PHYSICS FOR YOU JULY 18

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7 Volume 6 No. 7 July 08 Managing Editor Mahabir Singh Editor Anil Ahlawat Corporate Office: Plot 99, Sector 44 Institutional area, Gurgaon - 00 (H). Tel : info@mtg.in website : egd. Office: 406, Taj Apartment, Near Safdarjung Hospital, New Delhi CONTENTS Class Focus NEET / JEE Be JEE eady Monthly Tune Up 7 CBSE Drill Brain Map 46 Class Focus NEET / JEE 9 CBSE Drill 5 Be NEET eady 60 Monthly Tune Up 64 Competition Edge Physics Musing Problem Set 60 8 Success Story 68 JEE Advanced Solved Paper Physics Musing Solution Set 59 8 Crossword 84 Live Physics 85 Individual Subscription ates Subscribe online at Combined Subscription ates epeaters Class XII Class XI epeaters Class XII Class XI 9 months 5 months 7 months 9 months 5 months 7 months Mathematics Today PCM Chemistry Today PCB Physics For You PCMB Biology Today Send D.D/M.O in favour of MTG Learning Media (P) Ltd. Payments should be made directly to : MTG Learning Media (P) Ltd, Plot No. 99, Sector 44, Gurgaon - 00 (Haryana) We have not appointed any subscription agent. Printed and Published by Mahabir Singh on behalf of MTG Learning Media Pvt. Ltd. Printed at HT Media Ltd., B-, Sector-6, Noida, UP-007 and published at 406, Taj Apartment, ing oad, Near Safdarjung Hospital, New Delhi Editor : Anil Ahlawat eaders are adviced to make appropriate thorough enquiries before acting upon any advertisements published in this magazine. Focus/Infocus features are marketing incentives. MTG does not vouch or subscribe to the claims and representations made by advertisers. All disputes are subject to Delhi jurisdiction only. Copyright MTG Learning Media (P) Ltd. All rights reserved. eproduction in any form is prohibited. 7

8 PHYSICS MUSING Physics Musing was started in August 0 issue of Physics For You. The aim of Physics Musing is to augment the chances of bright students preparing for JEE (Main and Advanced) / NEET / AIIMS / JIPME with additional study material. In every issue of Physics For You, 0 challenging problems are proposed in various topics of JEE (Main and Advanced) / NEET. The detailed solutions of these problems will be published in next issue of Physics For You. The readers who have solved five or more problems may send their detailed solutions with their names and complete address. The names of those who send atleast five correct solutions will be published in the next issue. We hope that our readers will enrich their problem solving skills through Physics Musing and stand in better stead while facing the competitive exams. Set 60. The view in the figure is from above a plane mirror suspended by a thread connected to the centre of the mirror at point A. A scale is located 0.75 m (the distance from point A to point P) to the right of the centre of the mirror. Initially, the plane of the mirror is parallel to the side of the scale; and the angle of incidence of a light ray which is directed at the centre of the mirror is 0º. A small torque applied to the thread causes the mirror to turn.5 away from its initial position. The reflected ray then intersects the scale at point Q. The distance from point P to point Q on the scale is Normal to the mirror.5 in its intial position Initial Position of the mirror A m (a).00 m (b) 0.56 m (c).0 m (d) 0.86 m.. A mirror of length L moves horizontally as shown in the figure with a velocity v. The L V mirror is illuminated by a point P source of light P placed on the ground. The rate at which ground the length of the light spot on the ground increases is : (a) v (b) zero (c) v (d) v. A linear object AB is placed along the axis of a concave mirror. The object is moving towards the mirror with speed v. The speed of the image of the point A is 4v and the speed of the image of B is also P Q 4v. If centre of the line AB is at a distance L from the mirror then length of the object AB will be A B L (a) (b) 5 L (c) L (d) 4 L 4. Three infinite current carrying conductors are placed as shown in figure. Two wires carry same current while current in third wire is unknown. The three wires do not intersect with each other and all of them are in the plane of paper. Which of the following is correct about a point 'P' which is also in the same plane : P OP = x I I O Third wire (a) Magnetic field intensity at P is zero for all values of x. I (b) If the current in the third wire is (left to sinα right) then magnetic field will be zero at P for all values of x. I (c) If the current in the third wire is (right to sinα left) then magnetic field will be zero at P for all values of x. (d) None of these By Akhil Tewari, Author ank up Physics for JEE Main & Advanced, Professor, IITians PACE, Mumbai. 8

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10 5. Figure shows a system of three concentric metal shells A, B and C with radii a, a and a respectively. Shell B is a a A earthed and shell C is given a B C a charge Q. Now if shell C is connected to shell A, then the final charge on the shell B is equal to : (a) 4 Q 8Q 5Q Q (b) (c) (d) 7 6 A series parallel combination of batteries consisting of a large number N = 00 of identical cells, each with an internal resistances r = 0. W, is loaded with an external resistance = 0 W. The number n of parallel groups consisting of an equal number of cells connected in series, at which the external resistance generates the highest thermal power is (a) (b) (c) 4 (d) 6 7. The potential difference between the points P and Q in the adjoining circuit will be C P C ( CC 4 CC ) E (a) ( C+ C)( C + C4) C Q C CC E + 4 (b) CC ( C + C4) E ( CC CC 4) E ( CC CC 4) E (c) (d) ( C+ C)( C + C4) ( C+ C + C + C4) 8. There are two conducting spheres of radius a and b (b > a ) carrying equal and opposite charges. They are placed at a separation d (>>> a and b). The capacitance of system is 4πε0 (a) b a b d 4πε (b) 0 a d a b d 4πε (c) 0 4πε (d) a b d a b d Solution Senders of Physics Musing SET-59. Shiv Prasad Gupt, Mau. Kunal Mukherjee, Kolkata (WB). Akash Biswas, Kolkata (WB) 9. Find the potential difference V a V b between the points a and b shown in each part of the figure. (a) 5/ V (b) 9/ V (c) 4/ V (d) / V 0. A finite ladder circuit is constructed by connecting several sections of 6 mf, 8 mf capacitor combinations as shown in the figure. Circuit is terminated by a capacitor of capacitance C. Find the value of C, such that the equivalent capacitance of the ladder between the points A and B becomes independent of the number of sections in between? (a) 8 mf 9 G P A S E A Y (b) 0 mf (c) mf (d) 4 mf COSSWOD SOLUTION JUNE 08 6 A 7 M 5 A T O M I C C L O C T L 4 T O 7 P I C A L Y E A Winners. Sanket, Gulbargaa (Karnataka). M. Vallaba Gurunath, Virudhu Nagar A S 0 L 4 L S N L A S N C 5 E A I M P U L S E K U E 8 M 9 A D A S C S Y P I G B A L C C A X A F I 6 S T C O U N I M T C E 8 S T H E E T 0

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12 F CUS Class XI NEET/JEE09 Focus more to get high rank in NEET, JEE (Main and Advanced) by reading this column. This specially designed column is updated year after year by a panel of highly qualified teaching experts well-tuned to the requirements of these Entrance Tests. UNIT - : UNITS AND MEASUEMENT Physical quantities All the quantities in terms of which laws of physics can be described and which can be measured are known as physical quantities. There are two types of physical quantities : Fundamental quantities Derived quantities Fundamental quantities : Those physical quantities which do not depend upon any other quantity are known as fundamental quantities or base quantities. There are seven fundamental or base quantities in SI system. They are Length, Mass, Time, Electric current, Thermodynamic temperature, Amount of substance, Luminous intensity. Derived quantities : Those physical quantities which are derived from the fundamental quantities are known as derived quantities. e.g. Speed = Distance Time units Unit : Measurement of any physical quantity involves comparison with a certain basic arbitrarily chosen, internationally accepted reference standard known as unit. The result of a measurement of a physical quantity is expressed by a number (or numerical measure) accompanied by a unit. In general, Measure of a physical quantity = numerical value (n) unit (u) Characteristics of a Standard Unit A unit selected for measuring a physical quantity should fulfill the following requirements : It should be of suitable size. It should be well defined. It should be easily accessible. It should be easily reproducible at all places. It should not change with time. It should not change with change in its physical conditions like temperature, pressure etc. Fundamental and Derived Units Fundamental units : The units of fundamental or base quantities are known as fundamental or base units. Derived units : The units of derived quantities are known as derived units. System of Units A complete set of units having, both the base units and derived units is known as system of units. Different Types of System of Units CGS system : In this system centimetre, gram and second are the fundamental units of length, mass

13 and time respectively. It is a metric system of units. It is also known as Gaussian system of units FPS system : In this system foot, pound and second are the fundamental units of length, mass and time respectively. It is not a metric system of units. It is also known as British system of units. MKS system : In this system metre, kilogram and second are the fundamental units of length, mass and time respectively. It is also a metric system of units. International System of Units (SI) : The system of units which is at present internationally accepted for measurement is the Système Internationale d Unites (French for International System of Units), abbreviated as SI. The SI, with standard scheme of symbols, units and abbreviations, was developed and recommended by General Conference on Weights and Measures in 97 in France for international usage in scientific, technical, industrial and commercial work. It is based on the seven fundamental units or base units and two supplementary units. Seven base quantities, their units with definitions are as shown in the table. Base quantity SI Units Name Symbol Definition Length metre m The metre is the length of the path travelled by light in vacuum during a time interval of /99,79,458 of a second. Mass kilogram kg The kilogram is equal to the mass of the international prototype of the kilogram (a platinum-iridium alloy cylinder) kept at International Bureau of Weights and Measures, at Sevres, near Paris, France. Time second s The second is the duration of 9,9,6,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium- atom. Electric current ampere A The ampere is that constant current which, when flowing in two straight parallel conductors of infinite length and of negligible circular cross-section, and placed metre apart in vacuum, would produce between these conductors a force equal to 0 7 newton per metre of length. Thermodynamic temperature kelvin K The kelvin, is the fraction /7.6 of the thermodynamic temperature of the triple point of water. Amount of substance Luminous intensity mole mol The mole is the amount of substance of a system, which contains as many elementary entities as there are atoms in 0.0 kilogram of carbon-. candela cd The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency hertz and that has a radiant intensity in that direction of /68 watt per steradian. The two supplementary units in SI system are : adian (rad) : It is defined as the plane angle subtended at the centre of circle, by an arc of the circle equal in length to its radius. Steradian (sr) : It is defined as the solid angle subtended at the centre of a sphere by an area of the sphere equal to square of its radius. Advantages of SI The main advantages of SI over the other systems of units are the following : SI is a coherent system of units. SI is a rational system of units. SI is an absolute system of units. SI is a metric system.

14 accuracy and Precision of measuring instruments Accuracy : The accuracy of a measurement is a measure of how close the measured value is to the true value of the quantity. Precision : Precision tells us to what resolution or limit the quantity is measured by a measuring instrument. The accuracy in measurement may depend on several factors, including the limit or the resolution of the measuring instrument. Precision is determined by the least count of the measuring instrument. Smaller the least count, greater is the precision. errors in measurement Error : The result of every measurement by any measuring instrument contains some uncertainty. This uncertainty is known as error. Every calculated quantity which is based on measured values, also has an error The difference in the measured value and the true value of a quantity is known as error in measurement. In general, the errors in measurement can be broadly classified as : Systematic errors andom errors. Systematic errors : Systematic errors are those errors that tend to be in one direction, either positive or negative. Some of the sources of systematic errors are : Instrumental errors : These arise from the errors due to imperfect design or calibration of the measuring instrument, zero error in the instrument etc. Imperfection in experimental technique or procedure. Personal errors : These arise due to an individual s bias, lack of proper setting of the apparatus or individual s carelessness in taking observations without observing proper precautions, etc. Systematic errors can be minimised by improving experimental techniques, selecting better instruments and removing personal bias as far as possible. andom errors : andom errors are those errors, which occur irregularly and hence are random with respect to sign and size. These can arise due to random and unpredictable fluctuations in experimental conditions (e.g. unpredictable fluctuations in temperature, voltage supply, mechanical vibrations of experimental set-ups, etc), personal (unbiased) errors by the observer taking readings etc. Least count error : The smallest value that can be measured by the measuring instrument is known as its least count. All the readings or measured values are good only up to this value. The least count error is the error associated with the resolution of the instrument. Least count error belongs to the category of random errors but within a limited size; it occurs with both systematic and random errors. Using instruments of higher precision, improving experimental techniques, etc., we can reduce the least count error. Absolute Error, elative Error and Percentage Error Let a physical quantity a be measured n times. Let the measured values be a, a, a,... a n. The arithmetic mean of these values is a + a + a an Arithmetic mean, amean = n n amean = a n i i = Absolute error : The magnitude of the difference between the individual measurement and the true value of the quantity is known as the absolute error of the measurement. It is denoted by Da. The errors in the individual measured values from the true value are Da = a a mean Da = a a mean Da = a a mean Da n = a n a mean Da may be positive in certain cases and negative in some other cases. Absolute error Da will always be positive. Mean absolute error : It is the arithmetic mean of all the absolute errors. It is denoted by Da mean. a a a an amean = n 4

15 n amean = a n i i = The final value of measurement may be written as a = a mean ± Da mean a mean Da mean a a mean + Da mean This implies that any measurement of the physical quantity a is likely to lie between (a mean + Da mean ) and (a mean Da mean ). elative error or fractional error : It is defined as the ratio of mean absolute error to the mean value of the quantity measured. elative error or fractional error mean absolute error amean = = mean value amean Percentage error : When the relative error is expressed in percentage, it is known as percentage error. a Percentage error, δa = mean 00% amean Combination of Errors Addition : If X = A + B, then the maximum absolute error in X is DX = DA + DB When two quantities are added, the maximum absolute error in the final result is the sum of the absolute errors in the individual quantities. Subtraction : If X = A B then the maximum absolute error in X is DX = DA + DB When two quantities are subtracted, the maximum absolute error in the final result is the sum of the absolute errors in the individual quantities. Multiplication : If X = AB, then the maximum relative error in X is X A B = + X A B Maximum percentage error in X is X A B 00 = X A B When two quantities are multiplied, the maximum relative error in the final result is the sum of the relative errors in the individual quantities multiplied. Division : If X = A, then the maximum relative B error in X is X A B = + X A B Maximum percentage error in X is X A B 00 = X A B When two quantities are divided, the maximum relative error in the final result is the sum of the relative errors in the individual quantities divided. Power : If X = A k, then the maximum relative error in X is X = k A X A Maximum percentage error in X is X 00 = k A 00 X A The maximum relative error in a physical quantity raised to the power k is the k times the relative error in the individual quantity. In more general form if X = A B r s CD Maximum relative error in X is X p A X A q B C D = + + r + s B C D Maximum percentage error in X is X 00 = p A 00 + q B 00 X A B C D + r 00 + s 00 C D significant figures The reliable digits plus the first uncertain digit are known as significant digits or significant figures. ules to determine the Significant Figures ule : All non-zero digits are significant. e.g. has three significant figures. ule : All the zeros between two non-zero digits are significant, no matter where the decimal point is, if at all. e.g and 007 have five significant figures each. ule : If the number is less than, the zero(s) on the right of decimal point, but to the left of the first non-zero digit are not significant. e.g has two significant figures. ule 4 : The terminal or trailing zero(s) in a number without a decimal point are not significant. e.g. 00 has three significant figures. p q 5

16 ule 5 : The trailing zero(s) in a number with a decimal point are significant. e.g has four significant figures. Note : The power (or exponent) of 0 is irrelevant to the determination of significant figures. e.g has four significant figures. The change of units only changes the order of exponent but not the number of significant figures. e.g..600 m = cm = mm = km. All have four significant figures each. The digit 0 conventionally put on the left of a decimal for a number less than is never significant e.g. 0.5 has three significant figures. ounding Off The result of computation with approximate numbers, which contains more than one uncertain digit, should be rounded off. ules regarding rounding off by convention are as follows : ule : If the digit to be dropped is less than 5, then the preceding digit is left unchanged. e.g. 7. is rounded off to 7.. ule : If the digit to be dropped is more than 5, then the preceding digit is raised by one. e.g is rounded off to 6.8. ule : If the digit to be dropped is 5 followed by non-zero digit then the preceeding digit is raised by one. e.g is rounded off to 6.5. ule 4 : If the digit to be dropped is 5, then preceding digit is left unchanged, if it is even. e.g. 6.5 rounding off to 6.. ule 5 : If the digit to be dropped is 5, then the preceding digit is raised by one, if it is odd. e.g is rounded off to 4.8. ules for Arithmetic Operation with Significant Figures Addition/Subtraction : In addition or subtraction the final result should retain as many decimal places as are there in the number with the least decimal places. e.g.. m +.78 m +.46 m = 5.6 m. The final result should be rounded off to 5. m. Multiplication/Division : In multiplication or division, the final result should retain as many significant figures as are there in the original number with the least significant figures. e.g. Mass = 4.7 g, volume =.5 cm Density = Mass Volume 4. 7 g = = gcm 5. cm. The final result should be round off to.69 g cm. dimensional analysis Dimensions of a physical quantity The dimensions of a physical quantity are the powers (or exponents) to which the base quantities are raised to represent that quantity. Dimensions are denoted with square brackets [ ]. Length has the dimension [L], mass [M], time [T], electric current [A], thermodynamic temperature [K], luminous intensity [cd], and amount of substance [mol]. Using the square brackets [ ] around a quantity means that we are dealing with the dimensions of the quantity. Dimensional Formulae and Dimensional Equations Dimensional formula : The expression which shows how and which of the base quantities represent the dimensions of a physical quantity is known as the dimensional formula of the given physical quantity. Dimensional equation : An equation obtained by equating a physical quantity with its dimensional formula is known as the dimensional equation of the physical quantity. Application of Dimensional Analysis The main applications of dimensional analysis are the following : To check dimensional consistency of equation To deduce relation among the physical quantities To convert one system of unit into another system of unit. To check the dimensional consistency of equations : It is based on principle of homogeneity of dimensions which states that the equation is dimensionally correct if the dimensions of the various terms on either side of the equation are the same. Note : Only those physical quantities can be added or subtracted from each other which have the same dimensions. A dimensionally consistent equation need not be actually an exact (correct) equation, but a dimensionally wrong or inconsistent equation must be wrong. 6

17 To deduce relation among the physical quantities : If we know the dependence of the physical quantity on the other physical quantities, we can derive a relation among the physical quantities by using the principle of homogeneity of dimensions. To convert one system of unit into another system of unit : For this, we use the relation a b c M L T n = n M L T where M, L, T are fundamental units on one system; M, L, T are fundamental units on the other system, a, b, c are the dimensions of the quantity in mass, length and time, n is numerical value in one system and n is its numerical value in the other system. Note : This formula is valid only for absolute units and not for gravitational units. measuring instruments Vernier Callipers It is a device used to measure accurately upto 0 th of a millimetre. The vernier callipers is as shown in the figure. eading of Vernier callipers : Place the body between the jaws and the zero of vernier scale lies ahead of N th division of main scale. Then Main scale reading (MS) = N If n th division of vernier scale coincides with any division of main scale, then Vernier scale reading (VS) = n (VC) Total reading = MS + VS = N + n (VC) Zero error : When the jaws A and B touch each other and if the zero of the vernier scale does not coincide with the zero of the main scale, then the instrument has error called zero error. Zero error is always algebraically subtracted from the observed reading. (i) Positive zero error : Zero error is said to be positive if the zero of the vernier scale lies on the right of the zero of the main scale as shown in figure (i). Here, zero error = 0.00 cm + 5 VC (ii) Negative zero error : Zero error is said to be negative if the zero of the vernier scale lies on the left of the main scale as shown in figure (ii). Here, zero error = 0.00 cm (0 5) VC Vernier constant : It is the difference between values of one main scale division and one vernier scale division of vernier callipers. Let n vernier scale divisions (VSD) coincide with (n ) main scale divisions (MSD) \ n VSD = (n ) MSD n VSD= MSD n Vernier constant, VC = MSD VSD n = MSD MSD = n n MSD Value ofone main scale division VC = Total number ofdivisions on vernierscale Screw Gauge It works on the principle of micrometer screw. A screw gauge is as shown in figure. A B M S N H E 95 Linear (Pitch) scale Circular (Head) scale Pitch : It is defined as the linear distance moved by the screw forward or backward when one complete rotation is given to the circular cap. Distance moved onlinear scale Pitch of the screw = Number of rotations K 7

18 Least count of the screw gauge Pitchofthe screw = Total number ofdivisions onthecircular scale eading of a Screw Gauge : Place a wire between A and B, the edge of the cap lies ahead of N th division of linear scale. Then Linear scale reading (LS) = N If n th division of circular scale lies over reference line, then Circular scale reading (CS) = n (LC) Total reading = LS + CS = N + n (LC) Zero error : When the two studs A and B of the screw gauge are brought in contact and if the zero of the circular scale does not coincide with the reference line then the screw gauge has an error. This error is called zero error. (i) Positive zero error : Zero error is said to be positive if the zero of the circular scale lies below the reference line as shown in figure (i). Here, zero error = + LC (ii) Negative zero error : Zero error is said to be negative if the zero of the circular scale lies above the reference line as shown in figure (ii). Here, zero error = (98 00) LC = LC. The period of oscillation of a simple pendulum is T = π L/ g. Measured value of L is 0 cm known to mm accuracy and time for 00 oscillations of the pendulum is found to be 50 s using a wrist watch of s resolution. What is the accuracy in the determination of g? (a) % (b) % (c) 4% (d) 5%. Which of the following pairs have same dimensional formula for both the quantities? () Kinetic energy and torque () esistance and inductance () Young s modulus and pressure (a) () only (b) () only (c) () and () only (d) All of three. Which of the following systems of units is not based on units of mass, length and time alone? (a) SI (b) MKS (c) FPS (d) CGS abc 4. A physical quantity P = is determined by d measuring a, b, c and d separately with the percentage error of %, %, % and % respectively. Minimum amount of error is contributed by the measurement of (a) b (b) a (c) d (d) c 5. If dimensions of length are expressed as G x c y h z, where G, c and h are the universal gravitational constant, speed of light and Planck s constant respectively, then (a) x=, y=, z = (b) x=, y=, z = (c) x=, y=, z = (d) x=, y=, z = 6. S = A( e Bxt ), where S is speed, t is time and x is displacement. Then unit of B is (a) m s (b) m s (c) s (d) s 7. The dimensions of a a t in the equation P =, b bx where P is pressure, x is distance and t is time are (a) [M LT ] (b) [ML 0 T ] (c) [ML T ] (d) [M 0 LT ] 8. The number of significant figures in is (a) (b) 6 (c) 5 (d) 4 9. The mass of a block is 87. g and its volume is 5 cm. Its density upto correct significant figures is (a).488 g cm (b).5 g cm (c).48 g cm (d).4 g cm 8

19 0. A book with many printing errors contains four different formulae for the displacement y of a particle undergoing a certain periodic motion. t () y= asin T () y = asinvt () a t y = sin T a (4) y= ( a πt πt ) sin + cos T T where a is the maximum displacement of the particle, v is the speed of the particle, T is the time period of motion. Then dimensionally (a) and are wrong (b) and are wrong (c) and 4 are wrong (d) 4 and are wrong. In an experiment to measure the height of a bridge by dropping stone into water underneath, if the error in measurement of time is 0. s at the end of s, then the error in estimation of height of bridge will be (a) 0.49 m (b) 0.98 m (c).96 m (d). m. Which of the following units denotes the dimensions [ML /Q ] where Q denotes the electric charge? (a) henry (H) (b) H m (c) weber (Wb) (d) Wb m. The correct order in which the dimensions of time decreases in the following physical quantities is. Stefan s constant. Coefficient of volume expansion. Work done 4. Velocity gradient (a), 4,, (b),,, 4 (c) 4,,, (d),, 4, 4. The radius of a circle is. m. Area enclosed by it upto correct significant figures is (a) m (b) m (c) m (d) 4.68 m 5. When one metre, one kg and one minute are taken as fundamental units, the magnitude of a force is 6 units. What is the value of this force on CGS system? (a) 0 dyne (b) 0 5 dyne (c) 0 6 dyne (d) 0 7 dyne 6. The position x of a particle at time t is given by V x = ( e at ) a where V 0 is a constant and a > 0. The dimensions of V 0 and a are (a) [M 0 LT ] and [M 0 L 0 T ] (b) [M 0 LT 0 ] and [M 0 LT ] (c) [M 0 LT ] and [MLT ] (d) [M 0 LT ] and [M 0 LT] 7. In a measurement, the random error (a) can be decreased by increasing the number of readings and averaging them (b) can be decreased by changing the person who takes the reading (c) can be decreased by using new instrument (d) can be decreased by using a different method in taking the reading 8. A person observes that the full length of a train subtends an angle of 5 degrees. If the distance between the train and the person is km, the length of the train, calculated using parallax method, in metres is (a) 45 (b) 45p (c) 50p (d) A student measures the time period of 00 oscillations of a simple pendulum four times. The data set is 90 s, 9 s, 95 s and 9 s. If the minimum division in the measuring clock is s, then the reported mean time should be (a) 9 ± s (b) 9 ± 5.0 s (c) 9 ±.8 s (d) 9 ± s 0. In a slide callipers, (m + ) number of vernier divisions is equal to m number of smallest main scale divisions. If d unit is the magnitude of the smallest main scale division, then the magnitude of the vernier constant is d (a) ( m +) unit (b) d m unit (c) md ( m +) unit (d) ( m + ) d unit m SOLUTIONS L. (d): Here, T = π g Squaring both sides, we get L L T = 4π or g = 4π...(i) g T Take log and differentiate both sides of equation (i), we get g L T = g L T For maximum relative error, the individual errors should be added 9

20 g L = + T g L T Here, T t t = and T = n n T t = T t As errors in both L and t are the least count errors. g g = = = 0.05 The percentage error in g is g L T = + g L T 00 L = + T 00 L T = = 5%. (c) : Both kinetic energy and torque have same dimensional formula [ML T ]. esistance and inductance have different dimensional formulae. Dimensional formula of resistance is [ML T A ]. Dimensional formula of inductance is [ML T A ]. Both Young s modulus and pressure have same dimensional formula [ML T ].. (a) : International system (SI) is not based on units of mass, length and time alone. abc 4. (b): P = d P a b c d P 00 = a b c d 00 = % % % % = % + % + % + % The minimum amount of error is contributed by the measurement of a. 5. (b): Let L = kg x c y h z where k is a dimensionless constant of proportionality Equating dimensions on both sides, we get [M 0 LT 0 ] = [M L T ] x [LT ] y [ML T ] z = [M x + z L x + y + z T x y z ] Applying principle of homogeneity of dimensions, we get x + z = 0...(i) x + y + z =...(ii) x y z = 0...(iii) Solving equations (i), (ii) and (iii), we get x=, y=, z = 6. (a) : The power of exponent is always dimensionless. \ [Bxt] = [M 0 L 0 T 0 ] 0 [ B] = = = [ ML T ] [ xt] 0 [ MLT] Hence unit of B is m s. 7. (b): P = a t bx Dimensions of a = [T ], as t is subtracted from a. From P a t t = = bx bx [ t ] [ T ] 0 4 [] b = [] b = = [ M L T ] [ Px] [ ML T ][ L] a b = [ T ] = 0 [ ML T ] 0 4 [M LT ] 8. (c) : As per rules, number of significant figures in is (b): Density = mass 87. = =. 488 g cm volume 5 As mass has three significant figures and volume has two significant figures, therefore as per rule, density will have two significant figures. ounding off, we get, Density =.5 g cm 0. (b): Since LHS is displacement, so HS should have dimensions of displacement. Also argument of a trigonometric function should be dimensionless. In () argument is not dimensionless. In () argument is not dimensionless and a has not T dimension of displacement. Therefore () and () are dimensionally wrong.. (c) : From s= ut + at Here, s = h, u = 0, a = g h= 0 + gt = gt h = t h t ( g = constant) h = 98. = 96. m 0

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22 h h = 0. h 9. 6 m = h = = = 96. m (a) : Henry is unit of self-inductance (L). Induced emf, ε =L di dt or L ε dt ( W/ q) dt = = di di [ ][ ] [ ][ ] \ [ L ] = = = ML T T ML T T [ ][ ] [ ][ ] ML Q A Q QT Q \ henry (H) = [ML /Q ]. (a) : Stefan s coefficient Energy = Time Area (Temperature) 4 [ML T ] 0 4 = = [ ML T K ] 4 [T][L ][K] Coefficient of volume expansion, Change in volume = Original volume Temperature [L ] = = [ MLTK ] [L ][ K] Work done = Force Distance = [MLT ][L] = [ML T ] Velocity gradient = Velocity Distance [LT ] 0 0 = = [ T ] = [ M LT ] [L] The correct order is (a) :, 4,,. 4. (d): Area = πr = (. ) = m. 7 As per rule, the area will have three significant figures. ounding off, we get A = 4.68 m 5. (a) : M = kg M = g L = m L = cm T = min T = s n = 6 n =? The dimensional formula of force is [MLT ] \ a =, b =, c = a b c M L T n = n M L T = 6 kg m min g cm s 000 g = 00 cm 60 s 6 g cm s = = Hence, in the CGS system of units the value of given force is 0 dyne. 6. (a) : As a t is dimensionless [] a = [] t 0 0 [] a = = [ T ] = [ M LT ] [ T] [ V0 ] Also,[ x] = or [ V0 ] = [ x][] a [] a [V 0 ] = [L] [M 0 L 0 T ] = [M 0 LT ] 7. (a) : If the same measurement is repeated a number of times and the result is taken as the average value of all the observations taken, the random error can be reduced. π 8. (c) : θ = 5 = 5 rad, d = km 60 \ Length of train, l = q d π π = 5 km = km = 50πm (a) : Here, t = 90 s, t = 9 s, t = 95 s, t 4 = 9 s L.C. = s ti i Mean of the measurements, t = N t = = 9 s 4 t ti i Mean deviation = = = 5. s N 4 Since the least count of the instrument is s, so reported mean time = (9 ± ) s. m 0. (a) : (m + ) VSD = m MSD \ VSD= MSD m + Vernier constant = MSD VSD m = MSD + MSD m d = MSD= unit ( m + ) m +

23 Class XI Be JEE EADY with exclusive and brain storming MCQs Practicing these MCQs help to Strengthen your concepts and give you extra edge in your JEE Preparations. A neutron with velocity v strikes a stationary deuterium atom, its kinetic energy changes by a factor of (a) 5 (b) (c) 8 (d) 6 9. A statelllite of mass m revolves around the earth of radius at a height x from its surface. If g is acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is g g (a) (b) + x x / g (c) gx (d) + x. A rubber cord catapult has cross-sectional area 5 mm and initial length of rubber cord is 0 cm. It is stretched to 5 cm and then released to project a missile of mass 5 g. Taking Y rubber = N m. Velocity of projected missile is (a) 0 m s (b) 00 m s (c) 50 m s (d) 00 m s 4. A particle P is projected from a point O with an initial velocity of 60 m s at an angle 0 to the horizontal. At the same instant a second particle Q is projected in the opposite direction with initial speed 50 m s from a point level with O and 00 m from O. If the particle collide find the time when the collision occurs. y 60 m s 50 m s 0 a O x x A 00 m (a).s (b) 0.5s (c).5s (d).09s 5. In figure, a cyclic process ABCA of moles of an ideal P A gas is given. The temperature of the gas at B and C are 500 K and 000 K respectively. B C V If the work done on the gas in process CA is 500 J then find the net heat absorbed or released by an ideal gas. Take = 5/ J mol K. (a) 0 kj (b) 5 kj (c).5 kj (d).0 kj 6. Two parallel glass plates are held vertically at a small separation d and dipped in a liquid of surface tension T, angle of contact q = 0 and density r. The height of water that climbs up in the gap between the plates is given by

24 (a) T/drg (c) T/drg (b) T/ drg (d) None of these 7. A weightless ladder 0 ft long rests against a frictionless wall at an angle of 60 from the horizontal. A 50 pound man is 4 ft from the top of the the ladder. A horizontal force is needed to kept it from slipping. Choose the correct magnitude from the following. (a) 75 lb (b) 00 lb (c) 0 lb (d) 7. lb 8. The molecules of a given mass of a gas have root mean square speeds of 00 m s at 7 C and atmospheric pressure. The root mean square speeds (in m s ) of the molecules of the gas at 7 C and atmospheric pressure is (a) 00 (b) 00 (c) 400 (d) A spherical body of mass m and radius r is allowed to fall in a medium of viscosity h. The time in which the velocity of the body increases from zero to 0.6 time the terminal velocity (v) is called time constant t. Dimensionally t can be represented by mr 6πmrη (a) (b) 6πη g m (c) (d) None of these 6πηrv 0. Figure shows a tube having sound source at one end and observer at the other end. Source produces frequencies upto 0000 Hz. Find the frequencies at which person hears maximum intensity. Speed of sound is 400 m s. 0 cm 0 cm (a) 000 Hz (b) 500 Hz (c) 4000 Hz (d) 4500 Hz. The collision frequency of nitrogen molecule in a cylinder containing nitrogen molecule at.0 atm pressure and temperature 7 C. (Take radius of a nitrogen molecule is.0 Å) (a). 0 5 s (b) s (c) s (d). 0 5 s. A chain of length l is placed on a smooth spherical surface of radius with one of its ends fixed at the top of the sphere. What will be acceleration a of each element of the chain when its upper end is released? It is assumed that the length of chain l < p. g l g l (a) cos l (b) + l cos g l (c) + cos l (d). A train producing frequency of 640 Hz is moving towards point C with speed 7 km h. A person is sitting 7 m from point C as g l l cos 000 m Person 7 m shown in figure. Find the frequency heard by person if speed of sound is 0 m s. (a) 540 Hz (b) 660 Hz (c) 740 Hz (d) 590 Hz 4. Two Aluminium rods and a steel rod of equal crosssection area and equal length l 0 are joined rigidly side by side as shown in figure. Initially the rods are at 0 C. Find the length of the rod at the temperature q if Young s modulus of elasticity of the aluminium and steel are Y a and Y s respectively and coefficient of linear expansion of aluminium and steel are a a and a s respectively. Y Y (a) a a s s l0 α + α Aluminium + θ Ya + Ys Steel Ya a Ys s (d) l0 + α + α Aluminium θ Ya + Ys Ya a Ys s (c) l0 α α θ Ya Ys Y Y (d) a a s s l0 α α θ Ya Ys 5. Two bodies begin a free fall from the same height at a time interval of N s. If vertical separation between the two bodies is m after n second from the start of the first body, then n is equal to (a) nn (b) gn N N (c) + (d) gn gn 4 C 4

25 SOLUTIONS. (d): Let mass of neutron = m then mass of deuterium = m [ it has double nuclides thus has neutron]. Let initial velocity of neutron = v and final velocities of neutron and deuterium are v and v respectively. v v = 0 m m m v m v neutron deuterium Applying conservation of momentum mv + m(0) = mv + mv v = v + v...(i) applying conservation of energy mv = mv + ( mv ) v = v + v...(ii) from (i) and (ii), v = (v v ) + v v = v + 4v 4v v + v...(ii) 6v 4v v = 0 v v v = and v= Now fractional change in kinetic energy v K K mv mv v i f = = = 9 8 = Ki mv v 9. mv GmM GM (d): = g ( x) ( x) also = + + mv = + m GM ( x) ( + x) or mv = mg ( + x) ( + x) or / g g v = v = + x + x. (c) : Young s modulus of rubber, Y F l rubber = F = YA. l A l l On putting the values from question, F = = = 650 N Kinetic energy = Potential energy of rubber mv = F l F l v = = = 6500 m 5 0 = 5 0 = 50 m s 4. (d): If the particles collide they must be at the same point at the same time so as time is an important consideration we do not use the equation of the path. Let t be the time between projection and collision. For P we use O as origin and the x-axis along OA given x P = (60 cos 0 ) t; y P = (60 sin 0 ) t gt For Q we use A as origin and the x-axis along AO given x Q = (50 cos a) t, y Q = (50 sin a) t gt During collision x P + x Q = 00 t( cos α ) = 00...(i) Also, y P = y Q 0 = 50sinα sinα =...(ii) 5 4 Hence, from equation (ii), cos α= α= Therefore Q is projected at 6.9 to the horizontal. Then equation (i) gives, t( ) = 00 t = 09. s Therefore the particle collide.09 seconds after projection. 5. (a) : The change in internal P A energy during the cyclic process is zero. Hence, the heat supplied to the gas is equal to B C the work done by it. Hence, V DQ = W AB + W BC + W CA...(i) The work done during the process AB is zero W BC = P B (V C V B ) = n(t C T B ) = ( mol) (5/ J mol K ) (500 K) = 500 J As W CA = 500 J (given) \ DQ = (from (i)) DQ = 0 kj 6. (a) : The meniscus between the plates has cylindrical shape with radius r = d. The pressure just h inside the meniscus is A B T P0 T + P r = 0 d Now, P A = P B T or, P0 = P0 h g d T + ρ h = dρg 5

26 7. (d): AB is the ladder, let F be the horizontal force and W is the weight of man. Let N and N be normal reactions of ground and wall, respectively. Then for vertical equilibrium W = N...(i) For horizontal equilibrium N = F...(ii) Taking moments about A N (AB sin60 ) W(BC cos 60 ) = 0...(iii) Using (ii) and AB = 0 ft, BC = 4 ft, we get N B F 0 W 4 0 = 60 D W W C F = = N 0 5 = 50 5 = 0 = 7. lb 60 O A F 8. (a) : Here, let v rms = 00 m s at temperature T = 7 C = (7 + 7) K = 00 K and Pressure P = atm, Now for the same gas at temperature T = 7 C = (7 + 7) K = 400 K, P = atm, v rms =?, Using PV PV V P T 00 T = ; T V =. P T = 400 = M Again P rms and rms V v P M V v = = vrms V P \ = v V P rms P V vrms v = rms = ( 00) P V 00 v rms = m s 9. (d): None of the expressions has the dimension of time. 0. (c) : The sound wave bifurcates at the junction of the straight and the rectangular parts. The wave through the straight part travels a distance d = 0 cm and the wave through the rectangular part travels a distance d = 0 cm = 0 cm before they meet again and travel to the receiver. The path difference between the two waves received is, therefore Dd = d d = 0 cm 0 cm = 0 cm v 400 ms The wavelength of either wave is λ = =. υ υ For constructive interference, Dd = nl, where n is an integer.. or, d = n. v υ = nv 400 υ d = n= 000 n 0. 4 ft Thus, the frequencies within the specified range which cause maximum of intensity are 000 Hz, 000 Hz. (b): Mean free path λ = πd λ K T λ = = π Bd p p nk B ( T) (. 8 0 )( 90) λ= = (. 44)( 4. )( 0 ) ( ) K T v B rms = m = = 5. 0 m s \ collision frequency υ = v rms = = 46 0 λ s (d): Let m be the mass of the chain of length l. Consider an element of dl length dl of the chain at dq q an angle q with vertical, From figure, dl = dq; Force responsible for acceleration, df = (dm)g sin q; Mass of the element, dm = m l dl ; or dm = m. dθ l m df l d g mg = d θ ( sin θ) = sinθ θ l Net force on the chain can be obtained by integrating the above relation between 0 to a, we have α mg mg α mg F = sinθd θ= [ cos θ] = [ cos α] l l 0 l 0 mg = l cos l F g = = l Accelaration, a cos. m l. (b) 4. (a) 5. (c) : y = gn, y = gn ( N) y y = gn n N [ ( )] g = ( n N) N [ y y = m] N n = + gn 6

27 Class XI Units and Measurement Total Marks : 0 NEET / AIIMS Only One Option Correct Type. The dimensions of resistance are same as those of... where h is the Planck s constant, e is the charge. (a) h e h (c) e e (d) h e. The correct order in which the dimensions of length increase in the following physical quantities is. Permittivity of free space. esistance. Permeability of free space 4. Stress (a), 4,, (b),, 4, (c),,, 4 (d) 4,,,. If force F, area A and density D are taken as the fundamental quantities, the dimensions of Young s modulus Y will be (a) [F A D ] (b) [FA D ] (c) [FA D] (d) [FA D 0 ] 4. A wire has a mass (0. ± 0.00) g, radius (0.5 ± 0.005) mm and length (6 ± 0.06) cm. The maximum percentage error in the measurement of its density is (a) % (b) % (c) % (d) 4% 5. The SI unit of pressure gradient is (a) N m (b) N m (c) N m (d) N m 6. The length of a simple pendulum is about 00 cm known to have an accuracy of mm. Its period of oscillation is s determined by measuring the time Time Taken : 60 Min. for 00 oscillations using a clock of 0. s resolution. What is the accuracy in the determined value of g? (a) 0.% (b) 0.5% (c) 0.% (d) % 7. The circular scale of a screw gauge has 00 equal divisions. When it is given 4 complete rotations, it moves through mm. The least count of screw gauge is (a) cm (b) cm (c) 0.00 cm (d) cm 8. Young s modulus of steel is.9 0 N m. When expressed in CGS units of dyne cm, it will be equal to (a) (b).9 0 (c).9 0 (d) If P, Q, are physical quantities, having different dimensions, which of the following combinations can never be a meaningful quantity? (a) ( P Q) (b) PQ PQ (c) (d) ( Q ) 0. A book with many printing errors contains four different formulae for the displacement y of a particle undergoing a certain periodic motion. t () y= asin T () y = asinvt a () y t T a πt πt (4) y= ( a ) ( sin + cos T T ) 7

28 where a is the maximum displacement of the particle, v is the speed of the particle, T is the time period of motion. Then dimensionally (a) and are correct (b) and are correct (c) and 4 are correct (d) and 4 are correct. The dimensions of sb 4 (s = Stefan s constant and b = Wein s constant) are (a) [M 0 L 0 T 0 ] (b) [ML 4 T ] (c) [ML T] (d) [ML 6 T ]. The heat dissipated in a resistor can be obtained by the measurement of resistance, current and time. If the maximum error in the measurement of these quantities is %, % and % respectively, the maximum error in the determination of the dissipated heat is (a) 4% (b) 6% (c) 4 % (d) % Assertion & eason Type Directions : In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice as: (a) If both assertion and reason are true and reason is the correct explanation of assertion. (b) If both assertion and reason are true but reason is not the correct explanation of assertion. (c) If assertion is true but reason is false. (d) If both assertion and reason are false.. Assertion : S.I. units are logical and coherent. eason : S.I. system of units is a rationalised system. 4. Assertion : The dimensional formula for product of resistance and conductance is same as for dielectric constant. eason : Both have dimensions of time constant. 5. Assertion : If two quantities have same dimensions, then they must represent same physical quantities. eason : Work and torque have different dimensions. JEE MAIN / ADVANCED Only One Option Correct Type 6. If E = energy, G = gravitational constant, I = impulse and M = mass, then the dimensions of GIM are E same as that of (a) time (b) mass (c) length (d) force 7. A student measured the length of a rod and wrote it as.50 cm. Which instrument did he use to measure it? (a) A screw gauge having 50 divisions in the circular scale and pitch as mm. (b) A meter scale. (c) A vernier calliper where the 0 divisions in vernier scale matches with 9 divisions in main scale and main scale has 0 divisions in cm. (d) A screw gauge having 00 divisions in the circular scale and pitch as mm. 8. Which one of the following is dimensionally incorrect? (a) Capacitance C = [M L T 4 A ] (b) Magnetic field induction B = [ML 0 T A ] (c) Coefficient of self-induction L = [ML T A ] (d) Specific resistance r = [M L T A ] 9. In a particular system, the unit of length, mass and time are chosen to be 0 cm, 0 g and 0. s respectively. The unit of force in this system will be equivalent to (a) 0. N (b) N (c) 0 N (d) 00 N More than One Options Correct Type 0. In terms of potential difference V, electric current I, permittivity e 0, permeability µ 0 and speed of light c, the dimensionally correct equation(s) is(are) (a) µ 0 I = e 0 V (b) e 0 I = µ 0 V (c) I = e 0 cv (d) µ 0 ci = e 0 V. Planck s constant h, speed of light c and gravitational constant G are used to form a unit of length L and a unit of mass M. Then the correct option(s) is(are) (a) M c (b) M G (c) L h (d) L G. A student uses a simple pendulum of exactly m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of sec for this and records 40 seconds for 0 oscillations. For this observation, which of the following statement(s) is (are) true? (a) Error DT in measuring T, the time period, is 0.05 seconds. (b) Error DT in measuring T, the time period, is second. (c) Percentage error in the determination of g is 5%. (d) Percentage error in the determination of g is.5%. 8

29 . Let [e 0 ] denote of the permittivity of the vacuum, and [m 0 ] denote the permeability of the vacuum. If M = mass, L = length, T = time and I = electric current, then the dimensional formula of e 0 and m 0 are (a) [e 0 ] = M L T I (b) [e 0 ] = M L T 4 I (c) [m 0 ] = MLT I (d) [m 0 ] = ML T I. Numerical Value Type 4. A wire of length l = 6 ± 0.06 cm and radius r = 0.5 ± cm and mass m = 0. ± 0.00 g. Maximum percentage error in density is. 5. The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.0 cm and 5.5 cm of the main scale. The Vernier scale has 50 divisions equivalent to.45 cm. The 4 th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is cm. 6. The velocity of a transverse wave in a string is directly proportional to T and inversely proportional to µ. In a measurement, the mass applied at the end of string is.0 gm, length of string is m and mass of string is 5 gm. If possible error in measuring mass is 0. gm and that of length is mm, the percentage error in measurement of velocity is. Comprehension Type a The van der Waal s equation is P + V b T = V where P is pressure, V is molar volume and T is the temperature of the given sample of gas. is called molar gas constant, a and b are called van der Waal s constants. 7. The dimensional formula for a is same as that for (a) V (b) P (c) PV (d) T 8. The dimensional formula for ab/t is (a) [ML 5 T ] (b) [M 0 L T 0 ] (c) [ML T ] (d) [M 0 L 6 T 0 ] CHECK YOU PEFOMANCE No. of questions attempted No. of questions correct Marks scored in percentage Matrix Match Type 9. Match List I with List II and select the correct answer using the codes. List I List II A. Boltzmann constant P. [ML T ] B. Coefficient of Q. [ML T ] viscosity C. Planck constant. [MLT K ] D. Thermal conductivity S. [ML T K ] Codes: A B C D (a) P Q S (b) Q P S (c) S Q P (d) S P Q 0. Match List I with List II and select the correct answer using the codes. List I List II A. Physical quantities having same dimensions and same unit P. Torque, kinetic energy B. Physical quantities having same dimensions and different units C. Physical quantities having different dimensions and same unit D. Physical quantities having different dimensions and different units Codes: A B C D (a) T P Q,S (b) P,Q T S (c) P Q T,S (d) T Q,S P Q. Frequency, angular velocity. Potential energy, work S. Linear impulse, angular momentum T. None Keys are published in this issue. Search now! J If your score is > 80% Your preparation is going good, keep it up to get high score % Need more practice, try hard to score more next time. <60% Stress more on concepts and revise thoroughly. 9

30 . A teaspoonful of neutron star would weigh 6 billion tons Neutron stars contain some of the densest matter in the known universe. A neutron star is the remnants of a massive star that has run out of fuel. The dying star explodes in a supernova while its core collapses in on itself due to gravity, forming a super-dense neutron star. Astronomers measure the mind-bogglingly large masses of stars or galaxies in solar masses, with one solar mass equal to the Sun s mass ( x 0 0 kg. Typical neutron stars have a mass of up to three solar masses, which is crammed into a sphere with a radius of approximately ten kilometres, resulting in some of the densest matter in the known universe.. Mystery of microwave and liquids As per the latest researches worldwide, water in the liquid state has the characteristic to enable many new molecular interactions to develop. This helps in enhancing absorption of heat by food items. Due to this property, foods items like burgers become soft enough to be eaten after coming out of microwave ovens. 5 MIND BLOWING FACTS. Viscous fluids can flow at high speeds It is generally believed that viscous fluids can t flow fast enough like water a liquid with reduced viscous level. However, some scientists went on to prove that fluids like Ketchup can attain high speeds, too, if constantly sheered over a period of time till they attain momentum. Once thrust is achieved, the viscous forces dwindle down considerably and free movement is observed. 4. The Many World theory According to this theory, in addition to our universe, there are an infinite number of other universes. The theory was originally offered to resolve strange quantum interpretation of particles and their wave-particle dual nature and causality principle. In many world theory, you are not living in just one space; rather, there are infinite transcripts of you in other worlds that may happen to have a completely different behaviour. In multiple universe theory, each of your versions can have a completely different fate. The theory has many varieties, like the one that describes the whole universe as countless bubble universes constantly appearing and disappearing. This notion of the universe is rather opposed to the Big Bang theory that suggests the creation of the universe at.7 billion light years ago. 5. If you took out all the empty space in our atoms, the human race could fit in the volume of a sugar cube The atoms that make up the world around us seem solid, but are in fact over per cent empty space. An atom consists of a tiny, dense nucleus surrounded by a cloud of electrons, spread over a proportionately vast area. This is because as well as being particles, electrons act like waves. Electrons can only exist where the crests and troughs of these waves add up correctly. And instead of existing in one point, each electron s location is spread over a range of probabilities an orbital. They thus occupy a huge amount of space. 0

31 CLASS XI Chapterwise practice questions for CBSE Exams as per the latest pattern and marking scheme issued by CBSE for the academic session GENEAL INSTUCTIONS (i) All questions are compulsory. (ii) Q. no. to 5 are very short answer questions and carry mark each. (iii) Q. no. 6 to are short answer questions and carry marks each. (iv) Q. no. to 4 are also short answer questions and carry marks each. (v) Q. no. 5 to 7 are long answer questions and carry 5 marks each. (vi) Use log tables if necessary, use of calculator is not allowed. Time Allowed : hours Maximum Marks : 70 Physical World, Measurements and Kinematics SECTION A. Write the dimensions of : (i) Universal Gravitational constant and (ii) Coefficient of viscosity.. Is physics more of a philosophy or more of a mathematical science?. Name and define the SI unit of luminous intensity. 4. Two forces 5 and 0 kg wt are acting with an inclination of 0 between them. What is the angle which the resultant makes with 0 kg wt? 5. A stone tied at the end of string is whirled in a circle. If the string breaks, the stone flies away tangentially. Why? SECTION B 6. Discuss relation of physics to biology. 7. If force (F), velocity (V) and time (T) are taken as fundamental units, then find the dimensions of mass. 8. The shadow of a tower standing on a level plane is found to be 50 m longer when sun s altitude is 0 than when it is 60. Find the height of the tower m h Sun 9. Show that the area under velocity-time graph of a particle, in uniform motion gives the displacement of a particle in a given time. 0. A projectile is thrown with an initial velocity of xi + yj. The range of the projectile is twice the maximum height of the projectile. Calculate y x. O What will be the effect on maximum height of a projectile when its angle of projection is changed from 0 to 60, keeping the same initial velocity of projection?. A physical quantity p is related to four observations a, b, c and d as follows p = ab. c d

32 The percentage errors of measurement in a, b, c and d are %, %, 4% and %, respectively. What is the percentage error in the quantity p? If the value of p calculated using the above relation turns out to be 5.76, to what value should your round off the result?. Derive the relation between linear and angular velocity of a body in uniform circular motion. SECTION C. The displacement of a particle is zero at t = 0 and x at certain time t. It starts moving in the positive x-direction with a velocity which varies as v= k x, where k is a constant. Show that the velocity is proportional to time. 4. On a two-lane road, car A is travelling with a speed of 6 km h. Two cars B and C approach car A in opposite directions with a speed of 54 km h each. At a certain instant, when the distance AB is equal to AC, both being km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident? 5. Politics is the art of the possible. Similarly, Science is the art of the soluble. Explain this beautiful aphorism on the nature and practice of science. 6. The frequency depends upon (a) tension of the string, (b) length of the string, (c) linear mass density. Using dimensional analysis, derive an expression of frequency. 7. The following figure represents the velocity-time graph of a particle moving in a straight line. v (m s ) 0 A B 0 D C H G F E O 4 t(s) (i) Did the particle ever move with uniform velocity? (ii) Did the particle ever move with uniform acceleration? (iii) What is the distance traversed by the particle in 4 second? 8. Identify the types of motion in following cases : displacement O x C time B A t displacement O x D time E t distance O x time F t 9. Two ends of a train moving with a constant acceleration passes a certain point with velocities u and v. Show that the velocity with which the middle point of the train passes the same point is ( u + v ). O The velocity of a train increases at a constant rate a from 0 to v and then remains constant for some time interval and then finally decreases to zero at a constant rate b. If the total distance covered by the particle be x, then show that the total time taken x v will be t = + + v α β. ^ ^ ^ ^ 0. Let r () t = ti+ 4t j and r () t = 4t i + tj represent the positions of particles and respectively as function of time t; r () t and r () t are in m and t in s. What will be the relative speed of the two particles at the instant t = s?. In which of the following examples of motion, can the body be considered approximately a point object : (a) a railway carriage moving without jerks between two stations. (b) a monkey sitting on top of a man cycling smoothly on a circular track. (c) a spinning cricket ball that turns sharply on hitting the ground. (d) a tumbling beaker that has slipped off the edge of a table.. 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 8 km h while the other has the speed of 7 km h. The bird starts moving from first car towards the other and is moving with the speed of 6 km h and when the two cars were separated by 6 km. What is the total distance covered by the birds? What is the total displacement of the bird?. A particle starts from origin at t = 0 s with a velocity of 0 j m s and moves in the x y plane with a constant acceleration of ( 8 i + j) ms. (a) At what time is the x-coordinate of the particle 6 m? What is the y-coordinate of the particle at the time? (b) What is the speed of the particle at that instant?

33 4. State polygon law of vectors and prove it with the help of triangle law of vectors. SECTION D 5. State the rules for counting the number of significant figures in a measured quantity. O How is random error eliminated? What do you mean by (i) absolute error (ii) mean absolute error (iii) relative error and (iv) percentage error? 6. Draw and discuss the position-time graphs of two objects moving along a straight line, when their relative velocity is (i) zero (ii) positive (iii) negative. O A car accelerates from rest at a constant rate a for sometime after which it decelerates at constant rate b to come to rest. If the total time elapsed is t second, evaluate : (a) the maximum velocity reached and (b) the total distance travelled. 7. State parallelogram law of vector addition. Find analytically the magnitude and direction of resultant vector. Apply it to find the resultant when, (i) Two vectors are parallel to each other (ii) Two vectors are perpendicular to each other. O (a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by q(t) = tan ( v 0 y gt ) v0x (b) Show that the projection angle q 0 for a projectile launched from the origin is given by : q 0 = tan ( 4h ) m where the symbols have their usual meaning. SOLUTIONS. (i) Universal gravitational constant : [M L T ] (ii) Coefficient of viscosity : [M L T ]. Physics is more of a philosophy than a mathematical science because without a philosophical outlook, understanding and appreciation of physics is incomplete.. The SI unit of luminous intensity is candela (cd). One candela is defined as the luminous intensity, in a given direction, of a source that emits a monochromatic radiation of frequency hertz and has a radiant intensity in that direction of /68 watt per steradian. 4. Let q be the angle between vectors a and b and a be the angle between a and. B b q = 0 a O Here, a = 0 kg-wt, b = 5 kg-wt From the analysis of vector addition b sinθ 5 sin0 tan a = = a + b cosθ 0 + 5cos0 5sin( ) = 0 + 5cos( ) = 5 cos 0 0 5sin0 5 = = = 5 tan a = or a = 0 5. When a stone is going around a circular path, the instantaneous velocity of stone is acting tangentially to the circle. When the string breaks, the centripetal force stops to act. Due to inertia, the stone continue to move along the tangent to circular path. That is why, the stone flies off tangentially to the circular path. 6. Biology involves matter that is alive. The knowledge of physics has provided us a powerful tool to study the biological sciences. For example, electron microscope has made it possible to see even the structure of a cell. Similarly, optical microscope is widely used in the study of biology. 7. Let mass m F a V b T c or m = kf a V b T c...(i) where k is a dimensionless constant and a, b and c are the exponents. Writing dimensions on both sides, we get [ML 0 T 0 ] = [MLT ] a [LT ] b [T] c [ML 0 T 0 ] = [M a L a + b T a b + c ] Applying the principle of homogeneity of dimensions, we get a =... (ii) a A

34 a + b = 0... (iii) a b + c = 0... (iv) Solving eqns. (ii), (iii) and (iv), we get a =, b =, c = From eqn. (i), [m] = [FV T] 8. Here d = 50 m, q = 60, q = 0 d 50 h = = cotθ cot θ cot0 cot 60 \ h = = = 5 / = 5.7 = 4. m 9. For uniform acceleration, v t graph will be a straight line. Area under the curve BC is for time interval t = 0 to t is Area of BCOA = (AB + OC) OA = (v + v ) t = v av t = displacement ( v av = v + v ) Therefore area under v t graph gives displacement of the particle. 0. Here, ange of projectile = maximum height \ v sinθ v sin θ = g g sin q cos q = sin q ( sin q = sin q cos q) tan q = (i) As v cos q = x and v sin q = y y v \ x = sinθ v cos = tanθ θ = [from (i)] O Maximum height of a projection is given by h m = ( usin θ ) g where u is initial velocity and q is the angle of projection At q = 0 u sin 0 u hm = = g 4 g At q = 60 u sin 60 u hm = = g 4 g hm = h m So, maximum height at 60 will be three times the maximum height at 0. ab. p = c d p a b c d 00 = p 00 a b c d = / = = 4% Mean absolute error in p = 4% = 0.4 errors are in two significant digits. So the value of p should have two significant digits only. \ equired value of p is As the time rate of change of angular displacement is w = θ t Now, if the distance travelled by the object during time Dt is DS i.e., PP is DS, then v = S t but DS = Dq, therefore v = θ = w or v = w t. As v k x dx / dx =, = kx or = kdt dt / x Integrating both sides within the given limits, we get x dx t = k dt 0 / x 0 or / x x / 0 t 0 = k t or x = kt or x kt = Thus, v= k x = k kt k t = or v t 4. Velocity of car A = 6 km h = 0 m s ; Velocity of car B or C = 54 km h = 5 m s ; elative velocity of B w.r.t. A = 5 0 = 5 m s ; elative velocity of C w.r.t. A = = 5 m s ; As, AB = AC = km = 000 m O Dq P P Dq 4

35 000 Time available to B or C for crossing A = = 40 s 5 If car B accelerates with acceleration a, to cross A before car C does, then u = 5 m s, t = 40 s, S = 000 m, a =? Using, S = ut + at,wehave 000 = a 40 or = 800 a or a = m s 5. Nothing is impossible in politics and the politics is the art of possible. It is a well known fact that to win over votes, politicians make anything and everything possible even when they are least sure of the same. In politics, ministry may change overnight, but in science universal laws do not change overnight. Science is a systematised study of observations. A scientist patiently analyses these observations and comes out with certain laws e.g. Tycho Brahe worked for twenty long years to make observations on planetary motions. J. Kepler formulated his three famous laws of planetary motion from this huge reservoir of observations. Thus the statement that science is the art of the soluble means that a wide variety of physical processes are understood in terms of only a few basic concepts, i.e. there appears to be unity in diversity as if widely different phenomena are soluble and can be explained in terms of only a few fundamental laws. Newton s laws of gravitation are applicable throughout the universe. They are same for two small bodies as well as for the solar system. Whole of the universe can be dissolved into certain laws i.e. we can study the universe on the basis of a few laws. 6. Let the relation be : Frequency u = k L a f b m c Where k is a dimensionless constant L is the length, f is the tension and m is the linear mass density The dimensions of each quantity are : [L] = [M 0 L T 0 ] [ f ] = [M L T ] [m] = [M L T 0 ] [u] = [M 0 L 0 T ] [M 0 L 0 T ] = [M 0 L T 0 ] a [M L T ] b [M L T 0 ] c Thus, now equating the dimensions on both sides of the equation. so, b + c = 0; a + b c = 0; b = Thus, b =, c =, a = Thus the formula for frequency is υ= k f L m 7. (i) From given v t graph of the particle, the value of velocity, v is constant from A to B so particle moves with uniform velocity between the time interval t = s to t = s and also from time t = s to t = 4 s (from C to D). (ii) From time, t = 0 to t = s, graph is straight line with positive slope and from time t = s to t = s, slope is negative. So, the particle moved with constant acceleration between time interval t = 0 to t = s and t = s to t = s. (iii) Distance travelled = Area under v t graph = Area of triangle OAH + Area of rectangle HABG + Area of trapezium GBCF + Area of rectangle FCDE = 55 m 8. Graph A shows body is at rest. Graph B shows body has uniform motion, i.e., velocity is constant. Graph C shows uniform motion but negative velocity. Graph D is not a possible case, since x takes many values for same time. Graph E shows accelerated motion, since slope (velocity) increases. Graph F shows decelerated motion, since velocity decreases. 9. Let x be the total length of the train, V be the velocity of the middle point of the train while passing a certain point and a be the uniform acceleration of the train. Taking the motion of the train when middle point is passing from the given point, we have x u = u, v = V, s = Using, v = u + as, we have ax V = u + = u + ax (i) Taking the motion of train when the last end of train is passing from the given point, then u = u, v = v, a = a, s = x Now, we have, v = u + ax v u or ax = Putting this value in (i), we get v u u + v ( u + v ) V = u + = or V = 5

36 O The given motion can be represented by the velocitytime graph OABC as shown in figure. The portion OA represents the motion with constant acceleration a, the straight line AB parallel to time-axis represents the motion with uniform velocity v and the line BC represents the motion with constant deceleration b. a = slope of OA AD v = = \ OD = v OD OD α β= slopeof BC = BE v = EC EC \ EC = v β Now, total distance covered = Area under the graph OABC or x= AB+ OC AD [ ] = [ DE OD DE EC] AD [ AB = DE] = + + DE v v x v v v or = DE + + α β v α β x or v v DE = v α β Now, total time taken is given by t = OD + DE + EC v x v v = + v + v x v = + + α α β β v α β. ^ ^ 0. Here, r () t = ti+ 4t j, r () t = 4t i + tj Velocity, dr d v() t ( ti ^ ^ ^ ^ = = + 4t j) = i + 8t j dt dt dr d v() t ( 4t ^ ^ ^ ^ = = i + tj) = 8ti+ j dt dt The relative velocity of particle with respect to particle is v = v v = ( ^ i + 8t ^ j) ( 8t ^ i + ^ j) = ( ) ^ ^ 8t i + ( 8t ) j At t = s, v = ( 8) ^ i + ( 8 ) ^ j = ^ 5i + ^ 5 j elative speed of two particles at t = s, v = ( 5) + () 5 = = 5 m s. (a) The carriage can be considered a point object because the distance between two stations is very large as compared to the size of the railway carriage. ^ ^ (b) The monkey can be considered as a point object if the cyclist describes a circular track of very large radius because in that case the distance covered by the cyclist is quite large as compared to the size of monkey. The monkey can not be considered as a point object if the cyclist describes a circular track of small radius because in that case the distance covered by the cyclist is not very large as compared to the size of the monkey. (c) The spinning cricket ball can not be considered as a point object because the size of the spinning cricket ball is quite appreciable as compared to the distance through which the ball may turn on hitting the ground. (d) A beaker slipping off the edge of a table can not be considered as a point object because the size of the beaker is not negligible as compared to the height of the table.. Speed of first car = 8 km h - Speed of second car = 7 km h - \ elative speed of each car w.r.t each other = = 45 km h - Distance between the car = 6 km (given) \ Time of meeting the cars, Distance between the cars 6 4 t = = = h = 08. h elative speed of cars 45 5 Speed of the birds (v b ) = 6 km h - \ Total distance covered by the bird = v b t = = 8.8 km Total displacement of bird = distance covered by first car in 0.8 h on straight line. \ x = = 4.4 km. Here, u = 0. 0j ms at t = 0. dv a = = (. 80 i+ 0. j) ms dt So dv = (. 80 i+ 0. j) dt Integrating it within the limits of motion i.e. as time changes from 0 to t, velocity changes from u to v, we have v u= (. 80 i+ 0. j) t or v = u ti + 0. tj dr As v = or dr = vdt dt So, dr = ( u ti + 0. t jdt ) Integrating it within the conditions of motion i.e. as time changes from 0 to t, displacement is from 0 to r, we have r = ut+ t i+ t j or xi yj tj t + = i+ t j = t 40. i+ ( 0t+ t ) j 6

37 7

38 Here, we have, x = 4.0 t and y = 0 t + t \ t = (x/4) / (a) At x = 6 m; t = (6/4) / = s y = 0 + = 4 m (b) Velocity of the particle at time t is v = 0 j+ 80. ti + 0. tj When t = s, then, v = 0 j+ 80. i+ 0. j= 6 i+ 4 j Speed = v = =. 6 ms 4. Polygon law of vectors is the generalised form of triangle law of vectors and is used to add more than two vectors. According to this law : If a number of vectors can be represented both in magnitude and direction, by the sides of an open convex polygen taken in the same order, then their resultant is represented completely in magnitude and direction by the closing side of the polygon, taken in the opposite order. Let A, B, C, D be four vectors acting at the point O as shown in figure. These vectors can be represented both in magnitude and direction by the sides ab, bc, cd and de respectively of a polygon abcde, taken in the same order. e D O C A B a A If is the resultant of all these vectors, then, = A+ B+ C+ D= ab+ bc+ cd+ de = ac + cd + de ( as ab + bc = ac) = ad + de ( as ac + cd = ad) = ae ( as ad + de = ae) We find that is given by ae which is the closing side of the polygon taken in the opposite order. 5. efer to answer 96, page no. 5 (MTG CBSE Champion Class ). O efer to answer 7, page no. (MTG CBSE Champion Class ). 6. efer to answer 60, page no. 80 (MTG CBSE Champion Class ). b B c C D d O efer to answer 8, page no. 74 (MTG CBSE Champion Class ). 7. efer to answer 58, page no. 0 (MTG CBSE Champion Class ). O efer to answer 5, page no. 4 (MTG CBSE Champion Class ). Be an Engineer Specialising in Arts JNU will offer BTech, MS or MTech courses with arts as a specialisation stream Known for its liberal arts education, the Jawaharlal Nehru University (JNU), New Delhi will provide degree in engineering in the upcoming academic session. The new establishment at its campus, the School of Engineering would offer five-year dual degree wherein students would pursue BTech in an engineering discipline and the Master s in arts in any related subjects. For the first batch, starting this academic season, students can apply for BTech in Computer Science and Engineering, and Electronics and Communication Engineering only while BTech in civil engineering, mechanical engineering, chemical engineering, and environmental engineering will also be available for next batches. Apart from the new-age technical courses such as big data analytics, data mining, AI, algorithm, VLSI design, computer design etc, the students will also be exposed to the interdisciplinary subjects including Korean studies, environmental science, computational biology, computational finance, computational linguistic, management and entrepreneurship. We would give due weightage to social science in our engineering programmes. While students are free to choose a Master s in the engineering related subjects, we would expose students to an array of elective subjects. This would make the entire degree socially-relevant and interdisciplinary, said amesh Kumar Agarwal, dean, School of Engineering, JNU. From semester six onwards, students would chose two elective subjects as a minor course. The elective in management and entrepreneurship would be in line with the BTech-MBA trend and would make students have an edge over the upcoming entrepreneurship culture. He also suggested that Master s in computation finance is also structured based on the industry needs which would help students excel in the data-driven workplaces. JNU will take support from IIT (Indian Institute of Technology) faculty to teach some core subjects. The university will not hold any entrance test as the admissions will be done on the basis on Joint Seat Allocation Authority 08 (JoSAA) and JEE (Joint Entrance Examination) Mains score. Students can apply at university website - 8

39 F CUS Class XII NEET/JEE09 Focus more to get high rank in NEET, JEE (Main and Advanced) by reading this column. This specially designed column is updated year after year by a panel of highly qualified teaching experts well-tuned to the requirements of these Entrance Tests. UNIT - : ELECTOSTATICS The branch of physics, which deals with the study of charges at rest, the forces between the static charges, fields and potentials due to these charges is called electrostatics. Basic properties of charge Charge is transferable : If a charged body is put in contact with an uncharged body, the uncharged body becomes charged due to transfer of electrons from one body to the other. Charge is always associated with mass, i.e., charge can not exist without mass though mass can exist without charge. So, the presence of charge itself is a convincing proof of existence of mass. Quantization of charge : Total charge on a body is always an integral multiple of a basic unit of charge denoted by e and is given by q = ne where n is any integer, positive or negative and e = C. The basic unit of charge is the charge that an electron or proton carries. By convention the charge on electron is e ( C) and charge on proton is +e ( C). Additivity of charge : Total charge of a system is the algebraic sum (i.e. sum is taking into account with proper signs) of all individual charges in the system. Conservation of charge : Total charge of an isolated system remains unchanged with time. In other words, charge can neither be created nor be destroyed. Conservation of charge is found to hold good in all types of reactions either chemical or nuclear. Charge is invariant : Charge is independent of the frame of reference. Like charges repel each other while unlike charges attract each other. charging of a Body Charging a body means transfer of charges (i.e., electrons) from one body to the other. A body can be positively charged by losing some of its electrons and it can be negatively charged by gaining electrons. Methods of charging : A body can be charged by Friction Induction Conduction Charging by induction is preferred because one charged body can be used to charge any number of uncharged bodies without any loss of charge. If q be the source of charge, then charge induced on a body of dielectric constant K is given by q = q K 9

40 For metals, K = \ q = q i.e., charges induced are equal and opposite only in case of conductors. In general, magnitude of induced charge is less than that of inducing charge. coulomb s law It states that the electrostatic force of interaction (repulsion or attraction) between two electric charges q and q separated by a distance r, is directly proportional to the product of the charges and inversely proportional to the square of distance between them and acts along the straight line joining two charges. kqq 9 F =, where k = = 9 0 N m C r 4πε0 is the proportionality constant. e 0 = C N m is permittivity of free space. Coulomb s law in vector form qq ^ F = forceonq due to q = r 0 r 4πε where r is a unit vector in the direction from q to q F q r = r r q F F r O (a) r r^ q q q > 0 q r^ q q < 0 q F (b) F q F Similarly, qq ^ F = force on q due to q = r 4 0 r πε F = F elative permittivity or dielectric constant : The relative permittivity or the dielectric constant (e r or K) of a medium is defined as the ratio of the permittivity e of the medium to the permittivity e 0 of free space. ε ie.., εr or K = ε 0 The Coulomb s force between two charged particles in a medium is qq qq Fmedium = 4 r = πε 4πε 0εr r F Fmedium = air K principle of superposition According to principle of superposition, force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges. In a system of charges q, q... q n, the total force F on the charge q, due to all other charges, is then given by the vector sum of the forces F, F,..., F n. F = F + F F n qq ^ qq ^ qq n ^ = r r... r n πε r r r n q n qi ^ = r i 4πε0 i = r i continous charge distribution Linear charge density : Charge per unit length is called linear charge density. It is denoted by symbol l. λ= Charge Length ; Its SI unit is C m. Surface charge density : Charge per unit area is called surface charge density. It is denoted by symbol s. σ= Charge ; Its SI unit is C m. Area Volume charge density : Charge per unit volume is called volume charge density. It is denoted by symbol r. ρ= Charge Volume ; Its SI unit is C m. electric field The electric field at a point is defined as the force experienced by unit positive charge (called test charge) placed at that point. F Electric field, E = lim q 0 0 q0 where q 0 is the test charge. The direction of E is the same in the direction of F. 40

41 The electric field is sometimes called electric intensity or electric field intensity. The electric field is a vector quantity and its SI unit is N C or V m. Electric field due to a point charge q at a distant r from a charge is q E =. 4πε 0 r Electric field due to a system of charges is obtained by applying superposition principle. electric field lines An electric field line is a path, straight or curved such that tangent to it at any point gives the direction of electric field at that point. Properties of Electric Field Lines Electric field lines start from positive charges and end at negative charges. If there is a single charge, they may start or end at infinity. Two electric field lines can never cross each other. Electric field lines do not form closed loops. The electric field lines are closer where the electric field is stronger and the lines spread out where the electric field is weaker. In a uniform electric field, the electric field lines are equidistant, parallel straight lines. Electric field lines start or end normally on the surface of a conductor. Electric field lines do not exist inside a conductor. Electric field lines due to different charge configurations are as shown in the figure. +q q Electric dipole moment : It is a vector quantity whose magnitude is equal to product of the magnitude of either charge and distance between the charges. i.e. p = q a. By convention the direction of dipole moment is from negative charge to positive charge. The SI unit of electric dipole moment is C m. Electric field on axial line (end on position) of an electric dipole Electric field at a point on axial line of electric pr dipole is E = 4πε 0 ( r a ) where r is the distance of the point from the centre of the electric dipole p For r > > a, E = 4πε0r The direction of the electric field on axial line of an electric dipole is along the direction of the dipole moment (i.e. from q to q). Electric field on equatorial line (board on position) of an electric dipole Electric field at a point on equatorial line of electric p dipole is E = 4πε 0 ( r + a ) / p For r > > a, E = 4πε 0 r The direction of the electric field on equatorial line of the electric dipole is opposite to the direction of the dipole moment. (i.e. from q to q). The electric field at point P P due to an electric dipole is r p E = 4 r + cos θ q +q πε 0 a electric field due to a uniformly charged ring electric dipole +q q c It is a pair of two equal and opposite charges separated by a small distance. Electric field at a point on the axis of uniformly charged ring at a distance r from its centre is qr E = 4πε 0 ( r + a ) / where q is the charge on the ring and a is the radius of the ring. q For r >> a, E = 4πε 0 r At the centre of the ring, E = 0. 4

42 electric dipole in a uniform electric field When an electric dipole of dipole moment p is placed in a uniform electric field E, it will experience a torque and is given by τ= p E ; τ= pe sin θ where q is the angle between p and E. Torque acting on a dipole is maximum (t max = pe) when dipole is perpendicular to the field and minimum (t = 0) when dipole is parallel or antiparallel to the field. When a dipole is placed in a uniform electric field, it will experience only torque and the net force on the dipole is zero while when it is placed in a non uniform electric field, it will experience both torque and net force. potential energy of an electric dipole in a uniform electric field Potential energy of an electric dipole in a uniform electric field is U = pe(cosq cosq ) where q is the initial angle between p and E and q is the final angle between p and E. If the dipole is rotated from q = 0 (aligned parallel to E ) to q = q, then U = pe(cosq ) = pe ( cosq) If the dipole is rotated from q = 90 (aligned perpendicular to E ) to q = q, then U = pecosθ = p E electric flux Electric flux Df through an area element S in an electric field E is defined as φ= E S = E Scos θ where q is the angle between E and outward normal to the area element. It represents the number of electric field passing the area element. Electric flux is a scalar quantity. gauss s law It states that the total electric flux linked with a closed surface is /e 0 times the total charge enclosed by the closed surface. E ds q = S ε 0 where q is total charge enclosed by the closed surface S. The term q on the right side of Gauss s law includes the sum of all charges enclosed by the closed surface. The charges may be located anywhere inside closed the surface. The total electric flux through a closed surface is zero if no charge is enclosed by the surface. Gauss s law is true for any closed surface, regardless of its shape or size. The surface that we choose for the application of Gauss s law is called the Gaussian surface. In the situation when the surface is so chosen that there are some charges inside and some outside, the electric field (whose flux is calculated) is due to all the charges, both inside and outside the closed surface. However, the term (q) represents only the total charge inside the closed surface. Gauss s law is based on inverse square dependence on distance. applications of gauss s law Electric field due to thin infinitely long straight wire of uniform linear charge density l is λ E = πε0r where r is the perpendicular distance of the observation point from the wire. Electric field due to uniformly charged thin spherical shell of uniform surface charge density s and radius at a point distant r from the centre of the shell is given as follows : At a point outside the shell i.e., r > q E = 4πε 0 r At a point on the surface of the shell i.e., r = q E = 4πε 0 At a point inside the shell i.e., r < E = 0 Here, q = 4p E s The variation of E with r for a uniformly charged thin spherical shell is as shown in r the figure. O r= 4

43 Electric field due to a uniformly charged non conducting solid sphere of uniform volume charge density r and radius at a point distant r from the centre of the sphere is given as follows : At a point outside the sphere i.e., r > q E = 4πε r 0 At a point on the surface of the sphere i.e., r = q E = 4πε 0 At a point inside the sphere i.e., r < ρr qr E = = ε 4πε, 0 0 Here, q= 4 π ρ The variation of E with r for a uniform charged non conducting sphere is as shown in the figure. O r r = Electric field due to a infinite thin plane sheet of uniformly charge surface density s is E = σ ε 0 E is independent of r, distance of the point from sheet. Electric field due to two thin infinite parallel sheets of uniform surface density +s and s, is given as follows : At a point anywhere in the space between the two sheets E = σ ε 0 At point outside the sheets, E = 0. electric potential Electric potential at a point is defined as amount of work done in bringing a unit positive charge from infinity to that point. It is denoted by symbol V. V = W q Electric potential is a scalar quantity. The SI unit of potential is volt and its dimensional formula is [ML T A ]. Electric potential due a point charge q at a distance r from a charge is q V = 4 πε0 r E Electric potential due to system of charges : The electric potential at a point due to a system of charges is equal to the algebraic sum of the electric potentials due to individual charges at that point. q q q q n n qi V = r r r r 4 r πε = 0 n πε 0 i= i Electric potential at any point due to an electric dipole The electric potential at point P due to an electric dipole is V = πε 4 0 pcosθ r When the point P lies on the axial line of dipole i.e., q = 0. p V = 4πε0r When the point P lies on the equatorial line of the dipole, i.e., q = 90 \ V = 0. Electric potential due to a uniformly charged spherical shell of uniform surface charge density s and radius at a point distant r from the centre of the shell is given as follows : At a point outside the shell i.e., r > q V = 4πε0 r At a point on the surface of the shell i.e., r = q V = 4πε0 At a point inside the shell i.e., r > q V = 4πε 0 Here, q = 4p V s The variation of V with r for a uniformly charged thin spherical shell is shown in the O r r = figure. Electric potential due to a uniformly charged nonconducting solid sphere of uniform volume charge density r and radius at a point distant r from the centre of sphere is given as follows : At a point outside the sphere i.e., r > q V = πε r 4 0 At a point on the surface of the sphere i.e., r = q V = 4πε0 r P q +q a 4

44 At a point inside the sphere i.e., r < q( r ) V = 4πε 0 Here q= 4 π ρ equipotential surface An equipotential surface is a surface with a constant value of potential at all points on the surface. Properties of an Equipotential Surface Electric field lines are always perpendicular to an equipotential surface. Work done in moving an electric charge from one point to another on an equipotential surface is zero. Two equipotential surfaces can never intersect one another. elationship between E and V ; E = V where = ^ + ^ + ^ i j k x y z Negative sign shows that the direction of E is the direction of decreasing potential. electric potential energy Electric potential energy of a system of charges is the total amount of work done in bringing the various charges to their respective positions from infinitely large mutual separations. The SI unit of electrical potential energy is joule. Electric potential energy of a system of two charges is qq U = 4πε0 r where r is the distance between q and q. Electric potential energy of a system of n point charges qq j k U = 4πε0 all pairs rjk Note in this summation, we should include only one term for each pair of charges. conductors Those substances which can easily allow electricity to pass through them are called conductors. They have a large number of free charge carriers that are free to move inside the material. Metals, human and animal bodies and earth are conductors. Basic electrostatics properties of a conductor Inside a conductor, electric field is zero. At the surface of a charged conductor, electric field must be normal to the surface at every point. The interior of a conductor can have no excess charge in the static situation. Electric potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface. Electric field at the surface of a charged conductor E = σ ^ n ε 0 where s is the surface charge density and n is a unit vector normal to the surface in the outward direction. Electrostatic shielding : It is the phenomenon of protecting a certain region of space from external electric field. capacitor A condenser or a capacitor is a device that stores electric charge. It consists of two conductors separated by an insulator or dielectric. The two conductors carry equal and opposite charges ±Q. In an electrical circuit, a capacitor of fixed capacitance is represented by the symbol as shown in figure (a) while a capacitor of variable capacitance is represented by the symbol as shown in figure (b). (a) types of capacitors Parallel plate capacitor : It consists of two similar flat conducting plates, arranged parallel to one another, separated by a distance. Its capacitance is given by A C = ε 0 (when air is between the plates) d C = Kε 0 A (when dielectric is between the plates) d where A is area of each plate and d is separation between the two plates. When a dielectric slab of thickness t and dielectric constant K is introduced between the plates, then the capacitance of a parallel plate capacitor is given by A ε 0 C = d t K (b) 44

45 When a metallic conductor of thickness t is introduced between the plates, then capacitance of a parallel plate capacitor is given by A C = ε 0 d t Cylindrical capacitor : It consists of two co-axial cylinders of same length. Capacitance of an air filled cylindrical capacitor is πε0l C = b ln a where a and b are the inner and outer radii and L is the length. Spherical capacitor : It consists of two concentric spherical shells. Capacitance of an air filled spherical capacitor is ab C = 4πε0 b a where a and b are the inner and outer radii. Grouping of Capacitors Capacitors in series : For n capacitors connected in series, the equivalent capacitance C S is given by = CS C C Cn Capacitors in parallel : For n capacitors connected in parallel, the equivalent capacitance C P is given by C P = C + C C n When capacitors are connected in series, the charge through each capacitor is same. When capacitors are connected in parallel, the potential difference across each capacitor is same. energy stored in a capacitor Work done in charging a capacitor gets stored in the capacitor in the form of its electric potential energy and it is given by Q U = CV = QV = C The energy stored per unit volume in the electric field between the plates is called energy density (u). It is given by u= ε E 0 sharing of charges When two capacitors charged to different potentials are connected by a conducting wire, charge flows from the one at higher potential to the other at lower potential till their potentials become equal. The equal potential is called common potential. It is given by Total charge Q+ Q CV + CV V = = = Total capacity C+ C C+ C In sharing charges, there is absolutely no loss of charge. Some energy is, however, lost in the process in the form of heat etc which is given by U CC ( V V ) U = ( C + C ) effect of dielectric When a dielectric slab of dielectric constant K is introduced between the plates of a charged parallel plate capacitor and the charging battery remains connected, then Potential difference between the plates remains constant i.e., V = V 0 Capacitance C increases i.e., C = KC 0 Charge on a capacitor increases i.e., Q = KQ 0 Electric field between the plates remains unchanged i.e., E = E 0 Energy stored in a capacitor increases i.e., U = KU 0 When a dielectric slab of dielectric constant K is introduced between the plates of a charged parallel plate capacitor and the charging battery is disconnected, then Charge remains unchanged i.e., Q = Q 0 Capacitance increases i.e., C = KC 0 Potential difference between the plates decreases i.e., V = V0 K Electric field between the plates decreases i.e., E = E0 K Energy stored in the capacitor decreases i.e., U = U0 K where Q 0, C 0, V 0, E 0 and U 0 represents the charge, capacitance, potential difference, electric field and energy stored in the capacitor of a charged air filled parallel plate capacitor.. 45

46 KINEMATICS Uniform Circular Motion A particle moves along a circular path with a constant speed. Angular displacement, θ = s r ; Angular velocity, ω = θ t Also, ω = π ; Linear velocity, v = r ω T Centripetal acceleration, a = v = r ω O Linear acceleration, a = rα r v a θ r Non-Uniform Circular Motion Circular motion with variable speed esultant acceleration of the particle a= at + ar a r = v /r = radius acceleration a t = tangential acceleration Average speed total distance v av = total time Average velocity r vav = t For infinitesimally small change in time Instantaneous velocity r dr v = Lt = t 0 t dt Differentiating with respect to time Acceleration The time rate of change of velocity, dv a = dt Av. accn v aav = t Instantaneous speed d v = Lt t 0 t Speed The rate of distance covered with time is called speed, v distance = = total time d t Displacement The shortest distance between the two positions of a body in a particular direction. Differentiating with respect to time Velocity The rate of change of displacement displacement r v = = time t If a is constant then v = u + at v = ut + (/)at v = u + as S n = u + (a/)(n ) If a = f(t) a function of time t v= u+ f () t dt 0 t s= ut+ ( f ( tdtdt ) ) 0 Distance The actual path length covered by moving particle. est or Motion It is relative term and depends on frame of reference. Frame of reference A system consisting a set of coordinates and with reference to which observer describes any event. The observers are moving in a particular direction x at a relative velocity of v and each observer has their own set of coordinates (x, y, z) and (x, y, z ). x(m) v(m s ) a(m s ) 60 Constant slope means constant velocity Increasing slope 5 indicates increasing velocity Positive slope implies positive acceleration. 4 0 Basic Terminology Graphical epresentation 5 Distance Decreasing slope indicates decreasing velocity. 0 5 Time (in s) 0 0 MOTION Velocity 5 Horizontal line (zero slope) implies zero acceleration. Circular Motion Horizontal point (zero slope) implies zero velocity. Acceleration 5 elative Motion Flattening curve implies velocity becoming less negative. Negative slope implies negative acceleration. 0 0 Projectile Motion Negative slope indicates negative velocity. 5 5 Continuing zero slope implies object is at rest. 0 0 Projectile Fired Horizontally g Equation of trajectory : y = u x. Velocity after time t : v= u + g t β= tan gt u Time of flight : T = h g Horizontal range : = u Multiplication by a Scalar Quantity Multiplication of a vector by a scalar φ changes its magnitude by a factor of φ. h g elative Velocity According to vector law of addition rps = rps + rss dr dr dr PS = PS + SS dt dt dt So, v v v PS = PS + SS ie.., v v v actual = relative + reference vrelative = vactual vreference esolution of Vector Horizontal component of A, Ax = A cosθ Vertical component of A, Ay = A sinθ A= Ax + A y, tanθ= A y A x Projectile Fired at an Angle with Horizontal Position after time t : x = (u cos θ)t, y = (u sin θ)t gt Equation of trajectory : g y = xtanθ. x u cos θ Maximum height : H u sin θ = g u Time of flight : T = sinθ g Horizontal range : = u sinθ g Parallelogram Law For two coinitial vectors represented by the two adjacent side of a parallelogram, the diagonal of the parallelogram so formed will be the resultant. = A + B + ABcosθ Triangle Law If two vectors are represented by two sides of a triangle in same order, the resultant will be the third side but in opposite order. C = A + B Addition of Vectors VECTO Physical quantities having both magnitude and direction

47 . Four charges, +q, +q, q, q are placed on the circumference of a circle of radius r. What is the force on the charge +Q placed at the centre of the circle as shown in the figure? Qq (a) (b) 4πε 0 r 8Qq (c) (d) 4πε r 0 4πε 4πε 0 0 8Qq r Qq r. The figure shows two charged particles on x-axis. The particles are free to move. However, at one point, a third charged particle can be placed such that all three particles would be in equilibrium.. Is the third particle positively charged?. Is the point where the third particle is placed to the left, right or between the two charges? q q (a) negative, to their right (b) positive, between them (c) negative, to their left (d) positive, to their right. Three point charges +q, q and +q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = a, y = 0, z = 0), respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are (a) qa along + y direction (b) qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0) (c) qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0) (d) qa along + x direction 4. An electric dipole consists of two opposite charges each 0.05 mc separated by 0 mm. The dipole is placed in an uniform external electric field of 0 6 N C. The maximum torque exerted by the field on the dipole is (a) 6 0 N m (b) 0 N m (c) 5 0 N m (d).5 0 N m 5. A thin glass rod is bent into a semi circle of radius r. A charge +q is uniformly distributed along the upper half and a charge q is uniformly distributed along the lower half, as shown in the figure. The magnitude and direction of the electric field E produced at P, the centre of the circle, will be (a) 0 q (b) perpendicular to the line OP and επ 0 r directed downward q (c) perpendicular to the line OP and επr 0 directed downward q (d) along the axis OP επr 0 6. The total electric flux through a cube when a charge 8q is placed at one corner of the cube is ε (a) e 0 q (b) 0 (c) 4pe 0 q (d) q q ε 0 7. A non conducting sphere of radius a has a net charge +q uniformly distributed throughout its volume. A spherical conducting shell having inner and outer, radii b and c and net charge q is concentric with the sphere (see the figure). ead the following statements (i) The electric field at a distance r from the center qr of the sphere for r < a = 4πε0 a (ii) The electric field at distance r for a < r < b = 0 (iii) The electric field at distance r for b < r < c = 0 (iv) The charge on the inner surface of the spherical shell = q (v) The charge on the outer surface of the spherical shell = + q Which of the above statements are true? (a) (i), (ii) and (v) (b) (i), (iii) and (iv) (c) (ii), (iii) and (iv) (d) (ii), (iii) and (v) 8. In the uniform electric field of E = 0 4 N C, an electron is accelerated from rest. The velocity of the 48

48 electron (in m s ) when it has travelled a distance of 0 m is nearly (e/m of electron.8 0 C kg ) (a) (b) (c) (d) Positive and negative point charges of equal a a magnitude are kept at 00,, 00,, and respectively. The work done by the electric field when another positive point charge is moved from ( a, 0, 0) to (0, a, 0) is (a) positive (b) negative (c) zero (d) depends on the path connecting the initial and final positions. 0. A charge +q is placed at the origin O of X-Y axes as shown in the figure. The work done in taking a charge Q from A to B along the straight line AB is qq a b qq b a (a) 4πε 0 ab (b) 4πε 0 ab qq b qq a (c) 4πε a b (d) 4πε b b 0. The electric field and the potential of an electric dipole vary with distance r as (a) and r r (b) and r r (c) and (d) and r r r r. n identical droplets are charged to V volt each. If they coalesce to form a single drop, then its potential will be (a) n / V (b) n / V (c) nv (d) V/n. A small conducting sphere of radius r is lying concentrically inside a bigger hollow conducting sphere of radius. The bigger and smaller spheres are charged with Q and q (Q > q) and are insulated from each other. The potential difference between the spheres will be q Q Q q (a) 4πε 0 r (b) + 4πε0 r (c) q q 4πε0 r (d) 0 Q q 4πε0 r 4. A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral. Which one is correct statement? (a) A potential difference appears between the two cylinders when a charge density is given to the inner cylinder. (b) A potential difference appears between the two cylinders when a charge density is given to the outer cylinder. (c) No potential difference appears between the two cylinders when a uniform line charge is kept along the axis of the cylinders. (d) No potential difference appears between the two cylinders when same charge density is given to both the cylinders. 5. Consider a system of three charges q q q, and placed at points A, B and C, respectively, as shown in the figure. Take O to be the centre of the circle of radius and angle CAB = 60. Then q (a) the electric field at point O is directed along the negative x-axis 8πε0 (b) the potential energy of the system is zero (c) the magnitude of the force between the charges q at C and B is 54πε 0 q (d) the potential at point O is πε identical spheres of charge q and capacitance C each are combined to form a large sphere. The charge and capacitance of the large sphere is (a) 64q, C (b) 6q, 4C (c) 64q, 4C (d) 6q, 64C 7. A parallel plate capacitor of plate area A and plates separation distance d is charged by applying a potential V 0 between the plates. The dielectric constant of the medium between the plates is K. What is the uniform electric field E between the plates of the capacitor? (a) E = e 0 CV 0 / KA (b) E = V 0 / Kd (c) E = V 0 / KA (d) E = K V 0 d/ e 0 A 8. The capacitance of a parallel plate capacitor with air as dielectric is C. If a slab of dielectric constant K and of 49

49 the same thickness as the separation between the plates is introduced so as to fill (/4) th of the capacitor (shown in figure), then the new capacitance is C C (a) ( K + ) (b) ( K + ) 4 4 C (c) ( K +) (d) None of these 4 9. The number of ways one can arrange three identical capacitors to obtain distinct effective capacitances is (a) 8 (b) 6 (c) 4 (d) 0. A parallel plate capacitor is charged to a potential difference of 50 volts. It is then discharged through a resistance for seconds and its potential drops by 0 volts. Calculate the fraction of energy stored in the capacitance. (a) 0.4 (b) 0.5 (c) 0.50 (d) 0.64 SOLUTIONS. (b): Force on charge Q due to Qq charge q at A, F = 4 πε 0 r along AO Force on charge Q due to charge q at B, Qq F = along BO 4 πε 0 r Force on charge Q due to charge q at C, Q( q) F = along OC 4πε 0 r Force on charge Q due to charge q at D, Q( q) F4 = 4πε along OD 0 r F = F = F = F 4 = F The resultant force on charge Q is Fnet = ( F) + ( F) + ( F) ( F ) cos90 8Qq = F 8 = 4πε 0 r. (b): The third particle is positively charged because q and q will repel each other and the third charge will keep them in position. Therefore the third particle is positively charged. This has to be between two, nearer q. Without the third charge, the charges repel each other. To keep them in position if a +ve charge is kept between them, near to q, the force of attraction will be equal on both, if the distance from q is times the distance from q. (b) : O esultant electric dipole moment is along OC where C = (a, a, 0) p = p + p = ( qa) + ( qa) = qa 4. (d): Torque t = pe sinq For maximum torque, q = 90 \ t = pe = =.5 0 N m 5. (b) 6. (d): Since charge 8q is placed at one corner of the cube, it can be imagined to be placed at the centre of a large cube which can be formed using 8 similar cubes and arranging them as shown. Now 8q is at centre of the 8 cubes arranged to form a closed box. Total flux through the bigger cube = 8q ε 0 8q q Flux through one small cube = = 8 ε0 ε0 7. (b) (Gauss s law) 8. (d) : Starting from rest, the velocity of the electron when it has travelled distance s in a uniform electric field is v = ee as = m s 4 v = = m s 50

50 5

51 9. (c) : It can be seen that potential at the points both A and B are zero. When the charge is moved from A to B, work done by the electric field on the charge will be zero. q 0. (a) : Potential at point A is VA = 4πε0 a q Potential at point B is VB = 4πε0 b Work done in taking a charge Q from A to B is Qq Qq a b W = Q VB VA = b a = ( ) 4πε 4 ab 0 πε0. (d) q. (a) : Potential of droplet = 4πε0 r 4 4 / πr n = πr or r = rn For bigger drop, nq Vn = nq = V = V n / 4πε0 r 4πε / n. 0 n r Q q. (c) : Potential at A,VA = +, πε 4 0 Q q Potential at B, VB = + πε r. 4 0 The potential difference between the spheres is q q VB VA = r πε (a) : There will appear a potential difference between the two cylinders if there is an electric field in the space between them, since the potential difference is related to the electric field. 5. (c) : The electric field at O due to the two charges q/ will get cancelled. Electric field due to 60 q will be directed in ve x-axis. q q E = ( / ) = 4πε 0 6πε0 Potential energy of the system q q q q q = πε0 sin60 cos 60 The magnitude of the force between charges at B and C, q q F = = q 4πε 0 ( sin 60 ) 54 πε0 The electric potential at point O, q q q + V = = 0 4πε 0 6. (c) : 64 drops have formed a single drop of radius. Volume of large sphere = 64 Volume of small sphere 4 = 64 4 π πr = 4r Therefore charge of large sphere Q total = 64q and capacitance of large sphere, C = 4πε ; C = ( 4πε ). 4r C = 4C 0 0 E 7. (b): Dielectric constant, K = 0 E E0 V0 V0 E = = E0 = K Kd d 8. (b): If A be area of each plate and d is the distance A between the plates, then C = ε 0 d Kε0( A/ 4) Kε0A C = = d 4d ε0( A / 4) ε0a and C = = d 4d As C and C are in parallel. \ The new capacitance, C = C + C Kε0A ε0a C C = + = ( K + ) 4d 4d 4 9. (c) 0. (d): Initial energy stored in the capacitor, Ui = CV = C( 50) After s, when the potential drops by 0 V, the final potential is 40 V. Final energy stored in the capacitor, U f = C( 40) Fraction of energy stored is U C( 40) f = = U i 40 = C( 50) 5

52 CLASS XII Chapterwise practice questions for CBSE Exams as per the latest pattern and marking scheme issued by CBSE for the academic session GENEAL INSTUCTIONS (i) All questions are compulsory. (ii) Q. no. to 5 are very short answer questions and carry mark each. (iii) Q. no. 6 to are short answer questions and carry marks each. (iv) Q. no. to 4 are also short answer questions and carry marks each. (v) Q. no. 5 to 7 are long answer questions and carry 5 marks each. (vi) Use log tables if necessary, use of calculator is not allowed. Time Allowed : hours Maximum Marks : 70 SECTION A. If the electron s drift speed is so small, and the electron s charge is small, how can we still obtain large amount of current in a conductor?. Two conducting wires X and Y of same diameter but different materials are joined in series across a battery. If the number density of electrons in X is twice that in Y, find the ratio of drift velocity of electrons in the two wires.. Two identical cells, each of emf E, having negligible internal resistance, are connected in parallel with each other across an external resistance. What is the current through this resistance? 4. What are Ohmic and non-ohmic conductors? Give one example of each. 5. What is the power output of the source battery if it is short circuited? What is power dissipation inside the battery in that case? Current Electricity SECTION B 6. The variation of potential difference with length in case of two potentiometers A and B is given. Which of the two is more sensitive? 7. A cylindrical metallic wire is stretched to increase its length by 5%. Calculate the percentage change in its resistance. 8. Find the value of current in the circuit shown. 9. A dry cell of emf.6 V and internal resistance 0. W is connected to a resistor of resistance ohm. If the current drawn from the cell is A, (i) what is the voltage drop across? (ii) what is the rate of energy dissipation in the resistor? 5

53 0. Is the motion of a charge across junction momentum conserving? Why or why not?. Define relaxation time of electrons in a conductor. Explain how it varies with increase in temperature of a conductor. O Two metallic wires of the same material and same length but different cross-sectional areas are joined together (i) in series (ii) in parallel, to a source of emf. In which of the two wires will the drift velocity of electron be more in each of the two cases and why?. In a discharge tube, the number of hydrogen ions (i.e., protons) drifting across a cross-section per second is.0 0 8, while the number of electrons drifting in the opposite direction across another cross-section is per second. If the supply voltage is 0 V, what is the effective resistance of the tube? SECTION C. Let there be n resistors... n with max = max (... n ) and min = min {... n }. Show that when they are connected in parallel, the resultant resistance P < min and when they are connected in series, the resultant resistance S > max. Interpret the result physically. 4. Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter mm. Conductor B is a hollow tube of outer diameter mm and inner diameter mm. Find the ratio of resistance A to B. 5. Two cells of same emf E but internal resistance r and r are connected in series to an external resistor as shown in the figure. What should be the value of so that the potential difference across the terminals of the first cell becomes zero. O Two cells of voltage 0 V and V and internal resistances 0 W and 5 W respectively, are connected in parallel with the positive end of 0 V battery connected to negative pole of V battery as shown in the figure. Find the effective voltage and effective resistance of the combination. 6. Potential difference V across terminals of a cell were measured (in volts) against different current V V (in ampere) flowing through the cell. A graph was drawn which was a straight line ABC. Using the data given in the graph determine, (i) the e.m.f. and (ii) the internal resistance of the cell. 7. A cell of emf e and internal resistance r is connected across a variable external resistance. Plot graphs to show variation of (i) E with (ii) Terminal potential difference of the cell (V) with. 8. A number of identical cells, n, each of emf e, internal resistance r connected in series are charged by a d.c. source of emf e', using a resistor. (i) Draw the circuit arrangement. (ii) Deduce the expressions for (a) the charging current and (b) the potential difference across the combination of the cells. 9. Six equal resistors, each of value, are joined together as shown in the given figure. Calculate the equivalent resistance across AB. If a supply of emf e is connected across AB, compute the current through the arms DE and AB. 0. Answer the following : (a) Why are the connections between the resistors in a meter bridge made of thick copper strips? (b) Why is it generally preferred to obtain the balance point in the middle of the metre bridge wire? (c) Which material is used for the meter bridge wire and why?. Find the value of the unknown resistance X, in the following circuit, if no current flows through the section AO. Also calculate the current drawn by the circuit from the battery of emf 6 V and negligible internal resistance. 54

54 . Three cells are connected in parallel with their like poles connected together with wires of negligible resistance. If the emfs of the cells are V, V and 4 V respectively and their internal resistances are 4 W, W and W respectively, find the current through each cell. O State Kirchhoff s rules. Use these rules to determine the value of currents I, I and I in the circuit as shown V I I 0 I 60 I 4 V. In a meter bridge, the null point is found at a distance of 40 cm from A. If a resistance of W is connected in parallel with S, the null point occurs at 50.0 cm from A. Determine the values of and S. 4. AB is m long uniform wire of 0 W resistance. The other data are as shown in the circuit diagram. Calculate (i) potential gradient along AB (ii) length AO of the wire, when the galvanometer shows no deflection. SECTION D 5. (a) Derive the relation between current density j and potential difference V across a current carrying conductor of length l, area of cross-section A and the number density n of free electrons. (b) Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area m carrying a current of.5 A. [Assume that the number density of conduction electrons is m.] I A = 4V O.4 = 4V G B O (a) Two cells of emf e and e having internal resistances r and r respectively are connected in parallel as shown. Deduce the expressions of the equivalent emf a cell which can replace the combination between the points B and B. (b) In the figure shown, an ammeter A and a resistor of 4 W are connected to the terminals of the source. The emf of the source is V having an internal resistance of W. Calculate the voltmeter and ammeter readings. 6. Use Krichhoff s rules to obtain the balance condition in a Wheatstone bridge. Calculate the value of in the balance condition of the Wheatstone bridge, if the carbon resistor connected across the arm CD has the colour sequence red, red and orange, as shown in the figure. If now the resistances of the arms BC and CD are interchanged, to obtain the balance condition, another carbon resistor is connected in place of. What would now be the sequence of colour bands of the carbon resistor? O (a) State with the help of a circuit diagram, the working principle of a meter bridge. Obtain the expression used for determining the unknown resistance. (b) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? (c) Why is it considered important to obtain the balance point near the mid-point of the wire? 7. (a) State the working principle of a potentiometer. Draw a circuit diagram to compare emf of two primary cells. Drive the formula used. (b) Which material is used for potentiometer wire and why? (c) How can the sensitivity of a potentiometer be increased? O (a) In the figure a long uniform potentiometer wire AB is having a constant potential gradient 55

55 along its length. The null points for the two primary cells of emfs e and e connected in the manner shown are obtained at a distance of 0 cm and 00 cm from the end A. Find (i) e /e and (ii) position of null point for the cell e. (b) A potentiometer wire of length.0 m has a resistance of 5 W. It is connected to a 5 V battery in series with a resistance of 5 W. Determine the emf of the primary cell which gives a balance point at 60 cm. SOLUTIONS. Electric current in a conductor is I = nea v d. In a conductor the free electron density is of the order of 0 9 m. So, we obtain large amount of current in a conductor.. Since the wires are connected in series, current I through both is same. Therefore ratio of drift velocities vx I / nxea = X (Q I = nea v vy I / nyea d ) Y where, n X, n Y = respective electron densities A X, A Y = cross-sectional areas v X n = Y = ( Given : AX = AY, nx = ny) vy nx v X : v Y = :. So, current I = E 4. Ohmic conductors : Conductors which obey Ohm s law e.g., metallic conductors as their resistance is constant at constant temperature. Non-Ohmic conductors : Conductor which do not obey Ohm s law, e.g., p-n junction diode, thermistors etc. 5. When a battery is short circulated I = e/r. Power loss outside the battery, P output = I = 0 r Power loss inside the battery P inside = I r = ε r k 6. A = V A / l = V A > kb VB / l VB or k A > k B As, potential gradient of B is less than that of A, so, potentiometer B is more sensitive l = l + 5% of l = l + 00 l = l l or l =.05 l So =.05 =.05 (Q = r l A ) or =.05 or =.05 = = 0.5%. So, resistance of wire increases by 0.5%. 8. Circuit can be simplified I = I = 0 0 = 0. A = A and I = 5 60 = A I = 0. + = = 0.7 W Voltage drop across is V = I = 0.7 =.4 V ate of energy loss across the resistor = I =.8 J s 0. No, when an electron approaches a junction, in addition to the uniform E that it normally faces (which keep the drift velocity v d fixed), there are accumulation of charges on the surface of wires at the junction. These produce electric field. These fields alter direction of momentum.. elaxation time is the time between two consecutive collisions of a free electron in a conductor. With increase of temperature, relaxation time decreases due to more frequent collisions and increase in kinetic energy of electrons. O (i) In series, the current is same and using I = Anev d, v d = Ane, i.e., 56

56 v d l A for constant current, so v d is more in the wire with smaller cross-section area. (ii) In parallel, V is same and using v d = ee m.τ = ev ml τ, the drift velocity is same in both the wires.. Current due to protons I P = q 8 9 p = t s = 0.6 A Current due to electrons I e = q te = s = 0.4 A Total current in the discharge tube = I p + I e = 0.59 A Effective esistance = V 0 = = 88.5 ohms I When n resistors are connected in parallel, the effective resistance P is = P n min = min + min min P n Since min... min \ min > or min > P n P Due to one resistor of resistance min, there is only one route for the current in circuit. When n resistors are in parallel, there are additional (n ) routes for the current in circuit. When n resistors are in series, then S = n \ S > max Due to one resistor of resistance max, the current in circuit is more than the current due to n resistors in series of effective resistance S. So S > max. 4. For a solid wire of resistance A l = l, r = r, D = mm πd π() A = = mm 4 4 l l l A = ρ A = ρ π = 4ρ () / 4 π For hollow tube of resistance B, l = l, r = r ( D D A ) π( ) π = π = = mm l l l B ρ A ρ π = 4ρ A = : ( / 4) π B 5. Current in circuit, E, r E, r E+ E E I = r + r + I r+ r+ I As per question, let for the given value of, terminal potential difference across first cell is zero, i.e., E Ir = 0 or E E Ir ( r + r + ). r or = ( r+ r+ ) or r + r + = r or = r r O Applying Kirchhoff s junction rule, I = I + I or I = I I Applying Kirchhoff s loop rule to the upper loop containing cell of V, 5 W and, we get 5I I = or 5(I I) I = or 5I 5I I = (i) Applying Kirchhoff s loop rule to the outer loop containing cell of 0 V, 0 W and, we get 0 I + I = 0 (ii) Eqn. (ii) [Eqn. (i)] gives (0 I + I) (0 I 0 I I) = 0 4 or I + 0 I = 6 or 0 = + I (iii) Compare eqn. (iii) with e eff = I( + r eff ), we get e eff = V and r eff = 0 W 6. The slope of V-I graph provides external resistance = = = 5 A At zero current e = V =.4 volt At point B, V = e Ir 0.8 =.4 0.r 0.r = 0.6 or r = 5 W 7. (i) As the emf e of cell is independent of resistance of the circuit, so it does not change with increase in. (ii) Terminal potential difference of cell is V ε V = I = + r ε or V = + r/ When V = e r So when increases, decrease and hence V increases. 57

57 8. (i) Circuit arrangement is as shown in figure. + I d.c.source n cells r r r (ii) (a) Net emf of cells = ne Net internal resistance = nr ε nε So, charging current in the circuit is I = + nr (b) Potential difference across the combination of cells is V = ne + I nr or V = ne + ε nε + nr nr nε( + nr)+ ( ε nε) nr or V = + nr or V = n ε + n ε r+ n ε r n ε r or V = n ( ε + ε r ) + nr + nr 9. In the given circuit, one part is a balanced Wheatstone bridge. So no current will flow through the arm DE, so above circuit can be reduced to as shown below. \ Equivalent resistancebetween A and B is = A + + C eq D = + + = eq 4 = B eq = No current flows through the arm DE, whereas current flows through AB is ε ε I = = eq / or I = ε 0. (a) The resistivity of a copper wire is very low. Also, the connections are thick, so that the area of cross section is quite large and hence the resistance of the wires is almost negligible. (b) It is preferred to obtain the balance point in the middle of the meter bridge wire because it improves the sensitivity of the meter bridge and minimum error due to resistance of copper strips. (c) Constantan is used for meter bridge wire because its temperature coefficient of resistance is almost negligible due to which the resistance of the wire does not change with increase in temperature of the wire due to flow of current. F. Circuit can be rearranged as shown in the figure. As no current flows through the section AO, this is balanced Wheatstone bridge. P As =, = or X = 6 W Q S 4 X Equivalent resistance of the network is 9 = = = 6 Ω V 6 Current in the circuit is I = = = 6 A. The scheme of connections is shown in figure. Let I, I and I be the currents flowing through the three cells E, E and E. Applying Kirchhoff s junction law at the junction A, we get, I + I + I = 0 or I = (I + I )...(i) Applying Kirchhoff s loop law to the closed loop BE AE B and we get, 4I I + = 0 or 4I I = =... (ii) Applying the Kirchhoff s loop law to the closed loop BE AE B, we get 4I I + 4 = 0 or 4I I = 4 = or 4I + (I + I ) = (Using (i)) or 6I + I = or I + I =... (iii) Multiplying (iii) by and adding to (ii), we get (9 + 4)I = = or I = A From (iii), 6 7 I = I= = + = A From (i), 7 9 I = = A Note : Negative sign of currents shows that the actual direction is opposite to what has been taken in figure. O Two rules which are used to find the current in different branches of electric circuit are: (i) Kirchhoff s junction rule : It states that at any junction in an electrical circuit, sum of incoming currents is equal to sum of outgoing currents. (ii) Kirchhoff s loop rule: It states that in any closed loop in a circuit, algebraic sum of applied emf s and 58

58 potential drops across the resistors is equal to zero. 00 I S 5 V I I 0 60 P Q I 4 V Applying Kirchhoff s loop rule to the closed loop PSP, we get 0I + 00I 5 = 0...(i) Applying Kirchhoff s loop rule to the closed loop PQP, we get 0I + 60I 4 = 0...(ii) Applying Kirchhoff s first rule to the junction P I = I + I...(iii) Substituting the value of I in (i) and (ii) we get 0I + 0I 5 = 0...(iv) 80I + 0I 4 = 0...(v) Solving equations (iv) and (v), we get I= A, I = A= A From equation (iii), 55 I = I+ I = A= A When resistances and S are connected then balance point is found at a distance 40 cm from the zero end. 40 = S S = S = or... (i) When a resistance of W is connected in parallel with S then total resistance in the right gap is S S =...(ii) S + Since balance point is obtained at a distance of 50 cm from the zero end = or = S S 50...(iii) = S Dividing (i) by (iii), we get S = S = or S or S S = or S = S S I Putting the value of S in (ii), we get S = S or S = + S + or S + 4 = 6 or S = \ S = 6 W Putting the value of S in (i), we get 6 = = 6= 4Ω \ = 4 W and S = 6 W 4 4. (i) I = 50 A E = 4 volt 0Ω 4 V AB = I 0 = 0 =.6 V 50 A L = m Potential gradient k = V AB = 6. = 0.8 V m L (ii) Current I in the lower branch 4 I = = 4 E = 4 volt0 A Terminal potential between A P and Q, V PQ = = 0.8 V Balancing length V PQ = kl 0.8 = 0.8l l = meter. P.4 I I 0.6 B AB = 0Ω 4V G Q 5. efer to answer 5, page no. 7 (MTG CBSE Champion Class ). O (a) efer to answer 74, page no. 78 (MTG CBSE Champion Class ). (b) efer to answer 66(b), page no. 76 (MTG CBSE Champion Class ). 6. efer to answer 94, page no. 8 (MTG CBSE Champion Class ). O efer to answer 07, page no. 85 (MTG CBSE Champion Class ). 7. efer to answer 6, page no. 88 (MTG CBSE Champion Class ). O (a) efer to answer, page no. 87 (MTG CBSE Champion Class ). (b) efer to answer 8, page no. 87 (MTG CBSE Champion Class ). B 59

59 Be NEET EADY Practicing these MCQs help to Strengthen your concepts and give you extra edge in your NEET Preparations with Exclusive and brain storming MCQs. Two positively charged infinitely long wires (l > l ) are placed parallel to each other at x = a and x = +a as shown in figure. Find the x coordinate of a point at which electric field is zero. (a) (c) ( λ+ λ) ( λ λ ) a ( ) λ λ (b) x = -a ( λ λ) ( λ + λ ) ( ) ( λ + λ ) a (d) λ+ λ ( λ λ )a l y l O. A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage according to the law m r = av, where a = per volt. The same (but containing no dielectric) capacitor charged to a voltage 56 V is connected in parallel to the first non-linear uncharged capacitor. Determine the final voltage across the capacitors. (a) V (b) 0 V (c) 5 V (d) 0 V. A copper wire of length l and radius r is nickel plated till its final radius is r. If the resistivity of the copper and nickle are ρ c and ρ n, then find the equivalent resistance of wire. a x = a x (a) (c) ρc + ρ ρρ c n l πr n (b) ρ ρ c n l ρc + ρ n πr (d) ρc + ρ l ρρ c n πr ρc + ρ ρρ n π n c n r 4. The maximum intensity in Young s double slit experient is I 0. Distance between the slits is d = 5l, where l is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance D = 0d? (a) I 0 (b) I 0 (c) I 0 /4 (d) I 0 / 5. A hydrogen like atom of atomic number Z is in an excited state of quantum number n. It can emit a maximum energy photon of 04 ev. If it makes a transition to quantum state n, a photon of energy 40.8 ev is emitted. The value of n will be (a) (b) (c) (d) 4 6. The displacement current flows in the dielectric of a capacitor when the potential difference across its plate (a) is increasing with time (b) is decreasing with time (c) has assumed a constant value (d) both (a) and (b). 60

60 7. The half life of a particle of mass kg is 6.9 s and a stream of such particles is travelling with the kinetic energy of a particle being 0.05 ev. The fraction of particles which will decay when they travel a distance of m is (a) 0. (b) 0.0 (c) 0.00 (d) Two parallel vertical A C metallic rails AB and CD are separated by m. They are connected at the two ends by resistances and as shown in figure. A horizontal metallic bar of mass 0. kg slides without friction vertically down the rails under the B D action of gravity. There is a uniform horizontal magnetic field of 0.6 T perpendicular to the plane of the rails. It is observed that when the terminal velocity is attained, the power dissipated in and are 0.76 W and. W respectively. Find the terminal velocity of the bar. (a) m s (b) m s (c) 0.5 m s (d) 5 m s 9. If E 0 = 00 V, = 5 Ω, L = 0. H and C = 0 5 F and the frequency (u) is variable, then the current at u = 0 and L C u = will be respectively E (a) 0 A, 8 A (b) 8 A, 0 A (c) 8 A, 8 A (d) 0 A, 0 A 0. A thin prism P with angle 4 and made from glass of refractive index.54 is combined with another prism P made from glass of refractive index.7 to produce dispersion without deviation. What is the angle of the prism P? (a) 6 (b) (c) 4 (d) 9. A particle of charge mc and mass m = g starts y moving from origin at B t = 0 under an electric field x of 0 NC along x-axis E and magnetic field of 0 T z along the same axis with the velocity of v = 0 j m s as shown, the speed Burning needs air but why does a gust of wind blow off a lamp? When a lamp is lit, the oil vaporizes. The vapor rises because of the heat. The hot air also rises and draws in fresh air from below and from the sides. The flame of the lamp is formed because of burning of the mixture of the hydrocarbon vapor of the oil and air. A strong gust of wind disperses the vapor and the chemical reaction (combustion) would stop. We must know that the same gust of wind would help the fire of a burning wood or coal. This is because here the heat source is not in the air but on the coal or the wood that is burning which would not be dispersed by a wind. Why is the sun milder in the morning and evening compared to its afternoon brightness? To reach the earth the Sun rays have to pass through the atmosphere. The mass of the air column vertically upwards is much less than along the horizon. The thickness of the air column that the light ray has to go through is much greater. In other words, the sunlight has to pass through much greater amount of air to reach the Earth surface in the morning and evening. We know that air molecules, dust, aerosol etc., scatter light and also absorb energy. The increased absorption in the mornings and evenings makes the Sun less hot and less bright. The apparent change in the observed color of the Sun between mid-day and morning or evening is due to preferential scattering of different colors of the sunlight. The Sun appears orange at sunset or sunrise because most of the blue and green have been taken away by scattering during the long travel of sunlight through the atmosphere in the horizontal direction. On a rainy day why does a film of water condense inside the window panes? On a rainy day when the windows are closed the humidity inside the room rises. More so if there are many people in the room. This is because of exhaled breath and other sources of moisture in the room. Also, the temperature inside the room is usually slightly higher car is higher, because of the people or there may be a room heater on if it is winter. The rainwater cools the window panes. The inside air cools below the dew point as it comes in contact with the window panes and condenses. This completely blocks the view. 6

61 of the particle at the time of 0 s will be (a) 0 m s (b) 40 m s (c) 0 m s (d) None. A common emitter amplifier has a voltage gain of 50 and current gain is 5. The power gain of the amplifier is (a) 500 (b) 000 (c) 50 (d) 00. The combination of two bar magnets makes 0 oscillations per second in an oscillation magnetometer when like poles are tied together and oscillations per second when unlike poles are tied together. Find the ratio of the magnetic moments of the magnets. (a) (b) 0 (c) 9 (d) 0 4. In a black body radiation at certain temperature T, the wavelength having maximum intensity of radiation equals 9000 Å. When the temperature is increased form T to T the total radiation increases 6 times. The peak radiation at T is found to be capable of ejecting photoelectrons. The maximum kinetic energy of the photoelectrons is the same as the energy of photon that one gets when one of electrons in the M-shell of hydrogen atom jumps to L-shell. What is the work function of the metal? (a). ev (b) 0. ev (c) 0.88 ev (d) 6.6 ev 5. In the following circuit, the value of β is 00. Find I B, V CE, V BE and V BC, when I C =.5 ma. The transistor is in active cut off or saturation state. +0V 5.0 kω 0 kω C (a) V (b) 5 V (c) 7 V (d) V SOLUTIONS. (b): Let coordinates of P are (x, 0). The distance of point P from wire = (a + x) and from wire = (a x) E = E kλ kλ \ = x ( a+ x) ( a x) P x a a l (a x) = l (a + x) l a l x = l a + l x (l l )a = (l + l )x \ x = ( λ λ ) a ( λ+ λ) x = -a x = +a B E. (a) : From conservation of charge CV = (C + Cʹ) V f CV = (C + av f C) V f V f + V f V = 0 V f + V f 56 = 0 V f = volt. (c) : = ρ l A r Ni r Cu esistance of copper wire l l Cu = ρc π r ( A = πr ) esistance of nickle wire A Ni = π(r) πr = πr l Ni = ρn π r Both wires are connected in parallel. So equivalent CuNi ρρ c n l resistance = = Cu + Ni ρc + ρn πr 4. (d): Path difference x = yd D Here, y = d 5 = λ and D = 0d = 50l 5λ 5λ λ \ x = 50λ = 4 Corresponding phase difference f = π λ π = λ 4 I = I 0 cos φ = I 0 cos π 0 4 = I 5. (b): Let ground state energy (in ev) be E Then from the given condition E E n E = 04 ev or E = 04 ev 4n E 4n = 04 ev...(i) and E n E n = 40.8 ev E 4n E E n = 4n = 40.8 ev From equation (i) and (ii), 4 n = 5 n =...(ii) 4n dφ AdE 6. (d): I D = ε 0 = ε0 ; i D 0 whenever de 0 dt dt dt This will possible when potential difference across a capacitor varies. 6

62 7. (d) : Speed of charge particle, v = dn dt = Nl dn dt = ldt = l v = = K m = 0 m s 8. (a) : The rod will acquire terminal velocity only when magnetic force F M = BIl due to electromagnetic induction balances its weight, i.e., BIl = mg, i.e., I = = A 06. Now if ε is the emf induced in the rod, ε I = P = P + P so, ε = ( ) = 0.6 V (. 98/ ) Now as this ε is generated due to motion of rod with terminal velocity in the magnetic field, i.e., ε ε = Bv T l so v T = Bl = 06. = m s (d) 0. (b): In case of thin prism δ = (m )A, when two prisms are combined together, δ = δ + δ = (m )A + (mʹ )Aʹ For producing dispersion without deviation δ = 0, i.e., (mʹ )Aʹ = (m )A or Aʹ = (. 54 ) 4 (. 7 ) = So the angle of the other prism is and opposite to the first.. (b): As v = v x + v qe y = m t + (0) = (0) = 600 v = 40 m s. (c) : AC power gain is ratio of change in output power to the change in input power. AC power gain Change in output power = = Vc i Change in input power Vi i V = c ic V i = A v β ac i \ β ac = 5 A V = 50 b c b Now, AC power gain = A v β ac. (b) : u = u = and u = or, or, or, π π π MB I H ( M + M ) B I υ υ = M + M M M υ υ M M + υ υ + I = 50 5 = 50 H ( M M) B I + I H, where M > M = ( M+ M ) + ( M M ) = M ( M + M ) ( M M ) M = = 4. (c) : T 4 T = 6 T 4 = T l T = l T l = λ T = 4500 Å T hc λ = = 45 ev Now according to question maximum kinetic energy of electron, K max =.6 =.9 ev f = hc K max = 4.9 = 0.88 ev λ (d) : β = I IC B I B = I C β = 5. = 0.05 ma 00 Applying Kirchhoff s law to base emitter loop, V CE = V C I C C = 0 (.5 0 ) (5 0 ) = 7.5 V V BE = V B I B B = 0 ( ) (0 0 ) = 8.5 V \ V BC = (V BE V CE ) = ( ) = V 6

63 Class XII MONTHLY TUNE UP! PACTICE POBLEMS These practice problems enable you to self analyse your extent of understanding of specified chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer. Performance analysis table given at the end will help you to check your readiness. Electrostatics Total Marks : 0 NEET / AIIMS Only One Option Correct Type. A spherical metal plate of radius.5 cm is charged with 0 mc situated in air. Find mechanical force acting on one side of plate. If the same spherical plate carrying same charge were kept in acetone, what would be magnitude of mechanical force? (k acetone = 7) (a) 0 N (b) 48 N (c) 48 N (d) 84 N. An infinite sheet carrying a uniform surface charge density s lies on the xy-plane. The work done to ^ ^ ^ carry a charge from the point A = ai ( + j+ k) ^ ^ ^ to the point B = a( i+ j+ 6 k) (where, a is a constant with the dimension of length and e 0 is the permittivity of free space) is σaq (a) (b) σaq 5σaq (c) (d) σaq ε 0 ε ε 0 0 ε0. A 400 pf capacitor is charged by a 00 V supply. How much electrostatic energy is lost in the process of disconnecting from the supply and connecting another uncharged 400 pf capacitor? (a) 0 5 J (b) 0 6 J (c) 0 7 J (d) 0 4 J 4. A wire of length L(=0 cm) is bent into a semicircular arc. If the two equal halves of the arc, each to be uniformly charged with charges ± Q [ Q = 0 e 0 coulomb, where e 0 is the permittivity (in SI units) of free Time Taken : 60 Min. space, the net electric field at the centre O of the semicircular arc would be (a) (50 0 N C ) j y (b) (50 0 N C ) i (c) (5 0 N C ) j + ++ (d) (5 0 N C ) i + x 5. The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a sphere of radius a centred at the origin of the field will be given by (a) 4pe 0 Aa (b) Ae 0 a (c) 4pe 0 Aa (d) e 0 Aa 6. A parallel plate capacitor is connected to a battery. The plates are pulled apart with uniform speed v. If x is separation between plates, then the rate of change of the electrostatic energy of the condenser is proportional to (a) x (b) x (c) (d) x x 7. A simple pendulum has time period T. The bob is given negative charge and surface below it is given positive charge. The new time period will be (a) less than T (b) greater than T (c) equal to T (d) infinite 8. A charge of magnitude q is distributed uniformly on a solid sphere. The electric field at a point inside the sphere (r < ) varies as (a) /r (b) /r (c) r (d) r 64

64 9. An electric field is given by E = yi + xj ( ) NC. The work done in moving a C charge from r i j r i A = ( + ) mto B = ( 4 + j) m is (a) + 8 J (b) + 4 J (c) zero (d) J 0. An electron of mass m e, initially at rest, moves through a certain distance in a uniform electric field in time t. A proton of mass m p, also initially at rest, takes time t to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio of t /t is nearly equal to (a) (b) (m p /m e ) / (c) (m e /m p ) / (d) 86. Two point charges +q and q are held fixed at ( d, 0) and (d, 0) respectively of a x-y coordinate system. Then (a) the electric field E at all points on x-axis has the same direction (b) electric field at all points on y-axis is along x-axis (c) work has to be done in bringing a test charge from to the origin (d) the dipole moment is qd along the x-axis.. A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 0 V. The potential at the centre of the sphere is (a) zero (b) 0 V (c) same as at a point 5 cm away from the surface (d) same as at a point 5 cm away from the surface Assertion & eason Type Directions : In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice as: (a) If both assertion and reason are true and reason is the correct explanation of assertion. (b) If both assertion and reason are true but reason is not the correct explanation of assertion. (c) If assertion is true but reason is false. (d) If both assertion and reason are false.. Assertion : Gauss s law shows diversion when inverse square law is not obeyed. eason : Gauss s law is a consequence of conservation of charges. 4. Assertion : If a conductor is given charge then no excess inner charge appears. eason : Electric field inside conductor is zero. 5. Assertion : If an electron and proton possessing same kinetic energy enter an electric field in a perpendicular direction, the path of the electron is more curved than that of the proton. eason : Electron forms a larger curve due to its small mass. JEE MAIN / ADVANCED Only One Option Correct Type 6. A capacitor as shown in figure has square plates length l each and are inclined at an angle q with one another. For small value of q, capacitance is given by (a) (c) ε0l d ε l 0 d d θl d (b) ε 0l θ l d d θl d (d) ε 0l θ l + d d 7. A uniformly charged thin spherical shell of radius carries uniform surface charge density of s per unit area. It is made of two hemispherical shells, held together by pressing them with force F (see figure). F is proportional to F (a) ε σ (b) ε σ 0 0 σ σ (c) ε0 (d) ε0 8. A thin fixed ring of radius m has a positive charge 0 5 C uniformly distributed over it. A particle of mass 0.9 g and carrying a negative charge of 0 6 C is placed on the axis at distance cm from the centre of the ring. Show that the motion of the negatively charged particle is approximately simple harmonic. Calculate the time period of oscillation. (a) 0.5 s (b) 0.6 s (c) 0.7 s (d) 0.8 s 9. A charge q is distributed uniformly over the volume of a ball of radius. Assume the permittivity to be equal to unity. What is the electrostatic self energy of the ball? q (a) (b) q 5πε0 0πε0 q q (c) (d) 0πε 4πε 0 0 F 65

65 More than one Options Correct Type 0. A spherical symmetric charge system is centred at origin. Given, electric potential Q φ = ( r, 4 0) πε00 Q φ = ( r > 4 πε 0) 0r (a) within r = 0 total enclosed net charge is Q (b) electric field is discontinued at r = 0 (c) charge is only present at r = 0 (d) electrostatic energy is zero for r < 0.. A non-conducting solid sphere of radius is uniformly charged. The magnitude of the electric field due to the sphere at a distance r from its centre (a) increases as r increases, for r < (b) decreases as r increases, for 0 < r < (c) decreases as r increases, for < r < (d) is discontinuous at r =.. A dielectric slab of thickness d is inserted in a parallel plate capacitor whose negative plate is at x = 0 and positive plate is x = d. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from 0 to d, (a) the magnitude of the electric field remains the same (b) the direction of the electric field remains the same (c) the electric potential increases continuously (d) the electric potential increases at first, then decreases and again increases.. A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E and W denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted), and the work done on the system, in question, in the process of inserting the slab, then AV (a) Q = ε 0 AV (b) Q = ε 0 K d d V (c) E = Kd ε AV (d) W = 0 d K. Numerical Value Type 4. Four points charge + 8 mc, mc, mc, and + 8 mc 7 are fixed at the points m, m, + m and + 7 m respectively on the y-axis. A particle of mass kg and charge + 0. mc moves along the x-direction. Its speed at x = + is v 0. Find the least value of v 0 (in m s ) for which the particle is at the origin. Assume that space is gravity free. 9 Given = 9 0 N m C. 4πε0 5. Two fixed, equal, positive A +q charges, each of magnitude D O q C are located at points C A and B separated by a distance B +q of 6 m. An equal and opposite charge moves towards them along the line COD, the perpendicular bisector of the line AB. The moving charge, when it reaches the point C at a distance of 4 m from O, has a kinetic energy of 4 J. Calculate the distances (in m) of the farthest point D which the negative charge will reach before returning towards C. 6. A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of strength 8 π 0 5 Vm. When the field 7 is switched off, the drop is observed to fall with terminal velocity 0 m s. The magnitude of q is n e, then value of n is. Given g = 9.8 m s, viscosity of the air = N s m and the density of oil = 900 kg m GLIMPSE OF NEXT ISSUE... Focus NEET/JEE (XI) : Kinematics Focus NEET/JEE (XII) : Current Electricity Monthly Tune Up (XI) : Kinematics Monthly Tune Up (XII) : Current Electricity Brain Map : Electrostatics 66

66 Comprehension Type The nuclear charge (Ze) is non-uniformly distributed within a nucleus of radius. The charge density r(r) (charge per unit volume) is dependent only on the radial distance r from the centre of the nucleus as shown in figure. The electric field is only along the radial direction. 7. For a = 0, the value d (maximum value of r as shown in the figure) is Ze Ze 4Ze Ze (a) (b) (c) (d) 4π π π π 8. The electric field within the nucleus is generally observed to be linearly dependent on r. This implies (a) a = 0 (b) a = List I List II (A) Q, Q, Q, Q 4 all positive (P) +x (B) Q, Q positive; Q, Q 4 negative (Q) x (C) Q, Q 4 positive; Q, Q negative () +y (D) Q, Q positive; Q, Q 4 negative (S) y A B C D (a) P S Q (b) S Q P (c) P Q S (d) S Q P 0. The axis of a hollow cone as shown in figure is vertical. Its base radius is. It is kept in a uniform electric field E parallel to its axis. Match List I with List II and select the correct answer using the codes. (c) a = (d) a = Matrix Match Type 9. Four charges Q, Q, Q and Q 4 of same magnitude are fixed along the x-axis at x = a, a, +a and +a respectively. A positive charge q is placed on the positive y-axis at a distance b > 0. Four options of the signs of these charges are given in List I. The direction of the forces on the charge q is given in List II. Match List I with List II and select the correct answer using the codes. CHECK YOU PEFOMANCE No. of questions attempted No. of questions correct Marks scored in percentage List I List II (A) Magnitude of flux through base of cone (P) p E (B) Magnitude of flux (Q) π E through curved part of cone (C) Magnitude of flux () zero through curved part MNQP of cone (D) Net flux through the entire cone (S) non zero A B C D (a) (P,S) (P,S) (Q,S) (b) (Q,S) (P,S) (P,S) (,S) (c) (P,Q) (,S) (P,) (d) (P,) (,S) (Q,P) Q Keys are published in this issue. Search now! J If your score is > 80% Your preparation is going good, keep it up to get high score % Need more practice, try hard to score more next time. <60% Stress more on concepts and revise thoroughly. 67

67 Team Applauds Tejaswini Banbare for Cracking NEET 08 MTG : Why did you choose medical Entrance? Tejaswini : Because it was my dream to become a doctor and also because of my interest in biology. MTG : What exams have you appeared for and what is your rank in these exams? Tejaswini : NEET 08, AI 54. I scored 75/80 in physics, 60/80 in chemistry and 0/60 in Biology. MTG : Any other achievements? (Please mention the name of exams and rank.) Tejaswini : Gold medal in NSO and IMO. MTG : How did you prepare for NEET and other medical exams? Tejaswini : I first of all made sure that my concepts were clear and also worked to strengthen the basics of every subject. I personally believe that persistence and hard work along with good tuning with the teachers is necessary to crack NEET. MTG : What basic difference you found in various papers you cleared? Tejaswini : The basic difference is in the difficulty level, with AIIMS being more difficult than NEET. MTG : How many hours in a day did you study to prepare for the examination? Tejaswini : 6-7 hours daily MTG : On which topics and chapters you laid more stress in each subject? Tejaswini : For Biology, I never stressed on a particular topic. For me, every chapter was equally important. For Physics, motion in one and two dimensions, force, work energy and power, scalars and vectors are very important and play important role in all other chapters and electrodynamics. For Chemistry, I stressed more on inorganic section. MTG : How much time does one require for serious preparation for this exam? Tejaswini : To be on safer side years should be given for proper preparation of NEET. MTG : Any extra coaching? Tejaswini : Guidance by Bhargav aje Sir. MTG : Which subjects/topics you were strong/ weak at? Tejaswini : In Physics I was strong enough. But in Chemistry and Biology you feel that you are done but while solving, there would be confusion in many questions. MTG : Which Books/Magazines/Tutorial/Coaching Classes you followed? Tejaswini : Coachings - IIB (Bio), Creative Coaching Classes (Physics) Konale Coaching Classes (Chemistry) Books - MTG NEET guide, previous year questions by MTG. MTG : In your words what are the components of an ideal preparation plan? Tejaswini : For ideal preparation, study should be constant. It has to be precise. Study should be stress free, what is more important is the quality study not how much time you study. MTG : What role did the following play in your success : (a) Parents (b) Teachers Tejaswini : (a) Parents : Parents play a very important role. My parents have supported and motivated me throughout my preparation. They have given me every best possible thing. (b) Teachers : Teachers have always been supportive. They were very precise in their teaching and have cleared my doubts every time. 68

68 MTG : Your family background? Tejaswini : My mother and father both are teachers and my sister is a doctor. MTG : What mistake you think you shouldn t have made? Tejaswini : Initially, I was hesitant to ask doubts and did not interact with teachers. But I regret it because it is important to clear doubts otherwise you won t get to know your mistakes. MTG : How did you de-stress yourself during the preparation? Share your hobbies and how often could you pursue them. Tejaswini : My parents would help me to destress myself. My sister and parents would talk to me and motivate me whenever I was stressed. I like reading novels and story books. MTG : How have various MTG products like Explorer Books and Magazines helped you in your preparation? Tejaswini : MTG Books have helped me focus on every topic and the quality of questions was the same as NEET. MTG : Was this your first attempt? Tejaswini : Yes MTG : Had you not been selected then what would have been your future plan? Tejaswini : I would have given a second attempt. But if I had felt exhausted I would have gone for other medical courses like dental, BAMS. MTG : What do you think is the secret of your success? Tejaswini : Daily practice and stress less study are the key factors of my success. MTG : What advice would you like to gave our readers who are PMT aspirants? Tejaswini : Believe in yourself till the very end. Never get stressed, despite of the ups and downs. All the Best! ATTENTION COACHING INSTITUTES : a great offer from MTG CLASSOOM STUDY MATEIAL MTG Classroom Study Material for JEE (Main & Advanced), NEET and FOUNDATION MATEIAL for Class 6, 7, 8, 9, 0, & with YOU BAND NAME & COVE DESIGN. This study material will save you lots of money spent on teachers, typing, proof-reading and printing. Also, you will save enormous time. Normally, a good study material takes years to develop. But you can have the material printed with your logo delivered at your doorstep. educational publishing for JEE (Main & Advanced)/NEET/PMT... Order sample chapters on Phone/Fax/ . Phone : sales@mtg.in EXCELLENT CONTENT PAPE PINTING QUALITY 69

69 PAPE-I Section (Maximum Marks : 4) This section contains SIX (06) questions. Each question has FOU options for correct answer(s). ONE O MOE THAN ONE of these four option(s) is (are) correct option(s). For each question, choose the correct option(s) to answer the question. Answer to each question will be evaluated according to the following marking scheme: Full Marks : +4 If only (all) the correct option(s) is (are) chosen. Partial Marks : + If all the four options are correct but ONLY three options are chosen. Partial Marks : + If three or more options are correct but ONLY two options are chosen, both of which are correct options. Partial Marks : + If two or more options are correct but ONLY one option is chosen and it is a correct option. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : In all other cases. For Example : If first, third and fourth are the ONLY three correct options for a question with second option being an incorrect option; selecting only all the three correct options will result in +4 marks. Selecting only two of the three correct options (e.g. the first and fourth options), without selecting any incorrect option (second option in this case), will result in + marks. Selecting only one of the three correct options (either first or third or fourth option), without selecting any incorrect option (second option in this case), will result in + marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in - marks.. The potential energy of a particle of mass m at a distance r from a fixed point O is given by V(r) = kr/, where k is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius about the point O. If v is the speed of the particle and L is the magnitude of its angular momentum about O, which of the following statements is (are) true? 70 (a) v = k m (b) v = k m (d) L = mk. Consider a body of mass.0 kg at rest at the origin at time t = 0. A force F = (αt i + β j ) is applied on the body, where a =.0 N s and b =.0 N. The torque acting on the body about the origin at time t =.0 s is τ. Which of the following statements is (are) true? (a) τ = N m (b) The torque τ is in the direction of the unit vector +k (c) The velocity of the body at t = s is v = ( i + j ) m s (d) The magnitude of displacement of the body at t = s is m 6. A uniform capillary tube of inner radius r is dipped vertically into a beaker filled with water. The water rises to a height h in the capillary tube above the water surface in the beaker. The surface tension of water is s. The angle of contact between water and the wall of the capillary tube is q. Ignore the mass of water in the meniscus. Which of the following statements is (are) true? (a) For a given material of the capillary tube, h decreases with increase in r (b) For a given material of the capillary tube, h is independent of s (c) If this experiment is performed in a lift going up with a constant acceleration, then h decreases (d) h is proportional to contact angle q (c) L = mk

70 4. In the figure below, the switches S and S are closed simultaneously at t = 0 and a current starts to flow in the circuit. Both the batteries have the same magnitude of the electromotive force (emf) and the polarities are as indicated in the figure. Ignore mutual inductance between the inductors. The current I in the middle wire reaches its maximum magnitude I max at time t = t. Which of the following statements is (are) true? V S L V (a) Imax = V (b) Imax = 4 (c) τ= L ln (d) τ= L ln 5. Two infinitely long straight wires lie in the xy-plane along the lines x = ±. The wire located at x = + carries a constant current I and the wire located at x = carries a constant current I. A circular loop of radius is suspended with its centre at (0, 0, ) and in a plane parallel to the xy-plane. This loop carries a constant current I in the clockwise direction as seen from above the loop. The current in the wire is taken to be positive if it is in the + j direction. Which of the following statements regarding the magnetic field B is (are) true? (a) If I = I, then B cannot be equal to zero at the origin (0, 0, 0) (b) If I > 0 and I < 0, then B can be equal to zero at the origin (0, 0, 0) (c) If I < 0 and I > 0, then B can be equal to zero at the origin (0, 0, 0) (d) If I = I, then the z-component of the magnetic field at the centre of the loop is µ 0 I 6. One mole of a monatomic T ideal gas undergoes a II cyclic process as shown in the figure (where V is I III the volume and T is the IV temperature). Which of V the following statements is (are) true? I S L V (a) Process I is an isochoric process (b) In process II, gas absorbs heat (c) In process IV, gas releases heat (d) Processes I and III are not isobaric Section (Maximum Marks : 4) This section contains EIGHT (08) questions. The answer to each question is a NUMEICAL VALUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. 6.5, 7.00, -0., -.0, 0.7, -7.0) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer. Answer to each question will be evaluated according to the following marking scheme: Full Marks : + If ONLY the correct numerical value is entered as answer. Zero Marks : 0 In all other cases. 7. Two vectors A and B are defined as A = ai and B= a(cosωti + sin ωtj ), where a is a constant and w = p/6 rad s. If A+ B = A B at time t = t for the first time, the value of t, in seconds, is. 8. Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed.0 m s and the man behind walks at a speed.0 m s. A third man is standing at a height m above the same horizontal line such that all three men are in a vertical plane. The two walking men are blowing identical whistles which emit a sound of frequency 40 Hz. The speed of sound in air is 0 m s. At the instant, when the moving men are 0 m apart, the stationary man is equidistant from them. The frequency of beats in Hz, heard by the stationary man at this instant, is. 9. A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle 60 with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is ( )/ 0 s, then the height of the top of the inclined plane, in metres, is. Take g = 0 m s. 0. A spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is.0 N m and the mass of the block is.0 kg. Ignore the mass of the spring. Initially the spring is in an unstretched condition. Another block of mass.0 kg moving with a speed of.0 m s collides 7

71 elastically with the first block. The collision is such that the.0 kg block does not hit the wall. The distance, in metres, between the two blocks when the spring returns to its unstretched position for the first time after the collision is. m s kg kg. Three identical capacitors C, C and C have a capacitance of.0 mf each and they are uncharged initially. V 0 S They are connected in a circuit as C C shown in the figure S C and C is then filled completely with a dielectric material of relative permittivity e r. The cell electromotive force (emf) V 0 = 8 V. First the switch S is closed while the switch S is kept open. When the capacitor C is fully charged, S is opened and S is closed simultaneously. When all the capacitors reach equilibrium, the charge on C is found to be 5 mc. The value of e r =.. In the xy-plane, the region y y > 0 has a uniform magnetic field Bk B and the region y < 0 v 0 = p m s has another uniform x magnetic field Bk. B A positively charged particle is projected from the origin along the positive y-axis with speed v 0 = p m s at t = 0, as shown in the figure. Neglect gravity in this problem. Let t = T be the time when the particle crosses the x-axis from below for the first time. If B = 4B, the average speed of the particle, in m s, along the x-axis in the time interval T is.. Sunlight of intensity. kw m is incident normally on a thin convex lens of focal length 0 cm. Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light, in kw m, at a distance cm from the lens on the other side is. 4. Two conducting cylinders of equal length but different radii are connected in series between two heat baths kept at temperatures T = 00 K and T = 00 K, as shown in the figure. The radius of the bigger cylinder is twice that of the smaller one and the thermal conductivities of the materials of the smaller and the larger cylinders are K and K respectively. If the temperature at the junction of the two cylinders in the steady state is 00 K, then K /K =. Insulating material T K K T L Section (Maximum Marks : ) This section contains TWO (0) paragraphs. Based on each paragraph, there are TWO (0) questions. Each question has FOU options. ONLY ONE of these four options corresponds to the correct answer. For each question, choose the option corresponding to the correct answer. Answer to each question will be evaluated according to the following marking scheme: Full Marks : + If ONLY the correct option is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : In all other cases. Paragraph "X" In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, [E] and [B] stand for dimensions of electric and magnetic fields respectively, while [e 0 ] and [m 0 ] stand for dimensions of the permittivity and permeability of free space respectively. [L] and [T] are dimensions of length and time respectively. All the quantities are given in SI units. 5. The relation between [E] and [B] is (a) [E] = [B][L][T] (b) [E] = [B][L] [T] (c) [E] = [B][L][T] (d) [E] = [B][L] [T] 6. The relation between [e 0 ] and [m 0 ] is (a) [m 0 ] = [e 0 ][L] [T] (b) [m 0 ] = [e 0 ][L] [T] (c) [m 0 ] = [e 0 ] [L] [T] (d) [m 0 ] = [e 0 ] [L] [T] L 7

72 Paragraph "A" If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation z= x/y. If the errors in x, y and z are Dx, Dy and Dz, respectively, then x x x x y z± z = ± = ± y± y y x ± y 7. Consider the ratio r = ( a ) to be determined by ( + a) measuring a dimensionless quantity a. If the error in the measurement of a is Da (Da/a ), then what is the error Dr in determining r? a a (a) (b) ( + a) ( + a) a a a (c) (d) ( a ) ( a ) y The series expansion for ± y, to first power in 8. In an experiment the initial number of radioactive nuclei is 000. It is found that Dy/y, is (Dy/y). The relative errors in independent 000±40 nuclei decayed in the first.0 s. For variables are always added. So the error in z will be x y x, ln( + x) = x up to first power in x. The z = z + x y error Dl, in the determination of the decay The above derivation makes the assumption that constant l, in s, is Dx/x, Dy/y. Therefore, the higher powers of (a) 0.04 (b) 0.0 these quantities are neglected. (c) 0.0 (d) 0.0 PAPE-II Section (Maximum Marks : 4) This section contains SIX (06) questions. Each question has FOU options for correct answer(s). ONE O MOE THAN ONE of these four option(s) is (are) correct option(s). For each question, choose the correct option(s) to answer the question. Answer to each question will be evaluated according to the following marking scheme: Full Marks : +4 If only (all) the correct option(s) is (are) chosen. Partial Marks : + If all the four options are correct but ONLY three options are chosen. Partial Marks : + If three or more options are correct but ONLY two options are chosen, both of which are correct options. Partial Marks : + If two or more options are correct but ONLY one option is chosen and it is a correct option. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : In all other cases. For Example: If first, third and fourth are the ONLY three correct options for a question with second option being an incorrect option; selecting only all the three correct options will result in +4 marks. Selecting only two of the three correct options (e.g. the first and fourth options), without selecting any incorrect option (second option in this case), will result in + marks. Selecting only one of the three correct options (either first or third or fourth option),without selecting any incorrect option (second option in this case), will result in + marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in marks.. A particle of mass m is initially at rest at the origin. It is subjected to a force and starts moving along the x-axis. Its kinetic energy K changes with time as dk/dt = gt, where g is a positive constant of appropriate dimensions. Which of the following statements is (are) true? (a) The force applied on the particle is constant (b) The speed of the particle is proportional to time (c) The distance of the particle from the origin increases linearly with time (d) The force is conservative. Consider a thin square plate floating on a viscous liquid in a large tank. The height h of the liquid in the tank is much less than the width of the tank. The floating plate is pulled horizontally with a constant velocity u 0. Which of the following statements is (are) true? (a) The resistive force of liquid on the plate is inversely proportional to h (b) The resistive force of liquid on the plate is independent of the area of the plate (c) The tangential (shear) stress on the floor of the tank increases with u 0 (d) The tangential (shear) stress on the plate varies linearly with the viscosity h of the liquid 7

73 . An infinitely long thin non-conducting wire is parallel to the z-axis and carries a l z uniform line charge density l. It pierces a thin non-conducting P spherical shell of radius in such a way that the arc PQ subtends 0 O an angle 0 at the centre O of the spherical shell, as shown in Q the figure. The permittivity of free space is e 0. Which of the following statements is (are) true? (a) The electric flux through the shell is λ/ ε0 (b) The z-component of the electric field is zero at all the points on the surface of the shell (c) The electric flux through the shell is λ/ ε0 (d) The electric field is normal to the surface of the shell at all points 4. A wire is bent in the shape of a right angled triangle and is placed in front of a concave mirror of focal length f, as shown in the figure. Which of the figures shown in the four options qualitatively represent(s) the shape of the image of the bent wire? (These figures are not to scale.) a > 45 (a) a (b) 0 < a < 45 (c) a (d) 5. In a radioactive decay chain, 90Th nucleus decays to 8Pb nucleus. Let N a and N b be the number of a and b particles, respectively, emitted in this decay process. Which of the following statements is (are) true? (a) N a = 5 (b) N a = 6 (c) N b = (d) N b = 4 6. In an experiment to measure the speed of sound by a resonating air column, a tuning fork of frequency 500 Hz is used. The length of the air column is varied by changing the level of water in the resonance tube. Two successive resonances are heard at air columns of length 50.7 cm and 8.9 cm. Which of the following statements is (are) true? f 45 f (a) The speed of sound determined from this experiment is m s (b) The end correction in this experiment is 0.9 cm (c) The wavelength of the sound wave is 66.4 cm (d) The resonance at 50.7 cm corresponds to the fundamental harmonic Section (Maximum Marks : 4) This section contains EIGHT (08) questions. The answer to each question is a NUMEICAL VALUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded off to the second decimal place; e.g., 6.5, 7.00, 0.,.0, 0.7, 7.0) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer. Answer to each question will be evaluated according to the following marking scheme: Full Marks : + If ONLY the correct numerical value is entered as answer. Zero Marks : 0 In all other cases. 7. A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m = 0.4 kg is at rest on this surface. An impulse of.0 N s is applied to the block at time t = 0 so that it starts moving along the x-axis with a velocity v(t) = v 0 e t/t, where v 0 is a constant and t = 4 s. The displacement of the block, in metres, at t = t is. Take e = A ball is projected from the ground at an angle of 45 with the horizontal surface. It reaches a maximum height of 0 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 0 with the horizontal surface. The maximum height it reaches after the bounce, in metres, is. 9. A particle, of mass 0 kg and charge.0 C, is initially at rest. At time t = 0, the particle comes under the influence of an electric field Et () = E 0 sin wt i, where E 0 =.0 N C and w = 0 rad s. Consider the effect of only the electrical force on the particle. Then the maximum speed, in m s, attained by the particle at subsequent times is. 0. A moving coil galvanometer has 50 turns and each turn has an area 0 4 m. The magnetic field produced by the magnet inside the galvanometer is 0.0 T. The torsional constant of the suspension wire is 0 4 N m rad. When a current flows through 74

74 75

75 the galvanometer, a full scale deflection occurs if the coil rotates by 0. rad. The resistance of the coil of the galvanometer is 50 W. This galvanometer is to be converted into an ammeter capable of measuring current in the range A. For this purpose, a shunt resistance is to be added in parallel to the galvanometer. The value of this shunt resistance, in ohms, is.. A steel wire of diameter 0.5 mm and Young s modulus 0 N m carries a load of mass M. The length of the wire with the load is.0 m. A vernier scale with 0 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count.0 mm, is attached. The 0 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by. kg, the vernier scale division which coincides with a main scale division is. Take g = 0 m s and p =... One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 00 K and the universal gas constant = 8.0 J mol K, the decrease in its internal energy, in joule, is.. In a photoelectric experiment a parallel beam of monochromatic light with power of 00 W is incident on a perfectly absorbing cathode of work function 6.5 ev. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 00%. A potential difference of 500 V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F = n 0 4 N due to the impact of the electrons. The value of n is. Mass of the electron m e = 9 0 kg and.0 ev = J. 4. Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = to n = transition has energy 74.8 ev higher than the photon emitted in the n = to n = transition. The ionization energy of the hydrogen atom is.6 ev. The value of Z is. Section (Maximum Marks : ) This section contains FOU (04) questions. Each question has TWO (0) matching lists: LIST I and LIST II. FOU options are given representing matching of elements from LIST I and LIST II. ONLY ONE of these four options corresponds to a correct matching. For each question, choose the option corresponding to the correct matching. For each question, marks will be awarded according to the following marking scheme: Full Marks : + If ONLY the option corresponding to the correct matching is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : - In all other cases. 5. The electric field E is measured at a point P(0, 0, d) generated due to various charge distributions and the dependence of E on d is found to be different for different charge distributions. List-I contains different relations between E and d. List-II describes different electric charge distributions, along with their locations. Match the functions in List-I with the related charge distributions in List-II. List-I List-II P E is. A point charge Q at the origin independent of d Q. E. E S. E d d d. A small dipole with point charges Q at(0, 0, l) and Q at (0, 0, l). Take l d. An infinite line charge coincident with the x-axis, with uniform linear charge density l 4. Two infinite wires carrying uniform linear charge density parallel to the x- axis. The one along (y = 0, z = l) has a charge density + l and the one along (y = 0, z = l) has a charge density l. Take l d 5. Infinite plane charge coincident with the xy-plane with uniform surface charge density (a) P 5; Q, 4; ; S (b) P 5; Q ;, 4; S (c) P 5; Q ;, ; S 4 (d) P 4; Q, ; ; S 5 76

76 6. A planet of mass M, has two natural satellites with masses m and m. The radii of their circular orbits are and respectively. Ignore the gravitational force between the satellites. Define v, L, K and T to be, respectively, the orbital speed, angular momentum, kinetic energy and time period of revolution of satellite ; and v, L, K and T to be the corresponding quantities of satellite. Given m /m = and / = /4, match the ratios in List-I to the numbers in List-II. List-I List-II P. Q.. S. v v L L K K T T (a) P 4; Q ; ; S (b) P ; Q ; 4; S (c) P ; Q ; ; S 4 (d) P ; Q ; 4; S 7. One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the PV-diagram. P Among these four II P processes, one is 0 isobaric, one is IV III isochoric, one is I isothermal and one P 0 is adiabatic. Match the processes V V 0 V mentioned in 0 List-I with the corresponding statements in List-II. List-I List-II P. In process I. Work done by the gas is zero Q. In process II. Temperature of the gas remains unchanged. In process III. No heat is exchanged between the gas and its surroundings S. In process IV 4. Work done by the gas is 6P 0 V 0 (a) P 4; Q ; ; S (b) P ; Q ; ; S 4 (c) P ; Q 4; ; S (d) P ; Q 4; ; S 8. In the List-I below, four different paths of a particle are given as functions of time. In these functions, a and b are positive constants of appropriate dimensions and a b. In each case, the force acting on the particle is either zero or conservative. In List-II, five physical quantities of the particle are mentioned: p is the linear momentum, L is the angular momentum about the origin, K is the kinetic energy, U is the potential energy and E is the total energy. Match each path in List-I with those quantities in List-II, which are conserved for that path. P. Q.. S. List-I List-II rt ()= αti + βtj. p rt () = αcosωti + βsinωtj. L rt () = α(cosωti + sin ωtj ). K β rt ()= αti+ t j 4. U 5. E (a) P,,, 4, 5; Q, 5;,, 4, 5; S 5 (b) P,,, 4, 5; Q, 5;,, 4, 5; S, 5 (c) P,, 4; Q 5;,, 4; S, 5 (d) P,,, 5; Q, 5;,, 4, 5; S, 5 COMIC CAPSULE It s not the v f = v i + at that kills you, it s the F = m Dv Dt 77

77 SOLUTIONS Paper-I. (b,c) : Potential energy of a particle, V = kr Force acting on the particle, dv k F = = r = kr dr ( ) As particle is moving on a circular path of radius so F is the centripetal force. mv mv k F = or =k or v = m Now angular momentum of the particle about O, L = mv = m k m = mk. (a,c) : Here m = kg, F = ( αti + β j) = ( ti + j ) N As, F m dv t = = ( ti+ j) or v = ti + j dt dt ( ) 0 t v = i+ t j Also, t t t r = v dt = i+ t t j dt ; t r = i + j Velocity of the body at t = s, v i = + j i j ms = ( + ) ms Displacement of the body at t = s, s = r( t = s) r( t = 0s ) = i + j 0 i j 6 = ( + )m 6 s = + = Torque on the body at t = s, τ= r F = i + j i j 6 ( + ) = ( i j) + ( j i) = k k = k 6 6 τ= Nm. (a,c) : ise of water in capillary tube, h = σ cos θ ρrg For a given material of the capillary tube, h. r When lift is going up with a constant acceleration a then effective acceleration in the experiment will be a + g. σcos θ \ h = and h < h ρra ( + g) 4. (b,d) : Let I and I be the currents in both loops as shown in figure. L L I = (I I ) I V L t V V I = e e L [ ] [ t ] L t V I e = e L [ t ] For I max, di dt = 0 + V = L e L t V L e L t 0 L t L t I I V...(i) e e e L = t or = L t = ln L t = ln = t time when I is maximum. Using t in equation (i), I or I max max = 5. (a,b,d) : L L V e ln L e ln = [ L ] V V 4 = 4 x = I z O C(, 00, ) x = Magnetic field due to ring at origin = µ 0 I µ 0 = 8 ( I k ) 6 ( k ) Magnetic field at origin due to wires y 78

78 I I B B + = µ 0 0 µ π π k (Assume current I and I are along + j direction.) (a) If I = I, then I B O = µ 0 6 ( k) It cannot be zero. (b) If I > 0 and I < 0, B at origin due to wires will be along +k direction and B due to ring is along k direction. Hence net magnetic field at point O can be equal to zero. (c) If I < 0, I > 0, then net magnetic field at origin will be I I I B O = µ 0( + ) µ 0 + π 6 k. It cannot be zero. (d) As I = I, then B = B, so magnetic field along z-axis is only due to the ring I B = µ 0 in z direction. 6. (b,c,d) : (a) Process-I is not T isochoric as, V is decreasing. II (b) Process-II is isothermal expansion. I III DU = 0, W > 0, IV So, DQ > 0 V (c) Process-IV is isothermal compression, DU = 0, W < 0, DQ < 0 (d) Process-I and III are not isobaric because in isobaric process T V. 7. (.00) : A = ai and B = a(cosωti + sin ωtj ) + = + + ωt A B ( a acos ωti ) asinωtj = acos...(i) = ωt A B ( a acos ωti ) sinωtj = asin...(ii) Using equations (i) and (ii) in A+ B = A B ωt ωt a cos = asin tan ωt = ωt π = or π π t = ω π = t =.00 s 8. (5.00) : Apparent frequency at O due to source at A, υ υ A A = v = υ v cosθ cosθ cos q O m Observer (stationary) m m q 5 m 5 m q A m s B m s q cos q υa = 40 θ cos cosθ 0 (Using Binomial expansion) Apparent frequency at O due to source at B, υ B v = υ v + cosθ υb = 0 40 cosθ 0 + cos θ 40 0 (Using Binomial expansion) Beat frequency (u) = u A u B =40 cosθ 0 u = cos q From figure, cosθ= 5 \ υ= 5 = 500. Hz 9. (0.75) 0. (.09) : Time period of spring-block system m T = π = π = π s k Just after the collision speeds of blocks are shown in figure. v v kg kg Using momentum conservation principle, + 0 = v + v or, v + v = EXAM ALET...(i) The International Physics Olympiad (IPhO) is an annual competition in physics for secondary school students, aiming at promoting physics and the development of international contacts in physics education. It involves individual theoretical and experimental competitions in physics. The IPhO st edition took place in Warsaw, Poland, in 967, with 5 participating countries. Since then, IPhO has grown to worldwide coverage involving more than 85 countries and 400 competitors. The 49 th IPhO will be held in Lisbon, Portugal, from to 9 July 08. The event is organised by the Portuguese Physical Society, on behalf of the Portuguese Ministry of Education. The IPhO 08 is supported by the Government of Portugal and by public and private partners. The Portuguese Minister of Education, the IPhO 08 Organising Committee and the city of Lisbon address a warm welcome to all participants in this event. 79

79 As collision is elastic so 0 e = = v+ v v+ v v + v =...(ii) From eqns (i) and (ii), 4 v = ms, v = ms Spring returns to its unstretched position for the first time after T =πs So, required separation between the blocks = T v 4 09 = ( π) =. =. m. (.50) : When the switch S is closed and S is opened, and capacitor C becomes fully charged. Then charge on C = C V 0 = 8 mc = 8 mc When the switch S is closed and S is opened Applying loop rule CV0 q q q = 0 C εrc C q 5 q = 0 e r q εr q mf q ε r = 5 q mf CV 0 q As per question, CV 0 q = 5 mc \ q = 8mC.5 mc = mc So, ε r = = 5 d d. (.00) : Average speed along x-axis, v + t+ t mv0 mv0 We have, r = and r = qb qb Since B B = y 4 \ r = 4r Time spent by charged particle in B is t = πm qb Time spent by charged particle in B is t = πm qb Total distance along x-axis is d + d = r + r = (r + r ) = (5r ) = 0r Average speed = 0 r mv0 5t = qb qb πm v 0 r B B = m s ( v 0 = p m s ) r x. (0.00) : I 0 I 0 A 0 A 0 0 cm cm Let A 0 be the initial area covered by light and A 0 be the final area covered by light at cm. From figure, A0 = = A ; A = A since, I 0 A 0 = I 0 A 0 I0A0 I0 = = 00 I0 = 0 kw m A (4.00) : 00 K r K 00 K L L ate of heat flow is given by dq dt = ( T T ) L where = KA Since rate of heat flow is same, we can write = or = L L = ; K A π( r) = = = 4 KA KA K A πr r 00 K K 5. (c) : From Maxwell's equations, electric and magnetic fields are related as E = cb, where c is the velocity of electromagnetic wave [E] = [B][LT ] MONTHLY TUNE UP CLASS XI ANSWE KEY. (c). (a). (d) 4. (d) 5. (d) 6. (a) 7. (b) 8. (c) 9. (a) 0. (d). (b). (b). (b) 4. (c) 5. (d) 6. (a) 7. (c) 8. (c) 9. (a) 0. (a, c). (a,c,d). (a, c). (b, c) 4. (4.00) 5. (5.) 6. (.70) 7. (c) 8. (d) 9. (c) 0. (b) 80

80 6. (d) : We have, c = or [ c ] = µε 0 0 µε 0 0 [ LT ] = [m 0 ] = [e 0 ] [L] [T] [ µ 0][ ε 0] a r ( a) ( + a) 7. (b) : r = = + + a r ( a) ( + a) r a r = a a + r a( + a+ a) = + a r ( a)( + a) a ( a) a r = = ( a)( + a) ( + a) + ( a) 8. (c) : N = N 0 e lt lnn = lnn 0 lt Differentiation with respect to l ETUN TO BOADS AISES CLASS X HUDLE As mandatory exam returns after 009, the pass percentage has fallen even below the 009 figure. But there has been a sharp rise in the number of students scoring 90% and above. And the top is now crowded with 5 students sitting pretty at the first three ranks Four students tied at top slot with 499/500 score Prakhar Mittal imzhim Agarwal Nandini Garg Sreelakshmi G DPS, Gurgaon P Public School, Bijnor Scottish International School, Shamli Bhavan s Vidyalaya, Cochin High scorers across the country 90% and above 95% and above dn 0 N dλ Converting to error, N Nt = λ ; \ λ = = Paper-II. (a), (b), (d). (a), (c), (d). (a), (b) 4. (d) 5. (a), (c) 6. (a), (b), (c) 7. (6.0) 8. (0) 9. () 0. (5.56). (). (900). (4) 4. () 5. (b) 6. (b) 7. (c) 8. (a) For complete solution refer to MTG JEE ADVANCED EXPLOE 7,476 Navodayas outdo KVs in a reversal of Class XII results,,49 Foreign school students performance Fall in numbers and successful students Candidates appeared 7L 6L 5L 4L L Appeared Passed Pass %,787,88 98.%.7L 4.9L 6.6L 6.L Differently abled toppers Marks School Anushka Panda 489 Suncity, Gurgaon Sanya Uttam School, Gandhi 489 Ghaziabad Somya Deep JNV, Dhanpur, 484 Pradhan Orissa Institution-wise pass percentage JNV KVS Independent Central Tibetan School Govt-aided Govt Overall pass percentage South trumps rest of India once again Top regions in pass percentage Thiruvananthapuram 99.6 Chennai 97.4 Ajmer 9.9 Number of candidates placed in compartment,86,067(.45%) What happened in 009-The last mandatory Class X Board before the grading & CCE system came Candidates Overall pass ALL-INDIA TOPPE registered percentage Parnail Singh 8,4, % 494/500 (98.8%) Candidates appeared PWD candidates 8,5,400,68 90% and above 95% and above 4,9 47,8 8

81 SOLUTION SET-59. (c) : By work energy theorem, W ext + W spring = K.E. f K.E. i Let x, x be the initial and final extensions and v, v are initial and final velocities. Initially, Fx kx = mv...(i) andfinally, Fx kx = mv...(ii) In both cases, force applied is same, and velocity becomes maximum when F = kx (at equilibrium) (after which the mass will decelerate). x \ F = kx = (4k)x x = ( k = 4k)...(iii) 4 k v = v = m x max k Now, vmax = m x 4k x v = max m 4 k v = m x max v = v max. (a) : Since : F = a bx means particle will do SHM. At mean position ; F = 0 a x = b In the figure shown, C is the mean position and A and B are extreme positions. a x = v = A= b a a max and max ω. = b b b. (c) : Area = 0 cm ; Y = 8 0 t/cm A B B C 5t 5t t t lt lt 5 Total extension = t 60 t 00 t AY AY AY ( ) t ( ) = = = cm AY ^ ^ 4. SO = i + 4 j Component of velocity along SO ^ ^ i j vs = ^ i + ^ j + 4 = 5( ) 0 m s 5 5. ^ ^ ^ ^ i+ 4j v0 = 5i+ 5 j 5 5 = ( ) ms 0 5 υ= 00 = 05 Hz 0 0 rc = t / PA 0 P A in P P 0 0 T = P 0 T T = P rc t/ 0 t Pressure inside the surface, T T T P in = P P P r = c t = / t So, net inwards force, T TA F = P0A PinA= P A P A 0 t 0 = t Here volume between the plates V V = A t t = Putting the value of t, A AT F = = 4 ( 40 0 ) ( 70 0 ) = 45 N; V So force required to separate the plates is 45 N. 6. Initial pressure inside the bubble T Pi = P r Now a uniform surface charge is given to the bubble. The surface tension is a pulling force, which increases pressure inside the bubble by 4T r But the charges given to the surface will repel each other. So due to the charge given, pressure inside the bubble will decrease by σ ε0 So, final pressure inside the bubble T Pf = P r σ f ε 0 As the temperature of the gas inside the bubble is constant so, P i V i = P f V f 4T 4 4T 4 P0 + r P0 r f r = + r σ f 0 π π ε 7 ε Here Put r f = r, So,get σ = + Pr 0 T 0 4 r 8

82 7. Taking a small element at D, Nsinq = dmg N = dmg...(i) sinθ T sind Ncosq = dmw Tsind = dm(w + gcotq) But d is very small, sin d d md Tδ = l ( ω + g cot θ ) π T d l mdl g = ( ω π + cot θ) T mg ω = + cot θ π g (using (i)) 8. (a) As a rod AB moves, the point P will always lie on the circle. \ Its velocity will be along the circle as shown by v P in the figure. If the point P has to lie on the rod AB also then it should have component in x direction as v. \ v P sinq = v v P = vcosec q Here cosθ= x y A C P v = = v 5 5 P sinθ = 4 5 cosec θ = vp = v 4 v (b) ω= P 5v = 4 9. Least count = MSD VSD (Q MSD = mm) \ Least count = 0. mm Length of side of cube = M.S.. + V.S.. least count = = 0. mm v mass. 75 Density = = = volume ( 0. ) Using significant figures the correct answer would be d η η η 0. sinθ = 4 = and cosθ = d B B v b A A vb= vc θ v θ v C C F ext = 0, so linear momentum of the system remain conserved. Px = constant, P y = constant (along x-axis) mv = mv b cosq mv v= vbcosθ v = vb 4 η v...(i) For elastic collision vrel sepration e = = vcosθ= vb + v cosθ vrel approach v η v v v η or v cosθ = 4 + = η 4 η v= v v = 4 η + 4 η 4 η v( h ) = v (6 h ) v = v η 6 η (If > h then disc A recoil with v ) v = v ( h ) / (6 h ). espectively at smaller h, equal, or greater than MONTHLY TUNE UP CLASS XII ANSWE KEY. (b). (a). (b) 4. (d) 5. (c) 6. (d) 7. (a) 8. (c) 9. (c) 0. (b). (b). (b). (b) 4. (b) 5. (d) 6. (a) 7. (a) 8. (b) 9. (b) 0. (a,b,d). (a,c). (b,c). (a,c,d) 4. (.00) 5. (8.48) 6. (5.00) 7. (b) 8. (c) 9. (a) 0. (a) v c 8

83 Current Electricity CLASS XII eaders can send solutions at or post us with their complete address by 5 th of every month to get their names published in next issue ACOSS. The potential difference between the two terminals of a cell in an open circuit () 5. The magnitude of the drift velocity per unit electric field (8) 7. A device used for measuring the internal resistance of a cell () 0. The element that has negative temperature coefficient of resistivity (7). The reciprocal of resistance (). An alloy used for making wire bound standard resistors (8) DOWN. A type of resistor whose resistance depends on temperature (0). The colour of the third band of a coded resistor of resistance 500 MW (6) 4. The SI unit of conductance (7) 6. The physical quantity represented by the slope of V-I graph (I is on x-axis and V on y-axis) (0) 8. The carriers of electric current in a superconductor (9) 9. An example of a non-ohmic device (5). The resistance of an open key (8) 84

84 The Final Theory on the Origin of the Universe Stephen Hakwing last paper co-authored with European esearch Council grantee Thomas Hertog proposes a new cosmological theory His final theory on the origin of the universe has been published in the Journal of High Energy Physics. Hawking collaborated with Professor Thomas Hertog from KU Leuven on the paper which was submitted for publishing before Hawking s death in March this year. The paper s ideas are based on string theory, it predicts the universe is finite and that the universe s origin may, in fact, be far more simple than current theories relating to the Big Bang. Modern Big Bang theorists believe that the universe came into existence with a burst of inflation that started to occur just moments after the Big Bang itself. It is thought that once this inflation begins the quantum effects can keep it going indefinitely. The part of the universe that we can observe is just a hospitable pocket where inflation has ended. In their new paper, Hawking and Hertog say the eternal inflation model is wrong. New theory claims universe is finite. This is because Einstein s theory of general relativity breaks down on quantum scales. The problem with the usual account of eternal inflation is that it assumes an existing background universe that evolves according to Einstein s theory of general relativity and treats the quantum effects as small fluctuations around this, said Hertog. However, the dynamics of eternal inflation wipes out the separation between classical and quantum physics. As a consequence, Einstein s theory breaks down in eternal inflation. We predict that our universe, on the largest scales, is reasonably smooth and globally finite. So it is not a fractal structure, said Hawking. Their latest research doesn t disprove multiverses but reduces them to a much smaller range. Hawking s work paves way for future physics research. This means the multiverse theory will be able to be tested by a larger range of physicists in the future. Stephen Hawking s voice beamed to nearest black hole message from late British astrophysics giant Stephen Hawking was beamed towards A the nearest black hole on 5 June afternoon as his remains were laid to rest in London s Westminster Abbey. With celebrities and science enthusiasts from around the world in attendance, the ashes of the theoretical physicist were interred by the graves of fellow science greats Isaac Newton and Charles Darwin. A specially-written musical piece by Greek composer Vangelis featuring Hawking s famous synthesised voice was beamed into space by radio waves from a European Space Agency satellite dish in Spain. The ESA said the six-minute message, which is drawn from a speech Hawking gave about preserving the planet, was being transmitted towards the black hole A , which was discovered in 975 and is located,500 light years from Earth. This is a beautiful and symbolic gesture that creates a link between our father s presence on this planet, his wish to go into space and his explorations of the universe in his mind, said his daughter Lucy Hawking. It is a message of peace and hope, about unity and the need for us to live together in harmony on this planet, she said. The memorial stone placed on top of Hawking s grave included his most famous equation describing the entropy of a black hole. Here Lies What Was Mortal Of Stephen Hawking, read the words on the stone, which included an image of a black hole. Hawking, who had captured the imagination of millions around the world, died on March 4 at the age of

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91 egistered.n.i. No. 5780/99 Published on st of every month Postal egd. No. DL-SW-0/4047/8-0 Lic. No. U(SW)-0/08-0 to post without prepayment of postage N.D.P.S.O.N.D- on -rd same month