Cengage MECHANICS 1

,. iGHAPTER Units and Dimensions ',' \"\", ,\"', SystelUsoftfnh~···· Signifi~ant·Figutes. i...'j;,·1?:i~t%nSiohsoi\\a~hysj¢alQulUltity ElttorsiIlMeasuJ:ements Absolute Errdrs.\" •.·'j;,DltAehsi()ri~lJ:lq~nl~e PropagatioliOiCojllbinedE~ors . i» ··lJ~~§9fJ:)ill1el.lsiona[A!l:alysis 3.1

3.2 Physics for IIT-JEE: Mechanics I To measure a physical quantity, we need a standard known as Table 1 unit. For example, if length of some metal rod is measured to be 15 em. then em is the unit of length. 15 is thc numerical part. So M.K.S. c.G.s. F.P.S. S.I. Unit., System System =Physical Quantity Numerical Part x Unit (i) Length System (i) Length It is an extended (i) Length ft (foot) form of M.K.S. We have threc types of units: Fundamental units. Supplemen- m (meter) system. It includes tary units and Derived units as illustrated in Fig. 3.1 below. em (cen- timeter) four more fun- damental units Independent of Derived,from (ii) Mass (ii) Mass (ii) Mass each other and not fundamental kg (kilo- g (gram) (pound) (in addition to interconvertible gram) three basic units), units (iii) Time (iii) Time· (iii) Time which represent I Ul1itoflen'gth, mass s (second) s (second) s (second) fundamental quan- time~ temperature,' 1 titics in electricity, luminuous intensitY, magnetism, heat Unlts,'ofVc,lo,city, and light. Ciectric current. acceleration, Unit of, forcel work,etc, Table 2 'amount of substance, 1. plane angle A. Fundamental Quantities in S.I. System and Their Units 2.\" solid angle Sr. No. Physical Quantity Name of Unit Symbol of Fig. 3.1 Unit 1. Mass kilogram kg SYSTEMS OF UNITS 2. Length meter 3. Time second III It is a complete set of fundamental and derived units. We have 4. Temperature kelvin four types of systems of units. Generally, a system is named in 5. Luminous intensity candela s terms of fundamental, units on which it is based. 6. Electric current ampC'-rC'. K 7. Amount of substance mole Cd 1. M.K.S. system: In this system length, mass and time are taken as fundamental quantities. A 2. C.G.s. system: It is Gaussian system. In this also length, mol mass and time are taken as fundamental quantities. B. Supplementary Quantities in S.I. System and Their Units 3. F.P.S. system: In this also length, mass and time are taken as fundamental quantities. It is British Engineering system. Sr. No. Physical Quantity Name of Unit Symbol of Unit M.K.S. and C.G.S. systems are also called metric sys- tems or decimal systems, because mUltiples and submultiples -I.--------pranc-angle radian rad are related by powers of 10. Example: I km = 103 m 2. Solid angle steradian -Sf- - - - F.P.S. system is not used much nowadays because of • inconvenient multiplies and submultiples. The disadvantages of e.G.S. system is that many derived units in this system Advantage of S.l. system is that it assigns only one unit to become unnecessarily small.. various fonns of a particular physical quantity. For example, unit of all kinds of energy is J in this system. But in M. K.S. The main drawback of all the above systems is that they system: are confined to mechanies only. All the physical quantities Unit of mechanical energy is joule, that of heat energy is calorie, appearing in physics cannot be described by these systems. that of electric energy is Wh (watt hour), etc. So, we need such a system which takes eare of all the physical quantities appearing in physics. S.l. system is such a kind of DIMENSIONS OF A PHYSICAL QUANTITY system. These are the powers to which the fundamental units of mass, 4. S.l. system: It was introduced in 1971 by General Conference length and time have to be raised to represent a derived unit of on Weights and Measures. lt is also called as rationalised the physical quantity under consideration. Dimensions of any M.K.S. system because it is made by modifying the M.K.S. derived physical quantity can be represented in the form of fun- system. It is nothing but extended M.K.S. system. It is a damental units of mass, length and time. Knowing the units, comprehensive system (see Table 1). -dimensions can be easily written. This system contains seven fundamental units and two supplementary units as shown in Table 2. It also contains a To write the dimensions of a physical quantity, we use fol- large number of derived quantities, lowing symbols for mass, length and time: Mass - 1M]; Length -lLl; and Time - [T]:

DIMENSIONAL FORMULAE Units and Dimensions 3.3 Relations which express physical quantities in terms of appro- Convert 1 joule into erg. priate powers of fundamental units are known as dimensional Sol. louIe: 5.1. system, erg: C.O.S. system formulae. These formulae tell us about: Work = force x distance = mass x acceleration x length 1. Fundamental units involved to represent a quantity. = mass x -leng-th x length 2. The nature of their dependence, (time)' Obtain the dimensions of acceleration. Dimensions of work = IW] = [M' L2T \"J Sol. We know that acceleration: a = 1, b =2, c = -2. v sll s Now, a = - = - = -, (8 - distance, I - time) .'. using III = I[100_0g]' [100cmJ' = 107 raj = [[TLFI = IL \"r 1= IMoL']\"1 I g I em So, I joule = I07erg. So, the dimensions of acceleration arc () in mass, +1 in length and ~2 in time. Convert S4 kmh~l into ms~l. Some more examples: 1. Force: Force = mass x acceleration = 1M] x IL' T~21 Sol. Let v = 54 kmh~~~' = n2 ms~' = [M'L'T~21 [vi = LT~',a=O,h= I.c=-I 2. Momentum: Momentum = mass x Velocity fl, -- 54 [k- g]\" [k- mJ' [h-]<~'\" g ms = [MJ x [L'T\"] = [M'L'T~'] 1000= 54 x I x 1000 X [36001~' = 54 x = 15· 3. Work: Work = force x distance = [M' L' 7'~'21 x ILl ~ 3600 . = 1M' L'7'2] Hence, 54 kmh~' = 15 ms~1 USES OF DIMENSIONAL ANALYSIS 1. Conversion of units of a quantity from one system to another. 2. To check the accuracy of formulae. 3. Derivation of formulae. Conversion of Units of a Quantity from One System to To Check the Accuracy of Formulae Another The accuracy of the expression of any physical quantity can be Physical quantities can he converted from one system of units to checked by using the principle of homogeneity. According to another. Due to this conversion, the numerical part of physical this principle, dimensions of various quantities as a whole on quantity changes but the dimensions and the overall quantity both sides of an expression (related to a physical quantity) are remain the same, equal. Suppose a physical quantity has the dimensional formula Check the accuracy of the relation M{lLbrc . - u2 = 2as, where v and u are final and initial velocities, Let N! and N2 be the numerical values of a quantity in the a is acceleration and s is the distance. two systems of units, respectively. Sol. We have v2 - u2 = 2as In first system, Physical Quantity Q = N! Mll Lf~:: = N! VI In second system, Same Quantity Q = N2M~1 L~l~: = N2U2 Checking the dimensions on both sides, we get A physical quantity remains the same irrespective of the sys- LH.S = IL1'-'1' - ILr']' = rL2T~']_[L2T~2] = [L'1\" 2J tem of measurement, i.e., ~ R.H.S = [L'r'J[LJ = reT'J Q=N,U,=M~ • M~~~=M~L~~ Comparing LH.S. and R.H.S\" we find LH.S. ~ R.H.S. • N, -- N,<[~M~-'J\"[L-,'J\"[T-'J\" Hence, the formula is dimensionally correct. M, L2 12 Check whether the relation S = ut So, knowing the quantities on the right hand side the value of N2 can be obtained. + i1at' is dimensionally correct or not, where symbols have their usual meaning.

3.4 Physics for IIT-JEE: Mechanics I :2So, dimensions of == dimensions of P Sol. We have S = ut + ~at2, Checking the dimensions on both :::::? Dimensions of a == dimensions of P x dimensions of V2 2 sides, L.H.S. = [MOL' T0]' '* [al = [M' L -, 1'- 2 1 X [L3f = [M' L'T'I and +R.H.S. = [U- l l[T] [LT-'][T 2 ] Dimensions of b == dimensions of V = [MoL'Tol + [MoL'T o] = [Mol,'ro] =} [b] = [VI = IMoL3Tol Comparing the L.H.S. and R.H.S .. we find L.H.S. = R.H.S. Hence, the formula is dimensionally correct Unit of a == unit of px unit of \\1 2 = N x m6 = Nm4, \"7 m\" Unit of b \"unit of V = m3 Limitations Derive an expression fhr the time period of a simple pendulum of length l. 1. Sometimes an equation which is dimensionally correct. may Sol. Let t ex m(llbgc :::::> t = kmal\"f( not be correct actually. MOLor\"-' = M\" L\"[Lr -'1' Example: S= ut + at', l' = \"JT.. ::::} MOLorl = M(/L'H-CT ·,2e Vg Comparing the powers of M, Land T on both sides: a == 0, So, mere dimensional correctness does not indicate that an l>+c=0,-2c= I equation must be correct physically also. But if an equation '* {/ If= O. b = 112 and c = - j (2. Putting these values. we get f = kmo [!/2 =} t = k -, which is the required relation. is dimensionally wrong, then certainly it must be wrong, al- gg though it. may also have exceptions like: a D\" = u + :.;:(2\" - I). 2. Two different physical quantities may have same dimensional formula. For example, work and torque have same dimen- sions though they are different physical quantities. Note:. According to principle ()f homogeneity. the quanti- Useful Tips ties having same dimensions only be added or sllhlmeled. • All of the following havethe same dimensiomtl for- =So,dimensionally: L + L Land L - L = L. mula [MoLor\"]: Derivation Of Formulae Frequency, angular frequency, angular velocity and If we know the factors on which a quantity depends, then we velodty gradient. can derive its formula using the principle of homogeneity. • All of the following quantities are dimensionless: For a particle to move in a circular orbit Angle, solid angle, T-ratios, strains, Poisson's ratio, relative density, relative permittivity, refractive index uniformly, centripetal force is required which infact depends and relative permeability. upon mass (m), velocity (v) and radius (r) of the circle, Ex- • Following three, quantities have the Same dimen- press centripetal force in terms of these quantities. Sol. According to provided information, let F ex m(/vhr c . r-sional formula [M oL 2 2]: ::::} F = km(/vbr<' (i) Squa'l:e of'vefoCity~--gl:a\\;nanohal'poteIitial~ aiia IiHent' where k is a dimensionless constant of proportionality and a, b, heat. e arc the constant powers of m, v, r, respectively. • Following quanti tie's have the same dimensionless '* [M'L'r-'1 = IM\"(LT-1),'L'] r·-fornwla [M L 2 2]: '* [M'L'r- 2 1= 1M\" L/H,'T-\" I Work, energy, torque and heat. Now, using principle of homogeneity (i.e., comparing the • Foll~\\VinK have the same dimensional formula power of like quantities on both side), we have [MLT-· 1]: Momentum and impulse. a = I (ii) !J + c = I (iii) b = 2 (iv) • Both acceleration 'and gi·avitational field intensity have the same dimensional formula [Mo LT~2J. Using (ii), (iii) and (iv). we have (I = I, b = 2, C = -I Using these values in (i), we get F = kill I v2r~1 • Force\" weight, 'thrust and energy gradient hAve the mv2 same dimenSional formula [MLt-'} ::::} F = k--, which is the desired relation. • Eritropy:'gas conStant, Boltzmann constant and ther- r mal t:;,apacity have the, smile dimensions in ma~s, Find out the unit and dimensions of the length andtime. (p :2)constants a and b in the Van der Waal's equation + • Light yeat, radius of gyration and wavelength, have the same dimensioIull formula [MoLTo]. (V - b) = RT, where p is pressure, V is volume, R is gas constant and T is temperature. • Surface tension 'and sp'ring constant, have the same. dimensional formula [M L °T~2]. Sol. We can add and subtract only like quantities. • Planck's constant and' angular momentum have the ~ame dimensional fonnula [M L '1'-'].

It Rydberg constant and propagation constant have the Units and Dimensions 1.5 same dimensional formula [MoL -I rOj, SIGNIFICANT fIGURES \\9 Following quantities have the same dimensional for- mula [M L -'r,,2j: When we measure a physical quantity, generally the measured Pressure, stress, moduni ofelasticity and'energy den- value does not come out to be accurate. It may contain some sity, error. When the measured vulue is expressed as a number, then some digits which it contains are known reliably plus the first III It IS not possible for a qua'nUty, to have dimensions digit which is unreliable. but no units. For example: Let us measure the length of a glass plate using \\$ It is possible for a quantity t.o have linits but no di- a scale. Let this length lies somewhere between 2.6 and 2.7 mensions. cm, say 2.63 em. Here, digits 2 and 6 are reliable, but digit 3 is unreliable. Now, the reliable digits and first unreliable digit lit A quantity has same dimensions in different. systems arc known as signiticant figures or significant digits. Thus, the measurement 2.63 em contains three significant figores. of units. These significant figures arc desired when the observations of 1. Is light year a unit of time? any experiment have to be recorded and then to be used in cal- culations. Here, the knowledge of significant figures is helpful. 2. Docs magnitude of a quantity change with change in the unit of system'! Rules for Counting Significant Figures 3. All constants arc dimensionless. Comment. For a Number Greater Than 1 4. What is the difference between nm, mN and Nm? 1. All non-zero digits arc significant. There may be decimal 5. Name three physical quantities which have same eli men- point in between and location of decimal docs not maHer. Example: SiCHlS. a. 2357 has four significant figures h. 312 has three significant figures 6, Justify L + L = Land L ,- L = L, c. 325.23 has nve significant figures d. 32.523 has five significant. flgun:s 7. Displacement of a particle is given by: x = /\\2 sin 2 K t, where t denotes the time. What is the llnit 2. Ail zeros between two non-zero digits arc significant. Lo- cation of decimal does not matter. Example: of K'! a. 2307 has foul' signitlc~nt figures (p :2)S. In the equation + b. 320JJ3 has five significant ligures (V - h):::: constant, what is the c. 32.003 has five signit-icant figures unit of a? 3. If the number is without decim{tl part, then the terminal or trailing zeros are not significant. Example: 9. According to quantum mechanics light travels in the form a. 23500 has three significant figures h. 53000 has two significant figures of packets and energy associated with each packet is E = 4. Trailing zeros in the decimal part are significant. Example:· I?[, where h is Planck's constant and f is the frequency. a. 3.700 has four significant figures What is the dimensional formula of h? h. 2.50 has three significant figures 10. The velocity v (in cms---!) of a particle is given in terms Note: Iftrailing zeros come/rom some measurement then of time t (in sec) by the equation: v = of + .-I.i ~. What they may be significant as illustrated below, t -10 C Experiment 1: Let some length is measured to be 1500 mm. arc the dimensions of a, hand c? According to rule (3), it should have two significant figures. But we can write 1500 mIll = 1.500 m. Then, according to rule (4) 11. What is the dimensional formula of magnetic nux? it should have four significant figures. Now, the question arises: Docs thechangeofunitchange the number ofsignificant figures? 12. A force F is given by F = at + ht 2, where! is the lime. The answer is No. I~ fact, the measured length 1500 mm will What arc the dimensions of a and h? have four significant figures. Experiment 2: Now, let same length is measured again and this 13. Let us redefine I N as the force of attraction bet ween two lime the valLIe comes out to be 1.5 m. This measurement has two particles, each of mass I kg, separnted by 1 111. Then, what is the ne\\-\\' value of universal constant of gravitation? 14. ln a hypothetical new system ofmcasurcment, the gravita- tional force bet ween two particles, each of mass I kg, sepa- rated by 1 kIll is taken as a Linit offorce. Let liS call this new unit of force 'notwcn'. How many newton will be there in one 'notwen''? Given: G = 6.67x 10--.!1 Nm2kg\"·2. 15. Let us consider a new hypothetical system of measurement in which the unit of energy is called eluoj. Suppose, in this system, the gravitational force of attraction between two particles, each of mass 1 kg, separated by 1 km is taken as a unit of force. Then, how many joule arc contained in one eluoj? Given: G :;::; 6.67 x 10---! 1 Nm2kg-- 2 .

3.6 Physics for lIT-JEE: Mechanics I significant figures. We can write 1.5 01 = 15000101. Here, 1500 2. If digit to be dropped is more than 5, then the preceding digit mm will have two significant figures. is increased by one. Example: a. 7.86 after rounding off becomes 7.9 From the above experiments, we learn that only originally b. 5.937 after rounding off becomes 5.94 measured quantity will indicate the correct number afsignificant jigures. 3. If digit to be dropped is 5: a. If it is only 5 or 5 followed by zero, then the preceding So, in order to avoid confusion in counting the number of digit is raised by one if it is odd and left unchanged if it is significant figures, we usually express a measured quantity in even. Example: scientific notation. By this, the number of significant figures are i. 4.750 after rounding off becomes 4.8 clearly mentioned and do not change on changing the units. For ii. 4.75 after rounding off becomes 4.8 illustrations, see the table below. iii. 4.650 after rounding off becomes 4.6 iv. 4.65 after rounding off becomes 4.6 If original mea* If original mea- If original mea- b. If 5 is further followed by a non-zero digit, the preceding sured quantity is digit is raised by one. Example: sured quantity is 1.5 m sured quantity is i. 15.352 after rounding off becomes 15.4 1,500mm ii. 9.853 after rounding off hecomes 9.9 1.5 01 lSOcm ........~ 1,5000101 = 1.5xlO3 mm Note: In Intermeditltesteps during multistep calculations, = 1.500X!03 mm = 1.5x 102 em 150em we should/etain Olle digit more than tile. sigIJificallt dig- = 1.5 x Ht' km = 1.50x 103 mm its, and at the end of the calculation, roundoff to proper = 1.500 m All of the above significantjigurei;. = 1.500 x 102 Cm contain two sig- = 1.50 m = 1.500 x 10..3 km nificant figures. = 1.50xl02 cm = 1.50x 10. 3 km All of the above contain four sig- All of the above ~nifi.c.-a-n.t figures. contain three sig- nificant figures. -.~ For a Number Less Than 1 a. 0.05857 Round off to two significant figures: c. 5.07 x 106 b. 0.05837 Any zero to the right of a non-zero digit is significant. All zeros d. 5.0t X to' between decimal point and first non-zero digit arc not signif'icant. Example: Sol. b. 0.058 a. 0.059 d.5.0xI0\" a. 0.0074 has two significant figures c. 5.1 x 106 b. 0.00704 has three significant figures Significant Figures in CaLculations c. 0.007040 has four significant figures When we do the calculations using measured values, the result d. 0.07040 has four significant figures cannot be more accurate than any of the measured value. The ~ Write the number of significant ligures result must possess the accuracy,. level as that of original mea- in the following: b. 50.00 surements. So, to have proper accuracy in the final result we Becd d. 5.7 X 106 to follow some rules during different arithmetical operations, a. 0.053 f. 2400 c. 0.0500 h. 0.0305090 1. Addition and subtraction: The number of decimal places e. 5.70 x 106 in the final result of any of these operations has to be equal to the smallest number of decimal places in any of the terms g. 2400 kg involved in calculations. Example: a. SUlll of terms 2.29 and 62.7 is 64.99. After rounding off Sol. to one place of decimal it will become 65.0. a.2 b.4 c.3 d.2 e.3 f.2 g.4 h.6 b. Subtraction of62.7 from 82.29 gives 19.59. Afterrollnding off to one place of decimal it will become 19.6. RuLes for Rounding off the Uncertain Digits Note: ])uring the. subtraction of quantities of nearly When we do the calculations using measured values, the result equal magnitude, accuracy is almost destroyed, . e.g., may contain more than one uncertain digits which should be rounded off. The following rules arc used for rounding off: 328..,. 3.23=0.0$. Result 0.05 has only one significant jig- 1. If digit to be dropped is less than 5, thell the preceding digit lmiwhereasoriginal measurements have three significant remains unchanged. Example: a. 6.32 after rounding off becomes 6.3 .jigureseach. . b. 5.934 after rounding off becomes 5.93 So, it is advised to measure the difference directly in- stead of measuring the quantities first and then finding their difference.

Units and Dimensions 3.7 2. Multiplication and division: In these operations, the number Instrumental Errors of significant figures in the result is the same as the sniallest number of significant figures in any of the factors, Example: These errors are introduced due to improper designing and man- a. 1.3 x 1.2 = 1.56. After rounding off to two significant ufacturing defects of instrument. Often there may be zero error, figures it becomes 1.6. For example, a meter scale may be worn off at the end of zero mark, The instrumental errors can be reduced by 'Using more ac- 3500 .. . curate instruments and applying zero correction, when required. b. - - = 465.42. As 3500 has mInImum number of slg- 7.52 Personal Errors nificant figures, Le., two, so the quotient must have two These errors are introduced due to lack of proper care on part of significant figures. So, 465.42 = 470 (after rounding off). the observer. For example, lack of proper setting of the apparatus, c. Ifwe divide 3500 m by 7.52, 3500 In has four significant fjgures, then fjnal result should be 465 (after rounding off recording the reading without applying proper precautions and to three significant figures), so on. Subtract 2.5 x 104 from 3.9 x 105 with Random Errors due regard to significant figures. So!. Let x = 2.5 X 104 = 25, 000, )' = 3.9 X 105 = 3, 90, 000 The causes of such errors are not known precisely, Hence, it is not possible to eliminate the random errors, e.g\" same person booY - x = 3, 90, 000 - 25, ='3, 65, 000 = 3.65 X 105 repeating the same experiment may get different readings each time. These errors are also known as chance errors. = 3.6 x lOs (rounded off to one place of decimal) These are minimized by repeating the experiment and taking Calculate area enclosed by a circle of the arithmetic mean of a1l the observations, The mean value diameter 1.06 m to correct number of significant figures. should be close to the accurate value, 1.06 Gross Errors Sol. Here. r = _._..- = 0.53 m. These errors arise on account of shear carelessness of the ob- 2 server. For example: Area enclosed = m·2 = 3.14(0.53)2 = 0.882026 m' 1. Reading an instrument without setting it properly. 2. Taking the observations wrongly without caring for the = 0.882 m2 (rounded off to three significant figures) sources of errors. 3. Recording the observations wrongly. ERRORS IN MEASUREMENTS 4. Using wrong values of the observations in calculations. The difference in the true value and the measured value of a these errors can be minimized only if the observer is sincere quantity is called error, and mentally alert. One basic thing on which every branch of science depends is ABSOLUTE ERRORS mcasurement. There are always many factors which influence the measurement. These factors always introduce error (may be It is the magnitude of the difference between the true small, whatever be the level of accuracy), So,.oo measurement is perfect. We can only minimize the errors using best methods value and the measured value of a physical quantity. Let and technique,\\;, but we cannot eliminate them permanently. x I , X2, X3, ... , XII be n observations recorded corresponding Types of errors: We have three types of errors. 1. Systematic errors, 2. Random errors, and 3. Gross errors. to a physical quantity X, then mean value XII! is given by: Xl + X2 + X3 + ... + XII Systematic Errors . XII! = . As the correct value of X IS not Errors whose causes are known are called systematic errors, These errors can be minimized by applying some corrections. II These are of various types: known, so Xm is taken to be the absolute value or correct value, Errors due to External Factors The absolute error in various observations are given by These errors are due to fluctuation in atmospheric conditions like temperature, pressure, humidity, etc. 6xI = jXm -xlj, 6X2 = IXIII -x21, ... , L\\.xll = IXm -xul Errors due to Imperfection The arithmetic mean of these absolute errors is called mean These are introduced due to negligence of facts, e.g., error in absolute error. Let us denote it by L\\.xm. weighing of a body arising out of buoyancy is usually ignored. - ~Xl + 6.x2 + ... + L\\.xn = 1 £~--1 /',x;l. Then, /',x\", = - 11 n i,=! Now, the meawrement is likely to be in between Xm - 6x m and XI/! + 6x m , Therefore, the final result of the measured phys- ical quantity can be written as X = Xm ± L\\.xm

3.8 Physics for IIT-JEE: Mechanics I Relative Error and Percentage Error Error in Difference Relative error is defined as the ratio of mean ahsolute error to Maximum absolute error in difference o[two quantities is equal the mean value of the quantity measured, When it is multiplied to sum of the absoluteelTors in the individual quantities. Suppose by LOO, it becomes percentage error. x = a -b. mean absolute CITor 6.x m Let i..a ::::: absolute error in measurement of a, 6:.17 ::::: absolute Relative error = error in measurement of hand 6:.x ::::: absolute error in calCulation real value (mean value) ~XIl1 of x, i.e\" difference of a and h. Percentage error = - - x 100 ±(/l\"Thus, value of the maximum absolute error in x is given by /lX = + /lb). .Ym Repeated measurements of a certain quantity in an experiment gave the following values: 1.29, Error in Product 1.33, 1.34, 1.35, 1.32, 1.36, 1.30 and 1.33. Calculate the mean Maximum fractional crror or relative error in product of quan- tities is equal to sum of the fractional or relative errors in the value, mean absolute error, the relative error Hnd the per- individual quantities. Suppose x = a x h. centage error. ' Sol. Here, mean value: Let 6:.0 ::::: absolute error in measurement of a, 6:.b::::: absolute 1.29 + 1.33 + 1.34 + 1.35 + 1.32 + 1.36 + 1.30 + 1.33 error in mcasurement of hand 6:. x =,absolute error in calculation of x, i.c., product of a and h. 8 = 1.3275 = 1.33 (rounded off to two places of decimal). (/l\"So, thc maximum possihle value offraclional error in product Absolute errors in measurement arc: + .ot. quanll.tte.s .IS .glven by -/l~X = ± .-~. -/l~l!) /lXI = 11.33 - 1.291 = 0.04; /lX2 = 11.33 - 1.331 = 0.00; /lX, = 11.33 - 1.341 = OJ)I; /lX4 = 11.33 - J .351 = 0.02; x ab /lX5 = 11.33 - 1.321 = 0.01; /lX\" = IU3 - 1.361 = 0.03; Error in Division Maximum value of fractional or relative error in division of qua11lities is equal to slim of the fractional or relative errors in /lX7 = 11.33 - 1.301 = 0.03; /lXH = 11.33 - 1.331 = 0.00. (( Mean absolute error: the individual quantities. Suppose x = --. _ 0.04 + 0.00 + 0.01 + 0.02 + 0.01 + 0.03 + 0.03 + 0.00 h 6:.X/ll = -'\";----~-..-~......... '·---~8\"---- Let 6:.a :;;: absolute error in measurement of a, L\\b :=: ahsolute error in measurement ofb and /lX = absolute error in calculation of x, i.e., division of a and h. = 0.0175 (/l\"So, the maximum possible value of fractional error in division = 0.02 (rounded olT to two places of decimal). of. quanti.tie.s .IS .gIVen by ~/-lX.... = ± ..\",-,. + -/\".l,,h-- ) . Relative error = ±-6..:...x._m- = 0.02 = ±O.01503 = ±O.02 ±-- x (/ h .:rm 1.33 Error in Power of a Quantity (rounded off to two places of decimal) Percentage error = ±O.O 1503 x J00 = ± 1.503 = ± 1.5%. Fractional error or relative error in the quantity is equal to sum of fractional or relative error of the individual quantities multi plied PROPAGATION OF COMBINATION OF ERRORS by their powers, When we do calculations using measured values which them- As the error multiplies II times, therefore, in any formula, the selves contain error, definitely there will also be error in the final result. To calculate the net error in the tlnal result, we quantity with maximum power should be measured with highest should know how errors propagate in different mathematical operations. degree of accuracy, i.c., with least error. alii Consider a quantity x = ~\"'.-'\". bll Let L\\a :;;: absolute error in measurement of 0, 6:.b::::: absolute error in measurement of b ancl6:.x::::: absolute error in calculation of x. Then, fractional or relative error in x is given by Error in Summation [(/l\")~/lX-=± III --- +11 (/-lb)] Maximum absolute error in the sum of two quantities is equal to X {/ b sLIm of the absolute errors in the individual quantities. Suppose Note: Ifa set of experimental data is specified to II signif- icant figures, a result obtained by combining the data will x =a+b. also be valid to Il significant figures. When two or more (Jxpel'imentally obtained numbers are Let 6.a = absolute error in measurement of (I, 6:.b::::: ahsolute multiplied, the percentage uncertainly ofthe final result is equal to square root of the sum of the squares ofthe per- error in measurement of hand 6:.x::::: absolute error in calculation centage ullcertainties ofthe original numbers. of x, i.e., sum of a and b. ±(/l\"Thus, value of the maximum absolute error in x is given by /lX = + /lb).

Units and Dimensions 3.9 The initial and final temperatures of 3. A research worker takes 100 observations in an experi- ment. If he repeats the same experiment by taking 500 recoriled by an observer are (40.6 ± O.2YC and observations, how is the probable error affected? (78.3 ± O.3)\"C. Calculate the rise in temperature with 4. Which quantity in a given formula should he measured most accurately? Why? proper error limits. 5. A hody travels uniformly a distance of (13.8 ± 0.2) m in Sol. Here, 8, = (40.6 ± O.2)\"C, O2 = (78.3 ± O.3)'C a time (4.0 ± 0.3) s. Find the velocity of the body within Rise in temp: 0 = O2 .. 0, = 78.3 .. 40.6 = 37YC error limits and the percentage error. Error in Ii: ,,0 = ±(\"O, + ,,(2) = ±(O.2 + 0.3) = ±OS'C 6. Error in the measurement of radius of a sphere is l%. Hence, rise in temperature = (37.7 ± 0.5)\"C Find the error in the measurement of volume. The length and breadth of a rectangle 7. Given Ii, = 5.0 ± 0.2 Q, R, = 10.0 ± 0.1 Q. What is the total resistance in parallel with possible % enor? are (5.7 ± 0.1) em and (3.4 ± 0.2) cm. Calculate area of the rectangle with error limits. 8. The value of resistance is 10.845 ohm and the current is 3.23 ampere. On multiplying them, we get the potential Sol. Here, 1= (5.7 ± 0.1) em, b = (3.4 ± 0.2) em difference ~ 35.02935 V. What is the value of potential difference in terms of significant figures? Area: A = I x b = 5.7 x 3,4 = 19.38 em2 = 19 em2 9. The length of one rod is 2.53 cm and that of the other (rounding off to two significant figures) is 1.27 em. The least count of measuring instrument is \"A = ± (t:~ +0:\") = ± (ll:.!. + 0.2) 0.01 em. If the two rods are put together end to end, find A Ib 5.7 3.4 the combined length. = ± (0.3=-±.1.1~) = ±,~,48 10. A wire has a mass 0.3 ± 0.003 g, radius 0.5 ± 0.005 mm 5.7 x 3.4 and length 6 ± 0.06 cm. What is the maximum percentage 19.38 error 1n the measurement of density? 1,48 1,48 \"A =±--- x A = ±-- x 1938 11. The pressure on a square plate is measured by measuring 19.38 19.38' the force on the plate and the length of the sides of the = ±1,48 = ±['5 Ii'F (rounding orfta two significant figure,S) plate by using the formula P = If the maximum er- So. Area = (19.0 ± 1.5) em'- rors in the measurement of force and length arc 4% and The distance covered by a body in time 2%, respectively, then what is the maximum error in the (5.0 ± 0.6) s is (40.0 ± 0.4) m. Calculate the speed of the measurement of pressure? body. Also, determine the percentage error in the sped. 12. The density of a cube is measured by measuring its mass and the length of its sides. If the maximum errors in the Sol. Here, S = 40.0 ± 0,4 m and 1 = 5.0 ± 0.6 s measurement of mass and length are 3% and 2%, respec- tS tS)Speed I! = tively, then find the maximum error in the measurement = 45'0~.O0' = 8.0 illS , ( v = of the density of cube. As Vernier Callipers: Its Use to Measure Length, Internal \"V \"Ic.-=-\"·.+I' '··· and External Diameters and Depth of a Cylindrical ust Vessel \"SHere, = 0.4 m, .I' ~ 40.0 m, \"I ~ 0.6 s, 1 ~ 5.0 s. \"V..- = .0.,-4 + 0....6.. = 0.13 1. Description: Vernier callipers and screw gauge are used to \"Vv 40.0 5.0 measure the length of objects, thickness of wires, diameter of =} = 0.13 x 8.0 ~ 1.04 cylinders, etc. upto an accuracy of 1/1 oth or 1I100th of a mm. Hence, v = (8.0 :I: [,04) ms\" A Vernier Callipers consists of a main scale M graduated in c. ~Percentage error (~)v x 100) = 0.13 x I00 ~ 13%. cm and mm over which is an auxiliary scale (or vernier scale) V which can slide along the length of main scale M. The divisions Accuracy and Precision of the vernier scale being either slightly longer or slightly smaller than the divisions of the main scale. Accuracy tells us how close is the measured value to the true value. Precision indicates from which instrument, the measure- The main scale has two fixed jaws A and C as shown in Fig. 3.2 ment is taken. while B amI D are the jaws of vernier scale. The position of vernier scale is fixed with the hclp of screw S. In general, thc h---{COllcept Application Exercise 3.2 vernier scale has 10 divisions over a length of 9111111. 1. Which of the following length measurement is most pre- cise and why? a. 2.0 em b. 2.00 CI11 C. 2.000 em 2. In a number without decimal, what is the significance of ! zeros on the right of non-zero digits? i

3.10 Physics for IIT-JEE: Mechanics I We can measure the external diameter of an object, internal In general, if m vernier scale divisions are equal to (11. - 1) diameter of a hollow object or depth of some vessel using vernier main scale divisions, i.c., m = 11 - 1, then we have callipers. Diameter (external) of an object can be determined by placing the object in between the jaws A and B while internal Verm.er constant = least count= x ( I - ~11;1) diameter can be measured by inserting C and D in the object. (1 _ 1= 11_~~) x = MSD Fixed Movable nn If the pth vernier division coincides with anyone of the main Upper Upper scale division, then law Jaw Projecting Fraction to be added = £. J to d n Hence, the measurcd value oflength: L = complete main scale L..-'L..-,__-,='='=C7=___--,rO reading before zero of the VS. mark + £.. ~~~~~~~~~~~~~~14~p~~~Q 11 r ~B Fixed Movable 3. Zero error: In a correctly adjusted instrument, the zero oft.he Lower Lower vernier coincides with the zero of the main scale when the two Jaw jaws A and B are brought in contact. Ifit is not so, the instrument' Jaw has a zero error which may be positive or negative according as the zero of vernier scale lies to the right or left of the zero of the main scale when its two jaws are brought in close contact: with each other. Fig. 3.2 Screw Gauge: Its Use to Measure Thickness/Diameter of Thin Sheet or Wire While performing the experiment using a vernier callipers first of all the jaws of vernier scale, i.e., A and B touch while 1. Description: It works on the principle of micrometer screw straight edges of C and [) touch. If the instrument is free from and its diagram showing the constant is shown in the Fig. 3.3. zero error, then zero of main scale coincides with the zero of H is the linear scale or pitch scalc while E is the cap scale, vernier scale. A hollow cylindrical cap k is capable to rotate over H (hub or linear scale) when the screw is rotated. When zero of pitch After noticing whether the instrument is having zero error or scale coincides with zero of cap scale, then the instrument is not, holll the ubject whose external diameter is to be computed free from zero error. betweenjaws A and B as shown in the figure. Now, the most im- The wire whose diameter has to be measured is held portant task'is to measure the diameter, i.e., to take the readings between A and B. of vernier caliper. Reading of screw gauge is, To measure the depth of the vessel, a metallic strip E is con- the measured diameter = N + n x Least count nected to the back of M and vernier scale. When jaws A and B touch each other, the edge of E touches the edge of M. When where N is the division of linear scale beyond which the edge the vernier is separated from M, E moves outwards. of the cap lies, and n is the division of circular scale which 2. Vernier constant: The difference between one main scale lies over reference linc. If some zero error is there, subtract division and one vernier division is called the vernier constant it ti·om the above reading. or the least count of the vernier because it is the smallest length that can be measured accurately with its help. When the zero of circular scale advances beyond the reference line, the zero error is negative and if it is left behind To find the vernier constant (or least count): the reference line, the ?ero error is positive. In positive zero error, zero of the linear scale is not hidden from the circular a. Find the magnitude of the smallest division of the main scale. scale, while in negative zero error, zero of the\" linear scale is hidden from circular scale, h. Count the total number of divisions on the vernier scale. c. Slide the movable jaw that the zero mark (the first division) Moveable Stud Attached Circular Scale Ratchet Screw to Fine, Uniform Screw or Head Scale (to Avoid of the vernier scale coincides with any of the main scale Undue divisions. Reference (E) Pressure) d. Find the number of scale divisions which coincide with the Fixed-stud total number of vernier division. If m main scale divisions (MSD) are equal to n vernier scale divisions (VSD), then we can write m.x = ny, where.x is the length of 1 MSD and y is the length Jaw to Main Scale in Millimeter, Hold or Pitch Scale (lI) of 1 VSD. . = I VSD = x - y = x - m Object Now, Vermer -x U-shaped Frame of constant I MSD - n Stainless-steel =x (1-I;n) Fig. 3.3

Units and Dimensions 3.11 2. Pitch of the screw gauge: The pitch of screw is defined Cross~section of a Wire as the distance through which the screw moves forward or backward parallel to its axis when one complete rotation is R given to the circular cap, To find pitch: a. Rotate the circular scale H and coincide the zero mark Fig. 3.4 with the reference line. Notice the reading on the pitch scale. ,----~nc:ePtApplicationExertise 3.3 ll--..., b. Give four complete rotations to the circuJar scale and note the reading on the pitch scale again. 1. Each division on the main scale is 1 mIn. Out of the follow- c. Calculate the pitch of the screw by dividing the distance ing which of the vernier scales will give vernier constant moved by the number of rotations. equal to 0.1 mm? distance travellcd on the pitch scale a. 90 mm divided into 100 divisions Pitch of the screw = - - - b. 90 mm divided into lO divisions number of rotations e. 90 mm divided into 1000 divisions d. None of these Note: The pitch of a screw gauge is usually OS mm or lmm. 2. A vernier callipers has 20 divisions on the vernier scale 3. Least couut of the screw gauge: The least count of the which coincide with 19 on the main scale. The least count screw gauge is defined as the distance through which the screw moves backward or forward when the cap is rotated of the instrument is 0.1 mm. The main scale divisions are through on the circular scale. Note the number of divisions on the circular scale H. Then, of: pitch of the screw a. 0.5 mm b. I mm Least. count = 1 c.2mm no. of rotations on the circular scale d. --mm 4 4. Zero correction: In some instruments, when the jaws A and B are brought in contact without applying any undue pres- 3. The length of one solid thin cylinder is 2.25 em and that sure, the zero of the circular scale does not coincide with the of another cylinder is 1.31 em. The vernier constant of reference line, In some instruments, zero mark goes beyond the reference line, while in others it is left behind when the calliper is O.O! cm. If the two cylinders are put together two studs A and B are in contact with each other without undue pressure. end to end, the combined length will be expressed as: To find zero correction, count the number of divisions . a. (3.56 ± 0.(05) em on the circular scale that the zero mark has advanced beyond or is left behind the reference line. Multiply this number with b. (3.56 ± 0.0(1) cm the least count. This is the zero correction, If the zero of the circular scale has advanced beyond the reference line, the e. (3.56 ± 0.002) em zero correction is positive but if it is left behind the reference line, it is negative. d. (3.56 ± 0.10) em Remember zero above, add and zero below, subtract. 4. A screw gauge has 1.0 mm pitch and 200 divisions on the 5. Backlash error: With constant usc a little play always creeps circular scale.\"What is the least count of the instrument? in between the nut and the screw, due to which the screw does a. 5 x lO-3 mm not move forward or backward for a little distance when the b. 4 x 10.3 mm head is rotated. This error is called the backlash error, To c. 6 x 10-.3 mm avoid this, the screw is well greased and is always turned in the same direction while taking an observation. d. 2 x lO-2 mm 6. To read a screw gauge: A famous relation in physics relates mov- a. Find the zero correction. ing mass m to the rest mass Ino of a particle in terms of its speed v and the speed of light c. (This relation first arose as b. Find the least count. c. Insert the wire/sheet between the jaws A and B, Rotate screw R till it is held tightly without any pressure. Note the reading of main scale and circular scale. Then Diameter/thickness d = main scale reading in millimeter + circular scale reading x least count (Apply the zero correction)

3.12 Physics for IIT·JEE: Mechanics I a consequence of special theory of relativity due to Albert The wavelength associated with a moving Einstein.) A boy recalls the relation almost correctly but for- particle depends upon pth power of it') mass Ill, qth power of its velocity v and r'h power of Planck's constant h. Then, the . = mo correct sct of values of p, q, and r is (1 _ V 2)1/2 gets where to put the constant e. He writes, m a. p =1, q = -1, r =1 Guess where to put the missing c? b. P =1, q =1, r=1 Sol. According to the principle of homogeneity of dimensions, c. p=-1,q=-I,r=-1 powers of M, L, T on either side of the formula must be d. p = -l,q = -l,r =1 equal. For this, on R.B.S., the denominator (1 - V2)'/2 should Sol. d. Given).. = km\"vq fl. The dimensions of right hand side be dimensionless. Here, v2 is not dimensionless, So, there is something missing. Therefore, instead of (l - V 2)1/2 we should and left hand side terms should be equal. write (l- V2 /C2 )1/2, Hence, the correct formula would be m =(1---v-m2o-je-2),'/-2 So [MOL TO] = [Ml\"[Lr'Y'IMI}Y--'Y or IMo LTo] = IMP+\"]IL\"+2\"][T-Q-'] The velocity ofa body is given by the equa- tion v = ~ + el' + d1 3. Now, comparing powers of M, Land T, we get I The dimensional formula of b is +p I' = 0 (i), a. [MuLTo] b. [MLoT u] q + 21' = 1 (ii), - q - I' = 0 (iii) c. IMuLoT] d. [MLT- 1] After solving, p = -1, q = -1 and I' = l. Putting these values, we get Sol. a. In the equation, left hand side has the dimension of h velocity, Thus, from the principle of homogeneity each term A = k -\"---, which is' the required relation. in the right hand side should have the dimensions of velocity. mv [b] \" ','1- = Iv] or Ib] = IVI] = [LI If P represents radiation pressure, c rep~ The force F is given in terms of'timc t and resents speed of ligbtand Q rcpresents radiation striking unit area per second, then non-zero integers x, y, and z such displacement x by the equation that px QYCZ is dimensionless are F = A cos Bx + C sin DI. The dimensional formula of DIB a. x=l,y=l,z=-l is b. [MOl\"UT- 1] b. x = t,y = -t,z = 1 a. [MoLoTo] c. x = -t,y = 1,z = 1 c. [MuL -1 TO] d. [MoL1T-1] d. x = 1, y = 1, z = I Sol. d. In the given equation Bx and DI should be dimensionless. Sol. b. Dimensions of [~] [~:][HI = [x 'I and [DI = IT\"] co} = = Iveloeity] px QJe' = [M L -'T-'I'[MT 'V[IoT\" J'. The velocity of a body, which has fallen As it is dimensionless, so freely under gravity varies as gP h q , where g is the accelera~ IML \"'T\"2YIMT-'l''ILT-'1' = [MoLoTo] or tion due to gravity and h is the heigbt througb which it has fallen. The value of p and q are [MX'bL x \"T- 2x -\"\"\"1 = IMoLoTo] a. 11 b. 1 1 Comparing powers of M, Land T, we get 2 -2'2 2' x + y =0, -x + z =0, - 2x - 3Y - z =0 I1 d. 1 1 Solving, x = 1, y = - 1, z = 1. 2 c. 2'2 -2' Sol. c. Here, we are given v = kg Phq • Dimensions of left hand A physical quantity x is calculated from side and right hand side terms should be same. a 2 b3 .'. [MoLT\"'] = [L1\"2]\"[L]q = [U\"!''/T- 2p] the relation x = fJ' If percentage error ina, b, c and dare Comparing powers of Land T, l' + q = 1 and -21' = -1 c\",d 2%,1 %,3%, and 4%, respectively, what is the percentage Solving them, we get)J = 112 and q = Ij2 error in x'? Putting these values, we get v = kvlih, which is the required rclation.

Units and Dimensions 3.13 J-llx x 100=± [2lla- · - +llb3 - +llc- +1-ll-d BA x a b c 2d If X = and AX, AA and AB are the ~=±[2 x 2%+3 x 1%+3%+ x 4%J =±12% maximum absolute errors in X, A and B, respectively, then the maximum fractional error in X is given by a. AX = AA+AB b. AX = AA- AB The length and breadth of a field are mea- c. -A\"X-- = -A-A-- - AB sured as: I = (120 ± 2) m and b = (100 ± 5) m, respectively. ---~ X AB What is the area of the field? AX AA AB Sol. Here, ~A.:\". = ~I + ~b = ( I~o + I~o) = 0.0667 d. X=A\"+'S llA = 0.0667 x A Sol. d. When two quantities are divided, their maximum frac- Now, A = Ib = 120 x 100 = 12000 m2 tional or relative errors are added up. =} llA = 0.0667 x 12000 = 800A m2 Area of the field = A ± llA = 12000 ± 800.4 Hence: L';X = llA + -llB - - = (1.2 ± 0.08) x 104 m2 • X AB In an experiment of simple pendulum, A physical quantity is represented by time period measured was 50 s for 25 vibrations when the X = M aL b T-c. If percentage errors in the measurement of length nfthe simple pendulum was taken 100 cm~ !fthe least M, Land T are a, f3 and y, respectively, then total percent- count of stop watch is 0.1 s and that of meter scale is 0.1 cm, age error is calculate the maximum possible error in the measurement of value of g. a. (aa + f3b - ye) b. (aa + f3b + yc) c. (aa - f3b - yc) d. zero Sol. b. llx x llM x 100 + llL x 100 --- 100= (1-- b--· .. x ML Sol. The time period of a simple pendulum is given by Ii '-gT 4;r2[ 4;r21 llT l' = 2;r j/ or 1'2 = org = 1'2' + cT x 100=aa+bf3+cy As 4 and Jr are constants, maximum permissible error in g is Given: potential difference, . llg ll[ + -2ll-T V and current, I = (2 ± 0.2) A. The value of gIven by -- = -I l' resistance R in Q is ~g Here, III = 0.1 em, I = 1m = 100em, ll1' = 0.1 s, l' = 50 s. a. 4 ± 16.25% h.4±6.25% llg 0.1 (0.1) 0.1 (02.51) c. 4± 11)% d.4±8% g=T(xj+2 =106+ 50 llgO.IJor = =Sol. a. We know that R VII = 8/2 4 Q. - x 100 = [ T0O.1O + 25 x 100 = 0.1 + 0.4 = 0.5%. Percentage error in V is \"08.5 x 100 = 6.25. g =If X A x B and AX, AA and AB are Percentage error in I is \"0.22 x 100 = 10. maximum absolute errors in X, A and B, respectively, then the maximum relative error in X is given by Now, add up the percentage errors and get the answer. a. Ax = AA + AB h. AX = AA - AB =Given: resistance, Rl (8 ± 0.4) Q and =resistance, Rz (8 ± 0.6) Q. What is the net resistance when AX = 'AAA -BAB RJ and Rz are connected in series'! c,.X d. -A=X . _AA... +A-B a. (16 ± 0.4) Q b. (16 ± 0.6) Q X AB c. (16 ± 1.0) Q d. (16 ± 0.2) Q Sol. d. When two quantities are multiplied, their maximum rel- Sol. c. When resistances are connected in series, then net resis- ative errors rlre added up. tance: R = RI + R2 llX llA llB R = (8 + 8 ± 0.4 ± 0.6) Q = (16 ± 1.0) Q. Hence: X ='11 + B'

3.14 Physics for IlT-JEE: Mechanics I The focal length I of a mirror is given and 47th circular division coincides with the main scale. Find the curved surface area of the wire in cm2 to appropriate -I1 -1 + ~1,where and represent obj.ect and'Image 2:.)significant figure. (Usc JC = by = u u v (IIT-JEE,2004) v distances, respectively. a. AI Au Av Sol. Least count = II~; = 0,01 mm, I~=-+- uv Al =Avu +Avv Diameter =MSR + CSR =I mm + 47 (0,01) mm =1.47 mm, b. I 22 1.47 x S6 mm' = 2.58724 em2 , - Surface area = JC Dl = 7 x c. AI = -Au + -Av - -A-( u' -+- 'v-) = 2,6 em2 -I uv u+v d. -AI-I = -Au + -A-v + -A-u +. --A_.v_- In a Searle's experiment, the diameter of u v u+v u+v the wire as measured by a screw gauge ofleast count 0.001 em Sol.d. f= -u-v , -1f',1 = I',u +i\\-v +,I-',=(u=+'-v'-) is 0.050 cm. The length, measured by a scale of least count - , u+v u v u+v 0.1 em, is 110.0 em. When a weight of SO N is suspended =I-', u+ -I',+v - -I'-,u, , +I-', v- from tbe wire, the extension is measured to be 0.125 em by a u v u+v u+v micrometer ofleast count 0.001 cm. Find the maximum error in the measurement of Young's modulus of the material of Tbe following observations were taken the wire from these data. (IlT-JEE,2004) determinifi'g surface tension of water by capillary tube Sol. Maximum percentage error in Y is given by =metbod: Diameter of capillary, D 1.25 X 10-2 m and WL =rise of water in capillary, h 1.45 X 10-2 m. Taking Y=--x- = =g 9.80 I11S-2 and using the relation T (rghj2) x 103 IT lJ2 X Nm- l , what is the possible error in measurement of surface 4 tension T'! a.2.4% b.15% -I',Y) (I-',-D) +I',-x +I',-L Y\" D c. 1.6% d.O.15% ( -2 XL max Sol. c. Percentage error in T, = 2 (0,001) + (~.:.OO~) + (~) I', I' -/',%r +-I'%,g +-i-\\h%= -0,0-1 x 0,01 0,05 0,125 110 -%= 9,80 r gh L25 100 + = 0,0489, I' x 100 + -0,01 x lOa So, maximum percentage error = 4.89% 1.45 =0,8+0,1+0,7= 1.6% The side of a cube is measured by vernier B-..m If force, velocity and time are taken as callipers (10 divisions of a vernier scale coincide with 9 divi- fundamental quantities, find the dimensions of work. sions of main scale,'where 1 division of main scale is .1 mm). a. FVT b. FVT' The main scale reads 10 mm and first division of vernier scule coincides with the main scale. Mass of the cube is 2.736 g. c. FOVT-' d. FV 2T-' Find the density of the cube in appropriate significant Sol. a. W = M L2 T-' = [F1\"[V]\"T' figures. (IlT-JEE, 2005) =} ML2T- 2 = [MLT-'j\" [Lr-']bT\" Sol. Least count. of vernier calliper =} a = I, a + b = 2 =} b = I and c - 2a - b =-2 I division of 1~~~2: scale = ~ = 0,1 mm =} c=LSo.lWj=FVT, number of divisions in vernier scale 10 The side of cube = lO mm + I x 0.1 mm = 1.01 em. A screw gauge having 100 equal divisions mass 2,736 g , a pitch of length 1 mm is used to measure the diameter Now, densIty = --~ = 1 1 = 2.66gcm-· of a wire of length 5.6 cm. The main scale reading is 1 mm volume (LO 1)- em' (to correct significant figures)

Units and Dimensions 3.15 EXERCISES Solutions on page 3.26 c. Fractional error. 1. Write the number of significant figures in the following d. Percentage error. a. 0.053 b. 50.00 e. Express the result in terms of absolute eITor and per- c. 0.0500 d. 5.7x ]0\" centage error. f. 2400 e.5.70x]06 h. 0.0305090 12. a. Two plates have lengths measured as (1.9±0.3) m g. 2400 kg and (3.S±0.2) m. Calculate their combined length 2. Round off to two significant figures with error limits. h. The initial and Hna] temperatures of a liquid are mea- a. 0.05857 b. 0.05837 sured to be 67.7±(1.2°C and 76.3±0.3\"c. Calculate c. 5.07 x 106 d. 5.0] x ]0\" the rise in temperature with error limits. 3. With due regard of significant figures, add the following 13. The sides ofareetangle are (l 0.5 ±0.2) em and (5.2 ±0.1) cm. Calculate its perimeter with error limits. a. 953 and 0.324 b. 953 and 0.625 c. 953.0 and 0.324 d. 953.0 and 0.374 14. The length and breadth of a rectangle arc 5.7±O.1 cm and 4. With due regard of significant figures, subtract 3.4±0.2 em. Calculate area of the rectangle with error limits. a. 0.35 from 7 b. 0.65 from 7 15. A body travels uniformly a distance of 13.8 ± 0.2 m in a c. 0.35 from 7.0 d. 0.65 from 7.0 time 4.0 ± 0.3 s. Calculate its velocity with error limits. What is the percentage error in velocity? 5. A diamond weighs 3.71 g. It is put into a box weighing 1.4 kg. Find the total weight of the box and diamond to 16. The radius of a sphere is measured to be 2.1 ± 0.5 em. correct number of significant figures. Calculate its surface area with error limits. 6. a. Calculate the area enclosed by a circle of radius 17. Calculate the percentage error in specific resistance, 0.56 m to correct number of significant figures. p = Jrr2R/I, where r = radius of wire = 0.26 ± 0.02 h. Calculate the area enclosed by a circle of diameter cm, I = length of wire = 156.0 ± 0.1 em, R = resistance 1.12 m to correct number of significant figures. H.of wire = 64 ± 2 ohm. 7. a. Add 3.8 x 10 6 to 4.2 X 10-5 with due regard to 18. Time period of a pendulum is given by T = 2Jf significant figures. Length of pendulum = 20 cm and is measured upto 1 mm accuracy. Time period is about 0.6 s. The time of 100 b. Subtract 3.2 x 10-6 from 4.7 x 10-4 with due regard oscillations is measured with a watch of 1110 s resolution. to significant figures. What is the accuracy in the determination of g? c. Subtract 1.5 x lOJ from 4.8 x lO4 with due regard Solutions on page 3.27 to significant figures, 1. Which oflhe following is not measured in units of energy? 8. The length, breadth, and thickncss of a metal sheet are 4.234 m, 1.005 m, and 2.0 I em, respectively. Give the arca a. Couple x angle turned through b. Moment of inertia x (angular velocity)2 and volume of the sheet to correct number of signiticant figures. c. Force x distance d. Impulse x time 9. The diameter of a sphere is 3.34 m. Calculate its volume 2. The unit of surface tension may be expressed as with due regard to significant figures, a. joule metre b. newton metre 10. Solve with due regard to significant figures: c. joule mctre- 2 d. newton metre-2 5.42 x 0.6753 0.085 3. The equation of the stationary wave is 11. In an experiment, refractive index of glass was observed Y = 2A sin (-2)Jr.c-t) cos (2TJrX) to be 1.45, I.S6, 1.54, 1.44, 1.54, and 1.53. Calculate Which of the following statements is wrong? a. Mcan value of refractive index. a. The unit of ct is same as that of A. h. Mean absolute error. b. The unit of x is same as that of A.

3.16 Physics for IIT-JEE: Mechanics I c. The unit of 2nc/A is same as that of 2nxlAt. 13. Tn the reiation: ely = 2wsin(<vt + <Po), the dimensional +elt d. The unit of clJ... is same as that oLr/A. formula for (wI <Po) is 4. (2;)Given that: y = A sin [ (et - X»), where y and x a, MLT b. MLTo arc measured in the unit of length, Which of the following c. MLoT\" d. MOLoT\" statements is true? 14. A physical quantity depends upon five factors, all of which a. The unit of A is same as that of x and A, have dimensions. Then, method of dimensional analysis h. The unit of A is same as that of x but not of A. c. The unit of c is same as that. of 2][/)\". a. can be applied. b. cannot be applied. d. The unit of (et - x) is same as that of 2\"IA. c. depends upon factors involved. S. The dimensional representation of Planck's constant is d. both a. and c. identical to that of a. torque 15. A student when discussing the properties of a medium b. work (except vacuum) writes: Velocity of light in vacuum::: velocity of light in medium C. stress This formula is a. dimensionally correct d. angular momentum b. dimeii.sionally incorrect 6. The dimensional representation of latent heat is identical to that of c. numerically incorrect a. internal energy d. both (a) and (e) h. angular momentum c. gravitational potential 16. Given that T stands for lime period and I stands for the d. electric potential length of simple pendulum. If g is the acceleration due to gravity, then which of the following statements about the 7. The dimensions of shear modulus of rigidit.y are relation 1'2 :::: (1/g) is correct? a. M'L'T-2 b. M'L'T-' u. It is correct both dimensionally as well as numeri- c. M L'T2 d. M L\"-' T- 2 cally. b. It is correct neither dimensionally nor numerical!y. 8. Out of the following, the only pair that docs not have identical dimensions is c. It is correct dimensionally but not numerically. a. angular momentum and Planck's constant b. moment of inertia and moment of a force d. It is correct numerically but not dimensionally. c. work and torque 17. Suppose refractive index ')' is given as d. impulse and momentum Ii Il=i\\+--~ A' 9. The dimensions of self-induction are where A and 13 arc constants and A is wavelength, then a. MLT-~'A-2 b. ML2T-'k' dimensions of B arc same as that of d. ML 2T· 2A-' c. ML'r'A- 2 a. wavelength b. volume c. pressure d. area 10. The dimensional formula for latent heat is 18. A physical quantity x depends on quantit.ies y and z as fol- lows: x :::: Ay + 13 tan(C z), where A, iJ and C arc constants. a. MOL 2 T-- 2 b. Mcr-2 \\Vhich of the following do not have the same dimensions? c. M I/T~2 d. ML2T-' 11. Which of the following have same dimension? a. x and B b. C andz·~1 a. Thermal capacity c. yandBIA d. x and A b. Universal gas constant c. Boltzmann's constant 19. If I, and R denote inductance and resistance, respectively, d. All of the above then the dimensions of LIRare a. M'LoT\"Q' b. M°l.\"TQ\" 12. Dimensional representation of coefficient of friction is a. [ML2r21 b. IMLT~21 20. The best method of reducing random error is a. to changc the instrument used for measurement c. [Mo LOTol d. [MLT'] b. to take help of experienced observer

Units and Dimensions 3.17 c. to repeat the experiment many times and to take the 30. The etfective length of a simple pendulum'is the sum of average results the following three: Length of string, radius of boh, length of hook. d. none of the above In a simple pendulum experiment, the length of the string as measured by a metre scale is 92,0 em, The radius of the 21. A length is measured as 7.60 rn, This is the same as a. 7600mm b. 0.0076 mm bob combined· with the length of the hook as measured by a vernier callipers is 2.15 cm. The effective length of the c. 760 em d. 0.76 dm 22. Force F is given in terms of time t and distance x by pendulum is F = A sin Ct + B cos Dx. Then, dimensions of AlB and a. 92.0 ern b. 94.2 em C/Dare c. 94,15 em d. 94 em a. [MoL\"T°], [M\"L\"T\"'J (fr31. The frequency (n) of vibration of a string is given as b. [MLT'2], [MoL-'To] 11 = -1 ---, where l' is tension and 1 is the length of 21 III c. [M\"LoT°J, [MoLr'l vibrating string. Then, the dimensional formula for m is d. [M\"L'T-'], [MoLoTo] a. [MoL'T'] b. [MoLliTol c. [M' L~\"To] d. [MLoToJ 23. Which of the following pairs have identical dimensions? a. Momentum and force 32. In the relation y = rsin(J)f - kx), the dimensions of (vlk b. Pressure and surface tension c. Moment of force and angular momentum are d. Surface tension and surface energy a. [Mo LOTo] b. [MoL'T'1 c. [MoLoT'1 d. [MoL''!'o] 24. The dimensional formula of resistivity of a conductor is 33. Dimensions of 80/1.-0 are a. [M L 2T-- 2A-2] b.IML3 T- 3 A--21 a. [LT-'J b. [L1'-- 2 ] c. [L2T2] d. IL,2T'] c. [ML- 2 T- 2 A2 1 d. [M L21'--2 A' 'I 25. The dimensjonal formula for electric potential is: 34. Which of the following qu'antities has its unit as newton- second? a.IML'T--3 A--'1 b. [MLT- 3 A--'] a. Energy c. [ML'T-- 3K\"] d. none of these b. Torque 26. Which of the following pairs do not have identical eli men- c. Momentum siems? a. Pressure and stress d. Angular momentum h. Work and pressure energy c. Angular momentum and Planck's constant 35. If frequency (F), velocity (V) and density (lJ) are con- d. Moment of force and momentum sidered as fundamental units, the dimensional formula for momentum will be a. DV 1'2 b. DV2p' 27. The product (PV) has the dimensions c. lJ'V' F2 a. [M J.'-'T 2 ] b. [M'L 2T--'J 36. lfforee (F), acceleration (A) and time (1') be taken as the c. IM'L2T2] d. [M'L 2 T-- 3 ] fundamental physical tluantities, the dimensions of length on this s.ystem of units are 28. If the error in the measurement of momentum of a particle a. FAT' h. FAT c. FT d. AT' is + 100%, then the error in the measurement of kinetic 37. If the percentage errors of A, Band C arc a, hand c, energy is a.25% b.200% respectively, then the total percentage error in the product c. 300% d.400% ABC is 29. Choose the physical quantity that is different from others a. abc b.a+b+c in some respect. a. Moment of ineliia c . -1+ -1+ - -1 d. ab + be + ca h. Electric current c. Pressure energy abc d. Rate of change of velocity 38. Which of the following numbers has least number of sig- nificant figures? a. 0.80760 b. 0.80200 c. 0.08076 d. 80.267

3.18 Physics for IIT·JEE: Mechanics I 39. The dimensional formula for magnetizing lield H is 50. The time dep,endence of a physical quantity P is given by a. 1M\"C\" TO A] b. [Mo LT-' A] P = Poe--cw , where CI is a constant and t is time. Then, c. [M\"LTA-'] d. [MoL'T-'A] constant Q' is 40. The dimensions of intensity of a wave arc a. dimensionless b. has dimensions of T -2 a. IML2['3] b. [ML\"r- 3 j c. has dimensions of P d. has dimensions of y2 c. [M L-'T'] d. .[M'!}r'] 41. The dimensions of formula of capacitance is 51. Which of the following have got same dimensions? a. IM-' L -2TA2] b. [M-'L- 2T 3;\\2] 1. Latent heat 2. Energy per uni t mass c. [M-'C,2r 4 A2] d. [M\"L- 2T 2 A'] 3. Gravitational potential 4. Specific heaL 42. What are the dimensions of Gas constant? a.IMLT\"K-'] b. [MoCr- 2 K-'] a. 4 and 3 b. I and 4 c.IML2T,2K-'mo!\"] d. [MllL'T'K'] c. I,2and3 d. None of these. 43. If the order of magnitude of 499 is 2, then order of mag· 52. Of the following quantities which one has the dimensions nitude of 50 I will be different from the remaining three? L4 ~2 ~l d.3 I. Energy density 2. Force pcr unit area 44. The order of magnitude of 0.0070 I is 3. Product of charge per unit volume and voltage 4. Angular momentum per unit mass. a. -2 b. -1 c. 2 d. I a.1 b.2 c.3 d.4 45. The order of magnitude of 379 is 53. The frequency f of vibrations of a mass Tn. suspended a. I b. 2 c. 3 d.4 from a spring of spring constant k is given by f = Cnr' kY, 46. If X = a + b, the maximum percentage elTor in the mea- where C is a dimensionless constant. The values of x and yare, respectively surement of X will be b. a. (;';a + ll.~) x 100% 22 ah 1I h-b. c,~ a~)b) x 100% d. -2' 2 c. ( ;';a, + ~) x 100% 54. If C, the velocity of light, g the acceleration due to gravity and P the atmospheric pressure he the fundamental quan- a+b a+b tities in M.K.S. system, then the dirnensions of length will be same as that of d. (;';a x ;';b) x 100% a. Cig ab 47. Which of the following is the most precise instrument for c. PCg measuring length? 55. The quantities A and B are related by the relation a. Metre rod of least count 0.1 cm. AlB = 111, where III is the linear density and A is force. b. Vernier callipers of least count 0.01 cm. The dimensions of B will be c. Screw gauge of least count 0.00 J em. a. same as that of pressure d. Data is not sufficient to decide. 48. The numher of significant figures in 5.69 x 1015 kg is h. same as that of work a. I b. 2 c. 3 d. 18 c. same as that of momentum 49. The position x of a particle at time t is given by d. same as that of latent heat x = Va (l - e-al ), where Va is constant and a > O. The I. a dimensions of Vo and a are 56. The dimensions of 2\"EoE2 (EO = permittivity of free space a. MOLT-' and ]'-, b. MOLToandT\" and E = electric field) are c. M OLT-\"andLT-2 d. MOLT;' and T a. 1M L2T-'1 b. [MC'T'] d. [MLT-'I c. [M L 2r,21 57. A physical quantity X is representee! by X = (MY c-y Y'<:'). The maximum percentage errors in the measure- ment of M, Land T, respectively, arc a%, b%, and c%. The maximum percentage error in the measurement of X will be

. Units and Dimensions 3.19 a. (ax + by - cz)% 65. The length I, brcadth b and thickness t of a block of wood b. (ax - by - cz)% c. (ax + by + cz)% are are measured with the help of a meter scalc. The results d. (ax - by + cz)% after calculating the errors arc given as 'If;t;.;,58. The velocity of transverse wave in a string is v = where T is the tension in the string and m is mass per 1 = 15.12 ± 0.01 em, b = 10.15 ± 0.01 cm unit length. If T= 3.0 kgf, mass of string is 2.5 g and t = 5.28 ± 0.01 cm. Thc percentage error in volume upto length of string is 1.000 m, then the percentage error in the measurement of velocity is proper significant' figures is a. 0.36% b. 0.28% c. 0.48% d.O.64% 66. The number of particles crossing a unit area pcrpen- a.0.5 b.0.7 c. 2.3 d. 3.6 dicular to the X-axis in a unit time is given by n a - [2 _D(n -n= 2 1),_Whcrc nl and n2 are the number of 59. Write the dimensions of alb in the.relation P = ~' X2 - Xl where P is the pressure, x is the distance and t is the time. particles per unit volume at x = XI and x = X2. respec- tively, and D is the diffusion constant. The dimensions of Dare b, MLoT-' a. (MoLT- 2] b. (Mo I}T-4 ] d. MLT-2 c. MLoT' c. (MoL 2T- 2 ] d. (M oL 2 T- I ] b -x' 67. If E, M, J and G, respectively, denote energy, mass, 60. Writethedimensionsofax bintherelationE = - - - , angular momentum and gravitational constant, then at where E is the energy, x is the displacement and t is time. 2 ME5JG2 has the dI' menSl.OllS 0f' a. ML2T b. M- IL 2T I a. time b. angle d. MLT-2 C. mass d. length 61. If the velocity of light C, the universal gravitational con- 68. If L, R, C and V, respectively, represent inductance, re- stant G and Planck's constant h be chosen as fundamental units, the dimensions of mass in this system arc sistance, capacitance and potential difference, then the L a. hl/2CI/'G-I/2 dimensions of RCV are the same as that of C, hCG- I a. Charge b. 1/Charge c. Current d. llCurrent 4 69. The moment of inertia of a body rotating about a given 62, If the relation V = \": Pr , where the letters have their axis is 12.0 kgm2 in the S.l. system. What is the value of 8 nl the moment of inertia in a system of units in which the usual meanings, the dimensions of V are units oflength is 5 cm and the unit of mass is 10 g? a. MOL'To b, MOL'T-I 63. The length I, breadth b and thickness t of a block of wood a. 2.4 x 103 b. 6.0 X 103 c. 5.4 x 10' d. 4.8 X 105 were measured with the help of a measuring scale. The 70. If the velocity (V), accelcration (A) and force (F) are results with permissible errors are: 1 = 15.12 ± 0.01 em, taken as fundamental quantities instead of mass (M), length (L) and time (T), the dimensions of Young's mod- = =b 10.15 ± 0.01 cm and t 5.28 ± 0.01 cm. ulus (Y) would be The percentage error in volume upto proper significant figures is a. 0.28% b. 0.36% a. FA'V-4 b. FA 2 V'- S c, 0.48% d.0.64% d. FA'V·-2 64. The relative density of the material of a body is found 71. A gas bubble formed from an explosion under water os- cillates with a period proportional to padb E e, where P is by weighing it first in air and then in water. If the weight the static pressure, d is the density of water and E is the energy of explosion. Then, a, b, c are, respectively of the body in air is WI = 8.00 ± 0.05 N and weight in water is W2 = 6.00 ± 0.05 N, then the relative density, a. 1, 1, I I I -5 PI\" = ~ W'Ith the maX'Imum permlS. Sl'ble error I.S b. 3' 2' 6 WI- W, a. 4.00 ± 0.62% b. 4.00 ± 0.82% -5 1 I 1 -5 I - c. 4.00 ± 3.2% d. 4.00 ± 5.62% c. - '2' 3\" d. 2' 6 '3 6

3.20 Physics for IIT-JEE: Mechanics I 72. <The percentage error in the measurement of mass and 80. A quantity x is given by soL Ll V where 80 is p~lmittivity speed arc 2% and 3%. respectively. How much will be t;1 the maximum error in the estimation of K.E, obtained by of free space, L is length, Ll V is potential difference and measuring mass and speed? Llt is the time interval. The dimensional formula for x is u.5% b.l% c.8% d. 11% the same as that of a. resistance b. charge 73. An experiment measures quantities a, h, c and then X is c. voltage d. current a l / 2bz 81. Given that Y = a sin w! + ht + ct2cos wt, The unit of abc calculated from X = --, . If the percentage errors in c- is same as that of a, band care ± 1%, ±3%, and ±2%, respectively, then a. y b. yll c. (yltf d. (yll)3 the percentage error in X can be a. 12.5% b.7% c.1% d.4% 82. The potential energy of a particle varies with distance Ax'/2 74. The resistance of a metal is given by R = lV' where V x as U = -x-,-+-B, where A and B are constants, The is potential difference and I is the current, In a circuit dimensional formula for A x B is =the potential difference across resistance is V (8 ± 0.5) a. M'L'I'I'- 2 b. M'L\"/2 T-- 2 V and clIrrcntin resistance, I = (4 ± 0.2)A. What is the d. M'L 9/2 r· 2 value of resistance with its percentage error? a. (2 ± 5.6%) Q b. (2 ± 0.7%) Q 83. If x and a stand for distance, then for what value of n is (~. (2 ± 35%) Q d. (2 ± 11.25%) Q the given equation dimensionally correct? The equation 75. Which of the following product of e, h, fL, G (where fL is Jis dx = sin-oj x- the permeability) be taken so that the dimensions of the -Ja2-xlI a product are same as that of speed of light? a. Zero b. 2 c.. -2 d. I a. 1/e~2fL~'Go b. h2eGo/L 84. The specific resistance p of a circular wire of radius r, rr,.2R c. hOe2G~' fL d. hGe~2fLo resistance R and length I is given by p = -l~' Given: 76. Which of the following does not have the dimensions of velocity? (Given, So = permittivity of free space; I' = 0.24 ± 0.02 em, R = 30 ± I Q, and I = 4.80 ± fLO = permittivity of free space; v = frequency; A = wave- length; P = pressure; p = density; k= wave number; and 0.01 em, The percentage error in p is nearly w = angular frequency.) a. 7% b. 9% c. 13% d. 20% a. wk b. VA 85. Using mass (M), length (L), time (1') and electric Current (A) as fundamental quantities, the dimensions of permit- C. Ifd. tivity will be jE.'oflO a. [MLT--'A-'l b. [MLT~2A~21 77. The mass of a liquid flowing per second per unit area of 86. Assume that the mass m of the largest stone that can be cross section of a tube is proportional to px and vY , where moved by a flowing river depends upon the velocity vof P is the pressure difference and v is the velocity. Then, the water, its density p and the acceleration due to gravity the relation between x and y is g. Then, 111 is directly proportional to a. x ,- y b. x = -y b• V 4 d. v6 d. y = -x-\" c. y- = x 87. A spherical body of mass In and radius,. is allowed to fall in a medium of viscosity fl. The time in which the velocity 78. A physical quantity x is calculated from x = aJbC2' Cal- of the body increases from zero to 0,63 times the terminal velocity (v) is ca1led time constant (r), Dimensionally, r culate tht;: percentage error in measuring x when the per- can be represented by centage errors in measuring a, b, care 4,2, and 3 percent, respectively, a. 6n 1) b. ~;r1)) a.7% b.9% c. 11% d. 9.5% 79. In the formula X = 3YZ2, X and Z have the dimensions In d. None of these or tapacitance amI magnetic induction, respectively, The dimensions of Y in M,K,S, system are c. 6n 1}f'v a. M~3 L21'~2 A--4 b. ML ~2A 88. A student writes four different expressions for the dis- placement y in a periodic motion a\" a function of time t, C. M~3 L -'1\" A4 d. M~' c·2r' A4 a is amplitude and T is time period, which of the following can be correct?

Units and Dimensions 3.21 a. .)1=aT.sI2nr\"-t 3. ±0.01 em' b. ±O.I cm2 c. ±0.I1 cm2 d. ±0.2 cm' b. y = a sin Vt a.t 95. While measuring the acceleration due to gravity by a sim~ C.)I=~sm­ pIe pendulum, a student makes a positive error of 1% in a [2mT a 72d. y = sin T + cos T2m ] the length of the pendulum and a negative error of 3% in the value of time period. His percentage error in the measurement of g by the relation, g ~ 4,,2(111'2) will be 89. The relation tane ~ v'lrf!, givcs the angle of banking of the 3. 2% h. 4% c. 7% d. 10% cyclist going round the curve. Here, v is the speed of the cyclist, r is the radius of the curve and g is acceleration 96. While measuring acceleration due to gravity by a simple due to gravity. Which of the following statements about pendulum, a student makes a positive elTor of 2% in the the relation is true? length of the pendulum and a positive error of 1% in the a. It is both dimensionally as well as numerically COf- value of time period, His actual percentage error in the rect. measurement of the value of g will be b. It is neither dimensionally correct nor numerically correct. a. 3% b. 0% c. 4% d. 5% c. It is correct dimensionally but not numerically. 97. The relative density of a material is found by weighing d. It is correct numerically but not dimensionally. the body first in air and then in water. If the weight in air is (10.0 ± 0.1) gf and weight in water is (5.0 ± 0.1) gf, \"90. In the relation P = fie-·exz/KG, P is pressure, Z is distance, then the maximum permissible percentage error in relative eK is Boltzmann constant and is the temperature, The density is dimensional formula of fJ will be LI ~2 ~3 ~5 a. [Mo L2rO] b. [Ml L'r- 1] 98. Dimensional formula of a physical quantity x is r-d. [M°L 2 1] [M- 1l}T- 21. The errors in measuring the'quantities M, Land T, respectiv~ly, are 2%, 3% and 4%, The maximum 91. A liquid drop of density r, radius r and surface tension percentage of error that occurs in measuring the quuntity (J oscillates with time period T. Which of the following expressions for T2 is correct? x is a. 9 b. 10 c. 14 d. 19 prj per 99. The heat generated in a circuit is given by Q = /2 Rt, b. ,.3 a. where / is cllITent, R is resistance and t is time. If the d. None of these percentage errors in measuring 1, Rand tare 2%, I%, and c. 1%, respectively, then the maximum error in measuring P heat will be 92. A highly rigid cubical block A of small mass M and side L is fixed rigidly on the other cubical block of same di- a.2% h.3% c.4% d.6% mensions and of low modulus of rigidity.\" such that the lower face of A completely covers the upper face of B. 100. The internal and external diameters of a hollow cylinder The lower face of B is rigidly held on a horizontal surface. A small force F is applied perpendicular to one of the side are measured with the help of a vernier callipers, Their faces of A. After the force is withdrawn. block A executes small oscillations, the time period of which is given by values are 4.23 ± 0.01 em and 3.87 ,Ie 0.01 em, respec- tively. The thickness of the wall of the cylinder is 3. 0.36 ± 0.02 em b. 0.18 ± OJ)2 C111 a. 2,,~ryL c. 0.36 ± 0.01 em d. 0.18 ± OJ)] ern h. 2\" ej(MIJ/L) c. 2\"ejML/ry d. 2nejM/ryL Multiple CQrrect 93. The mass of a body is 20.000 g and its volume is 10.00 Answers iFMpe Solutions ()n page 3.32 cm3. If the measured values are expressed up to the cor- rect significant figures, the maximum error in the value of 1. Which of the following pairs have the same dimensions? density is 3. Torque and work 3. 0.001 g em,·3 b. 0.010 g cm-3 b. Angular momentum and Planck's cons.tant c. Energy and Young's modulils c. 0.100 g cm\"\" d. None of these d. Light year and wavelength 2. Which of the following pairs have different dimensions? 94. The length of a strip measured with a meter rod is 10.0 cm. Its width measured with a vernier callipers is LOO a. Frequency and angular velocity. em, The least count of the meter rod is 0, I cm and that of b. Tension and surf~lce tension. vernier callipers is 0.01 em. What will be the error in its area?

3.22 Physics for IIT-JEE: Mechanics I c. Density and energy density. a. x and y have the same dimensions d. Linear momentum and angular momentum. h. x and z have the same dirnensions 3. Pressure is dimensionally c. y and z have the same dimensions a. force per unit area d. None of the above three pairs have the same dimen- b. energy per unit volume c. momentum per unit area per second sions d. momentum per unit volume 10. The velocity, acceleration and force in two systems of units are related as under: iii. F' = C,~) F i. v' = ~2 v 4. Which of following pairs have the same dimensions? ii.a' = (afJ) a (L ::; inductance, C ::; capacitance. R = resistance) All the primed symbols belong to one system and un- a. ILi and CR b. LRandCR fJprinted ones belong to the other system. a and are L d. RCand I c. - and..,ILC - dimensionless constants, Which of the following is/are LC R correct? 5. Choose the correct statement(s) a. Length standards of the two systems arc related by: L' = (~:) L. a. A dimensionally correct equation must be correct. b. Mass standards of the two systems are related by: b. A dimensionally con'cet equation may be correct. m' = (a,lfJ2 ) m. c. A dimensionally incorrect equation must be incor- o. Time standards of the two systems are related by: rect. d. A dimensionally incorrect equation may be correct. 6. Which of the following pairs have the same dimensions? T' = (;,) T. h d. Momentum standards of the two systems are related a. - and magnetic flux (j,)by: P' = P. e II Assertion.Reasoning Solution~,on page 333 b. -- and electric flux TMpe e In the following questions, each question contains Statement C~ electrit flux and !l. I (Assertion) and Statement 11 (Reason). Each question has 4 80 choices (a), (b), (c), and (d) out of whieh ONLY ONE is Correct. d. electric flux and fLo I 7. The values of measurement of a physical quantity in 5 (a) Statement I is True, Statement Il is True; Statement II is a trials were found to be l.51, l.53, 1.53; 1.52 and 1.54. correct explanation for Statement 1. Then a. average absolute CITor is 0.01 (b) Statement I is True, Statement Il is True; Statement 11 is NOT a correct explanation for ~tatement 1. h. relative error is 0.01 (0) Statement I is True, Statement II is False. c. percentage error is 0,01 % d. percentage error is 1% (d) Statement I is False, Statement II is True. 8. If S and V are one main scale and one vermier scale and 1. Statement I: Mass, length and time are fundamental quanti- 11 - 1 divisions on the main scale are equivalent to n divi- ties, siems of the vernier, then Statement II: Mass, length and time are indeperident of one another. a. Least count is -S n 2. Statement I: Light year and wavelength have same dimen- sions, h. Vernier constant is S Statement II: Light year represents time while wavelength - represents distance. n L c. The same vernier constant can be used for circular 3. Statement I: The product RC and - have same dimensions. R verniers also ~Statement II: Both RC and have dimensions of time. d. The same vernier constant cannot be used for circular 4. Statement I: Unem momentum and impulse have same di- verniers mensions [MLr-'l EI Statement II: Impulse is equal to final momentum. 9. Consider three quantities: x= -,Y= - - - and B ,jILO£O \" - CR' Here, I is the length of a wire, C is the capaci- tance and R is a resistance, All other symbols have usual meanings, Then

Units and Dimensions 3.23 5. Statement I: Angular momentum ofan orbiting electron may presence or absence of certain factors-non dimensional con- be represented in terms of Planck's constant. stants or variables-cannot be identified by this method. So, ev- Statement II: Angular momentum of an orbiting electron is ery dimensionally correct relation cannot be taken as perfectly an integral multiple of ~. correct. 2n 6. If a kg, f3 meter and y second are the fundamental units, 6. Statement I: A (Angstrom) and AU are different units of 1 calorie can be expressed in new units as [1 cal = 4.2 1] length. a. a-' fi'y b. \",-' fi-'Y Statement II: A (Angstrom) is a small unit of length while c. 4.2,,-' fi d. 4.2\",-' fi--'Y' AU is a big unit of length. 7. Time period of oscillation of a drop depends on surface 7. Statement I: The number of significant figures in 0.001 is I tcn~ion (T, density of the liquid p and radius r. The while in 0.100 it is 3. Statement II: Zeros before a non-zero significant digit are relation is not counted while zeros after a non-zero significant digit arc !;:a. )p;2 b. counted, t: d·fJc. 8. Statement I: If error in measure'ment of mass is 2% and that r3 in measurement of velocity is 5%, then error in measurement 8. Energy of a S.H.M. is dependent on mass In, frequeney I of kinetic energy is 6%. Statement II: Error in kinetic energy is and amplitude A of oscillation. The relation is ~E = (~\"'- +2\"\"V). a. AMTI K In v CompreHensive c. Mf 2 A-2 d. MI2A2 Type Solutions em page 3.33 For Problems 9-11 Accuracy of measurement also lies in the way the result is ex- For Problems 1-5 pressed. The number of digits to which a value is to be expressed The van def Waal's equation of state for some gases can be is one digit more than number of sure numbers. Rules do exist expressed as to deal with number of digits after an operation is carried out on the given values, The error can be minimised by many trials and (p ;2)+ (V -- b) = NT llsing the correct methods and instruments, where P is the pressure, V is the molar volume and Tis the 9. If length and breadth arc measured as 4.234 and l.OS-m, the area of the rectangle is absolute temperature of the given sample of gas and a, band R are constants. 1. The dimensions of a are a. 4.4457 m2 b. 4.45 m2 a. ML'r·' b. ML'r' c. 4.446 m' d. 0.4446 m2 C. L3 d. [,6 10. The order of magnitude of 147 is 2. The dimensions of constant bare a. I b.2 c. 3 d.4 a. M L'T-2 b. MC-'T-2 11. Number of significant figures can reduce in c. L' d. L6 a. addition b. subtraction 3. Which of the following does not have the same dimen- d. division c. multiplication sional formula as that for RT? a. PV b. Pb a ab Matcliing Column Type c. V' d. V2 Solutions on p(lge 3.33 4. The dimensional representation of ablRT is a. M L'1'-2 b. MO [}T\" 1. If R is resistance, L is inductance, C is capacitance, H is latent heat and s is specific heat, then match the quantity c. ML-'r-2 d. None of these given in column I with the dimensions given in column II. 5. The dimensional formula of RT is same as that of a. energy b. force ~. Column i Column II .. c. specific heat d. latent heat i. LC n. L2r-2 .c .;-- LN-c For Problems 6-8 b; L21'~2 K\"l . Dimensional methods provide three major advantages in verifi- f--.ll. cation, derivation and changing the system of units. Any empiri- c.r'·' . cal formula that is derived based on this method has to be verified iii. H .. and proportionality constants found by experimental means. The r\"-sd. M'L,4 A-c4 .. iv. s

3.24 Physics for IIT·JEE: Mechanics I 2. There are four vernier scales; whose specifications are given (UT-JEE, 1988) in columli 1and the least count is given in column II. Match the 3. The dimensions of electrical conductivity are _ __ columns I and n with correct specification and corresponding (UT-JEE, 1997) 4. The equation of state for real gas is given by least count (8 = value of main scale division, 11 = number of (p + :,) x (v - b) = RT. The dimcnsions ofthe constant marks on vernier) a are _____, Columnl ColtImnlI . . ...•. (IIT·,JEE, 1997) j;·s =llTIm, 11\",10 .\" a. 0.05 oim ii.s=05mm,n=10 ................... Single Correct Answer Type iii. s = 0.5 mm, 11= 20' . i'V.s = 1 mm,lI = 100 . b,O.Oimm•.. .....• . .. (D1. The dimensions of 80E2(80: permittivity of free space, c.O..lmm .... . .......• E: electric field) are d.0.025 mm '.' •..•... > . 3. Match the columns a. MLT- 1 b. M L'T-2 c. ML-'T- 2 d. ML 2T-- 1 . - - - - - - .. - - - , - _ . _ - - - - - Column I . CoiumnII. '.' i. Backlash error a. Always subtracted (IIT·JEE, 2000) -'-ii. Zero'error f---'c---------~-- -- 2• A quantI.ty X·IS gI.ven by cO L\"-'V, where cO I.S the penm.t- b. Least count = I MSD--' 1 tivity of the frce space, L is a\"'Ilength, '\" V is a potential ...I . . ....•. .. difference and ~t is a time interval. The dimensional for- Iii; Vernier calHper~' . . .... mula (or X is the same as that of ··-.··.·.•·....VSD . c.May be \"'-veor+ve •.... liv. Error in Screw gauge d. Due to loose fittillgs \". • 4. Using significant figures, match the following a. resistalice b. charge c. voltage d. current Column I Column II (IIT-,JEE,2001) 3. A cube has a side of length 1.2 x to--' m. Calculate its i ..O.1~23-45---·--~-:~~·--~~~1 ii. 0:12100 cm volume, iii. 47.23 -;- 2.3 c. 1 a.1.7xlo- 6 m3 r - - - - - - - - -iv.3 X 108 d.2 b. 1.73 x 10--6 m' .------~'--------' c. 1.70 x 10-6 m' 5. Some physical quantities arc given in Column I and some (UT·JEE, 2003) possible SI units in which these quantities may be expressed m'd. 1.732x 10-6 are given in Column II. Match the physical quantities in 4. Pressure depends on distance as P = ~ exp ( - ~;). Column I with the units in Column II. r---.----Column where a, f3 are constants) z is distance, k is Boltzmann's' constant and () is temperature. The dimensions of f3 are - ......_!:_(J M, \":1.'- __ -C.-ol-umcn-II- -.'-.'.- - -. .----'.'__.'. _..__ a. MOLoTo b. M-IL-IT-' a. (volt) (coulomb)(metre) ' '.' ii.3RT/M b. (kilogram) (metre)3(SeCOnd)\"·2 c. MOL'To d. M- I LIT2 iii. F2 / q2 B2 iv. GM,/R, ......c. (meter)2 (second)~2 (IIT-JEE, 2004) ..---.----.--\"~--.-- 5. A wircoflength I =6±0.06cmandradius r =0.5 ± 0.00:; .....j ...(f\",,-ad) (volt)2(kg)--1 .' where G is universal gravitational constant; Me, mass ofthe cm has mass rn = 0,3 ± 0,003 g, Maximum percentage earth; M,n mass of sun; Re , radius of the earth; R, universal error in density is gas constant; T, absolute temperature; M, molar mass; F, a.4 b.2 c. 1 d.6.8 force; q, charge; B, magnetic field. (IIT-,JEE, 2007) (IIT-JEE, 2004) ~rdii\\les Solutions O1i page 3.~1 6. Which of the following set have different dimensions? Fill in the Blanks Type a. Pressure, Young's modulus, stress b. Emf, potential difference. electric potential I. Planck's constant has dimensions _.._~__.. c. Heat, work done, energy (IIT·,JEE, 1985) d. Dipole moment, electric flux, electric field 2. In thc formula X = 3 Y Z2, X and Z have dimensions of ca- (IIT·JEE 2005) pacitance and magnetic induction, respectively. The dimen- sions of Y in M,K,S. system are ___,______,_,_,

Units and Dimensions 3.25 7. In a screw gauge, the zero of main scale coincides with (:g11. If El , En and Em are the percentage errors in g, i.e\" fIfth division of circular scale in first figure. The circular x 100) for students I. II, and Ill, respectively, then divisions of screw gauge are 50. It moves 0.5 mm on main scale in one rotation. The diameter of the ball in second a. Iii =0 figure is b. E1 is minimum c. Ii, and En '~K I d•. En is maximum (IIT-JEE, 2008) R ~ Multiple Correct Answers Type Fig. 3.5 1. The dimensions of the quantities in one (or more) uf the following pairs are the same. Identify the pair(s). a. 2.25 mm b. 2.20 mm c. 1.20 mm d. 1.25 mm a. Torque and work b. Angular momentum and work (IIT-JEE, 2006) c. Energy and Young's modulus 8. A student performs an experiment for determination of d. Light year and wavelength (IIT-JEE, 1986) g [ = ::;:L ], L '\" I m. and he commits an error of I',L. 2. Let (EO] denote the dimensional formula of the permit- For T he takes the time of n osdl1ations with the stop tivity of the vacuum and Iflo) that of the permeability watch of least COllnt 6 T and he commits a human error 01'0.1 s. For which of the following data, the measurement of the vaellum. If M = mass. L = length, '[ = time and of g will be most accurate'? I = electric current, then a. [so] = M- I L-3 1'2[ a.I',L=0.5,I',1'=O.l,1I=20 b. [sol = M-I L -31'4[' b.I',L = 0.5, I', T = 0.1. 11 = 50 c. liLol = ML1'-2[-2 c• .6.L = O.5,.6.T = 0,01, 11 = 20 d. [flo I = ML 21'- I[ (IIT-JEE, 1998) d. I',L = 0.5,1',1' = 0.05, IL = 50 (IIT-JEE, 2006) 3. The S.l. unit of inductance, the henry can be written as 9. A student performs an experiment. to determine the a. weber/ampere h. volt-sec!amp Young's modulus of a wire, exactly 2 m long, by Searle's c. joule/(ampere)' d. ohm-second method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an 4. Which of the following pairs have the same dimensions? uncertainty of0.05 mm at a load of exactly 1.0 kg. The stu- dent also measures the diameter of the wire to be OA mm 3. Reynold numher and coefficient of friction with an uncertainty of 0.0 I mOl. Take g = 9.8 m/s2 (exact). The Young's modulus obtained from the reading is h. Curie and frequency of light wave c. Latent heat and gravitational potential. a. (2.0 ± 0.3) x lO\" Nm· 2 d. Planck's constant and torque (IIT-JEE, 1995) b. (2.0 ± 0.2) x lO\" Nm-2 Subjective c. (2.0±0.1) x 10\" Nm·-2 d. (2.0 ± 0.(5) x lO\" Nnc2 (UT-JEE, 2007) 1. Give the M.K.S. units for each of the following quantities, a. Young's modulus 10. Students I, II and III perform an experiment for measuring b. Magnetic induction the acceleration due to gravity (g) using a simple pendu- lum. They usc different lengths of the pendulum and/or c. Power of a lens (lIT·JEE,1980) record time for different number of oscillations. The ob- servations arc shown in the following table. Least count 2. A gas bubble, from an explosion under water, oscillates with for length = 0.1 em, Least count for time = 0.1 s. a period T proportional to padb E e, where p 1S the static pressure, d is the density of water and E is the total energy of the explosion. Find the values of a, hand c. (IIT-JEE, 1981) Shident Length of Number of Total Tim\"e f()r Time 3. Write the dimensions of the following in terms of mass, time, Pendulum Oscillations n Oscillations Period length and charge. (cm) (n) (s) (s) - - c- ; - - - - 6 : 4 . 6 - - - 8 - - - - · a. Magnetic flux h. Rigidity modulus 1cZi\"8.\"0 16.0 (m~JEE, 1982) II M.O 4 64.0 16.0 -!_II . _ -2-0.0- - - - -4 - - - 36.0 9.0 4. Match the physical quantities given in column I with dimen- sions expressed in terms of mass (M), length (L) tillle (T). and charge (Q) given in column II.

3.26 Physics for IIT-JEE: Mechanics I Column I Column II Column I Column II Angular momentum ML'T-' Torque ML2T-' Capacitance ohm-second Inductance M-' L-2T'Q2 Latent heat ML2Q-2 Magnetic induction coulomb (volt)-' Capacitance MLl T-'Q-2 Resistivity Ll T-2 Inductance coulomb'-joule -, (IIT-JEE, 1983) newton (ampere metre)-, volt-second (ampere)-l 6. The nth division of main scale coincides with (n + I)th divi- sions of vernier scale. Given one main scale divisi()n is equal 5. Column I gives three physical quantities. Select the appro- to 'a' units. Find the least count of the vernier. priate units for the choices given in Column II. Some of the (lIT-JEE, 2003) physical quantities may have more than one choice correct. (IIT-JEE, 1990) ANSWERS AND SOLUTIONS 1. a. 2 h.4 c. 3 d. 2 10. 5.42 x 0.6753 = 43.06030 = 43 e. 3 f. 2 g.4 h. 6 0.085 (up to two significant figures) 2. a. 0.059 h. 0.058 11. 1.45 + 1.56 + 1.54 + 1.44 + 1.54 + 1.53 d. 5.0x 106 3. f-Lm = 6 c.5.1x106 = 1.51 3. a. 953 h. 954 b. absolute error in each measurement c. 953.3 d. 953.4 t.1-'1 = 11.45 - 1.511 = 0.06, t.1-'2 = 11.56 - 1.511 = 0.05 t./LJ = 11.54 - 1.511 = 0.03, t.1-'4 = 11.44 - 1.511 = 0.07 4. a. 7 h. 6 c. 6.6 d. 6.4 5. 3.17 g = 0.00371 kg t./L5 = 11.54 - 1.511 = 0.03 1.4 + 0.00371 = 1.40371 kg ~ 1.4 kg [correct upto one place of decimal] t.1-'6 = 11.53 - 1.511 = 0.02 6. A = rrr2 = 22 = 0.9856 m2 ~ 0.99 m' I:l/J-m = 0.06 + 0.05 + 0.03 + 0.07 + 0.03 + 0.02 ](0.56)2 6 (two significant figures) = 60.26 = 0.0433 ~ 0.04 7. a. 3.8 x 10-6 + 4.2 x 10-5 = 0.38 X 10-5 + 4.2 X 10-5 = 4.58 X 10-5 ~ 4.6 X 10-5 c. t.l-'m = 0-.04 = 0.02649 = 0.03 Fractional error - - [upto one place of decimal] I-' 1.51 h. 4.7 x 10-4 - 3.2 x 10-6 = 4.7 x 10-4 - 0.032 X d. Percentage error = 3 % 10-4 e. I-' = 1.51 ± 0.04 or 1.51 ± 3% = 4.668 X 10-4 ~ 4.7 X 10--4 (one decimal place] c. 4.8 x 104 - 1.5 X 103 12. a. 1 = I, + 12 =(1.9±0.3) + (3.5±0.2) = (5.4 ± 0.5) m = 4.8 X 104 - 0.15 X 104 = 4.65 X 104 h. t.T = T2 - T, = (76.3±0.3) - (67.7±O.2) ~ 4.6 X 104 [one decimal place] = 8.6 ± O.5\"C 8. 1 = 4.234 m, h = 1.005 m, t = 2.01 cm = 0.0201 m 13. 1 = 10.5 ± 0.2 cm, b = 5.2 ± 0.1 em Area = 2[lb + ht + t) P = 21 + 2b = 31.4 cm, t.P = 2(t.1 + t.b) = 0.6 em = 2 (4.234 x 1.005 + 1.005 x 0.0201 + 0.0201 + 4.234] So, P = 31.4 ± 0.6 cm = 8.7209478 m2 ~ 8.72 m2 (three significant figures). 14. A = Ib = 5.7 x 3.4 = 19.38 cm2 ~ 19 em2 Volume = Ibt = 4.234 x 1.005 x 0.0201 (two significant figures) = 0.0855289 ml= 0.0855 m3 -t.A= -t.+1 -t..b => t.A = ( -0.1 + 0- .2) (19.38) 5.7 (three significant figures). Alb 3.4 9. V = 4 (3.34)3 = -4 x -22 (3-.34)3 = 19.5169 m3 t.A = 1.48 ~ 1.5 - -J[ 32 37 2 = 19.5m3 (two significant figures) (up to three significant figures) So, area = (19 ± 1.5) em'

Units and Dimensions 3.27 15. V= -S = -13.8 = -I ~ 3.4 6. c. Latcnt heat, L = heat. '[M--:Lc:'-r~--'2-] t4 - - . [LJ = 3.45 ms . mass [M] (two significant figures) r-= [L 2 2J I'>v = [ 0.2 + 04\".\"3\" ] 3.45 = 0.3087 ~ 0.31 work done 13.8 Gravitational potential,V = - - - - (two significant figures) mass. v = (3.4 ± 0.31) ms-' = =[M L'T-'J = - - =0.31 [V] [L'r'] [M] Percentage error 3.4 x 100 9.11 % ~ 9%. r7. d. ry = tangential stress = = F I A 22 eshearing strain xIL . 16. r =2.1 ±0.5cm =} A =4rrr' =4 x 7(2.1)' = 4 x 13.86 = 55.44 cm' ~ 55.4 em' I'> A I'>r =} I'>A = 2 x 0.5 x 55.44 = 26.4 8. b. [Moment of inertia] = [ML'] - = 2- -- Ar 2.1 [Moment of force] = [M L 2r-'] Area = 55.4 ± 26.4 cm' 9. . lei dl rrr2 R 22 (0.26)2 x 64 c. Induced e.m.f. = L- dt 17. P = -Z- = 7 x 156 ' I _ [e] [Wlq] = [ML'r- 2/AT) I'>P 2(0.02) 2 0.1 [,]- [dlldt] = [dlldt] [AT'] P = D.26 + 64 + 156 = [ML'r-'A-'] PI'>P 10. a.As Q =mL, x 100 = 18.57 ~ 18.6% L _ [QJ _ [ML'T-'] 18. r' = 1 ,1 [J-[m]- [M1 4rr'- g g =4rr -r' I'>g = 1'>1 0.121'>r = 2(_1_) = 1/10 = [MoL'T-'] II z+ r 20+ 600 600 11. d. All the four have same dimensions~ Le., [ML2r-'K-']. I'>g = x 100 = 0.833 ~ 0.8% 12, c, Coefficient of friction is a dimensionless quantity. g 13, d. Here, (wt + 1>0) is dimensionless because it is an ar- Objective Type gument of a trigonometric function. 1. d. Couple r x angle de = d W 14. b, If a quantity depends upon more than three factors, I each having dimensions, the method of dimensional 'i1w' = kinetic energy and Fdx = dW analysis cannot be applied. It is because applying prin- ciple of homogeneity will give only three equations. 2. c. Unit of surface tension is Nm-'. Also, 15. d. The formula can be written as Jrn-2 == Nmm-2 == Nm- l . velocity of light in vacuum = I -,--c''---c-c-''c--c--cc-- 3. d. Here, (2rretI A) as well as (2rrx I A) are dimensionless. velocity of light iu medium So, unit of et is same as that of A. Unit of x is same as This formula is dimensionally correct as both the [2:X]that of A. Also, [2:et] = sides are dimensionless. Numerically, this ratio is equal = MOLoro to refractive index which is > 1. Hence, the equation is [2rrx].Hence,[2Arre] = the option unit- numerically incorrect. In (d), :A: is . At' 16. c. The correct relation for time period of simple pendu- lum is T;;;:: 2rr(l/g)1/2, So, the given relation is numer- less. This is not the case with ciA. ically incorrect as the factor of 2rr is missing, But it is correct dimensionally. 4. b. Here, unit of y and Awill be same and that of x and . be same. -2r(r et - x) is dimenSI.Onless. Here, et A wIll - Ax A A velocity of light in vaccum 17. d. As J.L = , and are dimensionless. Unit of ct is same as that of A velocity of light in medium or x. hence, J.L is dimensionless. Thus, each term on the R.H.S. of given equation should be dimensionless, 5. d. Angular momentum, J = mvr B Le., A2 is dimensionless, i,e., B should have dimension This is same as that of Planck's constant. of A2, i.e., cm2, Le., area.

3.28 Physics f~r IIT-JEE: Mechanics I '*18. d. [x] = [Ay] = [B] [y] = [B/Al 32. b. y = r sinew! - kx) Also, [x] \"\" [AJ and [Cz] = dimensionless Here, wt == angle => W = .1- = T- I '* [CJ = [e'1. l' 19. b. Dimension of L / R is same as that of time. '*Similarly, kx = angle k = -I = L -I x 20. c. Random errors arc reduced by making large number w = T- I = LT- I of observations and taking mean of all the results. - -- k L-I 21. c. All the choices are equivalent but the answer must kOJ Or simply represents wave velocity possess three significant digits as significanl digits do not change on conversion from one system to another. (i! = 2nf = fA = v'wh.ere j' is frequency. So, appropriate choice is (c). - -2n-IA k 22. c. -A = -For-ce = [M0L0T0 J 33. d. We know that speed of c.m. wave is B Force '*I 11080 = 1 = [L -2 ,2 J - '*Ct = angle C = -an-gl-e = -I = T- I C = -- C2 7 JI1080 time T 34. c. Momentum == force x time == Ns '*Dx = angIe D -_ angle _- _I _- L-I 35. d. Momentum, p = InV = MLT' = M L -'L4 T- 4 T' distance L = DV4 F~' C = T-' = [MOLT \"J 36. d. AT2 = LT-2 X T2 = [MoLT o] - - D f.-I 37. h. In a product, percentage errors arc added up. 23. d. . . = -for-ce = M LT-2 = [MT -2 J 38. c. 0.08076 has least number of significant figures, i.e., Surface tension length L 4. B work done Surface energy = ,----;--- 39. a.H = - 1\"0 increase in area 40. b. Intensity of wave = ML'T-2 [MT- 2] energy MZ 2T-' = 1M L O T - 3 J -::c;-- = =-----= , .. L' area x time L'T .'. Surface tension and surface energy have same dimen- charge AT sions. 41. c. Capacitance, C = \"---.- . , . resistance x area potentIal ML'T-3A ·1 24. b. ReSIstIvIty = ---;----cc--- = [M- I L-2T 4 A'1 length ML'T-'A-' x L2 = [ML'y- 3A-'J '*42. C. PV = nRT PV ML\"'T-' x L3 - - - - -...- - - L R= - = liT - - -;-- - -x-K: c - - work mol 25. a. Electrie potential = - - = 1M L 2T- 2K- I mo]-II . charge = MI'T-' = [ML'T-'A- I ] '*43. d. 50 I = 0.501 X!03 order of magnitude of 50 I is -.-.:--- AT 3. 26. d. Moment of force = F x J. distance = [M L' T \"J 44. a.0.00701 = 0.701 x 10-2 Momentum = mass x velocity:;::: M 1..1'-1] .', Order of magnitude of 0.00701 is -2. 27. c. Pressure x Volume gives work = [M L'T--2J '*45. b. 379 = 3.79 x 10' Order of magnitude of 379 is 2. 28. c. Relation between EK and p is Ek = p2 . When p is 46. c, X = a + b,* !'J.X = !'J.a + !'J.b 2m !'J.X (!'J.a + !'J.b) Now, - x 100= x 100 doubled, Ek becomes four times. So, Ek increases by X a+b 300. =(~+~)XIOO 29. d. Rate of change of velocity is equal to acceleration a+b a+b which is a vector quantity and all others are scalar quan- thies. 47. c. Screw gauge has minimum least count of O.OO! em, hence it is most precise instrument. 30. h. (92.0 + 2.15) em = 94.15 em. Rounding off to first decimal place. we get 94.2 cm. 48. c. 5.69x 1015 kg has 3 significant figures as the power ~of 10 are not considered for significant figures. 31. e. n = 2I1 Vf;f; 49. a. Here, at is dimensionless m= -41T2-n2 = [MLr--'] = 1. M!.' 1J L2 x '[.2 . [+]'* a = ~ = = [T-I]

Units and Dimensions 3.29 x = Vo and Vo ==xa [LT- I ] = [MoLT-'I. Alternatively, So = 1 x q---,-q,='- and E = -F --. q a 4][ F r- 50. b. Here. ai' is a dimensionless. Therefore, a = .I, and 1, I q,q, F' F MLT-2 a has the dimensions of T- 2• I F,,'-coE = ----- x - = - = 51. c. Latent heat = -Q = -en-erg-y 2 . 8][ q2,,' L' m mass 57. c. X = M' L -\"T-z rnGravt.tatl.OnaI pot. = -U. \"'6XX x 100 = x (6AMi 'x 100) + Y (6LLX 100) 52. d. Energy density = -ener-gy = 1,MC IT-'] + Z ( T61' x 100) (Errors are always added) volume ' Force/area = M L -, T-2 6X [charge/volume] x [voltage] = -QI xW- .. --- x 100 = (ax + by + ez) per cent X vo. Q 58. fT [rn'g] 'I' = [nA,I'lig] '/' M d.v=y;= The dimensions of (4) are different. I It follows from here, 2 6v = ~ [6111 + 61 + 6M] i.e.. M LMT-' = M OL 2T-' Hence, correct choice is (d). v 2 III I M 53. d.! = Cm\" k\". Writing dimensions on both sides: I [0.1 0.001 O.lJ [MOLor-'] = MX[MLoT-'V = 2 3.0 + 1.000 + 2.5 [MoLoT-'] = [Mx +Yr'2y ] = I + 0.001 + 0.04] = 0.036 -[0.03 2 Comparing dimensions on both sides, we have ,. Percentage error in the measurement II = 0.036 x 100 = 3.6 0=x+yand-I=-2y,* y= 2,x=-2 59. b. Here, 'a' has the same dimensions as t2 T' 1\" T4 a = [1\"], b = Px = ML-~'T~-'l) = Ai Aliter. Remembering that frequency of oscillation of loaded spring is = [M- J 1'4] I ff = -I- - = _I(k)'/2 111 -' I ' T' m 27f . 27f Now, \"b- = -M--'T-4 = MLOr\" which gives x = -2I and y = 2I ' 60. b. Here, band x 2 = L' have same dimensions. x2 L2 AlsO,a = E x 1= (ML'T-')1' = M\"T' C' L'T-' a x b = [M-' L2T'] 54. d. - = ~, = [LI g LT-- A A force M LT-' Hence, correct answer is (b). 55. d. - = Ill, B = - = . .= 61. a. h = [M L2T-'], G = [M- I L'T-'], C = [LT-'] ,B m Imear densIly M L-' .'. h'I'G-'I'C'/2 .. B = [M oL 2 T-'] = M'/'LI\"\"'/2 x M'/'L -3/2 1\" x L'/21'-'/2 heat energy Ml}T-2 Latent heat = M mass = M LOTO = Mass = [M oL 2 r 2 ] Thus, B has same dimensions as that oflatent heat. b.V=~\"p:4\"= MI-'T-'L4 ' 56. b. ~eoE' is the expression for electrostatic energy den- 62 . 8 nl ML-'r-'L =MoL'T-' 2 63. b. Percentage error in volume sity, i.e.• the energy stored per unit volume in a parallel plate capacitor. 0.01 0.01 om I _, energy = - - x 100+ - - x 100+ -- x 100 .. -£oE = --- 15.12 10.15 5.28 2 volume = 0.07 +0.10 + 0.19 = 0.36 64. d . Relau.ve densl.ty, p, = W, W W[- 2

3.30 Physics for IIT-JEE: Mechanics I -;:-;::;::-8.0_0::-= = 4.00 On solving these equations, we get 8.00 - 6.00 -5 1 1 a =6- ' b=-2andc=-3 Ap, x 100= AW, x 100+ A(W, - W2) x lOa 72. c. Kinetic energy, E = 21mv' p,. W, W, - W2 2601 AE Am Av =0-.0-5x l 00 + 0.05 + 0.05 100 =5. 10 - x 100= - x 100+2- x 100 x Em v 8.00 2 .. p, = 4.00 ± 5.62% =2+2x3=8% 65. a. V = lbt a !/2};2 73. a. X = --3- AV At Ab At - x 100=- x 100+ - x 100+- x 100 c· V bt AX 1 Aa !;b '!;c - x 100=--- x 100+2- x 100+3- x 100 X 2 abc 0.01 0.01 0.01 = - - x 100 + - - x 100 + - x 100 15.12 10.15 5.28 = 21: x 1 + 2 x 3 + 3 x 2 = 12.5% =0.36% D(Il, - n,) V8 66. d.n = - - - - - 74. d. R = - = - = 2 ohm X2 - Xl I4 !;R !;V M =} r-'L-'xL =} D = [MOL 2r-'] - x 100=- x 100+- x 100 D= RV I L3 = '0\".85 x 100 + '04.2 x 100 = 11.25% EJ' [ML'Y-'][ML'Y-'f °0 0 67. b. M5C' = MS x [M 'L'r ']2 = M L r =} R = (2 ± 11.25)Q. This comes out to be dimensionless and angle is also 75. a. Here v = ealtb(l'C d Taking the dimensions dimensionless. MOLY-'A o = [AT'1a[ML'T-']b[MLy- 2A-'Y L =L current r-[M'\" L 3 2 ]d 68. d. --c;---c:=- = There will be 4 simultaneous equations by equating RCV thedimensionsofM,L, T andA. These are\" - 2c = 0, r (L dI) dI a - b - 2c - 2d = -I, b + c - d = 0 and 2b + c + df 3d = I dI Solving for 'a', 'b', 'c' and 'd' we get [AsRC = time constant T and pot. diff. V = L-] a=-2,b=l,c=-I,d=O dt Thus, v = e-'hll-'Co (L (T' )\"1)b69. d. \"2 = 11, (M,)a M, L2 T, 76. a.Here, w x k = T-' L -, which are not the dimensions of velocity. All others have got the dimensions of veloc- Dimensional formula of moment of inertia = 1M L 2To] ity. .. a = I, b = 2, c = 0 77. b. M 0< pXv)' =} ML\"-2y--\" At 11, = 12.0, M, = 1 kg, M2 = 10 g L, = 1 m, L2 = 5 em, T, =I s, T2 =1 s 11, _120 (!kg)' (~)2 (~)o - . 109 5cm I sec = 12 x (1000 g)' (100Cm)2 xl x = I, -x + y = -2 and -2x - y = -I 10 g 5 em From here, we get y = -1. Thus, x = ~y = 12 x 100 x 400 = 4.8 X lOs Ax !;a !;b 70. a. Let Y = [va Ab P1 78. d.Here, - x 100= - x 100+2- x 100 xa b r[ML-'T-'] = [LT-'l\"ILT-']b[MLT-2 + -1 !-;c x 100 M C\"\"IT--2 = MCLa+b+cT-a-2b-2c 2c .. c= l,a+b+c=-I,-a-2b-2c=-2 On solving, we get a = -4, b = 2 and c = 1 = [4 + 2 x 2 + 1/2 x 31% = 9.5% 71. c. Let T 0< pa db E' 79. c. Here. X = Q. QW Writing dimensions on both sides. V. But V = [Mo LOT] = [M C-·, T- 2 ]a[M L ~3]b[M Z}T-2 ], Q' F [MoL 01'1 = MlI+.fJ+'c L -.a-3b+2cT-2a-2c :.X=W:Now,Z=B= IL Thus, a + b + c=O, -a - 3b + 2c=O, --2a - 2c=0 X I Q' I'L' .. Y=-=--x-- 3Z' 3 W F'

Units and Dimensions 3.31 i.e., [~~] = [MoLoToJ Now, [K8J = [EnergyJ = lML2r 2J [K8J [M L2T-'J = [M Lr2] =} [ 1 = - - = q Ll.V q ct [ZJ [LJ = - x - = -- = Current V Ll.I Ll.I 2 H nee = [a] = [MLT- J = [ MOL2To J 81. d. y = a sin wi + bl + el' cos wi e ,1,11] [PJ [M L -I T- 2J =; =H...~re~bac= y' b = Vi i~' C = yl12 91. a. Let 1'2 = p\" rh (J' x 1'2 = (M L -3),,(Li'(MT-2), y il yl 12 (ylll' 82. b. Here. x 2 has the dimensions of L 2. B = [L 2] Also• ML 2T-- 2 = AL'/2 ML 7/ 2T- 2 =} a + c = 0, 3a + b = 0 and 2c = 2 = =Hence, a I, b 3 and c = -1 -L-' or A = .. AxB=ML\"/2T- 2 83. h. Here, xl! has the same dimensions as a 2. Thus, n ::::- 2 r2 pr3 = pr3(J-l = _ will make the expression dimensionally correct. 0.02 (J 84. d. Required percentage = 2 x 0.24 x lOa + 30 x [ fM]92. d. [17J = [MC- I T-2J 0.01 V VHence, -;;L = I [MJ 100 + - x 100 = 16.7 + 3.3 + 0.2 = 20% [M L-'T-2J[LJ = [TJ 4.80 85. c. By Coulomb's law F = -1- X qlq2 93. d. Maximum en'or in measuring mass = 0.001 g, be- -2- cause least count is 0.001 g. Similarly, maximum error 4][ co r in measuring volume is 0.0] em3 EO = 4n xQ,Fq2x ,.2' l'ak'mg d'ImenslO. OS Ll.p Ll.M il V 0.001 0.01 -=-+-=--+-- pM V 20.000 10.00 So = (AT)(AT) = [M-'L-3T4A2] ML'T ..2 x U = (5 X 10..5) + (I X 10..3) = 1.05 X 10-3 86. d. m ()( va ph g' =} ilp = (1.05 x 10 3) X p M LOTD ()( (LT-')a(M L -3)b(LT-2), = L05 X 10-3 x 20.000 = 0.002 gcm-' Comparing the powers of M, Land T and solving, 10.00 wegetb= l,c=3,a=6=} m()(v6 87,d. [ 6mnrry2] = [ MLM-L'T2--I ] = [L 3 T] 94. d. Ll.A = ( -Ll{.l + bLl.b) A As we have [ry] = [M C,' Tool J om)= ± ( -0.1 + - x (10.0 x 1.00) em2 [(~n;,rryr2] = [( ML;'~:~T-'f2] 10.0 1.00 [6:~rvJ = [ML-'T~LLT'] = 2 = ±0.02 x 10 = ±0.2 cm2 W'T J Ll.l Thus, none of the given expressions have the di- 95. c. Given that - x 100 = + 1% mensions of time. I 88. d. Since L.H.S. is displacement, so RH.S. should have Ll.T dimensions ofdisplacement. In (a), aT does not have the and \" - x 100 = -3% dimension of displacement. Also argument of a trigono- l' metric function should be dimensionless. In (b), argu- Percentage error in the measurement of ment is not dimensionless and in (c), al T does not g= [ -4n-2/] = 100 x -LIl./ - 2x -Ll.T x 100 1'2 l' have the dimensions of displacement. Hence, the COf- rect choice is (d). = [1% - 2 x (-3%)] = 7%. 89. a.Herc, [taneJ = [;;] = Mo~oTo.Also,inactualex­ viifI:.96. b. l' = 2n or 1'2 = 4n2l:.. g 2 L Ll.g LLl.L 2yLle.T- 4n T2; Ii\"=} g = = - Ll.g x 100= -Ll. L x 100-L2l. T- x 100 -- l' gL pression for the angle of banking of a road, there is no numerical factor involved. Therefore. the relation is both Ll.L ill' numerical1y and dimensionally correct. Actual % eITor in g = - L x 100 - 2 - x 100 l' 90. a. Argument ofexponential term must be dimensionless, =+2%-2x 1%=0%

3.32 Physics for IIT-JEE: Mechanics I 97. d. Relative density = Wa , p = -W\" = F] x[Areal= MLT~2 L2=[ML2T~2A~ll Wa-Ww [ iL AL tv where p is relative density, Wa weight in air and w is Dimensions of -h = [PlanCk'S constant] e charge .. -£>\"p = -£>,.-W\" - -£>\"w loss In weIght. P Wa W = [M~2;~I] = [M L2T~2A~'1 + -. £>\"p £>,.W\" £>\"w From Gauss theorem, electric flux;::; !l For maXUl1um error, - = - - So p Wa W 7. a., b., d. Mean value For ma~imum percentage error, £>\"p x lOa = -£>,.-W\" x lOa + -£>,.w x 100 I.S1 + 1.53 + 1.53 + 1.52 + 1.54 - p Wa tv =5 1.53 Given £>,. W\" = 0.1 gf and W\" = 10.0 gf Absolute enOfS are: (1.53 - 1.51 = 0.02), w = 10.0 - 5.0 = 5.0 gf £>,.w= £>,.W\" +£>,.Ww =0.1 +0.1 =0.2gf (1.53 - 1.53 = 0.00), (1.53 - 1.53 = 0.00), (1.53 - 1.52 = 0.01) and (1.54 - 1.53 = 0.01) -£>x\"p IOO= (0- .1) xIOO+ (0- .2) xlOO Mean absolute error 0.04 p 10.0 5.0 . 5 = 0.02 + 0.00 + 0.00 + 0-.0-1-+--0-.0-1- =1+4=5 5 = 0.008 \"\" 0.01 So, choice (a) is correct. omam Relative enur = - = 0.00653 \"\" 1.53 0.01 % Error = - - x 100 = 1% 1.53 99. d. Required error is 2 x 2% + 1% + I%. i.e.• 6%. 8. a., b., c. Least count = 1 MSD - I VSD = S - V 100. b. Subtract 3.87 from 4.23 and then divide by 2. (n - nV= S - I) S = ~ [.,' = (n - I) S] 11 n Multiple Correct It is also called vernier constant. . Answers type So, choices (a) and (b) are correct ~ Choice (d) is wrong and choice (c) is correct, since for all vernier scales similar approach can be used. r r.1. a., b., d. Torque = x F. Work = F Both have dimensions [M L2T~21. a., b., d. Unit of x unit of E . . Angular momentum and Planck's constant have same dimen- 9. = . . = umt of velocIty umt of B sions [ML 2r· I J. Because E = 11 B Light year and wavelength have same dimension [L). Y = _1_ = c -+ velocity of light 2. b., c., d. Frequency and angular velocity have same dimen- , j!to co sions IT~ll. unit of I length . Tension has dimension of force and surface tension Unit of z = = - - = veloClty unit. of RC time has dimension of (foree/length). Density has dimensions 10. a., b., c., d. Velocity = length/time, ace = lengthJ(timef -m-ass- ) and energy denS.Ity ha.s dlm.enslOns ( -en-er-gy) . ( volume volume (velocityf. , v2 v2 Angular momentum has dimensions of (linear momentum x => Length;::; - - - - , I.e., and L = L =- - acc. a' a distance). =} ~ = (~y (~) = (~y a~ =a3/~3 3. a., b., c. Pressure has dimensions [ML~ IT-21, force per area, energy per unit volume and momentum per unit area per F' F second have same dimension [ML\"'T~21. Now, m' = - and m = - 4. a., c. -L, C R and v1L7C7' all have dimensions of time [1']. a' a R m' F' a ] 1 1 5. b., d. A dimensionally correct equation mayor may not be :::} - = - - = - x -a~ =-- mFa' correct. For example, s = ut + al 2 is dimensionally correct, a~ a2~2 but not correct actually. A dimensionally incorrect equation may be correct also: v' v a Time::::::: veL/ace, i.e., T' = - and T = - For example, s = II + 2'(2n - I) is a correct equation, but a' a not correct dimensionally. 6. a., c. Dimensions of magnetic flux::::: magnet.ic field x area

Units and Dimensions 3.33 Momentum::;:;:: mass x velocity, i.e., =6. d. 1 cal 4.2 J = 4.2 kgm2s-2 = 112(a kg)(fJ m)2(y S)-2 P! = mlv', and P :::: mv '* '\" = 4.2a-'fJ-2 y 2 pI m'v' 0: 2 So, 1 cal = (4.2 ,,-1 fJ-·2 y 2) new units. -=--=---=- P 111 V ,,' fJ' fJ fJ3 7. c.LetTexaap'JrC Mil LOT = (MT-2),'(M L -3)\"1/ 1. a. Fundamental quantity is that quantity which does not '*Equating the powers of M: 0 = a + h b = -·a depend upon other quantities. Since mass, length and time '*L: 0 = -3h + c c = 3h are independent of one another, so they are fundamental '*T: I = -2a I 13 quantities. a = ---,h= -,C=- 2 22 2. c. Light year and wavelength both have same dimensions of '* T=kt: length. 8. d. E ex mafb A' 3. a. Both RC and ~. have dimensions of tilne. Both are the ML 2T- 2 =maT ·1>Lc time constants of RC circuits and LR circuits, respectively. .' a = I, b = 2, C = 2 4. c. Impulse is equal to change in momentum. Dimensionally For Problems 9-11 both are same. 9. b., 10. b., 11. b S. a. Angular momentum of an orbiting electron in nth orbit is Sol. h 9. b. A = lb = 4.4457 m 2 given by \"2-,.,.. One should retain only three significant figures. 6. a. A is equal to 10-10 m. whereas I AU is the distance 10. b. For x to be order of.ma.gmtude, 0.5 < '11O4x7 s 5 should be between sun and earth. satisfied. 7. a. Zeros before a non-zero significant digit are not counted 11. b. To find say difference between 20.17 and 20.15, we get, 0.02 (Le., one significant flgure only). while zeros after a non-zero significant digit are counted. t;.K (t;.m t;.v)8. d. - x 100 = - + 2- x 100 =2 + 2x5 = 12% K In v Comprehensive MatcHing' CoLumn [~me [~pe t)!i~,> \" • w 1. i. -+ c., ii. -+ d., iii. -+ a., iv. -+ h. For Problems 1-5 r0) [LC] = [~RC] = T2 1. a., 2. c., 3. c., 4. d., 5. a. (ii) fLR] = [~R2] = T[R2] = T [~r = T [~ Sol. FV2] [MLT-2L6) 1. a. [a) = [PV') = [ - A = ' = [ML'T-') (.!:I'.)2= T = T (M L2T.....2)2 [L'] 2. c. [b) = IV) = [L'] . /2( A'T 3. c. According to given equation, = M2L4'[-S A-4 PV + -a - Pb - ab RT (iii) [H] = [ -He-at] ML2T-' = L 2T- 2 -V2 = Mass V =---- M As dimensions of all the terms on L.H.S. must be equal Heat ..- - ] -M--L:-:2:T-::--2- temperature MK to dimensions of term RT on R.H.S., hence al V2 does not (iv) [s J = [ Mass x = have the same dimensional formula as that of RT. = L2T-'K-' '*4. d. [~~] = [liT) [~~] = [V') = [L6 ] 2. i. -)- c., ii. -+ a., iii. -+ d., iv. -+ b. We know that least count is given by sin, so 5. a. [RT] = [PV] = [ AI' 'V ] = [ML'T-2 ] = =0) least count sin = 1110 0.1 mm So dimensional formula of RT is same as that of energy. (ii) least count = sin = 0.511 0 = 0.05 mm For Problems 6-8 (iii) least count = sin = 0.5/20 = 0.025 mm 6. d., 7. c., 8. d (iv) least count = sin = 11100 = 0.01 mm Sol. 3. i. -+ d., ii. -+ a., c., iii. -+ b., iv. -)- c., d. Backlash enor is caused by loose fittings, wear and tear etc., in the screw mechanisms.



CHAPTER! Motion in One Dimension 4.1

4.2 Physics for IIT-JEE: Mechanics I FRAME OF REFERENCE poles. trees. etc. change their coordinates and we say that they are moving in train frame. Frame of reference is the frame in which an observer sits and makes the observations. Frame of reference is of two types: REST AND MOTION 1. Inertial frame of reference A body is said to be at rest when it does not change its position 2. Non-Inertial frame of reference with time; on the other hand, if the position of a body continually changes with time, it is said to be in motion. To visualise the A frame of reference which is either at rest or moving with state of rest or motion of a body leads us to the concept of constant velocity is known as inertial frame ofreference. A frame reference frame. Description of the state of a body requires a of reference moving with some acceleration is known as n011- second object. with respect to which the statc must be specified. inertial frame of reference (Fig. 4.1). oth~rwise it would have no sense. The inherent meaning of the above statement is that when we speak of a body in rest or in y motion we always say this in comparison with a second body, which is known as the reference body. To locate the position of z a body relative to the reference body, a system of coordinates fixed on the reference body is constructed. This is known as Fig. 4.1 reference frame. If two cars A and B move side by side in same To take the observations, we attach a Cartesian-coordinate system with the frame of reference. Cartesian-coordinate system direction with same speed, it would appear to the passengers of the cars that they arc mutually at rest with respect to each other. consists ofthrec mutually perpendicular axis x, y. and z, meeting Obviously relative to a reference frame on A, B is at rest. The reverse is also true. Absolute rest or absolute motion is unknown. at common point O. 0 is known as the origin. When a particle All motions are relative. moves in space, its position at an instant can be described with the help of its three position coordinates (x, y, z). which change Trajectory: The path followed by a point object during its with time during the motion of particle. At one time, one, two or all three position coordinates may change. Accordingly we motion is called its trajectory. Shape of path depends upon closer have three types of motions. reference frame. Motion in One Dimension Further, all the motions around us can he broadly divided into three types of motion viz., translatory motion, rotatory motion The motion of a particle is said to be in one dimension if only one and oscillatory motion, or a combination of them. out ofthe three coordinates specifying the position of the particle Translatory Motion changes with time, For example, motion along a straight line. When a body moyes in such a way that the linear distance COy- Motion in Two Dimensions ered by each particle of the body is same during the motion, then the motion is said to translatory or translation. The motion of a particle is said to be in two dimensions if any two out of the threc coordinates specifying the position of the A -------', object change with time. In such a mot.ion, the object moves in BO- ---\", , \"\" , a plane. For example, projectile motion, circular motion. \", '\" -0 Motion in Three Dimensions ... \" ...... The motion of an object is said to be in three dimensions if all the three coordinates specifying the position of the object \", DB'. change with respect to time. In such a motion, the object moves in a space. For example, random motion of a gas molecule. , -0 C There is no rule or restriction on the choice of a frame. We can C c' choose a frame of reference to describe the situation under study ..A:::-:::::::::::::::::::(;?::A' according to our convenience. For example, if we are in a train it is convenient to choose a frame attached to our compartment. ( ; ) .---------------------- B ff The coordinates of box kept on the bel1h do not change with time Fig. 4.2 and we say that the suitcase is at rest in train frame, but electric Translatory motion can again be of two types viz., curvi- linear or rectilinear, accordingly as the paths of every con- stituent pat1icles are similarly curved or straight line paths. Here it is important that the body does not change its orientation (Fig. 4.2).

Motion in One Dimension 4.3 Rotatory Motion /:).r = -;'2 - rl. Here -, is displacement vector from point When a body moves in such way that every constituent particle !.1r of it describes the same angular displacement about a particular axis of rotation then the mot.ion is said to bG rotatory or rotational I to 2. (Fig. 4.3). Displacement and Distance Axis of rotation Displacement is the vector drawn from initial position to final position and its magnitude is equal to the shortest distance be- tween the initial and final positions. To find displacement, we can use: s f fDisplacement, = vdt = +(vxi vy ] + v,k)dt Fig. 4.3 Distance is the length ofactual path travelled bv a body during Oscillatory Motion its motion in a given interval of time. To find distance, we can use: The motion in which an object repeats its motion about a fixed point either to and fro or back and forth is called oscillatory f fDistance, d = Ivl dt = {vf-;'-;;f+ v;dt motion. Such a motion always takes place within well defined boundaries which are called extreme points. • Distance is a scalar quantity and displacement is a vector quantity. POSITION VECTOR AND DISPLACEMENT VECTOR • Distance between a given set of initial and final positions can have infinite values but displacement is unique. Position Vector • Displacement can be negative, zero or positive, but dis- Let A is a point in space whose coordinates are (x, y, z), then tance is never negative. When a body returns to its initial we know that its position vector w.r.t. the origin of coordinate position of starting, its displacement is zero but distance or path length is not zero. rsystem is given by: = xi + y) + zk (Fig. 4.4). • Magnitude of displacement can never be greater than dis- J' tance, ~(X,);Z) • In uniform motion, displacement := distance, ,, (/ • Displacement is zero if particle returns to initial position, ,, .)' • Distance does not decrease with t.ime and never be a zero _ _o~:- --+'_-:~ x for a moving body. __________.. -:..-..IJ'\"/''' Z • Displacement can decrease with time, can be zero, or even negative if body is returning to its initial position, reaches the initial position and moves back from the initial position. So magnitude of displacement is not equal to the distance, however it can be so if the motion is along a straight line without change in dir~ction Fig. 4.4 Tn general distance?: displacement. Displacement Vector IJltlIflWW A particle is moving in a circle. What It is a vector joining the initial position of the particle to its final is its displacement when it covers (i) half the circle (ii) full position after an interval of time. Mathematically, it is equal to circle? the change in position vector (Fig. 4.5). Sol. (i) equal to diameter (ii) zero First we will discuss motion in one dimension ,. Sign convention: Any vector quantity directed towards right will be taken as positive and that directed towards left will be taken as negative (Fig. 4.6). Positive direction Negative direction f'ig .. 4.6 Fig. 4.5

4.4 Physics for IIT-JEE: Mechanics! Velocity (Instantaneous Velocity) fv;Magnitude of velocity (known as speed), v = + v? + v~. Time rate of change of position (x) or displacement (s )at any instant of time (t) is known as Instantaneous Velocity or simply Uniform Velocity Velocity at that instant of time. It is denoted by v (Fig. 4.7). A particle is said to be moving with uniform velocity if it covers equal displacements in equal intervals of time, however small . Displacement these intervals may be. Quantitatively, veloclty = --'-=::c.--- Uniform velocity is that velocity in which thc body continu- ously moves in the same direction with constant speed. TIme . (\"X)Mathematically, v = hm - = -dx or v = -dx 6HO \"t dt tit [alsov=(~:)] Voriable Velocity y A particle is said to be moving with variable velocity if it covers equal displacements in unequal intervals\".of time, or unequal The instantaneous displacements in equal intervals of time. Here velocity changes in either magnitude or direction or both. P, velocity\\1' is tangent to the,path at each point. Here\\7; and ~ Averoge Velocity ~ are the instantaneous It is the ratio ofthe total displacement (s) to the total time interval velocities at the points v, o Pi and Pz shown in the figure. (t) in which the displacement occurs. x s Total displacement =-= . Vav t Total time taken Fig. 4.7 Average velocity is a veclor quantity. Its direct.ion is same as Velocity is a vector quantity, it can be positlve, zem or neg- that of displacement. ative. According to our sign convention if the particle is going If at any time 11 position vector of the particle is \"I and at time towards right, velocity will be taken as positive and ifthe particle '2 '2 -t2 position is then for this interval, vav = J~ (Fig. 4.9). t2 - 11 is going towards left, velocity will be taken as negative. In general if the particle moves in space, then I' will change and time rate of change of position vector is known as velocity (Fig. 4.8). Thus v~ =til-' = -d [.x'i+y,}+zk'] The average velocity Vav between points PI and P2 dt dt has the same direction as the displacement D.r. tix ~ dy, tiz, =-l+-]+-k x dt tit dt Path => v= vxt + vy) + v.Je Fig. 4.9 To find average velocity we need to know only the total dis- dx placement from initial position to final position and need not where =-, consider the nature of motion between initial and final positions, Vx dt Physical meaning of average velocity: It is that uniform ve- locity with which if the object is made to move, it will cover tiy tiz the same displacement in a given time as it does with its actual v =-, Vz =- velocity in the same time. y tit dt Speed (Instantaneous Speed) )' The magnitude of the velocity at any instant of time is known -,., as Instantaneous Speed or simply speed at that instant of time. O~---+--,--,r, -+X Distance It is denoted by v. Quantitatively, speed = -ccc-- ___________, '.J,... ' ' z x Time Mathematically, it is the time rate at which the distance is z being travelled by the particle. Fig. 4.8 Note: Velocity is f{lso defined as. ti\"\"e rate of clutngeof \"- . . . . .~..ds d. lSPtaC~ment,v= dr'

Motion in One'Dimension 4.5 • Speed is a scalar quantity. It can never be negative. Features of Uniform Motion • Instantaneous speed is the speed of a particle at a particular • Uniform motion is a straight line motion with constant instant of time. velocity. • In uniform motion displacement and distance are equaL Uniform Speed • The average and instantaneous velocities have same values in uniform motion. A particle is said to be moving with unifonn speed if it covers equal distances in equal intervals of time, however small these • No net force is required for an object to be in uniform intervals olay be. motion. Uniform speed is the speed which always remains constant. • The velocity in uniform motion does not depend upon the Here the body may change its direction of motion. time interval. Variable Speed • The velocity in uniform motion is independent of choice of origin, A particle is said to be moving with variable speed if it covers equal distances in unequal intervals oftime or unequal distances Unit of velocity or speed: ms- 1 in SI system and cms- I in in equal intervals of time, COS system. Average Speed Dimensional formula of velocity or speed: [MoLT-I] It is the ratio of total distance (d) travelled by the particle to the IliiMIiRtHiiti A car travelled the first third of a dis· total time taken (t) in which this distance is travelled. tance d at a speed of 10 kmh-1, the second third at a speed of 20 kmh-1 and the last third at a speed of 60 kmh-1. De- termine the average speed of the car (Fig. 4.11). Total distance d • , ,dl3 dl3 dJ3 Vav = =- A 10 kmh-1 B 20kmh- 1 C 60 kmh-1 D . Total time taken t I, 12 I) If motion takes place in same direction then average speed and Fig. 4.11 average velocity are same. In Fig. 4.10, a particle goes from A to C. Distances, velocities Sol. Average speed = T-o~t-aTl~od-tias-lt-tainm-c-ee-ttar-ak.veen:l.le.d and time taken are shown. V, \\'2 C AB+BC+CD dj3+dj3+dj3 A SI B 82 II + 12 + 13 = d j3 d j3 d j3 10+20+60 I, I, d Fig. 4.10 = -d'---cd;----dc, = 18 kmjh SI . 30 + 60 + 180 II = - ; S, = V212 In@JiiwiHdl A ship moves due east at 12 kmh-I for VI 1 h and then turns exactly towards south to move for an hour at 5 kmh-I. Calculate its average speed and average velocity + +Vav =S-I -S-2 = VItI v,t, (1) for the given motion (Fig. 4.12). + +tl I, t, tl Sol. Time taken to complete the journey = 2 h SI +S2 (2) -SI +S-, Vj V2 +1. 12 km Iftl = 12 .= I, then v\" = -VI -2-V2: average speed • equal to IS '~'5km arithmetical mean of individual speeds. S, then v\" 2VtV2 . bkm , 2. If SI = S2 = = +- - - : average speed IS equal C VI V2 to harmonic mean of individual speeds. Fig. 4.12 UNIFORM MOTION Average speed = Total distance travelled 12 +5 Total time taken 2 A particle is said to be in unifonn motion if it covers equal displacements in equal intervals of time, however small these = 8.5 kmjh intervals may be. = =Total displacement travelled 13 Average velocity Total time taken 2 = 6.5 kmjhr

4.6 Physics for IIT-JEE: Mechanics I D'lrectl.Oo 0.f' averalge ve'DeIty: tan e = 152 a = lim (L'.V)· = dv 61-+0 6.t dt 8 = tan-I (152) = 22°37' S ofE vIn general, = vxl + vyi + v): )' ACCElERATED MOTION ()I)<CC------+--~ X The motion of an object is said to be accelerated motion if its ______, ':.-J,\" velocity changes with time. x Example: A vehicle moving on a crowded road. An accelerated motion is a kind of non-uniform motion. Average Acceleration Fig. 4.15 It is the ratio of total change in velocity to the total time taken vDifferentiating w.r.t. time, we get acceleration: in which this change in velocity takes place (Fig. 4.13). Total change in velocity L'. v a- = d-dvt = -ddt [v.,') + vy,j + v~, k'] =- aav= L'.t, Total time taken dvx ~ dvv ~ -I j -dvkz \" a,\"i + \" \"- a+ - ' +a =-> = ayj + azk v dt dt dt =} l+M ddvxt ' Gz = ddvtz dt'. dv\" where ax = Fig. 4.13 , ay = To find average aeeeleration, we need to know only the total Ja,,~-+- - -- - change in velocity from initial position to final position. We + a aJ. aiMagnitude of acceleratioIlj= need not consider how the motion takes place between these aInstantaneous acceleration at point Pis shown in Fig. 4.16: two points. Vector\": is tangent to the path; vector {{ points toward the con- If velocity of the'-p~aftiCIe at an instant t1 is VI and at instant 12 is V2l then the average acceleration is mathematically given by ecave side of the path, 0 :::: ~ 7T ~ v, . .. V2 - Vl 6.v (FIg. 4.14): a\" = -t2 --t, =L-'.t vr , r v, P, r P, v, P, r (b) Fig. 4.16 a~l' (a) • a = ddvt = dd2tx2 ' A'cccleratl.Oll.IS a1so dcfi ned as second Fig. 4.14 time derivative of position. dv Change in velocity = final velocity - initial velocity • Also, a = Vd-x. Acceleration (Instantaneous Acceleration) • Acceleration is a vector quantity. It can be positive, zero Time rate of ehange of velocity at any instant of time is known or negative. dv • Acceleration is in the direction of increasing velocity, e.g,. as Instantaneous Acceleration or simply Acceleration. a = - . let a particle is going from A to B. In both the cases the dt particle is going towards right, so velocity in both the cases is positive at any instant of time. In first case velocity is Actually, acceleration at an instant is defined as the limit of the increasing towards right, so acceleration is towards right average acceleration as the time interval At around that instant and positive. In second case, velocity is decreasing towards becomes infinitesimally small (Fig. 4.15). right (or increasing towards left), so acceleration is towards left and negative (Fig. 4.17). • Negative acceleration is also known as retardation or de- celeration. Deceleration and acceleration have opposite directions.

Motion in One Dimension 4.7 4 ms-1 point B is v. Let the particle covers a displacement s during this motion with constant acceleration a (Fig. 4.19). --<>.------<>-.-9-. case (1) case (II) AB 4 ms-- 1 2 m:( 1 - _ _ . . - - - - _ . .- 9 - . Constant accc,leration AB I~O I~f Fig. 4.17 A -.... -----i.>__----_e.. B Uniform Acceleration u 111.-----s-----~.1 v Fig. 4.19 An object is said to be moving with uniform acceleration if equal 1. Velocity-Time Relation change in velocity takes place in equal intervals of time, however small these intervals may be. V-U => + at rn uniform acceleration velocity changes with a uniform rate. -- a= t-O v = u (1) (2) Variable Acceleration A particle is said to be moving with variable acceleration ifits ve- 2. Displacement-Time Relation locity changcs equally in unequal intervals of time, or unequally in equal intervals of time. tV:iv= s =u-+2v Average Velocity in Uniformly Accelerated Motion s u+v Total displacement , 2 Generally, Average velocity = , but it the mo- +I 2 Total time taken tion is uniformly accelerated, then average :,ciocity can also be S = lit 'iat written as Initial velocity + Final velocity u + v 3. Displacement-Velocity Relation 1. 2 2 Average ve (lClly = (For constant acceleration) = vavt = (u + v) (v - u) :: v2 u2 - S , 2 a 2a A uniformly accelerated body in a =} v2 - u2 = 2as straight line has a velocity of' 2 ms- 1 at any time. After some (3) time its velocity becomes 4 ms-I . Find the average velocity. The above equations (I), (2), and (3) are known as equations u+v 2+4 _I ojmotiof1 for uniform acceleration. -2- Sol. v,\" = -2-= 3 ms A particle moving with uniform acceler- A car travelling at 20 ms- I takes a ation from A to B along a straight line has velocities VI and V2 at A and B, respectively. If C is the mid-point between A U-turn in 20 s without changiug its speed. What is the aver- and B then determine the velocity of the particle at C. age acceleration of the car? \"•, \"• •\"2 Sol. Initial velocity = VI = +20 mis, AC B Final velocity = V2 = -20 mls •x • f • change in velocity V2 - Vj Fig. 4.20 Average acceleratIOn = ._- = --- time taken t -20-20 , Sol. Let v be the velocity of the particle at C. Assume accel- = 20 = -2 mis' eration of the particle to be a and distance between A and B to be x. To find the velocity at point C. Consider the motion from A to C: ,x Apply v2 - u2 = 2as. we get v2 - vi = 20- 2 => v2 - vf = ax (1) --20 ms- l Applying the same equation from C to B, we get Fig. 4.18 vi - v2 x =2a- Negative sign indicates that average acceleration is towards 2 left during the time interval given. 2 V2-_ ax (2) Formulae for Uniformly Accelerated Motion in a rT;;r+;;rV2- Straight Line YFrom equation (I) and (2), we get v = Let a particle starts from point A at t = () and reaches at point B at t = t. Initial velocity -at point A is u and final velocity at Two trains P and Q moving along par- allel tracks with same uniform speed of 20 mls. The driver of train P decides to overtake train Q and accelerates his

4.8 Physics for IlT·JEE: Mechanics I train by 1 ms- 2• After 50 s, the train P crosses the engine of 4 the train Q. Find out what was the distance between the two D\" = 10 + 2[2 x 5 - l] =} D\" = 10 + 18 = 28 m trains initially provided the length of each train is 400 m. A particle starts from rest with a con- Sol. Let the initial distance between two train is x. =stant acceleration a I mis', Distance travelled by train P in 50 s: SF' = ut + I 2at2 = 20 x 50 + 2I x I x , = 2.250 m 1, Determine the velocity after 2 s. 50- 2, Calculate the distance travelled in 3 s, 3, Find the distance travelled in the third second, Distance travelled by train Q in 50 s: 4, If the particle was initially moving with a velocity of SQ =ut = 20 x 50= 1000m 5 mis, then find the distance travelled in the 3,d second, Now, SI' - SQ = 2250 - 1000 = 1,250 m. This distance Sol, must be equal to initial distance between the trains plus the sum of length of two trains. =} x + 800 = 1250 =} x = 450 m 1. We need velocity~time relation to solve this question. Displacement Travelled by a Particle in nth Second of Formula used, v = u + ai, So v = 0 + (1)(2) = 2 m/s its Motion in Uniformly Accelerated Motion 2, Here u = 0, I = 3 s. a = 1 mis', Let a particle starts from a point A at t= 0 and travels along the s = ut + 2I 'lI' = 0 x 3 + 21 x 1 X (3)2 = 4.5 m straight line ABC . .. DEI-: 3, D\" = u + a - 1) =0 + I x 3 - I) = 2.5 m 2(2n 2(2 At t = l.s the particle arrives at point B, at t = 2s the particle arrives at point C and in the last, the particle arrives at point 4. Here 1I = 5 mis, F at t = ns as shown in Fig, 4.21. OUf aim is to calculate the a I 2(211 2(2 displacement in the nth second (say D,,) which is equal to E F D\" = u+ - 1) = 5+ x 3 - 1) here. EMFa =7.5m A B CD Consider a particle initially moving with 0- - - - - - - -------8 • • [=2s t \"~x-,2 t~\" x---J I=x a velocity of 5 mls starts decelerating at a constant rate of 2 mls'. /=15 u Fig, 4,21 1. Determine the time at which the particle becomes station- ary, Let a = uniform acceleration of the particle, 2, Find the distance travelled in the 2nd second. u == initial velocity of object at point A Now, EF = AF - AE (I) Sol, Here u ~ 5 mis, a = -2 mis' A F is displacement travelled in t = n seconds, so 1, v = u + at =}. 0 = 5 - 21 =} t = 2.5 s AF = un + _I an2 2, D\" = u + a - 1) = 5 - 22 [2 x2 - 11 = 2m 2 2(2n A E is displacement travelled in t = n - 1 seconds, so ,------1 Concept Application Exercise 4.1 f-----, . I AE = u(11 - 1) + 2a(n - I)' Put the values of A F and A E in (l), we get 1. State the following statements as True or False: For ordinary terrestrial experiments; EF = Iun + I [urn - 1) + I - I)'] a. a child revolving in a giant wheel is a non-inertial ob- 2GlZ']- 2a(n server. b. a driver in a sports car moving with a constant high After simplifying, we get E F = u + ~(2n - 1) speed 0[200 kmlh on a straight road is in inertial frame. c. the pilot of an aeroplane whieh is taking off is a non- D\" = u+ a - 1) inertial observer. 2(2n d. a cyclist negotiating a sharp turn is an example of iner- tial frame. A particle having initial velocity 10 ms- i e. the guard of a train which is slowing down to stop at a station is an inertial frame of reference. is moving with constant acceleration 4 ms-2• The particle is 2. a. If velocity of a body is zero, does it mean that acceler- moving in a straight line, Find the distance travelled by the ation is also zero') (YeslNo) particle in 5th second of its motion, Sol, Given: u = 10 ms- i , a = 4 ms-'. n = 5 h. If acceleration of a body is zero, docs it mean that ve- Iocityis also zero? (Yes/No) T• D\" = u + 2a (211 - I), USll1g

Motion in One Dimension 4.9 c. If a body travels with a uniform acceleration ([I for time 11. The displacement of a body is given to be proportional to II and with uniform acceleration ([2 for a time 12, then square of time elapsed. What is the nature of the acceler- average acceleration is given by ation (constant or variable)? +{l! I} ([2t2 12. A car starts form rest and accelerates uniformly for 10 s to a = +11 12 . (Yes/No) a velocity of 8 m/s. It then runs at a constant velocity and is finally brought to rest in 64 m with a constant retardation. d. If a body starts from rest and moves with a uniform The total distance covered hy the car is 584 m. Find the acceleration, the displacements travelled hy the body value of acceleration, retardation and total time taken. in 1 s, 2 s, 3 s, ... , etc. are in the ratio of I : 4 : 9 ... , etc. (True/False) 13. A body covers 10111 in the 2nd second and 25 111 in 5th sec- ond of its motion. If the motion is uniformly accelerated, c. For a body moving with uniform acceleration, the dis- how far will it go in 7th second? placement travelled by thc body in successive seconds is in the ratio of 1 : 3 : 5 : 7... (True/False) 14. A body moving with uniform acceleration in a straight line describes 25 111 in 51h second and 33 m in 7 th second. 3, Say Yes or No: Find its initial velocity and acceleration. a. Can an object moving towards north have acceleration towards south? 15. Two trains, each of length 100 m, moving in opposite directions along parallel lines, meet each other with speeds p. Can an object reverse the direction of its motion even of 50 kmlh and 40 km/h. If their accelerations are 30 em/s2 and 20 emls', respectively, find the time, they will take to though it has constant acceleration? pass each other. c. Can an object reverse the direction of its acceleration USE OF DIFFERENTIATION AN DINTEGRATION even though it continues to move in the same direction? IN ONE-DIMENSIONAL MOnON d. Average speed is the magnitude of average velocity. e. At any instant of time, direction of change in velocity Let a particle is moving from point;\\ to B. Suppose the particle takes sorne finite time !:1t to cover a finite distance 6.x between and acceleration direction are different. points C and D (Fig. 4.23). It is not known that how the particle has travelled from C to D. Particle may have travelled with 4, Can a body have uniform velocity or with variable velocity. Acceleration mayor may not have been constant during the motion from C to D. But a. zero instantaneous velocity-and yet be accelerating? certainly we can write the average velocity from C to D. b. zero average speed but non-zero average velocity? Tt6X c. negative acceleration and yet be speeding up? d. magnitude of average velocity be equal to average Vav = speed? 5. A body covered a distance of 5 m along a semicircular path. Find the ratio of distance to displacement. 6. A particle travels tram point A to B on a circular path of radius IShr em. If the arc length AB be 10 cm, find the displacement. AB Fig. 4,22 ~ • ~A C D B • D.X • 7. A body moves at a speed of 100 mls for lOs and then 6t moves at a speed of 200 111/s for 20 s along the same Fig. 4,23 direction. The average speed is _ __ 8. A body moves in the southern direction for lOs at the speed Now what happens if 61 approaches zero, or in other words. of 10 m/s. It then starts moving in the eastern direction at !:1t is infinitesimally smalL When we say !:1t approaches zero, it the speed of 20 m/s for lOs. The magnitude of average does not mean that J':...t is equal to zero. It means that J':...I is very- velocity is __._,\"',._.__. The average speed is . ,_,_ _ _. The very close to zero. If Ell is very small. then distance covered total displacement will be _____. during this time will also be very small. So J':...x also approaches zero and then points Cand J) will lie very close to each other. 9. A body moves from (3, 2) to (7, 6). The initial and final Let us write, J':...t = dt when J':...t approaches zero and J':...x = dx position vectors are __ and . The displace- when £:\"x approaches zero. ment vector is .____ . 10, A car travelling at 108 kmlh has its speed reduced to 36 Then average velocity becomes Vav = 6X = d~< km/h after travelling a distance of 200 m. Find the retar- - dation (assumed uniform) and time taken for this process. 61 ell We can aIso wn·te Vav = lI.m (6-X- ) =e-lx. ~\\t\",,)O b.l dl

4.10 Physics for IIT-JEE: Mechanics I When b.t approaches zero then average velocity becomes instantaneous velocity. This we can explain as follows: Since points C and Dare very close to each other, we can neglect any u2 change in velocity from C to D , if there is any. We can assume At x = 0, 11 = u. This gives C3 = v2 that velocity remains constant for the motion from C to D. Then 2 -u 2 the average velocity becomes instantaneous velocity or simply - = ax + ::::} v2 = u2 + 2 ax (iii) 22 velocity, So v = Vuv when 6.t -+ O. or we can si.mpIy wn.te.Instantaneous veloci.ty as: Ii = -dx, The displacement of a body as a func- dt tion of time is given by s = (31 2 + 4t + 7 ) m. Calculate the It can be stated as follows: Time rate of change of position magnitude of its instantaneous velocity and acceleration at is equal to in.vtantaneous velocity or simply veloci1y. 1= 1 s. Similarly we can define instantaneous acceleration or simply acceleration as time rate of change of velocity. Sol. Given s = (3t 2 + 4t + 7) m. Velocity is the time rate of dv change of displacement. Cl=- ....,. ds d 3 1 2 + 4t + 7) = 6t + 4. So 1 v 1 = -dl = dt -( Also. dv dt a=v- At! = 1 s, v = 6 x 1 + 4 = 10 mls dx We have seen that by using differentiation, we can find veloc- Now acceleration is the rate of change of velocity: ity by differentiating position, and acceleration by differentiating _ I Ilal= d- v =-d(6t+4)=6m/s-3. dt dt velocity. We can do the reverse also. We can find velocity from acceleration and position from velocity. Here we will have to usc integration. Velocity is given by integration of acceleration with respect The acceleration ofa particle varies with to time. fv = adt +C time I seconds according to the relation a = 6t + 6 mls'. Find So, the velocity and position as a function of time. It is given that the particle starts from origin at t = 0 with velocity 2 ms-I. Position is given by integration of velocity with respect to f lv ftime. So Sol. We know that.l dv = adt -'} dv x = v dt + C l'= (6t + 6)dt Here Cis known as constant of integration. Its value depends upon the given conditions. Its value can be different for different => v - 2 = [6t2 + 6t]' => v = 3t 2 + 6t +2 conditions. 2 () Derivations of Equations of Motions by Calculus Method f f f fx I Let a particle starts moving with velocity u at time t = 0 along now dx = vdi => dx = (3t 2 + 6t + 2) dt a straight line. The particle has a constant acceleration a. () () Let at f = t, its velocity becomes v and it covers a displace- ment of s during this timc. => x = t' + 3t2 +2t v= fadt+Clorv=at+Cl At I = 0 a body is started from origin 'At t = O. v = u. This gives C1 = U with some initial velocity. The displacementx(m) of the body Putting CI, we get v = u + at varies with time 1(5) as x = -(2/3)t2 + 16t + 2. Find the ini- f fx = v cli + C2 => X = (ll + at) dt + C2 tial velocity of the body and also find how long does the body take to corne to rest? What is the acceleration of the body when it comes to rest? (i) Sol.vd=x -=d- ( --2r+3 16t+2) =--4t+16; dt dt 3 3 at t = 0, x = f u dt + f at dt + C2 = ut + ~at2 + C2 4 11 = -- x 0+ 16= 16m/s 3 At t = O. x = O. This gives C2 = 0 . V = O. we get, 4 + 16 = 0 => t =12 s Putt111g --t 3 . C2 • we get x = ut + 21at 2 (ii) Hence the body takes 12 s to come to rest (momentarily). Puttmg Nowa=dv = d (-~t+16)=-~m/s2 f fvdd-xv=a => dt dt 3 3 vdv= adx+C3 We sec that acceleration is constant, so when the body comes ,4 to rest its acceleration is - - m/s2 . 3

Motion in One Dimension 4.11 ,-----1\\ Concept Application Exercise 4.2 If-----. 1. The position x of a particle varies with time t according to -T11 Displacement s will .1 u be taken as + If the relation x = t3 + 3t2 + 2t. Find velocity and acceler- Dispiaccrncnts will ation as a function of time. be taken us·- H 2. The displacement of a particle along x-axis is given by If Inithll velocity will If Initial vcloeity will be taken as -I- u be taken as +11 x = 3 + 8t.+ 7t 2 Obtain its velocity and accelerfltion at Au Acceleration will Acceleration will be t = 2 s. be taken as -- g l',,',*,=im taken as g 3. The acceleration a in ms~2 of a particle is given by Ground Ground (l = 3t2 + 2t + 2, where t is the time. If the particle starts (ii) out with a velocity v = 2 ms·- I at t = 0, then find the velocity at the end of 2 s. --r -Displacements will 4. The displacement x of a particle along the x-axis at time t U be taken as -- fI II Initial velocity wl1l I.S gI.Ven byx = -alt + -(l2[ 2. FI' nd the aceeicra'tton 0f the be takcn as - Ii Acceleration \\vl11 be 23 particle. ~B,*,=1m taken as K 5. A particle moves along a straight line such that its displace- (iround .ment s at any time 1 is given by s = t 3 - 6t 2 + 31 + 4 m, (iii) t being in seconds. Find the velocity of the particle when Fig. 4.25 the acceleration is zero. ONE-DIMENSIONAL MOTION IN A VERTICAL When the motion takes place uncler the effect of gravitational LINE (MOTION UNDER GRAVITY) attractivefon;e only, the motion is known as Free Fall. Here free Sign convention: Any vector quantity directed upward will be fall docs not mean that the particle is falling down only. Even if taken as positive and directed downward will be taken as negative (Fig. 4.24). the particle is rising up or is momentarily at rest at highest point, jNegalive but if only gravitational force is acting on it, then motion will Direction be called as Free Fall. ll! equations of motion we replace a by - g (minus sign be- cause acceleration is always directed downward), weget: {;:::t~g~gt2} v2 = u2 - 2gs Posilive Some Formulae 1Direction 1. A body is dropped from a height H. What will be the time Fig. 4.24 taken by the body to reach the ground and vcloeity just before reaching the ground? According to this sign convention:' 11\"'0 1\". s will be taken as positive if final position lies above ini- tial position and negative if final position lies below initial II position. 1 2. velocity(initial or final) will be taken as positive ifitis upward and negative if it is downward, \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ 3. acceleration a is always taken to be - g, Ground Illustrations to lise sign convention: Look at three different Fig. 4.26 situations in Fig. 4.25 given below. In each of thc situations a Ans: T = /2;, V = ..j2g H pru-ticle is projected from a point A with initial velocity u and the particle reaches point B after some time, Proof: Let time taken to reach the ground is T and velocity on Earth attracts each and every' object towards its centre. As a reaching the ground is v, We have u = 0, s = - H, t = T result of this attraction a constant acceleration is produced in the objects lying close to the earth's surface. This acceleration is 2using s= ut 1 2, we have always directed downward. This acceleration is called accelera- tion dueto gravity. It is denoted by g. For tl,e bodies lying close gt to the earth's surface, value of g = 9.8 ms-- 2 and sometimes to -H=OXT-~gT2=}H=~gT2 =} T=j2; make the calculations easy we take g = to ms- 2 . Hence time taken to reach the ground, T = /2;

4.12 Physics for IIT-JEE: Mechanics I Note: Time offlight is illdepelldellt afmass, shape and Here time of flight is the time for which the particle remains in air. size ofthe bpdy ill vaCUUIn, But ill air iftwo. bodies ofsame Alternative method to find the time of flight: Consider the motion from A to A: 11lasSilllddifferellt size.ore fallell, then the bodyhavillg s = 0 (initial and final both points arc same), moreyolullle will take.lIlore time. . initial velocity:;;;: u, time taken:;;;: T Now using v = u- gt, we have 11 Using s = ut - 2,ft{2, we have 0 = uT - 2gT2 V=0--gT='_gJ2H 2u g T=- =} v = - ,j2g H, negative sign indicates that velocity is in g downward direction. (b) Magnitude of velocity on returning to the ground will be Hencc velocity on rcaching the ground: v = ,j2gH in down- same as that of initial velocity but direction wi1l be opposite. ward direction 2. A particle is projected vertically upward from ground with Proof: Let v is. the velocity on rcaching the ground. Then from the initial velocity u, previous formulae: a. Find the maximum height, H, the particle will attain and time 2 T that it will take to return to the ground (Fig, 4.27), v = -,j2gH = -/2g lI =} v = -1/, hence proved b. What will be velocity when the particle returns to the ground? \\ 2g , c. What will be the displacement and distance travelled by the particle during this time of whole motion, (c) Displacement travelled\", 0, distance travelled = 2H = u- Ii ~ A body is dropped from a height of80 m. Find the time taken by tbe body to reach tbe ground and =velocity on reaching the ground. Take g 10 m/s2 • Sol. Given u = 0, H = 80 m. ff-H /2 x 80 Time taken to reach the ground: T = - = ,I - - - = 4 s g V 10 VelocilY on reaching the ground: Fig. 4.27 v = ,j2g H = ,,12 x 10 x 80 - 40 mls u2 2u A ball thrown up is caught hy the Ans: (a) H = -, T = - 2g g Proof: Consider the motion from A to B: thrower after 4 s. How high does it go and with what ve- s = +H (final point lies above the initial point), initial veloc- locity was it thrown? Take g = 10 ms-2• = =ity u, final velocity v 0 2u Sol. Given T = 4 s, We know that T = - = = -Let tim.e taken to go from A to Bu = fl. Using v = u - gt g =} 0 u - gt, =} I, g =2u H=U~-~g(~y'2. =} 4 = - =} u 20 Illis 10 Usmg s = ut - I gt 2 u2 202 =} Also H = - = --=20m 2g 2 x ]() . u2 From the top of a huiiding 160 m bigh, =} H=- a ball is dropped and at the same time a stone is projected 2g vertically upward from the ground with a velocity of 40 ms-1 So . , .. H = -u2 (Fig. 4.28). Find wheu and where the ball and the stone will maxImum heIght attamed IS 2g meet. Take g = 10 ms-2• consider the return motion from B to A: s ~ - H (final point lies below the initial point) 1/ = 0 (at point B velocity is zero) Let time taken to go from B to A = t2' We have t: If;:t2= = =~ Here tJ is known as time of ascent and t2 is known as time of 160 III descent. We can see that Tfme of ascent = Time of descent :;;;: ~, g Total time of flight: T = t, + t2 = 2u Fig. 4.28 g

Motion in One Dimension 4.13 Sol, Let they meet after time t at height x trom ground. (l) u\"\" 29.4 !]lIs For ball: s = -(160 - x), u = 0, a = -g, then A -(160-x) =0 x t - '12gt' =} 160-x = '12gt2 = =For stone: s =.X, u -g, then x = 40t - 1 2 (2) 34.3 m 40 mis, a _gt 80m = =adding equations (1) and (2): 160 4t =} t 4 s 2 8- -r h put the value ofl in (1): 160-x= ~ x 10x4' C tH 2 h, =} x=80m Fig. 4.29 So they meet in the middle of the building at I = 4 s =} t = 8s A body is projected vertically upward. 1 If /1 and /2 be the times at which it is at a height h above Now from (2), h = '2(9.8)(8 - 5)'= 44.1 m. From figure H = 80 - 34.3 = 45.7 m the point of proje~tion while ascending and descending, re- So finally: h, = H - h = 45.7 - 44.1 = 1.6 m spectively. Tben prove that initial velocity of projection of ,..------i Concept Application Exercise 4,3 f-------, tbe body is ~g(tl + (2) and the. value of h is ~gtlt2' 2 2 Sol. Let the body is projected with initial velocity u and let at any time t the body is at height h from the point of projection. There will be two possible values of I which are I, and t2 as 1. i. State the following statements as True or False: a. A ball thrown vertically up takes more time to go up given in the question. than to come down. Now we can write h = ut - ~ gl' =} gl2 - 2ut + 2h = O. b. If a ball starts falling from the position Of rest, then it .. 2 travels a distance of 25 m during the third second of ThIS 18 a quadratic equation having two roots. its fall. Sum of roots: tl + t2 = - -2u =} -- c. A packet dropped from a rising balloon first moves g upwards and then moves downwards as observed by a 2h stationary observer on the ground, Product of roots: t! t2 = - d. In the absence of air resistance, all bodies fall on the g A balloon is at a height of 40 m and is surface of earth with the same rate. ii. Fill in the blanks ascending with a velocity of 10 ms-I. A bag of 5 kg weight a. When a body is thrown vertically upward, at high- is dropped from it. When will the body reach the surface of est point (both velocity and acceleration are the earth? Given g =10 mls'. zer%nly velocity is zer%nly acceleration is zero), Sol. Given s = -40 m, U = 10 mis, The initial velocity of the bag will be same as that of velocity of balloon. b. If air drag is not neglected, then which is greater: time Usm. g s = ut - 21. gr, , we have of ascent or time of descent? -40 = lOt - ~IOI' =} 512 - lOt - 40 = 0 c. A body is projected upward. Up to maximum height, =} (t-4)(5t+lO)=0 =} t=48,-28 time taken will be greater to travel _ _ _.(first Neglecting negative time, time taken by body to reach the half/second half). 2. A ball thrown up from the ground reaches a maximum =ground is t 4 s. height of 20 m. Find: a. its initial velocity. From a point A, 80 m above the ground, b. the time taken to reach the highest point. c. its velocity just before hitting the ground. a particle is projected vertically upwards with a velocity of d. its displacement between 0.5 sand 2.5 s. e. the time at which it is 15 m above the ground. 29.4 ms-I. Five seconds later another particle is dropped 3. A balloon is rising up with a velocity of 10 ms- 1 and a from a point B, 34.3 m vertically below A. Determine when and where one overtakes the other. Take g = 9.8 ms-2• bag is dropped from it when its height from the ground is 40 m. Calculate the time taken by the bag to reach the Sol. Let at time t, they reach at level C. Then ground. For A: -(h + 34.3) = 29.4t - ~gt2 (1) 4. A body is projected from the bottom of a smooth inclined plane with a velocity of 20 m/s. If it is just sufficient to and (2) carry it to the top in 4s, find the inclination and height of For B: -h = - '218 (t - 5)2 the plane. From equations (2) - (1) =} 34.3 = -29.41 + ~ g[t2 _(t 2 - 5)1 2 S. A ball is dropped from an elevator at an altitude of 200 m =} 7 = -6t + (2t - 5)5 (Fig. 4.30). How much time will the ball take to reach the ground if the elevator is

4.14 Physics for lIT-JEE: Mechanics I Elevator , Elevator-+- Elevator How toAnaLyse the Graphs and How to Draw the t o o ...,e ll =9.8ms-1 Graphs F r~o 'u=9.8ms-1 196m ,,I,t,. In one dimensional motion, generally, we corne across , position-time (or displacement-lime) graph, velocity-tirne ,, t, graph, acceleration-time graph, etc. J///v/m;;ll /T///T/T//T!//I /II/T/TI//lllII Whenever we draw a graph, we need an equation involving (i) (ii) (iii) the variables between which we have to draw the graph. For Fig. 4.30 e.g., to draw position-time graph we generally use the equation a. stationary? x = ut + 1 2 , to draw velocity-tinle graph we generally usc h. ascending with velocity 10 m/s? _([t c. descending with velocity 10 m/s? 2 the equation v = u + at, etc. Note that we can use these relations 6. A particle is projected vertically upwards. Prove that it will be at 3/4 of its greatest height at times whieh are in on/v when acceleration is constant. the ratio 1:3. Position-Time Graph of Various Types of Motions 7. A balloon rises from rest on the ground with constant of a Particle acceleration g/8. A stone is dropped from the balloon Partide is Stationary when the balloon has risen to a height of H. Find the time Let a particle be at some point P at time t= 0 which is at a distance taken by the stone to reach the ground. Xo from origin. Since the particle is stationary, so at any further time the particle will remain at point P. Hence position-time S. A parachutist after bailing out falls 50 m without friction. graph for a stationary particle is parallel to tiITle axis (Fig. 4.31). When parachute opens, it decelerates at 2 m/s2, He reaches the ground.with a speed of 3 m/s. At what height did he o bail out? Fig. 4.31 9. A ball is dropped from the top of the tower of height h. It covers a distance of hl2 in the last second of its motion, Particle is Moving with Canstant Velocity Towards Right How long docs the ball remain in air? Equation to be uscd: x = Xo + vt. Graph wi II be a straight line. 10. When a ball is thrown up, it reaches to a maximum height Let the particle be at some point P initially at time t = 0 which h travelling 5 m in the last second. Find the velocity with which the ball should be thrown up. is at a distance of Xo from origin. Since the patticle is moving towards right so its distance from origin goes on increasing. GRAPHS IN MOTION IN ONE DIMENSION Hence position-t.ime graph for a particle moving with constant velocity towards right will be a straight line inclined to time axis Graphical analysis is a very potential method of studying the making an acute angle a (Fig. 4.32). motion of a particle, Indeed the method of analyzing situations graphically can be effectively applied not only to motion but to Recall that tan 0' is slope of position-time graph which is any field. The variations of two quantities (related) with respect equal to velocity of the particle. to each other can be demonstrated by means of a graph. The greatest advantage of depicting variables by means of curves (I P lies in the fact that the whole situation can be understood at any instant. In other words, graphs reveal much more than .that Fig. 4.32 revealed by a table. In this section. we will study and interpret various types of graphs related to motion such as displacement- Particle is Moving with Constant Velocity Towards Left time graph, velocity-time graph, etc. Equation to be used: x = Xo ~ vt. Graph will be a straight line. For graphical representation, we require two coordin'l-te (ref- Let the particle be at some point p. at time t = 0 which is at a erence) ({xes, one variable being taken along one axis. The usual practice is to take the independent variable along x-axis and the distance of Xo from origin. Since the particle is moving towards dependent on along y-axis. In general cases involving time as left so first its distance from origin goes on decreasing and then one of the variables, time, being independent, is usually taken its distance from the origin goes on increasing in negative di- along x-axis. rection. Hence position-time graph for a particle moving with constant velocity towards left will be a straight line inclined to' Graphs play very important role in analyzing a motion. Some- time axis making an obtuse angle a (Fig. 4.33). times it becomes difficult to solve the problems analytically. But with the help of graphs we can solve the problems very easily and without much calculation.

Motion in One Dimension 4.15 Velocity-Time Graph of Various Types of Motions of a Particle op Particle is Moving with a Constant Velocity Since velocity is constant, so a = O. Fig. 4.33 Let at any time velocity of particle is u. Since velocity remains Here tan a, the slope. will be negative which indicates nega- ·constant, so at any time velocity remains same. Hence velocity- tive velocity. time graph for a particle moving with constant velocity is a Particle is Moving with Constant Acceleration Directed straight line parallel to time axis. Rightward (Positive Acceleration) Cu a0 Since acceleration is towards right so velocity increases in ----+u right direction. Hence slope of position-time graph goes on increasing. o Equation to be used: x = ut + ~at2 t Graph will be a curved line (parabolic) Fig. 4.36 VI = tanal. V2 = tanal Particle is Moving with a Constant Positive Acceleration It is clear from Fig. 4.34 that as time passes, slope goes on increasing. Equation to be used: v = u + at =>Now C¥z > al =} tanal > tanal V2 > VI As the time passes, velocity goes on increasing. Hence velocity-time graph for a particle moving with constant posi- SO velocity goes on increasing. tive acceleration is a straight line inclined to tillie axis making an acute angle a. Here tan a is the slope of velocity-time graph x (Fig. 4.37). 1'). --a-]---_. -v, v, Note that the slope of velocity-time graph is equal to accel- eration. I t, t, \\'~ > VI to ----+a Fig. 4.34 a tan (X Particle is Moving with Constant Acceleration Directed Fig. 4.37 Leftward (Negative Acceleration) Particle is Moving with a Constant Negative Since acceleration is towards left so velocity decreases in right Acceleration (Retardation) direction and increases in left direction. Equation to be used: v = u + at . be used: x = at + 2I at2 As the time passes, the velocity goes on decreasing, at time EquatIOn to t = to velocity becomes zero and after that it increases in nega- tive direction. Hence velocity-time graph for a particle moving Graph will be a curved line (parabola) with constant retardation is a straight line inclined to time axis making an obtuse angle Ci (Fig. 4.38). VI = tan a!, V2 = tana2 Since 0' is obtuse angle so slope, tan 0', becomes negative; It is clear from Fig. 4.35 that as time passes, slope goes on hence acce1eration is negative. decreasing. v Now (X2 < al ::::} tanaz < tanal ::::} V2 < VI SO velocity goes on decreasing. II x -'., v~ +-----a \"2 (~_ I I (f lana I, !~ Fig. 4.38 V2 <: \"1 I, Fig. 4.35

4.16 Physics for IIT-JEE: Mechanics I Particle is Moving with Increasing Acceleration Particle is Moving with Decreasing Acceleration at Constant Rate 0'2> 0'1 Acceleration-time graph for a particle moving with decreasing tan D:'2 > tan (Xl acceleration at constant rate is a straight line making an obtuse angle\" with time-axis (Fig. 4.43). az > Cl1 a -a, ti2 I I, I, {/2 > a, I, Fig. 4.43 Fig. 4.39 Graph will be straight line because acceleration is 'decreasing Particle is Moving with Decreasing Acceleration at constant rate. Let at t = 0 acceleration is ao. At some time t = to acceleration becomes zero and then it becomes negative, 0,'2 < 0'1 ::::} tanal < tanal Derivation of Equations of Uniformly Accelerated Motion from Velocity-Time Graph ,. Consider an object is moving with a uniform acceleration 'a' -a, a, along a straight line. The initial and final velocities of the object at time t = 0 and t = tare u and v, respectively. During time t, let I s be the distance travelled by object. In uniformly accelerated motion the velocity-time graph of an object -is a straight line, I, I, inclined to time axis. 1I.,<al =In Fig. 4.44. 0 A u =} 0 E = v and 0 D = A C = t \" Fig. 4.40 Velocity AcceLeration-time Graph of Various Types of v ---- ---- B Motions of a Particle u --------- (' Time Particle is Moving with Constant Acceleration Let at time t = 0, acceleration is Qo. Since acceleration is con- A ,,, stant, so acceleration at any further time will remain ao. Hence 'D acceleration-time graph for a particle moving with constant ac- o celeration is a straight line parallel to time axis (Fig. 4.41). Fig. 4.44 aOL\"-L----.I Fig. 4.41 First Equation of Motion Particle is Moving with Increasing Acceleration at In uniformly accelerated motion, the slope of velocity-time Constant Rate graph represents the acceleration of object. Acceleration-time graph for a particle moving with increasing acceleration at constant rate is a straight line making an acute i.e .• Acceleration = slope of graph AB = IBI CC = AE angle ex with time axis. Graph will be straight line because ac- AC celeration is increasing at constant rate (Fig. 4.42). OE-OA => v-u ~' = a=-- :;--a ---., AC t +=} v - u = at or v = II at Fig. 4.42 Second Equation of Motion In uniformly accelerated motion, the area included between velocity-time graph and time axis provides us the distance cov- ered by the object in a given interval of time. Therefore, S = arca of trapezium AIi D0 A 21: (OA + BD) x OD = 2I: (u + v) x t