CONTENTS 2. Co-ordinate geometry 5 3. Differentiation 7 4. Integration 9 5. Graphs 12 6. Algebra 12 7. Geometry 8. Types of vectors 9. Addition of two vectors 0 1 16 20 19 -2 N Trigonometry 17 19 Addition of more than two vectors 21 11. Subtraction of two vectors 21 12. Resolution of a vector 22 13. Multiplication & division by a scalar 26 14. Scalar products of two vectors 26 15. Vector products of two vectors 28 16. Exercise-I (Conceptual Questions) 32 17. Exercise-II (Previous Years Questions) 38 18. Exercise-III (Analytical Questions) 39 19. Exercise-IV (Assertion & Reason) 40 ss io n 10. Se A E Page 1. LL E BASIC MATHEMATICS USED IN PHYSICS & VECTORS S.No. NEET SYLLABUS Elementary concepts of differentiation and integration for describing motion. Scalar and vector quantities, Position and displacement vectors,general vectors and notation, equality of vectors, multiplication of vectors by a real number, addition and subtraction of vectors. Satyendranath Bose (1894-1974) Se ss io n Satyendranath Bose, born in Calcutta in 1894, is among the great Indian physicists who made a fundamental contribution to the advance of science in the twentieth century. An outstanding student throughout, Bose started his career in 1916 as a lecturer in physics in Calcutta University; five years later he joined Dacca University. Here in 1924, in a brilliant flash of insight, Bose gave a new derivation of Planck.s law, treating radiation as a gas of photons and employing new statistical methods of counting of photon states. He wrote a short paper on the subject and sent it to Einstein who immediately recognised its great significance, translated it in German and forwarded it for publication. Einstein then applied the same method to a gas of molecules. A -2 19 20 LL E N Albert Einstein, born in Ulm, Germany in 1879, is universally regarded as one of the greatest physicists of all time. His astonishing scientific career began with the publication of three path-breaking papers in 1905. In the first paper, he introduced the notion of light quanta (now called photons) and used it to explain the features of photoelectric effect that the classical wave theory of radiation could not account for. In the second paper, he developed a theory of Brownian motion that was confirmed experimentally a few years later and provided a convincing evidence of the atomic picture of matter. The third paper gave birth to the special theory of relativity that made Einstein a legend in his own life time. In the next decade, he explored the consequences of his new theory which included, among other things, the mass-energy equivalence enshrined in his famous equation E = mc2. He also created the general version of relativity (The General Theory of Relativity), which is the modern theory of gravitation. Some of Einstein's most significant later contributions are: the notion of stimulated emission introduced in an alternative derivation of Planck's blackbody radiation law, static model of the universe which started modern cosmology, quantum statistics of a gas of massive bosons, and a critical analysis of the foundations of quantum mechanics. The year 2005 was declared as International Year of Physics, in recognition of Einstein's monumental contribution to physics, in year 1905, describing revolutionary scientific ideas that have since influenced all of modern physics. The key new conceptual ingredient in Bose's work was that the particles were regarded as indistinguishable, a radical departure from the assumption that underlies the classical Maxwell-Boltzmann statistics. It was soon realised that the new Bose-Einstein statistics was applicable to particles with integers spins, and a new quantum statistics (Fermi -Dirac statistics) was needed for particles with half integers spins satisfying Pauli's exclusion principle. Particles with integers spins are now known as bosons in honour of Bose. An important consequence of Bose-Einstein statistics is that a gas of molecules below a certain temperature will undergo a phase transition to a state where a large fraction of atoms populate the same lowest energy state. Some seventy years were to pass before the pioneering ideas of Bose, developed further by Einstein, were dramatically confirmed in the observation of a new state of matter in a dilute gas of ultra cold alkali atoms - the Bose-Eintein condensate 0 ALBERT EINSTEIN (1879-1955) ALLEN Pre-Medical : Physics BASIC MATHEMATICS USED IN PHYSICS Mathematics is the supporting tool of Physics. Elementary knowledge of basic mathematics is useful in problem solving in Physics. In this chapter we study Elementary Algebra, Trigonometry, Coordinate Geometry and Calculus (differentiation and integration). 1. TRIGONOMETRY 1.1 Angle Consider a revolving line OP. P Suppose that it revolves in anticlockwise direction starting from its initial position OX . The angle is defined as the amount of revolution that the revolving line makes with its initial position. q O X The angle N From fig. the angle covered by the revolving line OP is q = ÐPOX also 1 right angle = p rad (radian) 2 20 1 right angle = 90° (degrees) -2 LL E 1° = 60' (minute) 1' = 60" (second) 19 is taken negative if it is covered in clockwise direction. 0 is taken positive if it is traced by the revolving line in anticlockwise direction and One radian is the angle subtended at the centre of a circle by an arc of the circle, whose length is equal to the » 57.3° n o 180 p ss io radius of the circle. 1 rad = To convert an angle from radian to degree multiply it by 180° p Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 Illustration 1. E q=1rad l=r Se p 180° A To convert an angle from degree to radian multiply it by r A circular arc is of length p cm. Find angle subtended by it at the centre in radian and degree. 6cm p cm q 6cm Solution 180° p 180° s p cm p = 30° q == =rad=30° As 1 rad = So q = ´ p 6 p r 6 cm 6 1 Pre-Medical : Physics ALLEN Illustration 2. When a clock shows 4 o'clock, how much angle do its minute and hour needles make? p 2p (1) 120° (2) rad (3) rad (4) 160° 3 3 Solution 12 11 1 10 From diagram angle q = 4 ´ 30° = 120° = 2p 9 rad 3 Ans. (1,3) 2 q 3 8 4 7 5 6 1.2 Trigonometrical ratios (or T ratios) perpendicular MP = hypotenuse OP cos q = tan q = perpendicular MP = base OM cot q = -2 Y' base OM = perpendicular MP 1 cos q 1 + tan2q = sec2q hypotenuse OP = perpendicular MP cot q = 1 tan q 1 + cot2q = cosec2q Illustrations Illustration 3. Given sin q = 3/5. Find all the other T-ratios, if q lies in the first quadrant. Solution 3 In D OMP , sin q = so MP = 3 and OP = 5 5 Q OM = (5)2 - (3)2 = OM 4 Now, cos q = = OP 5 cot q = 2 OM 4 = 3 MP 25 - 9 = 5 3 16 = 4 MP 3 tan q = = OM 4 sec q = P OP 5 = OM 4 O cosec q = OP 5 = MP 3 q M Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 sin2q + cos2q = 1 sec q = X M base OM = hypotenuse OP cosec q = Se A 1 sinq 90° ss io hypotenuse OP = base OM It can be easily proved that : sec q = cosec q = O n sin q = q X' 20 The three sides of a right angled triangle are connected to each other through six different ratios, called trigonometric ratios or simply T-ratios : P 19 LL E in anticlockwise direction.From P, draw perpendicular PM on OX. Then, side OP (in front of right angle) is called hypotenuse,side MP (in front of angle q) is called opposite side or perpendicular and side OM (making angle q with hypotenuse) is called adjacent side or base. 0 N Let two fixed lines XO X' and YOY' intersect at right angles t o e ach othe r at po int O. Then, (i) Point O is called origin. (ii) XOX' is known as X-axis and YOY' as Y-axis. (iii) Portions XOY, YOX', X'OY' and Y'OX are called I, II, III and IV quadrant respectively. Consider that the revolving line OP has traced out angle q (in I quadrant) Y E Pre-Medical : Physics ALLEN Table : The T-ratios of a few standard angles ranging from 00 to 1800 Angle (q) 0° 30° 45° 60° 90° 120° 135° 150° 180° sin q 0 1 2 1 3 2 1 3 2 1 1 2 0 cos q 1 3 2 1 1 2 0 tan q 0 1 3 2 2 1 3 – 2 1 2 – 3 ¥ 1 - 3 2 - 2 1 - –1 –1 0 3 (not defined) 1.3 Four Quadrants and ASTC Rule* 90° N IInd quadrant S in In second quadrant, only sinq and cosecq are positive. Ist quadrant All 180° 270° 20 * Remember as Add Sugar To Coffee or After School To College. -2 In fourth quadrant, only cosq and secq are positive Cos IVth quadrant 19 Tan IIIrd quadrant LL E In third quadrant, only tanq and cotq are positive. 0° 360° 0 In first quadrant, all trigonometric ratios are positive. 1.4 Trigonometrical Ratios of General Angles (Reduction Formulae) (ii) A sin(p+q)= –sinq Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 sin(2p-q)= –sinq E (iv) tan(2np+q)=tanq æ np ö + q÷ will remain same if n is even and sign of trigonometric Trigonometric function of an angle ç è 2 ø function will be according to value of that function in quadrant. sin(p–q) = + sinq (iii) cos(2np+q) = cosq ss io sin(2np +q) = sinq n Trigonometric function of an angle (2np + q) where n=0, 1, 2, 3,.... will be remain same. cos(p-q) = –cosq tan(p-q)= –tanq cos(p+q)= –cosq tan(p+q)= +tanq cos(2p-q)= +cosq tan(2p-q)= –tanq Se (i) æ np ö + q÷ will be changed into co-function if n is odd and sign of Trigonometric function of an angle ç è 2 ø trigonometric function will be according to value of that function in quadrant. æp ö sin ç + q÷ = + cos q è2 ø æp ö cos ç + q÷ = - sin q è2 ø æp ö tan ç + q÷ = - cot q è2 ø æp ö sin ç - q÷ = + cos q è2 ø æp ö cos ç - q÷ = + sin q è2 ø æp ö tan ç - q÷ = + cot q è2 ø Trigonometric function of an angle –q (negative angles) sin(–q) = –sinq cos(–q) = +cosq tan(–q) = –tanq 3 Pre-Medical : Physics ALLEN sin (90° + q ) = cos q cos (90° + q ) = –sin q sin (180° – q ) = sin q cos (180° – q ) = –cos q sin (– q ) = – sin q cos (– q ) = cos q sin (90° – q ) = cos q cos (90° – q ) = sin q tan (90° + q ) = –cot q tan (180° – q ) = –tan q tan (– q ) = – tan q tan (90° – q ) = cot q sin (180° + q ) = –sin q cos (180° + q ) = –cos q sin (270° – q ) = –cos q cos (270° – q ) = –sin q sin (270° + q ) = –cos q cos (270° + q ) = sin q sin (360° – q ) = –sin q cos (360° – q ) = cos q tan (180° + q ) = tan q tan (270° – q ) = cot q tan (270° + q ) = –cot q tan (360° – q ) = –tan q Illustrations Illustration 4. (ii) tan 210° = tan (180° + 30°) = tan 30° = 3 2 sin 300° = sin (270° + 30°) = – cos 30° = – cos 120° = cos (180° – 60°) = – cos60° = – 20 (iv) 3 1 2 n (iii) 1 LL E cos (–60°) = cos 60° = -2 1 2 (i) (iv) cos 120° 0 (iii) sin 300° N (ii) tan 210° 19 Find the value of (i) cos (–60°) Solution Find the values of : (i) tan (–30°) If sec q = (iii) sin 135° (iv) cos 150° (v) sin 270° (vi) cos 270° 5 p and 0 < q < . Find all the other T-ratios. 3 2 A 2. (ii) sin 120° Se 1. ss io BEGINNER'S BOX-1 sin (A + B) = sin A cos B + cos A sin B cos (A + B) = cos A cos B – sin A sin B sin ( A – B) = sin A cos B – cos A sin B cos (A – B) = cos A cos B + sin A sin B tan ( A + B ) = tan A + tan B 1 - tan A tan B sin 2 A = 2 sin A cos A tan 2 A = 2 tan A 1 - tan2 A tan ( A - B ) = cos 2 A = cos2 A – sin2 A cos 2 A = 2 cos2 A – 1 = 1 – 2sin2 A 1 + cos A = 2 cos2 4 tan A - tan B 1 + tan A tan B A A , 1 – cos A = 2sin2 2 2 Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 1.5 A few important trigonometric formulae E Pre-Medical : Physics ALLEN 1.6 Range of trigonometric functions As sin q = P H As cos q = B and B £ H H As P tan q = B and P £H so –1 £ sin q £ 1 so –1 £ cos q £ 1 so H q P B – ¥ < tan q < ¥ Remember : - a2 + b2 £ a sinq + b cosq £ a 2 + b2 1.7 Small Angle Approximation If q is small (say < 5°) then sinq » q, cosq » 1 & tanq » q. Here q must be in radians. Illustrations COORDINATE GEOMETRY 0 -2 (iii) cos1° » 1 19 2. p ö p p ö p æ p ö æ æ pö = sin ç » = tan ç ÷ » tan2° = tan ç 2° ´ (ii) ÷ ÷ ÷ è 180 ø 180 è è 90 ø 90 180° ø 180° ø LL E æ (i) sin1° = sin çè1° ´ N Illustration 5. Find the approximate values of (i) sin1° (ii) tan2° (iii) cos1° Solution 20 To specify the position of a point in space, we use right handed rectangular axes coordinate system. This system consists of (i) origin (ii) axis or axes. If a point is known to be on a given line or in a particular direction, only one coordinate is necessary to specify its position, if it is in a plane, two coordinates are required, if it is in space three coordinates are needed. n · Origin ss io This is any fixed point which is convenient to you. All measurements are taken w.r.t. this fixed point. · Axis or Axes Se Any fixed direction passing through origin and convenient to you can be taken as an axis. If the position of a point or position of all the points under consideration always happen to be in a particular direction, then only one axis is required. This is generally called the x-axis. If the positions of all the points under consideration are Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 A always in a plane, two perpendicular axes are required. These are generally called x and y-axis. If the points are distributed in a space, three perpendicular axes are taken which are called x, y and z-axis. E 2.1 Position of a point in xy plane The position of a point is specified by its distances from origin along (or parallel to) x and y-axis as shown in figure. Here x-coordinate and y-coordinate is called abscissa and ordinate respectively. y x y y origin (0,0) (x,y) x x 5 Pre-Medical : Physics ALLEN 2.2 Distance Formula The distance between two points (x1, y1) and (x2, y2) is given by d = Note : In space d = (x - x1 ) + ( y2 - y1 ) 2 2 2 (x 2 - x1 )2 + (y 2 - y 1 ) 2 + (z2 - z1 ) 2 y 2.3 Slope of a Line The slope of a line joining two points A(x1, y1) and B(x2, y2) is denoted by m and is given by Dy y 2 - y1 m= = = tan q [If both axes have identical scales] Dx x 2 - x1 B(x2, y2) y2 Dy y1 (x1,y1) q A Dx Here q is the angle made by line with positive x-axis. x1 Slope of a line is a quantitative measure of inclination. x2 x -2 For point (2, 14) find abscissa and ordinate. Also find distance from y and x-axis. Solution Abscissa = x-coordinate = 2 = distance from y-axis. 19 LL E Ordinate = y-coordinate = 14 = distance from x-axis. Illustration 7. 0 N Illustrations Illustration 6. 20 Find value of a if distance between the points (–9 cm, a cm) and (3 cm, 3cm) is 13 cm. Solution Þ 13 = éë3 - ( -9) ùû + [ 3 - a ] Þ 132 = 122 + (3–a)2 Þ (3–a)2 = 132 – 122 = 52 Þ (3–a) = ±5 Þ a = –2 cm or 8 cm (x - x1 ) + ( y2 - y1 ) 2 2 2 2 2 ss io n By using distance formula d = A Se Illustration 8. A dog wants to catch a cat. The dog follows the path whose equation is y–x = 0 while the cat follows the path whose equation is x2 + y2 = 8. The coordinates of possible points of catching the cat are : (1) (2, –2) (2) (2, 2) (3) (–2, 2) (4) (–2, –2) Solution Ans. (2,4) 1. BEGINNER'S BOX-2 2. Distance between two points ( 8, – 4) and (0, a) is 10. All the values are in the same unit of length. Find the positive value of a. Calculate the distance between two points (0, –1, 1) and (3, 3, 13). 3. Calculate slope of shown line y (0, 2) (3, 0) 6 x Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 Let catching point be (x1, y1) then, y1–x1=0 and x12 + y12 = 8 Therefore, 2x12 = 8 Þ x12 =4 Þ x1=±2; So possible points are (2, 2) and (–2, –2). E Pre-Medical : Physics ALLEN 3. DIFFERENTIATION 3.1 Function Constant: A quantity, whose value remains unchanged during mathematical operations, is called a constant quantity. The integers, fractions like p ,e etc are all constants. Variable: A quantity, which can take different values, is called a variable quantity. A variable is usually represented as x, y, z, etc. Function: A quantity y is called a function of a variable x, if corresponding to any given value of x, there exists a single definite value of y. The phrase 'y is function of x' is represented as y = f (x) For example, consider that y is a function of the variable x which is given by y = 3x 2 + 7x + 2 If x = 1, then y = 3 (1)2 + 7(1) + 2 = 12 and when x = 2, y = 3 (2)2 + 7(2) + 2 = 28 Therefore, when the value of variable x is changed, the value of the function y also changes but corresponding to each value of x, we get a single definite value of y. Hence, y = 3x2 + 7x + 2 represents a function of x. dy dx 3.2 Physical meaning of Ds Dt -2 average velocity of the body, vav = 0 The ratio of small change in the function y and the variable x is called the average rate of change of y w.r.t. x. For example, if a body covers a small distance Ds in small time Dt, then N (i) When D x ® 0 The limiting value of Dy Dx Lim is Dx ®0 Dy dy = Dx dx 20 (ii) Dv Dt LL E of the body, aav = 19 Also, if the velocity of a body changes by a small amount Dv in small time Dt, then average acceleration and instantaneous acceleration of the body If c = constant, Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 A 1. E Se 3.3 Theorems of differentiation Ds ds = Dt ®0 Dt dt Dv dv = a = Lim Dt ® 0 Dt dt v = Lim ss io Like wise, instantaneous velocity of the body n It is called the instantaneous rate of change of y w.r.t. x. The differentiation of a function w.r.t. a variable implies the instantaneous rate of change of the function w.r.t. that variable. d (c) = 0 dx 2. y = c u, where c is a constant and u is a function of x, du dy d = (cu) = c dx dx dx 3. y = u ± v ± w, where, u, v and w are functions of x, du dy d = ( u ± v ± w) = dx dx dx 4. y = u v where u and v are functions of x, dv du dy d +v = (uv) = u dx dx dx dx 5. u y = , where u and v are functions of x, v dy d æ uö = ç ÷= dx dx è v ø 6. y = xn, n real number, dy d n = (x ) = nxn–1 dx dx v dv dw ± dx ± dx du dv -u dx dx v2 7 Pre-Medical : Physics ALLEN Illustrations Illustration 9. Find dy , when dx (i) y = x (ii) y = x5 + x4 + 7 (iii) y = x2 + 4x –1/2 – 3x–2 Solution (i) y = dy d ( x) = dx dx x Þ d 1/2 (x ) dx 1 1 12 - 1 1 -1/ 2 x x = = 2 2 x 2 = dy d 5 d 5 d 4 d = (x + x4 +7) = (x ) + (x ) + (7) dx dx dx dx dx Þ (ii) y = x5 + x4 + 7 = = 5 x4 + 4x3 + 0 = 5x4 + 4x3 dy d 2 d 2 d d = (x + 4x–1/2 – 3x–2) = (x ) + (4x–1/2) – (3x–2) dx dx dx dx dx = d 2 d –1/2 d –2 (x ) + 4 (x ) –3 (x ) = 2x + 4 æç - 1 ö÷ x–3/2 –3(–2)x–3 dx dx dx è 2ø N (iii) y = x2 + 4x–1/2 – 3x–2 Þ (i) y = x7/2 (ii) y = x–3 (iii) y = x (v) y = 5x4 + 6x3/2 + 9x (vi) y = ax2 + bx + c (vii) y = 3x5 – 3x – ds . dt 19 dy for the following : dx (iv) y = x5 + x3 + 4x1/2 + 7 20 1 x Given s = t2 + 5t + 3, find 3. If s = ut + 4. The area of a blot of ink is growing such that after t seconds, its area is given by A = (3t 2+7) cm2 . Calculate the rate of increase of area at t=5 second. The area of a circle is given by A = p r2 , where r is the radius. Calculate the rate of increase of area w.r.t. radius. Obtain the differential coefficient of the following : 6. (i) (x – 1) (2x + 5) ss io (ii) Se ds 1 2 at , where u and a are constants. Obtain the value of . 2 dt A 5. n 2. 1 2x + 1 (iii) 3x + 4 4x + 5 (iv) x2 x3 + 1 3.4 Formulae for differential coefficients of trigonometric, logarithmic and exponential functions 8 • d (sin x) = cos x dx • d (cos x) = – sin x dx • d (tan x) = sec2 x dx • d (cot x) = – cosec2 x dx • d (sec x) = sec x tan x dx • d (cosec x) = – cosec x cot x dx • 1 d (logex) = dx x • d x (e ) = ex dx • d ( ax ) e = aeax dx Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 Find LL E 1. -2 BEGINNER'S BOX-3 0 = 2x – 2x–3/2 + 6x–3 E Pre-Medical : Physics ALLEN 3.5 Maximum and Minimum value of a Function y B Higher order derivatives are used to find the maximum and minimum values of a function. At the points of maxima and minima, first derivative (i.e. Maxima dy ) becomes zero. dx At point ‘A’ (minima) : A minima x d2 y As we see in figure, in the neighbourhood of A, slope increases so >0. dx2 dy d2 y = 0 and >0 dx dx2 Condition for minima : At point ‘B’ (maxima) : As we see in figure, in the neighbourhood of B, slope decreases so (3) 4 5 3 5 Ans. (3) 2 dy 1 1 d y = 0 Þ 5 ( 2x ) - 2 (1) + 0 = 0 Þ x = . Now at x= , 2 =10 which is positive dx 5 5 dx 20 For maximum/minimum value 2 4 æ1ö æ1ö 1 . Therefore y min = 5 ç ÷ - 2 ç ÷ + 1 = 5 5 è5ø è5ø n so y has minimum value at x = 4. (4) 0 2 5 -2 (2) LL E Solution 1 5 Illustrations 19 Illustration 10. The minimum value of y = 5x2 – 2x + 1 is N dy d2 y = 0 and <0 dx dx2 Condition for maxima : (1) d2 y <0 dx2 INTEGRATION ss io In integral calculus, the differential coefficient of a function is given. We are required to find the function. Integration is basically used for summation . å is used for summation of discrete values, while ò sign is used for continous function. Se If I is integration of f(x) with respect to x then I = ò f(x) dx [we can check A where c = an arbitrary constant Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 Let us proceed to obtain intergral of xn w.r.t. x. E dI = f(x) ] Q ò f'(x) dx = f(x) + c dx d n+1 (x ) = (n +1)xn dx Since the process of integration is the reverse process of differentiation, ò (n + 1) xn dx = x n +1 or (n + 1) ò x n dx = x n+1 Þ ò x n dx = x n +1 n +1 The above formula holds for all values of n, except n = –1. It is because, for n = –1, Q ò 1 d (loge x) = dx x x n dx = ò x -1dx = ò 1 dx x \ ò 1 dx = loge x x Similarly, the formulae for integration of some other functions can be obtained if we know the differential coefficients of various functions. 9 Pre-Medical : Physics x n +1 + c , Provided n ¹ – 1 n +1 1. ò x n dx = 2. ò sin x dx = - cos x + c (Q d (cos x)= – sin x) dx 3. ò cos x dx = sin x + c (Q d (sin x)= cos x) dx 4. ò 1 dx = loge x + c x (Q d 1 (loge x)= ) dx x 5. ò e x dx = e x + c (Q d x (e )= ex) dx Illustrations ò I= 1ö dx x ÷ø x2dx - ò cos xdx + ò 20 - cos x + 1 x2+1 x3 dx = – sin x + loge x + c = – sin x + loge x + c x 2 +1 3 n Solution 2 ss io æ ò çè x +1 xq q (p + q )/ q +c ò x dx = p + c = p + q x +1 q Illustration 12. Evaluate p p q 0 x -7 +1 1 -6 ò x dx = -7 + 1 + c = – 6 x + c (iii) -7 -2 x11/2 +1 2 13/2 dx = +c = + c (ii) x 11 13 +1 2 xp/q (p/q ¹ –1) (iii) LL E 11/2 (ii) x–7 19 Illustration 11. Integrate w.r.t. x. : (i) x11/2 Solution (i) ò x ALLEN Few basic formulae of integration Following are a few basic formulae of integration : N 4.1 BEGINNER'S BOX-4 ò x dx (v) ò 1ö æ ç x + ÷ dx è xø (ii) (vi) ò x ò -32 dx (iii) ò (3x -7 -1 + x ) dx 2 (iv) ò 1 ö æ ç x+ ÷ dx è xø æ a bö ç 2 + ÷ dx (a and b are constant) èx xø 4.2 Definite Integrals When a function is integrated between a lower limit and an upper limit, it is called a definite integral. d ( f(x) ) = f ' (x), then If dx ò f ' (x ) dx is called indefinite integral and ò b a f ' (x ) dx is called definite integral Here, a and b are called lower and upper limits of the variable x. After carrying out integration, the result is evaluated between upper and lower limits as explained below : 10 ò b a b f ' (x ) dx = f (x ) a = f(b) – f(a) Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 (i) 15 Se Evaluate the following integrals : A 1. E Pre-Medical : Physics ALLEN y 4.3 Area under a curve and definite integration y=F(x) Area of small shown darkly shaded element = ydx = f(x) dx If we sum up all areas between x=a and x= b then y b ò f ( x ) dx = shaded area between curve and x-axis. a x x=a dx Illustrations Illustration 13. x x=b 5 The integral ò x dx 2 is equal to 1 (2) 124 3 (4) 45 N Solution 5 5 1 3 (3) é x3 ù é 53 13 ù 125 1 124 = = - = x dx ê ú ê - ú= ò1 3û 3 3 3 ë 3 û1 ë 3 2 LL E BEGINNER'S BOX-5 (i) òR p (v) ò0 2 (ii) ò r2 r1 p sin x dx (vi) ò0 2 -k q 1 q2 x 2 dx v (iii) òu p cos x dx (vii) ò- ¥ Mv dv 2 p 2 ò0 (iv) x -1 2 dx 20 GMm dx x2 ¥ 19 Evaluate the following integrals cos x dx n 1. Ans. (2) 0 125 3 -2 (1) ss io 4.4 Average value of a continuous function in an interval A Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 E y av = Se Average value of a function y = f(x), over an interval a £ x £ b is given by b ò ydx a b b = ò dx ò ydx a b-a a Illustrations Velocity(m/s) Illustration 14. 20 The velocity-time graph of a car moving along a straight road is shown in figure. The average velocity of the car in first 25 seconds is (1) 20 m/s (2) 14 m/s (3) 10 m/s (4) 17.5 m/s Solution : 0 10 20 25 Time(s) Ans. (2) 25 Average velocity= ò vdt 0 25 - 0 = ù Area of v-t graph between t=0 to t = 25 s 1 éæ 25 + 10 ö 20) ú = 14 m/s = ( êçè ÷ 25 25 ë 2 ø û 11 Pre-Medical : Physics ALLEN Illustration 15. Determine the average value of y = 2x +3 in the interval 0 £ x £ 1. (1) 1 (2) 5 (3) 3 Solution (4) 4 Ans. (4) 1 y av = 1 1 é æ x2 ö ù = ò (2x + 3 ) dx = ê2 ç ÷ + 3x ú = 12 + 3 (1) - 02 - 3 (0 ) = 1 + 3 = 4 1-0 0 ë è 2ø û0 0 SOME STANDARD GRAPHS AND THEIR EQUATIONS y y x Straight line m=tanq q>90° y = mx + c q y 2 x = ky 2 x = -ky x O Parabola Parabola LL E 2 x O y y A a Circle 6. x x 2 y2 + =1 a2 b 2 Se b O Rectangular Hyperbola x 1 y xy = constant x O Ellipse x2+ y2 = a2 y ss io Parabola x Parabola y = -kx2 n y = kx O 20 y y x O x O x O m = –ve q q x y Straight line c O a y 1 Exponential Decay y=e –kx x a = semi major axis b = semi minor axis O x ALGEBRA 6.1 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. Equation ax2 + bx + c = 0 is the general quadratic equation. The general solution of the above quadratic equation or value of variable x is - b ± b2 - 4ac - b + b2 - 4ac - b - b2 - 4ac x1 = and x 2 = Þ 2a 2a 2a b c Sum of roots = x1 + x2 = - and product of roots = x1x2 = a a 2 For real roots discriminant b – 4ac ³ 0 and for imaginary roots b2 – 4ac < 0 x= 12 Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 O y y y = mx Straight line c c 0 y O x+ m = -2 = q y m=tanq q<90° m = +ve 19 m x -c A Straight line N 5. ò ydx E Pre-Medical : Physics ALLEN Illustrations Illustration 16. Solve the equation 2x2 + 5x – 12 = 0 Solution By comparison with the standard quadratic equation x= a = 2, b = 5 and c = –12 -5 ± 11 +6 -16 -5 ± 121 -5 ± (5)2 - 4 ´ 2 ´ (-12) = = , = 4 4 4 4 2´2 or x= 3 ,–4 2 Illustration 17. The speed (v) of a particle moving along a straight line is given by v = t2 + 3t – 4 where v is in m/s and t in seconds. Find time t at which the particle will momentarily come to rest. Solution When particle comes to rest, v = 0. Þ Neglect negative value of t, t= –3 ± 9 - 4(1) (–4) 2(1) Hence t = 1 s Illustration 18. Þ t = 1 or – 4 0 t2 + 3t – 4 = 0 N So we have (–2v)2 – 4(1) (100) ³ 0 Þ 4v2 – 400 ³ 0 Þ v2 – 100 ³ 0 Þ (v – 10) (v + 10) ³ 0 Þ v ³ 10 and v £ – 10 -2 n Hence possible positive values of v are v ³ 10 m/s. 20 For possible values of v, time t must be real so from b2 – 4ac ³ 0 19 Solution LL E The speed (v) and time (t) for an object moving along straight line are related as t2 + 100 = 2vt where v is in meter/second and t is in second. Find the possible positive values of v. 1. Solve for x : (i) 2. In quadratic equation ax2 + bx + c = 0, if discriminant is D = b2 – 4ac, then roots of the quadratic equation are : (choose the correct alternative) (1) Real and distinct, if D > 0 (2) Real and equal (i.e., repeated roots), if D = 0. Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 (ii) pqx2 – (p2 + q2) x + pq = 0 Se A 10x2 – 27x + 5 = 0 (3) Non-real (i.e. imaginary), if D < 0 E ss io BEGINNER'S BOX-6 (4) All of the above are correct 6.2 Binomial Expression : An algebraic expression containing two terms is called a binomial expression. æ 1ö For example (a+b), (a+b)3, (2x––3y)–1, ç x + ÷ etc. are binomial expressions. è yø Binomial Theorem (a + b)n = a n + na n -1 b1 + n(n - 1) n -2 2 a b + ......., , 2´1 (1 + x)n = 1 + nx + n ( n - 1) 2 x + ...... 2 ´1 Binomial Approximation If x is very small, compared to 1, then terms containing higher powers of x can be neglected so (1+x)n » 1 + nx 13 Pre-Medical : Physics Illustrations Illustration 19. Calculate ALLEN 0.99 Solution 0.99 = (1 – 0.01)1/2 » 1 – 1 (0.01) » 1 – 0.005 » 0.995 2 Illustration 20. The mass m of a body moving with a velocity v is given by m = m0 1- where m0 = rest mass of v2 c2 body = 10 kg and c = speed of light = 3 × 108 m/s. Find the value of m at v = 3 × 107 m/s. é æ 3 ´ 107 ö2 ù = 10 ê1 - ç ÷ ú 3 ´ 108 ø ú ëê è û -1/2 1 ù é = 10 ê1 ú ë 100 û é æ 1 öæ 1 ö ù 10 » 10 ê1 - ç - ÷ç = 10 + » 10.05 kg ÷ ú 200 ë è 2 øè 100 ø û LL E 6.3 Logarithm -1/2 m =logm-log n n • loge m = 2.303 log10m p a p+q a+b = then = q b p-q a-b n 6.4 Componendo and Dividendo Rule : If • log mn = n log m 20 • log 19 Common formulae : • log mn = log m+log n 0 -1/2 -2 æ v2 ö m = m0 ç 1 - 2 ÷ c ø è N Solution ss io 6.5 Arithmetic progression (AP) General form : a, a + d, a +2d, ..., a + (n–1)d. Here a = first term, d = common difference Se n n n [a+a+(n–1)d] = [2a+(n–1)d] = [Ist term + nth term] 2 2 2 Illustrations Illustration 21. Find the sum of given Arithmetic Progression 4 + 8 + 12 +.........+ 64 (1) 464 (2) 540 (3) 544 Solution Here a = 4, d = 4, n = 16 So, sum = (4) 646 Ans. (3) n 16 [First term + last term] = [4 + 64] = 8(68) = 544 2 2 Note : (i) Sum of first n natural numbers. (ii) Sum of square of first n natural numbers (iii) 14 Sum of cube of first n natural numbers n ( n + 1) n éë1 + n ùû = 2 2 n(n + 1) (2n + 1) Sn2 = 12 + 22 + 32 + .......+ n2 = 6 Sn = 1 + 2 + 3 +....+n = é n(n + 1) ù Sn3 = 1 + 2 + 3 + .......+ n = ê ú ë 2 û 3 3 3 3 2 Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 A Sum of n terms Sn = E Pre-Medical : Physics ALLEN BEGINNER'S BOX-7 1. Find sum of first 50 natural numbers. 2. Find 12 + 22 +.................+ 102. 6.6 Geometric Progression (GP) General form : a, ar, ar2,..., arn–1 Sum of n terms Sn = ( a 1 - rn Here a = first term, r = common ratio ) For 0 £ |r| < 1 Sum of ¥ term S ¥ = 1-r a (Q r < 1 \ r¥ ® 0) 1-r Illustration 22. Find the sum of given series 1 + 2 + 4 + 8 +..............+ 256 Here a = 1, r = 2, n = 9 (Q 256 = 28). So S9 = Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 Here, a = 1, r = E -2 n ss io (2) 1 A Solution : 1 1 1 + + + ....upto ¥ . 2 4 8 1 a So, S¥ = = 2 1-r (3) 2 Se Find 1 + Ans.[2] (1)(1 - 29 ) = 29 – 1 = 512 – 1 = 511 (1 - 2) Illustration 23. (1) ¥ (4) 513 19 (3) 512 20 Solution : (2) 511 LL E (1) 510 0 N Illustrations 1 1- 1 2 (4) 1.925 Ans.[3] =2 BEGINNER'S BOX-8 1 1 1 1 1 + - + + ........ ¥ . 2 4 8 16 32 1. Find 1 - 2. 1 1 é1 ù Find Fnet = GMm ê 2 + 2 + 2 + ..............up to ¥ ú . ër û 2r 4r 15 Pre-Medical : Physics 7. ALLEN GEOMETRY 7.1 Formulae for determination of area : 1. 2. Area of a square = (side)2 Area of rectangle = length × breadth 3. Area of a triangle = 1 (distance between parallel sides) × (sum of parallel sides) 2 Area enclosed by a circle = p r2 (r = radius) Area of trapezoid = 5. 6. 7. 8. 9. 10. Surface area of a sphere = 4p r2 Area of a parallelogram = base × height Area of curved surface of cylinder = 2prl Area of ellipse = p ab Surface area of a cube = 6(side)2 (r = radius) 11. Total surface area of cone = pr + prl where prl = pr r + h (r = radius and l = length) (a and b are semi major and semi minor axes respectively) 1. Volume of a rectangular slab = length × breadth × height = abt 2. Volume of a cube = (side) 4 3 pr 3 1 2 pr h 3 (r = radius and h is height) 22 = 3.14 ; 7 20 (r = radius and l is length) Note : p = b (r = radius) Volume of a cylinder = p r2l Volume of a cone = a n 5. Volume of a sphere = t 1 = 0.3182 » 0.3 . p ss io 4. LL E 3. 3 = lateral area 0 7.2 Formulae for determination of volume : 2 N 2 -2 2 19 4. 1 (base × height) 2 p2 = 9.8776 » 10 and A Illustration 24. Se Illustrations Calculate the area enclosed by shown ellipse Shaded area = Area of ellipse = pab Here a = 6 – 4 = 2 and b = 4 – 3 = 1 r Area = p × 2 × 1 = 2p units Illustration 25. 4 3 2 1 2 4 6 R=1m Calculate the volume of given disk. Solution Volume = pR2t = (3.14) (1)2 (10–3) = 3.14 × 10–3 m3 16 t=1mm Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 Solution E Pre-Medical : Physics ALLEN VECTORS Scalar Quantities A physical quantity which can be described completely by its magnitude only and does not require a direction is known as a scalar quantity. It obeys the ordinary rules of algebra. Ex : Distance, mass, time, speed, density, volume, temperature, electric current etc. Vector Quantities A physical quantity which requires magnitude and a particular direction, when it is expressed. Ex. : Displacement, velocity, acceleration, force etc. Vector quantities must obey the rules of vector algebra. B Equal Vectors Vectors which have equal magnitude and same direction are called equal vectors. r r A =B n A Anti–parallel Vectors : B ss io l A Those vectors which have opposite direction are called anti–parallel vector. B Negative (or Opposite) Vectors Vectors which have equal magnitude but opposite direction are called negative vectors of each other. uuur uuur AB and BA are negative vectors uuur uuur AB = -BA A Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 E l l Se Angle between two anti–parallel vectors is always 180° l 0 A 20 Angle between two parallel vectors is always 0° -2 Parallel Vectors :- Those vectors which have same direction are called parallel vectors. l A 19 l Q P LL E 1.1 Types of vector N A vector is represented by a line headed with an arrow. Its length is proportional to its magnitude. r A is a vector.. r uuur A = PQ r r Magnitude of A = A or A Co-initial vector Co-initial vectors are those vectors which have the same initial point. r r r In figure a,b and c are co-initial vectors. A B A B c b a Collinear Vectors : The vectors lying in the same line are known as collinear vectors. Angle between collinear vectors is either 0° or 180°. 17 Pre-Medical : Physics Example. (i) (iii) l l ALLEN (q = 0° ) (q = 180° ) ( q = 0° ) (q = 180° ) (ii) (iv) Coplanar Vectors Vectors located in the same plane are called coplanar vectors. Note :- Two vectors are always coplanar. ® ® a Concurrent vectors b Those vectors which pass through a common point are called concurrent vectors ® c r r r In figure a,b and c are concurrent vectors. r A Â = r A 0 Vector æ ö ç unit vector= Magnitude of the vector ÷ è ø 20 r Unit vector in the direction of A is 19 by Â ( Read as A cap or A hat or A caret). -2 Unit Vector A vector having unit magnitude is called unit vector. It is used to specify direction. A unit vector is represented LL E l Null or Zero Vector A vector having zero magnitude is called null vector. r r r Note : Sum of two vectors is always a vector so, ( A ) + ( - A ) = 0 r 0 is a zero vector or null vector.. N l r r ˆ = AA ˆ A = AA Y In an XYZ co-ordinate frame there are three unit vectors $i , $j and · Se A k$ , these are used to indicate X, Y and Z directions respectively. These three unit vectors are mutually perpendicular to each other. ^ j l di P These vectors are used in rotational motion to define rotational effects. Direction of these vectors is always along the axis of rotation in accordance with right hand screw rule or right hand thumb rule. uur r , Ex. : Infinitesimal angular displacement ( dq ) , Angular velocity ( w ) r r and Torque r Angular momentum ( J ) , Angular acceleration ( a ) ( t) 18 X ^ Polar Vector Vectors which have initial point or a point of application are called polar vectors. Axial Vector i k Z Ex. : Displacement, force etc. ^ Axis en em c a l s sp t initial point Q Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 Base Vectors ss io n A unit vector is used to specify the direction of a vector. E Pre-Medical : Physics ALLEN 1.2 Addition of two vectors Vector addition can be performed by using following methods (i) Graphical methods (ii) Analytical methods Addition of two vectors is quite different from simple algebraic sum of two numbers. l Triangle Law of Addition of Two Vectors If two vectors are represented by two sides of a triangle in same order then their sum or 'resultant vector' is given by the third side of the triangle taken in opposite order of the first two vectors. B vector addition B R= A+ B B a A A Step (i) A Step (ii) 2 19 C D B R= b n ss io B sin q A sin q and tanb = Þ R = A + B + 2AB cos q , tana = A + B cos q B + A cos q 2 B Bcos q A 20 · Parallelogram Law of Addition of Two Vectors : If two vectors are represented by two adjacent sides of a parallelogram which are directed away from their common point then their sum (i.e. resultant vector) is given by the diagonal of the parallelogram passing away through that common point. uuur uuur uuur r r r AB + AD = AC Þ A + B = R q a Bsin q B sin q A + B cos q -2 LL E r r Let direction of R make angle a with A : tan a = 0 N r Shift one vector B , without changing its direction, such that its tail coincide with head of the other vector r r r A . Now complete the triangle by drawing third side, directed from tail of A to head of B (it is in r r opposite order of A and B vectors). r r r Sum of two vectors is also called resultant vector of these two vectors. Resultant R = A + B r r r Length of R is the magnitude of vector sum i.e. A + B r r r \ R = A + B = (A + B cos q)2 + (B sin q)2 = A 2 + B 2 + 2AB cos q R q A A+ B a A B Illustrations Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 E Se A Illustration 26. Two forces of magnitudes 3N and 4N respectively are acting on a body. Calculate the resultant force if the angle between them is : (i) 0° (ii) 180° (iii) 90° Solution (i) (ii) q = 0°, both the forces are parallel, R = A + B \ Net force or resultant force R = 3 + 4 = 7N Direction of resultant is along both the forces q = 180°, both the forces are antiparallel,R = A~B \ Net force or resultant force R = 4 – 3 = 1N Direction of net force is along larger force i.e. along 4N. (iii) q = 90°, both the forces are perpendicular then R = A2 + B 2 + 2AB cos90° = A2 + B2 = 32 + 42 = 5N ; tan a = 3 4 Þ a = tan-1 ( 34 ) = 37° Magnitude of resultant is 5N which is acting at an angle of 37° from 4N force. 19 Pre-Medical : Physics ALLEN Illustration 27. Two vectors having equal magnitude of 5 units, have an angle of 60° between them. Find the magnitude of their resultant vector and its angle from one of the vectors. Solution 2 2 5 A = B = 5 unit and q = 60°; R = A + B + 2AB cos 60° = 5 3 unit R 3 o B sin 60 1 o o 2 60° tan a = = 3 = = tan 30 \ a = 30 o a 3 A + B cos 60 2 5 Illustration 28. r r A vector A and B make angles of 20° and 110° respectively with the X-axis. The magnitudes of these vectors are 5m and 12m respectively. Find their resultant vector. Solution r r R Angle between the A and B = 110°– 20° = 90° B N So R = A2 + B2 + 2AB cos90o = 52 + 122 = 13m r r Let angle of R from A is a -2 A C B ne dy 60° 10 dyne Se Two forces each numerically equal to 10 dynes are acting as shown in the figure, then find resultant of these two vectors. Solution The angle q between the two vectors is 120° and not 60°. 10 ss io Illustration 30. D n uuur uuur uuur Figure shows a parallelogram ABCD. Prove that AC + BD = 2BC Solution uuur uuur uuur uuur uuur uuur AC = AB + BC & BD = BC + CD [applying triangle law of vectors] uuur uuur uuur uuur uuur uuur uuur uuur uuur Now AC + BD = AB + BC + BC + CD = AB + 2BC + CD uuur uuur uuur uuur uuur uuur uuur uuur But CD = -AB \ AC + BD = AB + 2BC - AB = 2BC 20 Illustration 29. X-axis 19 LL E r æ 12 ö Þ a = tan-1 ç ÷ with vector A or (a + 20°) with X-axis è 5 ø A 0 ° 110 20° B sin q 12sin 90o 12 ´ 1 12 tan a = = = = A + B cos q 5 + 12cos 90o 5 + 12 ´ 0 5 l l l GOLDEN KEY POINTS r r r r A+B =B+A r r r r r r Vector addition is associative A + (B + C) = ( A + B) + C Resultant of two vectors will be maximum when they are parallel i.e. angle between them is zero. r r r r r r R max = A + B max == A 2 + B 2 + 2AB cos 0° = (A + B)2 = A + B or A + B max = A + B Vector addition is commutative l Resultant of two vectors will be minimum when they are antiparallel i.e. angle between them is 180°. r r R min = A + B = A 2 + B 2 + 2AB cos180° = (A - B) 2 = A ~ B (Larger — smaller) r r r r r r Þ R min = A ~ B Þ A + B min = A ~ B l l Resultant of two vectors of unequal magnitude can never be zero. If vectors are of unequal magnitude then minimum three coplanar vectors are required for zero resultant. 20 Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 A R = (10)2 + (10) 2 + 2(10)(10)(cos120°) = 100 + 100 - 100 = 10 dyne E Pre-Medical : Physics ALLEN l Resultant of two vectors of equal magnitude will be at their bisector. r r r r But if A > B then angle b > a If A = B q/2 R= A +B B q/2 l B R= b A+ B a A A r R will incline more towards the vector of bigger magnitude. r r If two vectors have equal magnitude i.e. A = B = a and angle between them is q then resultant will be along r r q the bisector of A and B and its magnitude is equal to 2a cos R=2acos q/2 2 B r r r q R = A + B = 2a cos 2 a a 120° =a 2 r r r r r i.e. If q = 120° then R = A + B = A = B = a q/2 Special Case : If q = 120° then R = 2a cos A N q/2 0 -2 1.3 Addition of More Than Two Vectors (Law of Polygon) 19 LL E If resultant of two unit vectors is another unit vector then the angle between them (q) = 120°. OR If the angle between two unit vectors (q) = 120°, then their resultant is another unit vector. C D C D R ss io A n B 20 If some vectors are represented by sides of a polygon in same order, then their resultant vector is represented r r r r r by the closing side of polygon in the opposite order. R = A + B + C + D A B GOLDEN KEY POINTS If n coplanar vectors of equal magnitude are arranged at equal angles of separation then their resultant is always zero. E B A 1.4 Subtraction of two vectors r r r r Let A and B are two vectors. Their difference i.e. A – B can be treated as r r sum of the vector A and vector ( -B ) . r r r r A - B = A + ( -B ) r r r To subtract B from A , reverse the direction of B and add to vector r A according to law of triangle. r r B sin q |A - B|= A 2 + B2 + 2AB cos ( p - q ) = A 2 + B 2 - 2AB cos q & tan a = A - B cos q r r where q is the angle between A and B . B q A a –B B E C D A- Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 l Se In a polygon if all the vectors taken in same order are such that the head of the last vector coincides with the tail of the first vector then their resultant is a null vector. r r r r r r A +B+C+D+E =0 A l –B 21 Pre-Medical : Physics ALLEN GOLDEN KEY POINTS l r r r r Vector subtraction does not follow commutative law i.e. A - B ¹ B - A r r r r r r Vector subtraction does not follow associative law i.e. A - B - C ¹ A - B - C l r r If two vectors have equal magnitude, i.e. A = B = a and q is the angle between them, then l ( ) ( ) r r q A - B = a 2 + a 2 - 2a 2 cos q = 2a sin 2 r r r r q = a i.e. A - B = A = B = a at q = 60° 2 If difference of two unit vectors is another unit vector then the angle between them is 60° or If two unit vectors are at angle of 60°, then their difference is also a unit vector. In physics whenever we want to calculate change in a vector quantity, we have to use vector subtraction. For r r r example, change in velocity, Dv = v 2 - v 1 Special case : If q = 60o then 2a sin l l N Illustrations 0 -2 19 12 + 12 + 2(1)(1) cos q = 12 + 12 - 2(1)(1) cos q n Illustration 32. r ˆ = ˆ - b| ˆ , then find the angle between ar and b If |aˆ + b||a Solution r ˆ = ˆ - b| ˆ Let angle between ar and b be q Q |aˆ + b||a \ Þ 2 + 2 cosq = 2 – 2 cos q Þ cos q = 0 \ q = 90° 20 LL E Illustrastion 31. The magnitude of pairs of displacement vectors are given. Which pairs of displacement vectors cannot be added to give a resultant vector of magnitude 13 cm? (1) 4 cm, 16 cm (2) 20 cm, 7 cm (3) 1 cm, 15 cm (4) 6 cm, 8 cm Solution Ans. (3) r r Resultant of two vectors A and B must satisfy A~ B £ R £ A + B If two forces act in opposite direction then their resultant is 10N and if they act mutually perpendicular then their resultant is 50N. Find the magnitudes of both the forces. 3. r r r r If A = 3$i + 2$j and B = 2$i + 3$j - k$ , then find a unit vector along A - B . 4. If magnitude of sum of two unit vectors is Se $ $ If a$ - b = 2 then calculate the value of a$ + 3 b . A 2. e j 2 then find the magnitude of subtraction of these unit vectors. 1.5 Resolution of vectors into rectangular components When a vector is splitted into components which are at right angles to each other then the components are called rectangular or orthogonal components of that vector. uuuur (i) Let vector ar = OA in X - Y plane, makes angle a from X-axis. Draw perpendiculars AB and AC from A on the X-axis and Y-axis respectively. uuuur (ii) The length OB is called projection of OA on X-axis or component of uuuur OA along X-axis and is represented by ax. Similarly OC is the projection uuuur of OA on Y-axis and is represented by ay. r uuuur uuur uuur According to law of vector addition. a = OA = OB + OC 22 Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 1. ss io BEGINNER'S BOX-9 E ALLEN Pre-Medical : Physics r Thus a has been resolved into two parts, one along OX and the other along OY, which are mutually perpendicular. In DOAB, OB = cos a OA Þ OB = OA cos a Þ ax = a cos a AB = sina Þ AB = OA sin a = OC Þ ay = a sin a \ ax =acosa and ay = asina OA uuur uuur If $i and $j denote unit vectors along OX and OY respectively then OB = a cos a $i and OC = a sin a $j uuuur uuur uuur So according to rule of vector addition OA = OB + OC Þ ar = a ˆi + a ˆj = a cos aˆi + a sin aˆj x y and ay r So |a|= a 2x + a 2y and tan a = ax 1.6 Rectangular Components of a Vector in Three Dimensions (i) 0 -2 D Also (OA)2 = (OE)2 + (EA)2 But (OE)2 = (OB)2 + (OD)2 and EA = OC X az E 20 r ˆ \ a = a xˆi + a y ˆj + a z k 19 LL E N (ii) uuuur r Consider a vector a represented by OA , as shown in figure. Consider O as origin and draw a rectangular parallelopiped with its three edges along the X, Y and Z axes. r r r r Vector a is the diagonal of the parallelopiped; its projections on X, Y and Z axis are a x , a y and a z r respectively. These are the three rectangular components of A . Y uuuur uuur uuur Using triangle law of vector addition OA = OE + EA C ay uuur uuur uuuur Using parallelogram law of vector addition OE = OB + OD A uuur uuur uuuur uuur uuuur uuur uuuur uuur uuuur uuur ( ) Q EA = OC \ OA = OB + OD + OC \ OA = OB + OD + EA a uuuur r uuur uuur uuuur ˆ ˆ ˆ ax B Now OA = a, OB = a x i, OC = a y j and OD = a z k O Z n \ (OA)2 = (OB)2 + (OD)2 + (OC)2 or a2 = a x2 + a y 2 + a z2 Þ a = a x 2 + a y 2 + a z 2 ...(i) Direction Cosines r Let a makes angle : a with X axis, b with Y axis and g with Z axis A az Þaz = a cos g • cos g = a ss io ax Þ ax = a cos a a a Þ ay = a cos b Se • cos a = ay • cos b = a Y b Z g a ax X Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 cos a, cos b and cos g are direction cosines of the vector. Putting the value of ax, ay and az in eq. (i) we get a2 = a2 cos2 a + a2 cos2 b + a2 cos2 g E Þ cos 2 a + cos 2 b + cos2 g = 1 Þ ( 1 - sin2 a ) + ( 1 - sin2 b ) + ( 1 - sin2 g ) = 1 Þ 3 - (sin2 a + sin2 b + sin2 g ) = 1 Þ sin2 a + sin2 b + sin2 g = 2 l l l l GOLDEN KEY POINTS A vector can be resolved into infinite number of components. $i $i $i For example 10$i = $i + $i + $i............10 times ; = + + ............20 times and so on. 2 2 2 Maximum number of rectangular components of a vector in a plane is two. But maximum number of rectangular components in space (3-dimensions) is three which are along X, Y and Z axes. A vector is independent of the orientation of axes but the components of that vector depend upon the orientation of axes. The component of a vector along its perpendicular direction is always zero. 23 Pre-Medical : Physics ALLEN Illustrations Illustration 33. r r If P = 3iˆ + 4ˆj + 12kˆ then find magnitude and the direction cosines of P . Solution r r Magnitude of P : P == P 2 + P 2 + P 2 = 32 + 42 + 122 = 169 = 13 x y z Direction cosines : cos a = P Px 3 4 P 12 = , cos b = y = , cos g = z = P 13 P 13 P 13 Illustration 34. Solution Y 2 2 2 a = ax + a y = 1 + 1 = 2 = b ay ax 1 1 = = Þ b = 45° Þ a = 45° & cos b = a a 2 2 Illustration 35. Horizontal component = 4 sin 60° = 2 3 N ss io Illustration 36. A force of 4N is inclined at an angle of 60° from the vertical. Find out its components along horizontal and vertical directions. Solution Vertical component = 4 cos 60° = 2N 20 Ay Ax A 1 1 1 1 1 1 = Þ a = cos -1 = Þ b = cos -1 Þ g = cos -1 ; cos b = ; cos g = z = A A A 3 3 3 3 3 3 n cos a = 19 LL E Given Ax = Ay = Az = 1 so A = A x2 + A y 2 + A z2 = 1 + 1 + 1 = 3 -2 r Find out the angle made by A = ˆi + ˆj + kˆ vector from X, Y and Z axes respectively. Solution X a N cos a = 0 2 = 45 45 ° ® a= ^ ° i+ ^ j Find the angle made by ( $i + $j ) vector from X and Y axes respectively. A 3 80 3 = = 40 3N A y = A sin 60° = 2 2 Illustration 38. ® A q = 60° Horizontal r r Determine that vector which when added to the resultant of P = 2iˆ + 7jˆ - 10kˆ and Q = ˆi + 2jˆ + 3kˆ gives a unit vector along X-axis. Solution r r r Resultant R = P + Q = (2$i + 7$j - 10k$ ) + ($i + 2$j + 3k$ ) = 3$i + 9$j - 7k$ r r But R + required vector = $i so required vector = ˆi - R = ˆi - ( 3iˆ + 9ˆj - 7kˆ ) = -2iˆ - 9ˆj + 7kˆ 24 Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 Now A Þ A = 80N 2 Vertical Q Ax = A cosq \ 40 = A cos 60° = Se A Illustration 37. A force is inclined at an angle of 60° from the horizontal. If the horizontal component of the force is 40N, calculate the vertical component. Solution E Pre-Medical : Physics ALLEN Illustration 39. Y B A 13 5° r r r Add vectors A,B and C which have equal magnitude of 50 units and are inclined at angles of 45°, 135° and 315° respectively from X-axis. Solution r r Angle between B and C is equal to 315°–135°= 180° They balance each other \ r so sum of these three is A i.e. 50 units at 45° from X-axis 45° 18 0° X C Illustration 40. C The sum of three vectors shown in figure, is zero. uuur uuur What is the magnitude of vector OB & OC ? B 45° O A = 10N OD = OC cos 45° and OE = OC sin 45° For zero resultant OE = OA or OC sin 45° = 10N 1 2 = OB Þ OB = 10 N O A = 10N 20 and OD = OB Þ OC cos 45° = OB Þ 10 2 ´ -2 2 uuur = 10N Þ OC = 10 2N 19 1 B 45° D LL E Þ OC ´ E C 0 uuur Resolve OC into two rectangular components. N Solution n Illustration 41. North 3N 5N ss io For shown situation, what will be the magnitude of minimum force in force is along east direction? A Solution East 4N South Se newton that can be applied in any direction so that the resultant 37° West Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 Let force be F so resultant is in east direction E r 4iˆ + 3jˆ + ( 5 cos 37°ˆi + 5 sin 37°ˆj ) + F = kiˆ N( j ) r r Þ 4iˆ + 3jˆ + 4iˆ + 3jˆ + F = kiˆ Þ 8iˆ + 6jˆ + F = kiˆ r 2 2 Þ F = ( k - 8 ) ˆi - 6jˆ Þ F = ( k - 8) + ( 6) Þ Fmin = 6N OR Fmin = y-components of existing forces æ 3ö = 3N + 5N sin37° = 3N +5N ç ÷ = 6N è 5ø 3N 5N 37° W E( i ) 4N Fmin S 25 Pre-Medical : Physics ALLEN BEGINNER'S BOX-10 2. r The x and y components of vector A are 4m and 6m respectively. The x and y components of vector r r r A + B are 10m and 9m respectively. For the vector B calculate the following (a) x and y components (b) length and (c) the angle it makes with x-axis $ Find the directional cosines of vector 5$i + 2$j + 6k . Also write the value of sum of squares of directional cosines 3. of this vector. r r r r If A = 6 $i – 6 $j + 5 k$ and B = $i + 2 $j – k$ , then find a unit vector parallel to the resultant of A & B . 1. e j 1.7 Multiplication and Division of a Vector by a Scalar r 1.8 Scalar Product of Two Vectors 0 N r A r r r If there is a vector A and a scalar K and if B = KA and C = where K > 0 then K (a) In multiplication of a vector by a scalar the magnitude becomes K times while the direction remains same. r r So that angle between A and B is zero. (b) In division of a vector by a scalar,the magnitude becomes (1/K) times and the direction remains same. So r r that angle between A and C is zero. GOLDEN KEY POINTS -2 19 20 LL E Definition : The scalar product (or dot product) of two vectors is defined as the product of their magnitudes with r r cosine of the angle between them. Thus if there are two vectors A and B having angle q between them then r r their scalar product is written as A × B = AB cos q r rr r Example : W = F.S Where F = force and S = displacement. l r r According to definition A × B = AB cos q q A.B = AB cos q A X Scalar product of two vectors will be maximum when cos q = max =1, i.e. q =0°, i.e, vectors are parallel. r r ( A × B )max = AB l r r Scalar product of two vectors will be zero when cos q = 0, i.e. q =90° so ( A × B ) = 0 if the scalar product of two nonzero vectors is zero then vectors are orthogonal or perpendicular to each other. l ˆ ˆ ˆ =1 × 1 × cos 90°=0 In case of orthogonal unit vectors ˆi,jˆ and kˆ : ˆi.jˆ = ˆj.k=k.i l The scalar product of a vector by itself is termed as self dot product and is given by r r r r r A.A = A A cos 0° = A 2 Þ A = A × A l ˆ ˆ = 1 ´ 1 ´ cos 0° = 1 In case of unit vector nˆ : n.n l r r ˆ ˆ = (A B + A B + A B ) ˆi + B ˆj + B k) In terms of components : A × B = (A xˆi + A y ˆj + A z k).(B x y z x x y y z z 26 ˆ ˆ =1 ˆ ˆ = ˆi .iˆ = ˆj.jˆ = k.k so n.n Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 l B Se A r r æ ö -1 A × B the angle between the vectors q = cos ç ÷ è AB ø ss io n l Dot product is always a scalar, which is positive if angle between the vectors is acute (i.e.q < 90°) and negative if angle between them is obtuse (i.e. 90° < q £ 180°). r r r r r r r r r r r Dot product is commutative A × B = B × A l Dot product is distributive A × ( B + C ) = A × B + A × C l E Pre-Medical : Physics ALLEN r r 1.9 Projection of A on B (i) (ii) r r r r r r æ A.B ö A.B r ˆ = A.B In scalar form : Projection of A on B=A cos q = A ç ÷= è AB ø B ® A q Acosq r r r r r æ A.B ö ˆ ˆ ˆ )B ˆ In vector form : Projection of A on B= ( A cos q ) B = ç B = ( A.B ÷ ç B ÷ è ø ® B Illustrations Illustration 42. Can scalar product be ever negative ? Solution Yes. Scalar product will be negative if q > 90°. r r r r \ When q > 90° then cosq is negative and P × Q will be negative. Q P × Q = PQ cos q N Illustration 43. r r r r If A = 4iˆ + njˆ - 2kˆ and B = 2iˆ + 3jˆ + kˆ , then find the value of n so that A ^ B Solution -2 Illustration 44. 19 LL E ˆ ˆ + 3ˆj + k) ˆ = 0 r (4 ´ 2) + (n ´ 3) + (-2 ´ 1) = 0 r 3n = – 6 r n = – 2 \ (4iˆ + njˆ - 2k).(2i 0 r r Dot product of two mutually perpendicular vectors is zero A × B = 0 \ cos q = ss io n 20 r r r r r r Three vectors A, B and C are such that A = B + C and their magnitudes are in ratio 5 : 4 : 3 respectively. r r [AIPMT (Mains) - 2008] Find angle between vector A and C Solution r r r r r r Given that : A = B + C Þ A – C = B r r r r r r r r By taking self dot product on both sides (A - C) . (A - C) = B.B Þ A2 + C2 – 2A.C = B2 r r Now let angle between A and C be q then A2 + C2 – 2AC cosq = B2 2 2 2 æ 3ö 18 3 A2 + C2 - B2 (5) + (3) - (4) = Þ q = cos–1 ç ÷ = 53° = = è 5ø 2 (5) (3) 30 5 2AC Se A OR r r r r r r r r Since 52 = 42 + 32 the vectors A, B and C with A = B + C make a triangle with angle between B and C Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 r r 3 as 90°. If q is the angle between A and C , then cosq = 5 E 1. 2. 3. 4. 5. \ q = 53° BEGINNER'S BOX-11 r r r r r r r r If a and b are two non collinear unit vectors and |a + b| = 3 , then find the value of (a - b) . (2a + b) r r If A = 4$i - 2$j + 4k$ and B = -4$i + 2$j + ak$ are perpendicular to each other then find value of a ? If vector (a$ + 2b$ ) is perpendicular to vector (5a$ - 4b$ ) , then find the angle between a$ and r r If A = 2$i - 2$j - k$ and B = $i + $j , then : r r r (a) Find angle between A and B . (b) Find the projection of resultant vector of A and r Find the vector components of a = 2$i + 3$j along the directions of $i + $j . b$ . r B on x-axis. 27 Pre-Medical : Physics ALLEN 1.10 Vector Product of Two Vectors Definition The vector product or cross product of two vectors is defined as a vector having magnitude equal to the product of their magnitudes with the sine of angle between them, and its direction is perpendicular to the plane r r containing both the vectors according to right hand screw rule or right hand thumb rule. If A and B are two r r r r r r vectors, then their vector product i.e. A ´ B is a vector C defined by C = A ´ B = AB sin qnˆ Right Hand Thumb Rule r r Place the vector A and B tail to tail. Now place stretched fingers r r and thumb of right hand perpendicular to the plane of A and B r such that the fingers are along the vector A . If the fingers are r now closed through smaller angle so as to go towards B , then r r r the thumb gives the direction of A ´ B i.e. C Ù C=A×B n Ù n A B q A×B 0 -2 19 LL E N Right Hand Screw Rule r r r r The direction of A ´ B i.e. C is perpendicular to the plane containing vectors A r r and B and towards the advancement of a right handed screw rotated from A (first r vector) to B (second vector) through the smaller angle between them. Thus, if a right r r handed screw whose axis is perpendicular to the plane formed by A and B is rotated r r from A to B through the smaller angle between them, then the direction of r r advancement of the screw gives the direction A ´ B . A B q ss io n 20 Examples of Vector Product r r r r r r (i) Torque : t = r ´ F (ii) Angular momentum : J = r ´ p r r r r r r (iii) Velocity : v = w ´ r (iv) Acceleration : a = a ´r r r r r r Here r is position vector and F,p,w and a are force, linear momentum, angular velocity and angular acceleration respectively. 1.11 Geometrical Meaning of Vector Product of Two Vectors Area of DPOQ = r r \ Area of parallelogram OPRQ = 2[area of D OPQ] = A ´ B Formulae to Find Area r r 1 r r If A and B are two adjacent sides of a triangle, then its area = A ´ B 2 r r r r If A and B are two adjacent sides of a parallelogram then its area = A ´ B r r 1 r r If A and B are diagonals of a parallelogram then its area = A ´ B 2 28 B = Area of parallelogram OPRQ base ´ height (OP) (NQ) 1 r r = = A´B 2 2 2 R q O N A P Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 = base × height Se A r r uuuur uuur (i) Consider two vectors A and B which are represented by OP and OQ and ÐPOQ = q (ii) Complete the parallelogram OPRQ. Join P with Q. Here OP = A and OQ = B. Draw QN ^ OP. r r (iii) Magnitude of cross product of A and B Q r r A ´ B = ABsinq = (OP) (OQ sin q) = (OP) (NQ) (QNQ=OQsinq) E Pre-Medical : Physics ALLEN GOLDEN KEY POINTS l Vector product of two vectors is always a vector perpendicular to the plane containing the two vectors, i.e., r r orthogonal (perpendicular) to both the vectors A and B l Vector product of two vectors is not commutative i.e. A ×B = C r r r r r r r r But A ´ B = B ´ A = AB sin q A´B ¹ B´A r r r r Note : A ´ B = -B ´ A r r r r i.e., in case of vectors A ´ B and B ´ A magnitudes are equal but B A B A directions are opposite [See the figure] r r r r -1 A ´ B ˆ Þ q = sin According to definition of vector product of two vectors A ´ B = AB sin qn AB The vector product of two vectors will be maximum when sin q = max. = 1, i.e., q =90° r r A ´ B max = AB sin 90o = AB LL E i.e. vector product is maximum if the vectors are orthogonal (perpendicular). The vector product of two non-zero vectors will be zero when sinq = 0, i.e. 19 l 0 l B × A = –C -2 l The vector product is distributive when the order of the vectors r r r r r r r is strictly maintained, i.e. A ´ ( B + C ) = A ´ B + A ´ C N l r r r r when q = 0° or 180° , |A ´ A|= 0 or |A ´ ( - A)| = 0 l 20 Therefore if the vector product of two non-zero vectors is zero, then the vectors are collinear. The self cross product, i.e., product of a vector by itself is a zero vector or a null vector. r r r r In case of unit vector n$ : n ˆ´n ˆ = 1 ´ 1 ´ sin 0° n ˆ = 0 so that ˆi ´ ˆi = 0, ˆj ´ ˆj = 0, kˆ ´ kˆ = 0 l In case of orthogonal unit vectors $i, $j and k$ ; according to right hand thumb rule A ˆj ´ ˆk = ˆi , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 l l Se ss io l ˆi ´ ˆj = kˆ , E n r r r i.e. A ´ A = (AA sin0o ) n ˆ =0 kˆ ´ ˆi = ˆj ˆi ˆj and ˆj ´ ˆi = -kˆ , kˆ ´ ˆj = -ˆi , kˆ j i ˆi ´ ˆk = -ˆj k r r In terms of components A ´ B = A x Ay ˆ A z = ˆi(A y B z - A z B y ) - ˆj(A x B z - A z B x ) + k(A x By - A y Bx ) Bx By Bz r r ˆ ´B ˆ r r A´B A Unit vector perpendicular to A as well as B is n̂ = ± r r = ± sin q A´B l r r r r r r r r r If A , B and C are coplanar, then A × ( B ´ C ) = 0 . [Q (B ´ C) is perpendicular to A ) r r r r r r r r Angle between ( A + B ) and ( A ´ B ) is 90° as A ´ B is perpendicular to plane containing A & B . l A scalar or a vector, cannot be divided by a vector. l Vectors of different types can be multiplied to generate new physical quantities which may be a scalar or a vector. If, in multiplication of two vectors, the generated physical quantity is a scalar, then their product is called scalar or dot product and if it is a vector, then their product is called vector or cross product. l 29 Pre-Medical : Physics ALLEN Illustrations Illustration 45. r ˆ and rr = (5iˆ - 3ˆj) , then calculate torque ( rt = rr ´ Fr ) . If F = (4iˆ - 10j) Solution r r Here r = 5 ˆi - 3 ˆj + 0kˆ and F = 4 ˆi - 10 ˆj + 0 kˆ ˆi ˆj kˆ r r r ˆ -50 + 12) = -38kˆ \ t = r ´ F = 5 -3 0 = ˆi(0 - 0) - ˆj(0 - 0) + k( 4 -10 0 Illustration 46. \ n̂ = ± 1 83 ˆ (7iˆ - 3jˆ - 5k) 0 20 r r \ A ´ B = 72 + ( -3)2 + ( -5)2 = 83 unit -2 ˆ -2 - 3) = 7iˆ - 3jˆ - 5kˆ 1 = ˆi(6 + 1) - ˆj(4 - 1) + k( 2 19 kˆ LL E ˆi ˆj r r A´B = 2 3 1 -1 N ˆ and (iˆ - ˆj + 2k) ˆ . Find a unit vector perpendicular to both the vectors (2iˆ + 3ˆj + k) Solution r r Let A = 2iˆ + 3jˆ + kˆ and B = ˆi - ˆj + 2kˆ r r r r A´B unit vector perpendicular to both A and B is n̂ = ± r r |A ´ B| Illustration 47. r r r If A = ˆi + 2 ˆj + 3kˆ , B = - ˆi + ˆj + 4 kˆ and C = 3 ˆi - 3 ˆj - 12kˆ , then find the angle between the vectors r r r r r A + B + C and A ´ B in degrees. ( ) n Solution ) ˆj kˆ 2 3 = 5iˆ - 7jˆ + 3kˆ 1 4 Se ˆi r r r r r r r Let P = A + B + C = 3iˆ - 5kˆ and Q = A ´ B = 1 -1 ss io ( 1. 2. 3. 4. 30 BEGINNER'S BOX-12 r r There are two vectors A = 3$i + $j and B = $j + 2k$ . For these two vectors – r r (a) If A & B are the adjacent sides of a parallalogram then find the magnitude of its area. r r (b) Find a unit vector which is perpendicular to both A & B . r r r r r r If 3 A ´ B = A × B , then find the angle between A and B . r r r r AB If the area of a triangle of sides A & B is equal to , then find the angle between A & B . 4 r r r r r r Find A . B if A = 2, B = 5, and A ´ B = 8 Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 A r r P × Q 15 - 15 r r = = 0 Þ q = 90° Angle between P & Q is given by cos q = PQ PQ E Pre-Medical : Physics ALLEN ANSWERS BEGINNER'S BOX-1 3 2 (v) –1 sinq = 4 , 5 cosq = cotq = 3 , 4 1 1. 2 (vi) 0 3 , 5 cosecq = tanq = 4 , 3 3. –2/3 (iii) 1 (iv) 5x4+3x2+2x–1/2 3. u + at 5. 2 pr 1. 2x - x 4 (iv) 3 (x + 1)2 (i) x16 +c 16 4. 2 (iv) a + b loge x + c x æ 1 1ö - ÷ è r2 r1 ø (ii) kq1q2 ç 1 (iii) M (v2 – u2) (iv) ¥ (v) 1 2 5 2 6 , , ;1 65 65 65 2. x2 + 2x + loge x + c 2 (vi) – 45m æ1ö (c) tan -1 ç ÷ è2ø (ii) – 2x –1/2 +c x -6 + loge x + c 2 GMm R 2. 2 ss io 1 (iii) – (4x + 5)2 x2 + loge x + c 2 r2 (a) 6m and 3m (b) BEGINNER'S BOX-5 1. 3 Se –1 2GMm BEGINNER'S BOX-10 (i) 4x + 3 (v) 3. 1 x2 2 (ii) – (2x + 1)2 (iii) – $i - $j + k$ LL E 2t + 5 30 cm2 s (i) 40N & 30N (vi) 2ax+b A Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATHIN PHY\01-THEORY.P65 E 1. –4 BEGINNER'S BOX-4 1. 2. Fnet = BEGINNER'S BOX-9 (ii) –3x (vii) 15x4 – 3+ 6. 2 3 1. 7 (i) x5/2 2 (v) 20x3 + 9x1/2 + 9 2. 4. 2. 385 1275 BEGINNER'S BOX-8 BEGINNER'S BOX-3 1. 1. 5 4 2. 13 2 2. (4) BEGINNER'S BOX-7 BEGINNER'S BOX-2 1. p q 5 1 , (ii) , q p 2 5 (i) 0 (iv) – (iii) -2 3 3 2 19 (ii) N 1 20 2. (i) - n 1. BEGINNER'S BOX-6 (vi) 1 (vii) 2 1. 3. 7 $i - 4 $j + 4k$ 9 BEGINNER'S BOX-11 1 2 2. 5 4. (a) 90° (b) 3 5. 3. 60° 5 $ $ ( i + j) 2 BEGINNER'S BOX-12 2$ 6$ 3 $ i- j+ k 7 7 7 1. (a) 7 units (b) 2. 30° r r 4. A . B = ± 6 3. 30° or 150° 31 Pre-Medical : Physics ALLEN Build Up Your Understanding EXERCISE-I (Conceptual Questions) 9. 1 and q lies in the first quadrant, the 5 value of cos q is : 6 (4) – 1 6 CALCULUS 4. The coordinates of a particle moving in XY–plane vary with time as x = 4t2; y = 2t. The locus of the particle is a :(2) Circle (4) No where Two particles A and B are moving in XY-plane. Their positions vary with time t according to relation : xA (t) = 3t, xB (t) = 6 yA (t) = t, yB (t) = 2 + 3t2 Distance between two particles at t = 1 is : (1) 5 10. (2) 3 (3) Parabola (4) Ellipse The slope of graph as shown in figure at points 1, 2 and 3 is m1, m2 and m3 respectively then (1) y = x 2 y 3 x (2) m1 < m2 < m3 (4) m1 = m3 > m2 (1) x–coordinate A 12. (3) Both x and y coordinates (4) Data insufficient. 7. (1) 0.2 cm/s (3) 0.6 cm/s (2) y–coordinate Magnitude of slope of the shown graph. y 12 (2) y = 2x (4) y = – x 4 The side of a square is increasing at the rate of 0.2 cm/s. The rate of increase of perimeter w.r.t. time is : Se A particle moves along the straight line y = 3x + 5. Which coordinate changes at a faster rate ? 11. (4) n 2 (1) m1 > m2 > m3 (3) m1 = m2 = m3 6. (3) y = –4x ss io 1 (3) 4 A particular straight line passes through origin and a point whose abscissa is double of ordinate of the point. The equation of such straight line is : LL E (1) Straight line (3) Both (1) and (2) (2) 0.4 cm/s (4) 0.8 cm/s Frequency f of a simple pendulum depends on its length l and acceleration g due to gravity according to the following equation f = 1 2p g . Graph l between which of the following quantities is a straight line ? x (1) First increases then decreases (2) First decreases then increases 32 (1) f on the ordinate and l on the abscissa (2) f on the ordinate and Öl on the abscissa (3) Increases (3) f2 on the ordinate and l on the abscissa (4) Decreases (4) f2 on the ordinate and 1/l on the abscissa Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATH IN PHY\02-EXERCISE..P65 (3) 1 (2) (2, 0) 0 5 (2) – 6 (1) (1, 0) -2 5 6 The equation of a curve is given as y = x 2 + 2 – 3x. The curve intersects the x-axis at N If tan q = (1) 5. 8. 20 3. GEOMETRY 19 TRIGONOMETRY 1. As q increases from 0° to 90°, the value of cos q :(1) Increases (2) Decreases (3) Remains constant (4) First decreases then increases. 2. The greatest value of the function –5 sinq + 12 cosq is (1) 12 (2) 13 (3) 7 (4) 17 E Pre-Medical : Physics ALLEN ALGEBRA (3) 5 4 (4) 4 3 In the given figure, each box represents a function machine. A function machine illustrates what it does with the input. Input (x) Square root Output (z) of the input Double the input and add three Which of the following statements is correct ? (1) z=2x+3 (2) z=2(x+3) (3) z = 2x + 3 (4) z = 2 ( x + 3) r | A| (3) n̂ = r A Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATH IN PHY\02-EXERCISE..P65 E 19. B (4) None of the above A physical quantity which has a direction : (1) must be a vector (2) may be a vector (3) must be a scalar (4) none of the above Which of the following physical quantities is an axial vector ? (1) displacement (2) force (3) velocity (4) torque 2 , then value of angle q is : (1) 30o r r (2) n̂ = A| A| The forces, which meet at one point but their lines of action do not lie in one plane, are called : (2) 45o (3) 60o (4) 75o r r The resultant of A and B makes an angle a with r r A and b with B , then : (1) a < b (2) a < b if A < B (3) a < b if A > B (4) a < b if A = B r r r r r Two vectors A and B are such that A + B = C and A 2 + B 2 = C 2. Which of the following statements, is correct ? r r (1) A is parallel to B r r (2) A is anti-parallel to B r r (3) A is perpendicular to B r r (4) A and B are equal in magnitude 24. 25. The minimum number of vectors of equal magnitude required to produce a zero resultant is : (1) non-coplanar and non-concurrent forces (1) 2 (2) 3 (2) coplanar and non-concurrent forces (3) non-coplanar and concurrent forces (3) 4 (4) more than 4 26. The direction of the angular velocity vector is along : How many minimum number of coplanar vectors having different magnitudes can be added to give zero resultant ? (1) Along the tangent of circular path (1) 2 (4) coplanar and concurrent forces 20. If R = 23. A 18. A 20 r A r n̂ = (1) | A| 17. q If n$ is a unit vector in the direction of the vector r A , then :- R Se 16. B LL E DEFINITION & TYPES OF VECTOR 15. Which of the following statements is false : (1) Mass, speed and energy are scalars (2) Momentum, force and torque are vectors (3) Distance is a scalar while displacement is a vector (4) A vector has only magnitude whereas as a scalar has both magnitude and direction 0 6 5 -2 (2) n 14. 8 7 19 (1) 1 1 1 + + + ....... ¥ is 4 16 64 N The sum of the series 1 + ss io 13. ADDITION & SUBTRACTION, MULTIPLICATION & DIVISION OF A VECTOR BY A SCALAR r r 21. Two vectors A and B lie in a plane, another vector r C lies outside this plane, then the resultant of these r r r three vectors i.e. A + B + C : (1) can be zero (2) cannot be zero r r (3) lies in the plane containing A & B r r (4) lies in the plane containing B & C r 22. In vector diagram shown in figure where ( R ) is the r r resultant of vectors ( A ) and ( B ). (2) Along the direction of radius vector (3) Opposite to the direction of radius vector (4) Along the axis of rotation 27. (2) 3 (3) 4 (4) 5 How many minimum number of vectors in different planes can be added to give zero resultant ? (1) 2 (2) 3 (3) 4 (4) 5 33 Pre-Medical : Physics (4) Infinite 32. Vector sum of two forces of 10N and 6N cannot be : 34. (1) 2 N and 2 N (2) 1 N and 1 N (3) 1 N and 3 N (4) 1 N and 4 N r r r If A + B = C and A + B = C, then the angle between r r A and B is : p (2) 4 35. p (1) 0 (3) (4) p 2 r r r The resultant of A & B is R1 . On reversing the r r vector B , the resultant becomes R2 . What is the value of R12 +R22 ? (1) A + B 2 (2) A – B 2 2 2 2 Which of the following sets of concurrent forces may be in equilibrium ? (1) F1 = 3N, F2 = 5N, F3 = 1N (2) F1 = 3N, F2 = 5N, F3 = 9N (3) F1 = 3N, F2 = 5N, F3 = 6N (4) F1 = 3N, F2 = 5N, F3 =15N 34 (2) 2 42. 1 (3) 3 RESOLUTION OF VECTOR (4) 2 5 If a unit vector is represented by 0.5$i - 0.8$j + ck$ , then the value of 'c' is : (1) 1 43. 44. (2) 0.11 1 1 , ,1 2 2 (2) 1 1 1 , , 2 2 2 (4) 0.39 1 2 1 2 , , 1 2 1 2 , , 1 2 1 2 The unit vector along ˆi + ˆj is : (1) k̂ 46. 0.01 (4) The direction cosines of a vector ˆi + ˆj + 2 kˆ are :- (3) 45. (3) r Vector P makes angles a, b & g with the X, Y and Z axes respectively, then sin2 a + sin2 b + sin2 g = (1) 0 (2) 1 (3) 2 (4) 3 Se 2 A 37. (1) (1) 2 (3) 2(A + B ) (4) 2(A – B ) r r r r Given that P + Q = P - Q . This can be true when : r r (1) P = Q r r (2) Q = 0 r r (3) Neither P nor Q is a null vector r r (4) P is perpendicular to Q 2 36. 41. LL E 33. (1) 4N (2) 8N (3) 12N (4) 2N Which of the following pair of forces will never give a resultant force of 2 N ? 40. (2) ˆi + ˆj (3) ˆi + ˆj 2 (4) ˆi + ˆj 2 The unit vector parallel to the resultant of the vectors r r A = 4iˆ + 3jˆ + 6kˆ and B = -ˆi + 3jˆ - 8kˆ is : (1) 1é ˆ ˆ 3i + 6 j - 2kˆ ù û 7ë (2) 1é ˆ ˆ 3i + 6 j + 2kˆ ù û 7ë (3) 1 é ˆ ˆ 3i + 6j + 2kˆ ù û 49 ë (4) 1 é ˆ ˆ 3i + 6j - 2kˆ ù û 49 ë Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATH IN PHY\02-EXERCISE..P65 The vector sum of the forces of 10 newton and 6 newton can be : (1) 2N (2) 8N (3) 18N (4) 20N 0 31. -2 (3) 4 39. N (2) 3 r 3 r A (2) A 2 r r (3) 2 A (4) 3 A What happens, when we multiply a vector by (– 2) ? (1) direction reverses and unit changes (2) direction reverses and magnitude is doubled (3) direction remains unchanged and unit changes (4) none of these Two vectors of equal magnitude have a resultant equal to either of them in magnitude . The angle between them is : (1) 60° (2) 90° (3) 105° (4) 120° If the sum of two unit vectors is a unit vector, then the magnitude of their difference is : (1) 19 (2) 3 (3) 4 (4) Infinite maximum number of rectangular into which a vector can be split in 38. 20 30. (1) 2 What is the components space ? (1) 2 ALLEN r r r r r r If vectors A and B are such that A + B = A = B , r r then A - B may be equated to n 29. What is the maximum number of components into which a vector can be split ? (1) 2 (2) 3 (3) 4 (4) Infinite What is the maximum number of rectangular components into which a vector can be split in its own plane ? ss io 28. E ALLENr (3) ˆi + ˆj + kˆ (4) ˆi + ˆj - kˆ (1) 0 55. 13 units and r R is : (1) 19 N (2) 13 N (3) 11 N (4) 5 N -1 æ 5 ö (1) cos ç ÷ è 12 ø -1 æ 5 ö (2) cos ç ÷ è 13 ø -1 æ 12 ö (3) cos ç ÷ è 13 ø -1 æ 2 ö (4) cos ç ÷ è 13 ø (4) cos (3/2) -1 r What is the angle between A and the resultant of r r ˆ and A - B ˆ ? A+B ) ( ) -1 æ A ö (2) tan ç ÷ èBø (1) 0° æBö (3) tan ç ÷ èAø -1 58. æA-Bö (4) tan ç ÷ èA+Bø -1 ( ) If n̂ = aiˆ + bjˆ is perpendicular to the vector ˆi + ˆj , then the value of a and b may be : (2) –2, 0 (3) 3, 0 (4) E 1 2 , - 1 (1) 300 (2) 600 (3) 90 (4) 180 59. (3) 6$i (4) 3$i - 4 $j ˆ N displaces an object through a A force (3iˆ + 2j) r r ˆ m. The work (W = F·S ) done is : distance (2iˆ - 3j) -1 æ A ö (3) sin ç ÷ èBø 60. (1) Are equal to each other. (2) Are equal to each other in magnitude. (3) Are not equal to each other in magnitude. 61. Aö -1 æ (2) cos ç - ÷ è Bø Aö -1 æ (4) sin ç - ÷ è Bø What is the component of (3iˆ + 4ˆj) along (iˆ + ˆj) ? (1) 1 ˆ ˆ j+i 2 (3) 5 ˆ ˆ j+i 2 0 The vector sum of two forces is perpendicular to their vector difference. In that case, the force : (4) Cannot be predicted. (1) zero (2) 12 J (3) 5 J (4) 13 J r r r r If P.Q = PQ, then angle between P and Q is : (1) 0° (2) 30° (3) 45° (4) 60° r r r The resultant of A and B is perpendicular to A . r r What is the angle between A and B ? -1 æ A ö (1) cos ç ÷ èBø 2 r r Given that A = B. What is the angle between (A+B) r r and (A–B) ? 0 53. (2) 7k$ Se (1) 1, 0 A Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATH IN PHY\02-EXERCISE..P65 52. 57. j r (1) 4$i + 3 j LL E ( e A vector perpendicular to 4$i - 3$j may be : 0 (3) sin (2/3) 56. -2 (2) tan-1 (2/3) 19 (1) tan-1 (3/2) N r The angle that the vector A = 2$i + 3$j makes with y-axis is : DOT PRODUCT 51. r If vectors P , (2) p/3 (3) p /2 (4) p /4 r r Q and R have magnitudes 5, 12 and r r r r P + Q = R , the angle between Q and Forces 3N, 4N and 12N act at a point in mutually perpendicular directions. The magnitude of the resultant force is : -1 50. r r r The magnitudes of vectors A , B and C are r r r respectively 12, 5 and 13 units and A + B = C , r r then the angle between A and B is : 20 49. (2) ĵ - kˆ 54. n 48. (1) ĵ + kˆ Pre-Medical : Physics ss io 47. r If A + B is a unit vector along x-axis and r r A = ˆi - ˆj + kˆ , then what is B ? ( ( ) ) (2) 3 ˆ ˆ j+i 2 ) (4) 7 ˆ ˆ j+i 2 ) ( ( r The vector B = 5iˆ + 2jˆ - Skˆ is perpendicular to the r vector A = 3iˆ + ˆj + 2kˆ if S = (1) 1 (3) 6.3 (2) 4.7 (4) 8.5 35 Pre-Medical : Physics ˆ ˆ (4) A.B r r r (1) A ´ B = 0 The angle between vectors $i + $j and $j + k$ is : e j e j (1) 900 (2) 1800 (3) 00 (4) 600 71. The angle between the two vectors ® Ù Ù Ù ® Ù Ù Ù (1) 4ˆj r Let A = ˆi A cos q + ˆj A sin q , be any vector. Another r r vector B which is normal to A is : (1) ˆiB cos q + ˆjB sin q (2) ˆiB sin q + ˆjB cos q (3) ˆiB sinq - ˆjB cosq (4) ˆiA cosq - ˆjA sinq (3) 1 67. (4) zero r A force F = ( 3iˆ + 4ˆj ) N acts on a body and displaces (1) 10J (2) 12J (3) 19J (4) 25J (1) (1) 3 (2) 4 (3) 5 (4) zero 73. (2) 1/2 (3) –1/2 (4) 1 (2) ˆi ´ ˆj = kˆ (4) ˆi ´ kˆ = -ˆi r r Two vectors P and Q are inclined to each other at angle q. Which of the following is the unit vector r r perpendicular to P and Q ? r r P´ Q (1) P.Q 74. ˆ P̂ ´ Q sin q r ˆ P̂ ´ Q P̂ ´ Q (3) (4) PQ sin q PQ sin q The magnitude of the vector product of two vectors 75. (1) Greater than AB (2) Less than AB (3) Equal to AB (4) Equal to zero r r r If P ´ Q = R , then which of the following statements ˆ is perpendicular to the If a vector (2iˆ + 3ˆj + 8k) (1) –1 ˆˆ i.i = 1 (3) ˆi.jˆ = 0 What is the projection of 3iˆ + 4kˆ on the y-axis ? ˆ , then the value of a is : vector (4ˆj - 4iˆ + a k) 36 ˆi, ˆj and kˆ are unit vectors along X, Y & Z axis (2) Se A by the force is : 69. If respectively, then tick the wrong statement : (2) 2 r r it by S = ( 3iˆ + 4jˆ ) m . The work done ( W = F × S ) 68. 72. (4) -4iˆ n (1) 3 (2) – (iˆ + ˆj) ˆ (3) (iˆ + k) r r The vector P = aiˆ + ajˆ + 3kˆ and Q = aiˆ - 2jˆ - kˆ are perpendicular to each other. The positive value of a is : r may be :- N (4) 45° r ss io 66. (3) 90° r r (3) A.B = 48 (4) A = 5 r A vector F1 is along the positive X-axis. If its vector LL E 65. (2) 180° A 1 = B 2 product with another vector F2 is zero then F2 A = 3 i + 4 j + 5 k and B = 3 i + 4 j - 5 k will be : (1) zero (2) 76. r r A and B may not be : is not true ? r r (1) R ^ P r r (2) R ^ Q r r r (3) R ^ (P + Q) r r r (4) R ^ (P ´ Q) Which of the following vector identities is false ? r r r r r r r r (2) P + Q = Q ´ P (1) P + Q = Q + P r r r r r r r r (3) P.Q = Q.P (4) P ´ Q ¹ Q ´ P Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATH IN PHY\02-EXERCISE..P65 r r (3) B.A 0 r ˆ (2) A.B -2 64. r r (1) A.B CROSS PRODUCT r r 70. If A = 3iˆ + 4ˆj and B = 6iˆ + 8jˆ and A and B are the r r magnitudes of A and B , then which of the following is not true ? 19 63. ALLEN r r What is the projection of A on B ? 20 62. E ) ( ) (1) 0 (2) A – B 2 2 (3) A + B + 2AB (4) none of these r r r r r r If A ´ B = 0 and B ´ C = 0 , then the angle between r r A and C may be : 2 78. (1) zero (3) 79. 82. (2) p 2 p 4 (1) p rad (3) 83. ) ˆ and 3iˆ form two sides of If the vectors ˆi + ˆj + k a triangle, then area of the triangle is : 3 unit 3 (3) unit 2 80. 84. (2) 2 3 unit (4) 3 2 unit r For a body, angular velocity w = ˆi - 2jˆ + 3kˆ and radius r r r r vector r = ˆi + ˆj + kˆ , then its velocity ( v = w ´ r ) is: (1) -5iˆ + 2jˆ + 3kˆ (2) -5iˆ + 2jˆ - 3kˆ (3) -5iˆ - 2jˆ + 3kˆ (4) -5iˆ - 2jˆ - 3kˆ (2) p rad 4 85. 86. (2) 5 3 unit (3) 10 3 unit (4) 20 3 unit r r (4) B = 0 r r If three vectors satisfy the relation A.B = 0 and r r r A.C = 0 , then A can be parallel to r r r r r r (1) C (2) B (3) B ´ C (4) B.C Se A ss io (1) 14 unit p rad 2 n 3iˆ + ˆj - 2kˆ and ˆi - 3jˆ + 4kˆ will be : ) r r A vector A points vertically upward and B points r r towards north. The vector product A ´ B is (1) zero (2) along west (3) along east (4) vertically downward r r r r r r If A ´ B = A.B , then the angle between A and B will be : (1) 30° (2) 45° (3) 60° (4) 75° r r Two vecto rs A an d B are such that r r r r A + B = A - B . Then select incorrect atternative r r r r r (1) A.B = 0 (2) A ´ B = 0 r r (3) A = 0 Area of a parallelogram, whose diagonals are ( (4) zero LL E 81. ) N (1) ( 2 (4) None ( r r r r The angle between vectors A ´ B and B ´ A is: -2 ( 0 r r r r What is the value of A + B · A ´ B ? 20 77. Pre-Medical : Physics 19 ALLEN ANSWER KEY Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATH IN PHY\02-EXERCISE..P65 EXERCISE-I (Conceptual Questions) E Que. Ans. Que. Ans. Que. Ans. Que. Ans. Que. Ans. Que. Ans. 1 2 2 2 3 1 4 3 5 2 6 2 7 2 8 3 9 1 10 1 11 4 12 4 13 4 14 3 15 4 16 1 17 2 18 4 19 3 20 4 21 2 22 2 23 3 24 3 25 1 26 2 27 3 28 4 29 1 30 2 31 2 32 4 33 4 34 1 35 3 36 2 37 3 38 4 39 2 40 4 41 2 42 2 43 3 44 3 45 3 46 1 47 2 48 2 49 2 50 1 51 4 52 3 53 2 54 3 55 3 56 2 57 1 58 1 59 2 60 61 4 62 2 63 4 64 3 65 3 66 1 67 4 68 4 69 3 70 3 71 4 72 4 73 2 74 1 75 4 76 2 77 1 78 1 79 3 80 1 81 2 82 1 83 2 84 2 85 3 86 3 4 37 Pre-Medical : Physics ALLEN AIPMT/NEET & AIIMS (2006-2018) EXERCISE-II (Previous Year Questions) AIPMT 2006 r r The vectors A an d B are su ch that r r r r r A + B = A - B . The angle between vectors A 4. directions indicated in the figure. Which of the following statements is true ? r and B is - (2) 60° (4) 45° d AIPMT 2007 r (1) b + er = r (3) d + cr = r r r r A +B , then the value of A . B 3 is : 2 2 +B + (3) (A2 + B2 + Re-AIPMT 2015 1/ 2 5. (2) A + B 1/2 (4) (A2+B2+AB)1/2 3 AB) AIIMS 2009 r r r th e to rque ( t = r ´ F) of a force If vectors r r wt wt A = cos wtiˆ + sin wtjˆ and B = cos ˆi + sin ˆj 2 2 are functions of time, then the value of t at which they are orthogonal to each other is : p (1) t = 0 (2) t = 4w p p (3) t = (4) t = 2w w r r F = -3iˆ + ˆj + 5kˆ acting at the point r = 7iˆ + 3jˆ + kˆ (2) 4iˆ + 4jˆ + 6kˆ (3) -14iˆ + 38ˆj - 16kˆ (4) -21iˆ + 3jˆ - 5kˆ NEET-I 2016 6. If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is :(1) 0° (2) 90° (3) 45° (4) 180° ANSWER KEY EXERCISE-II (Previous Year Questions) Que. Ans. 38 1 1 2 4 3 1 4 4 5 4 6 2 Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATH IN PHY\02-EXERCISE..P65 A Se ss io n (1) 14iˆ - 38ˆj + 16kˆ 20 Fin d r r (2) b + cr = f r r (4) d + er = f LL E 3. I J 3K AB f e N F (1) G A H r f r f c 0 r r If A ´ B = 2. (3) 75° -2 (1) 90° b a 19 1. AIPMT 2010 r r Six vectors, a through f have the magnitudes and E Pre-Medical : Physics ALLEN Check Your Understanding EXERCISE-III (Analytical Questions) The moon’s distance from the earth is 360000 km and its diameter subtends an angle of 42' at the eye of the observer. The diameter of the moon is (1) 4400 km (2) 1000 km (3) 3600 km (4) 8800 km If 3 cosq + 4 sinq = A sin(q + a), then values of A and a are -1 æ 3 ö (2) 5, sin ç ÷ è5ø -1 æ 4 ö (3) 7, sin ç ÷ è5ø (4) None of these the third vector is two having equal magnitude. Then the angles between vectors are : (1) 30o, 60o, 90o 9. 1 mv 2 . If the kinetic 2 energy of a particle moving along x-axis varies with x as K(x) = 9 – x2. then the region in which particle lies is : (1) x ³ 9 (2) –3 £ x £ 3 (3) 0 £ x £ 9 (4) –¥ < x < ¥ The sum of magnitudes of two forces acting at a point is 16N. If the resultant force is 8N and its direction is perpendicular to smaller force, then the forces are : (1) 6N & 10N (2) 8N & 8N (3) 4N & 12N (4) 2N & 14N At what angle must the two forces (x + y) and (x – y) æ P2 - Q2 ö (1) ç 2PQ ÷ è ø (3) 10. (4) 2 ) + y2 ? (1) q1 = q2 (3) q1 = 2q2 12. é - 2(x - y ) ù -1 é - (x + y ) ù cos -1 ê (1) cos ê ú 2 2 ú (2) 2 2 ë 2(x - y ) û ë x +y û 2 2 2 A 2 2 2 2 2 -1 é - (x + y ) ù -1 é (x - y ) ù (3) cos ê 2 2 ú (4) cos ê 2 2 ú ë x -y û ë x +y û (1) l cos (q/2) (2) q1 = q2 2 (4) None of the above n Square of the resultant of two forces of equal magnitude is equal to three times the product of their magnitude. The angle between them is : (1) 0° (2) 45° (3) 60° (4) 90° A unit radial vector r̂ makes angles of a = 30° relative to the x-axis, b = 60° relative to the y-axis, and g = 90° relative to the z-axis. The vector r̂ can be written as : (1) A vector of length l is turned through the angle q about its tail. What is the change in the position vector of its head ? 7. 20 q1 and q2 ? 11. P+Q P-Q r r r Given that P = Q = R. If P + Q = R then the angle r r r r r r between P & R is q1. If P+Q+R = 0 then the angle r r between P & R is q2. What is the relation between ss io (x act so that the resultant may be Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATH IN PHY\02-EXERCISE..P65 P Q Se 6. E (2) Q LL E with speed v is given by K = (4) 90o, 135o, 135o r r r r The resultant of two vectors P and Q is R . If Q is doubled then the new resultant vector is r perpendicular to ' P '. Then R is equal to : N 7 9 m/s (2) m/s (3) 4 m/s (4) 5 m/s 2 2 The kinetic energy of a particle of mass m moving (1) 5. (2) 45o, 45o, 90o (3) 45o, 60o, 90o If velocity of a particle is given by v = (2t + 3) m/s, then average velocity in interval 0 £ t £ 1s is : 4. 2 times that of either of the 0 3. -1 æ 3 ö (1) 5, cos ç ÷ è5ø r r r r Given that A + B + C = 0 . Out of these three vectors two are equal in magnitude and the magnitude of -2 2. 8. 19 1. (3) 1ˆ 3ˆ i+ j 2 2 (2) 2ˆ 1 ˆ i+ j 3 3 3 ˆ 1ˆ i+ j 2 2 (4) None of these (2) 2l sin (q/2) (4) l sin(q/2) (3) 2l cos (q/2) EXERCISE-III (Analytical Questions) ANSWER KEY Que. 1 2 3 4 5 6 7 8 9 10 11 12 Ans. 1 2 3 2 1 1 2 4 2 2 3 2 39 Pre-Medical : Physics ALLEN Target AIIMS EXERCISE-IV (Assertion & Reason) Directions for Assertion & Reason questions These questions consist of two statements each, printed as Assertion and Reason. While answering these Questions you are required to choose any one of the following four responses. (A) If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion. (B) If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion. (C) If Assertion is True but the Reason is False. (D) If both Assertion & Reason are false. e j A e j 20 10. (1) A 11. [AIIMS 2000] (2) B (3) C Assertion : Vector ($i + $j + k$ ) is perpendicular to ($i – 2 $j + k$ ) F I GH JK Reason : Two non-zero vectors are perpendicular if their dot product is zero. [AIIMS 2009] (1) A (2) B (3) C (4) D ANSWER KEY EXERCISE-IV (Assertion & Reason) 40 (4) D Que. 1 2 3 4 5 6 7 8 9 10 11 Ans. 3 3 3 4 1 1 3 4 4 4 1 Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATH IN PHY\02-EXERCISE..P65 5. n 4. [AIIMS 1998] (1) A (2) B (3) C (4) D Assertion : A physical quantity must be a vector, if magnitude as well as direction is associated with it. Reason : A physical quantity can be a scalar quantity, if it is associated with magnitude only. Se (1) A (2) B (3) C (4) D Assertion : A vector can have zero magnitude if one of its rectangular components is not zero. Reason : Scalar product of two vectors cannot be a negative quantity. (1) A (2) B (3) C (4) D Assertion : The angle between the two vectors $i + $j and $j + k$ is p radian. 3 r r Reason : Angle between two vectors A and B is r r -1 A. B given by q = cos . AB (1) A (2) B (3) C (4) D 9. ss io 3. -2 8. 0 7. Assertion : Distance is a scalar quantity. Reason : Distance is the length of path traversed. (1) A (2) B (3) C (4) D Assertion : The sum of squares of cosines of angles made by a vector with X, Y and Z axes is equal to unity. Reason : A vector making 45° with X-axis must have equal components along X and Y-axes. (1) A (2) B (3) C (4) D Assertion : Adding a scalar to a vector of the same dimensions is a meaningful algebraic operation. Reason : Displacement can be added to distance. (1) A (2) B (3) C (4) D Assertion : The dot product of one vector with another vector may be a scalar or a vector. Reason : If the product of two vectors is a vector quantity, then product is called a dot product. LL E 2. 6. 19 Assertion : If the initial and final positions coincide, the displacement is a null vector. Reason : A physical quantity cannot be called a vector, if its magnitude is zero. (1) A (2) B (3) C (4) D Assertion : A vector quantity is a quantity that has both magnitude and a direction and obeys the triangle law of addition and equivalently the parallelogram law of addition. Reason : The magnitude of the resultant vector of two given vectors can never be less than the magnitude of any of the given vector. (1) A (2) B (3) C (4) D Assertion : The direction of a zero (null) vector is indeterminate. r r r r r r Reason : We can have A ´ B = A.B with A ¹ 0 r r and B ¹ 0 . N 1. E E ss io n N 0 -2 19 20 LL E Se A Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATH IN PHY\02-EXERCISE..P65 ALLEN Pre-Medical : Physics IMPORTANT NOTES 41 42 E Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_01\01-BASIC MATH IN PHY\02-EXERCISE..P65 ss io n N 0 -2 19 20 LL E Se A Pre-Medical : Physics ALLEN IMPORTANT NOTES