Physics - Unit & Dimension, Basic Maths and Vector PDF

ALLEN Unit & Dimension, Basic Maths and Vector 1 UNIT & DIMENSION, BASIC MATHS AND VECTOR...

ALLEN Unit & Dimension, Basic Maths and Vector 1 UNIT & DIMENSION, BASIC MATHS AND VECTOR KEY CONCEPT PHYSICAL QUANTITIES AND UNITS Physical quantities : All quantities that can be measured are called physical quantities. eg. time, length, mass, force, work done, etc. In physics we study about physical quantities and their inter relationship. Measurement : Measurement is the comparison of a quantity with a standard of the same physical quantity. Classification : Physical quantities can be classified on the following bases : I. Based on their directional properties 1. Scalars:The physical quantities which have only magnitude but no direction are called scalar quantities. Ex. mass, density, volume, time, etc. 2. Vectors : The physical quantities which have both magnitude and direction and obey laws of vector algebra are called vector quantities. Ex. displacement, force, velocity, etc. II. Based on their dependency 1. Fundamental or base quantities : A set of physical quantities which are completely independent of each other and all other physical quantities can be expressed in terms of these physical quantities is called Set of Fundamental Quantities. 2. Derived quantities : The quantities which can be expressed in terms of the fundamental quantities are known as derived quantities. Ex. Speed (= distance/time), volume, acceleration, force, pressure, etc. Physical quantities can also be classified as dimensional and dimensionless quantities or constants and variables. Ex. Classify the quantities displacement, mass, force, time, speed, velocity, acceleration, moment of inertia, pressure and work under the following categories : (a) base and scalar (b) base and vector (c) derived and scalar (d) derived and vector Ans. (a) mass, time (b) displacement (c) speed, pressure, work (d) force, velocity, acceleration node06\B0B0-BA\Kota\JEE(Advanced)\Nurture\Phy\Sheet\Unit & Dimension, Basic Maths and Vector\Eng.p65 Units of Physical Quantities The chosen reference standard of measurement in multiples of which, a physical quantity is expressed is called the unit of that quantity. Four basic properties of units are : 1. They must be well defined. 2. They should be easily available and reproducible. 3. They should be invariable e.g. step as a unit of length is not invariable. 4. They should be accepted to all. System of Units : 1. FPS or British Engineering system : In this system length, mass and time are taken as fundamental quantities and their base units are foot (ft), pound (lb) and second (s) respectively. E 2 JEE-Physics ALLEN 2. CGS or Gaussian system : In this system the fundamental quantities are length, mass and time and their respective units are centimetre (cm), gram (g) and second (s). 3. MKS system : In this system also the fundamental quantities are length, mass and time but their fundamental units are metre (m), kilogram (kg) and second (s) respectively. 4. International system (SI) of units : This system is modification over the MKS system and so it is also known as Rationalised MKS system. Besides the three base units of MKS system four fundamental and two supplementary units are also included in this system. Classification of Units : The units of physical quantities can be classified as follows : 1. Fundamental or base units : The units of fundamental quantities are called base units. In SI there are seven base units. SI BASE QUANTITIES AND THEIR UNITS S.No. Physical quantity SI unit Symbol 1. Length metre m 2. Mass kilogram kg 3. Time second s 4. Temperature Kelvin K 5. Electric current ampere A 6. Luminous intensity candela cd 7. Amount of substance mole mol 2. Derived units : The units of derived quantities or the units that can be expressed in terms of the base units are called derived units unit of distance metre Ex. Unit of speed = = = ms-1 unit of time second Some derived units are named in honour of great scientists. Unit of force – newton (N) Unit of frequency – hertz (Hz) etc. node06\B0B0-BA\Kota\JEE(Advanced)\Nurture\Phy\Sheet\Unit & Dimension, Basic Maths and Vector\Eng.p65 UNITS OF SOME PHYSICAL QUANTITIES IN DIFFERENT SYSTEMS Type of Physical CGS MKS FPS Physical Quantity (Originated in (Originated in (Originated in Quantity France) France) Britain) Length cm m ft Fundamental Mass g kg lb Time s s s Force dyne newton(N) poundal Work or Derived erg joule(J) ft-poundal Energy Power erg/s watt(W) ft-poundal/s E  ALLEN Unit & Dimension, Basic Maths and Vector 3 Dimensions : Dimensions of a physical quantity are the powers (or exponents) to which the base quantities are raised to represent that quantity. To make it clear, consider the physical quantity force. As we shall learn later, force is equal to mass times acceleration. Acceleration is change in velocity divided by time interval. Velocity is length divided by time interval. Thus, force = mass × acceleration velocity length / time = mass × = mass × = mass × length × (time)–2 time time Thus, the dimensions of force are 1 in mass, 1 in length and –2 in time. The dimensions in all other base quantities are zero. 1. Dimensional formula : The physical quantity that is expressed in terms of the base quantities is enclosed in square brackets to remind that the equation is among the dimensions and not among the magnitudes. Thus above equation may be written as [force]= MLT –2. Such an expression for a physical quantity in terms of the base quantities is called the dimensional formula. Thus, the dimensional formula of force is MLT–2. The two versions given below are equivalent and are used interchangeably. (a) The dimensional formula of force is MLT–2. (b) The dimensions of force are 1 in mass, 1 in length and –2 in time. The dimensional formula of any physical quantity is that expression which represents how and which of the base quantities are included in that quantity. Ex. Dimensional formula of mass is [M1L0 T0] and that of speed (= distance/time) is [M0L1T–1] 2. Applications of dimensional analysis : (i) To convert a physical quantity from one system of units to the other : This is based on a fact that magnitude of a physical quantity remains same whatever system is used for measurement n2 = numerical value in II system i.e. magnitude = numeric value (n) × n 1 = numerical value in I system M1 = unit of mass in I system unit (u) = constant or n1u1 = n2u2 M2 = unit of mass in II system So if a quantity is represented by [MaLbTc] L1 = unit of length in I system L2 = unit of length in II system a b c é u1 ù é M1 ù é L1 ù é T1 ù T1 = unit of time in I system Then n 2 = n1 ê u ú = n1 ê M ú ê L ú ê T ú T2 = unit of time in II system ë 2û ë 2û ë 2û ë 2û node06\B0B0-BA\Kota\JEE(Advanced)\Nurture\Phy\Sheet\Unit & Dimension, Basic Maths and Vector\Eng.p65 Ex. 1m = 100 cm= 3.28 ft = 39.4 inch (SI) (CGS) (FPS) Ex. The acceleration due to gravity is 9.8 m s–2. Give its value in ft s–2 Sol. As 1m = 3.2 ft \ 9.8 m/s2 = 9.8 × 3.28 ft/s2 = 32.14 ft/s2 » 32 ft/s2 Ex. Convert 1 newton (SI unit of force) into dyne (CGS unit of force) Sol. The dimensional equation of force is [F] = [M1 L1 T–2] Therefore if n1, u1, and n2, u2 corresponds to SI & CGS units respectively, then 1 1 -2 -2 é M1 ù é L1 ù é T1 ù é kg ù é m ù é s ù n2 = n1 ê ú ê ú ê ú =1 ê úê ú ê ú = 1 × 1000 × 100 × 1= 105 M L ë 2û ë 2û ë 2û T ë g ûë cm ûë s û \ 1 newton = 105 dyne. E 4 JEE-Physics ALLEN Q. The value of Gravitational constant G in MKS system is 6.67 × 10–11 N–m2/kg2. What will be its value in CGS system ? Ans. 6.67 × 10–8 cm3/g s2 (ii) To check the dimensional correctness of a given physical relation : If in a given relation, the terms on both the sides have the same dimensions, then the relation is dimensionally correct. This is known as the principle of homogeneity of dimensions. L Ex. Check the accuracy of the relation T = 2 p for a simple pendulum using dimensional analysis. g Sol. The dimensions of LHS = the dimension of T = [M0 L0 T1] 12 æ dimensions of length ö The dimensions of RHS = ç ÷ (Q 2p is a dimensionless constant) è dimensions of acceleration ø 12 æ L ö =ç -2 ÷ = (T2)1/2 = [T] = [ M0 L0 T1] è LT ø Since the dimensions are same on both the sides, the relation is correct. (iii) To derive relationship between different physical quantities : Using the same principle of homogeneity of dimensions new relations among physical quantities can be derived if the dependent quantities are known. Ex. It is known that the time of revolution T of a satellite around the earth depends on the universal gravitational constant G, the mass of the earth M, and the radius of the circular orbit R. Obtain an expression for T using dimensional analysis. Sol. We have [T] = [G]a [M]b [R]c [M]0 [L]0 [T]1 = [M]–a [L]3a [T]–2a × [M]b × [L]c = [M]b–a [L]c+3a [T]–2a Comparing the exponents 1 For [T] : 1 = –2a Þ a = – 2 1 For [M] : 0 = b – a Þ b = a = – 2 3 For [L] : 0 = c + 3a Þ c = –3a = 2 node06\B0B0-BA\Kota\JEE(Advanced)\Nurture\Phy\Sheet\Unit & Dimension, Basic Maths and Vector\Eng.p65 R3 Putting the values we get T = G–1/2 M–1/2 R3/2 = GM R3 So the actual expression is T = 2 p GM Limitations of this method : (i) In Mechanics the formula for a physical quantity depending on more than three physical quantities cannot be derived. It can only be checked. (ii) This method can be used only if the dependency is of multiplication type. The formulae containing exponential, trigonometrical and logarithmic functions can't be derived using this method. Formulae containing more than one term which are added or subtracted like s = ut +at2 /2 also can't be derived. E  ALLEN Unit & Dimension, Basic Maths and Vector 5 (iii) The relation derived from this method gives no information about the dimensionless constants. (iv) If dimensions are given, physical quantity may not be unique as many physical quantities have the same dimensions. (v) It gives no information whether a physical quantity is a scalar or a vector. Units and Dimensions of Physical Quantities Quantity Common Symbol SI unit Dimension Displacement s METRE (m) L Mass m, M KILOGRAM (kg) M Time t SECOND (s) T Area A m2 L2 Volume V m3 L3 Density r kg/m3 M/L3 Velocity v, u m/s L/T Acceleration a m/s2 L/T2 Force F newton (N) ML/T2 Work W joule (J) (=N – m) ML2/T2 Energy E, U, K joule (J) ML2/T2 Power P watt (W) (=J/s) ML2/T3 Momentum p kg-m/s ML/T Gravitational constant G N-m2/kg2 L3/MT2 Angle q, j radian Angular velocity w radian/s T–1 Angular acceleration a radian/s2 T–2 Angular momentum L kg-m2/s ML2/T Moment of inertia I kg-m2 ML2 Torque t N-m ML2/T2 Angular frequency w radian/s T–1 Frequency v hertz (Hz) T–1 Period T s T Young's modulus Y N/m2 M/LT2 Bulk modulus B N/m2 M/LT2 h node06\B0B0-BA\Kota\JEE(Advanced)\Nurture\Phy\Sheet\Unit & Dimension, Basic Maths and Vector\Eng.p65 Shear modulus N/m2 M/LT2 Surface tension S N/m M/T2 Coefficient of viscosity h N-s/m2 M/LT Pressure P, p N/m2, Pa M/LT2 Wavelength l m L Intensity of wave I W/m2 M/T3 Temperature T KELVIN (K) K Specific heat capacity c J/kg–K L2/T2K Stefan's constant s W/m2–K4 M/T3K4 Heat Q J ML2/T2 Thermal conductivity K W/m–K ML/T3K E 6 JEE-Physics ALLEN Basic Mathematics used in physics Plane–angle It is measure of change in direction. If a line rotates in a plane about one of its ends, the other end sweeps an arc. Angle (q) between two orientation of the line is defined by ratio of the arc s length(s) to length of the line(r) q= radian r Angles measured in anticlockwise and clockwise directions are usually taken positive and negative respectively. Angle is measured in radians (rad) or degrees. One radian is the angle subtended at the centre of a circle by an arc of the circle, whose length is equal to the radius of the circle. 22 p rad= 180° p= 3.1415 = 7 1°= 60' (minute), 1' (minute) = 60" (sec) Example Write expression for circumference of a circle of radius 'r'. Solution s = (Total angle about a point) r = 2pr Trigonometrical ratios (or T–ratios) Let two fixed lines XOX' and YOY' intersecting at right angles to each other at point O. Point O is called origin. Line XOX' is known as x–axis and YOY' as y–axis. Regions XOY, YOX', X'OY' and Y'OX are called I, II, III and IV quadrant respectively. Consider a line OP making angle q with OX as shown. Line PM is perpendicular drawn from P on OX. In the right angled triangle OPM, side OP is called hypotenuse, the side OM adjacent to angle q is called base and the side PM opposite to angle q is called the perpendicular. Following ratios of the sides of a right angled triangle are known as trigonometrical ratios or T-ratio node06\B0B0-BA\Kota\JEE(Advanced)\Nurture\Phy\Sheet\Unit & Dimension, Basic Maths and Vector\Eng.p65 perpendicular MP base OM perpendicular MP sin q = = cos q = = tan q = = hypotenuse OP hypotenuse OP base OM base OM hypotenuse OP hypotenuse OP cot q = = sec q = = cosec q = = perpendicular MP base OM perpendicular MP 1 1 1 cosecq = sec q = cot q = sin q cos q tan q Some trigonometric identities sin2q + cos2q = 1 1 + tan2q = sec2q 1 + cot2q = cosec2q E  ALLEN Unit & Dimension, Basic Maths and Vector 7 Example 3 Given sin q =. Find all the other T–ratios, if q lies in the first quadrant. 5 Solution 3 In D OMP , sin q = So MP = 3 and OP = 5 Q OM = (5)2 - (3) 2 = 25 - 9 = 16 = 4 5 OM 4 MP 3 OM 4 Now cos q = = tan q = = cot q = = OP 5 OM 4 MP 3 OP 5 OP 5 sec q = = cosec q = = OM 4 MP 3 T–ratios of some commonly used angles p p p p 0 rad rad rad rad rad Angle (q) 6 4 3 2 0° 30° 45° 60° 90° 1 1 3 sin q 0 1 2 2 2 3 1 1 cos q 1 0 2 2 2 1 tan q 0 1 3 ¥ 3 sin (90°–q)=cosq sin (180°–q)=sinq sin (360°–q)=–sinq cos (90°–q)=sinq cos (180°–q)=–cosq cos (360°–q)= cosq tan (90°–q)=cotq tan (180°–q)=–tanq tan (360°–q)=–tanq sin (90°+q)=cosq sin (180°+q)=–sinq sin (–q)=-sinq cos (90°+q)=–sinq cos (180°+q)=–cosq cos (–q)=cosq tan (90°+q)=–cotq tan (180°+q)=tanq tan (–q)=–tanq When q is very small we can use following approximations : cos q » 1 node06\B0B0-BA\Kota\JEE(Advanced)\Nurture\Phy\Sheet\Unit & Dimension, Basic Maths and Vector\Eng.p65 sin q ; q ü ý If q is in radians tan q ; q þ tan q » sin q. In the given right angled triangle we have very commonly used T-ratios 3 4 3 sin 37° = cos37° = tan 37° = 53° 5 5 4 5 3 4 3 4 sin 53° = cos 53° = tan 53° = 5 5 3 37° 4 E 8 JEE-Physics ALLEN Example Find the value of (i) cos (–60°) (ii) tan 210° (iii) sin 300° (iv) cos 120° Solution 1 (i) cos (–60°) = cos 60° = 2 1 (ii) tan 210° = tan (180° + 30°) = tan 30° = 3 3 (iii) sin 300° = sin (270° + 30°) = – cos 30° = - 2 1 (iv) cos 120° = cos (180° – 60°) = – cos60° = – 2 VECTORS Precise description of laws of physics and physical phenomena requires expressing them in form of mathematical equations. In doing so we encounter several physical quantities, some of them have only magnitude and other have direction in addition to magnitude. Quantities of the former kind are referred as scalars and the latter as vectors and mathematical operations with vectors are collectively known as vector analysis. Vectors A vector has both magnitude and sense of direction, and follows triangle law of vector addition. For example, displacement, velocity, and force are vectors. r Vector quantities are usually denoted by putting an arrow over the corresponding letter, as A or ar. Sometimes in print work (books) vector quantities are usually denoted by boldface letters as A or a. r r Magnitude of a vector A is a positive scalar and written as A or A. Geometrical Representation of Vectors. A vector is represented by a directed straight line, having the magnitude and direction of the quantity represented by it. e.g. if we want to represent a force of 5 N acting 45° N of E (i) We choose direction coordinates. node06\B0B0-BA\Kota\JEE(Advanced)\Nurture\Phy\Sheet\Unit & Dimension, Basic Maths and Vector\Eng.p65 (ii) We choose a convenient scale like 1 cm º 1N (iii) We draw a line of length equal in magnitude and in the direction of vector to the chosen quantity. (iv) We put arrow in the direction of vector. AB Magnitude of vector: A B = 5N By definition magnitude of a vector quantity is scalar and is always positive. E  ALLEN Unit & Dimension, Basic Maths and Vector 9 3. TERMINOLOGY OF VECTORS Parallel vector: If two vectors have same direction, they are parallel to each other. They may be located anywhere in the space. Antiparallel vectors: When two vectors are in opposite direction they are said to be antiparallel vectors. Equality of vectors: When two vectors have equal magnitude and are in same direction and represent the same physical quantity, they are equal. r r i.e. a = b Thus when two parallel vectors have same magnitude they are equal. (Their initial point & terminal point may not be same) Negative of a vector: When a vector have equal magnitude and is in opposite direction, it is said to be negative vector of the former. r r r r i.e. a = - b or b = - a Thus when two antiparallel vectors have same magnitude they are negative of each other. Remark : Vector shifting is allowed without change in their direction. 2. Angle Between two Vectors It is the smaller angle formed when the initial points or the terminal points of the two vectors are brought together. It should be noted that 0º £ q £ 180º. node06\B0B0-BA\Kota\JEE(Advanced)\Nurture\Phy\Sheet\Unit & Dimension, Basic Maths and Vector\Eng.p65 3. Addition Of Vectors: Parallelogram law of addition: Steps: (i) Keep two vectors such that there tails coincide. (ii) Draw parallel vectors to both of them considering both of them as sides of a parallelogram. (iii) Then the diagonal drawn from the point where tails coincide represents the sum of two vectors, with its tail at point of coincidence of the two vectors. E 10 JEE-Physics ALLEN (i) (ii) (iii) r r AC = a + b Addition of more than two Vectors The triangle law can be extended to define addition of more than two vectors. Accordingly, if vectors to be added are drawn in head to tail fashion, resultant is defined by a vector drawn from the tail of the first vector to the head of the last vector. This is also known as the polygon rule for vector addition. r r r r Operation of addition of three vectors A, B and C and their resultant P are shown in figure. r r r r A + B+ C = P Here it is not necessary that three or more vectors and their resultant are coplanar. In fact, the vectors to be added and their resultant may be in different planes. However if all the vectors to be added are coplanar, their resultant must also be in the same plane containing the vectors. Subtraction of Vectors r A vector opposite in direction but equal in magnitude to another vector A is known as negative r r vector of A. It is written as - A. Addition of a vector and its negative vector results a vector of zero r magnitude, which is known as a null vector. A null vector is denoted by arrowed zero ( 0 ). The idea of negative vector explains operation of subtraction as addition of negative vector. Accordingly r r r to subtract a vector from another consider vectors A and B shown in the figure. To subtract B from r r r A , the negative vector – B is added to A according to the triangle law as shown in figure-II. node06\B0B0-BA\Kota\JEE(Advanced)\Nurture\Phy\Sheet\Unit & Dimension, Basic Maths and Vector\Eng.p65 r r uuur uuur r r uuur If two vectors a & b are represented by OA & OB then their sum a + b is a vector represented by OC , where OC is the diagonal of the parallelogram OACB. r r r r r r r r r r a + b = b + a (commutative) (a + b) + c = a + (b + c) (associativity) r r r r r r r r r r a+0=a=0+a a + ( -a ) = 0 = ( -a ) + a r r r r r r r r | a + b |£| a | + | b | | a - b | ³ || a | - | b || r r r r r r a±b = | a |2 + | b |2 ±2 | a || b | cos q where q is the angle between the vectors E  ALLEN Unit & Dimension, Basic Maths and Vector 11 Some Important Results : r r (1) If q = 0° Þ a || b r r r r then, | R | = | a | + | b | & | R | is maximum r r (2) If q = p Þ a anti || b r r r r then, | R | = | a | - | b | & | R | is minimum r r (3) If q = p/2 Þ a ^ b R= a 2 + b2 r r & tan a = b/a (a is angle made by R with a ) r r (4) | a | = | b | = a r | R | = 2acosq/2 & a = q/2 r r (5) If | a | = | b | = a & q = 120° r then | R | = a 4. Multiplication Of A Vector By A Scalar: r r r r If a is a vector & m is a scalar, then m a is a vector parallel to a whose modulus is ½m½ times that of a. r r This multiplication is called SCALAR MULTIPLICATION. If a and b are vectors & m, n are scalars, then: r r r m (a) = (a) m = ma r r r m (na) = n (ma) = (mn) a r r r (m + n) a = ma + na r r r m (a + b) = ma + mb Resolution of a Vector into Components Following laws of vector addition, a vector can be represented as a sum of two (in two-dimensional space) or three (in three-dimensional space) vectors each along predetermined directions. These directions are called axes and parts of the original vector along these axes are called components of the vector. UNIT VECTOR: node06\B0B0-BA\Kota\JEE(Advanced)\Nurture\Phy\Sheet\Unit & Dimension, Basic Maths and Vector\Eng.p65 A unit vector is a vector of magnitude of 1, with no units. Its only purpose is to point, i.e. to describe a direction in space. r A unit vector in direction of vector A is represented as  r A &  = r |A| r r r or A can be expressed in terms of a unit vector in its direction i.e. A = | A |  Unit Vectors along three coordinates axes:– unit vector along x-axis is î unit vector along y-axis is ĵ unit vector along z-axis is k̂ E 12 JEE-Physics ALLEN Cartesian components in two dimensions If a vector is resolved into its components along mutually perpendicular directions, the components are called Cartesian or rectangular components. r In figure is shown, a vector A resolved into its Cartesian components r r A x and A y along the x and y-axis. Magnitudes Ax and Ay of these components are given by the following equation. Ax=Acosq and Ay = Asinq r A = A x ˆi + A y ˆj 2 2 A = Ax + Ay Here î and ĵ are the unit vectors for x and y coordinates respectively. Mathematical operations e.g. addition, subtraction, differentiation and integration can be performed independently on these components. This is why in most of the problems use of Cartesian components becomes desirable. Cartesian components in three dimensions r A vector A resolved into its three Cartesian components one along each of the directions x, y, and z-axis is shown in the figure. r r r r A = A x + A y + A z = A x ˆi + A y ˆj + A z kˆ A = A 2x + A 2y + A 2z Product of Vectors In all physical situation, whose description involve product of two vectors, only two categories are observed. One category where product is also a vector involves multiplication of magnitudes of two vectors and sine of the angle between them, while the other category where product is a scalar involves multiplication of magnitudes of two vectors and cosine of the angle between them. Accordingly, we define two kinds of product operation. The former category is known as vector or cross product and the latter category as scalar or dot product. Scalar or dot product of two vectors r r node06\B0B0-BA\Kota\JEE(Advanced)\Nurture\Phy\Sheet\Unit & Dimension, Basic Maths and Vector\Eng.p65 The scalar product of two vectors A and B equals to the product of their magnitudes and the cosine of the angle q between them. r r A × B = ABcos q = OA × OB × cos q The above equation can also be written in the following ways. r r r r A × B = ( A cos q ) B = OP × OB A × B = A ( Bcos q ) = OA × OQ E  ALLEN Unit & Dimension, Basic Maths and Vector 13 Above two equations and figures, suggest a scalar product as product of magnitude of the one vector and magnitude of the component of another vector in the direction of the former vector. KEY POINTS r r r r Dot product of two vectors is commutative: A × B = B × A r r r r If two vectors are perpendicular, their dot product is zero. A × B = 0 , if A ^ B r r r r Dot product of a vector by itself is known as self-product. A × A = A 2 Þ A = A × A r r æ A×B ö -1 The angle between the vectors q = cos ç ÷ è AB ø r r (a) Component of A in direction of B r r r r r r r æ A.B ö r A |= ( A cos q ) ˆ = A ç r r ÷B B ˆ = ç r ö÷ B æ A.B ˆ = ( A.B ˆ )B ˆ | è A B ø è B ø r r r r r (b) Component of A perpendicular to B : A ^ = A - A| | Dot product of Cartesian unit vectors: ˆi × ˆi = ˆj × ˆj = kˆ × kˆ = 1 ˆi × ˆj = ˆj × kˆ = kˆ × ˆi = 0 r r If A = A x ˆi + A y ˆj + A z kˆ and B = Bx ˆi + By ˆj + Bz kˆ , their dot product is given by r r A × B = A x Bx + A y By + A z Bz Solved Examples 1. Two displacement vectors of same magnitude are arranged in the following manner (I) 1200 (II) 300 (III) 900 (IV) 450 Magnitude of resultant is maximum for node06\B0B0-BA\Kota\JEE(Advanced)\Nurture\Phy\Sheet\Unit & Dimension, Basic Maths and Vector\Eng.p65 (A) case I (B) case II (C) case III (D) case IV Ans. (B) r r Sol. Magnitude of Resultant of A and B = A 2 + B2 + 2ABcos q which is maximum for q = 30 0 r r r r 2. Two vectors P and Q are added, the magnitude of resultant is 15 units. If Q is reversed and added to P r r resultant has a magnitude 113 units. The resultant of P and a vector perpendicular to P and equal in r magnitude to Q has a magnitude (A) 13 units (B) 17 units (C) 19 units (D) 20 units Ans. (A) E 14 JEE-Physics ALLEN Sol. P2 + Q2 + 2PQcosq = 225...(i) P 5 P2 + Q2 – 2PQcosq = 113...(ii) R =1 By adding (i) & (ii) 2(P2 +Q2) = 338 Q Ö1 -Q 13 P2 + Q2 = 169 Þ P 2 + Q 2 = 13 P 3. Three forces are acting on a body to make it in equilibrium, which set can not do it? (A) 3 N, 3 N, 7 N (B) 2 N, 3 N, 6 N (C) 2 N, 1 N, 1 N (D) 8 N, 6 N, 1 N Ans. (A, B, D) Sol. They must form a triangle. (a + b ³ c) 4. Keeping one vector constant, if direction of other to be added in the first vector is changed continuously, r tip of the resultant vector describes a circle. In the following figure vector a is kept constant. When r r r r vector b added to ar changes its direction, the tip of the resultant vector r = a + b describes circle of r r radius b with its center at the tip of vector a. Maximum angle between vector a and the resultant r r r r = a + b is r q b a -1 æ b ö -1 æ b ö (A) tan çè ÷ø (B) tan çè 2 2 ÷ø (C) cos–1 (r/a) (D) cos–1 (a/r) r a -b Ans. (A,B,C) r r Sol. Angle between rr and b is maximum when rr is tangent to circle. r r r r r 5. ( ) If A = 2iˆ + ˆj + kˆ and B = 10iˆ + 5ˆj + 5kˆ , if the magnitude of component of B - A along A is 4 x. Then x will be. Ans. 6 r r node06\B0B0-BA\Kota\JEE(Advanced)\Nurture\Phy\Sheet\Unit & Dimension, Basic Maths and Vector\Eng.p65 ( Sol. r = B - A = 4 2iˆ + ˆj + kˆ ) r r r.A 4 ( 4 + 1 + 1) r cos q = = =4 6 A 6 x=6 r r 6. The component of A = ˆi + ˆj + 5kˆ perpendicular to B = 3iˆ + 4 ˆj is 4 ˆ 3 ˆ 8 6 4 ˆ 3 ˆ 8 ˆ 6 ˆ (A) - i+ j + 5kˆ (B) - ˆi - ˆj + 5kˆ (C) i- j + 5kˆ (D) + i- j + 5kˆ 25 25 25 25 25 25 25 25 Ans. (C) E  ALLEN Unit & Dimension, Basic Maths and Vector 15 A Sol. q B Acosq r r r æ A×B ö A|| = A cos q = A ç ÷ è AB ø r r A×B 3+ 4 7 = = = B 5 5 r 7 æ 3iˆ + 4ˆj ö 7 ˆ ˆ A|| = çç 5 è 5 ÷ø 25 ÷= ( 3i + 4 j ) r 21 ˆ 28 ˆ A|| = i+ j 25 25 r ( ) æ 21 è 25 28 ö A ^ = ˆi + ˆj + 5kˆ - ç ˆi + ˆj ÷ 25 ø 4 ˆ 3 ˆ = i- j + 5kˆ 25 25 ALGEBRA : SOME USEFUL FORMULAE Quadratic equation and its solution An algebraic equation of second order (highest power of the variable is equal to 2) is called a quadratic equation. General quadratic equation is ax2 + bx + c = 0. The general solution of the above quadratic - b ± b 2 - 4ac equation or value of variable x is x = 2a - b + b2 - 4ac - b - b 2 - 4ac Þ x1 = and x 2 = 2a 2a node06\B0B0-BA\Kota\JEE(Advanced)\Nurture\Phy\Sheet\Unit & Dimension, Basic Maths and Vector\Eng.p65 Example Solve 2x2 + 5x – 12 = 0 Solution By comparison with the standard quadratic equation a = 2, b = 5 and c = –12 -5 ± (5)2 - 4 ´ 2 ´ (-12) -5 ± 121 -5 ± 11 +6 -16 3 x= = = = , Þ x= ,–4 2´2 4 4 4 4 2 Binomial approximation In case, x is very small, then terms containing higher powers of x can be neglected. In such a case, (1 + x)n = 1 + nx Also (1 + x)–n = 1–nx and (1 – x)n =1–nx and (1 – x)–n = 1 + nx E 16 JEE-Physics ALLEN Exponential Expansion x 2 x3 x2 x3 e =1+x+ x + +........ and e = 1 – x + –x - +........ 2! 3! 2! 3! Componendo and dividendo theorem : p a p+q a+b If = then by componendo and dividendo theorem = q b p-q a -b Determinant a b -3 3 2 -4 D= = ad - bc , For example = 12 , = -6 c d -5 1 -3 3 a1 a2 a3 b1 b2 b3 b2 b3 b b b b D= = a1 - a 2 1 3 + a3 1 2 c1 c 2 c3 c 2 c3 c1 c 3 c1 c 2 Example + - + 5 4 3 1 6 2 6 2 1 2 1 6 =5 -4 +3 = 5 (9 – 48) – 4(18 – 42) + 3(16–7) = –72 8 9 7 9 7 8 7 8 9 Logarithm loge x = ln x (base e) log x = log10 x (base 10) m (a) Product formula log mn = log m + log n (b) Quotient formula log = logm – logn n (c) Power formula log mn = n log m GEOMETRY : SOME USEFUL FORMULAE Formulae for determination of area : Area of a square = (side)2 node06\B0B0-BA\Kota\JEE(Advanced)\Nurture\Phy\Sheet\Unit & Dimension, Basic Maths and Vector\Eng.p65 Area of rectangle = length × breadth Area of a triangle = (1/2) × base × height Area of a trapezoid =(1/2) × (distance between parallel sides) × (sum of parallel sides) Area enclosed by a circle = pr2 (where r = radius) Area of a sector a circle = 12 qr 2 (where r = is radius and q is angle subtended at a centre) Area of ellipse = p ab (where a and b are semi major and semi minor axis respectively) Surface area of a sphere = 4pr2 (where r = radius) Area of a parallelogram = base × height Area of curved surface of cylinder = 2prl (where r = radius and l = length) E  ALLEN Unit & Dimension, Basic Maths and Vector 17 Area of whole surface of cylinder = 2pr(r + l) (where l = length) Surface area of a cube = 6(side)2 Total surface area of a cone = pr2+prl [where prl = pr r 2 + h 2 = lateral area (h=height)] Formulae for determination of volume : Volume of a rectangular slab = length × breadth × height = abt Volume of a cube = (side)3 4 3 Volume of a sphere = p r (where r = radius) 3 Volume of a cylinder = p r2l (where r = radius and l = length) 1 2 Volume of a cone = p r h (where r = radius and h = height) 3 node06\B0B0-BA\Kota\JEE(Advanced)\Nurture\Phy\Sheet\Unit & Dimension, Basic Maths and Vector\Eng.p65 E 18 JEE-Physics ALLEN EXERCISE (S-1) Units & Dimensions 1. A particle is in a unidirectional potential field where the potential energy (U) of a particle depends on the x- coordinate given by Ux = k (1 – cos ax) & k and 'a' are constants. Find the physical dimensions of 'a' & k. UD0001 2. The equation for the speed of sound in a gas states that v = gk BT / m. Speed v is measured in m/s, g is a dimensionless constant, T is temperature in kelvin (K), and m is mass in kg. Find the SI unit for the Boltzmann constant, kB? UD0002 3. The time period (T) of a spring mass system depends upon mass (m) & spring constant (k) & length of the é Force ù spring (l) ê k =. Find the relation among T, m, l & k using dimensional method. ë length úû UD0003 4. The distance moved by a particle in time t from centre of a ring under the influence of its gravity is given by x = a sinwt, where a & w are constants. If w is found to depend on the radius of the ring (r), its mass (m) and universal gravitational constant (G). Using dimensional analysis find an expression for w in terms of r, m and G. UD0004 5. A satellite is orbiting around a planet. Its orbital velocity (v0) is found to depend upon (A) Radius of orbit (R) (B) Mass of planet (M) (C) Universal gravitation constant (G) node06\B0B0-BA\Kota\JEE(Advanced)\Nurture\Phy\Sheet\Unit & Dimension, Basic Maths and Vector\Eng.p65 Using dimensional analysis find an expression relating orbital velocity (v0) to the above physical quantities. UD0005 6. Assume that the largest stone of mass 'm' that can be moved by a flowing river depends upon the velocity of flow v, the density d & the acceleration due to gravity g. If 'm' varies as the Kth power of the velocity of flow, then find the value of K. UD0006 r ar 7. Given F = where symbols have their usual meaning. The dimensions of a is. t UD0007 E  ALLEN Unit & Dimension, Basic Maths and Vector 19 Addition of vectors 8. A block is applied two forces of magnitude 5N each. One force is acting towards East and the other acting along 60° North of East. The resultant of the two forces (in N) is of magnitude :- UD0008 9. Two forces act on a particle simultaneously as shown in the figure. Find net force in milli newton on the particle. [Dyne is the CGS unit of force] 100 dyne 60° 1.0 mili newton UD0009 10. The maximum and minimum magnitudes of the resultant of two forces are 35 N and 5 N respectively. Find the magnitude of resultant force when act orthogonally to each other. UD0010 11. Three forces of magnitudes 2 N, 3 N and 6 N act at corners of a cube along three sides as shown in figure. Find the resultant of these forces in N. 3N 2N 6N UD0011 Resolution of vectors and unit vector 12. The farm house shown in figure has rectangular shape and has sides parallel to the chosen x and y axes. The position vector of corner A is 125 m at 53° and corner C is 100 m at 37° from x axis. Find the length node06\B0B0-BA\Kota\JEE(Advanced)\Nurture\Phy\Sheet\Unit & Dimension, Basic Maths and Vector\Eng.p65 of the fencing required in meter. y A B D C 53° 37° x UD0012 13. Vector B has x, y and z components of 4.00, 6.00 and 3.00 units, respectively. Calculate the magnitude of B and the angles that B makes with the coordinates axes. UD0013 E 20 JEE-Physics