hb = 2g As both projectiles attained the same heights. sin245° Ua_ = uB 2 But — = (given) sin2 0=|-U Í-U = V2) VV2 or sin 0 = 1 or 0 = sin 1 í . Ha = u2A sin245° 2 g 4 or r1 = 3 r 6 56 • PHYSICS FOR you | MARCH 16 For projectile B uB sin20 Maximum height.. (c) : Let the angle of projection of B be 9.(i) . the potential energy of the body is U = 2 k(x + ý)2 = 2 k(x2 + ý2 + 2xý) = — kỉ? + — ky2 + — (2kxv) 2 2 2 = U— + U2 + 2jữlU (using (i) and (ii)) = w Ui + U2) \\\\\\\\\\\\\\\\ 4k8p ỉg .. \ Ha = Hb or or u2A sin245° 2 g sin20 sin245° 1 uB sin20 2 g („ l2 = A or sin20 = = uị \ uB .. 9. For projectile A Maximum height. (d): The dynamic lift on the wing is F = DPA = (P1 . we get 2g = 8a 2ạ ĩ tr ilkg 4g .05 X 105 Pa)(50 m2) = 2. So if we consider the dimensions of P. Dn = 5 m. P2 = 0. then The equation of motion of 4 kg block is 4g .2T = 4a and the equation of motion of 1 kg block is T .b 1 4b2 2 mass of the gas(m) molecular mass(M )of the gas \ m = mM =1 (12 + 32) = 22 g 15. (d) 16.• Pressure = Force. (b) : According to van der Waals equation for m moles of real gas í or P + P= m2a1 2 mRT (V -mb) = mRT m2a V -mb) V2 Comparing it with the given equation RT a ã P= we get. (b) : If a is the downward acceleration of 4 kg block..75 X 105 Pa)(50 m2) = (0.2g = 4a Adding eqns. (ii) by 2. A = 50 m2 F = (0.8 X 105 Pa -..(ii) ..11.75 X 105 Pa.P2)A where P1 and P2 are the pressures on the upper and bottom surfaces of the wing r espectively and A is the area of each surface.. If T is the tension in each part of string. (d) 17. the upward accelerati on of 1 kg block must be 2a.| Area 13. we get 2T .(i) . (i) and (iii). m = As m = 2V . Here. (a) : For the body thrown upwards u = 40 m s-1..(iii) or 2g 8 4 \ The acceleration of 1 kg block = 2a = 2 í —) = — upwards 12. then [constant] = [P] [MLT-2] [L2] = [ML-1T-2] [•. a = -g = -10 m s . (c) : The dimensions of the constant is equal to the dimensions of each term on the left hand side of the equation.1g = 1(2a) Multiplying eqn.8 X 105 Pa.5 X 105 N = 25 X 104 N 14.0. P1 = 0.. (b) : Let fo and fe be focal lengths of objective and eye piece respectively .1 m \ 10 = f or fo = 10fe . (d) : The potential energy per unit volume of the wire is 1 (Stress)2 1 S2 2 Young’s modulus 2 Y As stress. m = 10 g = 10 X 10-3 kg Velocit y of the bullet. then = fo ( + fe 1 = 1 m fe l D J 0..1 = fo + fe .. (a) : Here. we get n = 4 4 PHYSICS FORyou | MARCH '16 • Zi The body thrown upwards with velocity 40 m s-1 takes 4 seconds to reach the high est point. we get fo = 1 m and fe = 0.25 m 1 10 [1+21 _ 5 _ = 10 " 7" _ 5 _ = 14 19. For normal adjustment.1 m When the image is formed at least distance of distinct vision D (= 25 cm). Mass of the car..(ii) Solving eqns.(i) and 1. 18. m = 10 and L = 1. (b) : Let Vị and vs be the velocity of light wave and the velocity of soun d wave respectively. n = 10 As for ce on the car = rate of change of momentum of the bullets F = nmv = 10 X 10 X 10-3 X 500 N = 50 N The acceleration of the car is F 50 N -2 a = = = 0. So the body thrown upwards with velocity 80 m s-1 will take 8 seconds to reach the highest point. M = 2000 kg Mass of the bullet. L = fo + fe Here. S = force area ( F 1( A 1 1 V F2 J V A1 J But F1 = F2 (given) . m = o fe and length of the telescope. v = 500 m s-1 Number of bullets fired per second.1m 0.As Dn = u + a (2n -1) n 2 \ 5 = 40 . f Magnification of the telescope. (i) and (ii). Then Vị vs 1Ị Vị u = —— = or ị_ 1ị 1s 1s vs But Vị > vs \ Ằị > Ằs 21.— (2n -1) 2 -2 On solving. Hence distance travelled in 8th second is Dn = 80 -— (2 X 8 -1) = 5m n 2 Note : A body covers the same distance in the last second of its upward journey whatever be its velocity..025 ms 2 M 2000 kg 20.1m ( 1 + 0. u1 =( 214 = 16 . we get 4I1 + 2(I1 . i. Applying Kirchhoff’s second law to the closed loop ABDA. Point-wis e theory and íocmulao ÍOí last minute reviỉĩon. MTG's Objective series is created keeping just this insight in mind for Class XI & XII .8I2 = 0 or 6I1 = 10I2 or I1 =1012 = 512 .I2) . (Maỉn & Advanced). we get 4I1 + 8I2 ... or — I = 20 or I =---------------= — A 3 2 2 44 11 E • PHYSICS FOR you | MARCH 16 OBỊECTIVE BỐTÀNY f KEY PEATURES: 1 «11 problems 1100+ are new and ven-similar to what you will face in JEE.\ S1 = A .(i) S2 Aj As the two wires are of the same material. Á Wl-fcGAvailable at all leading book shops throughout the country For more information or for help in placing your order: Call 0124-6601200 or email info@mtg.. d = diameter) di 1 But = (given) 2 2 .20 = 0 or 4 I2 ^J + 812 = 20 (using(i)) 44 3 X 20 15. State w Level PETs & PUC alỉo. Y1 = y2 2 .. (d): The distribution of current in various branches is shown in figure.e.' u 'UJ = T or u1 : u2 = 16 : 1 22.. therefore their Youngs moduli are the same.(i) 6 3 Applying Kirchhoff’s second law to the closed loop ABCA.in Q *Application to read QR codes required Boost your Tundamentals with MTG's Objective series 700L0GY £ ÀỸ QBIẸCTIVE < 350 PHỸSICS ẨZr ? 695 ĩ 795 < 775 7 795 Deep knowledge and crystal clear understanding of tundamentals is key to success . Extremely useful for JEE .1 2 f S1' 2 f A2 J V S2 J V A1 J ''d21 V d1 J (d 14 V d1 J (using(i)) (where. Moment of inertia of the sphere about a diameter is Idia = 5 MR2 By theorem of parallel axes Moment of inertia of the sphere about a certain axis at a distance r from its ce ntre is I = Idia + Mr2 = 2 MR2 + Mr2 5 If k is the radius of gyration at that axis. J&KCET.2R2 = 3R2 or r = -R = V06R 5 5 5 f=N = N oe -ít ít = e T1/2 -t ( N N . we get 5( 15 I1 = I A 1= A 3 1 11 25 11 \ The current in 2 W = I. (i). AFMC. PHYSICS FORyou | MARCH '16 • 79 Substituting this value of I2 in eqn. NEET. AIPMT. thes e unique books ensure students get just the start they need. then I = Mk2 \ Mk2 = 2 MR2 + Mr2 5 or k2 = 2 R2 + r2 5 But k = R (given) R2 = 2 R2 + r or 5 (given) 2 R2 + r2 5 r = R2 . Scan now with your smartphone or tabtet Detailed Solutions to MCQs for clear understanding Additional inrormation for students for exams such as AIIMS. AMU. UP-CPMT. WB JEE. (c) : Let M be the mass of the sphere. Put together by MTGs renowned editorial team. BHU. HIGHLIGHTS: 5.255+ pages covering the latest syllabus of AIPMT and other entrance exams Check-Your-Grasp questions for self-assessment NCERT xtract from NCERT books Question Banks including questions from previous years test papers (solved) of various exams like AIIMS. etc. AMU.= e = e í = ln 2 5 1/2 At t =1/2 ^ 11 T-f ln2 = e 2 = = e .students preparing to compete in entrance exams. Odisha. Kerala PMT. -12 = 25 A -15 A =10 A 1 2 11 11 11 23. etc. AIPMT. UGETManipal. then velocity of efflux.5p X 10-7 s 2 6 = 0. t1 < t2 < t3.. B = 100 mT = 1o0 X10-3 T p(1.6 X 10 27 kg. As the water drains out. The time taken by the proton to traverse 90° (= p/2) arc is s (p /2)r pr pm v v 2v 2qB (using (i)) Here. (a) : If h is the initial height of liquid in drum above the small openi ng.6x 10 27 kg) t =2(1. (b) : Since the wall is perfectly reflecting. Thus. Er = E0 cos(-kz . h decreases. Due to it.e. (a) : When the proton is projected with velocity v at right angle to a unifo rm magnetic field B. there is no phase ch ange (Stokes’ law).(i) 4pen (AB)2 along AB and that due to charge at C is 1 (2 mC)(2 mC) F2 =■ along CB 4pe0 (BC)2 Let F be the resultant force of F1 and F2. 27. q = 1.05p X 10 6 s = 0. i. (a) : The fraction of nuclei which remain undecayed after time t is ln 2 Force on charge at B due to charge at A is 1 (2 mC)(2 mC) F1 = f = e 111/2 = 1 = 1 = elWi ^/2 25. then 1 (2 mC)(2 mC) F 4pe0 tan 9= 1 = 0 (AB)2 1 (2 mC)(2 mC) 4p^0 (BC)2 = (BC)2 (4cm)2 16 = = = 9 (AB)2 (3 cm)2 or 9 = tan 11 16 2 PHYSICS FOR you | MARCH 16 29.05p ms 26. This reduces the rate of drainage of water. 24. it follows a circular path whose radius r is given by mv .. Thus. as the material of the wall is optically inactive.6 X 10 19 C. (d): The intensity of light transmitted through the polarizing sheet = — \ The intensity of light which does not get transmitted = r -10 = I0 0 2 2 28.wt) 1 = E0 cos(kz + wt) 1 30..lnV2 r= qB r m or — = — v qB where m and q are its mass and charge respectively. If F makes an angle 9 with F2. The reflected wave differs from incident wave in only one aspe ct. (b) : The situation is shown in figure. Further.6 X10 19 C)(100 X10 3 T) il „_7 7 = X10 7 s = 0.. h ence v decreases. amplitude (E0) of the linearly polarized electromagnetic wave remains unchanged. (c) : As u = (30Ì + 40 )) ms 1 \ Uy = 40 m s-1 . as the drainage continues. a longer time is required to drain out the same volume of wa ter. it travels along-z axis. m = 1. v = yj2 gh. Then Vy = Uy + ayt or 0 = (40 m s-1) + (-1 m s-2) t or (1 m s-2) t = 40 m s-1 40 ms-1 or t = = 40 s 1ms 2 31. Ị V2 d. (a) : If P1 and T1 are the pressure and temperature of the gas before compre ssion. Y = 1. (c) : In this question. + J (0) đt — 0-----= 5----------ĩn-----= Yl J dt T T T 0 Jdt V =\l ^[t ]0/2 =J 40. then for an adi abatic compression or T1Y = T2y p (Y-1) = p( Y-1) P1 P2 (T \Y \p \(Y-1) T p V T1 / V P1 P or 1 \ p \(Y-1)/Y P2 Here. (c) : y°---.and F = (-ỏi-5j )N \ Fy = -5 N The y-component of the acceleration is Fy -5 N -2 ay = — =-----= -1ms 2 7 m 5 kg Let after time t the y-component of the velocity become zero. T1 = 290K. 2 = 8.5 P1 T2 = (290 K)(8)(1'5-1)/1'5) = (290 K)(8)0'5/1'5 = (290 K)(8)1/3 = 2(290 K) = 580 K 32. then the acceleration of proto n due to electric field E = eE = p = ma0 = — = a0 or E =-------west . P2 and T2 are corresponding quantities after compression. the proton moves from rest towards west' It is due t o a force on the proton by virtue of electric field along west' If m and e are the mass and charge of the proton.\T \=J —=^0 T ' T T \ V 2 ) V2 V V2 33." ° T/2 T >t From graph T V = V0 for 0 < t < T 2 V = 0 for —< t <T 2 The rms value of V is V = rms T T/2 T ỊV2 d. (c) : de Broglie wavelength associated with a charged particle of mass m and charge q accelerated through potential difference V is 1 = h yỊĩmqV For same V. the acceleration of proton due to magnetic field = 3 a0 . it is independent of mass.m e When the proton is projected towards north with a speed v0. r + 90° + r' = 180° or r' = 180° .90° . So.(ii) cos r The critical angle C is m2 sin C = — = tan r (using (ii)) mi or C = sin-1(tanr) 37. (c) : The gravitational force is a conservative force. so the work done by i t is independent of path. m = 1 kg . (b) : Here.j) Rarer medium (ịi2) Reíracted ray As the reflected and refracted rays are perpendicular to each other. Hence in both cases 12 — mu 2 = mgh where h is the height of the inclined plane..r .r) (using (i)) sin r = = tan r .. shows that the proton is experiencing forces due to electric field along west and magnetic field acting vertically downwards. Therefore. then by Snell’s law m1 sin i = m2 sin r' m2 sin i or = m1 sin r' But by law of reflection. Incident ray' Reílected ray Denser medium (p. Ui = U2 38. or u = ^2 gh Clearly..a0 = 2 a 0' The force on the proton due to magnetic field = ev0B = m(2a0) 2ma0 or B = downwards ev0 „ _ ma0 _ 2ma0 Thus E = 0westand B = 0down e ev0 34. Mass of water. i = r m2 sin r sin r \ w = sin(90°.. (a) : A stationary charge produces an electric field only in the space surro unding it' Thus the region surrounding a stationary electric dipole has electric field only ' PHYSICS FORyou | MARCH '16 • 81 36.(i) If m1 and m2 are the refractive indices of denser and rarer medium respectively. (a) : The situation is shown in figure.r = 90° . or 1 = Ịmaqa = \(4mp)(2e) = Ịs = Ằa mpqp (mp)(e) 1 1 1p : 1a= ^/2:1 35. it moves with an acc eleration 3a0 towards west. e.e. & Med. water). room). Order now! • Includes solved AIPMT 2015 paper . (a) : Let the bullet be fired with velocity u.) 24th ApnI Kerala PET 25th & 26th april Kerala PMT 27th & 28th april APEAMCET 29th april (Engg. T1 = 24. then by work-energy theorem K .4 K . (a) 40.T2) W T1 . 9th & 10th April (online) ViTEEE cn Ổ —1 > "b MGIMS 17th april AMU (Engg.) 1st June JipMER 5th June E • PHYSICS FOR you | MARCH 16 Lost-minute check on your AIPMT reơdiness HIGHLIGHTS: • 10 ModelTest Papers based on latest AIPMT • Last 13 years' solved test papers of AIPMT ĩ 500 MTG's AIPMT Explorer helps students self-assess their readiness for success in A IPMT.273 K) = 273 K (336 kJ)(24.4 K) = = 30 kJ 273 K 39. L = 336 kJ kg-1 Temperature of hot reservoir (i.4°C = 24.) AipMT 1st may COMED K 8th may Karnataka CET 4th & 5th may BiTSAT 14th to 28th may wb jee 17th may JEE Advanced 22nd may AiiMS 29th may AMU (Med.T2 T2 (336 kJ)(297.Kf=fx EXAM DATES 2016 JEE Main : 3rd April (offline). T2 = 0°C = 0 + 273 = 273 K The amount of heat extracted from water at 0°C to convert to ice at 0°C is Q2 = mL = (1 kg)(336 kJ kg 1) = 336 kJ The coefficient of performance (a) of a r efrigerator is or Q2 T2 Q2(T . Then its initial kinetic energy is 12 Kị = mu (where m is the mass of the bullet) i 2 On penetrating through a thickness X. students ca n easily measure their preparedness for success.Latent heat of ice.4 + 273 = 297.4 K Temperature of cold reservoir(i. the bullet loses 25% of its kinetic energy. Now its final kinetic energy is Kf = 75% of Kị = f 1 „2 ^ = 3 f 1 „2 ^ = I — mu I= I —mu I 100 ^ 2 ) 4 ^ 2 ) Iff is the resistive force offered by the block to the bullet. Attempting the tests put together by MTG’s experienced team of editors and e xperts strictly on the AIPMT pattern and matching difficulty levels. • Detailed Solutions for self-assessment and to practice time management Available at all leading book shops throughout the country. For more intormation or for help in placing your order: Call 0124-6601200 or email: info<a)mtg.in *Application to read QR codes required Scan now with your smartphone or táblet* PHYSICS FORyou | MARCH '16 • 83 or or 1 2 3 (1 2~]_f 2 412 J J 1( 1 2^_ f 4 12 J J Let s be the total thickness penetrated by the bullet into the block. Then by wo rk-energy theorem 1 mu2 _ fs ...(ii) Dividing eqn. (i) by eqn. (ii), we get 1x _— or s = 4 x 4 s _ 302 _ 151 ..(•) _ 300 _ 150 Since frequency speed of sound • _ V2 " u v! v2 (151 ^ or u2 _ u1 v2 = (300 Hz) 1 150J (using (i)) = 302 Hz Hence the number of beats heard per second = u2 - u = 302 - 300 = 2 41. (c) : The time period of vibration of the bar magnet in a uniform magnet ic field B is T _ 2p ...(i) MB where - and M are its moment of inertia and magnetic moment respectively. When the field strength is increased to 4 times of its earlier value, the new fi eld strength becomes B' = 4B and the new time period becomes T' _ 2p - _ 2p MB' M(4B) 44. (c) : The reddish appearance of the rising and the setting sun is due to sca ttering of light. c 11 B In the given circuit, capacitor connected between A and B is short circuited and the remaining two capacitors are in parallel. • The equivalent capacitance between X and Y is Ceq = C + C = 2C = 2(1 mF) = 2 mF 2p, MB T 2 (using (i)) But T = 3 s (given) 3s T'_ — _ 1.5s 2 42. (a) : The energy (E) and momentum (p) of photon are related as E E _ pc or p _ c „ ^ , hu But E = hu p _ — c 43. (b) : Let Vj and v2 be the speeds of sound at 27°C and at 31°C respectively. As v ^ •Ựt (in kelvin) • v2 _ T2 _ /273 + 31 _ Í3Õ4 ( 4 ^Ị1/2 •■ v1 _ T ~ 273 + 27 _ 300 _|1 + 300\ 1 ( 4 ^ _ 1 + I \ (by binomial theorem) 21300 J Form IV 1. Place of Publication New Delhi 2. Periodicity of its publication Monthly 3. Printer’s and Publisher’s Name Mahabir Singh Nationality Indian Address Physics For You, 406, Taj Apartment, New Delhi - 110029. 4. Editor’s Name Anil Ahlawat Nationality Indian Address Physics For You, 19, National Media Centre, Gurgaon Haryana - 122002 5. Name and address of Mahabir Singh individuals who own the 406, Taj Apartment, newspapers and partners or shareholders holding more than one percent of the tot al capital New Delhi I, Mahabir Singh, here by declare that particulars given above are true to the b est of my knowledge and belief. Mahabir Singh Publisher 54 • PHYSICS FOR you | MARCH 16 PHYSICS MUSING SOLUTION SET-31 1. (c) : Let at any time t, the displacement of first particle be S1 and that of second particle be S2. S = — at2 and S2 = u 1 t - — 2 1 a For S2 > S1 ( 1^ 1 2 2 2ut 2u u I t - — |>- at ^ t -^— + — < 0 V a) 2 a a2 1 a (u -Vu2 -2u)< t < )(u+ vu2 -2u) 2. Hence, the duration for which particle 2 remains ahead of particle 1 (u + Vu2 -2u)-(u--\/u“ 2 = 2tJŨ(Ũ-2) a (b) : According to work energy theorem, Loss in kinetic energy = Work done 1 2 1 2 \ 0 - mv = W„ + WT or WT = -W„ - mv 2 g 1 1 g 2 = -mg 2R -1 mv2 12 = -0.1 X10 X 2 X 0.1 - X 0.1 X 52 2 = - 0.2 - 0.1 X12.5 = -1.45 J 12 3. (a) : For ball A, mg(4h) = mvA ^ vA = 'ỉ8gh Similarly, for ball B, mgh = 1 mv2B ^ vb = ^2gh Since, the balls collide elastically, \ After collision, va = ^Ị2gh and vb =yf8gh Ratio of heights attained by ball A and B, i.e., hA = vA = 2Ẻ = 1 hB vị 8 gh 4 4. (d) : As L = mvr = constant. m mv2 m L2 L2 -3 r r m2 r2 m 5. 6. (b) : Note that cohesive force among mercury molecules is greater than adhesive force between glass and mercury molecules. Also, adhesive force between water an d glass molecules is greater than cohesive force among water molecules. Let the body be in motion for n seconds. If sn is the distance covered by the bo dy in n seconds, then 12 from s = v0t + — at , 1 2 sn = 2 an2 (as v0 = 0) If snth is the distance covered in the last second of its motion, then snth = v0 + 2(2n -1) = 2(2n -1) According to the given condition, _1 _ a^ ,. 1 (1 2 snth = 2sn or 2(2n - 1) = 212 an or 2(2n - 1) = n2 or n2 - 4n + 2 = 0 or n= 4 ±v(4)2 - 4(1)(2) 2 4 ±yỈ8 2 (Taking only the positive sign before V negative sign implies n < 1s) V8, n= 4 s = 3.4s Also, = 1 2 = s = — an = 2 y 2 X (4)(3.4)2 2 23m (v a = 4 ms 2) 7. Let O be the centre of the hemisphere and OY be the axis passing through the vertex of the cone. Volume of cone = nr 2h 3 Volume of hemisphere 23 = nr 3 V Densities are same, masses of the bodies will be proportional to their respect ive volumes. If the centre of gravity of the combined mass should lie on O, then, nr 2h 2 3 -— + -nr 33 PHYSICS FOR you | MARCH 16 (35 nr 2h2 2 4 3 ^ _ nr X 3 X 4 3 8 h2 3 r 2 or h _ rVã 8. If the temperature of surrounding increases by AT, the new length of rod becomes l' = l(1 + a AT) Due to change in length, moment of inertia of rod also changes and is given as J where n_ 1.3.9934 m ...2a 65GM 8a and 16GM GM ------1---8a 2a 20GM 8a 1 2 (-20GM 65GM mv > m\ + 2 1 8a 8a _ 145GM 3 I5GM ^v V 4a le" Vmin _ 2V a 10. Ml2 Ml2(1 + aAT )2 .2 or . or -----w_--------------w 33 or w = w(1 . thus its angular momentum remai ns constant during heating.3 X110 _ 2.2 or — .75 m c 4 X110 First overtone of closed organ pipe = — First overtone of open organ pipe _ -2v 2lo These two produce beats of frequency 2. x = 2a So the body will reach the smaller planet due to the planet’s gravitational field if it has sufficient energy to cross the point B(x = 2a). Vs _ 16GM GM ------1-------2a 10a . The distance (from the smaller planet) where the gravitational pulls of the two planets balance each other will be given by -GMm -G(16M )m x2 (10a .. where n_ 1.2 4lc 2lo 2lo 4lc ^ 3 X110 _ 2. the expressio ns for beat frequency are 3v 2v 2v 3v — -——_ 2. _——..0067 m or lo = 0.x )2 i.2.Vs) Now. i.2. \ 1pw_ I'p w where m' is the final angular velocity of rod after heating.2 Hz when sounded together.2 l0 lo ^ lo = 1. — _ 110 Hz 4lc 330 lc _ m _ 0. c 4lc Open organ pipe.e._ 2./_ Ml'2 p _ 3 As no external force or torque is acting on rod.2 a AT) [using binomial expansion for small a] Thus percentage change in angular velocity of rod due to heating can be given as Aw_ m-m X100% _2 a ATX100 % 9. The frequency in Mth mode of vibration of one end closed and an open org an pipe are given by Closed organ pipe.. _ (2n—l)v. 12 — mv > m(Vb .e. o 2l o Fundamental frequency of closed organ pipe.3. Asit Srivastava (UP) 6. vvhich has a mass of 1.5 Jupiter radius. has orbital period of 3. with radii between one to 1.___. The hills are likely tragments of the rugged uplands that have broken away and a re being oarried by the nitrogen glaoiers into Sputnik Planum. Researchers from Keele University in UK used the Wide Angle Search for P lanets-South (WASP-South) instrument -an array of eight cameras observing select ed regions of the Southern sky.. The newly discovered planets were designated WASP-119 b. These hills individually measure one to several kilometres aoross. courtesy gravitational waves .75 days. say scientists The Moon can affect how heavy the rain is. jumbled mountains on Sput nik PlanurrTs western border. Gaurav Sharma (New Delhi) 2.." said Coel Hellier from Keele University . are likely miniature versions of the larger. The orbital periods of the pl anets vary from 2. WASP-126 b.___. researchers said.__..5 days. Lower humi dity is less favourable for precipitation. the images sh ow.. Subrata Dutta (WB) 2.. vvhich are in the vast ioe plain named Sputnik Planum vvithin Plu to’s 'heart’. Sinoe water ioe is less dense than nitrogen-dominated ioe. Moon can affect rainfall.2 of the mass of Jupiter.. the air becomes warmer to the extent that it can make rain lighter. WASP-129 b. Now. WASP-126b orbits the brightest star of the five.____ J E • PHYSICS FOR you | MARcH 16 Five 'hot' Jupiter-like planets discovered t ‘cientists have discovered five new Jupiter-like planets that are similar in chara cteristics to our solar system’s biggest planet and orbit very close to their host stars. VVASP-119 b. Britant (Bihar) 5... and their masses range from 0. Thi s means that it can be a target for atmospheric characterisation..4 days and a much younger parent star. Its host star has a similar mass to t he Sun’s but appears to be muoh older based on its etteotive temperature and densi ty . and an orbital per iod of 2... Manmohan Krishna (Bihar) SET-30 1... Tsubasa Kohyama. we'll listen to the stars.___. has the longest orbital period. similar in size to Jupiter... WASP-129 b and WASP-133 b. a doctoral student. to study five stars showing planet-like transits in their light curve. soientists believe th ese vvater ioe hills are tloating in a sea of frozen nitrogen and move over time like ioebergs in Earth’s Arotio Ooean. deducing the c omposition and nature of the atmosphere from detailed study .. less massive than Jupiter. Samrat Gupta (WB) 4. Researchers at Washington Univ ersity in the US discovered that because the Moon causes the Earth’s atmosphere to bulge towards it when it is overhead. givin g an insight into the dwarf planet’s tasoinating and abundant geologioal aotivity .17 to 5.Solution Senders of Physics Musing SET-31 1. "This is the tirst study to connect the tidal force of the moon with rainfall. is a typical hot Jupiter.. WASP-126 b is the lowest-mass world found by researchers. Afrid Khan (UP) ___.. Parv Mehta (Haryana) 3. Anurag Mohanty (Odisha) 7. Arnab Jana (WB) 3. Its low surface gravity and a bright host star make the planet a good target for transmission spectroscopy.'Chains’ of the drit ting hills are tormed along the flow paths of the glaoiers. WASP-124 b. The hills. WASP-133 b has the shortest orbital period of the exoplanets detected by researchers. WASP-124 b..__.2 the ma ss of Jupiter. The change is small... Nasa craft spots 'tloating hills' in Pluto's heart Nasa’s New Horizons spacecraft has captured images of frozen nitrogen glaciers on Pluto carrying numerous 'tloating’ hills that may be tragments of water ioe.________________. according to a new study that could h elp improve weather forecasts and climate models.3 to 1. accounting for about 1% of the t otal variation in rainfall. said. I12 is the image of ỉị bỵ M2. "This detection means that the stars are no longer silent. What accounts for the diíĩerence? -Shiva Prasad (Kerala) Ans. The best questions and their Solutions will be printed in this column each month . will enable mankind to listen to the stars. Ql. soientists say. soientists trom the Laser Interterometer Gravita tional Wave Observatory (LIGO) said that they have tinally detected the elusive gravitational waves. The space between the slightlỵ open door and the wall is acting as a single s lit for waves. for a large bar magnet the value of magnetic field is negligible at the centre. I2 is the image of o bỵ M2. that gravitationally drew closer to each o ther. the magne tic field is verỵ small because the contributions of the nearest magnetic dipoles (tinỵ magnetic bars) is partiallỵ cancelled bỵ the rest of the dipoles because its con tribution is anti-parallel. Here last angle made bỵ mirrors with images + angle between the mirror = 2 X 157. the controversial to the trivial. ỵou can hear sounds Corning from the hallw aỵ. "For this binary black hole System. near the edges the magnetic fie ld is greater because there is a minor quantitỵ of dipole magnets to oppose the co ntribution of the nearest. vvhich Albert Einstein predioted a century ago. Q3. I21 is the image of I2 bỵ Mị. the team will tack le the questions. easy and tough. while revolutionising our understanding of the most violent events i n the universe. -Pritish Acharya 360° ——1 = 8 —1 = 7 lị is the image of o bỵ Mị. then total number of images íormed is____. We can divide a bar magnet into a large number of tinỵ bar magnets. Fro m the serious to the silly. each with the mass of about 30 Suns. Evaluate. It’s not that we just l ook up and see anymore.” Evans said. Analysis of the waves suggests they originated from a System of two black holes. For a point near the centre of the bar. Studying gravitational waves will push Einstein’s general theory of relativity to its limits. It’s a whole new sense.The landmark disoovery of the tirst direot evidenoe of gravitational waves or ri pples in spaoe-time. Magnetic fìeld at the centre of a bar magnet is verỵ weak. an assistant professor at MIT. Q2. and not just see them. On contrary. so sound is eíTectivelỵ diíĩracted bỵ the opening and spread throughout the room. the signal LIGO received of the black hole merger was played on spe akers for eager scientists. it made a distin ctive. Iftwo mirrors are kept at 45° to each other and a bodỵ is placed in the middle. You must have a direct line of sight to detect the light waves. whỵ? -Tanỉỵa Mondaỉ (WestBengal) Ans. like we always have — we actually can listen to the univer se now. Yet ỵou cannot see what is happening in the halbvaỵ. so virtuallỵ no diíĩraction for the light occu rs. Courtesy: The Times of India Live Physics YOU ẠSK WE ANSWER Do you have a question that you just carTt get answered? Use the vast expertise of our mtg team to get to the bottom of the question. according to researchers at Massachusetts Institute of Technolog y. the magnetic field as the superposition of the contributions of all small align ed magnets present in the bar. Sound waves have wavelengths larger than the slit width. Light wavelength s are much smaller than the slit width. Actuallỵ. If a classroom door is open slightlỵ. In a breakthrough announoement. the ripples in the fabric of space-time. The frequency of these waves that LIGO is designed to catch are actually i n the audible range for humans.5° + 45° . rising ‘whoooop!’ sound." said Matthew Evans. Accordingly. Required number of images = PHYSICS FOR YOU I MARCH '16 45' Readers can send their responses at editor@mtg. (7) 11. You pull the plates farther apart a small distance. It is an interferometer used to study fine spectrum lines. 4 ) 23. (4) 4. Thus. (6) 5.5° 112. Because energỵ stored is proportional to both charge and poten tial diíĩerence. (4) 24. so work is d one by you on the System of two plates when you pull them apart. (5) 2. the plates attract each other.5° 112. the cap acitance decreases. You charge a capacitor and then remove it from the batterỵ.5°U2) 67.in orpost us with complete addres s by25th of every month to win exciting prizes. Ans. (8) DOWN 1. the energỵ stored in the capacitor increases. with air between them. The capacitor cons ists of large movable plates. A unit of power identical to the watt but used for the reactive power of an alternating current. and the change in corresponding positional coordinate.5° 22. The absorption of one particle by a system.= 360° 22. A range of frequencies within specified limits used for definite purpose .5° 157. aluminium (11%) and traces o f other elements. A piece of ferromagnetic material that connects two or more magnetic cor es. (4) 12. (4. Because the electric field of large plates is independent of distance. A measure of the rate of decay of a periodic quantity. Winners'name with their valuable feedback will bepublished in next issue.5° 157. (7) 25. what happens to the charge on the capacitor? Is work do ne in pulling the plates apart? -Kushagra (Uttar Pradesh) Ans. pr oduced in very turbulent clouds. 3) 14.5° Stop because next angle > 180° It implies that final images íormed bỵ two mirrors will coincide. A mechanical device that prevents any sudden or oscillatory motion of a moving part of any piece of apparatus. (4. A line on a chart or graph joining points of equal temperature. Because the same charge is stored at a higher potential diíĩerence. A type of alloy containing magnesium (88%). (6) . (5) 15. the charge on the capacitor rema ins the same as the plates are pulled apart.5°(I1) 67. Roughly spherical ice particles. (6. much used before the introduction of accumulators. (7) 18. charges on the plates have nowhere to go. A primary cell. This energỵ must be transíe rred into the System from somewhere. The product of a component of momentum. Q4. usually a few millimeters in radius. Because the electric field is measure of the rate of change of potential with distance. The equipotential surface of the gravity potential that coincides with t he mean sea level. Because the capacitor is removed from the batterỵ. (6) 21. the electric field remains constant. 4) 20. ACROSS 3. Kinetic energy released in matter. the potential diíĩerence between the plates increases as the separation distance in creases. (3) 26. The ratio of the radiant flux incident on an object to the flux transmit ted. A mixture of isotopes of an element in proportions that differ from the natural isotopic composition. Chemistry Today. (6) 7. voice. Pick the combo best suited for your needs. About MTG's Magazines Períect for students who like to prepare at a steady pace.020 PHYSỈCS r” Hírtr 11 y|jị»gw> jg|^L BIỌLOGY Subscribe to MTG magazines today. A simple machine that converts rotational motion to linear motion. The transfer of matter such as water vapour or heat. other than pitch or intensity of a note prod uced by musical instrument. to build all-round command over key subjects. Practice steadily. (7) 13. ( 7) 9. month by month. A permanently electrified substance exhibiting electric charges of oppos ite sign at its extremities.ensure you prac tice bít by bit. (8) 17.mtg. Fill-in the Subscription Form at the bottom and mail it to us today. Mathematics Today & Biology Today . SI unit of magnetic flux. (6) 10. Since 1982.Physi cs For You. save'up to Rs 2. (5) PHYSICS FOR you | MARCH '16 (89 Now. MTG's magazines . (5) 16. The distinguishing quality. A small wave. A radar-like technique employing pulsed or continous-wave laser beam for remote sensing. (5) 19. etc. If in a rush. through the atmosph ere as a result of horizontal movement of air. log on t o www. The vascular layer of the eye lying between the retina and the sclera.6.in now to subscribe Online. A mixture of free electrons and ions or atomic nuclei. paced month by month. The elementary particle that mediates the strong interaction between qua rks. (9) Cut Here 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 L 8. 'On cover price of ỉ30/-each. with very-similar & model test papers Self-assessment tests for you to evaluate your readiness and conhdence for the b ig exams Content put together by a team . Did you know these magazines are the only source for solved test papers of all nati onal and State level engineering and medical college entrance exams? Trust of over 1 Crore readers. (5) 22. Our 2015 offers are here. 800 ísóvet 1. ? 600 for 1-yr. new studying techniques. trends & cha nges in syllabi All-round skill enhancement conhdence-building exercises. Gurgaon .500 (sáve ? 660) □ 3 yrs: ? 2.080) □ 2yrs:t 1. Should you want us to send you your copies by Courier or Regd. Sector 44.340) I Physics ■ Chemistry ■ Mathematics ■ Biology □ l yr:f 330 (sơve 130) Q 2 yrs: ? 600 Isáve ? 1201 □ 3yrs:t 775 (sáve f 305) Complete Postal Address:. 3-yr subs criptions respectively). Pos t instead. Plot 99. time management. Visit www.in.300 (save 12. even adv ice from past IIT/PMT toppers Bonus: exposure to competition at a global level. 2-yr. Call (0)8800255334/5 f or more info.122 003 (HR). addỉtional charges apply (f 240. MTG Learnmg Madia (P) Ltd.in to subscribe Online. E-mail info@mtq. Pin Code' Other Phone Email______ J Mobile &]□□□□□□□□□□ Enclose Demand Dratt favounng MTG Learmng Modia (P) Ltd. Note: Magazines are despatched by Book-Post on 4th of every month (each magazine separately).900 (sáve* 1. with questions from Intl.020) Q 3 yrs: ? 1.mtg. PHYSICS F0R You | MARCH 16 . Y ou can aiso pay Via Money Orders Mail this Subscription Form to Subscription Dep t. payable at New Delhi.comprising experts and members from MTG's well-experienced Editorial Board Stay up-to-date with important iníormation such as examination dates. Olympiads & Contests Please accept my subscription to: (Coníirm yovr choice by ticking the appropriate boxes) PCMB combo □ 1 yr:f 1000 (sáve 1440) SUBSCRIPTION FORM PCM combo □ 1 yr: ? 900 (saveĩ 180) PCB combo □ 1 yr: ? 900 (sáve ? 180) Individual magazines □ 2yrs:? 1.340) □ 3yrs:? 1.900 (sóveĩ 1. ? 450.500 (sá ve Ị 660) □ 2yrs:t 1.