Chapter2 Measurement of Physical quantities in Physics Introduction You have studied in the last chapter, the fundamentals of scientific study in relevance to the subject Physics, the science of physical phenomenon of nature. You can now proceed to frame the hypothesis/theory/law to explain any physical phenomenon of nature. But as we have already discussed that there is no meaning of words until unless our theory or law is based on certain physical parameters of study or may be called as physical quantities, those could be expressed and measured quantitatively. These quantities are the basis of experimental validation of the theory propounded. Now we shall try to understand the concept of physical quantities regarding their definition and their use in interpreting the laws of Physics. At the same time regarding measurement of these quantities, we shall glance over the different standards of measurements prevailing world wide to quantitatively measure physical quantities and how these standards make computations involving physical quantities make so simple. Unit One Now in this unit of about half an hour you will be able to learn a) Concept of fundamental and derived quantity. b) System of measurements in practice. c) System of measurements in S.I. system. d) System of units and conversion of units from one system to another. Concept of fundamental and derived quantity As we have discussed that to understand the seen phenomenon of nature theories/hypothesis/laws are framed. But there is no sense and meaning of words until unless the laws or postulates are based on measurable quantities. The laws of physics are expressed and experimentally validated in terms of these physical quantities. The laws expressed in terms of these physical quantities are invariant with space and time. For the purpose of Physics it is very essential that quantities should be defined clearly and precisely and should have the conceptual meaning for the postulated law. Among these are force, time, velocity, density, temperature, charge, magnetic susceptibility and numerous others. Now as per the practice all the physical quantities involved in the Physics today may be grouped in two sets. One set contains the physical quantities as a fundamental one and other set contains quantities derived from these fundamental one. For an example length and time may be taken as a fundamental quantities and the quantity velocity defined as the ratio of length upon time may be termed as a derived quantity. However which quantity will be taken as a fundamental and which one as derived depends on system to system. For example S.I. system of units takes length, mass and time as a fundamental quantity and the quantity force defined as the product of mass and acceleration (length/time 2 ) as a derived quantity. While the F.P.S. system of units and measurement takes force, length and time as a fundamental quantity and mass as a derived quantity. Now the intention of creating physical quantities is that physical laws are expressed in terms of these quantities and which are themselves evaluated in terms of numerical values. These values should be such that can be characterized as intrinsic to the experiment conducted and may be compared to the same quantity arrived in another set of experiment. Now the comparison of physical quantities can be done only when each physical quantity either fundamental or derived one is based on some standard of measurement. The standard of measurement is a set of standard where each fundamental quantity is assigned a specific unit value. The unit value to that quantity is such that it is invariant with space and time and is easily accessible. For example in the S.I. system of measurement the physical quantity mass has been given a unit value as one kilogram and defined as mass of the international standard body preserved at Severes, France. However it certainly involves the procedure to account the unit value of the fundamental quantity, where different objects having different values in the same set of conditions can be compared. For example by the procedure to obtain mass of a standard body, the spring balance in terms of stretch of spring may be used. The stretch of spring is directly proportional to the mass of the body. By comparing the different stretches of spring, the masses of different bodies, may be compared. So in final words the operational definition of fundamental quantity involves two steps, first choice of a standard and second the establishment of procedures for measuring the quantity in terms of standard so that a number and unit are determined as a measure of quantity. But very important aspect of choice of standard is that it should be accessible and invariant. For example we have selected our standard for length to measure the distance between two points as one meter, then by a comparison of this length with a second object three times in length as standard, we say that second bar has a length of three meters. However most quantities cannot be measured directly in comparison to standard and indirect approach using some involved procedure is required and also certain assumptions are made to ascertain the matter. For example measuring the time of sending and receiving the electromagnetic pulse with known speed, the distance can be measured as product of speed and one half of time interval. Here we have set half time of sending and receiving the signals as our standard of measurement and different distances may be compared with different times of observations. However the speed of pulse is to be determined through the other acclaimed procedure. Similarly, we use an indirect method to measure very small distances between atoms and molecules by particle scattering method. System of measurements in practice Now so far we have studied that quantities in physics are either fundamental one or derived one depending upon the system of measurement we are using and shall study the different systems of measurement prevailing and used all over the world. The fundamental quantities are defined in terms of a standard of measurement devised in that particular system of measurement and procedure to measure the quantity so that comparison of different objects may be done in terms of that quantity. The derived quantities are derived from these fundamental one and for the complete descriptions will have dimensions showing the fundamental quantities involved and units of these dimensions involved depending upon the system of measurement used. Now we shall discuss the different system of measurements prevailing all over the world and their merits and demerits. Imperial System of units measurement Before S.I. system of units adopted around the world, the British systems of English units and later Imperial system of units were used in Britain, the Commonwealth and the United States. The system came to be known as U.S. Customary units in the United States and is still in use there and in a few Caribbean countries. These various systems of measurement have at times been called footpoundsecond systems after the Imperial units for distance, weight and time. Many Imperial units remain in use in Britain despite the fact that it has officially switched to the SI system. Road signs are still in miles, yards, miles per hour, and so on. People tend to measure their own height in feet and inches and beer is sold in pints, to give just a few examples. Imperial units are used in many other places, for example, in many Commonwealth countries, which are considered metricated, land area is measured in acres and floor space in square feet, particularly for commercial transactions (rather than government statistics). Similarly, the imperial gallon is used in many countries that are considered metricated at gas/petrol stations, an example being the United Arab Emirates. Metric System of Measurement The metric system is a decimalised system of measurement based on the metre and the Kilogram. It exists in several variations, with different choices of base units, though these do not affect its daytoday use. Since the 1960s the International System of Units (SI), explained further below, is the internationally recognized standard metric system. Metric units of mass, length, and electricity are widely used around the world for both everyday and scientific purposes. The main advantage of the metric system is that it has a single base unit for each physical quantity. All other units are powers of ten or multiples of ten of this base unit. Unit conversions are always simple because they will be in the ratio of ten, one hundred, one thousand, etc. All lengths and distances, for example, are measured in meters, or thousandths of a metre (millimeters), or thousands of meters (kilometres), and so on. There is no profusion of different units with different conversion factors as in the Imperial system (e.g.inches, feet, yards, fathoms, rods). Multiples and submultiples are related to the fundamental unit by factors of powers of ten, so that one can convert by simply moving the decimal place: 1.234 metres is 1234 millimetres or 0.001234 kilometres. The use of fractions, such as 2/5 of a meter, is not prohibited, but uncommon. S.I system of measurement The International System of Units (abbreviated SI from the French language name Système International d'Unités) is the modern, revised form of the metric system. It is the world's most widely used system of units, both in everyday commerce and in science. The SI was developed in 1960 from the metrekilogramsecond (MKS) system, rather than the centimetregramsecond (CGS) system, which, in turn, had many variants. At its development the SI introduced several newly named fundamental units that were previously not a part of the metric system. The S.I. system of units has the following set of standards for measurement: Standard of Length The first standard of length measurement conceived was a bar of platinumiridium alloy kept at International Bureau of weights and measures near Paris. The distance between two lines engraved on gold plugs near the ends of the bar (when the bar was at 0.00 degree centigrade and supported in a certain mechanical system) was defined as one meter. But the main disadvantage of the meter bar is that it is not accurately producible at everywhere. In 1961 an atomic standard of length was accepted by International agreement. The wavelength of orange radiation emitted by atoms of Krypton (Kr 36 ) in electrical discharge was chosen. One meter is now defined to be 1,650,763.73 wavelengths of this light. The choice of atomic standard has offered a great advantage that there is enough precision in length measurements and all atoms generate light of same wavelength, at everywhere therefore accessible and invariant with respect to space and time. Standard of Time For the development of the time standard there are two different aspects for the purpose. One is for civil and another for scientific work according to the desired accuracy in the work. We define the time to know the duration between the start and end of events and classify them in such a sequence where precedence of start or end of one event can be compared with other by counting number of repetitions as the division of time phase. An oscillating pendulum, quartz crystal, oscillating spring or digital counter can be used for the purpose. Of the many repetitive phenomenon occurring in the nature, the rotation of earth on its own axis, which in time span is length of day, has been used as time standard and still the basis of determining civil time standard. One mean solar second being defined as 1/86,400 of a mean solar day and time expressed in terms of Earth’s rotation about its own axis is called universal time (UT) In 1956, The International Congress of Weight and measures redefined the second for the scientific purposes requiring high precision, in terms of the earth orbital motion about the sun and found it to be the fraction 1/31,556,925.9747 of the tropical year 1900 and time defined in terms of the earth orbital motion is called ephemeris time (ET). But the main disadvantage of the above time standards is that both UT and ET must be determined by astronomical observations extending over several weeks (for UT) or several years (for ET) and a secondary terrestrial clock, calibrated by the astronomical observations is needed for the purpose. Quartz crystal clocks, based on the electrically sustained natural periodic vibration of a quartz crystal serve as a time standard and have measured time with a maximum error of 0.02 sec in a year. One of the most common uses of a time standard is the determination of frequencies and for the purpose the atomic clocks using the periodic atomic vibrations as a standard have been developed which gives accurate time estimates to a accuracy of fraction of micro seconds and are invariant with space and time. As per the S.I. system of units the second (s) is the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the hyperfine levels of the ground state of the Cs 133 atoms. Standard of Mass The kilogram (kg) is the mass of the international standard body preserved at Severes, France. Standard of amount of Substance (Mole) The amount of substance that contains as many elementary entities (Avogadro number 6.02x10 23 ) like atoms if substance is monoatomic or molecules is called a mole. For example 0.012 kg of carbon12, called as one mole substance contains 10 0 . 6 23 x Nos of atoms of carbon12. Standard of Current The ampere (A) is the current in two very long parallel wires 1m apart that gives rise to a magnetic force of 10 0 . 2 7 − x N/m. Standard of Temperature The Kelvin (K) is 16 . 273 1 of the thermodynamic temperature of the triple point of water. Standard of Luminous Intensity The candela (cd) is the luminous intensity in the perpendicular direction of a surface of a area 000 , 600 1 sq meter of a black body at a temperature of freezing platinum at pressure of 1 atm. Units and their conversion We have studied earlier that all derived quantities are dependent upon the fundamental quantities and fundamental quantities on certain standards. The choice of standard units for these fundamental quantities determines the system of units for all physical quantities used all over the world. For example the M.K.S. system of units classifying fundamental quantities mass, length and time as Kilogram, meter, second. Therefore once the choice of the system of units has been made the derived or dependent quantities follow the same system of units and will have both number and unit in its notification. When such quantities are added, subtracted, multiplied, or divided in an algebraic equation the unit can be treated as an algebraic quantity. For example we wish to find out the distance traveled by a person traveling at a speed of 5 km per hour in five hours than Distance traveled = Speed x Time= h x h Km 5 0 . 5 =25 km Ex 1. 5 litre of benzene weigh more in summer or winter? (A) Summer (B) Winter (C) Equal in both (D) None of these Sol: Since volume increases and density decreases after rise in temperature so for a given volume of benzene say 5 litre it will weigh more in the winter for higher density. Ex 2. The SI unit of length is the meter. Suppose we adopt a new unit of length which equal to x meters the area 1m 2 expressed in terms of the new unit has a magnitude (A) x (B) x 2 (C)1/x (D)1/x 2 Sol: Area = 1m x 1m = (1 unit/x)(1 unit/x) = 1/x 2 unit 2 So magnitude=1/x 2 Ex3. Iftheunitoflengthismicrometer and the unit of time is microsecond,the unit of velocity will be (A) 100 m/s (B) 10 m/s (C) micrometer/s (D) 1 m/s Sol: Velocity=L/T = 1 m/s Ex 4. If the units of length and force are increased four times, then the units of energy will (A) Increase 8 times (B) increase 16 times (C) decrease 16 times (D) increase 4 times Sol: E 0 = F 0 d 0 En = 16 f 0 d 0 En = 16 E 0 Hence each unit of new energy dimension is 16 times each unit of energy in old dimension Ex 5. In a particular system, the unit of length, mass and time are chosen to be 10 cm.,10 gm, and0.1 s respectively. The unit of force in this system will be equivalent to (A) 1/10 N (B) 1 N (C) 10 N (D) 100 N Sol: unit of mass= 10 gm Length = 10 cm. Time = 0.10 s Unit force in the new system with dimensions M L T –2 have 10 (gm)(10 cm) (0.1 s) -2 And so equals to 1/10 N OR Since 1N = 1kg m.s –2 = 10 3 gm.(10 2 cm)( s -2 ) Hence each unit of force = 1N/10 Ex 6.What will be the unit of time in that system in which the unit of length is ‘metre’ unit of mass‘kg’ and unit of force ‘kg.wt.’? (A) 1√9.8 sec (B) (9.8) 2 sec (C)√9.8 sec (D) 9.8 sec Sol: Force F = MLT -2 kg (9.80 m/s 2 ) = kgmT -2 T = 1/√9.8 s OR F= M L T –2 T=√ ML/F = √ kg . m/ kg wt So T= √ kg . m / kgwt = 1/√ 9.8 sec Ex7. If the velocity of light c, acceleration due to gravity g , and the atmospheric pressure p are taken as the fundamental units , then the unit of mass will be (A) 1 kg (B) 81 kg (C) 9x10 18 kg (D) 81x10 34 kg Sol: M= c x g y p z = M z L x+y-z T –x –2y –2z ; so z=1, x=4, y=-3 M=pc 4 /g 3 =81x10 34 Kg Ex8 Is the time variation of position as shown in the fig is observed in nature? Explain. P o s i t i o n o f P a r t i c l e ( x ) T i m e i n S e c . ( t ) Sol:TheanswerisclearlyNObecause at any instant of time the particle can’t have two positions and time doesn’t decreases. Ex9. Thenormal durationof physicspractical periodinIndiancolleges is100minutes. Express this period in micro century. 1 micro century =10-6 x 100 year. Sol: T= 100 minutes =100/60x24x365x10 –4 = 1.9 Microcenturies; Ex 10. The SI and C.G.S. units of energy are joule and erg respectively. How many ergare equal to one joule. Sol: In M.K.S. system of units Energy =Kg m 2 /s 2 = 1 joule= 10 7 grams cm 2 /s 2 = 1 Erg in C.G.S. system Ex 11. Young’s modulus of steel is 19x10 10 N/m 2 . Express it in dyne/cm2. Here dyne is the C.G.S. unit of force. Sol: 19 x10 11 Dyne /cm 2 ; Ex 12. The density of a substance is 8 g/cm 3 . Now we have a new system where unit of length is 5 cm and unit of mass 20 g. Find the density in the new system. Sol:Density of substance = 8 g/cm 3 ; Unit of length = 5 cm ; Unit of mass = 20 g. Since dimension of density ρ= ML -3 T 0 ρ 2 /ρ 1 = (M 2 /M 1 ) (L 2 /L 1 ) -3 = 20 (5) -3 = 4/25; Density of substance in new system = 50 units Exercise 1 Q1.Iftheunitsofforce, energy and velocity in new system be 10 N, 5 J and 0.5 m s -1 respectively, find the units of mass, length and time in the new system. Which gives M=E/v 2 =5J/(0.5) 2 = 20 Kg; L=E/F=0.5 m; T=E/FV=1 sec Q2. The radius of proton is about 10 -9 microns and the radius of the universe is about 10 28 cm. Nameaphysical object whosesizeis approximatelyhalf waybetweenthesetwoona logarithmic scale. Q3 Assuming that the length of the day uniformly increases by 0.001s in a century. Calculate the cumulative effect on the measure of time over twenty centuries. Such a slowing down of the earth‘s rotation is indicated by observations of the occurrences of solar eclipses during this period. Q4. The unit of length convenient on the nuclear scale is fermi ; 1 f=10 -15 m. Nuclear sizes obey roughly the following empirical relation; r=r 0 A 10 where r is the radius of the nucleus. A its mass number and r 0 is a constant equal to 1.2 f. Find out whether mass density is nearly constant for different nuclei. Solutions Exercise 1 Ans1. Let M = F x E y v z and so x=0, y=1 and z=-2 Ans2: On logarithmic scale the exponent value= -15+(15+26)/2= -15 + 41/2=5.5 Hence the size of object=10 5.5 ≅ 1x10 6 ( Radius of moon) Ans3: Increase in length of day in 20 centuries= 20 x 0.001s Average increase in length of day = 0.001x 20/2 = 0.01s Cumulativeerror intimemeasurement =cumulativeerror indaylengthincreasein20 centuries = 20 x100 x 365 x 0.01/3600 h= 2.03 h Ans4:Mass density of nucleus=A(m p )/(4/3πr 3 ) ≅ 10 17 Kg/m 3 Introduction So far we have defined the physical quantities in terms of their interdependence and evolved procedures to measure the physical quantities. Now we shall try to understand the way the physical quantities appear in the laws of physics and the rules of mathematical manipulations followed. Eachphysical quantitymaybeattributedasetofdimensionsaccordingtothebaseunits involvedinthequantity and laws of physics involve mathematical manipulations of these physical quantities. Aseachstatement shouldfollowcertainrulesinvolvingthephysical quantities regarding their mathematical manipulations, is governed by the dimensional analysis. Beside that physical quantity regarding their representation numerically should follow certain rules so that degree of accuracy of the measurement could be ascertained at a glance of the presentation of measurement. It is governed by the rules of significant digits of measurement of physical quantities. These two are the scope of study now. So at the end of this unit of around one hour you will be able to learn • Dimensions of physical quantities and dimensional analysis. • Concept of significant digits for the presentation of measurement of quantities in Physics. Dimensions of Physical Quantities As we have stated earlier that physical quantities can be added, subtracted, multiplied or divided in the same manner as any other algebraic quantity. The addition or subtraction of quantities is meaningful only if the quantities have followed same standard of units and so are dimensionally homogeneous. The dimensional consistency is must for equation to be correct. The correctness of equation can be checked by comparing the physical dimensions of each term in the equation under discussion. A list of various physical quantities with their usual convention and dimensions are : S.No Quantity Symbol Physical formula S.I unit Dimension formula 1 Acceleration a a=∆v/∆t m/s 2 M 0 LT -2 2 Angular acceleration α α = ∆ω/∆t Rad/sec 2 M 0 L 0 T -2 3 Angular Displacement θ θ=arc/radius Radian M 0 L 0 T 0 4 Angular Momentum L L=m v r Kg m 2 /s M L 2 T -1 5 Angular Velocity ω ω=θ/t Rad/sec M 0 L 0 T -1 6 Area A l x b (Metre) 2 M 0 L 2 T 0 7 Capacitance C Q = CV Farad M -1 L -2 T 4 A 2 8 Charge q q=I t Coulomb M 0 L 0 T A 9 Current I ---- Ampereor A M 0 L 0 T 0 A 10 Density d d =M/V Kg/m 3 M L -3 T 0 11 Dielectric constant ε r ε r =ε/ε 0 ___ M 0 L 0 T 0 12 Displacement S ___ Metre or m M 0 LT 0 13 Electric dipole moment P P=q 2l Coulomb metre M 0 L T A 14 Energy KE/U KE=1/2mv 2 Joule M L 2 T -2 15 Force F F=ma Newton or N M L T -2 16 Force Constant K K=F/x N/m M L 0 T -2 17 Frequencyf f=1/T Hertz M 0 L 0 T -1 18 Gravitational Constant G G=F r 2 /m 1 m 2 N m 2 /kg 2 M -1 L 3 T -2 19 Heat Q Q=ms∆t Joule/calorie M L 2 T -2 20 Impulse - F x t N.sec M L T -1 21 Intensity of electrical field E E = F/q N/coul M L T -3 A -1 21 Intensity of electrical field E E = F/q N/coul M L T -3 A -1 22 Latent Heat L Q = mL Joule/kg M 0 L 2 T -2 23 Magnetic dipole moment M M=NIA Amp.m 2 M 0 L 2 A T 0 24 Magnetic flux φ E=dφ/dt Weber M L 2 T - 2 A -1 25 Magnetic intensity H B =µH A/m M L -1 T 0 A 26 Moment of Inertia I I=mr 2 Kg m 2 ML 2 T 0 27 Momentum P P = mv Kg m/s M L T -1 28 Mutual inductance L E=L.di/dt Henery M L 2 T -2 A -2 29 Power P P=W/t Watt M L 2 T -3 30 Pressure P P =F/A Pascal M L -1 T -2 31 Resistance R V=IR Ohm M L 2 T -3 A -2 32 Specific heat S Q=m s∆t Joule/kg.kelvi n M 0 L 2 T -2 θ -1 33 Strain ___ ∆l/l; ∆A/A; ∆v/v ___ M 0 L 0 T 0 34 Stress ___ F/A N/m 2 M 0 L 0 T -2 35 Surface Tension T F / l N/m M L 0 T -2 36 Temperature θ ___ Kelvin M 0 L 0 T 0 θ 1 37 Torque τ F.d N.m. M L 2 T -2 38 Universal gas constant R PV = nRT Joule/mol.k M L 2 T -2 θ -1 39 Velocity v v = ∆s/∆t m/s M 0 LT -1 40 Volume V V=l x b x h (Metre) 3 M 0 L 3 T 0 41 Work W F.d N.m M L 2 T -2 42 Young Modulus Y Y = (F/A)/ ∆l/l N/m 2 M L -1 T -2 Application of dimensional analysis 1. To find the unit of a physical Quantity. 2. To convert units of a physical Quantity from one system to another. 3. To check the dimensional correctness of a given relation. It is to be further noted that every dimensionally correct relation does not mean to have physically correct relationship. If we have some idea or can make an educated guess as to how one physical quantity relates to another we can use dimensions to derive the form of the equation. As an example, consider the equation for the period of pendulum bob. We might suppose that the period depends on the mass of the bob, the length of the pendulum and the acceleration due to gravity We can express this as T=m x l y g z . Where x, y and z are as yet undetermined indices. To find the values of x, y and z we convert the formula into its dimensions. On the left-hand side the dimension of the period is [T], the dimension of mass is [M], the dimension of the length of the pendulum is [L] and the dimension of g is [LT -2 ]. [T]=[M] x [L] y [LT -2 ] z . Equating left-hand indices with matching dimensions on the right-hand side. [M]: x=0 [L]: y+z=0 [T]: 2z=-1 From this we can deduce that z=1/2, while y=1/2 and x=1/2 Substituting these values into the original equation we obtain. T= k(l/g) 1/2 . Where k is a constant of proportionality. Compare this with the equation for the period of a pendulum g l T π 2 · The form of the equation is correct, but it cannot determine the constant of proportionality. Limitations of application of dimensional analysis 1. If the dimensions are given, physical quantity may not be unique as many physical quantities have same dimensions. 2. Since numerical constants have no dimensions, can’t be deduced by dimensional analysis. For example 1, etc. 3. The dimensional analysis can’t be used to derive relationship other than the product of power functions. However the dimensional correctness of the relation may be checked. For e.g. S=u t+1/2 a t 2 , y= a sin ω t 4. The method of dimension can’t be applied to derive relationship when a physical quantity depends on more than three quantities. For e.g. T=2 π√ I/(m g l) Ex1. Ifvelocity, forceandtimearetakentobefundamental quantitiesfinddimensional formula for quantity mass (A) V -1 FT -1 (B) V -1 FT (C)VF -1 T -1 (D)V -1 F -1 T Sol: M=K (V) x (F) y (T) z ; x= -1, y =1, z =1; M = K V -1 F T Ex2. Youmaynot knowintegration. But usingdimensional analysiscancheckorprove results .in the integral ∫dx/(2ax-x 2 ) 1/2 = a n sin -1 [x/a-1] The value of n should be (A) 1 (B) –1 (C) 0 (D) 1/2 Sol: The dimension of a is that of x and for dimensional consistency of the equation n should be equal to zero. Ex3. In the cauchy’s formula for the refractive index n = A+B/λ 2 the dimensions of A and B are (A) Both are dimensionless (B) A is dimension less , B has dimension M 0 L -2 T 0 (C)Aisdimensionless, Bhasdimensions M 0 L 2 T 0 (D)BothAandB have dimensions M 0 L 2 T 0 Sol: Since refractive index is dimensionless hence A should be dimensionless and B should have dimensions of Length 2 that is M 0 L 2 T 0 Ex4. A highly rigid cubical block A of small mass M and side L is fixed rigidly on the another cubical block B of same dimensions and of low modulus of rigidity ηsuch that lower face of A completely covers the upper face of B. The lower face of B is rigidly held on a horizontal surface. A small force F is applies perpendicular to one of the side face of A.After the force is withdrawn, block A executes small oscillation, the time period of which is given by (A) 2π√(mη L) (B) 2π√(Mη /L) (C) 2π√(ML/η ) (D) 2π√(M/Lη ) Sol: Check the dimensions of the left and right side quantities. Ex5.The frequency of oscillation of an object of mass m suspended by means of spring of force constant k is given by f = c m x ky, where c is a dimensionless constantthevalue of x and y are (A) x = ½, y = ½(B) x = - ½,y = ½ (C) x = ½, y = - ½ (D) x = - ½, y = - ½ Sol: Putting the dimensions on the two sides and equating the powers of the base quantities we get x= -1/2, y =1/2 Multiple Choice Type of Questions Ex1. Suppose a quantity x can be dimensionally represented in terms of M, L and T i.e.[X] = M a L b T c the quantity mass (A) Can always be dimensionally represented in terms ofL, T and x. (B) May be represented in terms of L,T and x if a = 0 (C) May be represented in terms of L, T and x if a = 0 (D) May be represented in terms of L, T and x if a ≠0 Sol: Any quantity may be represented in terms of other quantities as base quantities such that the quantities are dependent upon each other with a certain relationship. Here I the case the quantity M may be represented in terms of L, T, x if a≠ 0 so that M remains related with the rest of quantities. Ex2. Choose the correct statement (s) (A) A dimensionally correct equation may be correct (B) A dimensionally correct equation may be incorrect (C) A dimensionally incorrect equation must be correct (D) A dimensionally incorrect equation must be incorrect Sol:Adimensionallycorrect equationmaybecorrect or incorrect. But adimensionally incorrect equation can never be correct. Ex3. Choose the correct statement (s) (A) All quantities may be represented dimensionally in terms of the base quantities. (B) All basequantitiescannot berepresenteddimensionallyintermsof therest ofbase quantities. (C) The dimension of a base quantity in other base quantity is always zero. (D) The dimension of a derived quantity is never zero in any base quantity. Sol: ( A ), (B),( C ) Descriptive Type of Questions: Ex.1(a) Can a physical quantity have no unit and dimensions? If yes give an example. (b) Can a physical quantity have units without dimensions? If yes give an example. Ans41: (a) Strain has neither unit nor dimensions but it is a defined physical quantity. (b) Angle measured in radians is physical quantity, which has unit of radian but no dimension. Theanglemeasurement inaplanethat isradianandangleofasolidthat is steradian are supplementary units to supplement the physical quantities like angular displacement and angular velocity. No dimension has been assigned to these quantities. Ex2. Find the dimensional formulae of e 0 & m 0 (Where e 0 is the absolute permittivity and m 0 is the permeability of vacuum or free space respectively ). Sol: The dimensions ofε 0 = M -1 L -3 T 4 A 2 ;The dimensions of µ 0 = MLT -2 A -2 Ex3. Assumingthat thelargest massthat canbemovedbyaflowingriver dependson velocity offlow density ofriver water and on gravity, find out how the mass varies with velocity of flow. Sol: M=K(v) x (d) y (g) z ; x=6, y=1, z=-3 and so M= K v 6 d g –3 Ex4. The gas constant Rdepends uponpressure of the gas P, volume of the gas V, temperature of the gas T and number of moles n. Derive an expression for gas constant, R Sol: Let P= K V x n y R z T m , on solving fordimensions of base quantities M,L,T and n we get x=-1, y=1, z=1, T=1 and so PV=n RT Exercise 2 Matching Type of Questions: Q1. In the following table, there are two lists A and B, but the list B is not in order of list A. write down the list B in order of list A in each table. (a) Moment of inertia (i) Newton /Meter 2 (b) Surface tension(ii) kg/ (metre-s) (c) Angular acceleration(iii) kg – meter 2 (d) Coefficient of viscosity (iv) Newton/meter (e) Coefficient of elasticity (v) radian/s 2 (f) Momentum(vi) MLT -1 (g) Gravitational Constant(vii) ML 2 T -1 (h) Plank Constant (viii) M -1 L 3 T -2 Q2. ColumnIgivesthreephysical quantities. Select theappropriateunitsforthesefrom choices given in column II. Some of the physical quantities may have more than one choice : I II Capacitance Ohm second InductanceCoul 2 joul -1 Magnetic inductance Coulomb (Volt) 1 Newton(ampere-m) -1 Volt-sec (Ampere) -1 Q3. Match the physical quantities given in column I with dimensions expressed in column if in tabular form (a) Angular momentum(a) ML 2 T -2 (b) Latent heat(b) ML 2 Q -2 (c) Torque (c) ML 2 T -1 (d) Capacitance (d) ML 3 T -1 Q -2 (e) Inductance(e) M -1 L -2 T 2 Q 2 (f)Resistivity(f) L 2 T -2 Q4. The position of a particle at any time is given by S(t) = V 0 /a (1-e -at ), where a>0 andV 0 are constants. What are the dimensions of a and V 0 ? Q5. Theequationofawavetravelingalongthexaxisbyy=Ae[x/a-t/T]2,writethe dimensions of A , a and T. Q6. Suppose an attractive nuclear force acts between two protons which may be written as F = Ce-kt/r2 Write down the dimensional formula and appropriate SI units of C and K. Q7. Find the dimensional formula of L/R, where R is the resistance and L is the coefficient of self-inductance. Q8. Find out the dimensions of electrical conductivity. Q9. The equation of state of a real gas is given by [p+a/v 2 ](v-b) = RT, where p, v and Tare pressure , volume and temperature respectively and R is the universal gas constant.Find out the dimensions of the constant a in the above equation. Q10. The heat produced in a wire carrying an electric current depends on the current, the resistance and the time; assuming that the dependence is of the product of powers type, guess an equation between these quantities using dimensional analysis. The dimensional formula of resistance is ML 2 T -3 A -2 and heat is a form of energy. Q11. The frequency of vibration of a stream depends on the length L between the nodes, the tension F in the string and its mass per unit length M. Guess the expression for its frequency from dimensional Analysis. Q12. The kinetic energy E of a rotating body depends on its moment of inertia I and its angular speedw. AssumingtherelationtobeE=KIawbwhereKisadimensionless constant, find a and b. Moment of Inertia of the sphere about its diameter is 2/5Mr 2 . Q13(a)IntheformulaX=3YZ 2 , XandZhavedimensionsofcapacitanceandmagnetic induction respectively. What are the dimensions of Y in MKSQ system? (b) Calculate the dimension(s) of VCR/L, where V=supply voltage, C=capacitance, R=resistance and L=inductance. Q14. If velocity of light c, the gravitational constant G and plank constant h be chosen as the fundamental units, then find the dimensions of mass, length and time in the new system. Concept of Significant digits As we have seen every measurement pertains to the standard we are going to use and its numerical value is read from the calibrated scale based on that standard of measurement. The value of measurement contains two parts (i) One part with all digits read directly from the scale by the smallest subdivision of the scale called as certain digits of measurement (ii) and the second part contains doubtful digits at end corresponding to the eye judgment within the least subdivision of the scale. For example the length measured by the meter scale having least count of 1cm as 50.35 cm contains digits 5, 0 as certain 3 is doubtful and 5 is insignificant. The digits 5, 0, 3 are termed as significant digits and 5 as insignificant digit. There may be some confusion while dealing with the measurements containing zero at their end but rule is the same. As a general practice we report only the significant digits and magnitude of any physical quantity is represented by proper power of 10. For example if only 5 is significant in 500 cm then we report it as 5 x 10 2 cm, with 5 as a significant digit. If 5 and first zero that is two digits are significant then we write 5.0 x 10 2 cm. and if all the digits are significant we report it as 5.00 x 10 2 cm. If the integer part of the digit is zero then all the continuous zeros after the decimal are treated as insignificant digits as the number 0.00003 having first digit as zero then all the continuous digits are also insignificant. For this reason it is important to keep the trailing zeros to indicate the actual number of significant figures. However if first digit is nonzero as in 1.0005 then all zeros will be counted for significant digits containing five significant digits as 1,0,0,0 and 5 respectively. For numbers without decimal points, trailing zeros may or may not be significant. Thus, 400 indicate only one significant figure. To indicate that the trailing zeros are significant a decimal point must be added. For example, 400 has three significant figures, and has one significant figure. Exact numbers have an infinite number of significant digits. For example, if there are two oranges on a table, then the number of oranges is 2.000... . Defined numbers are also like this. For example, the number of centimeters per inch (2.54) has an infinite number of significant digits, as does the speed of light (299792458 m/s). Significant digits in arithmetical calculation As per internationally accepted practice for finding out the significant digits in the arithmeticalcalculationsay division or multiplication of two physical quantities following rule has been formulated for determination of significant: 1. In multiplication or division of two or more quantities the number of significant digits in the answer is equal to the number of significant digits in the quantity, which has least number of significantdigits.Thus 60.0/2.0 will have only two significant digits. The insignificant digits are dropped from the result if appear after the decimal point, and replaced to zero if appear to the left of the decimal point. The least significant digit is rounded according to the following rules: (A) If the digit next to one rounded is more than 5, then the digit to be rounded is increased by 1 (B) If next digit is less than 5 then rounding digit is left unchanged. (C) If the digit next to be rounded is equal to 5 then rounding digit is increased by one if it is odd otherwise left unchanged. For example, (2.80) (4.5039) = 12.61092 should be rounded off to 12.6 (three significant figures like 2.80). 2. For addition or subtraction of quantities all the numbers are written with the decimal point in one line up to the number, which has the first doubtful digit counted from left. All digits are written to that digit after rounding off at that level and addition/subtraction is performed. For example, 89.332 + 1.1 = 90.432 should be rounded to get 90.4 (the tenths place is the last significant place in 1.1). Ex1. Evaluate 25.2x1374/33.3 All the digits in this expression are significant. Sol: Expression has value =1039.7838 Since the expression has number with lowest number of significant digits as 3 with number 25.2 or 33.3hence the expression will also have three significant digits and number will be written as per rule 1.04x10 3 Ex2. Evaluate 24.36+0.0623+256.2 Sol:As per rule write the numbers with decimal point in a line and check where the first doubtful digit occurs between these numbers, which in our case occurs at 256.2. Now change the other numbers with proper rounding and than add as24.4 + 0.1 + 256.2 = 280.7 Ex3. GivenP=0.0030m; Q=2.40mandR=3000m. Findout thenumber of significant figures in P, Q, R respectively. Sol: No of significant figures in P=2(3,0) since the first digit is zero so all zeros after that are insignificant; No of significant figures in Q=3(2,4,0); No of significant figures in R=4(3,0,0,0) Ex4. The volume of one sphere is 1.76c.c. Findout the volume of 25suchspheres (according to idea of significant figures). Sol:The volume of 25 spheres=25 x1.76 = 44. Since the number 1.76 has three significant digits and so result should also be written to the three significant digits as 44.0 c.c, Exercise 3 Q1. The length breadth and thickness of a block are measured as 12.5 cm, 5.6 cm, and 0.32 cm respectively. Which measurement is least accurate? Q2. The radius of the circle is stated as 2.12 cm. Find out its area written as in appropriate number of significant digits. Q3. Roundoffthefollowingnumberstothreesignificant digits(a)15462(b)14.745(c) 14.750 (d) 14.650x10 12 . Hints and Solutions Exercise 2 Ans4:Thepower of eshouldbedimensionless sothedimensionof a=M 0 L 0 T -1 ; The dimension of V 0 =M 0 LT -1 ; Ans5:The dimension of a and T should be of such that x/a and t/T are dimensionless and dimensions of A should be of y. So the dimension of T= M 0 L 0 T; A= M 0 L T 0 ; a=M 0 L T 0 ; Ans6: The dimension of C = M L 1 T -2 ; The dimension of K=M 0 L 2 T -1 Ans7: The dimension of L=V/(dI/dt)=W/I 2 =ML 2 T -2 A -2 ; The dimension of R=W/I 2 t =M L 2 T -3 A -2 Ans8: Electrical conductivity=1/R= M -1 L -2 T 3 A 2 Ans9: a=pv 2 =ML 5 T -2; Ans10: H= I x R y t z which on solving for dimensions of two side for base quantities M, L, T and I we get H=I 2 R t Ans11: Let ν = K L x F y m z Put the dimensions of the quantities on two sides and solve for powers x,y, and z we get ν = K/L √ (F/m) Ans12: E.= K I x ω y Put the dimensions of E, I and ω and we get E=K I ω 2 Ans13: Y=X/3Z 2 ;Y= M -3 L -6 T 8 A 6 = M -3 L -6 T 2 Q 6 Ans14: M=√C h/G ; L=h 1/2 G 1/2 /C 3/2 , T= h 1/2 G 1/2 /C 5/2 Exercise 3 Ans1: The breadth of the block is only up to two significant digits hence supposed to be least accurate. Ans2: Area of circle=πR 2 = 14.112416 cm 2 =14.1 cm 2 (three significant figures) Ans3:(a) 1.55x 10 4 ; The last number has been rounded to five since digit next to 4 is six greater than 5 (b) 1.47 x 10 1 The last number has been rounded to 7 only since digit next to it is less than 5 (c)1.48x10 1 Thelast numberhasbeenroundedto8sincedigit next toit is5, anditselfodd number. (d) 1.46 x 10 13 The last number has been rounded to 6 only since digit next to it is equal to 5 and itself even number. Introduction Now let us continue our discussion about measurement about physical quantities. We know, that all sort of measurements are arrived at by taking measurements in some set of experiments of the physical quantities. The authenticity of the results of experiments is totally dependent upon the precision of the measurements taken, which itself dependent upon the certain factors like how the instrument is calibrated with respect to reference one, least sub division of instrument, skill of the person making measurements, secondary effect of environment or errors in the instrument etc. Although we try to make accurate measurements but it is also true that it is very difficult to arrive at the fictitious true value of measurement. One measurement of the same quantity taken many times will differ from each other. So it is a very difficult task to arrive the true value of the measurement. Now in this unit we shall try to understand the different causes of errors in measurement and will classify the errors according to their origination. Next we shall determine the standard procedures to arrive at the true value and mode of representing them for all purposes. Now in this next unit of around one hour you will be able to learn • Types of errors in measurements and level of uncertainty. • Presentation of magnitude of quantities in Physics. Errors and Uncertainty Errors are always the part of measurements and nothing can be done about. If a measurement is repeated, the values obtained will differ and none of the results can be preferred over the others. Although it is not possible to do anything about such error, it can be characterized. For instance, the repeated measurements may cluster tightly together or they may spread widely. This pattern can be analyzed systematically. When we measure something the measurement is meaningless without knowing the uncertainty in the measurement. This leads us to the idea of errors in measurement. Other factors such as the conditions under which the measurements are taken may also affect the uncertainty of the measurements. Thus when we report a measurement we must include the maximum and minimum errors in the measurement. For example, measuring the height of a person, the measure may be accurate to a scale of 1 mm. But depending on how the person being measured holds himself during the measurement we might be accurate in measuring to the nearest cm. Generally, errors can be divided into two broad and rough but useful classes: systematic and random. Systematic errors Systematic errors are errors, which tend to shift all measurements in a systematic way so their mean value is displaced. This may be due to such things as incorrect calibration of equipment, consistently improper use of equipment or failure to properly account for some effect. In a sense, a systematic error is rather like a blunder and large systematic errors can and must be eliminated in a good experiment. But small systematic errors will always be present. For instance, no instrument can ever be calibrated perfectly. Other sources of systematic errors are external effects which can change the results of the experiment, but for which the corrections are not well known. In science, the reasons why several independent confirmations of experimental results are often required (especially using different techniques) is because different apparatus at different places may be affected by different systematic effects. Aside from making mistakes (such as thinking one is using the x10 scale, and actually using the x100 scale), the reason why experiments sometimes yield results, which may be far outside the quoted errors, is because of systematic effects, which were not accounted for. Random errors Random errors are errors, which fluctuate from one measurement to the next. They yield results distributed about some mean value. They can occur for a variety of reasons. ♦ They may occur due to lack of sensitivity. For a sufficiently a small change an instrument may not be able to respond to it or to indicate it or the observer may not be able to discern it. ♦ They may occur due to noise. There may be extraneous disturbances that cannot be taken into account. ♦ They may be due to imprecise definition. For example taking measurements with a magnetic compass, the effect of improper leveling of instrument during observations. ♦ They may also occur due to statistical processes such as the roll of dice. Random errors displace measurements in an arbitrary direction whereas systematic errors displace measurements in a single direction. Some systematic error can be substantially eliminated (or properly taken into account). Random errors are unavoidable and must be lived with. So the systematic errors are to be removed from the measurements by rectification of the cause or my taking measurements by several instruments otherwise the results will remain shifted from the true value. However the random errors are very uncertain and it is very difficult to account for them in our measurements. Random errors will always remain in our measurements how precisely we have taken our measurements. By taking repetitive number of measurements and taking average of large number of measurements, the average value will be close to the true value. But still there is some uncertainty associated with the true value. The uncertainty associated with the mean value is determined by the standard deviation of the measurements as detailed below: Let x 1 x 2 x 3 x 4 x 5 …x N be the results of an experiment repeated N times then standard deviation is defined as ∑ · · − · N i i i N x x 1 2 1 ) ~ ( σ Where ∑ · xi N x 1 ~ is the average of all values ofx. It is supposed to be the best value of x derived from the experiments and the true value is likely to occur within a range ) ~ ( σ t x . However the interval of ) 96 . 1 ( ~ σ t x is quite often taken as the interval in which the true value lies with95%probability. Andif theinterval is chosentobe ) 3 ( ~ σ t x thanthe probability of occurrence of true value in that interval is 99 %. Probabilities of occurrence the true value in any range sayσ k x t ~ is given by ∫ ∫ ∞ + ∞ − + − dx x f dx x f x x ) ( ) ( ~ ~ σ σ But it is fully acceptable only if the numbers of experiments are large. In general the value of N should be greater than 8 for a good approximation. Fractional and percentage errors If∆ x is the error in the measurement in the value x then fractional and Percentage errors are defined as : Fractional error=∆ x/x Percentage error=∆ x/x100 % Propagation of errors (addition and subtraction) Let error in quantity x is t∆ x and error in quantity y is t∆ y then the error in x + y or x - y is t(∆ x+∆ y) that is the errors add. Prefixes and Magnitudes To make sense of the vast range over which physical quantities are measured, prefixes are used as a shortcut to writing the magnitude using scientific notation. Other prefixes which are commonly used but are not strictly part of the SI system. Q1.When a current of 2.5t 0.5 ampere flows through a wire, it develops a potential difference of 20t 1 volt. Find the resistance of the wire- (A) 6.0 t3 (B) 7.0 t2(C) 8.0 t2 (D) 9.0 t3 Ans1: R=V/I=8 Also ∆ R/R=∆ V/V-∆ I/I For max ∆ R/R all terms to be positiveand therefore ∆ R/R=∆ V/V + ∆ I/I=0.25; ∆ R=2 Ohm and so R= V/I t∆ R = 8.0 t2 Q2. In an experiment the value of two resistance were measured to be as given below R 1 = 5.0 t 0.2 ohm and R 2 = 10.0 ± 0.1 ohms. Find there combined resistance in (i) series (ii) parallel. Ans2: When resistance are in series R=(R 1 + R 2 ) ∆ R=∆ R 1 +∆ R 2 = t0.3 and R=15 t0.3 Ohm and when in parallel R=(R 1 R 2 )/(R 1 + R 2 ) ∆ R/R=∆ R 1 /R 1 +∆ R 2 /R 2 -∆ R 2 /(R 1 +R 2 )- ∆ R 2 /(R 1 +R 1 ) For max ∆ R all terms mustbe positive and ∆ R/R =7%; and R=3.3 t7% Q3. In an experiment to determine acceleration due to gravity by simple pendulum, a student commits 1%positiveerror inthemeasurement of lengthand3%negativeerror inthe measurement of time period. Find the percentage error in the value of measurement of g. Ans3: We have T=2π√L/g; or T 2 = K L / g 2 ln T= ln K +ln L- ln g; (2/T) dT=1/L dL-1/g dg;So actual percentage error in measurement of g value (dg/g) =7% Q4. A naval destroyer is testing five clocks. Exactly at noon as determined by the wwv signal on the successive days of a week the clocks read as follows Clock SunMon Tue Wed Thru Fri Sat A 12:36:40 12:36:56 12:37:12 12:37:2712:37:44 12:37:5912:38:14 B 11:59:59 12:00:02 11:59:57 12:00:07 12:00:02 11:59:56 12:00:03 C 15:50:45 15:51:43 15:52:4115:53:39 15:54:3715:55:35 15:56:33 D 12:03:59 12:02:52 12:01:4512:00:38 11:59:3111:58:24 11:57:17 E 12:03:59 12:02:49 12:01:5412:01:52 12:01:3212:01:22 12:01:12 Justify your choice. Ans4:The standard deviations of the five clocks are increasing in order of C, D, A, B, E. Hence the clocks with minimum standard deviation is the most consistent one. So the clocks placed in the same order may be kept in terms of consistency as good timekeepers. Q5. A wire has a mass 0.3±0.003 g, radius 0.5±0.005 mm and length 6±0.06 cm. Find out the maximum percentage error in the measurement of its density. Ans5:d=M/π r 2 L; For Max,∆ d/d all terms should be positive so Maximumerror =¹ ∆ M/M¹ +2/r¹∆ r¹ +¹ ∆ L/L¹ =0.04=4% Vernier Calliper The meter scale enables us to measure the length to the nearest millimeter only. Engineers and scientists need to measure much smaller distances accurately. For this a special type of scale called Vernier scale is used. Vernier Calliper The Vernier scale consists of a main scale graduated in centimeters and millimeters. On the Vernier scale 0.9 cm is divided into ten equal parts. The least count or the smallest reading which you can get with the instrument can be calculated as under: Least count = one main scale (MS) division one vernier scale (VS) division. = 1 mm 0.09 mm = 0.1 mm = 0.01 cm The least count of the vernier = 0.01 cm The Vernier calliper consists of a main scale fitted with a jaw at one end. Another jaw, containing the vernier scale, moves over the main scale. When the two jaws are in contact, the zero of the main scale and the zero of the vernier scale should coincide. If both the zeros do not coincide, there will be a positive or negative zero error. After calculating the least count place the object between the two jaws. Record the position of zero of the vernier scale on the main scale (3.2 cm in figure below). Principle of Vernier You will notice that one of the vernier scale divisions coincides with one of the main scale divisions. (In the illustration, 3 rd division on the vernier coincides with a MS division). Reading of the instrument = MS div + (coinciding VS div x L.C.) = 3.2 + (3 x 0.01) = 3.2 + 0.03 = 3.23 cm To measure the inner and outer diameter of a hollow cylinder or ring, inner and outer callipers are used. Take measurements by the two methods as shown in figure below. Micrometer ScrewGauge Micrometer screwgauge is another instrument used for measuring accurately the diameter of a thin wire or the thickness of a sheet of metal. It consists of a Ushaped frame fitted with a screwed spindle which is attached to a thimble. Screwgauge The screw has a known pitch such as 0.5 mm. Pitch of the screw is the distance moved by the spindle per revolution. Hence in this case, for one revolution of the screw the spindle moves forward or backward 0.5 mm. This movement of the spindle is shown on an engraved linear millimeter scale on the sleeve. On the thimble there is a circular scale which is divided into 50 or 100 equal parts. When the anvil and spindle end are brought in contact, the edge of the circular scale should be at the zero of the sleeve (linear scale) and the zero of the circular scale should be opposite to the datum line of the sleeve. If the zero is not coinciding with the datum line, there will be a positive or negative zero error as shown in figure below. Zero error in case of screw gauge While taking a reading, the thimble is turned until the wire is held firmly between the anvil and the spindle. The least count of the micrometer screw can be calculated using the formula given below: Least count = 0.01 mm Determination of Diameter of a Wire The wire whose thickness is to be determined is placed between the anvil and spindle end, the thimble is rotated till the wire is firmly held between the anvil and the spindle. The rachet is provided to avoid excessive pressure on the wire. It prevents the spindle from further movement. The thickness of the wire could be determined from the reading as shown in figure below. Reading = Linear scale reading + (Coinciding circular scale x Least count) = 2.5 mm + (46 x 0.01) = (2.5 + 0.46) mm = 2.96 mm Relationship in the Metric system of length 1 kilometer (km) = 10 3 m 1 centimeter (cm) = 10 -2 m 1 millimeter (mm) = 10 -3 m Q1. The pitch of a screw gauge is 1mm and there are 100 divisions on the circular scale. While measuring diameter of a wire the linear scale reads 1mm and 47 th division on the circular scale coincides with the reference line. The length of wire is 5.6 cm. Find the curved surface area (in cm 2 ) of the wire in appropriate number of significant digits. Q2. In a Searle’s experiment the diameter of the wire as measured by a screw gauge of least count 0.001 cm is 0.050 cm. The length measured by a scale of least count 0.1 cm is 110.0 cm. When a weight of 50 N is suspended from the wire the extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm. Find the maximum error in the measurement of Young’s modulus of the material of the wire from these data. Solution: Ans 1: Dia of wire=1+47/100=1.47 mm=0.147 cm; Length of wire =5.6 cm Hence curved surface area =πD L=2.6 cm 2 . Ans 2: Y=F.L/A.δ · F.L/πr 2 δ;On differentiating two sides and dividing with Y on two sides of the equation we get, dY/Y=dL/L-2/rdr-dδ /δ For maximum error all terms should be positive dY/Y=dL/L+2/rdr+dδ /δ =0.0489; dY=1.09x10 10 Motion in One Dimension Introduction We have studied till now that the laws of physics are expressed in terms of physical quantities. These quantities are either fundamental or derived one from these fundamental quantities so that law of nature could be best expressed in terms of involved quantities. These quantities are meaning less until unless we have set certain standards for quantifying the physical quantity. The set of standards and the units involved in totality denotes the true value of the physical quantity and then only can participate in mathematical manipulations of laws of physics. Beside that these physical quantities don’t take part in mathematical operations like ordinary numeric values but are assigned certain predefined properties according the role to play to best describe the fundamental law of nature and are so called as scalar, vector or tensor quantity. Now we shall start our expedition to understand the very fundamental aspect of physical observation of nature involving the motion of particles or rigid bodies. These all aspects of motion are covered in the study of classical mechanics; the oldest branch of Physics, which is further, subdivided namely Statics and Dynamics. (A) The statics is the branch of Physics that deals with the study of physical objects or system of objects that are at rest. (B) The Dynamics is the branch of Physics that deals with the study of physical objects or system of objects that are in motion. When we are not concerned with the cause of motion and limit our study to the involved parameters of particulate motion only then the dynamics of particles may be termed as Kinematics. As we have come across with two terms rest and motion and for the purpose of Physics it is not so easy to call the term rest and motion as we usually do with. The terms rest and motion denotes the state of motion of the object under consideration with respect to frame of reference of observer. We have stated earlier in our previous discussions that measurement of physical quantities depends upon the frame of reference of observer. For an example an object at rest with respect to the observer attached to a moving train, is in motion with respect to an observer attached to the Earth. Hence the complete description of a physical quantity desires a proper set of standard units and complete information of frame of reference for observance. The mechanics of motion of objects we are dealing here is a part of Classical mechanics, also known as Newtonian mechanics (1860), doctrine of Sir Alexander Newton. As physics is not a static tool for explaining all the phenomenon of nature but itself a developing one to face the challenges posed by the incident results. The Newtonian mechanics was developed for understanding the observations regarding motions of objects in the nature and are perceived through the naked eyes. It proves all experimental results when the speeds of objects are slow enough in comparison to the speed of light. However it is helpless in describing the collisions, decay and interactions of elementary particles of atom like proton, electron and neutron moving at high speed of the order of speed of light, regarding deviation from the results for the relative velocity of particles observed from the different frame of reference and prediction of position and velocity at a time for the high speed moving particles which is an essential requirement for describing any physical quantity. For explaining the above phenomenon the new theories like Einstein theory of relativity (1905) and Quantum mechanics (1925) been developed which satisfies all experimental results involving particles of small mass and high speed (v≅ c). These theories are considered as a more general theory and Newtonian mechanics is considered as a special case of application for particles moving with velocity very very less comparable to the light. We now return to the classical mechanics to study the slow motions, which can be perceived through our common sense without any intuitive effort. But before taking a leap for understanding laws of physics involved in motion of objects, we shall introduce the physical quantities involved in various types of motion and start our expedition with the simplest kind of motion that is motion in a single direction also called as one-dimensional motion. At the end of this chapter of around one hour, you will be able to learn Concept of point object Motion in One Dimension Distance Displacement Average Speed Average Velocity Instantaneous Velocity Average acceleration Instantaneous acceleration Motion with constant acceleration Motion with variable acceleration Time dependent acceleration Position dependent acceleration Velocity dependent acceleration Relative motion in one dimension Concept of point object When we think of a motion there may be different possibilities of motion either in the choice of path or the choice of the body itself. But to limit our discussion to the beginners we have simplified our choices. We start our discussion with motion of objects that have assumed physically zero dimensions, called particles or point objects. One tends to think of a particle as a tiny object, e.g., a piece of shot, but actually no size limit is implied by the word “particle”. If we are not interested in the rotational motion of an object, any object can be considered as a particle. For example, sometimes we consider the motion of earth around the sun. In this case we consider the motion of the center of the Earth in the circular path and ignore the rotational motion of the earth on its own axis then for our treatment the Earth may be considered as a particle. In some astronomical problems the solar system or even a whole galaxy is treated as a particle. In other words when the size of the object is very small in comparison to the distance it moves then the object may be treated as point object. There is a specific nature of point object is that all the points on the object undergo same displacement, hence the displacement of any of the points may be treated in the experiment. Hence in all the concept of point object has its significance with reference to the type of motion of the object is under consideration and displacement that undergo in comparison to its own dimensions. Motion in One Dimension 0 1 2 3 4 5 6 D i s t a n c e ( m i l e s ) T i m e 0 . 0 H r s . 0 1 2 3 4 5 6 D i s t a n c e ( m i l e s ) T i m e 0 . 1 0 H r s . 0 1 2 3 4 5 6 D i s t a n c e ( m i l e s ) T i m e 0 . 2 0 H r s . To describe the motion of the particle, we are now in a stage to develop the concepts of displacement, velocity and acceleration. In the general motion of a particle in three dimensions, these quantities are vectors, which have direction as well as magnitude. However at this stage we have confined our discussion to the movement of a particle in a straight linear path, with only two possible directions, distinguished by designating one positive and the other negative. A simple example of one-dimensional motion is a vehicle moving along a straight, aligned road. We can choose any convenient point on the vehicle for the location of the point mass for the discussion of motion. We shall now define various physical quantities associated with rectilinear motion of a point object and then try to understand the need of their development. Distance The distance is defined as the length of actual path the particle traverses in its motion. In our case when particle is restricted its motion in a one dimension then the distance is defined as the length of actual path the particle has traversed irrespective of its direction of motion. For example the vehicle traveling in one dimension in the fig say East-West direction, moving 6 miles East ward and 4 miles back west ward will be defined to have moved a distance 10 miles. Hence it is a scalar quantity and always positive and increasing with time. Its unit are (m) in MKS and (cm) in CGS system. Displacement The displacement of the particle or point object in general is defined as the actual displacement the particle has undergone with respect to its original position in the time interval under consideration. In other words it is the change in position vector of the object in a scheduled time. The displacement of particle is not concerned with the path it has elapsed and journey details. It is a unique value defined with a vector having direction as the movement from its initial to final position and magnitude equal to the straight distance between initial and ending positions.In our previous example for a motion in one dimension the displacement will be 2 miles east ward since the vehicle has traversed a total distance of 10 miles but has been displaced with net amount of 2 miles from it’s initial position say at origin. Hence it is a vector quantity and units are same as of distance. Average Speed The average speed of a particle is defined as the ratio of the total distance moved to the time taken up to that instant of motion. So it is simply the time rate at which the distance moved by the particle. So Average speed = total distance / total time For example, if the vehicle in the previous example moves a total distance 10 milesin what ever direction in 0.2 h, than its average speed is hour miles hours miles t d v / 50 2 . 0 10 ~ · · ∆ ∆ · which suggest us that the driver might have moved with this speed uniformly elapsing total distance of10 miles in the given time interval of 0.2 hours. It is therefore a scalar quantity having SI units of meters per second, and written as m/s. Test your understanding Ex1. A car covers a distance of 2 km in 2.5 minute. If it covers half of the distance with speed 40 km/hr, the rest distance it will cover with speed- (A) 56 km/hr (B) 60 km/hr (C) 50 km/hr (D) 48 km/hr Sol : Time taken to cover the first half distance t 1 =Distance traveled/Speed =(1/40)60 = 1.5min; Time taken to cover remaining half distance t 2 =T-t 1 = 1 min Speed of the car in the next half journey= Distance traveled/Time taken =1km/min=60km/hr (B) Ex2. Mark the wrong statement- (A) The time displacement graph of a particle cannot be parallel to the time axis. (B) The time displacement graph of a particle cannot be perpendicular to the time axis. (C) The time velocity graph of a particle cannot be perpendicular to the time axis (D) The area of the time velocity graph gives the displacement Sol : The time displacement graph of a particle can be parallel to the time axis for a particle at rest. (A) Ex3.The position vector of a particle is determined by the expression ∆ r = 3 t 2 i ^ + 4 t 2 j ^ + 7 k ^ . The distance traveled in the first 10s is- (A) 100 m (B) 150 m (C) 500 m (D) 300 m Sol: r = 3t 2 i^ +4t 2 j^ +7k^; v = dr/dt =dx/dt i^+dy/dt j^+dz/dt k^= 6ti^+8tj^ ; S = 0 ∫ 10 v dt = 0 ∫ 10 10t dt= 500 m Ex4.What is the displacement of the point of the wheel initially in contact with the ground when the wheel roles forward half a revolution? Take the radius of the wheel R and x-axis in forward direction. (A) R/√π 2 +4(B) R√π 2 +16(C) 2π R (D) π R Sol:Displacement in x direction ( x )= π R; Displacement in y direction (y) = 2R; Resultant displacement = √x 2 +y 2 = R√π 2 +4 Average Velocity The concept of velocity is similar to that of speed but differs in respect of that here net displacement is accounted in place of total distance moved by the particle. So average velocity is defined as the ratio of the net vector displacement of the particle up to that instant and the total time of motion up to that instant. Since the displacement is a vector quantity so the average velocity is also a vector because it includes the direction of motion. We have defined earlier the term displacement as a vector showing change in position of a particle with magnitude of direct distance measured between initial and final position and directed to the line joining the two points. Hence the average velocity is also a vector quantity, as a rate of change of displacement or in other words displacement per unit time, having the same direction as displacement. Now in the case of previous example the vehicle has elapsed a net displacement of 2 miles in 0.2 hours in the direction toward east so the average velocity in the given time frame may be given as ( ) Eastward hour miles hours miles t x v / 10 2 . 0 2 · · ∆ ∆ · Now consider the motion of the vehicle as discussed in the previous example with direction of motion as along x axis denoting the east ward direction and its position on the x axis at any instant (t) may be shown as in fig . The value of x depends on the unit chosen as the measure of distance and sign depends on its position relative to the origin O; if it is to the right, it is positive; to the left, negative. Suppose that our vehicle under consideration is at position x 1 at time t 1 and at point x 2 at time t 2 . The change in the position of the vehicle x 2 - x 1 is called the displacement in time t 2 - t 1 . It is customary and easy to use the Greek letter ∆ (capital delta) to indicate the change in a quantity. Thus the change in x as ∆ x The average velocity of the vehicle is defined to be the ratio of the displacement ∆ x and the interval ∆ t =t 2 –t 1 : It is to be noted that average velocity like displacement vector can be positive or negative according as displacement in the direction of positive/negative x-axis of the chosen coordinate system. Observation of figdepicts the displacement of particle as difference of ordinate on vertical axis in the said time interval. The line joining two points P 1 , P 2 is the hypotenuse of the triangle with sides ∆ x and ∆ t. The ratio is called the slope of the hypotenuse. In geometric terms it indicates the steepness of the line and as per definition it is a measure of average 1 ∆ t t t · − 2 1 t 2 t x 1 x 2 x m P 2 ∆ x x x · − 2 ( ) x t 1 1 P 2 P t 1 ∆ ∆ ∆ x t s l o p e v · · ν ( ) x t 2 2 0 velocity of the particle. The steeper the slope indicates that particle has traversed greater displacement in a given time interval and has got higher average velocity. Since in our case the vehicle has moved along the +ve x-axis therefore there is no doubt about the direction of velocity that is also along +ve x-axis. Test your understanding Ex5. A body covers a distance AB of 2 km with speed of 2.5 km/h, while going from A to B and comes back from B to A with speed 0.5 km/hr, his average speed will be- (A) 1.5 km/hr(B)0.83 km/hr (C) 1.2 km/hr(D) 1.8 km/hr Sol : V av =4/(2/2.5+2/0.5)=0.83 km/h Ex6. A particle moves with constant speed v along a regular hexagon ABCDEF in same order (i.e. A to B, B to C to D, D to E to F, F to A) then magnitude of average velocity for its motion from A to C is- (A) v(B) v/2(C) √ 3v/2(D) None of these Sol :AC = √3a; t = 2a/v; Av. Velocity = √3v/2 Ex7. A car travels for time t with a uniform velocity of 108 km/h on a straight road and then immediately reverses gear and travels for time t on the same road with a uniform velocity of 72 km/h. then the average velocity of the car in this time interval 2t is- (A) 90 km/h (B) 86 km/h (C) 36 km/h (D) 18 km/h. Sol : Average velocity V av =Total displacement /Total time of motion=(108+72)t/2t=90 Km/h Ex8. A person walks along an east-west street, and a graph of his displacement from home is shown in figure. His average velocity for the whole time interval is- (A) 0 m/s (B) 23 m/s (C) 8.4 m/s(D) None of the above Sol : (A) Since net displacement in whole time interval is zero. o t ( s ) 2 5 1 0 1 5 5 - 3 0 Q . N o . 6 4 S 4 0 Ex9. A point traveling along a straight line traverses one-third the distance with a velocity v 0 . The remaining part of the distance was covered with velocity v 1 for half the time and with velocity v 2 for the other half of the time. The mean velocity of the point averaged over the whole time of motion will be- (A) v 0 (v 1 +v 2 )/3(v 1 +v 2 +v 0 ) (B) 3v 0 (v 1 +v 2 )/(v 1 +v 2 +v 0 ) (C) v 0 (v 1 +v 2 )/(v 1 +v 2 +4v 0 ) (D) 3v 0 (v 1 +v 2 )/(v 1 +v 2 +4v 0 ) Sol : Time of motion of one third distance (T 1 ) =d/3 v 0 , Time of journey of rest 2/3 distance (T 2 ) = 4d/3(v 1 +v 2 ); V av =d/(T 1 +T 2 )= 3v 0 (v 1 +v 2 )/(v 1 +v 2 +v 0 ) Ex10. A table clock has its minute hand 4.0 cm long. The average velocity of the tip of the minute hand between 6.00 AM to 6.30 AM and 6.30 PM will respectively be- (in cm/s) (A) 4.4 x 10 -3 , 1.8 x 10 -4 (B) 1.8 x 10 -4 , 4.4 x 10 -3 (C) 8 x 10 -3 , 4.4 x 10 -3 (D) 4.4 x 10 -3 , 8 x 10 -4 Sol : Average velocity=displacement/time; Av. Velocity (6 AM to 6:30 AM)=8/30x60= 4.4 x 10 -3 Av. Velocity (6 AM to 6:30 PM)=8/12.5x3600=1.8x10 -4 Instantaneous Velocity At a first glance, it seems impossible to define the velocity of a particle at a single instant, i.e. , at a specific time t 1 , the particle is at a single specific position say x 1 then the question arises that if we were talking about velocity of a particle at a single point then as per definition, what about the displacement at that instant? It seems to be a paradox, which can be resolved when we realize motion as a whole at different instants and limit our ∆ t 1 ∆ t 2 ∆ t 3 P 1 P 2 t 1 t 2 ∆t 4 T a n g e n t a t p o i n t P 1 t ( s ) 0 discussion of motion to a infinitesimal time interval (∆ t) tending to zero then in the limiting condition displacement is termed as instantaneous velocity of the particle at that particular instant. The basic requirement for finding out instantaneous velocity of the particle is that one should have a complete mathematical algorithm available for position of particle at any instant and time. Figureshows x (t) curve indicating various sequence of time intervals ∆ t 1 , ∆ t 2 , ∆ t 3 , each one smaller than the previous one. For each time interval ∆ t, the average velocity is the slope of the dashed line appropriate for that interval. This figure shows that as the time intervals becomes smaller, the dashed lines get steeper but never incline more then the line tangent to the curve at point t 1 . We define the slope of this tangent line to be the instantaneous velocity at the time t 1 . The instantaneous velocity is then defined as the limit of the ratio ∆ x/∆ t as ∆ t approaches zero. In the limiting notation this is defined as derivative of x (t) with respect to t at time t 1 and is written as dx/dt and its value could be found by differential calculus. =Slope of line tangent to x (t) at time t=t 1 So instantaneous velocity of the particle at any instant say t dt dx t x v Limt · ∆ ∆ · → ∆ 0 Test your understanding Ex11. At an instant, the coordinates of a particle are x=at 2 , y=b t 2 and z=0,then its velocity at the instant t will be. (A) t√a 2 +b 2 (B) 2t√a 2 +b 2 (C) a 2 +b 2 (D) 2t 2 √a 2 +b 2 Sol : velocity component in x direction v x =dx/dt=2at Velocity component in y direction v y =dy/dt=2bt Resultant Velocity (v) =√ v x 2 +v y 2 =2t√a 2 +b 2 Ex12.The displacement of a body is given by x=√(a 2 -t 2 )+ t cos t 2 , where t is time and a is constant. Its velocity is. (A) a-t 2 -t sint 2 (B) 2 t/√a 2 -t 2 +cost 2 -t sin t 2 (C) -t/√a 2 -t 2 +cost 2 -2t 2 sint 2 (D) –a/(a 2 - t 2 )+cost 2 -t sint 2 Sol : Velocity component in x direction v=dx/dt=-t/√a 2 -t 2 +cost 2 -2t 2 sint 2 Ex 13. If the velocity–time diagram for the rectilinear motion of a particle is as shown in figure (representing half wave of a sine curve). Find the distance traveled by the particle in a time T/2 seconds. (A) T v max /π(B) V max T/3(C) V max T (D) V max T/2 Sol : S=2 0 ∫ Τ/ 2 V m sin 2π t/Tdt=V m T/π Average acceleration Average acceleration is defined as a quantity measuring the rate of change of instantaneous velocity in a certain time interval. The change in velocity vector may be either in magnitude or direction or both and change in either way will be termed as the average acceleration. Since instantaneous velocity is a vector quantity and we are interested in change in vector quantity so the acceleration can only be a vector quantity and will follow the laws of vector algebra. Since for a particle moving in one dimension will have velocity vector having orientation either on +ve or -ve x-axis therefore change in velocity vector will also be associated in the two possible directions. If the change in velocity vector is directed along the direction of velocity vector at the instant of starting point of motion under consideration than the resultant velocity vector gets added up and so the acceleration vector is termed as positive acceleration and if converse is true than the acceleration is designated as negative acceleration. So in final words average acceleration is the ratio of change in instantaneous velocity vector (∆ v) in a given time interval (∆ t). t v a av ∆ ∆ · The dimensions of acceleration in S.I unit system are meter per Second Square. v o t v m T / 2 Test your understanding Ex14. A particle is moving eastward with a velocity of 5 ms -1 in 10 s the velocity change to 5 ms -1 northward. The average acceleration in this time is- (A) Zero (B) 1/√2ms -2 towards north–west (C) 1/√2ms -2 towards north–east (D) 1/2ms -2 towards north–west Sol : Average acceleration=∆ v/∆ t=√50/10=1/√2ms -2 towards north–west Ex15.A rifle bullet loses 1/20 of its velocity in passing through a wooden plank. The least number of planks required stopping the bullet is- (A) 5 (B) 10 (C) 11 (D) 20 Sol : Let x 0 be the thickness of one plank and a be the retarding acceleration produced due to friction of the plank than x 0 =u 2 -v 2 /2a; x 0 =u 2 -(19u/20) 2 /2a; Number of planks=Total path length/thickness of one plank= u 2 /2ax 0 =400/(39)≈11 Instantaneous acceleration If the instantaneous velocity of the particle is varying continuously with position and time then instantaneous acceleration at that particular instant is defined as the change in velocity vector for a time interval ∆ t infinitesimally small tending to zero. It is to be noted that change in velocity vector is also a vector term and the law of vectors has to be followed. We shall see later on that such type of situation may be handled easily when vector is considered to be resolved into its components in the respective directions and the changes are also considered in the respective components directions individually. For the case of motion in one dimension the particle moves in a straight line with two possible directions of motion say either in +ve or -ve x-axis. Now if the particle moves such that velocity at every instant of motion is defined and there is no abrupt change in velocity vector than the instantaneous acceleration vector is just the rate of change in magnitude of velocity vector and direction will be in the direction of velocity vector if velocity time curve is increasing and opposite to the velocity vector if it is a decreasing curve. So if we plot graph between instantaneous velocity of the particle moving in one dimension and time then as per the definition the instantaneous acceleration is defined as slope of tangent line at that instant. t v t a t Lim ∆ ∆ · → ∆ 0 ) ( = Slope of v (t) curve The instantaneous acceleration is therefore derivative of velocity with respect of time and is written as dv/dt. Since velocity is also derivative of x with respect to t therefore acceleration is also referred as second derivative of x with respect to t. We shall study in the later stage how acceleration vector is useful in defining Newton’s second law of motion. Test your understanding Ex16. Relationship between the distance traveled by a body and the time is described by the equation S=A + Bt + Ct 2 + Dt 3 , where C=0.14 m/sec 2 , D= 0.01 m/sec 3 . In what time after motion begins, will acceleration be 1 m/sec 2 and what is the average acceleration during this time? (A) 12 seconds, 0.64 m/sec 2 (B) 6 seconds, 0.5 m/sec 2 (C) 3 seconds, 0.4 m/sec 2 (D) 8 seconds, 0.6 m/sec 2 . Sol : v=B+2C t+3 D t 2 ; a=2C+6 D t=1 at t=12 s a av ={v(12)-v(0) }/∆ t=0.64 m/s 2 Motion with constant acceleration along the axis of initial velocity vector Now let us consider the motion ofa particle moving with constant acceleration vector directed along the axis of the initial velocity vector. Since the acceleration vector is directed along the axis of initial velocity vector than it will lead to the change in magnitude of velocity vector. If the acceleration is positive or directed along the direction of velocity vector at any instant increases the magnitude, otherwise decreases the magnitude. Also if the direction of initial velocity vector is different from the acceleration vector than particle moves in the plane containing both the vectors in a parabolic trajectory, which we shall discuss in the next chapter. The motion of a particle with constant acceleration is quite common in nature. If air resistance is neglected than all the particles irrespective of their masses fall with constant acceleration in the influence of gravity that is attraction of the Earth. This acceleration due to gravity is designated by ‘g’ and has approximate value of 9.81 m/s 2 or 32.2 ft/s 2 For a particle moving with a constant acceleration in one dimension the velocity changes linearly with time. The positive acceleration means it adds the velocity vector and negative acceleration subtracts the velocity vector or in other words, for the case of one dimensional motion, the acceleration vector will be in the direction of velocity vector if it is positive other wise against the velocity vector. The both cases have been illustrated in Figand effect on motion are summarized here below: If velocity of the particle is v 0 at time t=0 then it’s value at time t as per definition is given by ( ) ( ) t a t v v + · 0 If the particle starts at x 0 at time t=0 and it’s position is x at time t then as per definition of displacement ∆ x is given by ( ) t x vav · ∆ Also ( ) 2 2 0 0 at t x v v v v t av + · + · ∆ · ( ) 2 2 0 t v a t x + · ∆ ( ) ( ) 2 2 0 0 t v x a t t x + + · a a a t t t - a t t t ( a ) ( b ) 0 0 We are sometimes interested in finding the final velocity of particle at a particular instant when its initial velocity, constant rate of acceleration and distance traveled or displacement is known to us. Since for constant acceleration 2 0v v v t av + · and t t x v v v t av 2 ) ( 0 + · · ∆ Now after eliminating t from the above equation yields x a t v v ∆ + · 2 0 2 2 If we are interested in finding out the displacement of particle in the n th second of its motion then ( ) 2 1 2 0 1 − + · − · − n a v S S S n n nth These equations are known as Newton’s equations of motion at constant acceleration. Graphical presentationof one-dimensional motionat constant acceleration (1) The graph between positions of particle at different instances vs. time is parabolic curve given by equation as shown in figfor +ve and -ve acceleration respectively. (2) The graph between velocities of particle with respect to time of x t P o s i t i v e C o n s t a n t a c c e t e r a t i o n x t N e g a t i v e C o n s t a n t a c c e t e r a t i o n O O v t O v I n i t i a l v e l o c i t y = ( - ) I ' v e I n i t i a l v e l o c i t y = 0 O v t v O v I n i t i a l v e l o c i t y = ( + ) I ' v e ( a ) ( b ) ( c ) t motion at constant +ve acceleration given by equation as shown in figure(a), (b), (c) for different initial velocities. (3) The graph between velocities of particle with respect to time of motion at constant -ve acceleration given by equation as shown in fig(a), (b), (c) for different initial velocities. The important and well-observed example of motion in one dimension at constant acceleration is the motion of a body under the influence of gravity. At normal heights above the earth surface the force/ acceleration remains constant and its magnitude denoted by g may be taken as 9.80 m/sec 2 . It is directed radially towards the earth center and for our purpose it may be assumed acting vertically downward that is towards the Earth surface and all the motions in the azimuth may be considered as linear motions without influenced by the air drag and all viscous resistance. Let us consider motion of a body falling freely under the influence of gravity. If we choose point of projection as origin and taking downward direction as positive, we have u=0 and a=g as body starts from rest that is falling freely under the influence of gravity only, hence as per Newton’s equation of motion at any instant ; These equations will yield to all unknowns at any instant of motion. It is to be noted that all the equations are vector equations and gives vector quantities on substitution of known parameters. g a + · v t O v I n i t i a l v e l o c i t y = ( + ) ' v e I n i t i a l v e l o c i t y = 0 O v t v O v I n i t i a l v e l o c i t y = ( - ) ' v e ( a ) ( c ) t s ( a ) t v ( b ) t a n θ · g θ t a ( c ) g t t g g ut S 2 2 2 1 2 1 · + · ( ) gt gt u t v · + · These equation are presented graphically as in figure 9 (a), (b), (c) Similarly if we consider motion of a body projected vertically upward under the influence of gravity and choosing point of projection as origin and vertical up direction as positive. We have g a − · t g ut S 2 2 1 − · ( ) gt u t v − · These equation are presented graphically as in figure (a), (b), (c) Test your understanding Ex17. An iron ball and a wooden ball of same radius are released from a height h in vacuum. The time taken by both of them to reach the ground is (A) Roughly equal(B) exactly equal (C) Unequal(D) equal only at equator Sol : (B) Since both have same initial velocity and fall with the same acceleration due to gravity (g) in the absence of air resistance. Ex18. A body moving with uniform acceleration describes 4m in 3 rd second and 12 m in the 5 th second, then distance described in next three second- (A) 100 m (B) 80 m (C) 60 m (D) 20 m Sol : Distance traveled in nth second S n =u+a/2(2n-1); a ( m / s ) 2 o t 2 u / g + v e - v e ( c ) ( a ) ( b ) 2 u / g t 0 u / g + v ( t ) 0 u / g 2 u / g + v e - v e s = u / 2 g 2 a=4 m/s 2 and u=-6 m/s V 5 =-6+4x5=14 and V 8 =-6+4x8=26 S=(V 5 + V 8 )t/2=60 m Ex19. A particle P is at the origin and starts with velocity 2i-4j (m/s) and constant acceleration (3i+5j) m/s 2 after it has traveled for 2 seconds. Its distance from the origin is- (A) 10 m(B) 10.2 m (C) 9.8 m (D) 11.7 m Sol : x = u x t+1/2 a x t 2 =10m; y= u y t+1/2 a y t 2 = 2; ¹ R ¹ =√ x 2 + y 2 =10.2 Ex20. A body moves from rest with constant acceleration then the variation of its K.E. with the distance (S) traveled is represented by: (A) Straight line (B) Parabola (C) Hyperbola (D) None of these Sol : K.E. = 1/2mv 2 = 1/2m(u 2 +2as) = m a s which is a equation of straight line. Ex21.A car accelerates from rest at a constant rate αfor sometime after which it decelerates at constant rate βto come to rest. If the total time elapsed is t sec, the maximum velocity of car will be- (A) α β /t (α +B)(B) (α +B) t/α b(C) α β /(α +B) t (D)α β t/(α +β ) Sol : v m = α t 1 ; v m = βt 2 ; t = t 1 + t 2 = v m /α+ v m /β; v m= α β t/(α +B) Ex22. A stone is dropped into a well and the sound of impact of stone on the water is heard after 2.056 sec of the release of stone from the top. If acceleration due to gravity is 980- cm/sec 2 and velocity of sound in air is 350 m/s, calculate the depth of the well (A) 1.96 m (B) 19.6 m (C) 6.91 m (D) 69.1 m Sol:Let dbe the depthofwell and t be the time taken to reach the stone on the water surface then d = 1/2gt 2 ; t = √2d/g; Also time taken by the sound of splash to reach the projector= d/vs Total timetakensincethereleaseof stonetosoundof splashtoreachtheprojector= d/vs+√2d/g = 2.05; d = 19.81m Ex23. A particle travels for 40 seconds under the influence of a constant force. If the distance traveled by the particle is s 1 in the first twenty seconds and s 2 in the next twenty seconds. Then (A) s 2 =s 1 (B) s 2 =2s 1 (C) s 2 =3s 1 (D) s 2 =4s 1 Sol : Let T be the total time of motion then distance moved in half of time period as say 20s is s 1 = ½ a (T/2) 2 =1/8a T 2 ; Total distance moved in time T is given as s 2 = ½ aT 2 Distance moved in second time interval of 20s (s 2 ) =1/2 aT 2 -1/8 aT 2 =3/8aT 2 And so s 2 /s 1 = 3 Ex24. A bullet fired into a fixed block of wood loses half its velocity after penetrating 60 cm. Before coming to rest it penetrates a further distance of (Assume constant frictional resistance) (A) 60 cm(B) 30 cm (C) 20 cm(D) 10 cm Sol : Let s be the distance moved by the bullet in first part of motion when loses half of its speed as u 2 /4 = u 2 – 2as 1 ; a = 3u 2 /8s 1 ; Let s 2 be the total distance moved up to the instant it comes to rest is given as s 2 = u 2 /2a = 4/3s 1 ; Distance moved in the second time interval (∆ s 2 )= s 2 -s 1 = s 1 /3 = 20 cm Ex25. A particle is projected vertically upwards and it reaches the maximum height H in time T seconds. The height of the particle at any time t will be (A) g(t-T) 2 (B) H-1/2g(t-T) 2 (C)1/2g(t-T) 2 (D) H-g(t-T) Sol : h= H-1/2g(t-T) 2 Ex26.At time t=0, an object is released from rest at the top of a tall building. At the time t 0 a second object is dropped from the same point, ignoring air resistance the time at which the objects have a vertical separation of l is- (A) t=l/gt 0 +t 0 /2 (B) t= l/gt 0 -t 0 /2 (C) t=l/gt 0 (D) none of the above Sol : Total displacement up to time t, ∆ S = l = ½ g t 0 2 + g t 0 (t- t 0 ) 2l/gt 0 = 2t-t 0 ; t = l/gt 0 +t 0 /2 Ex27. A steel ball is dropped from the roof of a building. A man standing in front of a 1 m high window in the building notices that the ball takes 0.1s to fall from the top to bottom of the window. The ball continues to fall and strikes the ground. On striking the ground, the ball gets rebounded with the same speed with which it hits the ground. If the ball appears at the bottom of the window 2 s after passing the bottom of the window on the way down, then the height of the building is (A) 12.40 m (B) 21.0 m (C) 24.0 m (D) 42.0 m Sol : Let u and v be the velocities of the ball at the instant it appears at the top and bottom of window during the downward journey. So {(u+v)/2}(0.1)=1; Or u+v=20 Also v=u+g(0.1) Or v=u+1; u=19/2 m/s Height above window (h 1 )= u 2 /20=(9.50)2/20 =4.50 m; Height below window h 3 = v+1/2g= 15.5 m; Total height of the building H = 4.50+1+15.50= 21.0 m Ex28. The vertical height of P above the ground is twice that of Q. A particle is projected downward with the speed of 9.8 m/s from P and simultaneously another particle is projected upward with the same speed of 9.8 m/s from Q both particles reach the ground simultaneously. The time taken to reach the ground is- (A) 3 sec (B) 4 sec (C) 5 sec (D) 6 sec. Sol : PG = 9.8t{1+0.5t}; QG = 9.8t{0.5t-1}; PG/QG = 2; t = 6 sec Ex29. A particle is projected vertically upwards from a point x on the ground. It takes a time t 1 to reach a point A at a height h above the ground. It continues to move and takes a time t 2 to reach the ground. The velocity of the particle at half the maximum height is (A) 2√2/g (t 1 + t 2 ) (B) g/2√2 (t 1 + t 2 ) (C) 2√2g/(t 1 + t 2 ) (D) (t 1 + t 2 )/ 2√2g Sol : Tm= (t 2 -t 1 )/2+t 1 ; Tm = (t 1 +t 2 )/2; u = g(t 1 +t 2 )/2; H = u 2 /2g = g 2 (1+t 2 )2/8g; V m = g(t 1 +t 2 )/2√2 Ex30. A juggler keeps on moving four balls in the air throwing the balls after regular intervals. When one ball leaves his hand (speed=20 m/s) the position of the other balls (height in meter) will be (take g=10 m/s 2 )- (A) 10, 20, 10(B) 15, 20, 15(C) 5, 10, 20(D) 5, 10, 20 Sol : The ball should be placed taking equal time interval of replacement and for that should be taking time interval 2/2=1 sec and respective positions of 15, 20,15m from the hand. Ex31. A man standing on the edge of a cliff throws a stone straight up with initial speed u and then throw another stone straight down with the same initial speed and from the same position. Find the ratio of the speed the stones would have attained when they hit the ground at the base of the cliff (A) √ 2:1(B) 1:√2 (C) 1:1 (D) 1:2 Sol : ( C ) The stones have the same velocity at the point of throw in the downward direction. Motion with variable acceleration So far we have discussed the motion of a particle moving in one dimension under constant acceleration. Now we shall extend the motion of a particle with variable acceleration in one dimension. Since motion is in one dimension then variation in acceleration is limited to change in magnitude only. We shall discuss the variation in three categories as: Time dependent acceleration Under this category of one-dimensional motion the acceleration of particle is a function of time or in other words it may be written as a=f (t). Then velocity may be obtained as integral off(t) with respect to dt and displacement may be obtained as further integral with respect to dt. ( ) t f dt dv a · · ( ) ( ) t F dt t f dt ds v · · · ∫ ( ) ∫ · dt t F s Position dependent acceleration Under this category of one-dimensional motion the acceleration of particle is dependent upon the displacement of particle from origin. Then Integrating both sides of equation results velocity as a function of displacement of particle and may be further integrated to find displacement as a function of time. ( ) x f dx dv v dt dv a · · · Velocity dependent acceleration Under this category of one-dimensional motion the acceleration of the particle is dependent upon the instantaneous velocity of particle. Then Integrating both sides of equation results velocity as a function of time, which may be further, integrated with respect to time to find displacement as a function of time. ( ) v f dt dv a · · Test your understanding Ex32. The acceleration (a) of moving particle varies with displacement (x) according to the following relation a=x 2 +3x, then correct relation between velocity and displacement is- (A) v=√x 3 +3x 2 +c (B) v=√2/3 x 3 +3x 2 +c ( C ) v=[2/3x 3 +3x 2 +c](D) v=2x+3where c is a constant Sol : a = x 2 +3x; a= vdv/dx = x 2 +3x; v = √2/3x 3 +3x 2 +c 1 Ex33. A particle moving along straight line has a velocity v ms -1 , when it cleared a distance of x metres. These two are connected by the relation v= √49+x. When its velocity is 1ms -1 , its acceleration (in ms -2 ) is- (A) 2 (B) 7 (C) 1 (D) 0.5 Sol : a= v dv/dx=0.5 m/s 2 Relative motion in one dimension As we have stated earlier that measurement of certain physical quantities like displacement, velocity and acceleration depends upon the reference frame of observation. The reference frame of observation may be defined as a frame from which observations are made. It may be either a frame fixed to the earth or a moving one. Now let us investigate how the measurements may differ when physical parameters like displacement, velocity and acceleration of the same moving object are measured from different reference frames. Now let us consider the simple one-dimensional motion of an object P. Initially at time t=0, the object P, earth frame and moving frame are all positioned at origin of the coordinate system as shown in figure. Now after a lapse of time ∆ t the object P has made a displacement a with respect to the frame attached to the earth and the moving frame has also elapsed a displacement b with respect to the earth frame. It is most worthy to note here that the displacement of the object P as measured by the moving frame will not be the same as measured by the earth frame, because of the displacement of the moving frame in the same interval and would be measured as a-b. It is now clear that the displacement and hence velocity measurements are dependent upon the frame of reference from where the physical quantities are observed. How ever the acceleration of the object as measured by the moving frame differs from the earth frame only when the moving frame is also accelerating. In other words the acceleration of an object is same as observed by the earth frame and a frame moving with constant velocity.For example if an object is accelerating with an acceleration a 1 with respect to the earth frame b a - b O x P o s i t i o n a t t = t + ∆ t y a P o s i t i o n a t t = 0 O ’ and at the same time the moving fame is also accelerating with an acceleration a 2 with respect to the earth frame than the acceleration of the object with respect to the moving frame will be given by a 1 -a 2 . The reference frames at rest or moving with constant velocity are termed as inertial reference frame and accelerating frames are known as non inertial frames. We shall discuss in detail the inertial and non inertial frames during the study of applications of Newton’s laws of motion, which are applicable in inertial reference frames where true value of acceleration is measured and inertial state of particle under observation is defined. However it is clear from the so far discussion that measured value of any physical quantity involving displacement depends upon the frame of reference of observation and for correct interpretation of physical quantity appropriate reference frame should be ascertained. Test your understanding Ex34. A boat P is moving at 40 km/hr and another boat Q is moving at 20 km/hr. Which one of the following is not a possible value for their relative velocity – (A) 10 km/hr. (B) 20 km/hr. (C) 30 km/hr. (D) 40 km/hr. Sol : (A) The relative velocity may have values v1-v2 to v1+v2. Ex35. Two trains along the same straight rails moving with constant velocities 60 km/hr and 30 km/hr towards each other. If at time t=0, the distance between them is 90 km, the time when they collide is- (A) 1 hr(B) 2 hr(C) 3 hr(D) 4 hr Sol : (A) Time of collision (t)= Initial Displacement/(Relative velocity of approach)= 1 hr Ex36. A boat moves relative to water with a velocity which is 1/n times the river flow velocity. At what angle to the stream direction must the boat to move for minimizing drifting? (A) π /2 (B) sin -1 (1/n) (C) π /2 + sin -1 (1/n) (D) π /2- sin -1 (1/n) Sol : Horizontal drift (D)= {v-v/n sinθ} t={v-v/n sinθ} L/( v/n cosθ sin θ= 1/n; θ=sin -1 (1/n); φ= π /2+sin -1 (1/n) Ex37. A thief walking slowly along a road sees a policeman at a perpendicular distance L from him and starts running at a constant speed u along the road. The policeman also starts running simultaneously with speed v always aiming at him. Find how soon the policeman will catch the thief- (A) t= v L/v 2 -u 2 (B) t= vL/v 2 -u 2 (C) t= uL/(v 2 -u 2 ) (D) t= v 2 /L 2 (v 2 -u 2 ) Sol : 0 ∫ Τ (u cosα -v)dt=l; u 0 ∫ Τ cosαdt- vT=l 0 ∫ Τ v cosαdt=uT ⇔ 0 ∫ Τ cos α dt=uT/v ; T=lv/(v 2 -u 2 ) Ex38. Two points move in the same straight line starting at the same moment from the same point. The first moves with constant velocity u and the second with constant acceleration f. During the time that elapses before second catches the first, the greatest distance between the particles is – (A) u/f(B) u 2 /2f(C) f/2u 2 (D) f/u 2 Sol : At max separation ft=u or t=u/f ; Xm = u 2 /f-1/2fu 2 /t 2 = u 2 /2f Ex39.A truck starts from rest with an acceleration of 1.5 m/s 2 while a car 150 m behind starts from rest with an acceleration of 2 m/s 2 . How long will it take before both the truck and car side by side? (A) 5.24 sec(B) 24.5 sec (C) 2.45 sec. (D) 52.4 sec. Sol : (B) 150= ½ (2-1.5) t 2 ; t=24.5 sec Ex40. Three points are located at the vertices of an equilateral triangle whose side equals a. They all start moving simultaneously with velocity v constant in magnitude. With the first point heading continuously for the second, the second for the third and the third for the first. How soon will the points converge? (A) 3v/a (B) 2a/3v (C) a/3v (D) a/v Sol : t = a/(v+vcos60)= 2a/3v Ex41. A motor cycle and a car start from rest at the same place at the same time and travel in the same direction .The cycle accelerates uniformly at 1m/s 2 up to a speed of 36 km/h and the car at 0.5 m/s 2 up to a speed of 54 km/h then the distance at which the car overtakes the cycle (A)100 m(B) 200 m(C) 300 m(D) 400 m Sol : Vcy = 10 m/s; Vca = 15 m/s; Time of acceleration t 1 =10 sec and t 2 =30 sec; Scy = 50 m t 2 – 40t+200=0; t=6 sec >30 sec; 50+10(t-10)=225+15(t-30); t=35 sec and Scy=Sca=300 m Ex42. A passenger is standing d m away from a bus. The bus begins to move with constant acceleration a. To catch the bus, the passenger runs at a constant speed v towards the bus. The minimum speed of the passenger so that he may catch the bus will be- (A) 2ad(B) √ ad (C) √ 2ad(D) ad Sol : d = v.t – 1/2at 2 ; at 2 - 2vt+2d = 0; t = (2vt√ 4v 2 – 8ad)/2a; t = v/at 1/a√v 2 -2ad for a real t; v 2 – 2ad>0; v min 2 = 2ad; v min = √2ad Ex43. A ship sails in still water at the rate of 5 m/s. it is sailing northwards in a river flowing eastwards with a velocity of 3 m/s. A monkey is climbing a vertical pole on the ship at the rate of 2 m/s. A person is walking along the bank of the river at the rate of 1 m/s. To him the monkey will appear climbing at the rate of- (A) √10 m/s(B) √11 m/s (C) √33 m/s (D) √45 m/s Sol : Velocity of monkey relative to man V rel = √29+4 = √33 and V rel = √41+4 = √45 Ex44. A string passes over a fixed pulley. Two boys P and Q of the same mass hang at the same height at each end. Both start to climb upward at the same time to reach the pulley. The velocity of P relative to the string is v and that of Q is 3v, then the time taken by P to reach the pulley is equal to (A) 1/3 rd of the time taken by Q (B) 3 times the time taken by Q (C) the time taken by Q (D) twice the time taken by Q Sol : (C) Velocity of Q relative to pulley = 3v-v= 2v↑; Velocity of P relative to pulley = v+v= 2v↑ Single Choice Type Objective Questions Q1. At a certain moment of time the angle between velocity vector v and the acceleration a of a particle, is greater than 90 0 . What can be inferred about its motion at the moment? (A) It is curvilinear and decelerated (B) it is rectilinear and accelerated. (C) It is curvilinear and accelerated (D) it is rectilinear and decelerated Q2. The motion of a particle of mass 1kg is confined to a plane and is determined by x=3t 2 , y=2t 3 where x and y are its coordinates at time t, then magnitude and direction of the force on the particle at t=1/2 second is- (A) 2 N at 30 0 with the x-axis (B) 6√2N at 45 0 with the x-axis (C) 6 N at 30 0 with the x-axis (D) 2N at 45 0 with the x-axis Q3. For a particle moving along a straight line, the displacement x depends on time t as x=α t 3 +β t 2 +γ t+δ . The ratio of its initial acceleration to its initial velocity depends- (A) Only on γ and β(B) only on αand γ(C) only on αand γ(D) only on α Q4. Which one of the following curves does not represent motion in one dimension? Q5. The length of second’s hand in a watch is 1cm. The change in velocity of its tip in 15 second is (A) Zero (B) π(30√2) cm/sec (C) π /30 cm/sec. (D) (π√ 2)/30 cm/sec. Q6. The modulus of the acceleration vector is constant. The trajectory of the particle is a/an- (A) Parabola (B) ellipse (C) hyperbola (D) circle v t v t x t v t Q7. A body A is thrown vertically upward with the initial velocity v 1 . Another body B is dropped from a height h. Find how the distance x between the bodies depends on the time t, if the bodies begin to move simultaneously. (A) x=h-v 1 t (B) x=(h-v 1 )t (C) x= h – v 1 /t (D) x= h/t-v 1 Q8. Which one of the following equations represents the motion of a body with finite constant acceleration in these equations y denotes the displacement of the body at time t and a, band c are the constant of the motion (A) y= a/t + bt(B) y=at(C) y=at + bt 2 (D) y=at + bt 2 + ct 3 Q9. The motion of particle is defined by x=a cos ω t and y=a sin ω t. The acceleration of particle is – (A) aω(B) a 2 ω(C) aω 2 (D) a 2 ω 2 Q10.The displacement- time relationship for a particle is given by x=a 0 + a 1 t + a 2 t 2 . The acceleration of the particle is- (A) a 0 (B) a 1 (C) a 2 (D) 2a 2 Q11. The acceleration vector of a particle is a constant. The trajectory of the particle is a/an- (A) parabola (B) ellipse (C) hyperbola (D) circle Q12. Which of the following distance- time graphs represents one-dimensional uniform motion? Q13. A horse rider is moving towards a big mirror with velocity v. The velocity of his image with respect to him is (A) 0 (B) 4v (C) 2v (D) v x t x t x t x t ( A ) ( B ) ( C ) ( D ) Q14. Figure shows the displacement- time graphs for two boys going home from the school. Which of the following statements is correct about their relative velocity- (A) first increases and then decreases (B) first decreases and then increases (C) is zero (D) is non zero but constant Q15.A particle is confined to move along the x-axis between reflecting walls at x=0 and x=a between these two limits, moves freely at constant velocity v. If the walls are perfectly reflecting than its displacement time graph is- Q16. The acceleration time graph of a particle moving along a straight line is as shown in figure. At what time the particle acquires its initial velocity? (A) 12 s(B) 5 s(C) 8 s (D)16 s x t A B t i m e 0 a a / v d i s p l a c e m e n t 2 a / v 3 a / v t i m e a / v 2 a / v 3 a / v 0 a d i s p l a c e m e n t a t i m e 0 a a / v d i s p l a c e m e n t 2 a / v 3 a / v t i m e 0 a a / v d i s p l a c e m e n t 2 a / v 3 a / v ( A ) ( B ) ( C ) ( D ) a ( m / s ) 2 1 0 4 t ( s ) o Q . 2 8 Q17. The displacement of a particle as a function of time is shown in fig. The fig indicates that- (A) The particle starts with a certain velocity, but the motion is retarded and finally the particle stops (B) The velocity of particle is constant throughout (C) The acceleration of the particle is constant throughout (D) The particle starts with a constant velocity, the motion isaccelerated and finally the particle moves with another constant velocity. Q18. A ship sailing south-east sees another ship which is steaming at the same rate as itself and which always appears to be in a direction due east and to be always coming nearer. Find the direction of motion of second vessel. (A) sails south-west(B) sails south-east (C) sails 30 0 south of west(D) sails 30 0 south of east Q19. An αparticle travels along the inside of straight hollow tube, 2.0 meter long, of a particle accelerator under uniform acceleration. How long is the particle in the tube, if it enters at a speed of 1000 m/s and leaves at 9000 m/s. (A) 4 x 10 -4 sec(B) 2 x 10 -7 sec(C) 4 x 10 -3 sec (D) 2 x 10 -6 sec Q20. The adjoining cure represents the velocity–time graph of a particle, its acceleration value along OA,AB and BC in metre/sec 2 are respectively- (A)1, 0, -0.5 (B)1, 0, 0.5 (C)1, 1, 0.5 (D)1, 0.5, 0 s ( m ) 4 t ( s ) o 2 0 Q . 2 9 0 1 0 2 0 3 0 V e l o c i t y ( m / s ) A B 1 0 5 t i m e ( s e c ) Q21. The v-t graph of a linear motion is shown in adjoining figure. The distance from origin after 8 seconds is (A) 18 meters (B) 16 meters(C) 8 meters(D) 6 meters Q22. The displacement-time graph of a moving particle is shown in the fig, the instantaneous velocity is negative at the point (A) D (B) F (C) C (D) E Q23. The graph between the displacement x and time t for a particle moving in a straight line is shown is figure. During the interval OA, AB, BC and CD the acceleration of the particle is- (A) +, 0, + , + (B) - , 0 , + , 0 (C) + , 0 , - , + (D) - , 0 , - , 0 Q24. A particle starts out at t=0 from the point x 0 =10 m with an initial velocity v 0 =15 m/s and a constant acceleration a = -5m/s 2 . Then its velocity – time graph is- 1 3 0 4 - 2 4 8 t ( s ) v ( m / s ) 5 7 y D C x T i m e D i s p l a c e m e n t x A D C B t o 0 0 2 4 6 ( s ) 1 0 2 0 v ( m / s ) t 0 0 2 4 6 ( s ) 1 0 2 0 v ( m / s ) t 0 0 2 4 6 ( s ) 1 0 2 0 v ( m / s ) t 0 0 2 4 6 ( s ) 1 0 2 0 v ( m / s ) t ( A ) ( B ) ( C ) ( D ) Q25. A particle starts out at t=0 from the point x 0 =10 m with an initial velocity v 0 = 15 m/s and a constant acceleration a= - 5m/s 2 . Then its displacement–time graph is – Q26. The displacement time graph for a one dimensional motion of a particle is shown in figure. Then the instantaneous velocity at t=20 sec is (A) 0.1 m/s (B) -0.1 m/s (C) -0.05 m/s (D) 1.0 m/s Q27. The acceleration versus time graph of a particle is as shown figure.The respective v-t graph of the particle is- v t O v O t v t O v t O ( A ) ( B ) ( C ) ( D ) Q28. The displacement- time graph of a moving particle with constant acceleration is shown in the figure. The velocity-time graph is given by- x ( m ) 4 0 3 0 2 0 1 0 0 0 2 6 4 t ( s ) x ( m ) 4 0 3 0 2 0 1 0 0 0 2 6 4 t ( s ) x ( m ) 4 0 3 0 2 0 1 0 0 0 2 6 4 t ( s ) x ( m ) 4 0 3 0 2 0 1 0 0 0 2 6 4 t ( s ) ( C ) ( D ) ( A ) ( B ) 2 0 3 0 4 0 t ( s ) o 6 0 1 S ( m ) 2 Q . 4 0 t O a 2 o s x ( m ) ( t / s ) 1 1 0 0 - 1 0 ( a ) 2 t ( s ) 1 0 0 - 1 0 ( b ) 2 t ( s ) v ( m / s ) v ( m / s ) 1 0 0 - 1 0 ( c ) 2 t ( s ) v ( m / s ) 1 0 0 ( d ) 2 t ( s ) v ( m / s ) 1 1 1 1 Q29. Two balls are dropped from the top of a high tower with a time interval t 0 second, where t 0 is smaller than the time taken by the ball to reach the floor, which is perfectly in elastic. The distance’s’ between the two balls, plotted against the time lapse t from the instant of dropping the second ball is represented by Q30. The graph below describes the motion of a ball rebounding from a horizontal surface being released from a point above the surface. The quantity represented on the y-axis is the ball’s – (A) Displacement (B) velocity (C) acceleration (D) momentum. Q31. The acceleration of a particle as a function of time is a= 1.5 t-0.15 t 2 . The particle starts motion from rest at a time t = 0 sec. Then the maximum velocity in the forward direction is. (A) 10 m/s (B) 25 m/s (C) 50 m/s (D) none of the above s t ( s ) o ( a ) s t ( s ) o ( b ) s t ( s ) o ( c ) s t ( s ) o ( d ) t ( s ) y o Q . 4 4 Q32. A balloon going upward with a velocity of 12 m/sec is at a height of 65 m from the Earth at any instant.Exactly at this instant a packet drops from it. How much time will the packet take in reaching the Earth? (g=10 m/sec 2 ) (A) 7.5 sec(B) 10 sec(C) 5 sec (D) None Q33. A body of mass 3 kg falls from the multistoried building 100 m high and buries itself2 meters deep in the sand. The time of penetration will be. (A) 9 sec (B) 0.9 sec(C) 0.09 sec (D)10 sec. Q34. A car moving with constant acceleration covers the distance between two points 60 m apart in 6 sec. Its speed as it passes the second point is 15 m/sec. At what prior distance from the first point was the car at rest? (A) 7.5 m (B) 15 m (C) 20 m (D) 25 m Q35. A body starts from rest with constant acceleration a. It’s velocity after n second is v. The displacement of body in last two seconds is- (A) 2v(n-1)/n(B) v(n-1)/n (C) v(n+1)/n (D) 2v(n+1)/n Q36. A rocket is fired vertically from the ground. It moves upward with a constant acceleration 10m/s 2 for 30 seconds after which the fuel is consumed. After what time from the instant of firing the rocket will attain the maximum height? Take g=10 m/s 2 (A) 30 s (B) 45 s(C) 60 s (D) 75 s Q37. A ball is thrown vertically upward with a velocity of 30 m/s. If the acceleration due to gravity is 10 m/s 2 , what will be the distance traveled by it in the last second of motion? (A) 5 m (B) 10 m(C)25 m(D) 30 m. Q38. A stone is released from an elevator going up with an acceleration a. The acceleration of the stone after the release is- (A) a upward (B) (g-a) upward (C) (g-a) downward (D) g downward Q39. The water falls at regular intervals from a tap 5 m above the ground. The third drop is leaving at instant when first one touches the ground. How far above the ground is the second drop at that instant? (A) 1.25 m(B) 2.50 m (C) 3.75 m (D) 4.00 m Q40. An object is thrown upward with a velocity u, then its displacement time graph is- Q41. A car moves with uniform acceleration along straight line PQR. Its speeds at P and R are 5 m/s and 25 m/s respectively. If PQ: QR=1:2, the ratio of the times taken by car to travel distance PQ and QR is- (A) 1:2 (B) 2:1(C) 1:1 (D) 1:5 Q42. The greatest acceleration or deceleration that a train may have is a. The minimum time in which the train can get from one station to the next at a distance s is- (A) √ s/a(B) √ 2s/a(C) 1/2√s/a(D) 2√s/a Q43. A particle moves with a constant acceleration such that in the successive time intervals t 1 , t 2 , t 3 its average velocities are v 1 , v 2 and v 3 . The ratio of v 1 -v 2 and v 2 -v 3 is (A) t 1 - t 2 : t 2 + t 3 (B) t 1 + t 2 : t 2 + t 3 (C) t 1 - t 2 : t 2 - t 1 (D) t 1 - t 2 : t 2 - t 3 Q44. A body is thrown up in a lift with a velocity u relative to the lift and the time of flight is found to be ‘t’. The acceleration with which the lift is moving up will be – (A) u-gt/t (B) u+gt/t (C) 2u-gt/t (D) 2u+gt/t Q45. A ball is dropped from a height of 20 m and rebounds with a velocity, which is 3/4 th of the velocity with which it hits the ground. What is the time interval between the first and second bounces (g=10 m/s 2 ) (A) 3 sec(B) 4 sec (C) 5 sec (D) 6 sec. s o u / g u / g 2 ( C ) s o u / g t u / g 2 ( B ) s o u / g t t u / g 2 ( D ) s o u / g t u / g 2 ( A ) E x 6 3 Q46. A pebble is thrown vertically upwards from bridge with an initial velocity of 4.9 m/s. It strikes the water after 2s. If acceleration due to gravity is 9.8 m/s 2 . The height of the bridge will be- (A) 4.9 m. (B) 19.6 m (C) 9.8 m (D) 24.5 m Q47. A man in a balloon rising vertically with an acceleration of 4.9 m/sec 2 , releases a ball 2 seconds after the balloon is let go from the ground. The greatest height above the ground reached by the ball is- (A) 14.7 m(B) 19.6 m (C) 9.8 m(D) 24.5 m Q48. A person standing on the floor of an elevator drops a coin. The coin reaches the floor of the elevator in a time t 1 , if the elevator is stationary and in time t 2 if it is moving with constant velocity. Then (A) t 1 =t 2 (B) t 1 t 2 (D) t 1 t 2 depending whether lift is going up or down Q49. A river has width 0.5 km and flows from west to East with a speed 30 km/hr. If a boatman starts sailing his boat at a speed 40 km/hr. normal to bank, the boat shall cross the river in time- (A) 0.6 minute (B) 0.75 minute (C) 0.45 minute (D) 3 minute Q50. Two trains take 3 seconds to pass one another when going in opposite direction but only 2.5 second if the speed of one is increased by 50%. The time one train would take to pass the other when going in the same direction at their original speed is – (A) 10 sec(B) 12 sec(C) 15 sec(D) 18 sec Q51.A steamer takes 12 days to reach from port A to B. Everyday only one steamer sets out from both the ports. How many steamers does each boat meet in the open sea (A) 12 (B) 13 (C) 23(D) 24 Q52. A body of mass 0.5 kg is found to be moving 30 m away from the starting point during the fourth second and 70 m towards the starting point during the ninth second of its motion. Identify the correct statements appropriate to the motion from the following. (A) The initial kinetic energy is 2500 J (B) It is uniformly accelerated motion (C) It is uniform acceleration to begin with and then uniform retardation (D) It will be at the starting point after 10 seconds. Q53. A freely falling object crosses T.V tower of height 102.9 m in three seconds. Find the height above the top of the tower from which it would have started falling. (A) 122.5 m (B) 102.9 m (C) 19.6 m (D) 82.3 m Q54. A boatman could row his boat with a speed 10 m/sec. He wants to take his boat from P to a point Q just opposite on the other bank of the river flowing at a speed 4 m/sec. He should row his boat (A) at right angle to the steam (B) at an angle of sin -1 (2/5) with PQ up the stream (C) at an angle of sin -1 (2/5) with PQ down the stream (D) at an angle cos -1 (2/5) with PQ down the stream Q55. A river is flowing from west to east at a speed of 5 meters/minute. A man on the south bank of the river capable of swimming at 10 meters/minute in still water wants to swim across the river in shortest time. He should swim in a direction- (A) Due North(B) 30 0 east of North (C) 30 0 west of North (D) 60 0 east of North Q56. Rahul hits a ball along the ground with a speed u in a direction, which makes an angle 30 0 with the line joining him and the fielder Rhodes. Rhodes runs to intercept the ball with a speed 2u/3. At what angle θ should he run to intercept the ball- (A) sin -1 √3/2 (B) sin -1 2/3 (C) sin -1 ¾ (D) sin -1 4/5 Q57. A bus moves over a straight level road with an acceleration a.A boy in the bus drops a ball outside. The acceleration of the ball with respect to the bus and the Earth respectively- (A) a and g (B) a+g and g-a (C) √a 2 +g 2 and g (D) √ a 2 +g 2 and a Q58. During a rainstorm, raindrops are observed to be striking the ground at an angle of θ with the vertical. A wind is blowing horizontal at the speed of5.0 m/s, the speed of raindrops is (A) 5 sinθ (B) 5/sinθ(C) 5 cosθ(D) 5/cosθ Q59. Two spheres of equal masses but radii R and 2R are allowed to fall in a liquid. The ratio of there terminal velocities is. (A) 1:4 (B) 1:2 (C) 1:1 (D) 2:1 Q60.A particles of mass m moves on the x-axis as follows: it starts from rest at t=0 from the point x=0 and comes to rest at t=1 at the point x=1. No other information is available about its motion at intermediate times (0