Volume 24 Managing Editor Mahabir Singh Editor Anil Ahlawat (BE, MBA) No January 2016 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 30 Core Concept 12 Thought Provoking Problems 22 PMT Practice Paper 25 JEE Accelerated Learning Series Brain Map 31 46 Ace Your Way CBSE XI 57 JEE Workouts 64 Ace Your Way CBSE XII 68 Exam Prep 2016 75 Physics Musing Solution Set 29 81 Live Physics You Ask We Answer 83 84 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 | january ‘16 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 30 single oPtion correct tyPe Two batteries of emf e1 and e2 having internal resistance r1 and r2 respectively are connected in series to an external resistance R Both the batteries are getting discharged The above described combination of these two batteries has to produce a weaker current than when any one of the batteries is connected to the same resistor For this requirement to be fulfilled e2 r r (a) e must not lie between and r2 + R r1 + R r r +R e (b) must not lie between and r1 + R r1 e1 r e r (c) must lie between and r2 + R e1 r1 + R e2 r +R r (d) must lie between and e1 r1 r1 + R A particle of charge per unit mass a is released from  origin with velocity v = v0i in a magnetic field  v0 B = − B0  k for x ≤ B0a  v0 and B = for x > B0a π   The x-coordinate of the particle at time t  >  3B0a  would be (a) π  v0  + v t− B0a  B0a  (b) π  v0  +v t − B0a  3B0a  (c) v0 v0  π  + t − B0a  3B0a  (d) v0 v0t + B0a Physics for you | January ‘16 An ideal choke takes a current of 10 A when connected to an ac supply of 125 V and 50 Hz A pure resistor under the same conditions takes a current of 12.5 A If the two are connected to an ac supply of 100 V and 40 Hz, then the current in series combination of above resistor and inductor is (a) 10/ A (c) 10 A (b) 12.5 A (d) 10 A When an athlete runs with some acceleration, he leans forward The line joining his centre of mass to his foot which is in contact with the ground makes an angle q with the vertical The coefficient of friction between his foot and the ground is m His foot does not slip on the ground His acceleration is (a) mg (b) mgtanq (c) gtanq (d) (gtanq)/m In a regular polygon of n sides, each corner is at a distance r from the centre Identical charges are placed at (n – 1) corners At the centre, the magnitude of intensity is E and the potential is V The ratio V/E is (a) rn (b) r(n – 1) (c) (n – 1)/r (d) r(n – 1)/n Solution Senders of Physics Musing set-29 Amatra Sen (WB) Manmohan Krishna (Bihar) Meena Chaturvedi (New Delhi) Naresh Chockalingam (Tamil Nadu) set-28 Azhar Qureshi (Bihar) Amatra Sen (WB) Preeti Puri (Haryana) subjective tyPe In figure, a long thin wire carrying a varying current I = I0 sin wt lies at a distance y above one edge of a rectangular wire loop of length L and width W lying in the X-Z plane What emf is induced in the loop? X I W y L Z Y A wire is wrapped N times over a solid sphere of mass m near its centre, which is placed on a smooth horizontal surface A horizontal magnetic field of  induction B is present Find the m R angular acceleration experienced  B by the sphere Assume that the I mass of the wire is negligible compared to the mass of the sphere A stone is dropped from a balloon going up with a uniform velocity of 5.0 m s–1 If the balloon was 50 m high when the stone was dropped, find its height when the stone hits the ground Take g = 10 m s–2 A body weighs 98 N on a spring balance at the north pole What will be its weight recorded on the same scale if it is shifted to the equator? Use g = GM/R2 = 9.8 m s–2 and the radius of the earth R = 6400 km 10 A wheel of radius r and moment of inertia I about its axis is fixed at the top of an inclined plane of inclination q as shown in figure A string is wrapped round the wheel and its free end supports a block of mass M which can slide on the plane Initially, the wheel is rotating at a speed w in a direction such that the block slides up the plane After some time, the wheel stops rotating How far will the block move before stopping?  M  nn 10 Physics for you | January ‘16 Superposition of Waves When two or more waves of similar kind, simultaneously arrive at a point then the resultant disturbance (or displacement in case of particles) at the point of meeting is given by a vector sum of the disturbances produced by each of the arriving waves After superposition, the waves pass through as if they did not encounter each other, hence there is no change in the properties of either of the arriving waves after superposition If the disturbances are produced along same line, vector sum becomes algebraic sum Let us see what we understand from this Consider two disturbances travelling in opposite directions meet each other as shown cm s–1 cm s–1 1cm cm cm cm cm cm cm We are going to draw the shape of the resultant waveform at (i) t = s (ii) t = 2.5 s (iii) t = s To this, imagine individual waves travelling as if they are not meeting each other and then we apply superposition t=2s cm cm cm t = 2.5 s t=4s cm s–1 cm cm cm s–1 cm cm Now, you see, that during superposition, the waves individually appear to loose their identity, but after superposition we can clearly see, they were always there! Applications of Superposition of waves • Interference • Standing waves • Beats Interference Before we begin with the topic, we need to understand what are coherent sources These are such sources for which the phase difference is independent of time This clearly is possible only if the frequency of both the waves are identical else (w1 – w2)t will come out to be time dependent expression Coming back to interference; when waves of similar kind from two or more coherent sources simultaneously arrive at a point then the resultant intensity at the point of superposition is different from the sum of intensity of the arriving waves and is dependent on the phase difference between the arriving waves which is directly dependent on path difference between the arriving waves To put this in simple words, let me put a simple statement, imagine two coherent sources of light made to interfere There might be a situation where light + light = darkness Isn't it opposite to our common sense? Yes, it is since we expect more bright light when two light sources are made to superimpose But this logic is true only for non-coherent sources Let us see this through an example Consider waves S1 from two coherent sources S and S meeting at a point P travelling different S2 distances Contributed By: Bishwajit Barnwal, Aakash Institute, Kolkata 12 physics for you | January ‘16 x1 x2 P The displacement equations for the two sources are say S1 : y1(x1, t) = A1sin(wt – kx1) S2 : y2(x2, t) = A2sin(wt – kx2) Clearly, the phase difference between them is Df = (wt – kx1) – (wt – kx2) = k(x2 – x1) 2p \ Df = (Dx ) l where Dx = x2 – x1 = path difference between the arriving waves Hence from phasor analysis of SHM, we now understand A2 that the resultant amplitude of  oscillation can be found out \ AR = A12 + A22 + A1 A2 cos(Df) where AR = resultant amplitude of oscillation Now we understand that A1 and A2 are fixed for wave but if we change the location of sources from the point P, x1 and x2 changes, hence Dx changes, hence Df changes Therefore AR becomes dependent on Dx Now, considering intensity of waves, as we know, I ∝ A2 2 \ from, AR = A1 + A2 + 2A1A2cos(Df) we have , I R = I1 + I2 + I1I2 cos(Df) where IR = intensity of the resultant wave due to arriving waves of intensities I1 and I2 Now extreme cases of interference may arise € Constructive interference: When the waves arrive such that crest of one wave coincides with the crest of the other, the waves are said to be in phase and in such case the amplitude and hence intensity of resultant wave is maximum A1  A1 + A2 = AR A2 This clearly is possible only if one wave is shifted with respect to the other wave by an integral multiple of complete wavelength (l)  2p  \ Dx = nl ⇒ Df =   (nl) = n(2p)  l  \ AR = A1 + A2 max IR max 14 = ( I1 + I2 )2 physics for you | January ‘16 A1 + AR = A1 – A2  AR A1 + € Destructive interference: If the waves arrive such that crest of one wave coincides with the trough of the other, the waves are said to be out of phase and in such case the resultant amplitude and hence intensity of wave is minimum A2 This clearly is possible only if one wave is shifted with respect to the other by an odd multiple of half wavelengths l \ Dx = (2n + 1) ⇒ Df = (2n + 1)p \ AR = A1 – A2 I Rmin = ( I1 − I2 )2 Hence, generalising, A1 – A2 ≤ AR ≤ A1 + A2 ( )2 ( )2 I1 − I2 ≤ I R ≤ I1 + I2 Special case: If the arriving waves are of identical amplitude and hence intensity, then A1 = A2 = A0 (say) I1 = I2 = I0 (say) \ AR = 2A0, IR = 4I0 max and AR max = 0, IR = Now, this is what I was talking about when I said light + light = darkness! In general at any point in such case,  Df   Df  AR = A0 cos  , I R = I0 cos2       Let us now see an application of interference Q1 Two coherent sources S1 and S2 are symmetriR cally placed with S1 S2 respect to centre at D 3 a separation 3l as R >> 3 shown D is a detector which can measure the resultant intensity at all points on circumference Find 10 Consider the a-decay of nucleus So, pe < pp, i.e., lesser momentum is associated with electron as compared to proton A → ZA−−24Y + 24 He Z X  238 → 234 92 U  90 Th + He 13 Here, d = 0.15 mm = 0.15 × 10–3 m = 15 × 10–5 m, l = 450 nm = 450 × 10–9 m = 4.5 × 10–7 m, D = 1.0 m (a) (i) Distance of the second bright fringe, nlD  lD  x2 =  xn =  d d  −7 × 4.5 × 10 × 1.0 × 4.5 = = × 10−2 −5 15 15 × 10 = 0.6 × 10–2 m = mm (ii) Distance of the second dark fringe, l D  3lD x2 =  xn = (2n − 1)  d 2d × 4.5 × 10−7 × 1.0 = = 4.5 mm × 15 × 10−5 lD (b) Fringe width β = d When screen is moved away, D increases, therefore width of the fringes increases but the angular separation (l/d) remains the same Before decay, neutron to proton ratio, 238 − 92 n/p = = 1.58 92 After decay, neutron to proton ratio, 234 − 90 n/p = = 1.60 90 Thus ratio increases OR 238 92U → 234 90Th + 2He + Q Energy released Q = Dmc2 = (mU – mTh – mHe)c2 = (238.05079 – 234.043630 – 4.002600)c2 = 0.00456 × 931.5 MeV Q = 4.25 MeV 11 Let m1 be the refractive index of lens m2 be the refractive index of medium 1  1 1 2 2 (i) (ii) (iii) (i) m1 > m2, it will act as a diverging lens (ii) m1 = m2, it will simply act as glass plate (iii) m1 < m2, it will act as converging lens 12 When a charged particle of charge q, mass m is accelerated under a potential difference V, let v be the velocity acquired by particle Then qV = mv or mv = 2qVm h h (i) de Broglie wavelength, l = = mv 2qVm or l ∝ qm \ q pmp e × 1837me le = = >1 lp qe me e × me So, le > lp, i.e., greater value of de Broglie wavelength is associated with electron as compared to proton (ii) Momentum of particle, p = mv = 2qVm or p ∝ qm \ 72 pe qe me me e = = × or or uc 103 × C >> 10–5 or C >> 10–8 F C >> 10–8 F or C = 0.01 mF 19 Here, umn = cRZ   ( n + p) − 1 , n2  where m = n + p, (p = 1, 2, 3, …) and R is Rydberg constant 1  n2 For p m1 > ma (b) : Two particles will meet at P After they will stick together, momentum m′v will remain same But charge is doubled, so radius is halved B y P (–r, r/2, 0) r x Finally, both move in dotted circle (d) : Volume of the balloon at any instant, when radius is r, V = pr 3 Time rate of change of volume, dr dV = pr dt dt Time rate of change of radius of balloon, dV dr = dt pr dt Flux through rubber band at the given instant, f = B (pr2) −df dr d = − (Bpr ) = −2 prB Induced emf, e = dt dt dt B dV  dV  =− = −2 prB   pr dt  2r dt As volume of the balloon is decreasing, dV is dt negative \ e= − (0.04) × (–100 × 10–6) = 20 mV × 10 × 10−2 (d) : As current at any instant in the circuit will be I = Idc + Iac = a + b sin wt 1/2  T I dt  1/2 ∫  1 T dt  So, Ieff =  = + ( sin w ) a b t   T  T ∫0   ∫0 dt  1/2 1 T  =  ∫ (a2 + 2ab sin wt + b2 sin2 wt ) dt  T  T T 1 As ∫ sin wt dt = and ∫ sin2 wt dt = T T 1/2   So, Ieff = a2 + b2     1 (c) : As = (m − 1)  −   R1 R2  f \ For equiconvex lens 3  1  =  − 1  +  10    R R  ⇒ R = 10 cm \ Focal length of left part 3  1  =  − 1  +  f1    ∞ R  f1 f2 ⇒ f1 = 20 cm f  Similarly, f2 = 20 cm 1  = (3 − 1)  −   f′ −10 10  −10 cm ⇒ f ′ = \ = + + = − + = − f eq f1 f ′ f 20 10 20 20 10 or feq = –10/3 cm Physics for you | January ‘16 81 (c) : (i) P.E = Mg I = 3.36(1 + 2t) × 10–2 A dI = × 3.36 × 10–2 A s–1 dt Magnetic induction at every point on the loop, mI B= pd Magnetic flux linked with loop at any instant, mI f = BA = ⋅ pr 2 pd df m0 r  dI  Induced emf, e = =   2d  dt  dt Induced current, m r  dI  e I= =   R R × 2d  dt  L = 0.5MgL (ii) P.E = Mg L = 0.5MgL 2 Mg L 2 (iii) P.E = Mg R = = MgL ≈ 0.2 MgL p p p p2 (iv) P.E = MgL ≈ 0.4 MgL p 2R  (v) P.E = Mg  R −   p  p−2 p−2 =  MgR = MgL  p  p 1.14 = MgL = 0.11 MgL p \ (i) = (ii) > (iv) > (iii) > (v) (a) : the weight of the removed cylinder is 15 N If a symmetrical hole was drilled on the other side, the uniform cylinder would have zero torque about P this implies that the torque due to excess weight (15 N) on the other side has caused instability Hence for equilibrium the torque of T must balance the torque due to excess weight \ T(2a) = (15)  a  3  or T = N We have considered torque about P so that torques of unknown forces N and f are zero −7 −3 −2 = p × 10 × (10 ) × × 3.36 × 10 8.4 × 10−4 × × = 5.024 × 10–11 A nn solution of December 2015 crossworD G O D T F 14 A C D R A S T 20 E S A a 2a/3 N P f (d) : As l / = v0t1 ⇒ t1 = l 3v0 t t l     2 Also = v1   + v2   2 2 4l ⇒ t2 = 3(v1 + v2 ) 3v (v + v ) l \ Mean velocity = = t1 + t v1 + v2 + 4v0 10 (a) : As the loop is very small, the distance of every point in its plane can be taken to be equal to d = m 82 I C Physics for you | January ‘16 23 B       C I 2R C L A Y C E L A L C R E R V I T I S F U S I V I 11 12 L E A H 15 C E S S L I O L O N I A 17 V U D I E M 18 C V T O K E S 21I 22 B R E D O O T S T R A N 24 E L E C T 26 E U I T L L U V I A L Y O M E T E R O G A M 10 Z T Y 13 D H Y N E D R L A T O M G R N F R A S P H A P P E R O R I G 16 J E E E T E R 19 F O U N D L C R R O N 25V I O L E T C H O G R A M Winners (December 2015) Akshat Puri (new Delhi) Karan Verma (UP) Harsh Gupta (Haryana) solution senders (november 2015) Mohammad naderi (iran, Khalkhal) Anuj sen (WB) Poonam shah (WB) A Discovered: A new Venus-like planet stronomers have discovered a new Venus-like rocky exoplanet 39 light years away , which may be cool enough to potentially host an atmosphere If it does, it is close enough that we could study that atmosphere in detail with the Hubble Space Telescope and future observatories like the Giant Magellan Telescope, researchers said “Our ultimate goal is to find a twin Earth, but along the way we’ve found a twin Venus,” said astronomer David Charbonneau of the Harvard-Smithsonian Center for Astrophysics in US.”We suspect it will have a Venus-like atmosphere too, and if it does we can’t wait to get a whiff,” he said The planet-GJ 1132b -orbits a red dwarf star only one-fifth the size of our Sun The star is also cooler than the Sun, emitting just 1/200th as much light GJ 1132b circles its star every 1.6 days at a distance of 1.4 million miles As a result, it is baked to a temperature of about 232 degrees Celsius Such temperatures would boil off any water the planet may have once held S L Lasers to make materials hotter than Sun’s core Witnessed: Black hole swallowing a star 300m light years away cientists have for the first time witnessed a black hole swallowing a star in a galaxy 300 million light years away and ejecting a flare of matter moving at nearly the speed of light The finding tracks the star -about the size of our Sun -as it shifts from its customary path, slips into the gravitational pull of a super massive black hole and is sucked in, said Sjoert van Velzen, a Hubble fellow at the Johns Hopkins University in US “It’s the first time we see everything from the stellar destruction followed by the launch of a conical outflow, also called a jet, and we watched it unfold over several months,” van Velzen said The first observation of the star being destroyed was made in December last year Researchers used radio telescopes to follow up as fast as possible They were just in time to catch the action By the time it was done, the team had data from satellites and ground-based telescopes that gathered X-ray, radio and optical signals, providing a stunning “multi-wavelength” portrait of this event It helped that the galaxy in question is closer to Earth than those studied previously in hopes of tracking a jet emerging after the destruction of a star asers could heat materials to temperatures hotter than the centre of the Sun in only 20 quadrillionths of a second, according to new research that could revolutionise energy production Physicists from Imperial College London have devised an extremely rapid heating mechanism that they believe could heat certain materials to ten million degrees in much less than a million millionth of a second The method, proposed for the first time, could be relevant to new avenues of research in thermonuclear fusion energy, where scientists are seeking to replicate the Sun’s ability to produce clean energy The heating would be about 100 times faster than rates currently seen in fusion experiments using the world’s most energetic laser system at the Lawrence Livermore National Laboratory in California Courtesy : The Times of India Physics for you | January ‘16 83 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 Is light visible or invisible? –Sajal Sengupta Ans Light is invisible until it hits an object It can only be visible to us whenever it get scattered or bounced off the dust particles in the intervening medium We can see the source of light like stars at night as they are too dim to light up much but that light does hit our eyes and therefore we see the light The above discussion on visibility of light is only confined to the electromagnetic radiation of wavelength from 400 nm to 700 nm which is called visible light, as these electromagnetic radiations gives the sensation of sight Q2 Why various colours seen in a film of soap? –Bidhan Banerjee Ans The colours in a soap film are caused by interference (internally and externally) of light waves The reason for various colours which are seen in a film of soap is due to interference of light reflecting off of two nearby surfaces, the outer surface and the inner surface or interference between the transmitted portions of the incident light In case of white light of range 400 nm to 700 nm, the interference enhances some wavelengths and suppresses other, which leads to the splitting of white light into visible colours The wavelength that got enhanced or suppressed depends only on thickness of the film and the refractive index of soap film Q3 It is said that while passing through turnings one should bend at maximum angle to have maximum speed, but in reality if one bends more than a specific value of angle, one falls down Why? –Suryakant Khilar (Odisha) 84 Physics for you | January ‘16 Ans A cyclist while going round a curve on a horizontal track has to bend himself a little from his vertical position in order to avoid overturning When he bends himself inward a component of the reaction of road provides him the necessary centripetal force for circular motion as tanq = v2 rg  2 −1 v or q = tan   , therefore, to take a safe turn  rg  cyclist should go slow and bend through a small angle q This balances the torque about the wheel contact patches generated by centrifugal force due to the turn with that of the gravitational force And the over bending of a cyclist on a turning road results a shift in the point of centre of gravity of the cyclist which increases the torque due to gravitational pull downward and it breaks the balanced condition Hence over bending results in a fall down of a cyclist Q4 Why does the earth rotate? –Apratarkya Banerjee Ans The reason for the rotation of earth and planets arise from the concept of formation of our solar system In the beginning solar system was a big cloud of dust and gases This cloud began to collapse, flattening into large disk, rotating faster and faster continuously, as they collapsed, their gravitational orbit set those dust and gases to spin As a result, the clumps of massive particle formed within that disk were going to naturally have some sort of rotation Due to conservation of angular momentum spinning went faster and faster for e.g., skaters speed up their rate of spin as they brought their arms closer to their body Also due to gravitational pull from all directions, the clump of massive particles became a round planet, our earth is also one of those clumps Inertia of motion keeps the planet spinning continuously on its axis unless any external agent resists this rotation Actually earth’s rotation is also affected by tidal pull of moon, it slows down at the rate of about millisecond per year Earth’s spinning was faster in the past At that time, a day was about 22 hour long only nn 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 numerical description of how far apart objects are (8) The point on a wave with the maximum value or upward displacement within a cycle (5) A term used to describe a large, circular plain (4) 10 Radiation, such as ultraviolet, able to produce a chemical change in exposed materials (7) 11 The angular distance of an object around or parallel to the horizon from a predefined zero point (7) 18 A stream of atomic nuclei that are observed to strike the Earth atmosphere with extremely high amounts of energy (6, 3) 20 A line on a map joining points at which the acceleration of free fall is constant (6) 24 The theoretical time reversal of a black hole, which arises as a valid solution in general relativity (5, 4) 25 A series or chain of craters (6) 26 A term used to describe a point directly underneath an celestial sphere (5) 27 The outer edge or border of a planet or other celestial body (4) 28 The chemical decomposition of materials into ions, excited atoms and molecules etc by ionizing radiation (10) down The surface of the Sun or other celestial body projected against the sky (4) The point of greatest separation of two stars, such as in a binary star system (8) An imaginary line in the sky traced by the Sun as it moves in its yearly path through the sky (8) A structural element that is capable of withstanding load primarily by resisting bending (4) A unit used to measure solar radiation and equivalent to 18 × 104 J m–2 (7) A mechanism that converts rotational motion to linear motion, and a torque to a linear force (5) An electrical device possessing reactance and selected for use because of that property (7) 12 An extension of string theory in which eleven dimensions are identified (1, 6) 13 The inner coat of the eye, having enormous number of light sensitive cells in the form of rods and cones (6) 14 The amount of potential energy stored in an elastic substance by means of elastic deformation (10) 15 A derived unit of energy, work or amount of heat in the international system of units (5) 16 A mechanically commulated electric motor powered from direct current (1, 1, 5) 17 The rate of mass flow per unit area (4, 4) 19 A region of space where the density of matter, or the curvature of space-time, becomes infinite and the concepts of space and time cease to have any meaning (11) 21 A term used to describe an exceptionally bright meteor (6) 22 Meteor, generally brighter than magnitude-4, which is about the same magnitude of planet Venus in the morning and evening sky (8) 23 Unit equivalent to coulomb per second (6)  Physics for you | january ‘16 85 86 Physics for you | january ‘16 ... 1)  −  , f R R  1 3  1  =  − 1? ??  +  = × =     10 10 10 10 f1    ? ?1   −30  −3 =  − 1? ??  −  = ×  = f2    10 20   10 × 20  10 0 8  1  =  − 1? ??  +  = f3   ... thrust will be (g = 10 m s–2) (a) 12 7.5 kg s? ?1 (b) 18 7.5 kg s? ?1 ? ?1 (c) 18 5.5 kg s (d) 13 7.5 kg s? ?1 30° (a) T2 = 75 N (b) T2 = 75 N (c) T1 = 25 N (d) T1 = 12 N 27 A car of mass 10 00 kg negotiates... × 10 –3 × 0 .1 = × 10 –4 m3 The final volume of the gas, V2 = V1 + DV = 2.4 × 10 –3 + × 10 –4 = 3.2 × 10 –3 m3 Let T2 be the final temperature of the gas, then P1V1 P2V2 PV = ⇒ T2 = 2 T1 T1 T2 P1V1