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Chapter One, , INTRODUCTION, , The National Curriculum Framework (NCF) – 2005 initiated a new phase, of curriculum revision. First, new syllabi for Science and Mathematics, for all stages of school education were developed. Based on these syllabi,, new textbooks were developed. As a part of this effort, Physics textbooks, for Classes XI and XII were published in 2006 and 2007, respectively., One of the major concerns expressed in NCF–2005 is regarding, Examination Reform., According to NCF–2005, “A good evaluation and examination, system can become an integral part of the learning process and, benefit both the learners themselves and the educational system, by giving credible feedback”., It further notes that,, “Education is concerned with preparing citizens for a, meaningful and productive life, and evaluation should be a way, of providing credible feedback on the extent to which we have, been successful in imparting such an education. Seen from this, perspective, current processes of evaluation, which measure and, assess a very limited range of faculties, are highly inadequate and, do not provide a complete picture of an individual’s abilitiy or, progress towards fulfilling the aims of education”., , 20/04/2018
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Exemplar Problems–Physics, The purpose of assessment is to determine the extent to which learning, has taken, on the one hand and to improve the teaching-learning process, and instructional materials, on the other. It should inter alia be able to, review the objectives that have been identified for different school stages, by gauging the extent to which the capabilities of learners have been, developed. Tests should be so designed that we must be able to gauge, what children have learnt, and their ability to use this knowledge for, problem-solving and application in the real world. In addition, they must, also be able to test the processes of thinking to gauge if the learner has, also learnt where to find information, how to use new information, and to, analyse and evaluate the same. The types of questions that are set for, assessment need to go beyond what is given in the book. Often children’s, learning is restricted as teachers do not accept their answers if they are, different from what is presented in the guidebooks. Designing good test, items and questions is an art, and teachers should spend time thinking, about and devising such questions., Observing on the current practices of the different boards of school, education in the country, the National Focus Group paper on, Examination Reform says:, “...Because the quality of question papers is low, they usually, call for rote memorisation and fail to test higher-order skills like, reasoning and analysis, let alone lateral thinking, creativity and, judgement”., It further advocates the inclusion of Multiple Choice Questions (MCQ)a type of question that has great untapped potential. It also notes the, limitation of testing through MCQ’s only. “While MCQ can more deeply, probe the level of conceptual understanding of students and gauge a, student’s mastery of subtleties, it cannot be the only kind of question in, any examination. MCQs work best in conjunction with some open-ended, essay questions in the second part of the paper, which tests expression, and the ability to formulate an argument using relevant facts.”, In order to address to the problem, the Department of Education in, Science and Mathematics undertook a programme, Development of, Exemplar Problems in Physics for Class XI during 2007-08. Problems, based on different chapters in textbook of Physics for Class XI published, by the NCERT has been developed. Problems have been classified broadly, into five categories:, 1. Multiple Choice Questions I (MCQ I): only one correct answer., 2. Multiple Choice Questions II (MCQII): may have one or more than, one correct answer., 3. Very Short Answer Questions (VSA): may be answered in one/two, sentences., 4. Short Answer Questions (SA): require some analytical/numerical, work., 5. Long Answer Questions (LA): require detailed analytical/numerical, solution., , 2, 20/04/2018
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Introduction, , Though most of the questions given in a particular chapter are based, on concepts covered in that chapter, some questions have been developed, which are based on concepts covered in more than one chapter., One of the major objectives of involving learners in solving problems, in teaching-learning process is to promote a more active learning, environment, improve student learning and also support young teachers, in their professional development during their early formative teaching, experiences. For this to be achieved, problem-solving based on good, question should form an integral part of teaching-learning process. Good, questions engage students in progressively deeper levels of thinking and, reasoning. It is envisaged that the questions presented through this book, would motivate teachers to design good questions. What makes a question, good? According to Robyn L. Miller et al. 1, Some characteristics of a good question are:, • stimulates students’ interest and curiosity., • helps students monitor their understanding., • offers students frequent opportunities to make conjectures and argue, about their validity., • draws on students’ prior knowledge, understanding, and/or, misunderstanding., • provides teachers a tool for frequent formative assessments of what, their students are learning., • supports teachers’ efforts to foster an active learning environment., , A NOTE, , TO, , STUDENTS, , A good number of problems have been provided in this book. Some are, easy, some are of average difficult level, some difficult and some problems, will challenge even the best amongst you. It is advised that you first, master the concepts covered in your textbook, solve the examples and, exercises provided in your textbook and then attempt to solve the, problems given in this book. There is no single prescription which can, help you in solving each and every problem in physics but still researches, in physics education show that most of the problems can be attempted, if you follow certain steps in a sequence. The following prescription due, to Dan Styer2 presents one such set of steps :, 1., Strategy design, , 2., , (a) Classify the problem by its method of solution., (b) Summarise the situation with a diagram., (c) Keep the goal in sight (perhaps by writing it down)., Execution tactics, (a) Work with symbols., (b) Keep packets of related variables together., , 1, 2, , http://www. math.cornell.edu/~ maria/mathfest_education preprint.pdf, http://www.oberlin.edu/physics/dstyer/SolvingProblems.html, , 3, 20/04/2018
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Exemplar Problems–Physics, (c) Be neat and organised., (d) Keep it simple., 3., , Answer checking, (a) Dimensionally consistent?, (b) Numerically reasonable (including sign)?, (c) Algebraically possible? (Example: no imaginary or infinite, answers), (d) Functionally reasonable? (Example: greater range with greater, initial speed), (e) Check special cases and symmetry., (f ) Report numbers with units specified and with reasonable, significant figures., , We would like to emphasise that the problems in this book should be, used to improve the quality of teaching-learning process of physics. Some, can be directly adopted for evaluation purpose but most of them should, be suitably adapted according to the time/marks assigned. Most of the, problems included under SA and LA can be used to generate more, problems of VSA or SA categories, respectively., , 4, 20/04/2018
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Chapter Two, , UNITS AND, MEASUREMENTS, , MCQ I, 2.1, , The number of significant figures in 0.06900 is, (a), (b), (c), (d), , 2.2, , The sum of the numbers 436.32, 227.2 and 0.301 in appropriate, significant figures is, (a), (b), (c), (d), , 2.3, , 5, 4, 2, 3, , 663.821, 664, 663.8, 663.82, , The mass and volume of a body are 4.237 g and 2.5 cm3,, respectively. The density of the material of the body in correct, significant figures is, , 20/04/2018
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Exemplar Problems–Physics, (a), (b), (c), (d), 2.4, , The numbers 2.745 and 2.735 on rounding off to 3 significant, figures will give, (a), (b), (c), (d), , 2.5, , and, and, and, and, , 2.74, 2.73, 2.73, 2.74, , 164 ± 3 cm2, 163.62 ± 2.6 cm2, 163.6 ± 2.6 cm2, 163.62 ± 3 cm2, , Which of the following pairs of physical quantities does not have, same dimensional formula?, (a), (b), (c), (d), , 2.7, , 2.75, 2.74, 2.75, 2.74, , The length and breadth of a rectangular sheet are 16.2 cm and, 10.1cm, respectively. The area of the sheet in appropriate significant, figures and error is, (a), (b), (c), (d), , 2.6, , 1.6048 g cm–3, 1.69 g cm–3, 1.7 g cm–3, 1.695 g cm–3, , Work and torque., Angular momentum and Planck’s constant., Tension and surface tension., Impulse and linear momentum., , Measure of two quantities along with the precision of respective, measuring instrument is, A = 2.5 m s–1 ± 0.5 m s–1, B = 0.10 s ± 0.01 s, The value of A B will be, (a), (b), (c), (d), , 2.8, , (0.25 ± 0.08) m, (0.25 ± 0.5) m, (0.25 ± 0.05) m, (0.25 ± 0.135) m, , You measure two quantities as A = 1.0 m ± 0.2 m, B = 2.0 m ± 0.2 m., We should report correct value for AB as:, (a), (b), (c), (d), , 1.4 m ± 0.4 m, 1.41m ± 0.15 m, 1.4m ± 0.3 m, 1.4m ± 0.2 m, , 6, 20/04/2018
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Units and Measurements, , 2.9, , Which of the following measurements is most precise?, (a), (b), (c), (d), , 2.10, , The mean length of an object is 5 cm. Which of the following, measurements is most accurate?, (a), (b), (c), (d), , 2.11, , 4.9 cm, 4.805 cm, 5.25 cm, 5.4 cm, , Young’s modulus of steel is 1.9 × 1011 N/m2. When expressed, in CGS units of dynes/cm2, it will be equal to (1N = 105 dyne,, 1m2 = 104 cm2), (a), (b), (c), (d), , 2.12, , 5.00 mm, 5.00 cm, 5.00 m, 5.00 km., , 1.9, 1.9, 1.9, 1.9, , ×, ×, ×, ×, , 10 10, 10 11, 10 12, 10 13, , If momentum (P ), area (A) and time (T ) are taken to be, fundamental quantities, then energy has the dimensional formula, (a), (b), (c), (d), , (P1, (P2, (P1, (P1, , A–1 T1), A1 T1), A–1/2 T1), A1/2 T–1), , MCQ II, 2.13, , On the basis of dimensions, decide which of the following relations, for the displacement of a particle undergoing simple harmonic, motion is not correct:, (a) y = a sin 2π t / T, (b) y = a sin vt., a, t , (c) y = sin , T, a , 2π t, 2π t , , − cos, (d) y = a 2 sin, , , T, T , , 2.14, , If P, Q, R are physical quantities, having different dimensions, which, of the following combinations can never be a meaningful quantity?, (a) (P – Q)/R, (b) PQ – R, , 7, 20/04/2018
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Exemplar Problems–Physics, (c) PQ/R, (d) (PR – Q2)/R, (e) (R + Q)/P, 2.15, , Photon is quantum of radiation with energy E = h ν where ν is, frequency and h is Planck’s constant. The dimensions of h are the, same as that of, (a), (b), (c), (d), , 2.16, , If Planck’s constant (h ) and speed of light in vacuum (c ) are taken, as two fundamental quantities, which one of the following can, in, addition, be taken to express length, mass and time in terms of, the three chosen fundamental quantities?, (a), (b), (c), (d), , 2.17, , Mass of electron (me ), Universal gravitational constant (G ), Charge of electron (e ), Mass of proton (mp ), , Which of the following ratios express pressure?, (a), (b), (c), (d), , 2.18, , Linear impulse, Angular impulse, Linear momentum, Angular momentum, , Force/ Area, Energy/ Volume, Energy/ Area, Force/ Volume, , Which of the following are not a unit of time?, (a), (b), (c), (d), , Second, Parsec, Year, Light year, , VSA, 2.19, , Why do we have different units for the same physical quantity?, , 2.20, , The radius of atom is of the order of 1 Å and radius of nucleus is, of the order of fermi. How many magnitudes higher is the volume, of atom as compared to the volume of nucleus?, , 2.21, , Name the device used for measuring the mass of atoms and, molecules., , 8, 20/04/2018
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Units and Measurements, , 2.22, , Express unified atomic mass unit in kg., , 2.23, , A function f (θ ) is defined as:, f (θ ) = 1–θ +, , θ2, 2!, , –, , θ3, 3!, , +, , θ4, 4!, , Why is it necessary for q to be a dimensionless quantity?, 2.24, , Why length, mass and time are chosen as base quantities, in mechanics?, , SA, 2.25, , (a) The earth-moon distance is about 60 earth radius. What will, be the diameter of the earth (approximately in degrees) as seen, from the moon?, (b) Moon is seen to be of (½)°diameter from the earth. What must, be the relative size compared to the earth?, (c) From parallax measurement, the sun is found to be at a, distance of about 400 times the earth-moon distance. Estimate, the ratio of sun-earth diameters., , 2.26, , Which of the following time measuring devices is most precise?, (a), (b), (c), (d), , A wall clock., A stop watch., A digital watch., An atomic clock., , Give reason for your answer., 2.27, , The distance of a galaxy is of the order of 1025 m. Calculate the, order of magnitude of time taken by light to reach us from, the galaxy., , 2.28, , The vernier scale of a travelling microscope has 50 divisions which, coincide with 49 main scale divisions. If each main scale division, is 0.5 mm, calculate the minimum inaccuracy in the measurement, of distance., , 2.29, , During a total solar eclipse the moon almost entirely covers the, sphere of the sun. Write the relation between the distances and, sizes of the sun and moon., , 2.30, , If the unit of force is 100 N, unit of length is 10 m and unit of time, is 100 s, what is the unit of mass in this system of units?, , 9, 20/04/2018
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Exemplar Problems–Physics, 2.31, , Give an example of, (a), (b), (c), (d), , 2.32, , a physical quantity which has a unit but no dimensions., a physical quantity which has neither unit nor dimensions., a constant which has a unit., a constant which has no unit., , Calculate the length of the arc of a circle of radius 31.0 cm which, subtends an angle of, , π, 6, , at the centre., , 2.33, , Calculate the solid angle subtended by the periphery of an area of, 1cm2 at a point situated symmetrically at a distance of 5 cm from, the area., , 2.34, , The displacement of a progressive wave is represented by, y = A sin(wt – k x ), where x is distance and t is time. Write the, dimensional formula of (i) ω and (ii) k., , 2.35, , Time for 20 oscillations of a pendulum is measured as t1= 39.6 s;, t2= 39.9 s; t3= 39.5 s. What is the precision in the measurements?, What is the accuracy of the measurement?, , LA, 2.36, , A new system of units is proposed in which unit of mass is α kg,, unit of length β m and unit of time γ s. How much will 5 J measure, in this new system?, , 2.37, , The volume of a liquid flowing out per second of a pipe of length l, and radius r is written by a student as, , v=, , π Pr 4, 8 ηl, , where P is the pressure difference between the two ends of the, pipe and η is coefficent of viscosity of the liquid having dimensional, formula ML–1 T–1., Check whether the equation is dimensionally correct., 2.38, , A physical quantity X is related to four measurable quantities a,, b, c and d as follows:, X = a2 b3 c5/2 d–2., The percentage error in the measurement of a, b, c and d are 1%,, 2%, 3% and 4%, respectively. What is the percentage error in, , 10, 20/04/2018
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Units and Measurements, , quantity X ? If the value of X calculated on the basis of the above, relation is 2.763, to what value should you round off the result., 2.39, , In the expression P = E l 2 m–5 G–2, E, m, l and G denote energy,, mass, angular momentum and gravitational constant, respectively., Show that P is a dimensionless quantity., , 2.40, , If velocity of light c, Planck’s constant h and gravitational contant, G are taken as fundamental quantities then express mass, length, and time in terms of dimensions of these quantities., , 2.41, , An artificial satellite is revolving around a planet of mass M and, radius R, in a circular orbit of radius r. From Kepler’s Third law, about the period of a satellite around a common central body,, square of the period of revolution T is proportional to the cube of, the radius of the orbit r. Show using dimensional analysis, that, , T =, , k r3, ,, R g, , where k is a dimensionless constant and g is acceleration due, to gravity., 2.42, , In an experiment to estimate the size of a molecule of oleic acid, 1 mL of oleic acid is dissolved in 19 mL of alcohol. Then 1 mL of, this solution is diluted to 20 mL by adding alcohol. Now 1 drop of, this diluted solution is placed on water in a shallow trough. The, solution spreads over the surface of water forming one molecule, thick layer. Now, lycopodium powder is sprinkled evenly over the, film and its diameter is measured. Knowing the volume of the drop, and area of the film we can calculate the thickness of the film which, will give us the size of oleic acid molecule., Read the passage carefully and answer the following questions:, (a) Why do we dissolve oleic acid in alcohol?, (b) What is the role of lycopodium powder?, (c) What would be the volume of oleic acid in each mL of solution, prepared?, (d) How will you calculate the volume of n drops of this solution, of oleic acid?, (e) What will be the volume of oleic acid in one drop of this, solution?, , 2.43, , (a) How many astronomical units (A.U.) make 1 parsec?, (b) Consider a sunlike star at a distance of 2 parsecs. When it is, seen through a telescope with 100 magnification, what should, be the angular size of the star? Sun appears to be (1/2)° from, , 11, 20/04/2018
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Exemplar Problems–Physics, the earth. Due to atmospheric fluctuations, eye can’t resolve, objects smaller than 1 arc minute., (c) Mars has approximately half of the earth’s diameter. When it, is closest to the earth it is at about 1/2 A.U. from the earth., Calculate what size it will appear when seen through the same, telescope., (Comment : This is to illustrate why a telescope can magnify planets but, not stars.), 2.44, , Einstein’s mass - energy relation emerging out of his famous, theory of relativity relates mass (m ) to energy (E ) as E = mc 2,, where c is speed of light in vacuum. At the nuclear level, the, magnitudes of energy are very small. The energy at nuclear, level is usually measured in MeV, where 1 MeV= 1.6×10–13J;, the masses are measured in unified atomic mass unit (u) where, 1u = 1.67 × 10–27 kg., (a) Show that the energy equivalent of 1 u is 931.5 MeV., (b) A student writes the relation as 1 u = 931.5 MeV. The teacher, points out that the relation is dimensionally incorrect. Write, the correct relation., , 12, 20/04/2018
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Chapter Three, , MOTION IN A, STRAIGHT LINE, , MCQ I, 3.1, , Among the four graphs (Fig. 3.1), there is only one graph for which, average velocity over the time intervel (0, T ) can vanish for a, suitably chosen T. Which one is it?, , (a), , (b), , (c), , (d), Fig. 3.1, 20/04/2018
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Exemplar Problems–Physics, 3.2, , A lift is coming from 8 t h floor and is just about to reach, 4 t h floor. Taking ground floor as origin and positive, direction upwards for all quantities, which one of the, following is correct?, (a) x < 0, v < 0, a > 0, (b) x > 0, v < 0, a < 0, (c) x > 0, v < 0, a > 0, (d) x > 0, v > 0, a < 0, , 3.3, , In one dimensional motion, instantaneous speed v satisfies, 0 ≤ v < v0., (a) The displacement in time T must always take non-negative, values., (b) The displacement x in time T satisfies – vo T < x < vo T., (c) The acceleration is always a non-negative number., (d) The motion has no turning points., , 3.4, , 3.5, , A vehicle travels half the distance L with speed V1 and the other, half with speed V2, then its average speed is, (a), , V1 + V2, 2, , (b), , 2V1 + V2, V1 + V2, , (c), , 2V1V2, V1 + V2, , (d), , L (V1 + V2 ), V1V2, , The displacement of a particle is given by x = (t – 2)2 where x is in, metres and t in seconds. The distance covered by the particle in, first 4 seconds is, (a) 4 m, (b) 8 m, (c) 12 m, (d) 16 m, , 3.6, , At a metro station, a girl walks up a stationary escalator in time t1., If she remains stationary on the escalator, then the escalator take, , 14, 20/04/2018
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Motion in a Straight Line, , her up in time t2. The time taken by her to walk up on the moving, escalator will be, (a) (t1 + t2)/2, (b) t1t2/(t2–t1), (c) t1t2/(t2+t1), (d) t1–t2, , 3.7, , The variation of quantity A with quantity B, plotted in, Fig. 3.2 describes the motion of a particle in a straight, line., , A, , MCQ II, , (a) Quantity B may represent time., (b) Quantity A is velocity if motion is uniform., (c) Quantity A is displacement if motion is uniform., , B, , (d) Quantity A is velocity if motion is uniformly, accelerated., 3.8, , A graph of x versus t is shown in Fig. 3.3. Choose, correct alternatives from below., (a) The particle was released from rest at t = 0., (b) At B, the acceleration a > 0., , Fig. 3.2, , x, B, A, , E, , C, , (c) At C, the velocity and the acceleration vanish., (d) Average velocity for the motion between A and D is, positive., (e) The speed at D exceeds that at E., 3.9, , D, , t, , Fig. 3.3, , For the one-dimensional motion, described by x = t–sint, (a) x (t) > 0 for all t > 0., (b) v (t) > 0 for all t > 0., (c) a (t) > 0 for all t > 0., (d) v (t) lies between 0 and 2., , 3.10, , A spring with one end attached to a mass and the other to a rigid, support is stretched and released., (a) Magnitude of acceleration, when just released is maximum., (b) Magnitude of acceleration, when at equilibrium position, is, maximum., (c) Speed is maximum when mass is at equilibrium position., (d) Magnitude of displacement is always maximum whenever speed, is minimum., , 15, 20/04/2018
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Exemplar Problems–Physics, A ball is bouncing elastically with a speed 1 m/s between walls of, a railway compartment of size 10 m in a direction perpendicular, to walls. The train is moving at a constant velocity of 10 m/s parallel, to the direction of motion of the ball. As seen from the ground,, , 3.11, , (a) the direction of motion of the ball changes every 10 seconds., (b) speed of ball changes every 10 seconds., (c) average speed of ball over any 20 second interval is fixed., (d) the acceleration of ball is the same as from the train., , VSA, 3.12, , Refer to the graphs in Fig 3.1. Match the following., Graph, , Characteristic, , (a), , (i) has v > 0 and a < 0 throughout., , (b), , (ii) has x > 0 throughout and has a point with, v = 0 and a point with a = 0., , (c), , (iii) has a point with zero displacement for t > 0., , (d), , (iv) has v < 0 and a > 0., , 3.13, , A uniformly moving cricket ball is turned back by hitting it with a, bat for a very short time interval. Show the variation of its, acceleration with time. (Take acceleration in the backward direction, as positive)., , 3.14, , Give examples of a one-dimensional motion where, (a) the particle moving along positive x-direction comes to rest, periodically and moves forward., (b) the particle moving along positive x-direction comes to rest, periodically and moves backward., , 3.15, , Give example of a motion where x > 0, v < 0, a > 0 at a particular, instant., , 3.16 An object falling through a fluid is observed to have acceleration, given by a = g – bv where g = gravitational acceleration and b is, constant. After a long time of release, it is observed to fall with, constant speed. What must be the value of constant speed?, , 16, 20/04/2018
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Motion in a Straight Line, , SA, 3.17, , A ball is dropped and its displacement vs time graph is, as shown Fig. 3.4 (displacement x is from ground and, all quantities are +ve upwards)., (a) Plot qualitatively velocity vs time graph., Fig. 3.4, , (b) Plot qualitatively acceleration vs time graph., 3.18, , A particle executes the motion described by x (t ) = x o (1 − e −γ t ) ; t ≥ 0 ,, x0 > 0., (a) Where does the particle start and with what velocity?, (b) Find maximum and minimum values of x (t), v (t), a (t). Show that, x (t) and a (t) increase with time and v (t) decreases with time., , 3.19, , A bird is tossing (flying to and fro) between two cars moving, towards each other on a straight road. One car has a speed of 18, m/h while the other has the speed of 27km/h. The bird starts, moving from first car towards the other and is moving with the, speed of 36km/h and when the two cars were separted by 36 km., What is the total distance covered by the bird? What is the total, displacement of the bird?, , 3.20, , A man runs across the roof-top of a tall building and jumps, horizontally with the hope of landing on the roof of the next, building which is of a lower height than the first. If his speed is 9, m/s, the (horizontal) distance between the two buildings is 10 m, and the height difference is 9 m, will he be able to land on the next, building ? (take g = 10 m/s2), , 3.21, , A ball is dropped from a building of height 45 m., v, Simultaneously another ball is thrown up with a speed vo, 40 m/s. Calculate the relative speed of the balls as a, function of time., , 3.22, , The velocity-displacement graph of a particle is shown, in Fig. 3.5., (a) Write the relation between v and x., (b) Obtain the relation between acceleration and, displacement and plot it., , xo, , O, , x, , Fig. 3.5, , 17, 20/04/2018
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Exemplar Problems–Physics, , LA, 3.23, , It is a common observation that rain clouds can be at about a, kilometre altitude above the ground., (a) If a rain drop falls from such a height freely under gravity, what, will be its speed? Also calculate in km/h. ( g = 10m/s2), (b) A typical rain drop is about 4mm diameter. Momentum is mass, x speed in magnitude. Estimate its momentum when it hits, ground., (c) Estimate the time required to flatten the drop., (d) Rate of change of momentum is force. Estimate how much force, such a drop would exert on you., (e) Estimate the order of magnitude force on umbrella. Typical, lateral separation between two rain drops is 5 cm., (Assume that umbrella is circular and has a diameter of 1m, and cloth is not pierced through !!), , 3.24, , A motor car moving at a speed of 72km/h can not come to a stop, in less than 3.0 s while for a truck this time interval is 5.0 s. On a, higway the car is behind the truck both moving at 72km/h. The, truck gives a signal that it is going to stop at emergency. At what, distance the car should be from the truck so that it does not bump, onto (collide with) the truck. Human response time is 0.5s., , (Comment : This is to illustrate why vehicles carry the message on the, rear side. “Keep safe Distance”), 3.25, , A monkey climbs up a slippery pole for 3 seconds and, subsequently slips for 3 seconds. Its velocity at time t is given, by v(t) = 2t (3-t); 0< t < 3 and v (t)=–(t–3)(6–t) for 3 < t < 6 s in, m/s. It repeats this cycle till it reaches the height of 20 m., (a) At what time is its velocity maximum?, (b) At what time is its average velocity maximum?, (c) At what time is its acceleration maximum in magnitude?, (d) How many cycles (counting fractions) are required to reach the, top?, , 3.26, , A man is standing on top of a building 100 m high. He throws two, balls vertically, one at t = 0 and other after a time interval (less, than 2 seconds). The later ball is thrown at a velocity of half the, first. The vertical gap between first and second ball is +15 m at, t = 2 s. The gap is found to remain constant. Calculate the velocity, with which the balls were thrown and the exact time interval, between their throw., , 18, 20/04/2018
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Chapter Four, , MOTION IN A PLANE, , MCQ I, 4.1, , The angle between A = ˆi + ˆj and B = ˆi − ˆj is, (a) 45° (b) 90° (c) –45° (d) 180°, , 4.2, , Which one of the following statements is true?, (a) A scalar quantity is the one that is conserved in a process., (b) A scalar quantity is the one that can never take negative values., (c) A scalar quantity is the one that does not vary from one point, to another in space., Y, (d) A scalar quantity has the same value for observers with, different orientations of the axes., u, , Figure 4.1 shows the orientation of two vectors u and v in the XY, plane., If u = a ˆi + b ˆj and, v = p ˆi + q ˆj, , v, , 4.3, , O, , X, , Fig. 4.1, , 20/04/2018
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Exemplar Problems–Physics, which of the following is correct?, (a), (b), (c), (d), 4.4, , The component of a vector r along X-axis will have maximum value, if, (a), (b), (c), (d), , 4.5, , Impulse, pressure and area, Impulse and area, Area and gravitational potential, Impulse and pressure, , In a two dimensional motion, instantaneous speed v0 is a positive, constant. Then which of the following are necessarily true?, (a), (b), (c), (d), , 4.8, , 60 m, 71 m, 100 m, 141 m, , Consider the quantities, pressure, power, energy, impulse,, gravitational potential, electrical charge, temperature, area. Out, of these, the only vector quantities are, (a), (b), (c), (d), , 4.7, , r is along positive Y-axis, r is along positive X-axis, r makes an angle of 45° with the X-axis, r is along negative Y-axis, , The horizontal range of a projectile fired at an angle of 15° is 50 m., If it is fired with the same speed at an angle of 45°, its range will, be, (a), (b), (c), (d), , 4.6, , a and p are positive while b and q are negative., a, p and b are positive while q is negative., a, q and b are positive while p is negative., a, b, p and q are all positive., , The average velocity is not zero at any time., Average acceleration must always vanish., Displacements in equal time intervals are equal., Equal path lengths are traversed in equal intervals., , In a two dimensional motion, instantaneous speed v0 is a positive, constant. Then which of the following are necessarily true?, (a) The acceleration of the particle is zero., (b) The acceleration of the particle is bounded., (c) The acceleration of the particle is necessarily in the plane of, motion., (d) The particle must be undergoing a uniform circular motion, , 20, 20/04/2018
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Motion in a Plane, , 4.9, , Three vectors A,B and C add up to zero. Find which is false., (a), (b), (c), (d), , (A×B)× C is not zero unless B,C are parallel, (A×B).C is not zero unless B,C are parallel, If A,B,C define a plane, (A×B), ×C is in that plane, 2, → C =A2+B2, (A×B).C=|A||B||C|→, , 4.10 It is found that |A+B|=|A|.This necessarily implies,, (a) B = 0, (b) A,B are antiparallel, (c) A,B are perpendicular, (d) A.B ≤ 0, , MCQ II, 4.11 Two particles are projected in air with speed vo at angles θ1 and θ2, (both acute) to the horizontal, respectively. If the height reached, by the first particle is greater than that of the second, then tick, the right choices, (a), (b), (c), (d), , angle of projection : q1 > q2, time of flight : T1 > T2, horizontal range : R1 > R2, total energy : U1 > U2., , 4.12 A particle slides down a frictionless parabolic, (y = x2) track (A – B – C) starting from rest at, point A (Fig. 4.2). Point B is at the vertex of, parabola and point C is at a height less than, that of point A. After C, the particle moves freely, in air as a projectile. If the particle reaches, highest point at P, then, (a), (b), (c), (d), , KE at P = KE at B, height at P = height at A, total energy at P = total energy at A, time of travel from A to B = time of travel from, B to P., , y, A, P, , vo, , C, -x2, , -x1, , B, (x = 0), , -xo, , x, , Fig. 4.2, , 4.13 Following are four differrent relations about displacement, velocity, and acceleration for the motion of a particle in general. Choose, the incorrect one (s) :, (a) v av =, , 1, [ v(t1 ) + v(t 2 )], 2, , (b) v av =, , r(t 2 ) − r(t1 ), t 2 − t1, , 21, 20/04/2018
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Exemplar Problems–Physics, , (c) r =, , 1, ( v(t 2 ) − v(t1 )) (t 2 − t1 ), 2, , (d) a av =, , v(t 2 ) − v(t1 ), t 2 − t1, , 4.14 For a particle performing uniform circular motion, choose the correct, statement(s) from the following:, (a), (b), (c), (d), , Magnitude of particle velocity (speed) remains constant., Particle velocity remains directed perpendicular to radius vector., Direction of acceleration keeps changing as particle moves., Angular momentum is constant in magnitude but direction, keeps changing., , 4.15 For two vectors A and B, A + B = A − B, , is always true when, , (a) A = B ≠ 0, (b) A ⊥ B, (c) A = B ≠ 0 and A and B are parallel or anti parallel, (d) when either A or B is zero., , VSA, , Q, R, , O, , Fig. 4.3, , P, , 4.16 A cyclist starts from centre O of a circular park of radius 1km and, moves along the path OPRQO as shown Fig. 4.3. If he maintains, constant speed of 10ms–1, what is his acceleration at point R in, magnitude and direction?, 4.17 A particle is projected in air at some angle to the horizontal,, moves along parabola as shown in Fig. 4.4, where x and y, indicate horizontal and vertical directions, respectively. Show, in the diagram, direction of velocity and acceleration at points, A, B and C., y, , B, C, H, A, x, Fig. 4.4, , 22, 20/04/2018
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Motion in a Plane, , 4.18 A ball is thrown from a roof top at an angle of 45° above the, horizontal. It hits the ground a few seconds later. At what point, during its motion, does the ball have, (a) greatest speed., (b) smallest speed., (c) greatest acceleration?, Explain, 4.19 A football is kicked into the air vertically upwards. What is its, (a) acceleration, and (b) velocity at the highest point?, 4.20 A, B and C are three non-collinear, non co-planar vectors.What, can you say about direction of A ×(B ×C)?, , SA, 4.21 A boy travelling in an open car moving on a levelled road with, constant speed tosses a ball vertically up in the air and catches it, back. Sketch the motion of the ball as observed by a boy standing, on the footpath. Give explanation to support your diagram., 4.22 A boy throws a ball in air at 60° to the horizontal along a road with, a speed of 10 m/s (36km/h). Another boy sitting in a passing by, car observes the ball. Sketch the motion of the ball as observed by, the boy in the car, if car has a speed of (18km/h). Give explanation, to support your diagram., 4.23 In dealing with motion of projectile in air, we ignore effect of air, resistance on motion. This gives trajectory as a parabola as you, have studied. What would the trajectory look like if air resistance, is included? Sketch such a trajectory and explain why you have, drawn it that way., 4.24 A fighter plane is flying horizontally at an altitude of 1.5 km with, speed 720 km/h. At what angle of sight (w.r.t. horizontal) when, the target is seen, should the pilot drop the bomb in order to, attack the target?, 4.25 (a) Earth can be thought of as a sphere of radius 6400 km. Any, object (or a person) is performing circular motion around the, axis of earth due to earth’s rotation (period 1 day). What is, acceleration of object on the surface of the earth (at equator), towards its centre? what is it at latitude θ ? How does these, accelerations compare with g = 9.8 m/s2?, , 23, 20/04/2018
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Exemplar Problems–Physics, (b) Earth also moves in circular orbit around sun once every year, with on orbital radius of 1.5 × 1011 m . What is the acceleration of, earth (or any object on the surface of the earth) towards the, centre of the sun? How does this acceleration compare with, g = 9.8 m/s2?, , V 2 4π 2 R , =, Hint : acceleration, , R, , T2 , , 4.26 Given below in column I are the relations between vectors a, b and, c and in column II are the orientations of a, b and c in the XY, plane. Match the relation in column I to correct orientations in, column II., Column I, , Column II, , (a) a + b = c, , (i), , (b) a – c = b, , (ii), , (c) b – a = c, , (d) a + b + c = 0, , (iii), , (iv), , 24, 20/04/2018
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Motion in a Plane, , 4.27 If A = 2 and B = 4 , then match the relations in column I with, the angle θ between A and B in column II., Column I, , Column II, , (a) A.B = 0, , (i) θ = 0, , (b) A.B = +8, , (ii) θ = 90°, , (c) A.B = 4, , (iii) θ = 180°, , (d) A.B = –8, , (iv) θ = 60°, , 4.28 If A = 2 and B = 4 , then match the relations in column I with, the angle θ between A and B in column II, Column I, , Column II, , (a), , A×B = 0, , (i) θ = 30°, , (b), , A×B = 8, , (ii) θ = 45°, , (c), , A ×B = 4, , (iii) θ = 90°, , (d), , A×B = 4 2, , (iv) θ = 0°, , LA, 4.29 A hill is 500 m high. Supplies are to be sent across the hill using a, canon that can hurl packets at a speed of 125 m/s over the hill., The canon is located at a distance of 800m from the foot of hill and, can be moved on the ground at a speed of 2 m/s; so that its distance, from the hill can be adjusted. What is the shortest time in which a, packet can reach on the ground across the hill ? Take g =10 m/s2., 4.30 A gun can fire shells with maximum speed v o and the, 2, maximum horizontal range that can be achieved is R = vo ., g, , vo, P, , q, , q, vo, , h, , R, , x, , T, , Fig 4.5, , 25, 20/04/2018
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Exemplar Problems–Physics, If a target farther away by distance ∆x (beyond R) has to, be hit with the same gun (Fig 4.5), show that it could be, achieved by raising the gun to a height at least, , ∆x , h = ∆x 1 +, , R , (Hint : This problem can be approached in two different ways:, (i) Refer to the diagram: target T is at horizontal distance, x = R + ∆x and below point of projection y = – h., (ii) From point P in the diagram: Projection at speed v o at an, angle θ below horizontal with height h and horizontal, range ∆x.), 4.31 A particle is projected in air at an angle, β to a surface which itself is inclined at, an angle α to the horizontal (Fig. 4.6)., (a) Find an expression of range on the, plane surface (distance on the plane, from the point of projection at which, particle will hit the surface)., (b) Time of flight., (c) β at which range will be maximum., , Fig. 4.6, , (Hint : This problem can be solved in two different ways:, (i) Point P at which particle hits the plane can be seen as, intersection of its trajectory (parobola) and straight line., Remember particle is projected at an angle (α + β ) w.r.t., horizontal., (ii) We can take x-direction along the plane and, y-direction perpendicular to the plane. In that case resolve g, (acceleration due to gravity) in two differrent components, gx, along the plane and gy perpendicular to the plane. Now the, problem can be solved as two independent motions in x and, y directions respectively with time as a common parameter.), P, L, , 4.32 A particle falling vertically from a height hits a plane surface inclined, to horizontal at an angle θ with speed vo and rebounds elastically, (Fig 4.7). Find the distance along the plane where if will hit, q, second time., , Fig 4.7, , 26, 20/04/2018
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Motion in a Plane, , (Hint: (i) After rebound, particle still has speed Vo to start., (ii) Work out angle particle speed has with horizontal after it, rebounds., (iii) Rest is similar to if particle is projected up the incline.), , 4.33 A girl riding a bicycle with a speed of 5 m/s towards north direction,, observes rain falling vertically down. If she increases her speed to, 10 m/s, rain appears to meet her at 45° to the vertical. What is the, speed of the rain? In what direction does rain fall as observed by a, ground based observer?, (Hint: Assume north to be î direction and vertically downward to, be − ˆj . Let the rain velocity vr be a ˆi + b ˆj . The velocity of rain as, observed by the girl is always vr − vgirl . Draw the vector diagram/s, for the information given and find a and b. You may draw all vectors, in the reference frame of ground based observer.), , 4.34 A river is flowing due east with a speed 3m/s. A, swimmer can swim in still water at a speed of 4 m/s, (Fig. 4.8)., (a) If swimmer starts swimming due north, what will, be his resultant velocity (magnitude and direction)?, (b) If he wants to start from point A on south bank and, reach opposite point B on north bank,, (a) which direction should he swim?, (b) what will be his resultant speed?, (c) From two different cases as mentioned in (a) and (b), above, in which case will he reach opposite bank in, shorter time?, , N, E, , B, 3m/s, , A, , Fig. 4.8, , 4.35 A cricket fielder can throw the cricket ball with a speed vo. If he, throws the ball while running with speed u at an angle θ to the, horizontal, find, (a) the effective angle to the horizontal at which the ball is projected, in air as seen by a spectator., (b) what will be time of flight?, (c) what is the distance (horizontal range) from the point of, projection at which the ball will land?, , 27, 20/04/2018
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Exemplar Problems–Physics, (d) find at which he should throw the ball that would maximise, the horizontal range as found in (iii)., (e) how does for maximum range change if u >vo, u = vo, u < vo?, o, , (f) how does in (v) compare with that for u = 0 (i.e.45 )?, 4.36 Motion in two dimensions, in a plane can be studied by expressing, position, velocity and acceleration as vectors in Cartesian, where, , co-ordinates, , are unit vector along, , x and y directions, respectively and Ax and Ay are corresponding, components of, , (Fig. 4.9). Motion can also be studied by, , expressing vectors in circular polar co-ordinates as, where, , , , , , , , and , , , , , , are unit, , vectors along direction in which ‘r’ and ‘ ’ are increasing., (a) Express, , in terms of, , ., , are unit vectors and are, perpendicular to each other., , (b) Show that both, , (c) Show that, , , , , , and, , where, = , , (d) For a particle moving along a spiral given by, , , where a = 1 (unit), find dimensions of ‘a’., Fig. 4.9, , (e) Find velocity and acceleration in polar vector, represention for particle moving along spiral, described in (d) above., 4.37 A man wants to reach from A to the opposite corner of the, square C (Fig. 4.10). The sides of the square are 100 m. A, central square of 50m × 50m is filled with sand. Outside, this square, he can walk at a speed 1 m/s. In the central, square, he can walk only at a speed of v m/s (v < 1). What is, smallest value of v for which he can reach faster via a straight, path through the sand than any path in the square outside, the sand?, , Fig. 4.10, , 28, 20/04/2018
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Chapter Five, , LAWS OF MOTION, , MCQ I, 5.1, , A ball is travelling with uniform translatory motion. This, means that, (a) it is at rest., (b) the path can be a straight line or circular and the ball travels, with uniform speed., (c) all parts of the ball have the same velocity (magnitude and, direction) and the velocity is constant., (d) the centre of the ball moves with constant velocity and the, ball spins about its centre uniformly., , 5.2, , A metre scale is moving with uniform velocity. This implies, (a) the force acting on the scale is zero, but a torque about the, centre of mass can act on the scale., (b) the force acting on the scale is zero and the torque acting, about centre of mass of the scale is also zero., , 20/04/2018
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Exemplar Problems–Physics, (c) the total force acting on it need not be zero but the torque on it, is zero., (d) neither the force nor the torque need to be zero., 5.3, , A cricket ball of mass 150 g has an initial velocity u = (3ˆi + 4 ˆj) m s−1, and a final velocity v = − (3ˆi + 4ˆj) m s−1 after being hit. The change, in momentum (final momentum-initial momentum) is (in kg m s1), (a) zero, (b) – (0.45ˆi + 0.6 ˆj), (c) – (0.9ˆi + 1.2ˆj), (d) – 5(ˆi + ˆj) ., , 5.4, , In the previous problem (5.3), the magnitude of the momentum, transferred during the hit is, (a) Zero (b) 0.75 kg m s–1 (c) 1.5 kg m s–1 (d) 14 kg m s–1., , 5.5, , Conservation of momentum in a collision between particles can, be understood from, (a), (b), (c), (d), , 5.6, , A hockey player is moving northward and suddenly turns westward, with the same speed to avoid an opponent. The force that acts on, the player is, (a), (b), (c), (d), , 5.7, , conservation of energy., Newton’s first law only., Newton’s second law only., both Newton’s second and third law., , frictional force along westward., muscle force along southward., frictional force along south-west., muscle force along south-west., , A body of mass 2kg travels according to the law x (t ) = pt + qt 2 + rt 3, where p = 3 m s−1 , q = 4 m s−2 and r = 5 m s−3 ., The force acting on the body at t = 2 seconds is, (a), (b), (c), (d), , 136 N, 134 N, 158 N, 68 N, , 30, 20/04/2018
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Laws of Motion, , 5.8, , A body with mass 5 kg is acted upon by a force F = ( –3ˆi + 4ˆj ) N. If, its initial velocity at t = 0 is v = ( 6ˆi -12 ˆj ) m s –1 , the time at which it, will just have a velocity along the y-axis is, (a) never, (b) 10 s, (c) 2 s, (d) 15 s, , 5.9, , A car of mass m starts from rest and acquires a velocity along east, v = v ˆi ( v > 0 ) in two seconds. Assuming the car moves with, uniform acceleration, the force exerted on the car is, (a), (b), , mv, eastward and is exerted by the car engine., 2, mv, eastward and is due to the friction on the tyres exerted by, 2, the road., , mv, eastward exerted due to the engine and, 2, overcomes the friction of the road., , (c) more than, , (d), , mv, exerted by the engine ., 2, , MCQ II, 5.10, , The motion of a particle of mass m is given by x = 0 for t < 0, s, x( t ) = A sin4p t for 0 < t <(1/4) s (A > o), and x = 0 for, t >(1/4) s. Which of the following statements is true?, (a) The force at t = (1/8) s on the particle is –16π2 A m., (b) The particle is acted upon by on impulse of magnitude, 4π2 A m at t = 0 s and t = (1/4) s., (c) The particle is not acted upon by any force., (d) The particle is not acted upon by a constant force., (e) There is no impulse acting on the particle., , 5.11 In Fig. 5.1, the co-efficient of friction between the floor, and the body B is 0.1. The co-efficient of friction between, the bodies B and A is 0.2. A force F is applied as shown, , 31, 20/04/2018
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Exemplar Problems–Physics, on B. The mass of A is m /2 and of B is m. Which of the, following statements are true?, (a), (b), (c), (d), (e), Fig. 5.1, , 5.12, , The bodies will move together if F = 0.25 mg., The body A will slip with respect to B if F = 0.5 mg., The bodies will move together if F = 0.5 mg., The bodies will be at rest if F = 0.1 mg., The maximum value of F for which the two bodies will move, together is 0.45 mg., , Mass m1 moves on a slope making an angle θ with the horizontal, and is attached to mass m2 by a string passing over a frictionless, pulley as shown in Fig. 5.2. The co-efficient of friction between m1, and the sloping surface is µ., Which of the following statements are true?, , m1, , m2, B, , (a) If m 2 > m 1 sin θ , the body will move up the plane., (b) If m 2 > m1 ( sin θ + µ cos θ ) , the body will move up the plane., , , , (c) If m 2 < m1 ( sin θ + µ cos θ ) , the body will move up the plane., Fig. 5.2, , (d) If m 2 < m 1 ( sin θ − µ cos θ ) , the body will move down the plane., 5.13, , In Fig. 5.3, a body A of mass m slides on plane inclined at angle, , θ1 to the horizontal and µ1 is the coefficent of friction between A, and the plane. A is connected by a light string passing over a, frictionless pulley to another body B, also of mass m, sliding on a, frictionless plane inclined at angle θ2 to the horizontal. Which of, the following statements are true?, (a) A will never move up the plane., (b) A will just start moving up the plane when, , µ=, , sin θ2 − sin θ1, ., cos θ1, , (c) For A to move up the plane, θ2 must always be greater, Fig. 5.3, , than θ1 ., (d) B will always slide down with constant speed., 5.14, , Two billiard balls A and B, each of mass 50g and moving in, opposite directions with speed of 5m s–1 each, collide and, rebound with the same speed. If the collision lasts for 10–3 s,, which of the following statements are true?, (a) The impulse imparted to each ball is 0.25 kg m s–1 and the force, on each ball is 250 N., , 32, 20/04/2018
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Laws of Motion, , (b) The impulse imparted to each ball is 0.25 kg m s–1 and the, force exerted on each ball is 25 × 10–5 N., (c) The impulse imparted to each ball is 0.5 Ns., (d) The impulse and the force on each ball are equal in magnitude, and opposite in direction., A body of mass 10kg is acted upon by two perpendicular forces,, 6N and 8N. The resultant acceleration of the body is, , 5.15, , 4, (a) 1 m s–2 at an angle of tan −1 w.r.t. 6N force., 3, 4, (b) 0.2 m s–2 at an angle of tan −1 w.r.t. 6N force., 3, 3, (c) 1 m s–2 at an angle of tan −1 w.r.t.8N force., 4, 3, (d) 0.2 m s–2 at an angle of tan −1 w.r.t.8N force., 4, , VSA, 5.16, , A girl riding a bicycle along a straight road with a speed of 5 m s–1, throws a stone of mass 0.5 kg which has a speed of 15 m s–1 with, respect to the ground along her direction of motion. The mass of, the girl and bicycle is 50 kg. Does the speed of the bicycle change, after the stone is thrown? What is the change in speed, if so?, , 5.17, , A person of mass 50 kg stands on a weighing scale on a lift. If the, lift is descending with a downward acceleration of, 9 m s–2, what would be the reading of the weighing scale?, (g = 10 m s–2 ), , 5.18, , The position time graph of a body of mass 2 kg is as given in, Fig. 5.4. What is the impulse on the body at t = 0 s and t = 4 s., , Fig. 5.4, , 33, 20/04/2018
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Exemplar Problems–Physics, 5.19, , A person driving a car suddenly applies the brakes on seeing a, child on the road ahead. If he is not wearing seat belt, he falls, forward and hits his head against the steering wheel. Why?, , 5.20, , The velocity of a body of mass 2 kg as a function of t is given by, v(t ) = 2t ˆi + t 2 ˆj . Find the momentum and the force acting on it, at, time t= 2s., , 5.21, , A block placed on a rough horizontal surface is pulled by a, horizontal force F. Let f be the force applied by the rough surface, on the block. Plot a graph of f versus F., , 5.22, , Why are porcelain objects wrapped in paper or straw before, packing for transportation?, , 5.23, , Why does a child feel more pain when she falls down on a hard, cement floor, than when she falls on the soft muddy ground in the, garden?, , 5.24, , A woman throws an object of mass 500 g with a speed of 25 m s1., (a) What is the impulse imparted to the object?, (b) If the object hits a wall and rebounds with half the original speed,, what is the change in momentum of the object?, , 5.25, , Why are mountain roads generally made, winding upwards rather than going straight up?, , SA, 5.26, , 34, , A mass of 2kg is suspended with thread AB, (Fig. 5.5). Thread CD of the same type is, attached to the other end of 2 kg mass. Lower, thread is pulled gradually, harder and, harder in the downward directon so as to, apply force on AB. Which of the threads will, break and why?, , 5.27, , In the above given problem if the lower thread is, pulled with a jerk, what happens?, , 5.28, , Two masses of 5 kg and 3 kg are suspended, with help of massless inextensible strings as, shown in Fig. 5.6. Calculate T 1 and T 2 when, whole system is going upwards with, acceleration = 2 m s 2 (use g = 9.8 m s –2)., , Fig. 5.5, , Fig. 5.6, , 20/04/2018
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Laws of Motion, , 5.29, , 5.30, , Block A of weight 100 N rests on a frictionless inclined, plane of slope angle 30° (Fig. 5.7). A flexible cord attached, to A passes over a frictonless pulley and is connected to, block B of weight W. Find the weight W for which the, system is in equilibrium., , NA, , °, , in, gs, , f, , m, , A block of mass M is held against a rough vertical wall by, pressing it with a finger. If the coefficient of friction between, the block and the wall is µ and the acceleration due to, gravity is g, calculate the minimum force required to be, applied by the finger to hold the block against the wall ?, , 5.31, , A 100 kg gun fires a ball of 1kg horizontally from a cliff of height, 500m. It falls on the ground at a distance of 400m from the bottom, of the cliff. Find the recoil velocity of the gun. (acceleration due to, gravity = 10 m s–2), , 5.32, , Figure 5.8 shows (x, t), (y, t ) diagram of a particle moving in, 2-dimensions., , 30, , F, , B, w, , 30°, mg cos 30°, , mg, Fig. 5.7, , x, , 1s, , 2s, , 3s, , t, , (b), , (a), Fig. 5.8, , If the particle has a mass of 500 g, find the force (direction and, magnitude) acting on the particle., 5.33, , A person in an elevator accelerating upwards with an acceleration, of 2 m s–2, tosses a coin vertically upwards with a speed of 20 m, s1. After how much time will the coin fall back into his hand?, ( g = 10 m s–2), , LA, 5.34, , There are three forces F1, F2 and F3 acting on a body, all acting on, a point P on the body. The body is found to move with uniform, speed., (a) Show that the forces are coplanar., (b) Show that the torque acting on the body about any point due, to these three forces is zero., , 35, 20/04/2018
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Exemplar Problems–Physics, 5.35, , When a body slides down from rest along a smooth inclined plane, making an angle of 45° with the horizontal, it takes time T. When, the same body slides down from rest along a rough inclined plane, making the same angle and through the same distance, it is seen, to take time pT, where p is some number greater than 1. Calculate, the co-efficient of friction between the body and the rough plane., , 5.36, , Figure 5.9 shows (v x ,t ),and(vy ,t ) diagrams for a body of unit, mass. Find the force as a function of time., , vx, (m s–1), , vy, (m s–1), , 2, , 2, , 1, , 1, 1s, , 2s, , O, , t, , (a), , 1s, , 2s, , 3s, , t, , (b), Fig. 5.9, , 5.37, , A racing car travels on a track (without banking) ABCDEFA, (Fig. 5.10). ABC is a circular arc of radius 2 R. CD and FA are, straight paths of length R and DEF is a circular arc of radius, R = 100 m. The co-effecient of friction on the road is µ = 0.1. The, maximum speed of the car is 50 m s–1. Find the minimum time, for completing one round., A, , F, , R, 90° R, O, , 2R, , E, D, C, , B, , Fig. 5.10, , 5.38, , The displacement vector of a particle of mass m is given by, r (t ) = ˆi A cos ωt + ˆj B sin ωt ., , (a) Show that the trajectory is an ellipse., (b) Show that F = −mω 2 r., , 36, 20/04/2018
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Laws of Motion, , 5.39, , A cricket bowler releases the ball in two different ways, (a) giving it only horizontal velocity, and, (b) giving it horizontal velocity and a small downward velocity., The speed vs at the time of release is the same. Both are released, at a height H from the ground. Which one will have greater speed, when the ball hits the ground? Neglect air resistance., , 5.40, , There are four forces acting at a point P produced by strings as, shown in Fig. 5.11, which is at rest. Find the forces F1 and F2 ., 2N, 1N, 45° 45°, 45°, 90°, , F1, , F2, , Fig. 5.11, , 5.41, , A rectangular box lies on a rough inclined surface. The co-efficient, of friction between the surface and the box is µ. Let the mass of the, box be m., (a) At what angle of inclination θ of the plane to the horizontal will, the box just start to slide down the plane?, (b) What is the force acting on the box down the plane, if the angle, of inclination of the plane is increased to α > θ ?, (c) What is the force needed to be applied upwards along the plane, to make the box either remain stationary or just move up with, uniform speed?, (d) What is the force needed to be applied upwards along the plane, to make the box move up the plane with acceleration a?, , 5.42, , A helicopter of mass 2000kg rises with a vertical acceleration of, 15 m s–2. The total mass of the crew and passengers is 500 kg., Give the magnitude and direction of the (g = 10 m s–2), (a) force on the floor of the helicopter by the crew and passengers., (b) action of the rotor of the helicopter on the surrounding air., (c) force on the helicopter due to the surrounding air., , 37, 20/04/2018
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Exemplar Problems–Physics, , Chapter Six, , WORK, ENERGY, AND POWER, , MCQ I, 6.1, , An electron and a proton are moving under the influence of mutual, forces. In calculating the change in the kinetic energy of the system, during motion, one ignores the magnetic force of one on another., This is because,, (a) the two magnetic forces are equal and opposite, so they produce, no net effect., (b) the magnetic forces do no work on each particle., (c) the magnetic forces do equal and opposite (but non-zero) work, on each particle., (d) the magenetic forces are necessarily negligible., , 6.2, , A proton is kept at rest. A positively charged particle is released, from rest at a distance d in its field. Consider two experiments;, one in which the charged particle is also a proton and in another,, a positron. In the same time t, the work done on the two moving, charged particles is, , 38, 20/04/2018
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Work, Energy and Power, , (a) same as the same force law is involved in the two experiments., (b) less for the case of a positron, as the positron moves away, more rapidly and the force on it weakens., (c) more for the case of a positron, as the positron moves away a, larger distance., (d) same as the work done by charged particle on the stationary proton., 6.3, , A man squatting on the ground gets straight up and stand. The, force of reaction of ground on the man during the process is, (a), (b), (c), (d), , 6.4, , A bicyclist comes to a skidding stop in 10 m. During this process,, the force on the bicycle due to the road is 200N and is directly, opposed to the motion. The work done by the cycle on the road is, (a), (b), (c), (d), , 6.5, , Kinetic energy., Potential energy., Total mechanical energy., Total linear momentum., , During inelastic collision between two bodies, which of the following, quantities always remain conserved?, (a), (b), (c), (d), , 6.7, , + 2000J, – 200J, zero, – 20,000J, , A body is falling freely under the action of gravity alone in vacuum., Which of the following quantities remain constant during the fall?, (a), (b), (c), (d), , 6.6, , constant and equal to mg in magnitude., constant and greater than mg in magnitude., variable but always greater than mg., at first greater than mg, and later becomes equal to mg., , Total kinetic energy., Total mechanical energy., Total linear momentum., Speed of each body., , Two inclined frictionless tracks, one gradual and the other, steep meet at A from where two stones are allowed to, slide down from rest, one on each track as shown in Fig., 6.1., Which of the following statement is correct?, (a) Both the stones reach the bottom at the same time, but not with the same speed., (b) Both the stones reach the bottom with the same speed, and stone I reaches the bottom earlier than stone II., , Fig. 6.1, , 39, 20/04/2018
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Exemplar Problems–Physics, (c) Both the stones reach the bottom with the same speed and, stone II reaches the bottom earlier than stone I., (d) Both the stones reach the bottom at different times and with, different speeds., 6.8, , The potential energy function for a particle executing linear SHM, 1 2, kx where k is the force constant of the, 2, oscillator (Fig. 6.2). For k = 0.5N/m, the graph of V(x) versus x is, shown in the figure. A particle of total energy E turns back when, , is given by V ( x ) =, , it reaches x = ±x m . If V and K indicate the P.E. and K.E.,, respectively of the particle at x = +xm, then which of the following is, correct?, (a), (b), (c), (d), , Fig. 6.2, , 6.9, , V = O,, V = E,, V < E,, V = O,, , K=E, K=O, K=O, K < E., , Two identical ball bearings in contact with each other and resting, on a frictionless table are hit head-on by another ball bearing of the, same mass moving initially with a speed V as shown in Fig. 6.3., , Fig. 6.3, , If the collision is elastic, which of the following (Fig. 6.4) is a possible, result after collision?, , (a), , (b), , (c), , (d), Fig. 6.4, , 40, 20/04/2018
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Work, Energy and Power, , 6.10, , A body of mass 0.5 kg travels in a straight line with velocity v = a, x3/2 where a = 5 m–1/2s–1. The work done by the net force during its, displacement from x = 0 to x = 2 m is, (a), (b), (c), (d), , 6.11, , 1.5 J, 50 J, 10 J, 100 J, , A body is moving unidirectionally under the influence of a source of, constant power supplying energy. Which of the diagrams shown in, Fig. 6.5 correctly shows the displacement-time curve for its motion?, , (a), , (b), , (c), , (d), Fig. 6.5, , 6.12, , Which of the diagrams shown in Fig. 6.6 most closely shows the, variation in kinetic energy of the earth as it moves once around, the sun in its elliptical orbit?, , (a), , (b), K.E, , t, , (c), , Fig. 6.6, , (d), , 41, 20/04/2018
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Exemplar Problems–Physics, 6.13, , Which of the diagrams shown in Fig. 6.7 represents variation of, total mechanical energy of a pendulum oscillating in air as, function of time?, E, , E, , t, t, , (a), , (b), , E, , E, , t, , t, , (d), , (c), Fig. 6.7, , 6.14, , A mass of 5 kg is moving along a circular path of radius 1 m. If, the mass moves with 300 revolutions per minute, its kinetic energy, would be, (a), (b), (c), (d), , 6.15, , 250π2, 100π2, 5π2, 0, , A raindrop falling from a height h above ground, attains a near terminal, velocity when it has fallen through a height (3/4)h. Which of the, diagrams shown in Fig. 6.8 correctly shows the change in kinetic and, potential energy of the drop during its fall up to the ground?, h, , h, , PE, , PE, , h/4, , KE, , KE, , t, (a), , t, (b), , 42, 20/04/2018
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Work, Energy and Power, , h, , PE, h, , KE, PE, , KE, t, , (c), , (d), , t, , Fig. 6.8, , 6.16, , In a shotput event an athlete throws the shotput of mass 10 kg, with an initial speed of 1m s –1 at 45° from a height 1.5 m above, ground. Assuming air resistance to be negligible and acceleration, due to gravity to be 10 m s –2 , the kinetic energy of the shotput, when it just reaches the ground will be, (a), (b), (c), (d), , 6.17, , 2.5 J, 5.0 J, 52.5 J, 155.0 J, , Which of the diagrams in Fig. 6.9 correctly shows the change in, kinetic energy of an iron sphere falling freely in a lake having, sufficient depth to impart it a terminal velocity?, , K.E, , K.E, , depth, (a), , depth, (b), , K.E, , K.E, , depth, , depth, (c), , Fig. 6.9, , (d), , 43, 20/04/2018
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Exemplar Problems–Physics, 6.18, , A cricket ball of mass 150 g moving with a speed of 126 km/h hits, at the middle of the bat, held firmly at its position by the batsman., The ball moves straight back to the bowler after hitting the bat., Assuming that collision between ball and bat is completely elastic, and the two remain in contact for 0.001s, the force that the batsman, had to apply to hold the bat firmly at its place would be, (a), (b), (c), (d), , 10.5 N, 21 N, 1.05 ×104 N, 2.1 × 104 N, , MCQ II, 6.19, , A man, of mass m, standing at the bottom of the staircase, of height, L climbs it and stands at its top., (a) Work done by all forces on man is equal to the rise in potential, energy mgL., (b) Work done by all forces on man is zero., (c) Work done by the gravitational force on man is mgL., (d) The reaction force from a step does not do work because the, point of application of the force does not move while the force, exists., , 6.20, , A bullet of mass m fired at 30° to the horizontal leaves the barrel of, the gun with a velocity v. The bullet hits a soft target at a height h, above the ground while it is moving downward and emerges out, with half the kinetic energy it had before hitting the target., Which of the following statements are correct in respect of bullet, after it emerges out of the target?, (a) The velocity of the bullet will be reduced to half its initial, value., (b) The velocity of the bullet will be more than half of its earlier, velocity., (c) The bullet will continue to move along the same parabolic, path., (d) The bullet will move in a different parabolic path., (e) The bullet will fall vertically downward after hitting the target., (f) The internal energy of the particles of the target will increase., , 6.21, , Two blocks M1 and M2 having equal mass are free to move on a, horizontal frictionless surface. M2 is attached to a massless spring, as shown in Fig. 6.10. Iniially M2 is at rest and M1 is moving toward, M2 with speed v and collides head-on with M2., (a) While spring is fully compressed all the KE of M 1, is stored as PE of spring., , 44, 20/04/2018
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Work, Energy and Power, , (b) While spring is fully compressed the system, momentum is not conserved, though final, momentum is equal to initial momentum., (c) If spring is massless, the final state of the M1 is, state of rest., (d) If the surface on which blocks are moving has, friction, then collision cannot be elastic., , Fig. 6.10, , VSA, 6.22, , A rough inclined plane is placed on a cart moving with a constant, velocity u on horizontal ground. A block of mass M rests on the, incline. Is any work done by force of friction between the block, and incline? Is there then a dissipation of energy?, , 6.23, , Why is electrical power required at all when the elevator is, descending? Why should there be a limit on the number of, passengers in this case?, , 6.24, , A body is being raised to a height h from the surface of earth., What is the sign of work done by, (a) applied force, (b) gravitational force?, , 6.25, , Calculate the work done by a car against gravity in moving along, a straight horizontal road. The mass of the car is 400 kg and the, distance moved is 2m., , 6.26, , A body falls towards earth in air. Will its total mechanical energy, be conserved during the fall? Justify., , 6.27, , A body is moved along a closed loop. Is the work done in moving, the body necessarily zero? If not, state the condition under which, work done over a closed path is always zero., , 6.28, , In an elastic collision of two billiard balls, which of the following, quantities remain conserved during the short time of collision of, the balls (i.e., when they are in contact)., (a) Kinetic energy., (b) Total linear momentum?, Give reason for your answer in each case., , 6.29, , Calculate the power of a crane in watts, which lifts a mass of, 100 kg to a height of 10 m in 20s., , 45, 20/04/2018
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Exemplar Problems–Physics, 6.30, , The average work done by a human heart while it beats once is 0.5, J. Calculate the power used by heart if it beats 72 times in a minute., , 6.31, , Give example of a situation in which an applied force does not, result in a change in kinetic energy., , 6.32, , Two bodies of unequal mass are moving in the same direction, with equal kinetic energy. The two bodies are brought to rest by, applying retarding force of same magnitude. How would the, distance moved by them before coming to rest compare?, , 6.33, , A bob of mass m suspended by a light string of length L is whirled, into a vertical circle as shown in Fig. 6.11. What will be the, trajectory of the particle if the string is cut at, , C, m, , x, B, , (a) Point B?, (b) Point C?, (c) Point X?, , L, , A, , Fig. 6.11, , SA, 6.34, , 6.35, , A graph of potential energy V ( x ), verses x is shown in Fig. 6.12., A particle of energy E0 is executing, motion in it. Draw graph of velocity, and kinetic energy versus x for one, complete cycle AFA., , V(x), A, , F, Eo, , B, D, C, Fig. 6.12, , x, , A ball of mass m, moving with a speed, 2v0, collides inelastically (e > 0) with an identical ball at rest. Show, that, (a) For head-on collision, both the balls move forward., (b) For a general collision, the angle between the two velocities of, scattered balls is less than 90°., , 6.36, , Consider a one-dimensional motion of a particle with total energy, E. There are four regions A, B, C and D in which the relation, between potential energy V, kinetic energy (K) and total energy, E is as given below:, Region A : V > E, Region B : V < E, Region C : K > E, Region D : V > K, State with reason in each case whether a particle can be found in, the given region or not., , 46, 20/04/2018
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Work, Energy and Power, , 6.37, , The bob A of a pendulum released from horizontal to, the vertical hits another bob B of the same mass at, rest on a table as shown in Fig. 6.13., , A, m, 1m, , If the length of the pendulum is 1m, calculate, (a) the height to which bob A will rise after collision., (b) the speed with which bob B starts moving., Neglect the size of the bobs and assume the collision, to be elastic., 6.38, , B, , m, , Fig. 6.13, , A raindrop of mass 1.00 g falling from a height of 1 km hits the, ground with a speed of 50 m s–1. Calculate, (a) the loss of P.E. of the drop., (b) the gain in K.E. of the drop., (c) Is the gain in K.E. equal to loss of P.E.? If not why., Take g = 10 m s-2, , 6.39, , Two pendulums with identical bobs and lengths are suspended, from a common support such that in rest position the two bobs, are in contact (Fig. 6.14). One of the bobs is released after being, displaced by 10o so that it collides elastically head-on with the, other bob., (a) Describe the motion of two bobs., (b) Draw a graph showing variation in energy of either pendulum, with time, for 0 ≤ t ≤ 2T , where T is the period of each, pendulum., , 6.40, , Suppose the average mass of raindrops is 3.0 × 10-5kg and their, average terminal velocity 9 m s-1. Calculate the energy transferred, by rain to each square metre of the surface at a place which receives, 100 cm of rain in a year., , 6.41, , An engine is attached to a wagon through a shock absorber of length, 1.5m. The system with a total mass of 50,000 kg is moving with a, speed of 36 km h-1 when the brakes are applied to bring it to rest. In, the process of the system being brought to rest, the spring of the, shock absorber gets compressed by 1.0 m. If 90% of energy of the, wagon is lost due to friction, calculate the spring constant., , 6.42, , An adult weighing 600N raises the centre of gravity of his body by, 0.25 m while taking each step of 1 m length in jogging. If he jogs, for 6 km, calculate the energy utilised by him in jogging assuming, that there is no energy loss due to friction of ground and air., Assuming that the body of the adult is capable of converting 10%, of energy intake in the form of food, calculate the energy equivalents, , Fig. 6.14, , 47, 20/04/2018
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Exemplar Problems–Physics, of food that would be required to compensate energy utilised for, jogging., 6.43, , On complete combustion a litre of petrol gives off heat equivalent, to 3×107 J. In a test drive a car weighing 1200 kg. including the, mass of driver, runs 15 km per litre while moving with a uniform, speed on a straight track. Assuming that friction offered by the, road surface and air to be uniform, calculate the force of friction, acting on the car during the test drive, if the efficiency of the car, engine were 0.5., , LA, 6.44, , F, m, , A block of mass 1 kg is pushed up a surface inclined to horizontal, at an angle of 30° by a force of 10 N parallel to the inclined, surface (Fig. 6.15).The coefficient of friction between block and, the incline is 0.1. If the block is pushed up by 10 m along the, incline, calulate, (a), (b), (c), (d), (e), , 30o, , Fig. 6.15, , 6.45, , work done against gravity, work done against force of friction, increase in potential energy, increase in kinetic energy, work done by applied force., , A curved surface is shown in Fig. 6.16. The portion BCD is free, of friction. There are three spherical balls of identical radii and, masses. Balls are released from rest one by one from A which is, at a slightly greater height than C., , A, , C, D, B, Fig. 6.16, , With the surface AB, ball 1 has large enough friction to cause, rolling down without slipping; ball 2 has a small friction and ball, 3 has a negligible friction., (a) For which balls is total mechanical energy conserved?, (b) Which ball (s) can reach D?, (c) For balls which do not reach D, which of the balls can reach, back A?, , 48, 20/04/2018
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Work, Energy and Power, , 6.46, , A rocket accelerates straight up by ejecting gas downwards. In a, small time interval ∆t, it ejects a gas of mass ∆m at a relative, speed u. Calculate KE of the entire system at t + ∆t and t and, , ( 2 ) ∆m u, , show that the device that ejects gas does work = 1, , 2, , in, , this time interval (neglect gravity)., 6.47 Two identical steel cubes (masses 50g, side 1cm) collide, head-on face to face with a speed of 10cm/s each. Find, the maximum compression of each. Young’s modulus for, steel = Y= 2 × 10 11 N/m 2., 6.48, , A baloon filled with helium rises against gravity increasing its, potential energy. The speed of the baloon also increases as it, rises. How do you reconcile this with the law of conservation of, mechanical energy? You can neglect viscous drag of air and, assume that density of air is constant., , 49, 20/04/2018
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Exemplar Problems–Physics, , Chapter Seven, , SYSTEM OF, PARTICLES AND, ROTATIONAL MOTION, MCQ I, 7.1, , For which of the following does the centre of mass lie outside the, body ?, (a), (b), (c), (d), , 7.2, , Which of the following points is the likely position of the centre of, mass of the system shown in Fig. 7.1?, (a), (b), (c), (d), , 7.3, , A pencil, A shotput, A dice, A bangle, , A, B, C, D, , A particle of mass m is moving in yz-plane with a uniform velocity v, with its trajectory running parallel to +ve y-axis and intersecting, , Hollow, sphere, Air, R/2, A, B, C, , R/2, , D, , Sand, Fig. 7.1, , 50, 20/04/2018
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System of Particles and, Rotational Motion, , z-axis at z = a (Fig. 7.2). The change in its angular momentum about, the origin as it bounces elastically from a wall at y = constant is:, (a), (b), (c), (d), 7.4, , When a disc rotates with uniform angular velocity, which of the, following is not true?, (a), (b), (c), (d), , 7.5, , The sense of rotation remains same., The orientation of the axis of rotation remains same., The speed of rotation is non-zero and remains same., The angular acceleration is non-zero and remains same., , increased, decreased, the same, changed in unpredicted manner., , Fig. 7.3, , In problem 7.5, the CM of the plate is now in the following quadrant, of x-y plane,, (a), (b), (c), (d), , 7.7, , Fig. 7.2, , A uniform square plate has a small piece Q of an irregular, shape removed and glued to the centre of the plate leaving, a hole behind (Fig. 7.3). The moment of inertia about the, z-axis is then, (a), (b), (c), (d), , 7.6, , mva êx, 2mva êx, ymv êx, 2ymv êx, , I, II, III, IV, , The density of a non-uniform rod of length 1m is given by, ρ (x) = a(1+bx 2), where a and b are constants and o ≤ x ≤ 1 ., The centre of mass of the rod will be at, (a), , 3(2 + b ), 4 (3 + b ), , 4 (2 + b ), (b) 3(3 + b ), (c), , 3(3 + b ), 4 (2 + b ), , (d), , 4 (3 + b ), 3(2 + b ), , 51, 20/04/2018
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Exemplar Problems–Physics, 7.8, , A Merry-go-round, made of a ring-like platform of radius R and, mass M, is revolving with angular speed ω . A person of mass M is, standing on it. At one instant, the person jumps off the round,, radially away from the centre of the round (as seen from the round)., The speed of the round afterwards is, (a) 2ω, , (b) ω, , (c), , ω, 2, , (d) 0, , MCQ II, 7.9, , Choose the correct alternatives:, (a) For a general rotational motion, angular momentum L and, angular velocity ω need not be parallel., (b) For a rotational motion about a fixed axis, angular momentum, L and angular velocity ω are always parallel., (c) For a general translational motion , momentum p and velocity, v are always parallel., (d) For a general translational motion, acceleration a and velocity, v are always parallel., , 7.10 Figure 7.4 shows two identical particles 1 and 2, each of mass m,, moving in opposite directions with same speed v along parallel lines., At a particular instant, r1 and r2 are their respective position vectors, drawn from point A which is in the plane of the parallel lines ., Choose the correct options:, , A, r1, , d1, , (a) Angular momentum l1 of particle 1 about A is l1 = mvd1, 1, , d2, r2, , v, , v, , (b) Angular momentum l2 of particle 2 about A is l2 = mvr2, (c) Total angular momentum of the system about A is, , l = mv(r1 + r2 ), 2, , (d) Total angular momentum of the system about A is l = mv (d 2 − d1 ) ⊗, , Fig. 7.4, , represents a unit vector coming out of the page., ⊗ represents a unit vector going into the page., , 7.11 The net external torque on a system of particles about an axis is, zero. Which of the following are compatible with it ?, (a), (b), (c), (d), , The forces may be acting radially from a point on the axis., The forces may be acting on the axis of rotation., The forces may be acting parallel to the axis of rotation., The torque caused by some forces may be equal and opposite, to that caused by other forces., , 52, 20/04/2018
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System of Particles and, Rotational Motion, , 7.12 Figure 7.5 shows a lamina in x-y plane. Two axes z and, z ′ pass perpendicular to its plane. A force F acts in the, plane of lamina at point P as shown. Which of the following, are true? (The point P is closer to z′-axis than the z-axis.), , z, , z, F, P, , ˆ., (a) Torque τ caused by F about z axis is along -k, ˆ., (b) Torque τ′ caused by F about z′ axis is along -k, (c) Torque τ caused by F about z axis is greater in magnitude, than that about z axis., (d) Total torque is given be τ = τ + τ′, τ′., , Fig. 7.5, , 7.13 With reference to Fig. 7.6 of a cube of edge a and, mass m, state whether the following are true or false., (O is the centre of the cube.), (a) The moment of inertia of cube about z-axis is, Iz = Ix + Iy, (b) The moment of inertia of cube about z ′ is, m a2, 2, (c) The moment of inertia of cube about z″ is, I 'z = I z +, , = Iz +, , ma, 2, , 2, , (d) Ix = Iy, , Fig. 7.6, , VSA, 7.14 The centre of gravity of a body on the earth coincides with its centre, of mass for a ‘small’ object whereas for an ‘extended’ object it may, not. What is the qualitative meaning of ‘small’ and ‘extended’ in, this regard?, For which of the following the two coincides? A building, a pond, a, lake, a mountain?, 7.15 Why does a solid sphere have smaller moment of inertia than a, hollow cylinder of same mass and radius, about an axis passing, through their axes of symmetry?, 7.16 The variation of angular position θ , of a point on a rotating, rigid body, with time t is shown in Fig. 7.7. Is the body rotating, clock-wise or anti-clockwise?, , Fig. 7.7, , 53, 20/04/2018
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Exemplar Problems–Physics, 7.17 A uniform cube of mass m and side a is placed on a frictionless, horizontal surface. A vertical force F is applied to the edge as shown, in Fig. 7.8. Match the following (most appropriate choice):, (a) mg/4 < F < mg / 2, (b) F > mg/2, (c) F > mg, (d) F = mg/4, , Fig. 7.8, , (i) Cube will move up., (ii) Cube will not exhibit motion., (iii) Cube will begin to rotate and, slip at A., (iv) Normal reaction effectively at, a/3 from A, no motion., , 7.18 A uniform sphere of mass m and radius R is placed on a rough, horizontal surface (Fig. 7.9). The sphere is struck horizontally at a, height h from the floor. Match the following:, (a) h = R/2, (b) h = R, (c) h = 3R/2, (d) h = 7R/5, , Fig. 7.9, , (i) Sphere rolls without slipping with a, constant velocity and no loss of energy., (ii) Sphere spins clockwise, loses energy by, friction., (iii) Sphere spins anti-clockwise, loses energy, by friction., (iv) Sphere has only a translational motion,, looses energy by friction., , SA, 7.19 The vector sum of a system of non-collinear forces acting on a rigid, body is given to be non-zero. If the vector sum of all the torques due, to the system of forces about a certain point is found to be zero,, does this mean that it is necessarily zero about any arbitrary point?, 7.20 A wheel in uniform motion about an axis passing through its centre, and perpendicular to its plane is considered to be in mechanical, (translational plus rotational) equilibrium because no net external, force or torque is required to sustain its motion. However, the, particles that constitute the wheel do experience a centripetal, acceleration directed towards the centre. How do you reconcile this, fact with the wheel being in equilibrium?, How would you set a half-wheel into uniform motion about an, axis passing through the centre of mass of the wheel and, perpendicular to its plane? Will you require external forces to, sustain the motion?, 7.21 A door is hinged at one end and is free to rotate about a vertical, axis (Fig. 7.10). Does its weight cause any torque about this axis?, Give reason for your answer., Fig. 7.10, , 54, 20/04/2018
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System of Particles and, Rotational Motion, , 7.22 (n-1) equal point masses each of mass m are placed at the vertices, of a regular n-polygon. The vacant vertex has a position vector a, with respect to the centre of the polygon. Find the position vector of, centre of mass., , LA, 7.23 Find the centre of mass of a uniform (a) half-disc, (b) quarter-disc., 7.24 Two discs of moments of inertia I1 and I 2 about their respective, axes (normal to the disc and passing through the centre), and, rotating with angular speed ω1 and ω 2 are brought into contact, face to face with their axes of rotation coincident., (a) Does the law of conservation of angular momentum apply to, the situation? why?, (b) Find the angular speed of the two-disc system., (c) Calculate the loss in kinetic energy of the system in the process., (d) Account for this loss., 7.25 A disc of radius R is rotating with an angular speed ωo about a, horizontal axis. It is placed on a horizontal table. The coefficient of, kinetic friction is µk., (a) What was the velocity of its centre of mass before being brought, in contact with the table?, (b) What happens to the linear velocity of a point on its rim when, placed in contact with the table?, (c) What happens to the linear speed of the centre of mass when, disc is placed in contact with the table?, (d) Which force is responsible for the effects in (b) and (c)., (e) What condition should be satisfied for rolling to begin?, (f) Calculate the time taken for the rolling to begin., 7.26 Two cylindrical hollow drums of radii R and 2R, and of a common, height h, are rotating with angular velocities ω (anti-clockwise) and, ω (clockwise), respectively. Their axes, fixed are parallel and in a, horizontal plane separated by (3R + δ ) . They are now brought in, contact (δ → 0) ., (a) Show the frictional forces just after contact., (b) Identify forces and torques external to the system just after, contact., (c) What would be the ratio of final angular velocities when friction, ceases?, , 55, 20/04/2018
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Exemplar Problems–Physics, 7.27 A uniform square plate S (side c) and a uniform rectangular plate, R (sides b, a) have identical areas and masses (Fig. 7.11)., , Fig. 7.11, , Show that, , (i ) I xR / I xS < 1; (ii ) I yR / I yS > 1; (iii ) I zR / I zS > 1., 7.28 A uniform disc of radius R, is resting on a table on its rim.The, coefficient of friction between disc and table is µ (Fig 7.12). Now the, disc is pulled with a force F as shown in the figure. What is the, maximum value of F for which the disc rolls without slipping ?, , Fig. 7.12, , 56, 20/04/2018
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Chapter Eight, , GRAVITATION, , MCQ I, 8.1, , The earth is an approximate sphere. If the interior contained matter, which is not of the same density everywhere, then on the surface, of the earth, the acceleration due to gravity, (a) will be directed towards the centre but not the same everywhere., (b) will have the same value everywhere but not directed towards, the centre., (c) will be same everywhere in magnitude directed towards the, centre., (d) cannot be zero at any point., , 8.2, , As observed from earth, the sun appears to move in an, approximate circular orbit. For the motion of another planet like, mercury as observed from earth, this would, (a) be similarly true., (b) not be true because the force between earth and mercury is, not inverse square law., , 20/04/2018
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Exemplar Problems–Physics, (c) not be true because the major gravitational force on mercury, is due to sun., (d) not be true because mercury is influenced by forces other than, gravitational forces., 8.3, , Different points in earth are at slightly different distances from, the sun and hence experience different forces due to gravitation., For a rigid body, we know that if various forces act at various, points in it, the resultant motion is as if a net force acts on, the c.m. (centre of mass) causing translation and a net torque, at the c.m. causing rotation around an axis through the c.m., For the earth-sun system (approximating the earth as a, uniform density sphere), (a) the torque is zero., (b) the torque causes the earth to spin., (c) the rigid body result is not applicable since the earth is not, even approximately a rigid body., (d) the torque causes the earth to move around the sun., , 8.4, , Satellites orbiting the earth have finite life and sometimes debris, of satellites fall to the earth. This is because,, (a) the solar cells and batteries in satellites run out., (b) the laws of gravitation predict a trajectory spiralling inwards., (c) of viscous forces causing the speed of satellite and hence height, to gradually decrease., (d) of collisions with other satellites., , 8.5, , Both earth and moon are subject to the gravitational force of the, sun. As observed from the sun, the orbit of the moon, (a) will be elliptical., (b) will not be strictly elliptical because the total gravitational force, on it is not central., (c) is not elliptical but will necessarily be a closed curve., (d) deviates considerably from being elliptical due to influence of, planets other than earth., , 8.6, , In our solar system, the inter-planetary region has chunks of, matter (much smaller in size compared to planets) called asteroids., They, (a) will not move around the sun since they have very small masses, compared to sun., (b) will move in an irregular way because of their small masses, and will drift away into outer space., (c) will move around the sun in closed orbits but not obey, Kepler’s laws., (d) will move in orbits like planets and obey Kepler’s laws., , 58, 20/04/2018
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Gravitation, , 8.7, , Choose the wrong option., (a) Inertial mass is a measure of difficulty of accelerating a body, by an external force whereas the gravitational mass is relevant, in determining the gravitational force on it by an external mass., (b) That the gravitational mass and inertial mass are equal is an, experimental result., (c) That the acceleration due to gravity on earth is the same for all, bodies is due to the equality of gravitational mass and inertial mass., (d) Gravitational mass of a particle like proton can depend on the, presence of neighouring heavy objects but the inertial mass, cannot., , 8.8, , Particles of masses 2M, m and M are respectively at points A, B, and C with AB = ½ (BC). m is much-much smaller than M and at, time t = 0, they are all at rest (Fig. 8.1)., At subsequent times before any collision takes place:, A, 2M, , C, M, , B, m, Fig. 8.1, , (a), (b), (c), (d), , m will remain at rest., m will move towards M., m will move towards 2M., m will have oscillatory motion., , MCQ II, 8.9, , Which of the following options are correct?, (a) Acceleration due to gravity decreases with increasing altitude., (b) Acceleration due to gravity increases with increasing depth, (assume the earth to be a sphere of uniform density)., (c) Acceleration due to gravity increases with increasing latitude., (d) Acceleration due to gravity is independent of the mass of the, earth., , 8.10, , If the law of gravitation, instead of being inverse-square law,, becomes an inverse-cube law(a) planets will not have elliptic orbits., (b) circular orbits of planets is not possible., (c) projectile motion of a stone thrown by hand on the surface of, the earth will be approximately parabolic., (d) there will be no gravitational force inside a spherical shell of, uniform density., , 59, 20/04/2018
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Exemplar Problems–Physics, 8.11, , If the mass of sun were ten times smaller and gravitational constant, G were ten times larger in magnitudes(a), (b), (c), (d), , 8.12, , If the sun and the planets carried huge amounts of opposite charges,, (a), (b), (c), (d), , 8.13, , all three of Kepler’s laws would still be valid., only the third law will be valid., the second law will not change., the first law will still be valid., , There have been suggestions that the value of the gravitational, constant G becomes smaller when considered over very large, time period (in billions of years) in the future. If that happens,, for our earth,, (a), (b), (c), (d), , 8.14, , walking on ground would became more difficult., the acceleration due to gravity on earth will not change., raindrops will fall much faster., airplanes will have to travel much faster., , nothing will change., we will become hotter after billions of years., we will be going around but not strictly in closed orbits., after sufficiently long time we will leave the solar system., , Supposing Newton’s law of gravitation for gravitation forces, F1 and F2 between two masses m1 and m2 at positions r1 and r2 read, n, , F1 – F2 = –, , r12, m m , GM 0 2 1 2 2 where M0 is a constant of dimension, 3, r12, M0 , , of mass, r12 = r1 – r2 and n is a number. In such a case,, (a) the acceleration due to gravity on earth will be different for, different objects., (b) none of the three laws of Kepler will be valid., (c) only the third law will become invalid., (d) for n negative, an object lighter than water will sink in water., 8.15, , Which of the following are true?, (a) A polar satellite goes around the earth’s pole in northsouth direction., (b) A geostationary satellite goes around the earth in east-west, direction., (c) A geostationary satellite goes around the earth in west-east, direction., (d) A polar satellite goes around the earth in east-west direction., , 60, 20/04/2018
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Gravitation, , 8.16, , The centre of mass of an extended body on the surface of the earth, and its centre of gravity, (a), (b), (c), (d), (e), , are always at the same point for any size of the body., are always at the same point only for spherical bodies., can never be at the same point., is close to each other for objects, say of sizes less than 100 m., both can change if the object is taken deep inside the earth., , VSA, 8.17, , Molecules in air in the atmosphere are attracted by gravitational, force of the earth. Explain why all of them do not fall into the earth, just like an apple falling from a tree., , 8.18, , Give one example each of central force and non-central force., , 8.19, , Draw areal velocity versus time graph for mars., , 8.20, , What is the direction of areal velocity of the earth around the sun?, , 8.21, , How is the gravitational force between two point masses affected, when they are dipped in water keeping the separation between, them the same?, , 8.22, , Is it possibe for a body to have inertia but no weight?, , 8.23, , We can shield a charge from electric fields by putting it inside a, hollow conductor. Can we shield a body from the gravitational, influence of nearby matter by putting it inside a hollow sphere or, by some other means?, , 8.24, , An astronaut inside a small spaceship orbiting around the earth, cannot detect gravity. If the space station orbiting around the earth, has a large size, can he hope to detect gravity?, , 8.25, , The gravitational force between a hollow spherical shell (of radius, R and uniform density) and a point mass is F. Show the nature of, F vs r graph where r is the distance of the point from the centre of, the hollow spherical shell of uniform density., , 8.26, , Out of aphelion and perihelion, where is the speed of the earth, more and why ?, , 8.27, , What is the angle between the equatorial plane and the orbital, plane of, (a) Polar satellite?, (b) Geostationary satellite?, , 61, 20/04/2018
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Exemplar Problems–Physics, , SA, 8.28, , Mean solar day is the time interval between two successive noon, when sun passes through zenith point (meridian)., Sidereal day is the time interval between two successive transit of, a distant star through the zenith point (meridian)., By drawing appropriate diagram showing earth’s spin and orbital, motion, show that mean solar day is four minutes longer than the, sidereal day. In other words, distant stars would rise 4 minutes, early every successive day., (Hint: you may assume circular orbit for the earth)., , 8.29, , Two identical heavy spheres are separated by a distance 10 times, their radius. Will an object placed at the mid point of the line joining, their centres be in stable equilibrium or unstable equilibrium?, Give reason for your answer., , 8.30, , Show the nature of the following graph for a satellite orbiting the, earth., (a) KE vs orbital radius R, (b) PE vs orbital radius R, (c) TE vs orbital radius R., , 8.31, , Earth, , Shown are several curves (Fig. 8.2). Explain with reason, which, ones amongst them can be possible trajectories traced by a, projectile (neglect air friction)., , Earth, Earth, , (a), , (b), , (c), , Earth, Earth, , Earth, , (e), , (d), , (f ), , Fig. 8.2, , 62, 20/04/2018
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Gravitation, , 8.32, , An object of mass m is raised from the surface of the earth to a, height equal to the radius of the earth, that is, taken from a distance, R to 2R from the centre of the earth. What is the gain in its potential, energy?, , 8.33, , A mass m is placed at P a distance h along the normal through the, centre O of a thin circular ring of mass M and radius r (Fig. 8.3)., , r, o, , h, , P, m, , M, , Fig. 8.3, , If the mass is removed further away such that OP becomes 2h, by, what factor the force of gravitation will decrease, if, h=r?, , LA, 8.34, , A star like the sun has several bodies moving around it at different, distances. Consider that all of them are moving in circular orbits., Let r be the distance of the body from the centre of the star and let, its linear velocity be v, angular velocity ω, kinetic energy K,, gravitational potential energy U, total energy E and angular, momentum l. As the radius r of the orbit increases, determine, which of the above quantities increase and which ones decrease., , 8.35, , Six point masses of mass m each are at the vertices of a regular, hexagon of side l. Calculate the force on any of the masses., , 8.36, , A satellite is to be placed in equatorial geostationary orbit around, earth for communication., (a) Calculate height of such a satellite., (b) Find out the minimum number of satellites that are needed to, cover entire earth so that at least one satellites is visible from, any point on the equator., , [M = 6 × 1024 kg, R = 6400 km, T = 24h, G = 6.67 × 10–11 SI units], , 63, 20/04/2018
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Exemplar Problems–Physics, 8.37, , Earth’s orbit is an ellipse with eccentricity 0.0167. Thus,, earth’s distance from the sun and speed as it moves around, the sun varies from day to day. This means that the length, of the solar day is not constant through the year. Assume, that earth’s spin axis is normal to its orbital plane and find, out the length of the shortest and the longest day. A day, should be taken from noon to noon. Does this explain, variation of length of the day during the year?, , 8.38, , A satellite is in an elliptic orbit around the earth with, aphelion of 6R and perihelion of 2 R where R= 6400 km is, the radius of the earth. Find eccentricity of the orbit. Find, the velocity of the satellite at apogee and perigee. What, should be done if this satellite has to be transferred to a, circular orbit of radius 6R ?, [G = 6.67 × 10–11 SI units and M = 6 × 1024 kg], , 64, 20/04/2018
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Chapter Nine, , MECHANICAL, PROPERTIES OF, SOLIDS, MCQ I, 9.1, , Modulus of rigidity of ideal liquids is, (a), (b), (c), (d), , 9.2, , The maximum load a wire can withstand without breaking, when, its length is reduced to half of its original length, will, (a), (b), (c), (d), , 9.3, , infinity., zero., unity., some finite small non-zero constant value., , be double., be half., be four times., remain same., , The temperature of a wire is doubled. The Young’s modulus of, elasticity, (a) will also double., (b) will become four times., , 20/04/2018
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Exemplar Problems–Physics, ( c ) will remain same., (d) will decrease., 9.4, , A spring is stretched by applying a load to its free end. The strain, produced in the spring is, (a), (b), (c), (d), , 9.5, , volumetric., shear., longitudinal and shear., longitudinal., , A rigid bar of mass M is supported symmetrically by three wires, each of length l . Those at each end are of copper and the middle, one is of iron. The ratio of their diameters, if each is to have the, same tension, is equal to, (a) Ycopper/Yiron, , (b), , (c), , (d), 9.6, , Yiron, Ycopper, 2, Y iron, 2, Y copper, , Yiron ., Ycopper, , A mild steel wire of length 2L and cross-sectional area A is, stretched, well within elastic limit, horizontally between two pillars, (Fig. 9.1). A mass m is suspended from the mid point of the wire., Strain in the wire is, 2L, , x, m, , Fig. 9.1, , x2, (a), 2 L2, (b), , x, L, , 66, 20/04/2018
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Mechanical Properties of Solids, , x2, (c), L, (d), 9.7, , x2 ., 2L, , A rectangular frame is to be suspended symmetrically by two, strings of equal length on two supports (Fig. 9.2). It can be done, in one of the following three ways;, , (a), , (b), , (c), , Fig. 9.2, , The tension in the strings will be, (a) the same in all cases., (b) least in (a)., (c) least in (b)., (d) least in (c)., 9.8, , Consider two cylindrical rods of identical dimensions, one of rubber, and the other of steel. Both the rods are fixed rigidly at one end to, the roof. A mass M is attached to each of the free ends at the centre, of the rods., (a) Both the rods will elongate but there shall be no perceptible, change in shape., (b) The steel rod will elongate and change shape but the rubber, rod will only elongate., (c) The steel rod will elongate without any perceptible change in, shape, but the rubber rod will elongate and the shape of the, bottom edge will change to an ellipse., (d) The steel rod will elongate, without any perceptible change in, shape, but the rubber rod will elongate with the shape of the, bottom edge tapered to a tip at the centre., , 67, 20/04/2018
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Exemplar Problems–Physics, , MCQ II, 9.9, , The stress-strain graphs for two materials are shown in Fig.9.3, (assume same scale)., , Fig. 9.3, , (a) Material (ii) is more elastic than material (i) and hence material, (ii) is more brittle., (b) Material (i) and (ii) have the same elasticity and the same, brittleness., (c) Material (ii) is elastic over a larger region of strain as compared to (i)., (d) Material (ii) is more brittle than material (i)., 9.10, , A wire is suspended from the ceiling and stretched under the action, of a weight F suspended from its other end. The force exerted by, the ceiling on it is equal and opposite to the weight., (a), (b), (c), (d), , 9.11, , Tensile stress at any cross section A of the wire is F/A., Tensile stress at any cross section is zero., Tensile stress at any cross section A of the wire is 2F/A., Tension at any cross section A of the wire is F., , A rod of length l and negligible mass is suspended at its two ends, by two wires of steel (wire A) and aluminium (wire B) of equal, lengths (Fig. 9.4). The cross-sectional areas of wires A and B are, 1.0 mm2 and 2.0 mm2, respectively., , (Y Al, , = 70 × 109 Nm −2 and Ysteel = 200 × 109 Nm –2 ), , A, , B, , steel, , Al, , m, , Fig. 9.4, , 68, 20/04/2018
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Mechanical Properties of Solids, , (a) Mass m should be suspended close to wire A to have equal, stresses in both the wires., (b) Mass m should be suspended close to B to have equal stresses, in both the wires., (c) Mass m should be suspended at the middle of the wires to, have equal stresses in both the wires., (d) Mass m should be suspended close to wire A to have equal, strain in both wires., 9.12, , For an ideal liquid, (a), (b), (c), (d), , 9.13, , the bulk modulus is infinite., the bulk modulus is zero., the shear modulus is infinite., the shear modulus is zero., , A copper and a steel wire of the same diameter are connected end, to end. A deforming force F is applied to this composite wire which, causes a total elongation of 1cm. The two wires will have, (a), (b), (c), (d), , the same stress., different stress., the same strain., different strain., , VSA, 9.14, , The Young’s modulus for steel is much more than that for rubber., For the same longitudinal strain, which one will have greater, tensile stress?, , 9.15, , Is stress a vector quantity ?, , 9.16, , Identical springs of steel and copper are equally stretched. On, which, more work will have to be done?, , 9.17, , What is the Young’s modulus for a perfect rigid body ?, , 9.18, , What is the Bulk modulus for a perfect rigid body?, , SA, 9.19, , A wire of length L and radius r is clamped rigidly at one end. When, the other end of the wire is pulled by a force f, its length increases, by l. Another wire of the same material of length 2L and radius 2r,, is pulled by a force 2f. Find the increase in length of this wire., , 69, 20/04/2018
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Exemplar Problems–Physics, 9.20, , A steel rod (Y = 2.0 × 1011 Nm–2; and α = 10–50 C–1) of length 1 m, and area of cross-section 1 cm2 is heated from 0°C to 200°C, without, being allowed to extend or bend. What is the tension produced in, the rod?, , 9.21, , To what depth must a rubber ball be taken in deep sea so that its, volume is decreased by 0.1%. (The bulk modulus of rubber is, 9.8×108 N m–2; and the density of sea water is 103 kg m–3.), , 9.22, , A truck is pulling a car out of a ditch by means of a steel cable, that is 9.1 m long and has a radius of 5 mm. When the car just, begins to move, the tension in the cable is 800 N. How much has, the cable stretched? (Young’s modulus for steel is 2 × 1011 Nm–2.), , 9.23, , Two identical solid balls, one of ivory and the other of wet-clay,, are dropped from the same height on the floor. Which one will rise, to a greater height after striking the floor and why?, , LA, 9.24, , Consider a long steel bar under a tensile stress due to forces F, acting at the edges along the length of the bar (Fig. 9.5). Consider, a plane making an angle θ with the length. What are the tensile, and shearing stresses on this plane?, a, , F, , F, , a', , Fig. 9.5, , (a) For what angle is the tensile stress a maximum?, (b) For what angle is the shearing stress a maximum?, 9.25, , (a) A steel wire of mass µ per unit length with a circular cross, section has a radius of 0.1 cm. The wire is of length 10 m when, measured lying horizontal, and hangs from a hook on the wall., A mass of 25 kg is hung from the free end of the wire. Assuming, the wire to be uniform and lateral strains << longitudinal, strains, find the extension in the length of the wire. The density, of steel is 7860 kg m–3 (Young’s modules Y=2×1011 Nm–2)., (b) If the yield strength of steel is 2.5×108 Nm–2, what is the maximum, weight that can be hung at the lower end of the wire?, , 70, 20/04/2018
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Mechanical Properties of Solids, , 9.26, , A steel rod of length 2l, cross sectional area A and mass M is set, rotating in a horizontal plane about an axis passing through the, centre. If Y is the Young’s modulus for steel, find the extension in, the length of the rod. (Assume the rod is uniform.), , 9.27, , An equilateral triangle ABC is formed by two Cu rods AB and BC, and one Al rod. It is heated in such a way that temperature of, each rod increases by ∆T. Find change in the angle ABC. [Coeff. of, linear expansion for Cu is α1 ,Coeff. of linear expansion for, Al is α 2 ], , 9.28, , In nature, the failure of structural members usually result from, large torque because of twisting or bending rather than due to, tensile or compressive strains. This process of structural, breakdown is called buckling and in cases of tall cylindrical, structures like trees, the torque is caused by its own weight bending, the structure. Thus the vertical through the centre of gravity does, not fall within the base. The elastic torque caused because of this, Y πr 4, . Y is, 4R, the Young’s modulus, r is the radius of the trunk and R is the, radius of curvature of the bent surface along the height of the tree, containing the centre of gravity (the neutral surface). Estimate the, critical height of a tree for a given radius of the trunk., bending about the central axis of the tree is given by, , 9.29, , A stone of mass m is tied to an elastic string of negligble mass and, spring constant k. The unstretched length of the string is L and, has negligible mass. The other end of the string is fixed to a nail at, a point P. Initially the stone is at the same level as the point P. The, stone is dropped vertically from point P., (a) Find the distance y from the top when the mass comes to rest, for an instant, for the first time., (b) What is the maximum velocity attained by the stone in this, drop?, (c) What shall be the nature of the motion after the stone has, reached its lowest point?, , 71, 20/04/2018
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Exemplar Problems–Physics, , Chapter Ten, , MECHANICAL, PROPERTIES OF, FLUIDS, , MCQ I, 10.1, , A tall cylinder is filled with viscous oil. A round pebble is dropped, from the top with zero initial velocity. From the plot shown in, Fig. 10.1, indicate the one that represents the velocity (v) of the, pebble as a function of time (t)., , v, , v, , v, , v, , t, , t, , t, , t, , (a), , (b), , (c), , (d), , Fig. 10.1, , 72, 20/04/2018
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Mechanical Properties of Fluids, , 10.2, , Which of the following diagrams (Fig. 10.2) does not represent a, streamline flow?, , Fig. 10.2, , 10.3, , Along a streamline, (a) the velocity of a fluid particle remains constant., (b) the velocity of all fluid particles crossing a given position is, constant., (c) the velocity of all fluid particles at a given instant is constant., (d) the speed of a fluid particle remains constant., , 10.4, , An ideal fluid flows through a pipe of circular cross-section made, of two sections with diameters 2.5 cm and 3.75 cm. The ratio of, the velocities in the two pipes is, (a) 9:4, (b) 3:2, , 10.5, , (c), , 3: 2, , (d), , 2: 3, , The angle of contact at the interface of water-glass is 0°,, Ethylalcohol-glass is 0°, Mercury-glass is 140° and Methyliodideglass is 30°. A glass capillary is put in a trough containing one of, these four liquids. It is observed that the meniscus is convex. The, liquid in the trough is, (a), (b), (c), (d), , water, ethylalcohol, mercury, methyliodide., , MCQ II, 10.6, , For a surface molecule, (a) the net force on it is zero., (b) there is a net downward force., , 73, 20/04/2018
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Exemplar Problems–Physics, , (c) the potential energy is less than that of a molecule inside., (d) the potential energy is more than that of a molecule inside., 10.7, , Pressure is a scalar quantity because, (a), (b), (c), (d), , 10.8, , it is the ratio of force to area and both force and area are vectors., it is the ratio of the magnitude of the force to area., it is the ratio of the component of the force normal to the area., it does not depend on the size of the area chosen., , A wooden block with a coin placed on its top, floats in water as, shown in Fig.10.3., The distance l and h are shown in the figure. After some time the, coin falls into the water. Then, (a) l decreases., (b) h decreases., (c) l increases., (d) h increase., , Fig. 10.3, , 10.9, , With increase in temperature, the viscosity of, (a), (b), (c), (d), , gases decreases., liquids increases., gases increases., liquids decreases., , 10.10 Streamline flow is more likely for liquids with, (a), (b), (c), (d), , high density., high viscosity., low density., low viscosity., , VSA, 10.11 Is viscosity a vector?, 10.12 Is surface tension a vector?, 10.13 Iceberg floats in water with part of it submerged. What is the fraction, of the volume of iceberg submerged if the density of ice is ρi =, 0.917 g cm–3?, 10.14 A vessel filled with water is kept on a weighing pan and the, scale adjusted to zero. A block of mass M and density ρ is, suspended by a massless spring of spring constant k. This block, is submerged inside into the water in the vessel. What is the, reading of the scale?, , 74, 20/04/2018
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Mechanical Properties of Fluids, , 10.15 A cubical block of density ρ is floating on the surface of water., Out of its height L, fraction x is submerged in water. The vessel is, in an elevator accelerating upward with acceleration a . What is, the fraction immersed?, , SA, 10.16 The sap in trees, which consists mainly of water in summer, rises, in a system of capillaries of radius r = 2.5×10–5 m. The surface, tension of sap is T = 7.28×10–2 Nm–1 and the angle of contact is 0°., Does surface tension alone account for the supply of water to the, top of all trees?, 10.17 The free surface of oil in a tanker, at rest, is horizontal. If the tanker, starts accelerating the free surface will be titled by an angle θ. If, the acceleration is a m s–2, what will be the slope of the free surface?., 10.18 Two mercury droplets of radii 0.1 cm. and 0.2 cm. collapse into, one single drop. What amount of energy is released? The surface, tension of mercury T= 435.5 × 10–3 N m–1., 10.19 If a drop of liquid breaks into smaller droplets, it results in lowering, of temperature of the droplets. Let a drop of radius R, break into N, small droplets each of radius r. Estimate the drop in temperature., 10.20 The sufrace tension and vapour pressure of water at 20°C is, 7.28×10–2 Nm–1 and 2.33×103 Pa, respectively. What is the radius, of the smallest spherical water droplet which can form without, evaporating at 20°C?, , LA, 10.21 (a) Pressure decreases as one ascends the atmosphere. If the, density of air is ρ, what is the change in pressure dp over a, differential height dh?, (b) Considering the pressure p to be proportional to the density,, find the pressure p at a height h if the pressure on the surface, of the earth is p0., (c) If p0 = 1.03×105 N m–2, ρ0 = 1.29 kg m–3 and g = 9.8 m s–2, at, what height will the pressure drop to (1/10) the value at the, surface of the earth?, (d) This model of the atmosphere works for relatively small distances., Identify the underlying assumption that limits the model., 10.22 Surface tension is exhibited by liquids due to force of attraction, between molecules of the liquid. The surface tension decreases, , 75, 20/04/2018
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Exemplar Problems–Physics, , with increase in temperature and vanishes at boiling point. Given, that the latent heat of vaporisation for water Lv = 540 k cal kg–1,, the mechanical equivalent of heat J = 4.2 J cal–1, density of water, ρ w = 103 kg l–1, Avagadro’s No NA = 6.0 × 1026 k mole –1 and the, molecular weight of water MA = 18 kg for 1 k mole., (a) estimate the energy required for one molecule of water to, evaporate., (b) show that the inter–molecular distance for water is, 1/3, , M, 1 , d= A ×, , N A ρw , , and find its value., , (c) 1 g of water in the vapor state at 1 atm occupies 1601cm3., Estimate the intermolecular distance at boiling point, in the, vapour state., (d) During vaporisation a molecule overcomes a force F, assumed, constant, to go from an inter-molecular distance d to d ′ ., Estimate the value of F., (e) Calculate F/d, which is a measure of the surface tension., 10.23 A hot air balloon is a sphere of radius 8 m. The air inside is at a, temperature of 60°C. How large a mass can the balloon lift when, the outside temperature is 20°C? (Assume air is an ideal gas,, R = 8.314 J mole–1K-1, 1 atm. = 1.013×105 Pa; the membrane, tension is 5 N m–1.), , 76, 20/04/2018
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Chapter Eleven, , THERMAL, PROPERTIES OF, MATTER, MCQ I, 11.1, , A bimetallic strip is made of aluminium and steel (α Al > α steel ) ., On heating, the strip will, (a), (b), (c), (d), , 11.2, , A uniform metallic rod rotates about its perpendicular bisector, with constant angular speed. If it is heated uniformly to raise its, temperature slightly, (a), (b), (c), (d), , 11.3, , remain straight., get twisted., will bend with aluminium on concave side., will bend with steel on concave side., , its speed of rotation increases., its speed of rotation decreases., its speed of rotation remains same., its speed increases because its moment of inertia increases., , The graph between two temperature scales A and B is shown in, Fig. 11.1. Between upper fixed point and lower fixed point there, , 20/04/2018
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Exemplar Problems–Physics, are 150 equal division on scale A and 100 on scale B., The relationship for conversion between the two scales, is given by, (a), , t A − 180, t, = B, 100, 150, , (b), , t A − 30, t, = B, 150, 100, , (c), , t B − 180, t, = A, 150, 100, , (d), , t B − 40, t, = A, 100, 180, Fig. 11.1, , 11.4, , An aluminium sphere is dipped into water. Which of the following, is true?, (a) Buoyancy will be less in water at 0°C than that in water at, 4°C., (b) Buoyancy will be more in water at 0°C than that in water at, 4°C., (c) Buoyancy in water at 0°C will be same as that in water at, 4°C., (d) Buoyancy may be more or less in water at 4°C depending on, the radius of the sphere., , 11.5, , As the temperature is increased, the time period of a pendulum, (a) increases as its effective length increases even though its centre, of mass still remains at the centre of the bob., (b) decreases as its effective length increases even though its centre, of mass still remains at the centre of the bob., (c) increases as its effective length increases due to shifting of, centre of mass below the centre of the bob., (d) decreases as its effective length remains same but the centre, of mass shifts above the centre of the bob., , 11.6, , Heat is associated with, (a) kinetic energy of random motion of molecules., (b) kinetic energy of orderly motion of molecules., (c) total kinetic energy of random and orderly motion of, molecules., (d) kinetic energy of random motion in some cases and kinetic, energy of orderly motion in other., , 78, 20/04/2018
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Thermal Properties of Matter, , 11.7, , The radius of a metal sphere at room temperature T is R, and the, coefficient of linear expansion of the metal is α . The sphere is, heated a little by a temperature ∆T so that its new temperature is, T + ∆T . The increase in the volume of the sphere is approximately, (a) 2 π R α ∆ T, (b) π R 2 α ∆ T, (c) 4π R 3α ∆T /3, (d) 4π R 3α ∆T, , 11.8, , A sphere, a cube and a thin circular plate, all of same material, and same mass are initially heated to same high temperature., (a), (b), (c), (d), , Plate will cool fastest and cube the slowest, Sphere will cool fastest and cube the slowest, Plate will cool fastest and sphere the slowest, Cube will cool fastest and plate the slowest., , MCQ II, 11.9, , Mark the correct options:, (a) A system X is in thermal equilibrium with Y but not with Z., System Y and Z may be in thermal equilibrium with each, other., (b) A system X is in thermal equilibrium with Y but not with Z., Systems Y and Z are not in thermal equilibrium with each, other., (c) A system X is neither in thermal equilibrium with Y nor with, Z. The systems Y and Z must be in thermal equilibrium with, each other., (d) A system X is neither in thermal equilibrium with Y nor with, Z. The system Y and Z may be in thermal equilibrium with, each other., , 11.10, , ‘Gulab Jamuns’ (assumed to be spherical) are to be heated in an, oven. They are available in two sizes, one twice bigger (in radius), than the other. Pizzas (assumed to be discs) are also to be heated, in oven. They are also in two sizes, one twice big (in radius) than, the other. All four are put together to be heated to oven, temperature. Choose the correct option from the following:, (a), (b), (c), (d), , Both size gulab jamuns will get heated in the same time., Smaller gulab jamuns are heated before bigger ones., Smaller pizzas are heated before bigger ones., Bigger pizzas are heated before smaller ones., , 79, 20/04/2018
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Exemplar Problems–Physics, 11.11, , Refer to the plot of temperature versus time (Fig. 11.2) showing, the changes in the state of ice on heating (not to scale)., , E, Which of the following is correct?, , A, O, , D, , C, , o, , Temperature ( C), , 100, , B, tm, , time (min), , (a) The region AB represents ice and water in thermal, equilibrium., (b) At B water starts boiling., (c) At C all the water gets converted into steam., (d) C to D represents water and steam in equilibrium at, boiling point., , Fig. 11.2, , 11.12, , A glass full of hot milk is poured on the table. It begins to cool, gradually. Which of the following is correct?, (a) The rate of cooling is constant till milk attains the, temperature of the surrounding., (b) The temperature of milk falls off exponentially with time., (c) While cooling, there is a flow of heat from milk to the, surrounding as well as from surrounding to the milk but, the net flow of heat is from milk to the surounding and that, is why it cools., (d) All three phenomenon, conduction, convection and radiation, are responsible for the loss of heat from milk to the, surroundings., , VSA, 11.13, , Is the bulb of a thermometer made of diathermic or adiabatic wall?, , 11.14, , A student records the initial length l, change in temperature, ∆T and change in length ∆l of a rod as follows:, S.No., , l (m), , ∆T ( C), , ∆l (m), , 1., , 2, , 10, , 4 × 10− 4, , 2., , 1, , 10, , 4 × 10− 4, , 3., , 2, , 20, , 2 × 10− 4, , 4., , 3, , 10, , 6 × 10 − 4, , If the first observation is correct, what can you say about, observations 2, 3 and 4., , 80, 20/04/2018
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Thermal Properties of Matter, , 11.15, , Why does a metal bar appear hotter than a wooden bar at the, same temperature? Equivalently it also appears cooler than, wooden bar if they are both colder than room temperature., , 11.16, , Calculate the temperature which has same numeral value on, celsius and Fahrenheit scale., , 11.17, , These days people use steel utensils with copper bottom. This is, supposed to be good for uniform heating of food. Explain this, effect using the fact that copper is the better conductor., , SA, 11.18, , Find out the increase in moment of inertia I of a uniform rod, (coefficient of linear expansion α ) about its perpendicular, bisector when its temperature is slightly increased by ∆T., , 11.19, , During summers in India, one of the common practice to keep, cool is to make ice balls of crushed ice, dip it in flavoured sugar, syrup and sip it. For this a stick is inserted into crushed ice and, is squeezed in the palm to make it into the ball. Equivalently in, winter, in those areas where it snows, people make snow balls, and throw around. Explain the formation of ball out of crushed, ice or snow in the light of P–T diagram of water., , 11.20, , 100 g of water is supercooled to –10°C. At this point, due to, some disturbance mechanised or otherwise some of it suddenly, freezes to ice. What will be the temperature of the resultant, mixture and how much mass would freeze?, S w = 1cal/g/o C and L w Fusion = 80cal/g , , 11.21, , One day in the morning, Ramesh filled up 1/3 bucket of, hot water from geyser, to take bath. Remaining 2/3 was to, be filled by cold water (at room temperature) to bring, mixture to a comfortable temperature. Suddenly Ramesh, had to attend to something which would take some times,, say 5-10 minutes before he could take bath. Now he had, two options: (i) fill the remaining bucket completely by cold, water and then attend to the work, (ii) first attend to the, work and fill the remaining bucket just before taking bath., Which option do you think would have kept water warmer ?, Explain., , 81, 20/04/2018
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Exemplar Problems–Physics, , LA, 11.22, , We would like to prepare a scale whose length does not change, with temperature. It is proposed to prepare a unit scale of this, type whose length remains, say 10 cm. We can use a bimetallic, strip made of brass and iron each of different length whose length, (both components) would change in such a way that difference, between their lengths remain constant. If αiron = 1.2 × 10 −5 / K and, , αbrass = 1.8 × 10−5 / K , what should we take as length of each strip?, 11.23, , We would like to make a vessel whose volume does not change, with temperature (take a hint from the problem above). We can, use brass and iron ( βvbrass = 6 × 10–5/K and βviron = 3.55 ×10–5/, K) to create a volume of 100 cc. How do you think you can, achieve this., , 11.24, , Calculate the stress developed inside a tooth cavity filled with, copper when hot tea at temperature of 57oC is drunk. You can, take body (tooth) temperature to be 37o C and α = 1.7 × 10-5/oC,, bulk modulus for copper = 140 × 10 9 N/m2., , 11.25, , A rail track made of steel having length 10 m is clamped on a, raillway line at its two ends (Fig 11.3). On a summer day due to, rise in temperature by 20° C , it is deformed as shown in figure., Find x (displacement of the centre) if α steel = 1.2 × 10-5 /°C., , Fig. 11.3, , 11.26, , A thin rod having length L0 at 0°C and coefficient of linear, expansion α has its two ends maintained at temperatures θ1 and, θ2, respectively. Find its new length., , 11.27, , According to Stefan’s law of radiation, a black body radiates, energy σT 4 from its unit surface area every second where T is the, surface temperature of the black body and σ = 5.67 × 10–8 W/, m2K4 is known as Stefan’s constant. A nuclear weapon may be, thought of as a ball of radius 0.5 m. When detonated, it reaches, temperature of 106K and can be treated as a black body., (a) Estimate the power it radiates., (b) If surrounding has water at 30 ° C , how much water can 10%, of the energy produced evaporate in 1 s?, Sw = 4186.0 J/kg K and L v = 22.6 × 105 J/kg , , (c) If all this energy U is in the form of radiation, corresponding, momentum is p = U/c. How much momentum per unit time, does it impart on unit area at a distance of 1 km?, , 82, 20/04/2018
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Chapter Twelve, , THERMODYNAMICS, , MCQ I, 12.1, , An ideal gas undergoes four different processes from the, same initial state (Fig. 12.1). Four processes are adiabatic,, isothermal, isobaric and isochoric. Out of 1, 2, 3 and 4 which, one is adiabatic., P, (a), (b), (c), (d), , 12.2, , 4, 3, 2, 1, , 4, 3, 2, 1, V, , Fig. 12.1, , If an average person jogs, hse produces 14.5 × 103 cal/min. This is, removed by the evaporation of sweat. The amount of sweat evaporated, per minute (assuming 1 kg requires 580 × 103 cal for evaparation) is, (a), (b), (c), (d), , 0.25, 2.25, 0.05, 0.20, , kg, kg, kg, kg, , 20/04/2018
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Exemplar Problems–Physics, 12.3, , Consider P-V diagram for an ideal gas shown in Fig 12.2., , P, , 1, P=, , Con s ta n t, V, , 2, V, Fig. 12.2, , Out of the following diagrams (Fig. 12.3), which represents, the T-P diagram?, TT, , T, , 2, , 2, , 1, , 1, (i), , P, , P, , (ii), , T, , T, , 1, , 1, , 2, , 2, , P, , P, , (iii), , (iv), Fig. 12.3, , (a) (iv), (b) (ii), (c) (iii), (d) (i), 12.4, , An ideal gas undergoes cyclic process ABCDA as shown in given, P-V diagram (Fig. 12.4)., The amount of work done by, the gas is, (a) 6PoVo, (b) –2 PoVo, (c) + 2 PoVo, (d) + 4 PoVo, , Fig 12.4, , 84, 20/04/2018
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Thermodynamics, , 12.5, , Consider two containers A and B containing identical gases at, the same pressure, volume and temperature. The gas in container, A is compressed to half of its original volume isothermally while, the gas in container B is compressed to half of its original value, adiabatically. The ratio of final pressure of gas in B to that of gas, in A is, (a) 2γ −1, γ −1, , 1, (b) , 2, , 12.6, , 1 , (c) , , 1 − γ , , 2, , 1 , (d) , , γ −1, , 2, , Three copper blocks of masses M1, M2 and M3 kg respectively are, brought into thermal contact till they reach equilibrium. Before, contact, they were at T1, T2, T3 (T1 > T2 > T3 ). Assuming there is no, heat loss to the surroundings, the equilibrium temprature T is, (s is specific heat of copper), (a) T =, , T1 + T2 + T3, 3, , (b) T =, , M1T1 + M 2 T2 + M 3 T3, M1 + M 2 + M 3, , (c) T =, , M1T1 + M 2 T2 + M 3 T3, 3 ( M1 + M 2 + M 3 ), , (d) T =, , M1T1s + M 2 T2s + M 3T3s, M1 + M 2 + M 3, , MCQ II, 12.7, , Which of the processes described below are irreversible?, (a) The increase in temprature of an iron rod by hammering it., (b) A gas in a small cantainer at a temprature T1 is brought in, contact with a big reservoir at a higher temprature T2 which, increases the temprature of the gas., (c) A quasi-static isothermal expansion of an ideal gas in cylinder, fitted with a frictionless piston., , 85, 20/04/2018
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Exemplar Problems–Physics, (d) An ideal gas is enclosed in a piston cylinder arrangement, with adiabatic walls. A weight W is added to the piston,, resulting in compression of gas., 12.8, , An ideal gas undergoes isothermal process from some initial state, i to final state f. Choose the correct alternatives., (a), (b), (c), (d), , P, , dU = 0, dQ= 0, dQ = dU, dQ = dW, , I, , 12.9, , IV, A, , II, B, , III, , V, Fig. 12.5, , 12.10, , Figure 12.5 shows the P-V diagram of an ideal gas undergoing a, change of state from A to B. Four different parts I, II, III and IV as, shown in the figure may lead to the same change of state., (a) Change in internal energy is same in IV and III cases, but not in I, and II., (b) Change in internal energy is same in all the four cases., (c) Work done is maximum in case I, (d) Work done is minimum in case II., Consider a cycle followed by an engine (Fig. 12.6), 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 to mechanical energy in such a, process, which is not possible., (b) mechanical energy is completely converted to heat in this, process,which is not possible., (c) curves representing two adiabatic processes don’t intersect., (d) curves representing an adiabatic process and an isothermal, process don’t intersect., , Fig. 12.6, , 12.11, , Consider a heat engine as shown in, Fig. 12.7. Q1 and Q2 are heat added to heat, bath T1 and heat taken from T2 in one cycle, of engine. W is the mechanical work done, on the engine., If W > 0, then possibilities are:, (a), (b), (c), (d), , Q1 > Q2 > 0, Q2 > Q1 > 0, Q2 < Q1 < 0, Q1 < 0, Q2 > 0, , Fig .12.7, , 86, 20/04/2018
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Thermodynamics, , VSA, 12.12, , Can a system be heated and its temperature remains constant?, , 12.13, , A system goes from P to Q by two different paths in the P-V, diagram as shown in Fig. 12.8. Heat given to the system in, path 1 is 1000 J. The work done by the system along path 1 is, more than path 2 by 100 J. What is the heat exchanged by the, system in path 2?, , 12.14, , If a refrigerator’s door is kept open, will the room become cool or, hot? Explain., , 12.15, , Is it possible to increase the temperature of a gas without adding, heat to it? Explain., , 12.16, , Air pressure in a car tyre increases during driving. Explain., , Fig. 12.8, , SA, 12.17, , Consider a Carnot’s cycle operating between T1 = 500 K and, T2=300K producing 1 k J of mechanical work per cycle. Find the, heat transferred to the engine by the reservoirs., , 12.18, , A person of mass 60 kg wants to lose 5kg by going up and down, a 10m high stairs. Assume he burns twice as much fat while, going up than coming down. If 1 kg of fat is burnt on expending, 7000 kilo calories, how many times must he go up and down to, reduce his weight by 5 kg?, , 12.19, , Consider a cycle tyre being filled with air by a pump. Let V be the, volume of the tyre (fixed) and at each stroke of the pump ∆V ( V ), of air is transferred to the tube adiabatically. What is the work, done when the pressure in the tube is increased from P1 to P2?, , 12.20, , In a refrigerator one removes heat from a lower temperature, and deposits to the surroundings at a higher temperature. In, this process, mechanical work has to be done, which is provided, by an electric motor. If the motor is of 1kW power, and heat is, transferred from –3°C to 27°C, find the heat taken out of the, refrigerator per second assuming its efficiency is 50% of a, perfect engine., , 87, 20/04/2018
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Exemplar Problems–Physics, 12.21, , If the co-efficient of performance of a refrigerator is 5 and operates, at the room temperature (27 °C), find the temperature inside the, refrigerator., , 12.22, , The initial state of a certain gas is (Pi, Vi, Ti). It undergoes, expansion till its volume becoms Vf . Consider the following two, cases:, (a) the expansion takes place at constant temperature., (b) the expansion takes place at constant pressure., Plot the P-V diagram for each case. In which of the two cases, is, the work done by the gas more?, , LA, P, , 12.23, 1(P1, V1, T1), , Consider a P-V diagram in which the path followed by one mole, of perfect gas in a cylindrical container is shown in Fig. 12.9., , 1/2, , PV = constant (a) Find the work done when the gas is taken from state 1 to, 2(P2, V2, T2), , state 2., (b) What is the ratio of temperature T1/T2, if V2 = 2V1?, , V1, , (c) Given the internal energy for one mole of gas at temperature T, is (3/2) RT, find the heat supplied to the gas when it is taken, from state 1 to 2, with V2 = 2V1., , V2 V, Fig. 12.9, , 12.24, , A to B : volume constant, B to C : adiabatic, C to D : volume constant, D to A : adiabatic, , P, C, , B, , A cycle followed by an engine (made of one mole of perfect gas in, a cylinder with a piston) is shown in Fig. 12.10., , VC = VD = 2VA = 2VB, A, , VA=VB, , D, , VC=VD, Fig. 12.10, , V, , (a) In which part of the cycle heat is supplied to the engine from, outside?, (b) In which part of the cycle heat is being given to the, surrounding by the engine?, (c) What is the work done by the engine in one cycle? Write your, answer in term of PA, PB, VA., (d) What is the efficiency of the engine?, 3, [ γ = 5 3 for the gas], ( Cv = R for one mole), 2, , 88, 20/04/2018
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Thermodynamics, , 12.25, , A cycle followed by an engine (made of one mole of an ideal gas, in a cylinder with a piston) is shown in Fig. 12.11. Find heat, exchanged by the engine, with the surroundings for each section, of the cycle. (Cv = (3/2) R), AB, BC, CD, DA, , :, :, :, :, , constant volume, constant pressure, adiabatic, constant pressure, , 12.26, , Consider that an ideal gas (n moles) is expanding in a process, given by P = f ( V ), which passes through a point (V0, P0 ). Show, that the gas is absorbing heat at (P0, V0 ) if the slope of the curve, P = f (V ) is larger than the slope of the adiabat passing, through (P0, V0 )., , 12.27, , Consider one mole of perfect gas in a cylinder of unit cross section, with a piston attached (Fig. 12.12). A spring (spring constant k), is attached (unstretched length L ) to the piston and to the bottom, of the cylinder. Initially the spring is unstretched and the gas is, in equilibrium. A certain amount of heat Q is supplied to the gas, causing an increase of volume from Vo to V1., (a) What is the initial pressure of the system?, (b) What is the final pressure of the system?, (c) Using the first law of thermodynamics, write down a relation, between Q, Pa, V, Vo and k., , Fig. 12.11, , Fig. 12.12, , 89, 20/04/2018
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Exemplar Problems–Physics, , Chapter Thirteen, , KINETIC THEORY, , MCQ I, 13.1, , A cubic vessel (with faces horizontal + vertical) contains an ideal, gas at NTP. The vessel is being carried by a rocket which is moving, at a speed of 500m s–1 in vertical direction. The pressure of the, gas inside the vessel as observed by us on the ground, (a) remains the same because 500m s−1 is very much smaller, than vrms of the gas., (b) remains the same because motion of the vessel as a whole, does not affect the relative motion of the gas molecules and, the walls., (c) will increase by a factor equal to ( v 2rms + (500)2 ) / v, , 2, rms, , where, , vrms was the original mean square velocity of the gas., (d) will be different on the top wall and bottom wall of the vessel., 13.2, , 1 mole of an ideal gas is contained in a cubical volume V,, ABCDEFGH at 300 K (Fig. 13.1). One face of the cube (EFGH) is, made up of a material which totally absorbs any gas molecule, , 90, 20/04/2018
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Kinetic Theory, , incident on it. At any given time,, (a) the pressure on EFGH would be zero., (b) the pressure on all the faces will the equal., (c) the pressure of EFGH would be double the pressure, on ABCD., (d) the pressure on EFGH would be half that on ABCD., 13.3, , Boyle’s law is applicable for an, (a), (b), (c), (d), , 13.4, , Fig. 13.1, , adiabatic process., isothermal process., isobaric process., isochoric process., , A cylinder containing an ideal gas is in vertical position and has, a piston of mass M that is able to move up or down without friction, (Fig. 13.2). If the temperature is increased,, , Fig. 13.2, , (a) both p and V of the gas will change., (b) only p will increase according to Charle’s law., (c) V will change but not p., (d) p will change but not V., 13.5, , Volume versus temperature graphs for a given mass of an V, ideal gas are shown in Fig. 13.3 at two different values of (l), constant pressure. What can be inferred about relation 40, 30, between P 1 & P 2?, (a) P1 > P2, (b) P1 = P2, (c) P1 < P2, (d) data is insufficient., , P2, , P1, , 20, 10, , 100 200 300 400500, T (K), Fig.13.3, , 91, 20/04/2018
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Exemplar Problems–Physics, 13.6, , 1 mole of H2 gas is contained in a box of volume V = 1.00 m3 at, T = 300K. The gas is heated to a temperature of T = 3000K and, the gas gets converted to a gas of hydrogen atoms. The final, pressure would be (considering all gases to be ideal), (a), (b), (c), (d), , 13.7, , same as the pressure initially., 2 times the pressure initially., 10 times the pressure initially., 20 times the pressure initially., , A vessel of volume V contains a mixture of 1 mole of Hydrogen, and 1 mole of Oxygen (both considered as ideal). Let f1(v)dv, denote, the fraction of molecules with speed between v and (v + dv) with, f2 (v)dv, similarly for oxygen. Then, (a) f1(v ) + f 2 (v ) = f (v ) obeys the Maxwell’s distribution law., (b) f1(v), f2 (v) will obey the Maxwell’s distribution law separately., (c) Neither f1 (v), nor f2 (v) will obey the Maxwell’s distribution, law., (d) f2 (v) and f1 (v) will be the same., , 13.8, , An inflated rubber balloon contains one mole of an ideal gas,, has a pressure p, volume V and temperature T. If the temperature, rises to 1.1 T, and the volume is increaset to 1.05 V, the final, pressure will be, (a), (b), (c), (d), , 1.1 p, p, less than p, between p and 1.1., , MCQ II, 13.9, , ABCDEFGH is a hollow cube made of an insulator (Fig. 13.4)., Face ABCD has positve charge on it. Inside the cube, we have, ionized hydrogen., The usual kinetic theory expression for pressure, (a) will be valid., (b) will not be valid since the ions would experience forces other, than due to collisions with the walls., (c) will not be valid since collisions with walls would not be elastic., (d) will not be valid because isotropy is lost., , Fig. 13.4, , 13.10, , Diatomic molecules like hydrogen have energies due to both, translational as well as rotational motion. From the equation in, kinetic theory pV =, , 2, E , E is, 3, , 92, 20/04/2018
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Kinetic Theory, , (a) the total energy per unit volume., (b) only the translational part of energy because rotational energy, is very small compared to the translational energy., (c) only the translational part of the energy because during, collisions with the wall pressure relates to change in linear, momentum., (d) the translational part of the energy because rotational energies, of molecules can be of either sign and its average over all the, molecules is zero., 13.11, , In a diatomic molecule, the rotational energy at a given, temperature, (a), (b), (c), (d), , 13.12, , obeys Maxwell’s distribution., have the same value for all molecules., equals the translational kinetic energy for each molecule., is (2/3)rd the translational kinetic energy for each molecule., , Which of the following diagrams (Fig. 13.5) depicts ideal gas, behaviour?, , P = const, , P, , T = const, , V, , T, , V, , (b), , (a), , P, , PV, , V = const, , T, , T, , (c), , (d), Fig. 13.5, , 13.13, , When an ideal gas is compressed adiabatically, its temperature, rises: the molecules on the average have more kinetic energy than, before. The kinetic energy increases,, (a) because of collisions with moving parts of the wall only., (b) because of collisions with the entire wall., , 93, 20/04/2018
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Exemplar Problems–Physics, (c) because the molecules gets accelerated in their motion inside, the volume., (d) because of redistribution of energy amongst the molecules., , VSA, 13.14, , Calculate the number of atoms in 39.4 g gold. Molar mass of, gold is 197g mole–1., , 13.15, , The volume of a given mass of a gas at 27°C, 1 atm is 100 cc., What will be its volume at 327°C?, , 13.16, , The molecules of a given mass of a gas have root mean square, speeds of 100 m s −1 at 27°C and 1.00 atmospheric pressure., What will be the root mean square speeds of the molecules of the, gas at 127°C and 2.0 atmospheric pressure?, , 13.17, , Two molecules of a gas have speeds of 9 × 10 6 m s −1 and, , 1 × 106 m s−1 , respectively. What is the root mean square speed of, these molecules., 13.18, , A gas mixture consists of 2.0 moles of oxygen and 4.0 moles of, neon at temperature T. Neglecting all vibrational modes, calculate, the total internal energy of the system. (Oxygen has two rotational, modes.), , 13.19, , Calculate the ratio of the mean free paths of the molecules of two, gases having molecular diameters 1 A and 2 A . The gases may, be considered under identical conditions of temperature, pressure, and volume., , SA, 13.20, V1, µ1, p1, , V2, µ 2, p2, , The container shown in Fig. 13.6 has two chambers, separated, by a partition, of volumes V1 = 2.0 litre and V2= 3.0 litre. The, chambers contain µ1 = 4.0 and µ2 = 5.0 moles of a gas at, pressures p1 = 1.00 atm and p2 = 2.00 atm. Calculate the pressure, after the partition is removed and the mixture attains equilibrium., , ,, , Fig 13.6, , 13.21, , A gas mixture consists of molecules of types A, B and C with, masses m A > m B > m C . Rank the three types of molecules in, decreasing order of (a) average K.E., (b) rms speeds., , 94, 20/04/2018
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Kinetic Theory, , 13.22, , We have 0.5 g of hydrogen gas in a cubic chamber of size 3cm, kept at NTP. The gas in the chamber is compressed keeping the, temperature constant till a final pressure of 100 atm. Is one, justified in assuming the ideal gas law, in the final state?, o, , (Hydrogen molecules can be consider as spheres of radius 1 A )., 13.23, , When air is pumped into a cycle tyre the volume and pressure of, the air in the tyre both are increased. What about Boyle’s law in, this case?, , 13.24, , A ballon has 5.0 g mole of helium at 7°C. Calculate, (a) the number of atoms of helium in the balloon,, (b) the total internal energy of the system., , 13.25, , Calculate the number of degrees of freedom of molecules of, hydrogen in 1 cc of hydrogen gas at NTP., , 13.26, , An insulated container containing monoatomic gas of molar mass, m is moving with a velocity vo . If the container is suddenly, stopped, find the change in temperature., , LA, 13.27, , Explain why, (a) there is no atmosphere on moon., (b) there is fall in temperature with altitude., , 13.28, , Consider an ideal gas with following distribution of speeds., Speed (m/s), , % of molecules, , 200, , 10, , 400, , 20, , 600, , 40, , 800, , 20, , 1000, , 10, , (i) Calculate Vrms and hence T. (m = 3.0 × 10−26 kg), (ii) If all the molecules with speed 1000 m/s escape from the, system, calculate new Vrms and hence T., , 95, 20/04/2018
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Exemplar Problems–Physics, 13.29, , Ten small planes are flying at a speed of 150 km/h in total, darkness in an air space that is 20 × 20 × 1.5 km3 in volume. You, are in one of the planes, flying at random within this space with, no way of knowing where the other planes are. On the average, about how long a time will elapse between near collision with, your plane. Assume for this rough computation that a saftey, region around the plane can be approximated by a sphere of, radius 10m., , 13.30, , A box of 1.00m3 is filled with nitrogen at 1.50 atm at 300K. The, box has a hole of an area 0.010 mm2. How much time is required, for the pressure to reduce by 0.10 atm, if the pressure outside is, 1 atm., , 13.31, , Consider a rectangular block of wood moving with a velocity v0, in a gas at temperature T and mass density ρ. Assume the velocity, is along x-axis and the area of cross-section of the block, perpendicular to v0 is A. Show that the drag force on the block is, kT, 4ρ Av0, , where m is the mass of the gas molecule., m, , 96, 20/04/2018
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Chapter Fourteen, , OSCILLATIONS, , MCQ I, 14.1, , The displacement of a particle is represented by the equation, , π, , y = 3 cos − 2ωt ., 4, , The motion of the particle is, (a), (b), (c), (d), 14.2, , simple harmonic with period 2p/w., simple harmonic with period π/ω., periodic but not simple harmonic., non-periodic., , The displacement of a particle is represented by the equation, , y = sin3 ωt . The motion is, (a), (b), (c), (d), , non-periodic., periodic but not simple harmonic., simple harmonic with period 2π/ω., simple harmonic with period π/ω., , 20/04/2018
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Exemplar Problems–Physics, 14.3, , The relation between acceleration and displacement of four, particles are given below:, (a), (b), (c), (d), , ax = + 2x., ax = + 2x2., ax = – 2x2., ax = – 2x., , Which one of the particles is executing simple harmonic motion?, 14.4, , Motion of an oscillating liquid column in a U-tube is, (a) periodic but not simple harmonic., (b) non-periodic., (c) simple harmonic and time period is independent of the density, of the liquid., (d) simple harmonic and time-period is directly proportional to, the density of the liquid., , 14.5, , A particle is acted simultaneously by mutually perpendicular, simple hormonic motions x = a cos ωt and y = a sin ωt . The, trajectory of motion of the particle will be, (a), (b), (c), (d), , 14.6, , an ellipse., a parabola., a circle., a straight line., , The displacement of a particle varies with time according to the, relation, y = a sin ωt + b cos ωt., (a) The motion is oscillatory but not S.H.M., (b) The motion is S.H.M. with amplitude a + b., (c) The motion is S.H.M. with amplitude a2 + b2., (d) The motion is S.H.M. with amplitude, , 14.7, , 2, 2, a +b ., , Four pendulums A, B, C and D are suspended from the same, , Fig. 14.1, , 98, 20/04/2018
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Oscillations, , elastic support as shown in Fig. 14.1. A and C are of the same, length, while B is smaller than A and D is larger than A. If A is, given a transverse displacement,, (a), (b), (c), (d), 14.8, , D will vibrate with maximum amplitude., C will vibrate with maximum amplitude., B will vibrate with maximum amplitude., All the four will oscillate with equal amplitude., , Figure 14.2. shows the circular motion of a particle. The radius, of the circle, the period, sense of revolution and the initial position, are indicated on the figure. The simple harmonic motion of the, x-projection of the radius vector of the rotating particle P is, 2π t , ., (a) x (t) = B sin , 30 , πt , (b) x (t) = B cos , ., 15 , , y, P (t = 0), T = 30s, B, o, , x, , πt π , + ., (c) x (t) = B sin , 15 2 , πt π , + ., (d) x (t) = B cos , 15 2 , , 14.9, , Fig. 14.2, , The equation of motion of a particle is x = a cos (α t )2., The motion is, (a), (b), (c), (d), , periodic but not oscillatory., periodic and oscillatory., oscillatory but not periodic., neither periodic nor oscillatory., , 14.10 A particle executing S.H.M. has a maximum speed of 30 cm/s, and a maximum acceleration of 60 cm/s2. The period of, oscillation is, (a) π s., (b), , π, , s., 2, (c) 2π s., (d), , π, t, , s., , 14.11 When a mass m is connected individually to two springs S1 and, S2, the oscillation frequencies are ν 1 and ν 2. If the same mass is, , 99, 20/04/2018
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Exemplar Problems–Physics, attached to the two springs as shown in Fig. 14.3, the oscillation, frequency would be, (a) ν 1 + ν 2., (b), , 2, , 2, , ν1 + ν 2 ., −1, , 1, 1 , ., (c) +, ν1 ν 2 , Fig. 14.3, , (d), , 2, 2, ν1 − ν 2 ., , MCQ II, 14.12 The rotation of earth about its axis is, (a), (b), (c), (d), , periodic motion., simple harmonic motion., periodic but not simple harmonic motion., non-periodic motion., , 14.13 Motion of a ball bearing inside a smooth curved bowl, when, released from a point slightly above the lower point is, (a), (b), (c), (d), , simple harmonic motion., non-periodic motion., periodic motion., periodic but not S.H.M., , displacement, , 14.14 Displacement vs. time curve for a particle executing S.H.M. is, shown in Fig. 14.4. Choose the correct statements., , 0, , 1, , 2, , 3, , 4, , 5, , Fig. 14.4, , 6, , 7 time (s), , (a) Phase of the oscillator is same at t = 0 s and, t = 2 s., (b) Phase of the oscillator is same at t = 2 s and, t = 6 s., (c) Phase of the oscillator is same at t = 1 s and, t = 7 s., (d) Phase of the oscillator is same at t = 1 s and, t = 5 s., , 14.15 Which of the following statements is/are true for a simple, harmonic oscillator?, (a) Force acting is directly proportional to displacement from the, mean position and opposite to it., , 100, 20/04/2018
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Oscillations, , (b) Motion is periodic., (c) Acceleration of the oscillator is constant., (d) The velocity is periodic., , (a) The force is zero at t =, , 3T, ., 4, , (b) The acceleration is maximum at t =, (c) The velocity is maximum at t =, , 4T, ., 4, , T, ., 4, , (d) The P.E. is equal to K.E. of oscillation at t =, 14.17, , displacement, , 14.16 The displacement time graph of a particle executing S.H.M. is, shown in Fig. 14.5. Which of the following statement is/are true?, , 0, , 2T/4, T/4, , 3T/4 T, , 5T/4 time (s), , Fig.14.5, , T, ., 2, , A body is performing S.H.M. Then its, (a) average total energy per cycle is equal to its maximum kinetic, energy., (b) average kinetic energy per cycle is equal to half of its, maximum kinetic energy., (c) mean velocity over a complete cycle is equal to, , 2, , π, , times of its, , maximum velocity., (d) root mean square velocity is, , 1, times of its maximum velocity., 2, , 14.18 A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart (Fig. 14.6). Take the direction from A to B, as the + ve direction and choose the correct statements., , Fig. 14.6, , (a) The sign of velocity, acceleration and force on the particle when, it is 3 cm away from A going towards B are positive., (b) The sign of velocity of the particle at C going towards O is, negative., (c) The sign of velocity, acceleration and force on the particle when, it is 4 cm away from B going towards A are negative., (d) The sign of acceleration and force on the particle when it is at, point B is negative., , 101, 20/04/2018
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Exemplar Problems–Physics, , VSA, 14.19 Displacement versus time curve for a particle executing S.H.M., is shown in Fig. 14.7. Identify the points marked at which (i), velocity of the oscillator is zero, (ii) speed of the oscillator is, maximum., , Fig. 14.7, , 14.20 Two identical springs of spring constant K are attached to a block, of mass m and to fixed supports as shown in Fig. 14.8. When the, mass is displaced from equilllibrium position by a distance x, towards right, find the restoring force, , Fig. 14.8, , 14.21 What are the two basic characteristics of a simple harmonic, motion?, 14.22 When will the motion of a simple pendulum be simple harmonic?, 14.23 What is the ratio of maxmimum acceleration to the maximum, velocity of a simple harmonic oscillator?, 14.24 What is the ratio between the distance travelled by the oscillator, y, in one time period and amplitude?, A , , 14.25 In Fig. 14.9, what will be the sign of, the velocity of the point P′ , which is, the projection of the velocity of the, reference particle P . P is moving in, a circle of radius R in anticlockwise, direction., , P, , , , t+, o, , P1, , Fig. 14.9, , 102, 20/04/2018, , A, , x
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Oscillations, , 14.26 Show that for a particle executing S.H.M, velocity and, displacement have a phase difference of π/2., 14.27 Draw a graph to show the variation of P.E., K.E. and total energy, of a simple harmonic oscillator with displacement., 14.28 The length of a second’s pendulum on the surface of Earth is, 1m. What will be the length of a second’s pendulum on the moon?, , SA, , Inextensible, string, , 14.29 Find the time period of mass M when displaced from its, equilibrium positon and then released for the system shown in, Fig 14.10., 14.30 Show that the motion of a particle represented by, y = sin ω t – cos ω t is simple harmonic with a period of 2π/ω., , Fig. 14.10, , 14.31 Find the displacement of a simple harmonic oscillator at which, its P.E. is half of the maximum energy of the oscillator., 14.32, , A body of mass m is situated in a potential field U(x) = U0 (1-cos αx ), when U0 and α are constants. Find the time period of small, oscillations., , 14.33 A mass of 2 kg is attached to the spring of spring constant, 50 Nm–1. The block is pulled to a distance of 5cm from its, equilibrium position at x = 0 on a horizontal frictionless surface, from rest at t = 0. Write the expression for its displacement at, anytime t., 14.34 Consider a pair of identical pendulums, which oscillate with equal, amplitude independently such that when one pendulum is at, its extreme position making an angle of 2° to the right with the, vertical, the other pendulum makes an angle of 1° to the left of, the vertical. What is the phase difference between the pendulums?, , LA, 14.35 A person normally weighing 50 kg stands on a massless platform, which oscillates up and down harmonically at a frequency of, 2.0 s–1 and an amplitude 5.0 cm. A weighing machine on the, platform gives the persons weight against time., (a) Will there be any change in weight of the body, during the, oscillation?, , 103, 20/04/2018
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Exemplar Problems–Physics, (b) If answer to part (a) is yes, what will be the maximum and, minimum reading in the machine and at which position?, 14.36 A body of mass m is attached to one end of a massless spring, which is suspended vertically from a fixed point. The mass is, held in hand so that the spring is neither stretched nor, compressed. Suddenly the support of the hand is removed. The, lowest position attained by the mass during oscillation is 4cm, below the point, where it was held in hand., (a) What is the amplitude of oscillation?, (b) Find the frequency of oscillation?, 14.37 A cylindrical log of wood of height h and area of cross-section A, floats in water. It is pressed and then released. Show that the log, would execute S.H.M. with a time period., , T = 2π, , m, Aρ g, , where m is mass of the body and ρ is density of the liquid., 14.38 One end of a V-tube containing mercury is connected to a suction, pump and the other end to atmosphere. The two arms of the, tube are inclined to horizontal at an angle of 45° each. A small, pressure difference is created between two columns when the, suction pump is removed. Will the column of mercury in V-tube, execute simple harmonic motion? Neglect capillary and viscous, forces.Find the time period of oscillation., 14.39 A tunnel is dug through the centre of the Earth. Show that a, body of mass ‘m’ when dropped from rest from one end of the, tunnel will execute simple harmonic motion., 14.40 A simple pendulum of time, period 1s and length l is hung, from a fixed support at O,, such that the bob is at a, distance H vertically above A, on the ground (Fig. 14.11)., The amplitude is θo . The, string snaps at θ = θ0 /2. Find, the time taken by the bob to, hit the ground. Also find, distance from A where bob, hits the ground. Assume θ0, to be small so that, , sin θ0, , θ0 and cos θ0, , 1., , , OP=l, , P, H, A, , Fig. 14.11, , 104, 20/04/2018
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Chapter Fifteen, , WAVES, , MCQ I, 15.1, , Water waves produced by a motor boat sailing in water are, (a), (b), (c), (d), , 15.2, , neither longitudinal nor transverse., both longitudinal and transverse., only longitudinal., only transverse., , Sound waves of wavelength λ travelling in a medium with a speed, of v m/s enter into another medium where its speed is 2v m/s., Wavelength of sound waves in the second medium is, (a) λ, , λ, (b), , 2, , (c) 2λ, (d) 4 λ, , 20/04/2018
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Exemplar Problems–Physics, 15.3, , Speed of sound wave in air, (a), (b), (c), (d), , 15.4, , Change in temperature of the medium changes, (a), (b), (c), (d), , 15.5, , frequency of sound waves., amplitude of sound waves., wavelength of sound waves., loudness of sound waves., , With propagation of longitudinal waves through a medium, the, quantity transmitted is, (a), (b), (c), (d), , 15.6, , is independent of temperature., increases with pressure., increases with increase in humidity., decreases with increase in humidity., , matter., energy., energy and matter., energy, matter and momentum., , Which of the following statements are true for wave motion?, (a) Mechanical transverse waves can propagate through all, mediums., (b) Longitudinal waves can propagate through solids only., (c) Mechanical transverse waves can propagate through solids, only., (d) Longitudinal waves can propagate through vacuum., , 15.7, , A sound wave is passing through air column in the form of, compression and rarefaction. In consecutive compressions and, rarefactions,, (a), (b), (c), (d), , 15.8, , density remains constant., Boyle’s law is obeyed., bulk modulus of air oscillates., there is no transfer of heat., , Equation of a plane progressive wave is given by, , x, y = 0.6 sin 2π t − . On reflection from a denser medium its, , 2, amplitude becomes 2/3 of the amplitude of the incident wave., The equation of the reflected wave is, x, , (a) y = 0.6 sin 2π t + , , 2, , 106, 20/04/2018
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Waves, , x, , (b) y = − 0.4 sin 2π t + , , 2, x, , (c) y = 0.4 sin 2π t + , , 2, x, , (d) y = −0.4 sin 2π t − ., , 2, 15.9, , A string of mass 2.5 kg is under a tension of 200 N. The length of, the stretched string is 20.0 m. If the transverse jerk is struck at, one end of the string, the disturbance will reach the other end in, (a), (b), (c), (d), , 15.10, , one second, 0.5 second, 2 seconds, data given is insufficient., , A train whistling at constant frequency is moving towards a, station at a constant speed V. The train goes past a stationary, observer on the station. The frequency n ′ of the sound as heard, by the observer is plotted as a function of time t (Fig 15.1) . Identify, the expected curve., n, , n, , t, t, , (b), , (a), , n, n, , t, , t, , (c), , (d), Fig 15.1, , 107, 20/04/2018
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Exemplar Problems–Physics, , MCQ II, 15.11, , A transverse harmonic wave on a string is described by y (x,t) =, 3.0 sin (36t + 0.018x + π/4), where x and y are in cm and t is in s. The positive direction of x is, from left to right., (a), (b), (c), (d), , 15.12, , The wave is travelling from right to left., The speed of the wave is 20m/s., Frequency of the wave is 5.7 Hz., The least distance between two successive crests in the wave, is 2.5 cm., , The displacement of a string is given by, y (x,t) = 0.06 sin (2πx/3) cos (120πt), where x and y are in m and t in s. The length of the string is 1.5m, and its mass is 3.0 × 10−2 kg ., (a) It represents a progressive wave of frequency 60Hz., (b) It represents a stationary wave of frequency 60Hz., (c) It is the result of superposition of two waves of wavelength 3, m, frequency 60Hz each travelling with a speed of 180 m/s, in opposite direction., (d) Amplitude of this wave is constant., , 15.13, , Speed of sound waves in a fluid depends upon, (a), (b), (c), (d), , 15.14, , During propagation of a plane progressive mechanical wave, (a), (b), (c), (d), , 15.15, , directty on density of the medium., square of Bulk modulus of the medium., inversly on the square root of density., directly on the square root of bulk modulus of the medium., , all the particles are vibrating in the same phase., amplitude of all the particles is equal., particles of the medium executes S.H.M., wave velocity depends upon the nature of the medium., , The transverse displacement of a string (clamped at its both ends), is given by y (x,t) = 0.06 sin (2πx/3) cos (120πt)., All the points on the string between two consecutive nodes, vibrate with, (a), (b), (c), (d), , same frequency, same phase, same energy, different amplitude., , 108, 20/04/2018
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Waves, , 15.16, , A train, standing in a station yard, blows a whistle of frequency, 400 Hz in still air. The wind starts blowing in the direction from, the yard to the station with a speed of 10m/s. Given that the, speed of sound in still air is 340m/s,, (a) the frequency of sound as heard by an observer standing on, the platform is 400Hz., (b) the speed of sound for the observer standing on the platform, is 350m/s., (c) the frequency of sound as heard by the observer standing, on the platform will increase., (d) the frequency of sound as heard by the observer standing, on the platform will decrease., , 15.17, , Which of the following statements are true for a stationary wave?, (a) Every particle has a fixed amplitude which is different from, the amplitude of its nearest particle., (b) All the particles cross their mean position at the same time., (c) All the particles are oscillating with same amplitude., (d) There is no net transfer of energy across any plane., (e) There are some particles which are always at rest., , VSA, 15.18, , A sonometer wire is vibrating in resonance with a tuning fork., Keeping the tension applied same, the length of the wire is, doubled. Under what conditions would the tuning fork still be is, resonance with the wire?, , 15.19, , An organ pipe of length L open at both ends is found to vibrate, in its first harmonic when sounded with a tuning fork of 480 Hz., What should be the length of a pipe closed at one end, so that it, also vibrates in its first harmonic with the same tuning fork?, , 15.20, , A tuning fork A, marked 512 Hz, produces 5 beats per second,, where sounded with another unmarked tuning fork B. If B is, loaded with wax the number of beats is again 5 per second. What, is the frequency of the tuning fork B when not loaded?, , 15.21, , The displacement of an elastic wave is given by the function, y = 3 sin ωt + 4 cos ωt., where y is in cm and t is in second. Calculate the resultant, amplitude., , 15.22, , A sitar wire is replaced by another wire of same length and, material but of three times the earlier radius. If the tension in the, wire remains the same, by what factor will the frequency change?, , 109, 20/04/2018
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Exemplar Problems–Physics, 15.23, , At what temperatures (in oC) will the speed of sound in air be, 3 times its value at OoC?, , 15.24, , When two waves of almost equal frequencies n 1 and n 2 reach at, a point simultaneously, what is the time interval between, successive maxima?, , SA, 15.25, , A steel wire has a length of 12 m and a mass of 2.10 kg. What will, be the speed of a transverse wave on this wire when a tension of, 2.06 × 104N is applied?, , 15.26, , A pipe 20 cm long is closed at one end. Which harmonic mode of, the pipe is resonantly excited by a source of 1237.5 Hz ?(sound, velocity in air = 330 m s–1), , 15.27, , A train standing at the outer signal of a railway station blows a, whistle of frequency 400 Hz still air. The train begins to move, with a speed of 10 m s-1 towards the platform. What is the, frequency of the sound for an observer standing on the platform?, (sound velocity in air = 330 m s–1), , 15.28, , The wave pattern on a stretched string is shown in, Fig. 15.2. Interpret what kind of wave this is and find its, wavelength., , Fig. 15.2, , 110, 20/04/2018
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Waves, , 15.29, , The pattern of standing waves formed on a stretched string at, two instants of time are shown in Fig. 15.3. The velocity of two, waves superimposing to form stationary waves is 360 ms–1 and, their frequencies are 256 Hz., , Fig. 15.3, , (a) Calculate the time at which the second curve is plotted., (b) Mark nodes and antinodes on the curve., (c) Calculate the distance between A ′ and C ′ ., 15.30, , A tuning fork vibrating with a frequency of 512Hz is, kept close to the open end of a tube filled with water, (Fig. 15.4). The water level in the tube is gradually, lowered. When the water level is 17cm below the open, end, maximum intensity of sound is heard. If the room, temperature is 20°C, calculate, (a) speed of sound in air at room temperature, (b) speed of sound in air at 0°C, (c) if the water in the tube is replaced with mercury,, will there be any difference in your observations?, , 15.31, , Fig. 15.4, , Show that when a string fixed at its two ends vibrates in 1 loop,, 2 loops, 3 loops and 4 loops, the frequencies are in the ratio, 1:2:3:4., , LA, 15.32, , The earth has a radius of 6400 km. The inner core of 1000 km, radius is solid. Outside it, there is a region from 1000 km to a, radius of 3500 km which is in molten state. Then again from, 3500 km to 6400 km the earth is solid. Only longitudinal (P), waves can travel inside a liquid. Assume that the P wave has a, speed of 8 km s–1 in solid parts and of 5 km s–1 in liquid parts of, the earth. An earthquake occurs at some place close to the surface, of the earth. Calculate the time after which it will be recorded in a, seismometer at a diametrically opposite point on the earth if wave, travels along diameter?, , 111, 20/04/2018
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Exemplar Problems–Physics, 15.33, , If c is r.m.s. speed of molecules in a gas and v is the speed of, sound waves in the gas, show that c/v is constant and independent, of temperature for all diatomic gases., , 15.34, , Given below are some functions of x and t to represent the, displacement of an elastic wave., (a), (b), (c), (d), , y = 5 cos (4x ) sin (20t ), y = 4 sin (5x – t/2) + 3 cos (5x – t/2), y = 10 cos [(252 – 250) πt ] cos [(252+250)πt ], y = 100 cos (100πt + 0.5x ), , State which of these represent, (a) a travelling wave along –x direction, (b) a stationary wave, (c) beats, (d) a travelling wave along +x direction., Given reasons for your answers., , 15.35, , In the given progressive wave, y = 5 sin (100πt – 0.4πx ), where y and x are in m, t is in s. What is the, (a), (b), (c), (d), (e), , 15.36, , amplitude, wave length, frequency, wave velocity, particle velocity amplitude., , For the harmonic travelling wave y = 2 cos 2π (10t–0.0080x + 3.5), where x and y are in cm and t is second. What is the phase, difference between the oscillatory motion at two points separated, by a distance of, (a) 4 m, (b) 0.5 m, , λ, (c), , 2, , 3λ, (at a given instant of time), 4, (e) What is the phase difference between the oscillation of a, particle located at x = 100cm, at t = T s and t = 5 s?, , (d), , 112, 20/04/2018
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ANSWERS, , Chapter, , 2, , 2.1, , (b), , 2.2, , (b), , 2.3, , (c), , 2.4, , (d), , 2.5, , (a), , 2.6, , (c), , 2.7, , (a), , 2.8, , (d), , 2.9, , (a), , 2.10, , (a), , 2.11, , (c), , 2.12, , (d), , 2.13, , (b), (c), , 2.14, , (a), (e), , 2.15, , (b), (d), , 2.16, , (a), (b), (d), , 2.17, , (a), (b), , 2.18, , (b), (d), , 2.19, , Because, bodies differ in order of magnitude significantly in respect, to the same physical quantity. For example, interatomic distances, are of the order of angstroms, inter-city distances are of the order of, km, and interstellar distances are of the order of light year., , 20/04/2018
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Exemplar Problems–Physics, 2.20, , 1015, , 2.21, , Mass spectrograph, , 2.22, , 1 u = 1.67 × 10–27 kg, , 2.23, , Since f (θ ) is a sum of different powers of θ, it has to be, dimensionless, , 2.24, , Because all other quantities of mechanics can be expressed in, terms of length, mass and time through simple relations., , 2.25, , (a ) θ =, , RE, 1, =, rad 1o, 60R E 60, , ∴ Diameter of the earth as seen from the moon is about 2°., (b) At earth-moon distance, moon is seen as (1/2)° diameter and, earth is seen as 2° diameter. Hence, diameter of earth is 4 times, the diameter of moon., , Dearth, =4, Dmoon, , (c), , rsun, = 400, rmoon, , (Here r stands for distance, and D for diameter.), Sun and moon both appear to be of the same angular diameter as, seen from the earth., , ∴, , Dsun Dmoon, =, rsun, rmoon, , ∴, , Dsun, = 400, Dmoon, , But, , D earth, D, = 4 ∴ sun = 100 ., Dmoon, Dearth, , 2.26, , An atomic clock is the most precise time measuring device because, atomic oscillations are repeated with a precision of 1s in 1013 s., , 2.27, , 3 × 1016 s, , 2.28, , 0.01 mm, , 114, 20/04/2018
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Answers, , 2.29, , θ = ( π Rs 2 / R e s 2 )( π Rm 2 / R em 2 ), , , Rs, R, = es, Rm Rem, , 2.30, , 105 kg, , 2.31, , (a), (b), (c), (d), , 2.32, , θ=, , 2.33, , 4 × 10, , 2.34, , Dimensional formula of ω =T, –1, Dimensional formula of k = L, , 2.35, , (a), , Angle or solid angle, Relative density, etc., Planck’s constant, universal gravitational constant, etc., Raynold number, , l, 3.14, l = r θ l = 31 ×, cm = 16.3 cm, r, 6, – 2, , steradian, –1, , Precision is given by the least count of the instrument., For 20 oscillations, precision = 0.1 s, For 1 oscillation, precision = 0.005 s., , (b), , Average time t =, , Period =, , 39.6 + 39.9 + 39.5, s = 39.6s, 3, , 39.6, = 1.98 s, 20, , Max. observed error = (1.995 –1.980)s = 0.015s., 2.36, , 2, , –2, , Since energy has dimensions of ML T , 1J in new units becomes, , γ 2 / αβ2 J. Hence 5 J becomes 5γ 2 / αβ2 ., 2.37, , ρr 4, . Therefore, the, ηl, dimensions of the right hand side comes out to be, , The dimensional part in the expression is, , [ML–1 T –2 ][L4 ] [L3 ], =, , which is volume upon time. Hence, the, [T], [ML–1 T –1 ][L], formula is dimensionally correct., 2.38, , The fractional error in X is, , dX 2da 3db 2.5dc 2d(d), =, +, +, +, X, a, b, c, d, = 0.235, , 0.24, , 115, 20/04/2018
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Exemplar Problems–Physics, , Since the error is in first decimal, hence the result should be, rounded off as 2.8., 2.39, , Since E, l and G have dimensional formulas:, , E → ML2 T –2, l → ML2 T –1, G → L3 M –1 T –2, 2, , Hence, P = E l m–5 G–2 will have dimensions:, , [ 2 –2 ][M 2 L4 T –2 ][ M2 T 4 ], [ P ] = ML T, [ M5 ][L6 ], = M0 L0 T 0, Thus, P is dimensionless., 2.40, , M, L, T, in terms of new units become, , M→, 2.41, , ch, ,L→, G, , Given T 2 α r 3, , hG, ,T →, c3, , hG, c5, , T α r 3 / 2 . T is also function of g and, , R T α gx Ry, x, , ∴ [ Lo Mo T1 ] = [ L3/2 Mo T o ][ L1Mo T -2 ] [ L1Mo T o ], 3, For L, 0 = + x + y, 2, 1, For T, 1 = 0 − 2x x = −, 2, 3 1, Therefore, 0 = − + y y = −1, 2 2, k r3, Thus, T = k r 3 / 2 g −1/ 2 R −1 =, R g, 2.42, , y, , (a) Because oleic acid dissolves in alcohol but does not disssolve, in water., (b) When lycopodium powder is spread on water, it spreads on the, entire surface. When a drop of the prepared solution is dropped, on water, oleic acid does not dissolve in water, it spreads on, the water surface pushing the lycopodium powder away to clear, a circular area where the drop falls. This allows measuring, the area where oleic acid spreads., , 116, 20/04/2018
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Answers, , (c), , 1, 1, 1, =, mL ×, mL, 20, 20 400, , (d) By means of a burette and measuring cylinder and measuring, the number of drops., (e) If n drops of the solution make 1 mL, the volume of oleic acid, in one drop will be (1/400)n mL., 2.43, , (a) By definition of parsec, , 1A.U. , ∴ 1 parsec = , , 1arc sec , 1 deg = 3600 arc sec, , π, ∴ 1 arcsec = 3600 × 180 radians, 3600 × 180 A.U. = 206265 A.U., ∴ 1 parsec =, ≈ 2 × 105 A.U., π, (b), , At 1 A.U. distance, sun is (1/2°) in diameter., 1/ 2, degree in diameter = 15, Therefore, at 1 parsec, star is, 2 × 105, × 10-5 arcmin., With 100 magnification, it should look 15 × 10-3 arcmin. However,, due to atmospheric fluctuations, it will still look of about 1 arcmin., It can’t be magnified using telescope., , Dmars 1, Dearth, 1, [from Answer 2.25 (c)], = ,, =, Dearth 2, Dsun, 400, D, 1, ∴ mars =, ., Dsun, 800, , (c), , At 1 A.U. sun is seen as 1/2 degree in diameter, and mars will, be seen as 1/1600 degree in diameter., At 1/2 A.U, mars will be seen as 1/800 degree in diameter. With, 60, 100 magnification mars will be seen as 1/8 degree =, = 7.5, 8, arcmin., This is larger than resolution limit due to atmospheric, fluctuations. Hence, it looks magnified., 2.44, , (a) Since 1 u = 1.67 × 10–27 kg, its energy equivalent is 1.67×10–27 c2, in SI units. When converted to eV and MeV, it turns out to be, 1 u ≡ 931.5 MeV., (b) 1 u × c2 = 931.5 MeV., , 117, 20/04/2018
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Exemplar Problems–Physics, , Chapter, 3.1, , (b), , 3.2, , (a), , 3.3, , (b), , 3.4, , (c), , 3.5, , (b), , 3.6, , (c), , 3.7, , (a), (c), (d), , 3.8, , (a), (c), (e), , 3.9, , (a), (d), , 3.10, , (a), (c), , 3.11, , (b), (c), (d), , 3.12, , (a) (iii), (b) (ii), (c) iv, (d) (i), , 3, , 3.13, , 3.14, , (i) x (t) = t - sin t, (ii) x (t) = sin t, , 3.15, , x(t) = A + B e − γt ; A > B, γ > 0 are suitably chosen positive constants., , 3.16, , v = g/b, , 3.17, , The ball is released and is falling under gravity. Acceleration is –g ,, except for the short time intervals in which the ball collides with, , a, 0, , t, , -g, , 118, 20/04/2018
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Answers, , ground, and when the impulsive force acts and produces a large, acceleration., 3.18, , (a) x = 0, v = γ x o, , 3.19, , Relative speed of cars = 45km/h, time required to meet, , =, , 36 km, = 0.80h, 45 km/h, , Thus, distance covered by the bird = 36 km/h × 0.8h = 28.8 km., 3.20, , Suppose that the fall of 9 m will take time t. Hence, , y − yo = voy, , Since voy = 0,, , t =, , 9m/s, , gt 2, −, 2, 9m, , 2(y − yo ), 2×9m, →, = 1.8 ≈ 1.34 seconds., g, 10 m/s2, , In this time, the distance moved horizontally is, x-xo = voxt = 9 m/s × 1.34s = 12.06 m., , 10m, , Yes-he will land., 3.21, , Both are free falling. Hence, there is no acceleration of one w.r.t., another. Therefore, relative speed remains constant (=40 m/s )., a, , 3.22, , v = (-vo/xo) x + vo, a = (vo/xo)2 x - vo2/xo, The variation of a with x is shown in the figure. It is a straight line, with a positive slope and a negative intercept., , 3.23, , (a) v =, (b), , 2 gh = 2 × 10 × 1000 = 141m/s = 510 km/h., 4π 3, 4π, m =, r ρ=, (2 × 10−3 )3 (103 ) = 3.4 × 10−5 kg., 3, 3, P = mv ≈ 4.7 × 10−3 kg m/s ≈ 5 × 10−3 kg m/s., , (c), , Diameter ≈ 4mm, , 0, , x, , ∆t ≈ d / v = 28 µ s ≈ 30 µ s, , ∆P 4.7 × 10−3, =, ≈ 168N ≈ 1.7 × 102 N., ∆t, 28 × 10−6, , (d), , F =, , (e), , Area of cross-section = π d 2 / 4 ≈ 0.8m 2 ., , 119, 20/04/2018
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Exemplar Problems–Physics, , With average separation of 5 cm, no. of drops that will fall almost, simultaneously is, , 0.8m 2, ≈ 320., (5 × 10−2 )2, , Net force ≈ 54000 N (Practically drops are damped by air viscosity)., 3.24, , Car behind the truck, , 20, = 4ms –2, 5, 20, Regardation of car =, ms –2, 3, Let the truck be at a distance x from the car when breaks are applied, , Regardation of truck =, , Distance of truck from A at t > 0.5 s is, Distance of car from A is, , x + 20t – 2t2., , 10 + 20(t – 0.5) –, , 10, (t – 0.5)2 ., 3, , If the two meet, x + 20t – 2t2 = 10 + 20t – 10 –, x =–, , 10 2 10, 10, t +, t – 0.25 ×, ., 3, 3, 3, , 4 2 10, 5, t +, t– ., 3, 3, 6, , To find xmin,, , dx, 8, 10, =– t+, =0, 3, 3, dt, 10 5, = s., which gives tmin=, 8, 4, 2, , Therefore, xmin= –, , 45, 10 5 5 5, × – = ., +, 34, 3 4 6 4, , Therefore, x > 1.25m., Second method: This method does not require the use of calculus., If the car is behind the truck,, Vcar = 20 – (20/3)(t – 0.5) for t > 0.5 s as car declerate only after 0.5 s., Vtruck = 20 – 4t, Find t from equating the two or from velocity vs time graph. This, yields t = 5/4 s., In this time truck would travel truck,, Struck= 20(5/4) – (1/2)(4)(5/4)2 = 21.875m, , 120, 20/04/2018
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Answers, , and car would travel, S car = 20(0.5) + 20(5/4 – 0.5) –, 2, , 1 , 5, , (20 / 3) × – 0.5 = 23.125m, 2, 4, , Thus Scar – Struck = 1.25m., If the car maintains this distance initially, its speed after 1.25s will, he always less than that of truck and hence collision never occurs., 3.25, , (a) (3/2)s,, , 3.26, , v1=20 m/s, v2 = 10m/s, time difference = 1s., , (b) (9/4)s,, , (c) 0,3s,, , Chapter, , (d) 6 cycles., , 4, , 4.1, , (b), , 4.2, , (d), , 4.3, , (b), , 4.4, , (b), , 4.5, , (c), , 4.6, , (b), , 4.7, , (d), , 4.8, , (c), , 4.9, , (c), , 4.10, , (b), , 4.11, , (a), (b), , 4.12, , (c), , 4.13, , (a), (c), , 4.14, , (a), (b), (c), , 4.15, , (b), (d), , 4.16, , v2, in the direction RO., R, , 4.17, , The students may discuss with their teachers and find answer., , 4.18, , (a) Just before it hits the ground., , 121, 20/04/2018
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Exemplar Problems–Physics, (b) At the highest point reached., (c) a = g = constant., 4.19, , acceleration – g., velocity – zero., , 4.20, , Since B× C is perpendicular to plane of B and C, cross product of, any vector will lie in the plane of B and C., vo, , 4.21, v (speed, tossed), , , , u(car speed), , 0, , For a ground-based observer, the ball is a projectile with speed vo, and the angle of projection θ with horizontal in as shown above., 4.22, , (a), , (b), , Since the speed of car matches with the horizontal speed of the, projectile, boy sitting in the car will see only vertical component of, motion as shown in Fig (b)., 4.23, , Due to air resistance, particle energy as, , y, , well as horizontal component of velocity, keep on decreasing making the fall steeper, than rise as shown in the figure., , 0, , 1, 2H, H, , φ = tan −1 = tan −1 , g, R, vo, , 4.24, , R = vo, , 4.25, , Acceleration, , x, , gH , = 23°12 ', 2 , , v 2 4π 2 R, =, R, T2, , 122, 20/04/2018
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Answers, , 4.26, , (a) matches with (iv), (b) matches with (iii), (c) matches with (i), (d) matches with (ii), , 4.27, , (a) matches with (ii), (b) matches with (i), (c) matches with (iv), (d) matches with (iii), , 4.28, , (a) matches with (iv), , (b) matches with (iii), (c) matches with (i), (d) matches with (ii), 4.29, , The minimm vertical velocity required for crossing the hill is given by, , v ⊥2 ≥ 2gh = 10,000, v⊥ > 100 m/s, As canon can haul packets with a speed of 125m/s, so the, maximum value of horizontal velocity, v, , will be, , v = 1252 − 1002 = 75 m / s, The time taken to reach the top of the hill with velocity v ⊥ is given by, , 1, gT 2 = h T = 10 s., 2, , In 10s the horizontal distance covered = 750 m., So cannon has to be moved through a distance of 50 m on the, ground., So total time taken (shortest) by the packet to reach ground, across the hill =, 4.31, , (a) L =, (b), (c), , 4.32, , 50, s + 10s + 10s = 45 s., 2, , 2vo 2 sin β cos(α + β ), g cos2 α, 2v o sin β, T =, g cos α, , β=, , π, 4, , −, , α, 2, , Av 02, sin θ, g, , 123, 20/04/2018
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Answers, , Time taken to go from A to C via this path, = Toutside =, , AR + RC, s, 1, , = 2 752 + 252 s, = 2 × 25 10 s, , 1, , , , For Tsand < Toutside, 50 2 +1 < 2 × 25 10, v, , 1, ⇒ +1< 5, v, ⇒, , 1, < 5 -1 or v >, v, , 1, ≈ 0.81 m/s., 5 -1, , Chapter, 5.1, , (c), , 5.2, , (b), , 5.3, , (c), , 5.4, , (c), , 5.5, , (d), , 5.6, , (c), , 5.7, , (a), , 5.8, , (b), , 5.9, , (b), , 5.10, , (a), (b) and (d), , 5.11, , (a), (b), (d) and (e), , 5, , 5.12, , (b) and (d), , 5.13, , (b), (c), , 5.14, , (c), (d), , 5.15, , (a), (c), , 5.16, , Yes, due to the principle of conservation of momentum., Initial momentum = 50.5 × 5 kg m s–1, , 125, 20/04/2018
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Exemplar Problems–Physics, Final momentum = (50 v + 0.5 × 15) kg m s–1, v, 5.17, , = 4.9 m s-1, change in speed = 0.1 m s –1, , Let R be the reading of the scale, in newtons., Effective downward acceleration =, , 50 g − R, =g, 50, , R = 5g = 50N. (The weighing scale will show 5 kg)., 5.18, 5.19, , Zero; −, , 3, kg m s-1, 2, , The only retarding force that acts on him, if he is not using a seat, belt comes from the friction exerted by the seat. This is not enough, to prevent him from moving forward when the vehicle is brought to, a sudden halt., , 5.20, , p = 8ˆi + 8ˆj, F = (4ˆi + 8ˆj)N, , 5.21, , f = F until the block is stationary., f remains constant if F increases beyond this, point and the block starts moving., , f, F, , 5.22, , In transportation, the vehicle say a truck, may need to halt suddenly., To bring a fragile material, like porcelain object to a sudden halt, means applying a large force and this is likely to damage the object., If it is wrapped up in say, straw, the object can travel some distance, as the straw is soft before coming to a halt. The force needed to, achieve this is less, thus reducing the possibility of damage., , 5.23, , The body of the child is brought to a sudden halt when she/he falls, on a cement floor. The mud floor yields and the body travels some, distance before it comes to rest , which takes some time. This means, the force which brings the child to rest is less for the fall on a mud, floor, as the change in momentum is brought about over a longer, period., (b) 18.75 kg m s–1, , 5.24, , (a) 12.5 N s, , 5.25, , f = µ R = µ mg cosθ is the force of friction, if θ is angle made by the, slope. If θ is small, force of friction is high and there is less chance of, skidding. The road straight up would have a larger slope., , 5.26, , AB, because force on the upper thread will be equal to sum of the, weight of the body and the applied force., , 126, 20/04/2018
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Answers, 5.27, , If the force is large and sudden, thread CD breaks because as CD is, jerked, the pull is not transmitted to AB instantaneously, (transmission depends on the elastic properties of the body)., Therefore, before the mass moves, CD breaks., , 5.28, , T1 = 94.4 N, T2 = 35.4 N, , 5.29, , W = 50 N, , 5.30, , If F is the force of the finger on the book, F = N, the normal reaction, of the wall on the book. The minimum upward frictional force needed, to ensure that the book does not fall is Mg. The frictional force = µN., Thus, minimum value of F =, , 5.31, , 0.4 m s–1, , 5.32, , x = t, y = t 2, , Mg, , µ, , ., , a x = 0, a y = 2 m s −1, F = 0.5×2 = 1N. along y-axis., , 2V, 2 × 20 40 10, =, =, =, = 3.33 s., g + a 10 + 2 12, 3, , 5.33, , t=, , 5.34, , (a) Since the body is moving with no acceleration, the sum of the, forces is zero F1 + F2 + F3 = 0 . Let F1 , F2 , F3 be the three forces, passing through a point. Let F1 and F2 be in the plane A (one can, always draw a plane having two intersecting lines such that the, two lines lie on the plane). Then F1 + F2 must be in the plane A., , F1, O, , F2, , P, F3, , Since F3 = – ( F1 + F2 ) , F3 is also in the plane A., (b) Consider the torque of the forces about P. Since all the forces, pass through P, the torque is zero. Now consider torque about, another point 0. Then torque about 0 is, Torque = OP × ( F1 + F2 + F3 ), Since F1 + F2 + F3 = 0 , torque = 0, 5.35, , General case, , s=, , 1 2, at t = 2s / a, 2, , Smooth case, , g, Acceleration a = g sin θ =, ∴t1 =, , 2, , 2 2s / g, , 127, 20/04/2018
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Answers, , x 2 y2, +, =1, A2 B 2, , x = A cos ωt , y = B sin ωt , , 5.39, , For (a), , 1 2, v z = gH, 2, , vz = 2gH, , vs 2 + vz 2 = vs 2 + 2 gH, , Speed at ground =, , 1, , 2, , , , For (b) also mvs + mgH is the total energy of the ball when it, 2, , hits the ground., So the speed would be the same for both (a) and (b)., , 5.40, , F2 =, F1 +, , F1 =, 5.41, , 5.42, , F3 + F4, , =, , 2, F3, 2, , =, , F4, , F4 − F3, 2, , 2 +1, 3 N, =, 2, 2, , 2, =, , 1, 2, , N, , (a), , θ = tan −1 µ, , (b), , mg sin α − µ mg cos α, , (c), , mg ( sin α + µ cos α ), , (d), , mg ( sin θ + µ cos θ ) + ma., , (a) F - (500 ×10) = (500 × 15) or F = 12.5 × 10 3 N, where F is the, upward reaction of the floor and is equal to the force downwards on, the floor, by Newton’s 3rd law of motion, (b) R - (2500 × 10) = (2500 × 15) or R = 6.25 × 104 N, action of the air, on the system, upwards. The action of the rotor on the surrounding, air is 6.25 × 104 N downwards., (c) Force on the helicopter due to the air = 6.25 × 104 N upwards., , 129, 20/04/2018
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Exemplar Problems–Physics, , Chapter, , 6, , 6.1, , (b), , 6.2, , (c), , 6.3, , (d), , 6.4, , (c), , 6.5, , (c), , 6.6, , (c), , 6.7, , (c), , 6.8, , (b), , 6.9, , (b), , 6.10, , (b), , 6.11, , (b) as displacement α t 3 / 2, , 6.12, , (d), , 6.13, , (d), , 6.14, , (a), , 6.15, , (b), , 6.16, , (d), , 6.17, , (b), , 6.18, , (c), , 6.19, , (b), (d), , 6.20, , (b), (d), (f), , 6.21, , (c), , 6.22, , Yes, No., , 6.23, , To prevent elevator from falling freely under gravity., , 6.24, , (a) Positive, (b) Negative, , 6.25, , Work done against gravity in moving along horizontal road is zero ., , 6.26, , No, because resistive force of air also acts on the body which is a, non-conservative force. So the gain in KE would be smaller than the, loss in PE., , 6.27, , No, work done over each closed path is necessarily zero only if all, the forces acting on the system are conservative., , 130, 20/04/2018
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Answers, , 6.28, , (b) Total linear momentum., While balls are in contact, there may be deformation which means, elastic potential energy which came from part of KE. Momentum is, always conserved., , 6.29, 6.30, , mgh 100 × 9.8 × 10, =, W = 490W, T, 20, ∆E 0.5 × 72, = 0.6 watts, P=, =, ∆t, 60, , Power =, , 6.31, , A charged particle moving in an uniform magnetic field., , 6.32, , Work done = change in KE, Both bodies had same KE and hence same amount of work is needed, to be done. Since force aplied is same, they would come to rest within, the same distance., , 6.33, , (a) Straight line: vertical, downward, (b), , Parabolic path with vertex at C., , (c) Parabolic path with vertex higher than C., 6.34, KE, Eo, , C, , Velocity, B, , D, , C, , D, , B, Eo, , O, , F, , F, X, C, , 6.35, , x, , D, , (a) For head on collission:, Conservation of momentum ⇒ 2mv0 = mv1 + mv2, Or 2v0 = v1 + v2, , v -v, and e = 2 1 ⇒ v2 = v1+ 2v0e, 2v, , ∴, , 2v1 = 2v00– 2ev0, , ∴, , v1 = v0(1 – e), , Since e <1 ⇒ v1 has the same sign as v0, therefore the ball, moves on after collission., (b) Conservation of momentum ⇒ p = p1+ p2, But KE is lost, , , , p 2 p 2, p2, > 2 + 2, 2m 2m 2m, , 131, 20/04/2018
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Exemplar Problems–Physics, , p1, p, , p, , p2, , , before, , after, , p1, , p2, , ∴ p2 > p12 + p22, Thus p, p1 and p2 are related as shown in the figure., , θ is acute (less than 90°) ( p 2 = p12 + p 22 would give θ = 90°), 6.36, , Region A : No, as KE will become negative., Region B : Yes, total energy can be greater than PE for non zero K.E., Region C : Yes, KE can be greater than total energy if its PE is negative., Region D : Yes, as PE can be greater than KE., , 6.37, , (a) Ball A transfers its entire momentum to the ball on the table, and does not rise at all., (b) v = 2gh = 4.42m/s, , 6.38, , (a) Loss of PE = mgh = 1 ×10, (b) Gain in KE =, , –3, , ×10 ×10, , –3, , = 10J, , 1, 1, mv 2 = × 10 –3 × 2500 = 1.25J, 2, 2, , (c) No, because a part of PE is used up in doing work against the, viscous drag of air., 6.39, , (b), , E1, E0, , E2, E0, , T/4 3T/4, , 6.40, , t, 5T/4 7T/4, , T/4 3T/4, , t, 5T/4 7T/4, , m = 3.0 × 10 – 5 kg ρ = 10 – 3 kg/m2 v = 9 m/s, A = 1m2, , h = 100 cm ⇒ n = 1m3, , M = ρ v = 10–3 kg, E =, , 1, 1, Mv 2 = ×103 ×(9)2 = 4.05 ×104 J ., 2, 2, , 132, 20/04/2018
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Answers, , 6.41, , KE =, , 1, 1, mv 2 ≅ × 5 × 10 4 × 102, 2, 2, 5, , = 2.5 × 10 J., 10% of this is stored in the spring., , 1 2, kx = 2.5 ×104, 2, x=1m, 4, , k = 5 × 10 N/m., 6.42, , In 6 km there are 6000 steps., ∴ E = 6000 (mg)h, = 6000 × 600 × 0.25, 5, , = 9 × 10 J., This is 10 % of intake., ∴, 6.43, , 6, , Intake energy = 10 E = 9 ×10 J., 7, , With 0.5 efficiency, 1 litre generates 1.5 × 10 J, which is used for 15, km drive., ∴ F d = 1.5 ×107J. with d = 15000 m, ∴ F = 1000 N : force of friction., , 6.44, , (a) Wg = mg sinθ d = 1×10 × 0.5 ×10 = 50 J., (b) Wf = µ mg cosθ d = 0.1 ×10 × 0.866 × 10 = 8.66 J., (c) ∆ U = mgh = 1 ×10 × 5 = 50 J, (d) a = {F - (mg sinθ + µ mg cosθ )} = [10 - 5.87 ], = 4.13 m/s2, v = u + at or v2 = u2 + 2ad, , ∆K =, , 1, 1, mv 2 – mu 2 = mad = 41.3 J, 2, 2, , (e) W = F d = 100 J, 6.45, , (a) Energy is conserved for balls 1 and 3., (b) Ball 1 acquires rotational energy, ball 2 loses energy by friction., They cannot cross at C. Ball 3 can cross over., , 133, 20/04/2018
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Exemplar Problems–Physics, , (c) Ball 1, 2 turn back before reaching C. Because of loss of energy,, ball 2 cannot reach back to A. Ball 1 has a rotational motion in, “wrong” sense when it reaches B. It cannot roll back to A,, because of kinetic friction., 6.46, , 1, 1, ( M − ∆m ) (v + ∆v )2 + ∆m (v − u )2, 2, 2, rocket, gas, 1, 1, = Mv 2 + Mv ∆v − ∆mvu + ∆mu 2, 2, 2, 1, (KE )t = Mv 2, 2, ( KE )t +∆t =, , ( KE )t +∆t − ( KE )t = ( M ∆v − ∆mu )v +, , 1, 1, ∆ m u 2 = ∆m u 2 = W, 2, 2, , (By Work - Energy theorem), , Mdv dm , , =, ( u ) ( M ∆v − ∆m u ) = 0, dt, dt, , , , , , Since , , 6.47, , F, ∆L, =Y, A, L, , Hooke’s law :, , where A is the surface area and L is length of the side of the cube. If, k is spring or compression constant, then F = k ∆ L, , A, = YL, L, 1, 2, –4, Initial KE = 2 × mv =5 ×10 J, ∴ k= Y, , 2, , Final PE = 2 ×, , ∴, 6.48, , ∆L =, , KE, k, , 1, k ( ∆L )2, 2, =, , KE, YL, , 5 ×10 –4, =1.58 × 10–7m, 11, 2 ×10, × 0.1, , =, , Let m , V, ρH e denote respectively the mass, volume and density of, helium baloon and ρair be density of air, Volume V of baloon displaces volume V of air., , (, , ), , So, V ρair − ρHe g = m a, , (1), , Integrating with respect to t,, , V ( ρair − ρHe ) gt = m v, , , , 1, 2, 1, 1 V2, V 2 ρ −ρ, mv 2 = m 2 ( ρair − ρHe ) g 2t 2 =, air, He, 2, 2 m, 2m, , (, , 2, , ) gt, , 2 2, , 134, 20/04/2018, , (2)
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Answers, , If the baloon rises to a height h, from s = ut +, , s = ut +, , 1 2, 1, at ,we get h = at 2, 2, 2, , 1 V ( ρair − ρhe ) 2, 1 2, 1, gt, at ,we get h = at 2 =, m, 2, 2, 2, , (3), , From Eqs. (3) and (2),, , 1, 1, , mv 2 = V ( ρa − ρHe ) g , V ( ρa − ρHe ) gt 2 , 2, 2m, , , = V ( ρa – ρHe ) gh, Rearranging the terms,, , , , 1, mv 2 + V ρHe gh = V ρair hg, 2, KEbaloon PEbaloon change in PE of air ., , So, as the baloon goes up, an equal volume of air comes down,, increase in PE and KE of the baloon is at the cost of PE of air [which, comes down]., , Chapter 7, 7.1, , (d), , 7.2, , (c), , 7.3, , ˆy and, after reflection from the wall, the, The initial velocity is vi = v e, final velocity is v f = −v eˆ y . The trajectory is described as, , r = y eˆy + a eˆz . Hence the change in angular momentum is, ˆ x . Hence the answer is (b)., r × m ( v f − vi ) = 2mvae, 7.4, , (d), , 7.5, , (b), , 7.6, , (c), , 7.7, , When b →0, the density becomes uniform and hence the centre of, mass is at x = 0.5. Only option (a) tends to 0.5 as b → 0., , 7.8, , (b) ω, , 7.9, , (a), (c), , 7.10, , (a), (d), , 7.11, , All are true., , 135, 20/04/2018
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Exemplar Problems–Physics, , 7.12, , ˆ., (a) False, it is along k, (b) True, (c) True, (d) False, there is no sense in adding torques about 2 different, axes., , 7.13, , (a) False, perpendicular axis theorem is applicable only to a lamina., (b) True, (c) False, z and z” are not parallel axes., (d) True., , 7.14, , When the vertical height of the object is very small as compared to, earth’s radius, we call the object small, otherwise it is extended., (a) Building and pond are small objects., (b) A deep lake and a mountain are examples of extended objects., , 7.15, , I = mi ri 2 . All the mass in a cylinder lies at distance R from the, axis of symmetry but most of the mass of a solid sphere lies at a, smaller distance than R., , 7.16, , Positive slope indicates anticlockwise rotation which is traditionally, taken as positive., , 7.17, , (a) ii, (b) iii, (c) i, (d) iv, , 7.18, , (a) iii,, , 7.19, , No. Given Fi = 0, , (b) iv, , (c) ii (d) i., , i, , The sum of torques about a certain point ‘0, , ri × Fi = 0, i, The sum of torques about any other point O′,, , ( ri – a ) × Fi = ri × Fi – a × Fi, i, i, i, Here, the second term need not vanish., 7.20, , The centripetal acceleration in a wheel arise due to the internal, elastic forces which in pairs cancel each other; being part of a, symmetrical system., , 136, 20/04/2018
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Answers, , (b), , F'', , F ′ = F = F ′′ where F and F ′′ and external forces through support., F, , Fnet = 0, External torque = F × 3R, anticlockwise., (c), , Let ω1 and ω2 be final angular velocities (anticlockwise and, F', , clockwise respectively), , F, , Finally there will be no friction., Hence, R ω1 = 2 R ω2 , 7.27, , ω1, =2, ω2, , (i) Area of square = area of rectangle ⇒ c2 = ab, 2, I xR I yR b 2 a 2 ab , ×, = 2 × 2 = 2 =1, I xS I yS c, c, c , I yR I xR, I yR, >, , >1, (i) and (ii), I, I, I, , yS, , xS, , yS, , I xR, < 1., and, I xS, I zr – I ZS ∝ ( a 2 + b 2 − 2c 2 ), , (iii), , = a 2 + b 2 − 2ab > 0, , ∴ ( I zR − I zS ) > 0, ∴, 7.28, , I zR, > 1., I zS, , Let the accelaration of the centre of mass of disc be ‘a’, then, Ma = F-f, , (1), , The angular accelaration of the disc is α = a/R. (if there is no sliding)., Then, , 1, 2, MR α = Rf, 2, , , (2), , ⇒ Ma = 2f, Thus, f = F/3. Since there is no sliding,, ⇒ f ≤ µmg, , F ≤ 3 µ Mg., , 139, 20/04/2018
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Exemplar Problems–Physics, , Chapter, , 8, , 8.1, , (d), , 8.2, , (c), , 8.3, , (a), , 8.4, , (c), , 8.5, , (b), , 8.6, , (d), , 8.7, , (d), , 8.8, , (c), , 8.9, , (a), (c), , 8.10, , (a), (c), , 8.11, , (a), (c), (d), , 8.12, , (c), (d), , 8.13, , (c), (d), , 8.14, , (a), (c), (d), , 8.15, , (a), (c), , 8.16, , (d), , 8.17, , Molecules experience the vertically downward force due to gravity, just like an apple falling from a tree. Due to thermal motion, which, is random, their velocity is not in the vertical direction. The downward, force of gravity causes the density of air in the atmosphere close to, earth higher than the density as we go up., , 8.18, , Central force; gravitational force of a point mass, electrostatic force, due to a point charge., Non-central force: spin-dependent nuclear forces, magnetic force, between two current carrying loops., , 8.19, , Areal, velocity, t, , 8.20, , It is normal to the plane containing the earth and the sun as areal, velocity, , ∆A 1, = r × v∆t ., ∆t, 2, , 140, 20/04/2018
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Answers, , 8.21, , It remains same as the gravitational force is independent of the, medium separating the masses., , 8.22, , Yes, a body will always have mass but the gravitational force on it, can be zero; for example, when it is kept at the centre of the earth., , 8.23, 8.24, , No., Yes, if the size of the spaceship is large enough for him to detect the, variation in g., , F, , 8.25, 0, , 8.26, 8.27, , Rr, , At perihelion because the earth has to cover greater linear distance, to keep the areal velocity constant., o, , (a) 90, , o, , (b) 0, , 8.28, , Every day the earth advances in the orbit by approximately 1 o. Then,, it will have to rotate by 361° (which we define as 1 day) to have sun, at zenith point again. Since 361° corresponds to 24 hours; extra 1°, corresponds to approximately 4 minute [3 min 59 seconds]., , 8.29, , Consider moving the mass at the middle by a small amount, h to the right. Then the forces on it are:, , right and, , GMm, ( R − h )2 to the, , 10R, M, , GMm, ( R + h )2 to the left. The first is larger than the, , R, , m, h, , M, , second. Hence the net force will also be towards the right. Hence, the equilibrium is unstable., 8.30, , KE, , R, , 8.31, , The trajectory of a particle under gravitational force of the earth, will be a conic section (for motion outside the earth) with the centre, of the earth as a focus. Only (c) meets this requirement., , 141, 20/04/2018
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Exemplar Problems–Physics, 8.32, , mgR/2., , 8.33, , Only the horizontal component (i.e. along the line joining m and O), will survive. The horizontal component of the force on any point on, the ring changes by a factor:, , 2r, , , ( 2, 2 3/2 , 4r + r ) , =, 8.34, , µ, , , ( 2, 2 3/2 , r +r ) , , 4 2., 5 5, , As r increases:, , U = −, , , GMm , increases., r , , , vc =, , , GM , decreases., r , , 1 , vc, × 3 decreases., r r 2, , ω=, , K decreases because v increases., E increases because U = 2K and U < O, l increases because mvr ∝, 8.35, , r., , AB = C, A, , F, , 3, = 3l, (AC) = 2 AG = 2.l ., 2, AD = AH + HJ + JD, , H, , l, l, = +l +, 2, 2, , G, , E, , = 2l., J, , AE = AC = 3l , AF = l, Force along AD due to m at F and B, , 1 1, = Gm 2 2 + Gm 2, l 2, , 1 1 Gm, l 2 2 = l 2, , D, , C, , 2, , Force along AD due to masses at E and C, , = Gm 2, , 1, Gm 2, cos, +, cos(30°), (, ), 30, 3l 2, 3l 2, , 142, 20/04/2018, , B
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Answers, Gm 2, 3l 2, , =, , 3=, , Gm 2, 3l 2, , ., , Force due to mass M at D, , Gm 2, ., 4l 2, , =, , ∴ Total Force =, , 8.36, , Gm 2, l2, , GMT 2 , (a) r = , , 4 π2 , h=, , 1, , GMT 2, 4 π2, , 1, 1, , 1 + 3 + 4 ., , , , 3, , 1, , 3, , –R, , = 4.23x107 – 6.4x106, , = 3.59×107 m., (b) θ = cos –1, = cos –1, = cos –1, , R, R+h, 1, 1+ h r, 1, 1+5.61, , , E, , = 81o18 ′, 2θ=162o 36', 360o, 2.21; Hence minimum number = 3., 2θ, 8.37, , Angular momentem and areal velocity are constant as earth orbits, the sun., 2, 2, At perigee rp ω p = ra ωa at apogee., , If ‘a’ is the semi-major axis of earth’s orbit, then r p = a (1 - e ) and, , ra = a (1 + e ) ., , , ∴, , ωp, , 2, , 1 + e , e = 0.0167, = , , ωa 1 - e , , ωp, ωa, , = 1.0691, , 143, 20/04/2018
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Exemplar Problems–Physics, Let ω be angular speed which is geometric mean of ωp and ωa and, corresponds to mean solar day,, , ω ω , p , = 1.0691, ω ωa , ωp, ω, , =, = 1.034., ω, ωa, If ω corresponds to 1° per day, (mean angular speed), then, , ω p = 1.034 per day and ω a = 0.967 per day. Since 361° = 14hrs:, mean solar day, we get 361.034 which corresponds to 24 hrs 8.14″, (8.1″ longer) and 360.967° corresponds to 23 hrs 59 min 52″, (7.9″ smaller)., This does not explain the actual variation of the length of the day, during the year., 8.38, , ra = a (1 + e ) = 6 R, rp = a (1 − e ) = 2R, , e =, , 1, 2, , Conservation of angular momentum:, angular momentum at perigee = angular momentum at apogee, , ∴ mv prp = mv a ra, va 1, = ., vp 3, Conservation of Energy:, ∴, , Energy at perigee = Energy at apogee, , 1, GMm 1, GMm, mv p 2 −, = mv a 2 −, 2, 2, rp, ra, 1, , ∴ v p 2 1 − = −2GM, 9, , , 1 1, 2GM − , rp ra , vp =, 2, va, , 1, −, v p , , , , 1 1 , r − r = 2GM, p , a, , 1/ 2, , 1 1 , r − r , p , a, 1/ 2, , 2GM 1 1 , R 2 − 6 , =, , 1, , , , 1 − , , , 9, , , 144, 20/04/2018
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Answers, , 1/ 2, , 2 / 3 GM , =, , 8/9 R , v p = 6.85km/s, , For r = 6R , vc =, , =, , 3 GM, = 6.85km/s, 4 R, , v a = 2.28km/s., , GM, = 3.23km/s., 6R, , Hence to transfer to a circular orbit at apogee, we have to boost, the velocity by ∆ = (3.23 – 2.28) = 0.95 km/s. This can be done by, suitably firing rockets from the satellite., , Chapter, 9.1, , (b), , 9.2, , (d), , 9.3, , (d), , 9.4, , (c), , 9.5, , (b), , 9.6, , (a), , 9.7, , (c), , 9.8, , (d), , 9.9, , (c), (d), , 9.10, , (a), (d), , 9.11, , (b), (d), , 9.12, , (a), (d), , 9.13, , (a), (d), , 9.14, , Steel, , 9.15, , No, , 9.16, , Copper, , 9.17, , Infinite, , 9.18, , Infinite, , 9, , 145, 20/04/2018
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Exemplar Problems–Physics, 9.19, , Let Y be the Young’s modulus of the material. Then, , f /πr 2, l /L, Let the increase in length of the second wire be l′ . Then, 2f, 4π r 2 = Y, l ′ / 2L, Y =, , Or, l ′ =, 9.20, , l πr 2, 2f, 1 2f, ×, 2L = l, 2L =, 2, L, f, 4π r 2, Y 4π r, , Because of the increase in temperature the increase in length per, unit length of the rod is, , ∆l, = α∆T = 10−5 × 2 × 10−2 = 2 × 10 –3, l0, Let the compressive tension on the rod be T and the cross sectional, area be a, then, , T /a, =Y, ∆l / l 0, ∴T = Y, , ∆l, × a = 2 × 1011 × 2 × 10−3 × 10 –4, lo, , = 4 × 10 4 N, 9.21, , Let the depth be h, then the pressure is, P = ρ gh = 103 × 9.8 × h, Now, , P, =B, ∆V / V, , ∴P = B, , ∴h =, , 9.22, , ∆V, = 9.8 × 108 × 0.1 × 10–2, V, , 9.8 × 108 × 0.1 × 10 –2, = 102 m, 9.8 × 103, , Let the increase in length be ∆l , then, , 800, = 2 × 1011, –6, (π × 25 × 10 )/ ( ∆l / 9.1), ∴ ∆l =, , 9.1 × 800, m, π × 25 × 10 –6 × 2 × 1011, , 0.5 × 10 –3 m, , 146, 20/04/2018
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Answers, , 9.23, , As the ivory ball is more elastic than the wet-clay ball, it will tend to, retain its shape instantaneously after the collision. Hence, there will, be a large energy and momentum transfer compared to the wet clay, ball. Thus, the ivory ball will rise higher after the collision., , 9.24, , Let the cross sectional area of the bar be A. Consider the equilibrium, of the plane aa ′ . A force F must be acting on this plane making an, angle, , π, , 2, , − θ with the normal ON. Resolving F into components, along, , the plane and normal to the plane, , FP = F cos θ, N, , FN = F sin θ, , FN, , a, , /2, , Let the area of the face aa ′ be A ′ , then, , O, , – , , F, , FP, a , , A, = sin θ, A′, A, ∴ A′ =, sin θ, , F sin θ F, = sin2 θ and the shearing stress, A′, A, F, F sin 2θ, = cos θ sin θ =, . Maximum tensile stress is, A, 2A, , The tensile stress T =, , F cos θ, A′, when θ = π, Z =, , 2 and maximum shearing stress when 2θ = π / 2 or, , θ = π /4 ., 9.25, , (a) Consider an element dx at a distance x from the load (x = 0). If T, (x) and T (x + dx) are tensions on the two cross sections a distance, dx apart, then, , T (x + dx) – T(x) = µgdx (where µ is the mass/length), dT, dx = µ gdx, dx, T ( x ) = µ gx + C, , At x = 0, T (0) = Mg C = Mg, , ∴ T ( x ) = µ gx + Mg, Let the length dx at x increase by dr, then, , 147, 20/04/2018
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Exemplar Problems–Physics, , T(x), , A =Y, dx, dr, 1, T(x), or,, =, dx YA, 1 L, r =, ( µ gx + Mg )dx, YA 0, dr, , dx, , L, , 1 µ gx 2, , + Mgx , , YA 2, 0, 1 mgl, , =, + MgL , YA 2, , , =, , x, x=0, Mg, , (m is the mass of the wire), 9, –2, A = π × (10 –3 )2 m 2 , Y = 200 ×10 Nm, , m = π ×(10 –3 )2 ×10 × 7860kg, r =, , 2 × 10, , 11, , 1, π × 786 ×10 –7 ×10 ×10, , + 25 ×10 ×10 , –6, , × π × 10, , 2, , , = [196.5×10 –6 +3.98×10 –3 ] 4×10-3 m, (b) The maximum tension would be at x = L., T= µ gL+Mg = (m+M)g, The yield force, , = 250 ×106 × π × (10–3 )2 = 250 × π N, At yield, , (m + M)g = 250 × π, , m = π × (10 –3 )2 ×10 × 7860 << M ∴ Mg, Hence, M =, 9.26, , 250 × π, = 25 × π, 10, , 250 × π, , 75kg., , Consider an element at r of width dr. Let T (r) and T (r+dr) be the tensions at, the two edges., – T (r+dr) + T (r) = µω2rdr where µ is the mass/length, dT, –, dr = µω 2rdr, dr, , 148, 20/04/2018
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Exemplar Problems–Physics, , dθ, , 3, = 2α1∆t × (1/2 ) − α 2 ∆t, 2, , = (α1 − α 2 ) ∆t, Or, dθ =, 9.28, , 2(α1 − α 2 )∆t, 3, , When the tree is about to buckle, , Wd =, , Y πr 4, 4R, , 1, h from the, 2, , h, then the centre of gravity is at a height l, , If R, ground., , From ∆ ABC, , 1 , ( R − d )2 + h , 2 , , R2, , If d, , 2, , R, , R2, , R 2 − 2Rd +, , ∴d =, , h2, 8R, , 1 2, h, 4, , h, , d A, R, , h/2, W, , If w 0 is the weight/volume, 4, , r dC, , B, , 2, , Y πr, h, = w0 (π r 2h ), 4R, 8R, 1/3, , 2Y , w , o , , h, 9.29, , r 2/3, , (a) Till the stone drops through a length L it will be in free fall. After, that the elasticity of the string will force it to a SHM. Let the stone, come to rest instantaneously at y., The loss in P.E. of the stone is the P.E. stored in the stretched string., , 1, k (y − L )2, 2, 1 2, 1 2, Or, mgy = ky − kyL + kL, 2, 2, mgy =, , Or,, , 1 2, 1, ky − (kL + mg )y + kL2 = 0, 2, 2, , y=, , (kL + mg ) ± (kL + mg )2 − k 2 L2, k, , 150, 20/04/2018
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Answers, , =, , (kL + mg ) ± 2mgkL + m 2 g 2, k, , Retain the positive sign., , ∴y =, , (kL + mg ) + 2mgkL + m 2 g 2, k, , (b) The maximum velocity is attained when the body passes, through, the “equilibrium, position” i.e. when the instantaneous acceleration, is zero. That is mg - kx = 0 where x is the extension from L:, , mg = kx, Let the velecity be v. Then, , 1, 1, mv 2 + kx 2 = mg( L + x ), 2, 2, 1, 1, mv 2 = mg ( L + x ) − kx 2, 2, 2, Now mg = kx, , x =, , mg, k, , 1, mg 1 m 2 g 2, , ∴ mv 2 = mg L +, − k, 2, k2, k 2, , = mgL +, , m 2g2, k, , −, , 1 m 2 g2, 2, , k, , 1, 1 m 2g2, mv 2 = mgL +, 2, 2 k, , ∴ v 2 = 2gL + mg 2 / k, v = (2 gL + mg 2 / k )1/ 2, (c) Consider the particle at an instantaneous position y. Then, , md 2y, = mg − k (y − L ), dt 2, , , d 2y k, + (y − L ) − g = 0, dt 2 m, , Make a transformation of variables: z =, , k, (y − L ) − g, m, , 151, 20/04/2018
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Exemplar Problems–Physics, 2, Then d z + k z = 0, 2, , m, , dt, , ∴z = A cos(ωt + φ ) where ω =, , m, y = L + g + A ′ cos(ωt + φ ), k , , , k, m, , Thus the stone performs SHM with angular frequency ω about, the point, , y0 = L +, , m, g, k, , Chapter 10, 10.1, , (c), , 10.2, , (d), , 10.3, , (b), , 10.4, , (a), , 10.5, , (c), , 10.6, , (a), (d), , 10.7, , (c), (d), , 10.8, , (a), (b), , 10.9, , (c), (d), , 10.10 (b), (c), 10.11 No., 10.12 No., 10.13 Let the volume of the iceberg be V. The weight of the iceberg is ρi Vg., If x is the fraction submerged, then the volume of water displaced is, xV. The buoyant force is ρ wxVg where ρ w is the density of water., , ρi Vg = ρw xVg, ∴x =, , ρi, = 0.917, ρw, , 10.14 Let x be the compression on the spring. As the block is in equilibrium, Mg – (kx + ρwVg) = 0, , 152, 20/04/2018
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Answers, , where ρw is the density of water and V is the volume of the block. The, reading in the pan is the force applied by the water on the pan i.e.,, mvessel + mwater + ρwVg., Since the scale has been adjusted to zero without the block, the new, reading is ρwVg., 10.15 Let the density of water be ρw., Then ρaL3 + ρ L3g = ρw xL3 (g + a), , ∴x =, , ρ, ρw, , Thus, the fraction of the block submerged is independent of any, acceleration, whether gravity or elevator., 10.16 The height to which the sap will rise is, , h =, , 2T cos 0°, 2(7.2 × 10 −2 ), =, 3, ρ gr, 10 × 9.8 × 2.5 × 10 −5, , 0.6m, , This is the maximum height to which the sap can rise due to surface, tension. Since many trees have heights much more than this,, capillary action alone cannot account for the rise of water in all, trees., 10.17 If the tanker acclerates in the positive x direction, then the water, will bulge at the back of the tanker. The free surface will be such, that the tangential force on any fluid parcel is zero., , Consider a parcel at the surface, of unit volume. The forces on the, fluid are, , ˆ, −ρ g y, , and, , ˆ, − ρa x, , 153, 20/04/2018
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Exemplar Problems–Physics, The component of the weight along the surface is ρ g sin θ, The component of the acceleration force along the surface is, , ρ a cos θ, , ∴ ρ g sin θ = ρ a cos θ, Hence, tan θ =a/g, 10.18 Let v1 and v2 be the volume of the droplets and v of the resulting, drop., Then v = v1+v2, , r 3 = r13 + r23 = ( 0.001 + 0.008 ) cm 3 = 0.009 cm3, ∴r, , 0.21cm, , (, , ∴ ∆U = 4π T r 2 − (r12 + r2 2 ), , ), , = 4π × 435.5 × 10−3 ( 0.212 − 0.05 ) × 10−4 J, −32 × 10–7 J, 10.19 R 3 = Nr 3, R, r = 1/ 3, N, , ∆U = 4π T (R 2 − Nr 2 ), Suppose all this energy is released at the cost of lowering the, temperature. If s is the specific heat then the change in temperature, would be,, , ∆θ =, , ∆U 4π T ( R 2 − Nr 2 ), =, ,where ρ is the density., 4, ms, π R 3 ρs, 3, , ∴ ∆θ =, , =, , 3T 1 r 2, , − 3 N, , ρs R R, , 3T 1 r 2 R 3, −, ρ s R R 3r 3, , 3T 1 1 , = ρs − , R r , , , 10.20 The drop will evaporate if the water pressure is more than the vapour, pressure. The membrane pressure (water), , p=, , 2T, = 2.33 × 103 Pa, r, , ∴r =, , 2T, 2(7.28 × 10 −2 ), =, = 6.25 × 10-5 m, p, 2.33 × 10 3, , 154, 20/04/2018
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Exemplar Problems–Physics, , 10.22 (a) 1 kg of water requires Lv k cal, , ∴ MA kg of water requires MALv k cal, Since there are NA molecules in MA kg of water the energy required for 1, molecule to evaporate is, , u=, , M A Lv, J, NA, 90, , 18 × 540 × 4.2 × 103, =, J, 6 × 10 26, = 90 × 18 × 4.2 × 10–23 J, 6.8 × 10–20 J, (b) Consider the water molecules to be points at a distance d from, each other., , M, , A, NA molecules occupy ρ l, w, , M, , A, Thus, the volume around one molecule is N ρ l, A w, , The volume around one molecule is d3 = (MA/NA ρw), 1/ 3, , 1/3, , MA , ∴d = , , N A ρw , , 18, , , =, , 6 × 10 26 × 10 3 , 1/ 3, , = ( 30 × 10−30 ), , m, , 3.1 × 10−10 m, , (c) 1 kg of vapour occupies 1601 × 10-3 m3., , ∴ 18 kg of vapour occupies 18 × 1601 × 10–3 m3, ⇒ 6 × 1026 molecules occupies 18 × 1601 × 10 –3 m3, , ∴ 1 molecule occupies, , 18 × 1601 × 10-3, m3, 6×1026, , If d ′ is the inter molecular distance, then, , d ′3 = ( 3 × 1601 × 10-29)m3, , ∴ d ′ = (30 × 1601)1/3 × 10-10 m, = 36.3 × 10-10 m, , 156, 20/04/2018
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Answers, , (d) F ( d ′ -d) = u F =, , (e), , F /d =, , 6.8 × 10−20, u, =, = 0.2048 × 10−10 N, d '− d (36.3 − 3.1) × 10−10, , 0.2048 × 10−10, = 0.066N m –1 = 6.6 × 10−2 N m –1, 3.1 × 10−10, , 10.23 Let the pressure inside the balloon be Pi and the outside pressure be Po, Pi – Po =, , 2γ, r, , Considering the air to be an ideal gas, Pi V= ni R Ti where V is the volume of the air inside the balloon, ni is, the number of moles inside and Ti is the temperature inside, and, Po V= no R To where V is the volume of the air displaced and no is the, number of moles displaced and To is the temperature outside., , Pi V, Mi, =, where Mi is the mass of air inside and MA is the molar, R Ti, MA, P V, Mo, mass of air and no = o =, where Mo is the mass of air outside, R To, MA, , ni =, , that has been displaced. If W is the load it can raise, then, W + Mi g = Mo g, ⇒ W= Mo g – Mi g, Air is 21% O2 and 79% N2, ∴ Molar mass of air MA = 0.21×32 + 0.79×28 = 28.84 g., , M A V Po Pi , −, , g, R To Ti , 4, 0.02884 × π × 83 × 9.8 1.013 × 105 1.013 × 10 5, 2×5 , −, −, , , 3, =, , 293, 333, 8 × 313 N, , 8.314, 3, 4, 0.02884 × π × 8, 1 , 1, 3, × 1.013 × 105 , −, × 9.8N, 8.314, 293, 333, , , W =, , = 3044.2 N., , 157, 20/04/2018
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Exemplar Problems–Physics, , Chapter, 11.1, , (d), , 11.2, , (b), , 11.3, , (b), , 11.4, , (a), , 11.5, , (a), , 11.6, , (a), , 11.7, , (d), Original volume Vo =, , 11, , 4, π R3, 3, , Coeff of linear expansion = α, , ∴ Coeff of volume expansion, 1 dV, ∴, = 3α, V dT, dV = 3V α dT, 11.8, , (c), , 11.9, , (b), (d), , 3α, , 4 π R 3α ∆ T, , 11.10 (b), 11.11 (a), (d), 11.12 (b), (c), (d), 11.13 Diathermic, 11.14 2 and 3 are wrong, 4th is correct., 11.15 Due to difference in conductivity, metals having high conductivity, compared to wood. On touch with a finger, heat from the surrounding, flows faster to the finger from metals and so one feels the heat., Similarly, when one touches a cold metal the heat from the finger, flows away to the surroundings faster., 11.16 −40 C = −40 F, 11.17 Since Cu has a high conductivity, compard to steel, the junction of Cu and, steel gets heated quickly but steel does, not conduct as quickly, thereby allowing, food inside to get heated uniformly., , Steel, Cu, Junction, Flame, , 158, 20/04/2018
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Answers, 1, Ml 2, 12, 1, 1, 1, 1, I' =, M (l + ∆l )2 =, Ml 2 +, 2 Ml ∆l +, M ( ∆l )2 α, 12, 12, 12, 12, 1, ≈I+, Ml 2 2α∆T, 12, , 11.18 I =, , = I + 2 I α ∆T, ∴ ∆ I = 2α I ∆ T, 11.19 Refer to the P.T diagram of water and double headed, arrow. Increasing pressure at 0°C and 1 atm takes, ice into liquid state and decreasing pressure in liquid, state at 0°C and 1 atm takes water to ice state., When crushed ice is squeezed, some of it melts. filling, up gap between ice flakes. Upon releasing pressure,, this water freezes binding all ice flakes making the, , P, (atm), Liquid, , 218, solid, 1, 0.006, 220, , ball more stable., , 0.01, , gas, , 374, , T (K), , 11.20 Resultant mixture reaches 0oC. 12.5 g of ice and rest is water., 11.21 The first option would have kept water warmer because according, to Newton’s law of cooling, the rate of loss of heat is directly, proportional to the difference of temperature of the body and the, surrounding and in the first case the temperature difference is less,, so rate of loss of heat will be less., 11.22 l iron − l, brass, = 10 cm at all tempertature, , ∴ l iron (1 + αiron ∆t ) − l, (1 + α, ∆t ) = 10cm, brass, brass, l iron αiron = l brass αbrass, ∴, , ∴, , l iron, l brass, , =, , 1.8 3, =, 1.2 2, , 1, = 10cm l brass = 20 cm, l, 2 brass, , and l ° iron = 30 cm, , 159, 20/04/2018
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Answers, , Integrating, , ∴ L = L o + L o α θ1 +, , (θ2 − θ1 ), Lo, , α x dx o, , 1, , , = L o 1 + α (θ 2 + θ1 ) as, , , 2, , L0, , 1, , 0 xdx = 2 L 0, 2, , Method II, If temperature of the rod varies linearly, we can assume average, , 1, (θ1 + θ2 ) and hence new length, 2, 1, L = L o 1 + α (θ2 + θ1 ) , 2, , , , temperature to be, , 11.27 (i) 1.8 × 1017 J/S, , (ii) 7 × 109 kg, , (iii) 47.7 N/m2., , Chapter, 12.1, , 12, , (c) adiabatic, A is isobaric process, D is isochoric. Of B and C, B has the smaller, slope (magnitude) hence is isothermal. Remaing process is, adiabatic., , 12.2, , (a), , 12.3, , (c), , 12.4, , (b), , 12.5, , (a), , 12.6, , (b), , 12.7, , (a), (b) and (d)., , 12.8, , (a), (d), , 12.9, , (b), (c), , 12.10, , (a), (c), , 12.11, , (a), (c), , 161, 20/04/2018
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Exemplar Problems–Physics, , 12.12, , If the system does work against the surroundings so that it, compensates for the heat supplied, the temperature can remain, constant., , 12.13, , U p − U Q = W.D. in path 1 on the system + 1000 J, = W.D. in path 2 on the system + Q, , Q = (–100 + 1000)J = 900 J, 12.14, , Here heat removed is less than the heat supplied and hence the, room, including the refrigerator (which is not insulated from the, room) becomes hotter., , 12.15, , Yes. When the gas undergoes adiabatic compression, its, temperature increases., dQ = dU + dW, As dQ = 0 (adiabatic process), so dU = -dW, In compression, work is done on the system So, dW = -ve, , dU = + ve, So internal energy of the gas increases, i.e. its temperature increases., 12.16, , During driving, temperature of the gas increases while its volume, remains constant., So according to Charle’s law, at constant V, P α T., Therefore, pressure of gas increases., , 12.17, , Q T2 3, =, = , Q1 − Q2 = 103 J, Q1 T1 5, 5, 3, Q1 1 − = 103 J Q1 = × 102 J = 2500 J, Q 2 = 1500J, 2, , 5, , 12.18, , 5 × 7000 × 10 3 × 4.2 J = 60 × 15 × 10 × N, , N=, , 21 × 7 × 106 147, =, × 103 = 16.3 × 103 times., 900, 9, , 162, 20/04/2018
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Exemplar Problems–Physics, ∆W = 2 A V1 ( 2 − 1) = 2RT1 ( 2 − 1), , ∆Q = (7 / 2)RT1 ( 2 − 1), 12.24 (a), , A to B, , (b), , C to D, , (c), , WAB =, , B, , p dV, , = 0; WCD = 0 ., , A, , Similarly. W BC, , C, C, dV, V −r +1 , = pdV = k r = k, , −R + 1 , V, B, B, , =, Similarly, W DA =, , VC, , VB, , 1, ( Pc Vc − PB VB ), 1−γ, , 1, ( PAV A − PD VD ), 1− γ, γ, , V , , Now PC = PB B = 2−γ PB, VC , Similarly, PD = PA 2-γ, Totat work done = WBC +WDA, , (d), , =, , 1, PB VB ( 2 −γ +1 − 1) − PA V A ( 2 −γ +1 − 1), 1−γ , , =, , 1, (21−γ − 1)( PB − PA )V A, 1−γ, , =, , 2/3, , 3, 1, 1 − , ( PB − PA ) V A, 2, 2, , , Heat supplied during process A, B, dQAB = dUAB, , 3, 3, QAB = nR(TB − TA ) = (PB − PA )VA, 2, 2, Net Work done 1 2 3 , = 1 −, , Efficiency =, Heat Supplied 2 , 12.25 Q AB = U AB =, , 3, 3, R(TB − T A ) = V A ( PB − PA ), 2, 2, , QBc = U BC + WBC, , 164, 20/04/2018
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Exemplar Problems–Physics, ∆Q = Pa (V − Vo ) +, , 1, k (V − Vo )2 + CV (T − To ), 2, , where To = Pa Vo/R,, T = [Pa+(R/A)-(V-Vo)]V/R, , Chapter, 13.1, , 13, , (b), Comment for discussion: This brings in concepts of relative motion, and that when collision takes place, it is the relative velocity which, changes., , 13.2, , (d), Comment for discussion: In the ideal case that we normally consider,, each collision transfers twice the magnitude of its normal, momentum. On the face EFGH, it transfers only half of that., , 13.3, , (b), , 13.4, , (c) This is a constant pressure, , 13.5, , (a), , 13.6, , (d), , ( p = Mg / A ) arrangement., , Comment for discussion: The usual statement for the perfect gas law, somehow emphasizes molecules. If a gas exists in atomic form, (perfectly possible) or a combination of atomic and molecular form,, the law is not clearly stated., 13.7, , (b), Comment: In a mixture, the average kinetic energy are equating., Hence, distribution in velocity are quite different., , 13.8, , (d), Comment for discussion: In this chapter, one has discussed constant, pressure and constant volume situations but in real life there are, many situations where both change. If the surfaces were rigid, p, would rise to 1.1 p. However, as the pressure rises, V also rises such, that pv finally is 1.1 RT with p final > p and Vfinal > V. Hence (d)., , 13.9, , (b),(d), , 13.10 (c), 13.11 (a), (d), , 166, 20/04/2018
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Answers, , ( ), , Comment : The equation <K.E. of translation> = 3 2 RT , <Rotational, energy> = RT is taught. The fact that the distribution of the two is, independent of each other is not emphasized. They are, independently Maxwellian., 13.12 (a), (c), 13.13 (a), Comment : Conceptually, it is not often clear to the students that, elastic collisions with a moving object leads to change in its energy., 13.14 ∴ Molar mass of gold is 197 g mole–1, the number of atoms = 6.0 ×, 1023, , ∴ No. of atoms in 39.4g =, , 6.0 × 1023 × 39.4, = 1.2 × 1023, 197, , 13.15 Keeping P constant, we have, , V2 =, 13.16, , V1T2 100 × 600, =, = 200cc, T1, 300, , P1V1 P2V2, =, T1, T2, V1 P2T1 2 × 300 3, =, =, =, V2 P1T2, 400, 2, P1 =, , 1 M −2, 1 M −2, c1 ; P2 =, c2, 3 V1, 3 V2, , ∴ c 22 = c12 ×, , V2 P2, ×, V1 P1, , = (100)2 ×, c2 =, , 2, ×2, 3, , 200, m s–1, 3, , 13.17 vrms =, , v12 + v 22, 2, , =, , (9 × 106 )2 + (1 × 106 )2, 2, , =, , (81 + 1) × 1012, = 41 × 106 m s −1., 2, , 13.18 O2 has 5 degrees of freedom. Therfore, energy per mole =, , 5, RT, 2, , ∴ For 2 moles of O2, energy = 5RT, 3, Neon has 3 degrees of freedom ∴ Energy per mole = RT, 2, ∴ For 4 mole of neon, energy = 4 × 3 RT = 6RT, 2, ∴ Total energy = 11RT., , 167, 20/04/2018
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Answers, , 13.22 We have 0.25 × 6 × 1023 molecules, each of volume 10–30m3., Molecular volume = 2.5 × 10–7m3, Supposing Ideal gas law is valid., Final volume =, , Vin, (3)3 × 10−6, =, ≈ 2.7 × 10 −7 m 3, 100, 100, , which is about the molecular volume. Hence, intermolecular forces, cannot be neglected. Therfore the ideal gas situation does not hold., 13.23 When air is pumped, more molecules are pumped in. Boyle’s law is, stated for situation where number of molecules remain constant., 13.24 µ = 5.0, T = 280K, No of atoms = µ N A = 5.0 × 6.02 × 1023, = 30 × 1023, Average kinetic energy per molecule =, , ∴ Total internal energy =, , 3, kT, 2, , 3, kT × N, 2, 3, = × 30 × 1023 × 1.38 × 10−23 × 280, 2, = 1.74 × 104 J, , 13.25 Volume occupied by 1gram mole of gas at NTP = 22400cc, , ∴ Number of molecules in 1cc of hydrogen, , =, , 6.023 × 1023, = 2.688 × 1019, 22400, , As each diatomic molecule has 5 degrees of freedom,, hydrogen being diatomic also has 5 degrees of freedom, , ∴ Total no of degrees of freedom = 5 × 2.688 × 1019, = 1.344 × 1020, 13.26 Loss in K.E of the gas = ∆E =, , 1, (mn )v o 2, 2, , where n = no: of moles., If its temperature changes by ∆T , then, , n, , 3, 1, R ∆T = mn vo 2 ., 2, 2, , ∴ ∆T =, , mvo 2, 3R, , 169, 20/04/2018
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Exemplar Problems–Physics, , 13.27 The moon has small gravitational force and hence the escape velocity, is small. As the moon is in the proximity of the Earth as seen from the, Sun, the moon has the same amount of heat per unit area as that of, the Earth. The air molecules have large range of speeds. Even though, the rms speed of the air molecules is smaller than the escape velocity, on the moon, a significant number of molecules have speed greater, than escape velocity and they escape. Now rest of the molecules, arrange the speed distribution for the equilibrium temperature. Again, a significant number of molecules escape as their speeds exceed escape, speed. Hence, over a long time the moon has lost most of its, atmosphere., At 300 K, , Vrms =, , 3kT, =, m, , 3 × 1.38 × 10−23 × 300, = 1.7 km/s, 7.3 × 10−26, , Vesc for moon = 4.6 km/s, (b), , As the molecules move higher their potential energy, , increases and hence kinetic energy decreases and hence, temperature reduces., At greater height more volume is available and gas expands and, hence some cooling takes place., 13.28 (This problem is designed to give an idea about cooling by evaporation), (i), , n i vi 2, V 2rms =, , i, , ni, , 10 × (200)2 + 20 × (400)2 + 40 × (600)2 + 20 × (800)2 + 10 × (1000)2, 100, 10 × 1002 × (1 × 4 + 2 × 16 + 4 × 36 + 2 × 64 + 1 × 100), =, 100, , =, , = 1000 × (4 + 32 + 144 + 128 + 100) = 408 × 1000m 2/s2, , ∴ vrms = 639m/s, 1, 3, mv 2rms = kT, 2, 2, , ∴T =, , 1 mv 2rms 1 3.0 × 10−26 × 4.08 × 105, = ×, 3 k, 3, 1.38 × 10−23, , = 2.96 × 102 K = 296K, , 170, 20/04/2018
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Answers, , 2, (ii) V rms =, , 10 × (200)2 + 20 × (400)2 + 40 × (600)2 + 20 × (800)2, 90, 10 × 1002 × (1 × 4 + 2 × 16 + 4 × 36 + 2 × 64), 90, 308, = 10000 ×, = 342 × 1000 m 2/s 2, 9, , =, , vrms = 584m/s, , T =, 13.29 Time t =, , t=, =, , λ, , v, 1, , d = diameter and n = number density, 2π d 2n, , λ=, n=, , 1 mV 2rms, = 248K, k, 3, , 10, N, =, = 0.0167 km −3, V 20 × 20 × 1.5, 1, , 2π d 2 ( N / V ) × v, 1, 1.414 × 3.14 × (20)2 × 0.0167 × 10−3 × 150, , = 225 h, 13.30 V1x = speed of molecule inside the box along x direction, n1 = number of molecules per unit volume, In time ∆ t, particles moving along the wall will collide if they are, within (V1x ∆t ) distance. Let a = area of the wall. No. of particles, colliding in time ∆t =, wall)., , 1, n i (Vix ∆t )a (factor of 1/2 due to motion towards, 2, , In general, gas is in equilibrium as the wall is very large as compared, to hole., , ∴ V12 x + V12y + V12z = V 2rms, , V 2rms, 3, 1, 3, 3kT, mV 2rms = kT V 2rms =, m, 2, 2, kT, ∴ V 21x =, m, ∴ V12x =, , ∴ No. of particles colliding in time ∆t =, , 1, kT, n1, ∆t a . If particles, 2, m, , 171, 20/04/2018
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Exemplar Problems–Physics, , collide along hole, they move out. Similarly outer particles colliding, along hole will move in., , ∴ Net particle flow in time ∆t =, , 1, kT, (n1 − n 2 ), ∆t a as temperature, 2, m, , is same in and out., , PV, pV = µ RT µ =, µ N A PN A RT, n=, =, V, RT, After some time τ pressure changes to p1′ inside, ∴ n1′ =, , P1′N A, RT, 1, 2, , n1V − n1′V = no. of particle gone out = (n1 − n 2 ), P1N A, P ′N, N, 1, V − 1 A V = ( P1 − P2 ) A, 2, RT, RT, RT, ′, P − P1 V m, ∴τ = 2 1, , P1 − P2 a kT, , ∴, , kT, τa, m, , kT, τa, m, , 46.7 × 10−27, 1.5 − 1.4 5 × 1.00, = 2 , , 1.5 − 1.0 0.01 × 10−6 1.38 × 10−23 × 300, = 1.38 ×105 s, 13.31 n = no. of molecules per unit volume, vrms = rms speed of gas molecules, When block is moving with speed vo, relative speed of molecules w.r.t., front face = v + vo, Coming head on, momentum transferred to block per collission, = 2m (v+vo), where m = mass of molecule., No. of collission in time ∆t =, , 1, (v + v o )n ∆tA , where A = area of cross, 2, , section of block and factor of 1/2 appears due to particles moving, towards block., , ∴ Momentum transferred in time ∆t = m (v + vo )2 nA ∆t from front, surface, Similarly momentum transferred in time ∆t = m (v − vo )2 nA ∆t from, back surface, 2, ∴ Net force (drag force) = mnA (v + v o ) − (v − v o ) from front, , = mnA (4vvo) = (4mnAv)vo, , = (4 ρ Av )v o, , 172, 20/04/2018
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Answers, , We also have, , 1, 1, mv 2 = kT, 2, 2, , Therefore, v =, , (v - is the velocity along x-axis), , kT, ., m, , Thus drag = 4 ρ A, , kT, v0 ., m, , Chapter, 14.1, , (b), , 14.2, , (b), , 14.3, , (d), , 14.4, , (c), , 14.5, , (c), , 14.6, , (d), , 14.7, , (b), , 14.8, , (a), , 14.9, , (c), , 14, , 14.10 (a), 14.11 (b), 14.12 (a), (c), 14.13 (a), (c), 14.14 (d), (b), 14.15 (a), (b), (d), 14.16 (a), (b), (c), 14.17 (a), (b) (d), 14.18 (a), (c), (d), 14.19 (i) (A),(C),(E),(G), , (ii) (B), (D), (F), (H), , 14.20 2kx towards left., 14.21 (a) Acceleration is directly proportional to displacement., (b) Acceleration is directed opposite to displacement., 14.22 When the bob of the pendulum is displaced from the mean position, so that sinθ ≅ θ, , 173, 20/04/2018
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Exemplar Problems–Physics, 14.23 + ω, 14.24 Four, 14.25 -ve, , 14.27, , 1, 1, 14.28 lm = l E = m, 6, 6, 14.29 If mass m moves down by h, then the spring extends by 2h (because, each side expands by h). The tension along the string and spring is, the same., In equilibrium, mg = 2 (k. 2h), where k is the spring constant., On pulling the mass down by x,, , F = mg - 2k ( 2h + 2x ), = – 4kx, , m, 4k, , So. T = 2π, 14.30 y =, 14.31, , 2 sin (ωt − π / 4) ; T = 2π / ω, , A, 2, , 14.32 U = U o (1 − cos α x ), , F=, , –dU –d, =, (U o – U o cos a x ), dx, dx, , = –U oα sin α x, −U o αα x, , ( for small α x ,sin α x α x ), , = –U oα 2 x, We know that F = –kx, So, k = U oα 2, , T = 2π, , m, U oα 2, , 174, 20/04/2018
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Answers, , 14.33 x = 5 sin 5t., 14.34 θ1 = θo sin ( ωt + δ1 ), , θ2 = θo sin (ωt + δ 2 ), For the first, θ = 2°, ∴ sin (ωt + δ1 ) = 1, For the 2nd, θ = −1°, ∴ sin (ωt + δ 2 ) = −1/ 2, , ∴ ωt + δ1 = 90°, ωt + δ 2 = −30°, ∴ δ1 − δ 2 = 120°, 14.35 (a) Yes., (b) Maximum weight = Mg +MAω2, , = 50 × 9.8 + 50 ×, , 5, × ( 2π × 2 )2, 100, , = 490 + 400 = 890N., Minimum weight = Mg –MAω2, , = 50 × 9.8 − 50 ×, , 5, × ( 2π × 2 )2, 100, , = 490–400, = 90 N., Maximum weight is at the topmost position,, Minimum weight is at the lowermost position., 14.36 (a) 2cm (b) 2.8 s–1, 14.37 Let the log be pressed and let the vertical displacement at the, equilibrium position be xo., At equilibrium, mg = Buoyant force, , = Ax o ρ g, When it is displaced by a further displacement x, the buoyant force, is A( x o + x )ρ g., Net restoning force, = Buoyant force – weight, = A( x o + x )ρ g – mg, , = ( A ρ g )x . i.e. proportional to x., ∴ T = 2π, , m, Aρg, , 175, 20/04/2018
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Exemplar Problems–Physics, 14.38 Consider the liquid in the length dx. It’s mass is A ρdx at a height x., PE = A ρdx gx, dx, , The PE of the left column, h1 x, , h1, , =, , A ρ gxdx, , l, , l, , 45°, , h2, 45°, , o, , x2, = Aρg, 2, , h1, , = Aρg, o, , h12 A ρ gl 2 sin 2 45°, =, 2, 2, , Similarly, P.E. of the right column = A ρ g, , h 22 A ρ gl 2 sin 2 45°, =, 2, 2, , h1 = h2 = l sin 45° where l is the length of the liquid in one arm of the, tube., Total P.E. = A ρ gh 2 = A ρ gl 2 sin2 45° =, , A ρ gl 2, 2, , If the change in liquid level along the tube in left side in y, then, length of the liquid in left side is l–y and in the right side is l + y., Total P.E. = A ρ g(l – y )2 sin 2 45° + A ρ g(l + y )2 sin2 45°, Change in PE = (PE)f – (PE)i, , Aρg, 2, Aρg, =, 2, =, , (l – y )2 + (l + y )2 − l 2 , , , l 2 + y 2 − 2 ly + l 2 + y 2 + 2 ly − l 2 , , , , = A ρ g y 2 + l 2 , Change in K.E. =, , 1, A ρ 2ly 2, 2, , Change in total energy = 0, , ∆( P .E ) + ∆( K .E ) = 0, A ρ g l 2 + y 2 + A ρly 2 = 0, Differentiating both sides w.r.t. time,, , 176, 20/04/2018
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Answers, , Chapter, 15.1, , (b), , 15.2, , (c), , 15.3, , (c), , 15.4, , (c), , 15.5, , (b), , 15.6, , (c), , 15.7, , (d), , 15.8, , (b), , 15.9, , (b), , 15, , 15.10 (c), 15.11 (a), (b), (c), 15.12 (b), (c), 15.13 (c), (d), 15.14 (b), (c), (d), 15.15 (a), (b), (d), 15.16 (a), (b), 15.17 (a), (b), (d), (e), 15.18 Wire of twice the length vibrates in its second harmonic. Thus if the, tuning fork resonaters at L, it will resonate at 2L., 15.19 L/2 as λ is constant., 15.20 517 Hz., 15.21 5cm, 15.22 1/3. Since frequency α, , 1, m = πr 2ρ, m, , 15.23 2184oC, since C α T, , 1, 15.24 n − n, 1, 2, , , , 15.25 343 m s–1. n =, , , , 1, 2l, , T , , m, , , , v, , , , = 412.5 with v = 330 m/s , 15.26 3nd harmonic since n o =, , 4l, , , , , c , , c − v , , 15.27 412.5Hz n ' = n , , 179, 20/04/2018
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Exemplar Problems–Physics, , 15.28 Stationary waves; 20cm, 15.29 (a) 9.8 × 10-4s. (b) Nodes-A, B, C, D, E. Antinodes-A1, C1. (c) 1.41m., 15.30 (a) 348.16 ms-1, (b) 336 m/s, (c) Resonance will be observed at 17cm length of air column, only, intensity of sound heard may be greater due to more complete, reflection of the sound waves at the mercury surface., 15.31 From the relation, ν =, , nv, , the result follows., 2L, , 6400 − 3500 2500 1000 × 2, +, +, 8, 5, 8 , , 15.32 t = , , , = 1975 s., = 32 minute 55 second., 15.33 c =, , c, =, v, , 3P, , ρ, 3, , γ, , =, , 3RT, ,v =, M, , and γ =, , γP, γ RT, =, M, ρ, , 7, for diatomic gases., 5, , 15.34 (a) (ii), (b) (iv), (c) (iii), (d) (i)., 15.35 (a) 5m, (b) 5m, (c) 50Hz, (d) 250ms-1, (e) 500π ms-1., 15.36 (a) 6.4π radian, (b) 0.8π radian, (c) π radian, (d) 3π /2 radian, (e), 80π radian., , 180, 20/04/2018
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SAMPLE, QUESTION PAPERS, DESIGN, , OF, , QUESTION PAPER, , PHYSICS, CLASS XI, TIME: 3 HRS., , MAX. MARKS: 70, , The weightage or the distribution of marks over different dimensions of the question, paper shall be as follows:, A., , Weightage to content/subject units, , Sr. No., , Marks, , 1., , Physical world and measurement, , 03, , 2., , Kinematics, , 10, , 3., , Laws of motion, , 10, , 4., , Work, Energy and Power, , 06, , 5., , Motion of system of particles and rigid body, , 06, , 6., , Gravitation, , 05, , 7., , Properties of bulk matter, , 10, , 8., , Thermodynamics, , 05, , 9., , Behaviour of perfect gas and kinetic theory of gases, , 05, , Oscillation and waves, , 10, , Total, , 70, , 10., , B., , Unit, , Weightage to form of questions, , Sr. No., , Form of Question, , Marks for each No. of Question Total Marks, question, , 1., , Long Answer Type (1A), , 5, , 3, , 15, , 2., , Short Answer (SAI), , 3, , 09, , 27, , 3., , Short Answer (SAII)/, , 2, , 10, , 20, , 1, , 08, , 08, , –, , 30, , 70, , Multiple Choice Question (MCQ), 4., , Very Short Answer (VSA)/, Multiple Choice Question (MCQ), Total, , 20/04/2018
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Exemplar Problems–Physics, 1 Mark quesiton may be Very Short Answer (VSA) type or Multiple choice Quesition, with only one option correct., 2 Mark question may be Short Answer (SAII) or Multiple choice Question with more, than one option correct., C., , Scheme of options, 1. There will be no over all option., 2. Internal choices (either for type) on a very selective basis has been given in some, questions., , D., , Weightage to difficulty levels of questions, Sr. No., , Estimated difficulty level, , Percentage, , 1., , Easy, , 15, , 2., , Average, , 70, , 3., , Difficult, , 15, , 182, 20/04/2018
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—, , Total, 8 (8), , 2 (2), , Oscillation & Waves, , Properties of bulk of Matter, , VII, , —, , X, , Gravitation, , VI, , 1 (1), , 1 (1), , Motion of System of Particles, & Rigid body, , V, , 1 (1), , Behavior of Perfect gas &, Kinetic Theory of gases, , Work, Energy and Power, , IV, , 1 (1), , IX, , Laws of Motion, , III, , 1 (1), , —, , Kinematics, , II, , 1(1), , VSA (1 Mark), , VIII Thermodynamics, , Physical Work and, Measurement, , I, , Topic, , 20 (10), , —, , 4 (2), , 2 (1), , 2 (1), , 2 (1), , 2 (1), , 2 (1), , —, , 4 (2), , 2 (1), , SAI (2 Marks), , 27 (9), , 3 (1), , —, , 3 (1), , 3 (1), , 3 (1), , 3 (1), , 3 (1), , 9 (3), , —, , —, , SA II (3 Marks), , Sample Paper 1, Blue Print, , 15 (3), , 5 (1), , —, , —, , 5 (1), , —, , —, , —, , —, , 5 (1), , —, , LA (5 Marks), , 70 (30), , 10 (4), , 5 (3), , 5 (2), , 10 (3), , 5 (2), , 6 (3), , 6 (3), , 10 (4), , 10 (4), , 3 (2), , Total, , Sample Question Papers, , 183, , 20/04/2018
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Exemplar Problems–Physics, , SAMPLE PAPER I, PHYSICS – XI, Time : Three Hours, , Max. Marks : 70, , General Instructions, (a), , All questions are compulsory., , (b), , There are 30 questions in total. Questions 1 to 8 carry one mark each, questions 9 to 18, carry two marks each, questions 19 to 27 carry three marks each and questions 28 to, 30 carry five marks each., , (c), , 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 three questions of five marks each., You have to attempt only one of the given choices in such questions., , (d), , Use of calculators is not permitted., , (e), , You may use the following physical constants wherever necessary :, c = 3 × 108ms-1, h = 6.6 × 10-34Js, µo = 4π × 10–7 TmA–1, Boltzmann constant k = 1.38 × 1023 JK-1, Avogadro’s number NA = 6.023 × 1023/mole, , 1., , If momentum (P), area (A) and time (T) are taken to be fundamental quantities, then, energy has the dimensional formula, (a) [P1 A-1 T1], (b) [P2 A1 T1], (c) [P1 A-1/2 T1], (d) [P1 A1/2 T-1], , 2., , The average velocity of a particle is equal to its instantaneous velocity. What is the, nature of its motion?, , 3., , A Force of F = ( 6i - 3 j ) N acts on a mass of 2kg. Find the magnitude of acceleration., , 4., , The work done by a body against friction always results in, (a), (b), (c), (d), , loss of kinetic energy, loss of potential energy, gain of kinetic energy, gain of potential energy., , 184, 20/04/2018
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Sample Question Papers, 5., , Which of the following points is the likely position of the centre of, mass of the system shown in Fig. 1., (a), , A, , (b), , B, , (c), , C, , (d), , D, , Hollow sphere, Air, , R/2, , Fig. 1, , A, B, C, , R/2, , D, Sand, , 6., , Two molecules of a gas have speeds 9 × 106 m/s and 1.0 × 106 m/s,, respectively. What is the r.m.s. speed?, , 7., , A particle in S.H.M has displacement x given by x = 3 cos (5πt+π) where x is in metres, and t in seconds. Where is the particle at t = 0 and t = 1/2 s?, , 8., , When the displacement of a particle in S.H.M. is one-fourth of the amplitude, what, fraction of the total energy is the kinetic energy?, , 9., , The displacement of a progressive wave is represented by, y = A sin(ωt – kx), where x is distance and t is time., Write the dimensional formula of (i) ω and (ii) k., , 10. 100 g of water is supercooled to –10oC. At this point, due to some disturbance mechanised, or otherwise some of it suddenly freezes to ice. What will be the temperature of the, resultant mixture and how much mass would freeze?, S w = 1cal/g/o C and Lw Fusion = 80cal/g , , OR, One day in the morning to take bath, I filled up 1/3 bucket of hot water from geyser., Remaining 2/3 was to be filled by cold water (at room temperature) to bring the mixture, to a comfortable temperature. Suddenly I had to attend to some work which would take,, say 5-10 minutes before I can take bath. Now I had two options: (i) fill the remaining, bucket completely by cold water and then attend to the work; (ii) first attend to the work, and fill the remaining bucket just before taking bath. Which option do you think would, have kept water warmer? Explain., 11. Prove the following;, For two angles of projection ‘θ’ and (90-θ) (with horizontal) with same velocity ‘V’, (a) range is the same,, (b) heights are in the ratio: tan2 θ:1., 12. What is meant by ‘escape velocity’? Obtain an expression for escape velocity of an object, projected from the surface of the earth., , 185, 20/04/2018
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Exemplar Problems–Physics, 13., , A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 km/h. At, what angle of sight (w.r.t. horizontal) when the target is seen, should the pilot drop, the bomb in order to hit the target?, , 14., , A sphere of radius R rolls without slipping on a horizontal road., A, B, C and D are four points on the vertical line through the, point of contact ‘A’ (Fig.2). What are the translational velocities, of particles at points A, B, C, D? The velocity of the centre of, mass is Vcm., , 15., , Fig. 2, , A thermodynamic system is taken from an original state D to an intermediate state E, by the linear process shown in the Fig. 3., Its volume is then reduced to the original value from, E to F by an isobaric process. Calculate the total work, done by the gas from D to E to F., , 16., , A flask contains Argon and Chlorine in the ratio 2:1, by mass. The temperature of the mixture is 37°C., Obtain the ratio of (i) average kinetic energy per, molecule and (ii) root mean square speed Vrms of the, molecules of the two gases. Atomic mass of argon =, 39.9u; molecular mass of chlorine = 70.9u., , Fig. 3, -17, , kg in Brownian, , 17., , Calculate the root mean square speed of smoke particles of mass 5 × 10, motion in air at NTP?, , 18., , A ball with a speed of 9m/s strikes another identical ball at rest such that after collision, the direction of each ball makes an angle 30° with the original direction. Find the speed, of two balls after collision. Is the kinetic energy conserved in this collision process?, , 19., , Derive a relation for the maximum velocity with which a car can safely negotiate a, circular turn of radius r on a road banked at an angle θ, given that the coefficient of, friction between the car types and the road is µ., , 20., , Give reasons for the following:–, (a) A circketer moves his hands backwards while holding a catch., (b) It is easier to pull a lawn mower than to push it., (c) A carpet is beaten with a stick to remove the dust from it., , 21., , A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s–2. The crew, and the passengers weight 300 kg. Give the magnitude and direction of the, (a) force on the floor by the crew and passengers,, (b) action of the rotor of the helicopter on the surrounding air,, (c) force on the helicopter due to the surrounding air., , 22., , A woman pushes a trunk on a railway platform which has a rough surface. She applies a force of 100 N over a distance of 10 m. Thereafter, she gets progressively tired, , 186, 20/04/2018
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Sample Question Papers, and her applied force reduces linearly with distance to 50 N. The total distance through, which the trunk has been moved is 20 m. Plot the force applied by the woman and the, frictional force, which is 50 N. Calculate the work done by the two forces over 20m., 23., , Derive equations of motion for a rigid body rotating with constant angular acceleration ‘α’ and initial angular velocity ωo., , 24., , Derive an expression for the kinetic energy and potential energy of a statellite orbiting, around a planet. A satellite of mass 200kg revolves around a planet of mass 5 × 1030, kg in a circular orbit 6.6 × 106 m radius. Calculate the B.E. of the satellite., G = 6.6 ×10-11 Nm 2 / kg2 ., , 25., , State and prove Bernoulli’s theroem., , 26., , Consider a cycle tyre being filled with air by a pump. Let V be the volume of the tyre, (fixed) and at each stroke of the pump ∆V ( V ) of air is transferred to the tube, adiabatically. What is the work done when the pressure in the tube is increased from, P1 to P2?, OR, In a refrigerator one removes heat from a lower temperature and deposits to the surroundings at a higher temperature. In this process, mechanical work has to be done,, which is provided by an electric motor. If the motor is of 1kW power, and heat is transferred from -3o C to 27o C , find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine., , 27., , Show that when a string fixed at its two ends vibrates in 1 loop, 2 loops, 3 loops and 4, loops, the frequencies are in the ratio 1:2:3:4., , 28., , (a) Define coefficient of viscosity and write its SI unit., (b) Define terminal velocity and find an expression for the terminal velocity in case of, a sphere falling through a viscous liquid., OR, The stress-strain graph for a metal wire is shown in Fig. 4. The wire returns to its, original state O along the curve EFO when it is gradually unloaded. Point B corresponds to the fracture of the wire., (i), , Upto what point of the curve is Hooke’s law obeyed?, , (ii), , Which point on the curve corresponds to the elastic, limit or yield point of the wire?, , (iii), , Indicate the elastic and plastic regions of the stressstrain graph., , (iv), , Describe what happens when the wire is loaded up, to a stress corresponding to the point A on the graph, and then unloaded gradually. In particular explain, the dotted curve., , Stress, C, A, , B, , E, F, , O, , O , , Strain, , Fig. 4, , 187, 20/04/2018
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Exemplar Problems–Physics, (v), , 29., , What is peculiar about the portion of the stress-strain graph from C to B? Upto, what stress can the wire be subjected without causing fracture?, , It is a common observation that rain clouds can be at about a kilometre altitude above, the ground., (a) If a rain drop falls from such a height freely under gravity, what will be its speed?, Also calculate in km/h. (g = 10m/s2)., (b) A typical rain drop is about 4mm diameter. Estimate its momentum if it hits you., (c) Estimate the time required to flatten the drop i.e. time between first contact and, the last contact., (d) Estimate how much force such a drop would exert on you., (e) Estimate to the order of magnitude force on an umbrella. Typical lateral separation, between two rain drops is 5 cm., (Assume that the umbrella cloth is not pierced through !!), OR, A cricket fielder can throw the cricket ball with a speed vo. If he throws the ball while, running with speed u at an angle θ to the horizontal, find, (i) The effective angle to the horizontal at which the ball is projected in air as seen by, a spectator., (ii) What will be time of flight?, (iii) What is the distance (horizontal range) from the point of projection at which the, ball will land?, (iv) Find θ at which he should throw the ball that would maximise the horizontal, range as found in (iii)., (v) How does θ for maximum range change if u >vo,, , 30., , u = vo,, , u < vo?, , (a) Show that in S.H.M., acceleration is directy proportional to its displacement at a, given instant, (b) A cylindrical log of wood of height h and area of cross-section A floats in water. It is, pressed and then released. Show that the log would execute S.H.M. with a time, period,, , m, Aρg, where m is mass of the body and ρ is density of the liquid, T = 2π, , OR, , 188, 20/04/2018
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Sample Question Papers, A progressive wave represented by, y = 5 sin (100πt-0.4πx), where y and x are in m, t is in s. What is the, (a), (b), (c), (d), (e), , amplitude, wave length, frequency, wave velocity, magnitude of particle velocity., , 189, 20/04/2018
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Exemplar Problems–Physics, , SAMPLE PAPER I, SOLUTIONS AND MARKING SCHEME, 1., , (d), , (1), , 2., , Uniform motion, , (1), , 3., , a = ( 3ˆi − 1.5 ˆj ) m/s2 ; a = 3.35m/s2, , 4., , (a), , (1), , 5., , (c), , (1), , 6., , 6.4 × 106 m/s, , 7., , –3m; O m., , 8., , K .E . 15, =, 16, E, , 9., , (i) [ M °L °T –1 ] , (ii) [ M °L–1T °], , (½)+(½), , (Formula, , (1), , ½ , Result ½) (1), (½)+(½), , (Formula, , (1), , ½ , Ratio ½) (2), 1 + 1 (2), , 10. Resultant mixture reaches 0oC. 12.5 g of ice and rest of water., , (1+1), , OR, The first option would have kept water warmer because according to Newtons law of, cooling the rate of loss of heat is directly proportional to the difference of temperature of, the body and the surrounding. In the first case the temperature difference is less so, rate of loss of heat will be less., , 11. Proof R1 = R2 and, , h1 tan 2 θ, =, h2, 1, , (2), , 1 + 1 (2), , 12. The minimum velocity of prejection of an object so that it just escapes the gravitational, force of the planet from which its is projected., , (1), , 1, GMm, mv 2 =, or v =, Re, 2, , (1), , 2GM, Re, , 13. Let the time taken by the bomb to hit the target be t., , 190, 20/04/2018
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Sample Question Papers, , 1500 =, , 1 2, gt, 2, , Or, t = 300 = 17.32 s, , (1), , Horizontal distance covered by the bomb = 17.32 × v, = 17.32 × 200 = 3464m., , ∴ tan θ =, , 1500, 3464, , or θ = tan –1 0.43, 14., , (1), , v A = v CM – ωR = 0, vCM = ωR, v B = v CM –, , vC = vCM +, , (½), , ωR, 2, , ωR, 2, , = v CM / 2, , (½), , 3, vCM, 2, , (½), , =, , v D = v CM + ω R = 2v CM, , (½), , 15. Work done = area under the P-V curve, =, , 1, (300)(30) = 450J, 2, , 16. Since Argon and Chlorine are both at the same temperature, the ratio of their, 1, average K.E. per molecule is 1:1, 2, 3, MV 2rms=K.E. per molecule = kT., 2, , ∴, , Vrms ( Argon), Vrms (Chlorine ), , =, , (1), (1), , (½), (½), , 70.9, M (Cl ), =, 39.9, M ( Ar ), , = 1.77 = 1.33, , (½+½), , 191, 20/04/2018
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Exemplar Problems–Physics, , 17., , PV =, , 1, m, 2, mVrms, =, RT, 3, M, , Vrms =, , (1), , 3 ( Nk ) T, 3kT, =, Nµ, µ, , (½), , 3 × 1.38 × 10−23 × 273, 5 × 10 –17, , (½), , =, , = 15 × 10–3 m s–1, = 1.5 cm s–1, 18., , (1), , m × 9 = mv1 cos 30° + mv 2 cos 30°, , (½), , 0 = mv1 sin 30° – mv 2 sin 30°, , (½), , v1 + v2 = 6 3, v1 = v 2, , v1 = v2 = 3 3 m s –1, Tf – Ti =, , 1, 1, 2, m ( 3 3 ) × 2 – m × 92 = –13.5m joule, 2, 2, , (½), , m is mass of either balls. So, K.E. is not conserved., , (½), , rg ( tanθ + µ ), 1 – µ tan θ, , 19., , Diagram, derivation of relation V =, , 20., , (a) He does so to increase the time taken for the catch. Since F = Ma = M, , (1) + (2), , dv, ,, dt, therefore increasing the time for the catch reduces the impact of force by the, , (1), , ball on the hands., (b) As seen from the figure, when the lawn mower is pulled by force F at θ ° to the, horizontal, the horizontal component F cos θ causes translatory motion of the, lawn mower while the vertical component cancels the weight of the lawn mower., , (1), , If the lawn mower is pushed by a force F at θ° to the horizontal, the horizontal, component is again F cos θ, while the, vertical component F sin θ adds on to the, weight mg, making it move difficult to, push the lawn mower., , F, F, , F, F, , F, F, , mg, (Pull), , mg, , 192, 20/04/2018
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Sample Question Papers, (c) By Newton’s law of inertia, when the carpet is beaten by the stick, it suddenly, moves forward but the dust particles tend to remain at their original positions, at rest, so they fall down under gravity., 21., , 22., , (a) 7.5 × 103 N, downwards, (b) 3.25 × 104N, downwards, (c) 3.25 × 104N, upwards, , (1), , (1+1+1), , Work done by the women = 1750 J, Work done by the frictional force = –1000 J, (1+1+1), , 23., , ω f = ωi + α t ;θ = ωi t +, , 24., , Derivation of K.E. =, , P.E. = –, Vb = –, , 1 2 2, α t ; ω f = ωi2 + 2αθ, 2, (1+1+1), , GMm, ,, 2r, , GMm, r, , 1, 1 GMm, mv 2 = –, = –5 ×1015 J, 2, 2 r, , (1+1+1), , 25., , Statement and proof of Bernoulli’s theorem., , 26., , P (V + ∆v )γ = ( P + ∆p )V γ, , ∆v , ∆p , P 1 + γ, = P 1 +, , , , V , , P , , γ, , (1 + 2), , (1), , ∆v ∆p dv V, =, ;, =, V, P dp γ p, P2, , W.D. =, , , P, , 1, , P2, , P dv =, , V, , P γ p dp =, P, 1, , ( P2 − P1 ), γ, , V, , (2), , OR, , η =1−, , 270, 1, =, 300 10, , Efficiency of refrigerator = 0.5η =, , (1), , 1, 20, , (1), , If Q is the heat/s transferred at higher temperture, , 193, 20/04/2018
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Exemplar Problems–Physics, , then, , W, 1, =, or Q = 20W = 20µKJ, Q 20, , and heat removed from lower temperture = 19 kJ., 27., , From the relation, ν =, , nv, , the result follows., 2L, 1 1 1 1, + + + +1, 2 2 2 2, , Calculation of ratio of frequencies:, 28., , (1), , (a) Coefficent of viscosity for a fliud is defined as the ratio of shearing stress, to the strain rate:, , µ=, , F / A Fl, =, v /l, vA, , (1), , SI unit of viscosity is poiseiulle (Pl), , (½), , (b) Terminal velocity is the constant maximum velocity attained by a body, falling through a viscous fluid when viscous force nullifies the net downward, force., (1), Derivation of v T =, , 2 2 ( ρ – ρL ), r, 9, η, , (2½), OR, , (i) Upto point P., (1+1+1+1+1), (ii) Point E, (iii) Elastic region : O to E, Plastic region : E to B, (iv) Strain increases proporional to the load upto P. Beyond P, it increases by an, increasingly greater amount for a given increase in the load. Beyond the elastic limit E, it does not retrace the curve backward. The wire is unloaded but, returns along the dotted line AO ′ . Point O ′ , corresponding to zero load which, implies a permanent strain in the wire., (v) From C to B, strain increases even if the wire is being unloaded and at B it, fractures. Stress upto that corresponding to C can be applied without causing fracture., 29., , (a) v = 2 gh = 2 × 10 × 1000 = 141m/s = 510 km/h, (b), , (1+1+1+1+1), , 4π 3, 4π, r ρ=, (2 × 10 −3 )3 (103 ) = 3.4 × 10 −5 kg, 3, 3, P = mv ≈ 4.7 × 10 − 3 kg m/s ≈ 5 × 10 − 3 kg m/s, , m =, , 194, 20/04/2018
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Exemplar Problems–Physics, So, acceleration of a body executing S.H.M. is directly proportional to the displacement, of the particle from the mean position at that instant., (b) Let the block be pressed and let the vertical displacement at the equilibrium position, be xo., At equilibrium, mg = Buoyant force, = A.x o .ρ .g, , When it is displaced by a further displacement x, the buoyant force is A( x o + x )ρ g, Net restoring force, = Buoyant force — weight, = A( x o + x )ρ g — mg, = ( A ρ g )x . i.e. proportional to x., , ∴ T = 2π, , m, Aρg, OR, , (a) 5m, , (b) 5m, , (c) 50Hz, , (d) 250ms, , -1, , (e) 500π ms-1., , (1+1+1+1+1), , 196, 20/04/2018
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Sample Paper II, Blue Print, , Topic, , VSA (1 Mark), , SAI (2 Marks), , SA II (3 Marks), , LA (5 Marks), , Total, , 1(1), , 1(2), , —, , —, , 3, , I, , Physical Word and, Measurement, , II, , Kinematics, , —, , 2(2), , 2(3), , —, , 10, , III, , Laws of Motion, , —, , 1(2), , 1(3), , 1(5), , 10, , IV, , Work, Energy and Power, , 1(1), , 1(2), , 1 (3), , —, , 6, , V, , Motion of System of Particles, & Rigid Body, , —, , 2 (3), , —, , 6, , 1(1), , 2(2), , —, , —, , 5, , VII Propertics of Bulk Modules, Bulk Matter, , 2(1), , —, , 1(3), , 1(5), , 10, , VIII Thermodynamics, , 2(1), , —, , 1(3), , —, , 5, , IX, , Behavior of Perfect Gas &, Kinetic Theory of Gases, , 1(1), , 2(2), , —, , —, , 5, , X, , Oscillations & Wave, , —, , 1(2), , 1(3), , 1(5), , 10, , VI, , Gravitation, , 70, , Sample Question Papers, , Total, , —, , 197, 20/04/2018
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Exemplar Problems–Physics, , SAMPLE PAPER II, Time : Three Hours, , Max. Marks : 70, , (a), , All questions are compulsory., , (b), , There are 30 questions in total. Questions 1 to 8 carry one mark each, questions 9 to 18, carry two marks each, questions 19 to 27 carry three marks each and questions 28 to, 30 carry five marks each., , (c), , There is no overall choice., , (d), , Use of calculators is not permitted., , (e), , You may use the following physical constants wherever necessary :, c = 3 × 108ms-1, h = 6.6 × 10-34Js, µo = 4π × 10–7 TmA–1, Boltzmann constant k = 1.38 × 1023 JK-1, Avogadro’s number NA = 6.023 × 1023/mole, , 1., , Modulus of rigidity of liquids is, (a) infinity; (b) zero; (c) unity; (d) some finite small non-zero constant value., , 2., , If all other parameters except the one mentioned in each of the options below be the, same for two objects, in which case (s) they would have the same kinetic energy?, (a) Mass of object A is two times that of B., (b) Volume of object A is half that of B., (c) Object A if falling freely while object B is moving upward with the same speed at any, given point of time., (d) Object A is moving horizontally with a constant speed while object B is falling freely., , 3., , If the sun and the planets carried huge amounts of opposite charges,, (a) all three of Kepler’s laws would still be valid., (b) only the third law will be valid., (c) the second law will not change., (d) the first law will still be valid., , 198, 20/04/2018
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Sample Question Papers, 4., , Which of the following pairs of physical quantities does not have the same dimensional, formula?, (a) Work and torque., (b) Angular momentum and Planck’s constant., (c) Tension and surface tension., (d) Impulse and linear momentum., , 5., , An ideal gas undergoes four different processes from same initial state (Fig.1). Four, processes are adiabatic, isothermal, isobaric and isochoric. Out of A, B, C, and D, which, one is adiabatic?, P, (a) (B), , A, , (b) (A), , B, C, , (c) (C), D, , (d) (D), Fig. 1, , V, , 6., , Why do two layers of a cloth of equal thickness provide warmer covering than a single, layer of cloth of double the thickness?, , 7., , Volume versus temperature graphs for a given mass of an ideal gas are shown in Fig 2 at, two different values of constant pressure. What can be inferred about relations between, P1 & P2?, , 8., , 9., , (a), , P1 > P2, , (b), , P1 = P2, , (c), , P1 < P2, , (d), , data is insufficient., , Along a streamline, , V, (l), 40, 30, , P2, , P1, , 20, 10, , 100 200 300 400500, T (K), Fig. 2, , (a), , the velocity of a fluid particle remains constant., , (b), , the velocity of all fluid particles crossing a given position is constant., , (c), , the velocity of all fluid particles at a given instant is constant., , (d), , the speed of a fluid particle remains constant., , State Newton’s third law of motion and use it to deduce the principle of conservation of, linear momentum., , 199, 20/04/2018
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Exemplar Problems–Physics, x, B, , 10. A graph of x v/s t is shown in Fig. 3. Choose correct, alternatives from below., (2), , A, , C, , E, , (a) The particle was released from rest at t = 0., (b) At B, the acceleration a > 0., (c) At C, the velocity and the acceleration vanish., , D, , O, Fig. 3, , (d) Average velocity for the motion between A and D is, positive., (e) The speed at D exceeds that at E., 11. A vehicle travels half the distance L with speed V1 and the other half with speed V2, then its average speed is, (a), , V1 + V2, 2, , (b), , 2V1 + V2, V1 + V2, , 2V1V2, (c) V + V, 1, 2, (d), 12., , L (V1 + V2 ), V1V2, Which of the diagrams shown in Fig. 4 most closely shows the variation in kinetic, energy of the earth as it moves once around the sun in its elliptical orbit?, , (a), , (b), K.E, , t, , (c), , (d), Fig. 4, , 200, 20/04/2018, , t
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Sample Question Papers, , 13., , The vernier scale of a travelling microscope has 50 divisions which coincide with 49, main scale divisions. If each main scale division is 0.5 mm, calculate the minimum, inaccuracy in the measurement of distance., , 14. A vessel contains two monatomic gases in the ratio 1:1 by mass. The temperature of the, mixture is 27°C. If their atomic masses are in the ratio 7:4, what is the (i) average kinetic, energy per molecule (ii) r.m.s. speed of the atoms of the gases., 15. A 500kg satellite is in a circular orbit of radius Re about the earth. How much energy is, required to transfer it to a circular orbit of radius 4Re? What are the changes in the, kinetic and potential energy? ( R e = 6.37 × 106 m, g = 9.8 × m s –2 ,), 16. A pipe of 17 cm length, closed at one end, is found to resonate with a 1.5 kHz source., (a) Which harmonic of the pipe resonate with the above source? (b) Will resonance with, the same source be observed if the pipe is open at both ends? Justify your answer. (Speed, of sound in air = 340 m s-1), 17. Show that the average kinetic energy of a molecule of an ideal gas is directly propotional, to the absolute temperature of the gas., 18., , Obtain an expression for the acceleration due to gravity at a depth h below the surface, of the earth., , ˆ where r is in metres and t in, 19. The position of a particle is given by r = 6t ˆi + 4t 2 ˆj + 10k, seconds., (a) Find the velocity and acceleration as a function of time., (b) Find the magnitude and direction of the velocity at t = 2s., 20., , A river is flowing due east with a speed 3m/s. A swimmer can, swim in still water at a speed of 4 m/s (Fig. 5)., (a) If swimmer starts swimming due north, what will be his, resultant velocity (magnitude and direction)?, , N, E, , B, 3m/s, , (b) If he wants to start from point A on south bank and reach, opposite point B on north bank,, (i) which direction should he swim?, , A, , (ii) what will be his resultant speed?, (c) From two different cases as mentioned in (a) and (b) above,, in which case will he reach opposite bank in shorter time?, , Fig. 5, , 21. (a) A raindrop of mass 1 g falls from rest, from a height of 1 km and hits the ground with, a speed of 50 m s-1., (i) What are the final K.E. of the drop and its initial P.E.?, (ii) How do you account for the difference between the two?, (Take g = 10ms–2)., , 201, 20/04/2018
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Exemplar Problems–Physics, , (b) Two identical ball bearings in contact with each other and resting on a frictionless, table are hit head-on by another ball bearing of the same mass moving initially, with a speed V as shown in Fig. 6., , Fig. 6, , If the collision is elastic, which of the following (Fig. 7) is a possible result, after collision?, 1, , 1, , (a), , 2, , 3, , (c), V/2, , V=0, , V /3, 1, , 2, , 3, , V /1, , V /2, , V /3, , (d), (b), , Fig. 7, , 22. Explain why:, (a) It is easier to pull a hand cart than to push it., (b) Figure 8 shows (x, t), (y,t) diagrams of a particle moving in 2-dimensions., If the particle has a mass of 500 g, find the force (direction and magnitude) acting, on the particle., , x, , 1s, , 2s, , 3s, , t, , Fig. 8, , 202, 20/04/2018
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Sample Question Papers, , 23., , (a) State parallel axis and perpendicular axis theorem., (b) Find the moment of inertia of a sphere about a tangent to the sphere, given the, moment of inertia of the sphere about any of its diameters to be 2MR2/5, where M, is the mass of the sphere and R is the radius of the sphere., , 24. A 3m long ladder weighing 20 kg leans on a frictionless wall. Its feet rest on the floor, 1 m from the wall Find the reaction forces of the wall and the floor. (3), 25. A fully loaded Boeing aircraft has a mass of 3.3×105 kg. Its total wing area is 500 m2., It is in level flight with a speed of 960km/h. (a) Estimate the pressure difference, between the lower and upper surfaces of the wings. (b) Estimate the fractional, increase in the speed of the air on the upper surface of the wing relative to the lower, surface., (The density of air ρ = 1.2 kg m –3 ), 26. Explain briefly the working principle of a refrigerator and obtain an expression for its, coefficient of performance., 27. Derive an expression for the apparent frequency of the sound heard by a listener when, source of sound and the listener both move in the same direction., 28. (a) Show that for small amplitudes the motion of a simple pendulum is simple harmonic,, hence obtain an expression for its time period., (b) Consider a pair of identical pendulums, which oscillate independently such that, when one pendulum is at its extreme position making an angle of 2° to the right with, the vertical, the other pendulum is at its extreme position making an angle of 1° to, the left of the vertical. What is the phase difference between the pendulums?, 29. (a) What is capillary rise? Derive an expression for the height to which a liquid rises in, a capillary tube of radius r., (b) Why small drops of a liquid are always spherical in shape., 30. (a) Derive an expression for the maximum safe speed for a car on a banked track,, inclined at angle α to the horizontal. µ is the cofficient of friction between the tracks, and the tyres., (b) A 100 kg gun fires a ball of 1kg from a cliff of height 500 m. It falls on the ground at, a distance of 400m from the bottom of the cliff. Find the recoil velocity of the gun., (acceleration due to gravity = 10 m s-2), , 203, 20/04/2018
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Exemplar Problems–Physics, , SAMPLE PAPER II, SOLUTIONS AND MARKING SCHEME, 1., , (b), , (1), , 2., , (e), , (1), , 3., , (c), , (1), , 4., , (c), , (1), , 5., , (c), , 6., , Air enclosed between two layers of cloth prevents the transmission of heat from, our body to outside., , (1), , 7., , (a), , (1), , 8., , (b), , (2), , 9., , Statement, , (1), , dp1, dp, d, = − 2 or, ( p1 + p2 ) = 0, dt, dt, dt, , (½), , p1 + p2 = constant ., , (½), , 10. (a), (c), (e), , (2), , 11. (c), , (2), , 12. (d), , (2), , 13. 0.01 mm, , (2), , 14. (i) 1:1, (ii) 1.32:1, , (2), , 15., , (1), , ∆E = 11.75 ×109 J, ∆KE = -11.75 ×109 J, , (½), , ∆PE = -23.475 ×109 J, , (½), , 204, 20/04/2018
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Sample Question Papers, , 16. (a), , n × 340 × 102, = 500n , where n is the harmonic for a closed pipe. Closed pipe vibrates, 4 × 17, , in 3rd harmonic with source of 1.5 KHz., , (1), , (b) For a pipe open at both ends,, n × 340 × 102, = 103 n where n is the harmonic. No integral value of n is possible for, 2 × 17, 1.5KHz. So answer is No., (1), , 17., , P =, , 1 MC 2, 3 V, , PV =, , (½), , 1, 2, M C 2 = K .E, 3, 3, , (½), , PV = nRT, , (½), , K.E ∝ T, , (½), , 18. AP = h, g′ =, , (½), GM ′, , ( Re, , 2, , −h), , (½), , A, h, , 4, 3, M ′ = π ( Re − h ) ρ, 3, , P, , h , , g ′ = g 1 −, , R, , e , , 19. (a) v = 6 i + 8t j, , Re, , (½), , (½), (1+1+1), , a =8j, (b) v = 6i + 16 j or v = 36 + 256 = 19.8m/s ., v makes an angle of tan-1 (8/3) with x-axis., 20. (i) 5 m / s at 37° to N., , (3), , (ii) (a) tan −1 ( 3 / 7 ) of N, (b) 7 m/s, (iii) in case (i) he reaches the opposite bank in shortest time., , 205, 20/04/2018
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Exemplar Problems–Physics, , 21. (i) (a) 1.25 J , 10J, (b) Difference is due to the work done by viscous force of air, (ii) (b), , (3), , 22. (a), , (1), F, , F, , F, F, , F, F, mg, , mg, (Pull), , FSinθ reduces the downward force in the case of pull., (b) x = t , y = t 2, , a x = 0, a y = 2 m s −1, F = 0.5×2 = 1N. along y-axis, , 23. (a) Statement of parallel axis theorem, , (b), , 7, MR 2 (Using parallel axis), 5, , Statement of perpendicular axis theorem, , (2), , (1), , (1), (1), , 24. Let F1 and F2 be the reaction forces of the wall and the floor respectively., N–W = 0, , (3), , F–F1 = 0 (½), , 2 2 F1 – (1/ 2 ) W = 0, , (½), , W = N = 20 × 9.8 N = 196 N, F = F1 = w / 4 2 = 34.6N, F2 =, , F 2 + N 2 = 199.0N, , (½), (½), , 206, 20/04/2018
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Sample Question Papers, The force F2 makes an angle α with the horizontal, , tan α = N / F = 4 2, α = tan –1 4 2, , (½), , 25. (a) The weight of the Boeing aircraft is balanced by the upwards force due to the pressure, difference:, , ∆P × A = 3.3 × 105 kg × 9.8m s –2, , (½), , ∆P = (3.3 × 105 kg × 9.8m s –2 )/500m 2, , (½), , = 6.5×103Nm–2, (b) The pressure difference between the lower and upper surfaces of the wing is, ∆P = ( ρ / 2 ) (v 22 – v12 ), , (½), , where v2 is the speed of air over the upper surface and v1 is the speed under the, bottom surface., v 2 – v1 =, , 2∆p, ρ ( v 2 + v1 ), , (½), , v av = ( v1 + v 2 ) /2 = 960km/h = 267m s –1, , (v 2, , – v1 ) / v av = ∆P / ρ v 2av ≅ 0.08, , (½), , (½), , 26. (a) Principle of reverse heat engine, , (1), , (b), , (1), Source T2, , W = Q1 Q1 System, , Sink T2, , 207, 20/04/2018
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Notes, , 20/04/2018
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Notes, , 20/04/2018
