tạp chí physics for you số tháng 11 năm 2015

Volume 23 Managing Editor Mahabir Singh Editor Anil Ahlawat (BE, MBA) No 11 November 2015 Corporate Office: Plot 99, Sector 44 Institutional area, Gurgaon -122 003 (HR) Tel : 0124-4951200 e-mail : info@mtg.in website : www.mtg.in CONTENTS Regd Office: 406, Taj Apartment, Near Safdarjung Hospital, New Delhi - 110029 Physics Musing Problem Set 28 JEE Workouts 12 Core Concept 18 Physics Musing Solution Set 27 21 Exam Prep 2016 23 JEE Accelerated Learning Series Brain Map 31 46 JEE Advanced Practice Paper 2016 58 Ace Your Way CBSE XI 64 Thought Provoking Problems 71 Ace Your Way CBSE XII 74 You Ask We Answer 82 Live Physics 83 Crossword 85 subscribe online at www.mtg.in individual subscription rates combined subscription rates yr yrs yrs yr yrs yrs Mathematics Today 330 600 775 PCM 900 1500 1900 Chemistry Today 330 600 775 PCB 900 1500 1900 Physics For You 330 600 775 PCMB 1000 1800 2300 Biology Today 330 600 775 Send D.D/M.O in favour of MTG Learning Media (P) Ltd Payments should be made directly to : MTG Learning Media (P) Ltd, Plot No 99, Sector 44, Gurgaon - 122003 (Haryana) We have not appointed any subscription agent Owned, Printed and Published by Mahabir Singh from 406, Taj Apartment, New Delhi - 29 and printed by Personal Graphics and Advertisers (P) Ltd., Okhla Industrial Area, Phase-II, New Delhi Readers are adviced to make appropriate thorough enquiries before acting upon any advertisements published in this magazine Focus/Infocus features are marketing incentives MTG does not vouch or subscribe to the claims and representations made by advertisers All disputes are subject to Delhi jurisdiction only Editor : Anil Ahlawat Copyright© MTG Learning Media (P) Ltd All rights reserved Reproduction in any form is prohibited Physics for you | November ‘15 P PHYSICS MUSING hysics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh The aim of Physics Musing is to augment the chances of bright students preparing for JEE (Main and Advanced) / AIIMS / Other PMTs with additional study material In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / various PMTs The detailed solutions of these problems will be published in next issue of Physics For You The readers who have solved five or more problems may send their detailed solutions with their names and complete address The names of those who send atleast five correct solutions will be published in the next issue We hope that our readers will enrich their problem solving skills through “Physics Musing” and stand in better stead while facing the competitive exams single oPtion correct tyPe A metal ring of initial radius r and cross-sectional area A is fitted onto a wooden disc of radius R > r If Young’s modulus of the metal is Y, then the tension in the ring is (a) AYR (b) Yr AR r (c) AY (R − r ) r (d) Y (R − r ) Ar A piece of pure gold (r = 19.3 g cm–3) is suspected to be hollow from inside It weighs 38.250 g in air and 33.865 g in water The volume of the hollow portion in gold is (a) 1.982 cm3 (b) 2.403 cm3 (c) 3.825 cm3 (d) 4.385 cm3 A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats g It is moving with speed v and is suddenly brought to rest Assuming no heat is lost to the surroundings, its temperature increases by (a) ( g − 1) Mv 2( g + 2)R (b) ( g −1) Mv 2 gR (c) gMv 2R (d) ( g −1) Mv 2R In the figure shown, there are 10 cells each of emf e and internal resistance r The current through resistance R is (a) zero (b) e/r (c) 3e/r (d) 4e/r Physics for you | november ‘15 A rectangular loop with a sliding connector of length m is situated in a uniform magnetic field of T perpendicular to the plane of loop Resistance of connector is W Two resistances of W and W are connected as shown in the figure The external force required to keep the connector moving with a constant velocity m s–1 is (a) N (b) N (c) N (d) N A thin lens of refractive index 1.5 and focal length in air 20 cm is placed inside a large container containing two immiscible liquids as shown in figure If an object is placed at an infinite distance close to principal axis, the distance between two images will be (a) 25 cm (b) 40 cm (c) 65 cm (d) 85 cm Solution Senders of Physics Musing set-27 Manmohan Krishna (Bihar) Anubhav Jana (WB) Shiekh Md Shakeel Hassan (Assam) Swati Shah (Rajasthan) set-26 Md Samim Jahin (Assam) Deep Anand Basumatary (Assam) Harsimran Singh (Punjab) Sayantan Bhanja (WB) The figure shows several equipotential lines Comparing between points A and B, choose the best possible statement (a) The electric field has a greater magnitude at point A and is directed to left (b) The electric field has a greater magnitude at point A and is directed to right (c) The electric field has a greater magnitude at point B and is directed to left (d) The electric field has a greater magnitude at point B and is directed to right A light wire AB of length 10 cm can slide on a vertical frame as shown in figure There is a film of soap solution trapped between the frame and the wire Find the mass of the load W that should be suspended from the wire to keep it in equilibrium Neglect friction Surface tension of soap solution is 25 dyne cm–1 (Take g = 10 m s–2) (a) 0.25 g (b) 0.50 g (c) 2.50 g (d) 5.00 g ParagraPh tyPe Read the given paragraph and answer question number and 10 Consider the situation shown in figure in which a block A of mass kg is placed over a block B of mass kg The combination of the blocks are placed on an inclined plane of inclination 37° with horizontal The system is released from rest (Take g = 10 m s–2 and sin 37° = 0.6) The coefficient of friction between block B and inclined plane is 0.4 and in between the two blocks is 0.5 Then (a) Both blocks will move but block A will slide over the blocks B (b) Both blocks will move together (c) None of them will move (d) Only block A will move 10 The frictional force acting between the blocks will be (a) N (b) 6.4 N (c) N (d) zero ATTENTION COACHING INSTITUTES: nn CLASSROOM STUDY MATERIAL a great offer from MTG MTG � LENT EXCEL Y � IT QUAL TENT ER � PAP TING � PRIN � CON 10 offers “Classroom Study Material” for J E E ( M a i n & A d va n c e d ) , A I P M T a n d FOUNDATION MATERIAL for Class 7, 8, 9, 10, 11 & 12 with YOUR BRAND NAME & COVER DESIGN This study material will save you lots of money spent on teachers, typing, proof-reading and printing Also, you will save enormous time Normally, a good study material takes years to develop But you can have the material printed with your logo delivered at your doorstep Profit from associating with MTG Brand – the most popular name in educational publishing for JEE (Main & Advanced)/AIPMT/PMT Order sample chapters on Phone/Fax/e-mail Phone : 0124-4951200 09312680856, 09717933372 e-mail : sales@mtg.in | www.mtg.in Physics for you | november ‘15 Your logo here one or more oPtions correct tyPe questions Assume ABCDEF to be a regular hexagon Choose the correct E D  statements    + DB + BE = (a) ED   F C FE = (b)  BC = 2FE (c) AD   A B (d) DC = − AF A spring mass system is hanging from the ceiling of an elevator in equilibrium as shown in figure The elevator suddenly starts accelerating upwards with acceleration a, consider all the statements in the reference frame of elevator and choose the correct one(s) (a) The frequency of oscillation is k m k 2π m ma k m ( g + a) (c) The amplitude of resulting SHM is k (d) Maximum speed of block during oscillation is  m  k a   (b) The amplitude of the resulting SHM is An ideal gas has molar heat capacity at constant pressure CP = 5R The gas is kept in a cylindrical vessel fitted with a piston which is free to move Mass of the frictionless piston is kg Initial volume of the gas is 0.0027 m3 and cross-section area of the piston is 0.09 m2 The initial temperature of the gas is 300 K 12 Physics for you | NOVember ‘15 class-Xi Atmospheric pressure P0 = 1.05 × 105 N m–2 An amount of 2.5 × 104 J of heat energy is supplied to the gas, then (a) Initial pressure of the gas is 1.06 × 105 N m–2 (b) Final temperature of the gas is 1000 K (c) Final pressure of the gas is 1.06 × 105 N m–2 (d) Work done by gas is 9.94 × 103 J A man has fallen into a ditch of width d and two of his friends are slowly pulling him out using a light rope and two fixed pulleys as shown in figure Assume both the friends d apply forces of equal magnitude Choose the correct statements (a) The force exerted by both the friends decreases as the man moves up (b) T h e f o r c e ap p l i e d by e a c h f r i e n d i s mg d + 4h2 when the man is at depth h 4h (c) The force exerted by both the friends increases as the man moves up mg d + h2 (d) The force applied by each friend is h when the man is at depth h Two balls are thrown from an inclined plane at angle of projection a with the plane, one up the incline and other down the incline as shown in figure (Here, T stands for total time of flight) Which of the following are correct? v sin2 α (a) h1 = h2 = g cos q 2v0 sin α (b) T1 = T2 = g cos q (c) R2 – R1 = g(sinq) T12 (d) vt = vt Two identical buggies move one after other due to inertia (without friction) with the same velocity v0 A man of mass m rides the rear buggy At a certain moment, the man jumps into the front buggy with a velocity u relative to his buggy If mass of each buggy is equal to M and velocity of buggies after jumping of man are vrear and vfront Then m (a) vrear = v0 + u m+M m u (b) vrear = v0 − m+M mM (c) vfront = v0 + u (m + M )2 mM u (d) vfront = v0 − (m + M )2 A spherical body of radius R rolls on a horizontal surface with linear velocity v Let L1 and L2 be the magnitudes of angular momenta of the body about centre of mass and point of contact P respectively Then (here K is the radius of gyration about its geometrical axis) (a) L2 = 2L1 if radius of gyration K = R (b) L2 = 2L1 for all cases (c) L2 > 2L1 if radius of gyration K < R (d) L2 > 2L1 if radius of gyration K > R Two solid spheres A and B of equal volumes but of different densities dA and dB are connected by a string They are fully immersed in a fluid of density dF They get arranged A into an equilibrium state as shown in the figure with a tension in the string B The arrangement is possible only if (a) dA < dF (b) dB > dF (c) dA > dF (d) dA + dB = 2dF A body of mass m is attached to a spring of spring constant k which hangs from the ceiling of an elevator at rest in equilibrium Now the elevator starts accelerating upwards with its acceleration varying with time as a = pt + q, where p and q are positive constants In the frame of elevator, (a) The block will perform SHM for all value of p and q (b) The block will not perform SHM in general for all value of p and q except p = (c) The block will perform SHM provided for all value of p and q except p = (d) The velocity of the block will vary simple harmonically for all value of p and q 10 A string of mass m is fixed at both its ends The fundamental mode of string is excited and it has an angular frequency w and the maximum displacement amplitude A Then (a) The maximum kinetic energy of the string is EK = mA2 w2 (b) The maximum kinetic energy of the string is EK = mA2 w2 (c) The mean kinetic energy of the string averaged over one periodic time is < EK > = mA2 w2 (d) The mean kinetic energy of the string averaged over one periodic time is < EK > = mA2 w2 11 A bottle is kept on the ground as shown in the figure The bottle can be modelled as having two cylindrical zones The lower zone of the bottle has a cross-sectional radius of R and is filled with honey of density 2r The upper zone of the bottle is filled with the water of density r and has a crosssectional radius R The height of the lower zone is H while that of the upper zone is 2H If now the honey and the water parts are mixed together to form a homogeneous solution, then (Assume that total volume does not change) (a) The pressure inside the bottle at the base will remain unaltered (b) The normal reaction on the bottle from the ground will remain unaltered (c) The pressure inside the bottle at the base will 1 increase by an amount   rgH 2 (d) The pressure inside the bottle at the base will 1 decrease by an amount   rgH 4 Physics for you | November ‘15 13 12 A particle moving with kinetic energy J makes an elastic collision (head-on) with a stationary particle which has twice its mass During impact (a) The minimum kinetic energy of system is J (b) The maximum elastic potential energy of the system is J (c) Momentum and total energy are conserved at every instant (d) The ratio of kinetic energy to potential energy of the system first decreases and then increases 13 Two blocks A and B each of mass m are connected by a massless spring of natural length L and spring constant k The blocks are initially resting on a smooth horizontal floor with the spring at its natural length as shown in figure A third identical block C, moving on the floor with a speed v along the line joining A and B, collides with A Then (a) The maximum compression of the spring is v m/k (b) The maximum compression of the spring is v m/2k (c) The kinetic energy of A-B system at maximum compression of the spring is zero (d) The kinetic energy of A-B system at maximum compression of the spring is mv2/4 14 Three planets of same density and with radii R1, R2 and R3 such that R1 = 2R2 = 3R3, have gravitational fields on the surfaces E1, E2, E3 and escape velocities v1, v2, v3 respectively Then E1 E1 = (a) (b) E = E2 v1 v1 = =2 (c) (d) v3 v2 15 Water is flowing smoothly through a closed pipe system At one point A, the speed of the water is 3.0 m s–1 while at another point B, 1.0 m higher, the speed is 4.0 m s–1 The pressure at A is 20 kPa when the water is flowing and 18 kPa when the water flow stops Then (a) the pressure at B when water is flowing is 6.7 kPa (b) the pressure at B when water is flowing is 8.2 kPa (c) the pressure at B when water stops flowing is 10.2 kPa (d) the pressure at B when water stops flowing is 8.2 kPa 14 Physics for you | NOVember ‘15 solutions (a, b,c, a regular  d):As  ABCDEF   is hexagon, \ FE = BC , AD = 2FE , DC = − AF Also, from triangle law addition,  of vector     ED + DB + BE = (a, b, d) : As it is a isochronous system k \ υ= 2π m ma From the reference frame of elevator, A = k k ma m vmax = wA = = a m k k 5R 3R , CV = 2 ∆W n(CP − CV ) = = 1− = 5 ∆Q nC p (a, c) : CP = 2∆Q = 2× × 104 = 104 J 5 Pressure is constant and equal to mg × 10 P = P0 + = 1.05 × 105 + A 0.09 = 1.06 × 105 N m–2 T (b, c) : From figure, h sin q =   d2 d/2 h2 + mg As man moves slowly 2T sin q = mg mg T= 2sin q As man moves upward, q becomes small \ sin q decreases ⇒ T increases \ ∆W = mg T= 2×h T h 2 d mg d + 4h h + = 4h (a, b, c) : Maximum height of projectile on an (v sin a)2 inclined plane, h1 max = = h2 max g cos q ⇒ (a) is correct Time of flight 2v sin a T1 = = T2 ⇒ (b) is correct g cos q where a = angle of projection from inclined plane q = angle of inclination of surface R1 = (v0 cos a)T1 − g sin q T12 (Range upward the inclined plane) R2 = (v0 cos a)T2 + g sin q T22 (Range downward the inclined plane) ⇒ (R2 – R1) = g sin q T12 ⇒ (c) is correct vt and vt are the velocities of the particles at their maximum height Let the particles reach their maximum heights at time t1 and t2 respectively Hence, = (v0 sin a) – (g cos q) t1 v sin a ⇒ t1 = g cos q v sin a Similarly, t2 = g cos q Hence, t2 = t1 Hence, vt = v0 cos a – (g sin q) t1 vt = v0 cos a + (g sin q) t2 ⇒ vt ≠ vt (b, c) : As no external force is applied to the system v0 v v0 u+v v0 Conserving linear momentum of man and rear buggy, (M + m)v0 = Mv + m(v + u) mu = vrear ⇒ v = v0 − M +m Conserving linear momentum of man and front buggy, m(u + v) + Mv0 = (M + m)v′ mu   m  u + v0 − + Mv0 = (M + m)v ′  M + m  Mmu = ( M + m)v ′ M +m Mmu v ′ = v0 + = vfront ( M + m)2 (a, d) (a, b, d) : Let V be the volume of each sphere and T is the tension in the string For the string to be taut, dFVg > dAVg or dF > dA (a) is correct and dBVg > dFVg or dB > dF ( M + m)v0 + dFVg (b) is correct For an equilibrium dFVg + dFVg + T = T + dAVg + dBVg or dA + dB = 2dF (d) is correct (c, d) : In the frame of elevator d2x mg + ma − kx = m dt 2 d x k m( g + a)  ⇒ = − x −  m k   dt or d2x dt =− k m( g + pt + q)  x−   m k  There is a term involving t on R.H.S., this does not represent S.H.M unless p = Differentiating with respect to time d 2v k mp  d3x k  dx mp  = − − or = − v −   m  dt k  m k  dt dt Thus the velocity of the block will vary simple harmonically 10 (a, d) : Let the displacement of the string be given by πx y( x , t ) = A sin cos(wt + d) L where d is a phase factor So the transverse velocity is given by πx ∂y v( x , t ) = = − wA sin sin(wt + d) ∂y L The maximum kinetic energy is equal to the string’s total energy of oscillation Note that all points of the string achieve their maximum kinetic energy at the same instant of time, where y = for all x Since m dm = mdx where m =   is the mass per unit L length of the uniform string The maximum kinetic energy,   ∂y   EK = max imum of  ∫   dm    ∂t    L  ∂y   = maximum of  ∫ m   dx     ∂y    A T dAVg T dFVg B dBVg  ∂y  The maximum value of   , occurs when  ∂y  sin (wt + d) = L m πx Hence EK = A2 w2 ∫ sin2 dx L Physics for you | NOVember ‘15 15 where M is the mass of the Sun, m is the mass of the body, r is the distance of closest approach and v is the velocity at O Since the angular momentum of the body about the Sun will remain conserved, therefore, (mv0)d = mvr ⇒ v0d = vr (ii) From (i), and (ii), we get GMm  v0d  mv02 = − + m r 2  r   GMm GMm  ⇒ + − −  r r     v 2d  GM   Solving, we get, r =  +  GM  − 1 v02       Let m be the mass of the planet  GMm GMm  Mv +  − −  0.08 0.08  where v is the velocity of M at point Q = Here, r = (0.06)2 + (0.08)2 = 0.1 m \  v2  = 2Gm  − 08   Putting the values of m and G, we get v = 6.5 × 10–5 m s–1 (b) Since the gravitational force due to two masses at point Q is zero So, at point Q, there is no acceleration of the body When the body is at point P, then there are two forces of equal magnitude along PA and PB The magnitude of either force is GMm F= r The resultant force along PQ is given by FR = Fcosq + Fcosq = 2Fcosq, 0.06 0.06 where, cosq = = = 0.6 r 0.1 ⇒ FR = 1.2F = = 1.2 GMm r2 1.2 × 6.67 × 10−11 × 0.1 × 6.4 = 5.12 × (0.1) 10–9 Applying the conservation of angular momentum at the perigee and apogee, we get, mvprp = mvara v p a + c or = = (i) va rp a − c Using conservation of total mechanical energy, we get, GMm GMm mv 2p − = mva2 − rp 2 1 1 ⇒ v 2p − va2 = 2GM  −   rp  r Putting v p = va a , we get, rp va = (using (i) and e = Applying principle of conservation of energy, KEi + PEi = KEf + PEf 1 GMm + mv02 = mv − 2 r Physics for you | November ‘15 .(ii) GM  a − c  GM  − e  =   a a + c  a  + e  N Hence, acceleration = 5.12 × 10 –9 N/0.1 kg = 5.12 × 10–8 m s–2 72 .(i) Similarly, vp = GM a c ) a 1 + e  1 − e    Due to gravitational attractive force, the velocity of mass m will increase and mass 2m will decrease Suppose at minimum separation r, velocity of both particles is v According to conservation of linear momentum, 2mu + = mv + 2mv ⇒ v = 2u/3 (i) According to conservation of energy, m Gm(2m) (2m) u2 − d Gm(2m) 1 = mv + (2m)v − r 2 2m d where u Let mass of given semicircle be m Consider a small element of length rdq as shown in figure The mass of this element is y  d O r c2 is the mass of the photon dI = Gm πr r2 .(i) dq (using (i)) dq By symmetry, the net field along y-axis is zero The component along negative x-axis is dIsinq The resultant field intensity is given by π Gm I = ∫ dI sin q = ∫ I= Gm πr πr sin q dq = [− cos π + cos 0] =  ^ \ I = − (2Gm / πr ) i 2Gm πr Gm πr c c  GM  = 1− λ ′ λ  Rc  Since acceleration due to gravity varies near the earth surface as  2h  g (h) = g 1 −  R  m Gravitational field intensity at centre O due to this Gdm u or  GM  λGM λ ′ = λ 1 +  or | λ ′ −λ |=  Rc  Rc x m m (rdq) = dq πr π element is dI = Rc  GM  or λ ′ = λ 1 −   Rc  GM If < < 1, then using binomial theorem Rc h1 dm = GM −1 (rd) hu GM  hu  , R  c  or u′ = u − 2Gm2   2Gm2 mu − = m  u − d r (using (i)) 3  ⇒ r= 1 u2  −    d 6Gm  hu′ = hu − [− cos q]0π along negative x -axis If hu is the energy of the photon neglecting the effect of gravitational attraction and hu′ is the energy after the red shift, then h m h2 From figure h1 > h2, so W1 will be lesser than W2 h h and W2 – W1 = mg2 – mg1 = mg  −  R R GM h or W2 − W1 = 2m R2 R   GM As g = and (h1 − h2 ) = h   R 2mhG   or W2 − W1 = ×  πR  r 3  R3 3   M = r × πR    \ W2 − W1 = πrGhm nn Our greatest weakness lies in giving up The most certain way to succeed is always to try just one more time -Thomas A Edison Physics for you | November ‘15 73 Series CHAPTERWISE PRACTICE PAPER : Semiconductor Electronics - materials devices and simple circuits | Communication Systems Time Allowed : hours Maximum Marks : 70 GENERAL INSTRUCTIONS (i) All questions are compulsory There are 26 questions in all (ii) This question paper has five sections: Section A, Section B, Section C, Section D and Section E (iii) Section A contains five questions of one mark each, Section B contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each (iv) There is no overall choice However, an internal choice has been provided in one question of two marks, one question of three marks and all the three questions of five marks weightage You have to attempt only one of the choices in such questions section-A Define the term critical frequency in relation to sky wave propagation of electromagnetic waves Name two factors on which electrical conductivity of a pure semiconductor at a given temperature depends Give one example each of a system that uses the (i) Skywave (ii) Space wave mode of propagation What happens to the width of depleting layer of a p-n junction when it is (a) forward biased (b) reverse biased? If the base region of a transistor is made large as compared to a usual transistor, how does it affect (a) the collector current (b) current gain of this transistor ? section-b Why is communication using line of sight mode limited to frequencies above 40 MHz? A semiconductor is known to have an electron concentration of × 1013 per cm3 and a hole concentration of × 1012 per cm3 74 Physics for you | November ‘15 (i) Is the semiconductor n-type or p-type? (ii) What is the resistivity of the sample if the electron mobility is 23,000 cm2 V–1 s–1 and hole mobility is 100 cm2V–1s–1? A diode detector, with a load resistance R = 250 kW in parallel with a capacitor C = 100 pF, is used to detect an amplitude modulated carrier Find the highest modulation frequency that can be detected without excessive distortion The output of an AND gate is connected to both the inputs of a NAND gate Draw the logic circuit of this combination of gates and write its truth table 10 In half-wave rectification, what is the output frequency if the input frequency is 50 Hz ? What is the output frequency of a full-wave rectifier for the same input frequency? OR A Zener of power rating W is to be used as a voltage regulator If Zener has a breakdown of V and it has to regulate voltage which fluctuates between V and V, what should be the value of Rs for safe operation as per the given figure? section-c 11 What is a carrier wave? Why high frequency carrier waves are employed for transmission? 12 From the output characteristics of common emitter circuit shown in figure, calculate the values of bac and bdc of the transistor when VCE is 10 V and IC = 4.0 mA 13 Draw a block diagram for a simple amplitude modulator and explain briefly how amplitude is modulation achieved 14 State the factor, which controls (i) wavelength of light and (ii) intensity of light, emitted by a LED Write two advantages of LED over incandescent lamp 15 Write the symbol and truth table of an AND gate Explain how this gate is realised in practices by using two diodes 16 Consider the circuit arrangement shown in figure (i) for studying input and output characteristics of n-p-n transistor in CE configuration 76 Physics for you | November ‘15 Select the values of RB and RC for a transistor whose VBE = 0.7 V, so that the transistor is operating at point Q as shown in the characteristics shown in figure (ii) Given that the input impedance of the transistor is very small and VCC = VBB = 16 V, also find the voltage gain and power gain of circuit making appropriate assumptions OR If each diode in given figure has a forward bias resistance of 25 W and infinite resistance in reverse bias, what will be the values of the current I1, I2, I3 and I4? A C E I1 G I4 125  I3 125  I2 125  25  5V B D F H 17 In the circuit shown in figure, when the input voltage of the base resistance is 10 V, VBE is zero and VCE is also zero Find the values of IB, IC and b 18 Suppose a n-type wafer is created by doping Si crystal having × 1028 atoms per m3 with 1ppm concentration of As On the surface 200 ppm concentration of Boron is added to create p region in this wafer Considering ni = 1.5 × 1016 m–3, (i) Calculate the densities of the charge carriers in the n and p regions (ii) Comment which charge carriers would contribute largely for the reverse saturation current when diode is reverse biased 19 Write the truth table for the circuits given in figure consisting of NOR gates only Identify the logic operations (OR, AND, NOT) performed by the two circuits 20 An amplitude modulated wave is as shown in figure Calculate (i) the percentage modulation, (ii) peak carrier voltage and, (iii) peak value of information voltage V 20 V 100 V t 21 An n-p-n transistor is connected in CE configuration in which collector supply is V and the voltage-drop across the load resistor of 800 W connected in the collector circuit is 0.8 V If the current amplification factor is 25, determine collector-emitter voltage and base current If the internal resistance of the transistor is 200 W, calculate the voltage-gain and power-gain 22 An amplitude modulated wave is represented as cm(t) = 5(1 + 0.6 cos 6280 t) sin 211 × 104 t, volts (i) What are the minimum and maximum amplitudes of the A.M wave? (ii) What frequency components are contained in the modulated wave? (iii) What are the amplitudes of the components? section-D 23 Two students namely Shobit and Amit were asked to take up a project on efficient lighting for road ways, cycle paths and bus lanes They found LED is the best source for the above said reasons They collected the information from various sources and submitted the project about its working, advantages and its applications by presenting with a good working model (a) By seeing these two students, what kind of qualities you want to adopt from them? (b) Explain LED with neat diagram and draw its symbol section-e 24 Explain briefly, with the help of circuit diagram, how V-I characteristics of a p-n junction diode are obtained in (i) forward bias, and (ii) reverse bias Draw the shape of the curves obtained OR Discuss common emitter amplifier, using n-p-n transistor Find its current gain, voltage gain and power gain 25 Draw the energy band diagrams of p-type and n-type semiconductors Explain with a circuit diagram the working of full-wave rectifier OR What are energy bands? How are these formed? Distinguish between a conductor, an insulator and a semiconductor on the basis of energy band diagram 26 Draw a labelled circuit diagram of a common emitter amplifier using a p-n-p transistor Explain how the input and output voltages are out of phase by 180° for a common-emitter transistor amplifier Define the term voltage gain and write an expression for it OR Explain briefly, with the help of a circuit diagram, how a p-n junction works as a half wave rectifier Explain with the help of a circuit diagram how a Zener diode works as a dc voltage regulator Draw its I-V characteristics solutions Critical frequency is the highest frequency of the radiowaves which when sent normally towards the given layer of ionosphere gets reflected from ionosphere and returns to the earth It is given by uc = (Nmax)1/2 where Nmax is the maximum number density of electrons in the given layer of ionosphere (a) Width of the forbidden gap (Eg) (b) Intrinsic charge carrier concentration (ni) (i) Short wave broadcast (ii) Television transmission (a) Decreases (b) Increases When the base region becomes large, most of the charge carriers coming from emitter will be neutralized in the base by the electron-hole combination Due to this (a) collector current decreases and as a result of this (b) current gain also reduces At frequencies above 40 MHz, communication is limited to line-of-sight communication At these frequencies, the sizes of transmitting and receiving antennas are relatively smaller and can be placed at heights of many wavelengths above the ground During line of sight communication, the waves coming directly from transmitting antenna towards receiving antenna get blocked at some point by the curvature of the earth Physics for you | November ‘15 77 (i) Given ne = ×1013 cm–3, nh = × 1012 cm–3 Since ne > nh, the semiconductor is n-type (ii) As me = 23,000 cm2 V –1s–1, mh = 100 cm2 V –1s–1 s = e(neme + nhmh) = (1.6 × 10–19)[(8 × 1013) (23,000) + (5 × 1012) × 100] –1 –1 = 0.29448 W cm 1 Thus, ρ = = = 3.396 W cm s 0.29448 W−1 cm −1 Here, R = 250 kW = 250 × 103 W = 2.5 × 105 W, C = 100 pF = 100 × 10–12 F = × 10–10 F For satisfactory detection, > RC uc or uc < 1 = = 40 kHz RC (2.5 × 10 W)(10−10 F) Thus, the highest modulation frequency that can be detected with permissible distortion is 40 kHz The logic circuit for the given combination of gates is shown in the figure Y = A.B A B AND Gate Y OR Given P = W, Vz = V Vi(max) = V P Iz = = = 0.2 A Vz For safe operation, Vi(max) − Vz − Rs = = = = 10 W Iz 0.2 0.2 11 A carrier wave is an electromagnetic wave of high frequency and of constant amplitude, which is employed to carry the audio signals from one location to other on the surface of earth For the transmission of audio signals, the high frequency carrier waves are used, because the effective power radiated by a longer wavelength NAND Gate Truth table A B Y′ = A · B Y = Y ′ Y ′ 0 1 1 0 1 1 l base-band signal should be small as P ∝   λ Thus, for a good transmission, we need high powers and hence we need high frequency carrier waves to carry the base-band signal (message signal) 12 Consider any two characteristics for two values of IB which lie above and below the given value of IC Here IC = 4.0 mA, therefore we select the two characteristics for IB = 20 mA and 30 mA 10 In half wave rectification, only one ripple is obtained per cycle in the output Output frequency of a half wave rectifier = input frequency = 50 Hz In full wave rectification, two ripples are obtained per cycle in the output Output frequency = × input frequency = × 50 = 100 Hz 78 Physics for you | November ‘15 From the graph, at VCE = 10 V, DIB = (30 – 20) mA = 10 mA DIC = (4.5 – 3.0) mA = 1.5 mA DIC 1.5 mA = = 150 Therefore, bac = DIB 10 mΑ At VCE = 10 V, either (i) IB = 30 mA and IC = 4.5 mA or (ii) IB = 20 mA and IC = mA Therefore 4.5 mA I = 150 For (i), bdc = C = IB 30 mA mA IC = = 150 For (ii), bdc = IB 20 mΑ 13 The block diagram is shown here (ii) Truth table Bandpass AM x(t) Square y(t) m(t) wave filter + law device Am sin mt centred at c c(t) (Modulating Ac sin ct 2(t) Bx ( t ) + Cx signal) (Carrier wave) The modulating signal is superposed on carrier wave of high frequency Let the signal produced be x(t) = Amsinwmt + Acsinwct (i) The resultant wave so obtained is sent to square law device which produces wave y(t) = Bx(t) + Cx2(t) (ii) where B and C are arbitrary constants From equations (i) and (ii), we get C y(t) = BAmsinwmt + BAc sinwct + (Am+ A2c ) C C − Am cos 2wmt − Ac2 cos 2wct 2 + CAmAccos(wc – wm)t – CAmAccos(wc + wm)t C ( A + Ac2 ) and m sinusoidal waves of frequencies wm, 2wm, wc, 2wc , (wc – wm) and (wc + wm) This is finally sent to bandpass filter which rejects dc and sinusoids of frequencies wm , 2wm and 2wc allows the frequencies wc, wc – wm and wc + wm The output of bandpass filter is an amplitude modulated wave In this equation, there is dc term 14 (i) Wavelength of light emitted depends on the nature of semiconductor (ii) Intensity of light emitted depends on the forward current Advantages of LED over incandescent lamp are as follows (a) low operational voltage and less power (b) fast action and no warm-up time required B B Y=A.B 0 1 (iii) Realisation of AND gate +5V +5V (a) When A = 0, B = Both the diodes D1 and D2 conduct Potential at Y is zero as most of potential falls across the resistance R (b) When A = 1, B = Diode D1 not conduct but D2 conducts Voltage of V drop across R Now potential at Y is (c) When A = 0, B = D1 conducts, D2 not conduct Voltage drop of V across R Potential at Y is (d) When A = 1, B = D1 and D2 both not conduct, no current flow through R Hence output Y = 16 From the output characteristics at point Q, VCE = V and IC = mA VCC = ICRC + VCE 16 − V − VCE , RC = = kW RC = CC IC × 10−3 Since, VBB = IBRB + VBE 16 − 0.7 RB = = 510 kW 30 × 10−6 I × 10−3 = 133 Now, b = C = I B 30 × 10−6 R × 103 Voltage gain, AV = b C = 133 × = 0.52 RB 510 × 103 R Power gain, AP = b × AV = b2 C RB 15 (i) The symbol of AND gate is as follows A A 0 Y = (133)2 × × 103 510 × 103 = 69 Physics for you | November ‘15 79 OR I3 is zero as the diode in that branch is reverse biased Resistance in the branch AB and EF are each (25 + 125) W = 150 W A C E I1 G 17 I4 125  I3 125  I2 125  25  5V B D F H As AB and EF are identical parallel branches, their 150 = 75 W effective resistance is \ Net resistance in the circuit = (75 + 25) W = 100 W = 0.05 A \ Current I1 = 100 As resistances of AB and EF are equal, 0.05 \ I2 = I4 = = 0.025 A Given Vi = 10 V, RB = 400 kW = 400 × 103 W RC = kW = × 103 W, VBE = VCE = 0, VCC = 10 V As Vi – VBE = RBIB \ 10 – = (400 × 103)IB 10 IB = = 25 × 10−6 A = 25 mA 400 × 10 and VCC – VCE = ICRC 10 IC = = 3.33 × 10−3 = 3.33 mA × 103 I 3.33 × 10−3 b= C = = 133 IB 25 × 10−6 18 (i) For n-type region, ne = ND = × × 1028 = × 1022 m–3 10    ppm =    106  As nenh = ni2, 80 Physics for you | November ‘15 n (1.5 × 1016 m−3 )2 nh = i = = 0.45 × 1010 m–3 ne × 1022 m−3 For p-type region, 200 25 –3 nh = NA = × × 1028 = × 10 m 10 ni2 (1.5 × 1016 m −3 )2 = Now, ne = = 2.25 × 107 m–3 nh × 1025 m −3 (ii) The minority carrier holes of n-region wafer (nh = 0.45 × 1010 m–3) would contribute more to reverse saturation current than minority carrier electrons of p–region wafer (ne = 2.25 × 107 m–3) when p – n junction is reverse biased 19 Boolean expression for logic circuit (a) Y=A A Here the given NOR gate with A short circuit input is acting as NOT gate Its truth table is Boolean expression for logic circuit (b) Y So by De-Morgan’s theorem ( ) Y = A + B = A⋅B = A⋅B Hence, the logic circuit acts like AND gate Its truth table is A 0 1 B 1 Y 0 20 According to diagram, Vmax = V 100 20 = 50 V , Vmin = = 10 V 2 20 V 100 V t (i) Percentage modulation, P = m × 100 − Vmin V × 100 P = max Vmax + Vmin where m = modulation index 50 − 10 = × 100 = × 100 = 66.7% 50 + 10 (ii) Peak carrier voltage, Vc = Vmax + Vmin 50 + 10 = = 30 V (iii) Peak information voltage = Vm = mVc = × 30 = 20 V 21 Here, VCC = V, RC = 800 W, ICRC = 0.8 V, b = 25, ri = 200 W 0.8 V = 1.0 × 10−3 A As ICRC = 0.8 V, IC = 800 W Further, VCE = VCC – ICRC = V – 0.8 V = 7.2 V I I 1.0 × 10−3 Since b = C , I B = C = A IB b 25 = 0.04 × 10–3 A = 40 mA  800 W  R  Voltage gain, AV = b  out  = 25   = 100  Rin   200 W  (As Rout = RC = 800 W and Rin = ri = 200 W) Power gain, AP = bAV = 25(100) = 2500 22 Given the A.M wave, cm(t) = 5(1 + 0.6 cos 6280 t) sin 211 × 104 t, volts Comparing with the standard A.M wave, cm(t) = Ac(1 + m cos wm t) sin wc t, we get Ac = V, m = 0.6 w 6280 Modulating frequency, fm = m = = kHz 2π 2π w 211 × 104 Carrier frequency, fc = c = = 336 kHz 2π 2π (i) Minimum amplitude of A.M wave = Ac – mAc = – 0.6 × = V Maximum amplitude of A.M wave = Ac + mAc = + 0.6 × = V (ii) Frequency components of the A.M wave are fc – fm, fc, fc + fm i.e., 336 – 1,336,336 + or 335 kHz, 336 kHz, 337 kHz (iii) The amplitudes of the three components are mAc mA × × , Ac , c i.e., , 5, 2 2 or 1.5 V, V, 1.5 V 23 (a) Initiative, curiosity, scientific awareness, community service (b) Light emitting diode (LED) is a junction diode made of gallium arsenide or indium phosphide in which when holeelectron pairs recombine at forward biased p-n junction, energy is released in the form of light The principle on which LED works is spontaneous emission of radiation, when an electron jumps from higher energy level to lower energy level in a semiconductor atom At forward biased p-n junction, free electrons of n-type combine with holes of p-type semiconductor As free electrons lie in conduction band and holes lie in valence band, so electron falls from the higher to lower energy level containing holes and the energy is released in the form of radiation The energy of radiation emitted by LED is equal to or less than the forbidden energy band gap Eg of the semiconductor used, and is given hc by hu = = Eg λ The frequency of emitted radiation in LED, thus depends upon the band gap energy Eg of the semiconductor used The intensity of emitted radiation in LED depends upon the forward current flowing through the LED More the forward current flowing through the LED, more the hole-electron pairs combine at forward biased p-n junction, releasing more number of photons and hence larger will be the intensity of emitted radiation 24 Refer point 9.3(4,5) page no 588 (MTG Excel in Physics) OR Refer point 9.4(8) page no 595 (MTG Excel in Physics) 25 Refer point 9.2, (2 (a, b)) page no 585 and point 9.3(6(ii)) page no 588 (MTG Excel in Physics) OR Refer point 9.1 (3,4,5,6) page no 583 (MTG Excel in Physics) 26 Refer point 9.4 (9) page no 595 (MTG Excel in Physics) OR Refer point 9.3(6(i), 7(i)) page no 589 (MTG Excel in Physics) nn Physics for you | November ‘15 81 Y U ASK WE ANSWER Do you have a question that you just can’t get answered? Use the vast expertise of our mtg team to get to the bottom of the question From the serious to the silly, the controversial to the trivial, the team will tackle the questions, easy and tough The best questions and their solutions will be printed in this column each month Q1 When you load up a plastic shopping bag with groceries and then carry the bag by the loops at the top of the bag, why will the loops initially withstand the load but then, several minutes later, begin to stretch, perhaps to the point of tearing? – Akash Jain (UP) Ans If you suspended a load from the lower end of a spring hanging from a ceiling, the spring will stretch by a certain amount and then stay stretched Plastic, which consists of polymers, is different If you suspend a load from the lower end of a plastic strip, the strip will initially stretch like the spring but thereafter it will gradually stretch more in what is called viscoelastic creep The mechanism of this creep can vary from polymer to polymer The polymer consists of many long and entangled molecules When the polymer is put under load, these molecules gradually disentangle somewhat because very are pulled in the direction of the load The gradual reorientation of the molecules allows the plastic to gradually stretch Q2 What causes the tides? Why most shore locations have two high tides per day but others have only one? – Sahil Verma (Delhi) Ans The primary cause of tides is the Moon’s gravitational pull on Earth’s oceans even though that force is not strong enough to lift the water Because the force varies over Earth’s surface (strongest on the side facing the Moon, weakest on the opposite side), the force reshapes the water distribution by stretching it parallel to the line connecting Earth and the Moon The stretching 82 Physics for you | November ‘15 produces two bulges in the water distribution, one on the side facing the Moon and one on the opposite side If Earth did not rotate, then a shore location in the bulge facing the Moon would have high water (high tide) all day and a shore location in the opposite bulge would as well However, Earth’s rotation means that a shore location rotates through both bulges in about a day and will thus have two high-water intervals The bulges are not exactly positional on a line through Earth and the Moon because the water motion encounters friction within the water and against shorelines The friction delays the water’s response to the stretching by the Moon So, the high-water point in a port city may occur an hour or more after the Moon is highest in the sky Another complicating factor is that the gravitational force from the Sun also tends to stretch the water distribution However, the solar effect is, roughly less than half the lunar effect Although the Sun is much larger than the Moon, it is also much farther from Earth During New Moon and Full Moon, the Sun and Moon are aligned and their tidal effects sum to give larger tides called spring tides When the directions to the Sun and Moon are separated by 90°, the sum gives neap tides Because of these various complications some shore locations can have only one noticeable tide per day Q3 When an airplane flies somewhat overhead and close enough to be heard, lower your head by stooping to the ground Why does the frequency of the airplane noise increases as you lower your head ? – Shreyas (Haryana) Ans The sound you hear consists of the sound coming directly to you from the airplane and the sound that reflects to you from the ground The two sets of sound waves undergo interference at your ears and you hear primarily the waves that constructively interfere, they reinforce one another rather than cancel one another The height above the ground at which constructive interference occurs depends on the wavelength, greater wavelength requires greater height As you lower your head, you move down to the heights at which shorter wavelengths or higher frequencies, undergoes constructive interference So, as you stoop, the sound you hear increases in frequency  to d jo in tl y s a w a rd e of ry ve 2015 wa e disco nald for th s P ri z o s l D a e c m b M o e N v B he Arthur trinos d u n e a n a t s a jit a te th e laurea shows Takaaki K utrinos, th s, which e n n o f ti o a ld u ill re o c tu c s o that ike na neutrino ameleon-l le physics ring the ch le in partic zz ademy of u c p A g h in is By uncove l Swed ng-stand ya lo o R a e d e th lv , had so e cosmos ch rasp of th ctions, su alter our g uclear rea n a in d d e id te in a a a s o rem les cre Sciences cule partic the neutrin s is s a e in w d m a it c t re e a a d th For d Neutrinos hers prove the stars s sun and an researc c e laureate ri e th m d A n l a as in the ti eutrinos e n article un th p f o l g a n s c lli d ti e e in k r disp hypoth are three to anothe 56 There one kind m o real in 19 fr s te s a massle ey oscill inos were showed th that neutr n o ti o n long-held T NASA plans to set up Mars colonies by 2030 Live Physics ize 2015 Nobel Pr e in P h y s ic s N ISRO will launch India’s own satellite navigation: IRNSS T he Global Positioning System used by almost all countries in the world is controlled by USA Govt American monopoly on satellite based navigation is all set to end now, as Indian Space Research Organisation will launch India’s own satellite based navigation next year Christened as Indian Regional Navigation Satellite System or IRNSS, this navigation system would be controlled by Indian Govt Scientists at ISRO started a series of meetings with various location and navigation device manufacturers, mobile phone companies and global information system (GIS) technology developers from all over the world, to explain them the advantages and benefits of using IRNSS As per reports coming in, ISRO will offer two types of services via IRNSS; Standard Positioning Service (SPS), which would be available for all users, mobile phones, e-commerce services, digital services and Restricted Service (RS),which would be exclusively for defense users ASA is planning to have humans living on Mars in the next few decades Moving to have Earth independent colonies on the Red Planet will be the end point of years of research, the agency has said, but it plans for that to be complete by the 2030s NASA laid out the plans in a large document: “NASA’s Journey to Mars -Pioneering Next Steps in Space Exploration” The document lays out the three stages of NASA’s plan to get to Mars The first is Earth Reliant; the second is Proving Ground, where the operations will be tested out in deep space, but in an environment that allows humans to get back to Earth in days NASA hopes that those two first stages allow it to get to the Earth independent stage It sees the Earth independent colonies as being a global achievement that marks a transition in humanity’s expansion as we go to Mars not just to visit, but to stay While there, humans will live and work within habitats that support human life for years, with only routine maintenance They’ll harvest resources to create fuel, water, oxygen and building materials and use advanced communication systems to send information back with only a 20-minute delay NASA has expressed such interest before, most recently proposing to send a small greenhouse to the planet in order to experiment with cultivating plant life, something that would be essential to establishing a permanent colony in the future Although Mars does not currently seem to be a great habitat for existing life It is still possible, things may be living beneath the surface something that can only be explored effectively by humans, not robots Physics for you | November ‘15 83 84 Physics for you | November ‘15 Readers can send their responses at editor@mtg.in or post us with complete address by 25th of every month to win exciting prizes Winners' name with their valuable feedback will be published in next issue across A region under the influence of some physical agency (5) An abnormal transient electrical disturbance in a conductor (5) Rotating part of electric motor or generator (8) A penlike device attached to a VDU by which information may be input to a computer (5, 3) A manually operated device by which people communicate with computer (8) A mechanical device that prevents any sudden or oscillatory motion of a moving part of any piece of apparatus (7) 13 A device that allows microwave radiation to pass in one direction, while absorbing it in the reverse direction (8) 15 The converse of compression (11) 17 A type of circuit having two stable states (8) 19 A thread like body, particularly the conductor of metal or carbon in an incandescent lamp (8) 22 X-rays of long wavelength produced when electrons are accelerated by voltages of 25 kV or less (5, 4) 24 A conductor or group of conductors used for the transmission and/or distribution of electrical power (4) 25 A device incorporated in an electrical or other indicating instrument to provide the necessary damping (6) down A system of wires or waveguides that conveys radio frequency power between a radio aerial and a transmitter or reciever, with minimum loss (6) A planoconcave lens placed between the objective and eyepiece in a telescope to increase magnification (6, 4) Process of boiling (10) The adjustment of tuning of the notes of a keyboard instrument to give a near diatonic scale for all keys (11) Cut Here 11 12 10 13 14 15 16 18 17 20 19 21 22 23 24 25 10 The science concerned with the production, properties and propagation of sound wave (9) 11 A device used for stabilizing voltage, consisting of a sensitive metallic resistor whose resistance increases with temperature (9) 12 To bring neutrons into thermal equilibrium with surrounding (10) 14 The number of hydrogen (or equivalent) atoms that an atom will combine with or displace (7) 16 The study of the production and effects of very low temperatures (10) 18 A unit of work or energy equal to one watt operating for one hour (4, 4) 20 An acronym for allied submarine detection investigation committee (5) 21 The disappearance of signals for short time due to variations in the height and density of ionisation of ionosphere (6) 23 A missile or space vehicle powered by ejecting gas (6)  Physics for you | November ‘15 85 86 Physics for you | November ‘15 [...]... gravitational force are as follows : • • • • 32 force and Coulomb force and gravitational force follow the same inverse square law Coulomb force can be attractive or repulsive while gravitational force is always attractive Coulomb force between the two charges depends on the medium between two charges while gravitational force is independent of the medium between the two bodies The ratio of coulomb force to... coulomb force to the gravitational force between two protons at same distance apart is e2 = 1.3 × 1036 4pe 0Gm pm p Physics for you | november ‘15 Superposition theorem : The interaction between any two charges is independent of the presence of all other charges Electrical force is a vector quantity therefore, the net force on any one charge is the vector sum of all the forces exerted on it due to each... surface charge density (assumed uniform) ^ n is normal unit vector E r E /20 r Q is total charge on the ring x = distance of point on the axis from centre of the ring Electric field is always along the axis E Emax R Physics for you | november ‘15 x 2 35  s^ E= n e0 f f (i) For r ≥ R,  kQ ^ E=  2r |r | f f f (ii) For r < R, E=0 f (i) For r ≥ R,  kQ ^ E=  2r |r | (ii) For r < R,  kQ  E= 3r R ^ R... or r2 = r1 = × 1000 = 800 kg m −3 5 5 10 (c) : Gravitational force remains constant on the falling spherical ball It is represented by straight line P The viscous force (F = 6 phrv) increases as the velocity increases with time Hence, it is represented by curve Q Net force = gravitational force – viscous force As viscous force increases, net force decreases and finally becomes zero Then the body falls... the point Q is total charge = s4pR2 Physics for you | november ‘15 37 Uniformly charged kQ solid non-conducting For r ≥ R, V = R sphere kQ(3R 2 − r 2) For r ≤ R, V = 2R 3 r = (3R 2 − r 2) 6e 0 f f f f f f Infinite line charge Not defined f f Infinite nonconducting thin sheet Not defined Infinite charged conducting thin sheet Not defined f f f f SELF CHECK 6 A uniformly charged solid sphere of radius... dipole is placed in a uniform electric field, 4pe 0 (r 2 + a 2)3/2 it will experience only torque and the net force on where q is the charge on the ring and a is the radius of the dipole will be zero while when it is placed in a the ring non uniform electric field, it will experience both 1 q At very large distance i.e r >> a, E = 2 torque and net force 4pe 0 r List of formula for electric field intensity... greater than the pseudo force, so the  balloon deflects towards right airVg  HeVa T 20 (air– He)Vg airVa HeVg  (air– He)Va  T Physics for you | november ‘15 \ tan q = (rair − rHe )Va a = (rair − rHe )Vg g Force exerted by a liquid on a vertical surface Let us assume, that the wall of a dam of width w stops a liquid of height h from flowing It obviously is experiencing force due to the liquid’s... + ++ + + + + ++ + + (b) + + ++ + + (d) (JEE Main 2015) 36 Physics for you | november ‘15 r E kQ/R2 Sphere acts like a point charge placed at centre for points outside the sphere  E is always along radial direction Q is the total charge (= s4pR2) f /0 n is the unit vector perpendicular to the surface f f E s is the surface charge density (assumed uniform) R r E kQ/R 2 R r 4 A thin disc of radius b... Electric potential due to various charge distributions are given in table below : Name /Type Point charge Ring (uniform/ non-uniform charge distribution) Formula V= kq r kQ at centre , R kQ V= , along the axis R2 + x 2 V= Uniformly charged kQ hollow conducting/ For r ≥ R , V = r non-conducting/solid For r ≤ R, V = kQ conducting sphere R Note f f f f f f f Graph q is source charge r is the distance of the point... (hA)g = rlVsub g = weight of displaced liquid Physics for you | november ‘15 19 Note : This result is applicable only if the container is vertically unaccelerated, else we need to replace g with geff in the result Now, let us see what would the condition of floatation for the object be For equilibrium, Fup = mg ⇒ rl(Ah)g = rs(AH)g h rs ⇒ = ≤ 1 [ h ≤ H for floatation] H rl \ If rs ≤ rl, the object ... any form is prohibited Physics for you | November ‘15 P PHYSICS MUSING hysics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh The aim of Physics. .. the axis E Emax R Physics for you | november ‘15 x 35  s^ E= n e0 f f (i) For r ≥ R,  kQ ^ E=  2r |r | f f f (ii) For r < R, E=0 f (i) For r ≥ R,  kQ ^ E=  2r |r | (ii) For r < R,  kQ ... gravitational force are as follows : • • • • 32 force and Coulomb force and gravitational force follow the same inverse square law Coulomb force can be attractive or repulsive while gravitational force