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2 Mathematics In Physics, , 0.1 Introduction., Mathematics is the language of physics. It becomes very easier to describe, understand and apply the, physical principles, if we have a good knowledge of mathematics., For example : Newton’s law of gravitation states that every body in this universe attracts every other body, with a force which is directly proportional to the product of their masses and is inversely proportional to the, square of the distance between them., m 1m 2, , y, , r, , 2, , or F =, , Gm 1 m 2, r2, , m, , This law can be expressed by a single mathematical relationship F , , ca, , de, , The techniques of mathematics such as algebra, trigonometry, calculus, graph and logarithm can be used, to make predictions from the basic equation., , eA, , If we are poor at grammar and vocabulary, it would be difficult for us to communicate our feelings,, similarly for better understanding and expressing of physics the basic knowledge of mathematics is must., , nc, , In this introductory chapter we will learn some fundamental mathematics., , ie, , 0.2 Algebra., , Sc, , (1) Quadratic equation : An equation of second degree is called a quadratic equation. Standard, , A, 'S, , quadratic equation ax 2 + bx + c = 0, , JH, , Here a is called the coefficient of x2, b is called the coefficient of x and c is a constant term, x is the variable, whose value (roots of the equation) are to be determined, Roots of the equation are : x =, , − b  b 2 − 4 ac, 2a, , This formula can be written as, x=, , Note, , :❑, , − Coefficient of x  (Coefficient of x ) 2 − 4 (Coefficient of x 2 )  (Constant term), 2(Coefficient of x 2 ), , If  and  be the roots of the quadratic equation then, , Sum of roots  +  = –, , c, b, and product of roots =, a, a, , Ex-1. Solve the equation 10 x 2 − 27 x + 5 = 0, Sol:, , By comparing the given equation with standard equation a = 10, b = – 27, and c = 5, , x=, , − (−27 )  (−27 ) 2 − 4  10  5 27  23, − b  b 2 − 4 ac, =, =, 2a, 2  10, 20

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Mathematics In Physics 3,  x1 =, , 27 − 23 1, 27 + 23 5, =, =, and x 2 =, 20, 5, 20, 2, ,  Roots of the equation are, , 5, 1, and ., 5, 2, , (2) Binomial theorem : If n is any number positive, negative or fraction and x is any real number, such, that x < 1 i.e. x lies between – 1 and + 1 then according to binomial theorem, (1 + x )n = 1 + nx +, , n(n − 1) 2 n(n − 1)(n − 2) 3, x +, x + ....., 2!, 3!, , Here 2 ! (Factorial 2) = 2  1, 3 ! (Factorial 3) = 3  2  1 and 4 ! (Factorial 4) = 4  3  2  1, , Note, , :❑, , If |x| << 1 then only the first two terms are significant. It is so because the values of, , m, , y, , second and the higher order terms being very very small, can be neglected. So the expression, can be written as, , de, , (1 + x)n = 1 + nx, , ca, , (1 + x)–n = 1 – nx, , eA, , (1 – x)n = 1 – nx, , nc, , (1 – x)–n = 1 + nx, , (1001)1/3 = (1000 + 1)1/3 = 10(1 + 0.001)1/3, , Sc, , Sol:, , ie, , Ex-2. Evaluate (1001)1/3 upto six places of decimal., , n(n − 1) 2, x + ......, 2!, , JH, , x = 0.001 and n = 1/3, , A, 'S, , By comparing the given equation with standard equation (1 + x )n = 1 + nx +, ,  10 (1 + 0 . 001 )1 / 3, , , , 1 1, , 2,  − 1   (.001 ), , , 3, 3, 1, 1, , , , , = 10 1 + (0 . 001 ) +, + ....  = 10 1 + 0 . 00033 − (0 .000001 ) + .... , , , 3, 2!, 9, , , , , , , , = 10[1.0003301 ] = 10 .003301 (Approx.), , Ex-3., , The value of acceleration due to gravity (g) at a height h above the surface of earth is given by, , g' =, , gR 2, (R + h) 2, , . If h  R then, , h, , (a) g' = g 1 − , R, , 2, , Sol: (b), , 2h , , (b) g' = g 1 −, , R , , 2, , 1, h,  R , ,  , g' = g,  = g,  = 1 + , R, +, h, 1, +, h, /, R, R, , , ,  , , , −2, , h, , (c) g' = g 1 + , R, , , 2h , , (d) g' = g 1 +, , R , , , 2, , , h (−2) (−3)  h , = g 1 + (−2) +,   + ....... , R, 2!  R , , , , h, 2h , , g' = g1 −, .),  ( if h  R then by neglecting higher power of, R, R , 

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4 Mathematics In Physics, (3) Arithmetic progression : It is a sequence of numbers which are arranged in increasing order and, having a constant difference between them., Example : 1, 3, 5, 7, 9, 11, 13, ……, , or, , 2, 4, 6, 8, 10, 12, ….., , In general arithmetic progression can be written as a0, a1, a2, a3, a4, a5 ……., (i) nth term of arithmetic progression an = a0 + (n − 1)d, a0 = First term, n = Number of terms, d = Common difference = (a1 – a0) or (a2 – a1) or (a3 – a2), (ii) Sum of arithmetic progression S n =, , n, 2a 0 + (n − 1)d  = n a 0 + an , 2, 2, , Ex-4. Find the sum of series 7 + 10 + 13 + 16 + 19 + 22 + 25, n, a 0 + a n  = 7 7 + 25  = 112, 2, 2, , [As n = 7; a0 = 7; an = a7 = 25], , y, , Sn =, , m, , Sol:, , 4, 8, 16, 32, 64, 128 ……, , or, , 5, 10, 20, 40, 80, ……., , ca, , Example :, , de, , (4) Geometric progression : It is a sequence of numbers in which every term is obtained by, multiplying the previous term by a constant quantity. This constant quantity is called the common ratio., , a(1 − r n ), 1−r, , Sn =, , a (r n − 1), r −1, , nc, , Sn =, , if r < 1, , A, 'S, , Sc, , ie, , (i) Sum of ‘n’ terms of G.P., , eA, , In general geometric progression can be written as a, ar, ar2, ar3, ar4, …., Here a = first term, r = common ratio, , S =, , JH, , (ii) Sum of infinite terms of G.P., , S =, , Ex-5. Find the sum of series Q = 2q +, Sol:, , if r > 1, , a, 1−r, , a, r −1, , if r < 1, if r > 1, , q q, q, + +, + ......, 3 9 27, , q q, q, , , + ...... , Above equation can be written as Q = q + q + + +, 3, 9, 27, , , , , ,  q , 3, 5, By using the formula of sum of infinite terms of G.P. Q = q + , =q+ q= q, 2, 2, 1 − 1 , , 3 , , (5) Some common formulae of algebra, (i) (a + b)2 = a2 + b2 + 2ab, (ii) (a – b)2 = a2 + b2 – 2ab, (iii) (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca, (iv) (a + b) (a – b) = a2 – b2

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Mathematics In Physics 5, , (v) (a + b)3 = a3 + b3 + 3ab(a + b), (vi) (a – b)3 = a3 – b3 – 3ab(a – b), (vii) (a + b)2 – (a – b)2 = 4ab, (viii) (a + b)2 + (a – b)2 = 2(a2 + b2), (ix) a 3 − b 3 = (a − b)(a 2 + b 2 + ab), (x) a 3 + b 3 = (a + b)(a 2 + b 2 − ab), (6) Componendo and dividendo method : If, , a+b c+d, a c, =, =, then, a−b c−d, b d, , 0.3 Trigonometry., Arc, AB S, =, =, (formula true for radian only), Radius OA r, , y, , B, , O, , de, , unit of angle is radian or degree, , A, , r, , ca, , relation between radian and degree :, , S, , , , m, , Angle ( ) =, , eA, , 2 radian = 360o; 1 radian = 57.3o, , Base, BC, =, Hypotenuse, AC, , Sc, , cos  =, , cosec =, , A, 'S, , Perpendicu lar AB, =, Hypotenuse, AC, , JH, , sin  =, , ie, , nc, , (1) Trigonometric ratio : In right angled triangle ABC, the largest side AC, which is opposite to the, right angle is called hypotenuse, and if angle considered is  , then side opposite to , AB, will be termed, as perpendicular and BC is called the base of the triangle., , Perpendicu lar AB, tan  =, =, Base, BC, , sec  =, , Hypotenuse, AC, =, Perpendicu lar, AB, , A, , Hypotenuse, AC, =, Base, BC, B, , Base, BC, cot  =, =, Perpendicu lar AB, , 90o, , , , C, , (2) Value of trigonometric ratio of standard angles, Angle, , 0o, , 30o, , 45o, , 60o, , 90o, , 120o, , 135o, , 150o, , 180o, , 270o, , 360o, , sin, , 0, , 1/2, , 1/2, , 3/2, , 1, , 3/2, , 1/ 2, , 1/2, , 0, , –1, , 0, , cos, , 1, , 3/2, , 1/2, , 1/2, , 0, , – 1/2, , –, 1/2, , –, 3/2, , –1, , 0, , 1, , tan, , 0, , 1/3, , 1, , 3, , , , – 3, , –1, , –, 1/3, , 0, , –, , 0, , (3) Important points :, (i) Value of sin or cos lies between – 1 and +1, however tan and cot can have any real value., (ii) Value of sec and cosec can not be numerically less than one., (iii) (90o – ) will lie in first quadrant

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6 Mathematics In Physics, (90o + ) will lie in second quadrant, Second quadrant, , First quadrant, , (Only sin and cosec are, positive), , (All T-ratio positive), , (180o – ) will lie in second quadrant, (180o + ) will lie in third quadrant, (270o + ) and (0o – ) will lie in fourth quadrant., (4) Fundamental trigonometrical relation, (i) tan  =, , sin , cos , , (ii) cosec  =, , (v) sin 2  + cos 2  = 1, , 1, sin , , Third quadrant, , Fourth quadrant, , (Only tan and cot are, positive), , (Only cos and sec are, positive), , (iii) sec  =, , 1, cos , , (iv) cot  =, , 1, tan , , (vii) cosec 2 − cot 2  = 1, , (vi) sec 2  − tan 2  = 1, , y, , (5) T-Ratios of allied angles : The angles whose sum or difference with angle  is zero or a multiple of, 90° are called angle allied to ., (i), , sin(− ) = − sin, , cos(− ) = cos , , (ii), , sin(90 o −  ) = cos , , cos(90 o −  ) = sin, , (iii), , sin(90 o +  ) = cos , , cos(90 o +  ) = − sin, , (iv), , sin(180 o −  ) = sin, , cos(180 o −  ) = − cos , , tan( 180 o −  ) = − tan , , (v), , sin(180 o +  ) = − sin, , cos(180 o +  ) = − cos , , tan( 180 o +  ) = tan , , (vi), , sin(270 o −  ) = − cos , , cos(270 o −  ) = − sin, , tan( 270 o −  ) = cot , , (vii), , sin(270 o +  ) = − cos , , cos(270 o +  ) = sin, , tan( 270 o +  ) = − cot , , (viii), , sin(360 o −  ) = − sin, , cos(360 o −  ) = cos , , tan( 360 o −  ) = − tan , , (ix), , sin(360 o +  ) = sin, , cos(360 o +  ) = cos , , tan( 360 o +  ) = tan , , :❑, , m, de, , ca, , eA, , nc, , ie, , Sc, A, 'S, , tan( 90 o −  ) = cot , tan( 90 o +  ) = − cot , , JH, , Note, , tan( − ) = − tan , , Angle ( 2n +  ) lies in first quadrant, if  in an acute angle. Similarly (2n −  ) will lie, , in fourth quadrant. Where n = 0, 1, 2, 3, 4, ❑ Angle (− ) is presumed always lie in fourth quadrant, whatever the value of  ., ❑ If parent angle is 90° or 270° then sin change to cos  , tan  change to cot  and sec  change, to cos ec ., ❑ If parent angle is 180° or 360° then no change in trigonometric function, , Ex-6., , Find the values of (i) cos(−60 o ) (ii) tan 210 o (iii) sin 300 o (iv) cos 120 o (v) sin(−1485 o ), , Sol:, , (i) cos(−60 o ) = cos 60 o =, , 1, 2, , (ii) tan( 210 o ) = tan( 180 o + 30 o ) = tan 30 o =, , 1, 3, , (iii) sin(300 o ) = sin(360 o − 60 o ) = − sin 60 o =, , − 3, 2

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Mathematics In Physics 7, (iv) cos(120 o ) = cos(90 o + 30 o ) = − sin 30 o =, , −1, 2, , (v) sin(−1485 o ) = − sin(3  360 o + 45 o ) = − sin 45 o = −, , 1, 2, , (6) Addition formulae, (i) sin( A + B) = sin A cos B + cos A sin B, (ii) cos( A + B) = cos A cos B − sin A sin B, (iii) tan( A + B) =, , tan A + tan B, 1 − tan A tan B, , Putting B = A in these formulae, we get, (iv) sin 2 A = 2 sin A cos A, , Ex-7., , m, , 2 tan A, 1 − tan 2 A, , de, , (vi) tan 2 A =, , 1, 2, , (c), , 1, , 3, , nc, , (b), , eA, , 3, 2, , ca, , If A = 60 o then value of sin 2 A will be, (a), , y, , (v) cos 2 A = cos 2 A − sin 2 A = 1 − 2 sin 2 A = 2 cos 2 A − 1, , (7) Difference formulae, , A, 'S, , (i) sin( A − B) = sin A cos B − cos A sin B, , 3 1, 3,  =, 2 2, 2, , Sc, , ie, , sin 2 A = 2 sin A cos A = 2 sin 60 cos 60 = 2 , , Sol: (a), , (iii) tan( A − B) =, , JH, , (ii) cos( A − B) = cos A cos B + sin A sin B, tan A − tan B, 1 + tan A tan B, , (8) Transformation formulae, , sin( A + B) + sin( A − B) = 2 sin A cos B, cos( A − B) − cos( A + B) = 2 sin A sin B, sin( A + B) − sin( A − B) = 2 cos A sin B, cos( A − B) + cos( A + B) = 2 cos A cos B, If we put ( A + B) = C and ( A − B) = D then on adding and subtracting, we get, A=, , C−D, C+D, and B =, 2, 2, , Putting these values in the above equation we get, (i) sin C + sin D = 2 sin, , C+D, C−D, cos, 2, 2, , (d), , 1, 2

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8 Mathematics In Physics, (ii) cos C + cos D = 2 cos, , C+D, C−D, cos, 2, 2, , (iii) sin C − sin D = 2 cos, , C+D, C−D, sin, 2, 2, , (iv) cos C − cos D = −2 sin, , C+D, C−D, sin, 2, 2, , (9) The sine and cosine formulae for a triangle : In a triangle ABC of sides a, b, c and angles A, B, and C, the following formulae hold good., (i), , a, b, c, =, =, sin A sin B sin C, , (ii) a 2 = b 2 + c 2 − 2bc cos A, , m, , y, , (iii) b 2 = c 2 + a 2 − 2ca cos B, , de, , (iv) c 2 = a 2 + b 2 − 2ab cos C, , ca, , (v) Area of a triangle ABC = S (S − a)(S − b)(S − c) ; where, S = (a + b + c) / 3, , eA, , 0.4 Logarithm., , nc, , Logarithm of a number with respect to a given base is the power to which the base must be raised to, represent that number., , Sc, , ie, , If a x = N then log a N = x, , Here x is called the logarithm of N to the base a., , A, 'S, , There are two system of logarithm : Logarithm to the base 10 are called common logarithms where as, logarithms to the base e are called natural logarithm. They are written as ln., , JH, , Conversion of natural log into common log :, , log e x = 2.3026 log 10 x, , Important formulae of logarithm :, (i) log a (mn ) = log a m + log a n (Product formula), m , (ii) log a   = log a m − log a n (Quotient formula), n, , (iii) log a m n = n log a m (Power formula), (iv) log a m = log b m log a b (Base change formula), , Note, , : ❑ Antilogarithm is the reverse process of logarithm i.e., the number whose logarithm is x is, called antilogarithm of x. If log n = x then n = antilog of x, , 0.5 Graphs., A graph is a line, straight or curved which shows the variation of one quantity w.r.t. other, which are, interrelated with each other.

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Mathematics In Physics 9, , In a relation of two quantities, the quantity which is made to alter at will, is called the independent, variable and the other quantity which varies as a result of this change is called the dependent variable., Conventionally, in any graph, the independent variable (i.e. cause) is represented along x-axis and dependent, variable (i.e. effect) is represented along y-axis., For example, we want to depict V = IR graphically, in which R is a constant called resistance, V is the, applied voltage (cause) and I (effect) is the resulting current. We will represent voltage on x-axis and current on, y-axis., Some important graphs for various equations, Y, , Y, , , , , y = mx + c, , X, , O, , m, , y, , O, , C, , y = mx, , X, , m = tan = slope of line with x-axis, , ca, , de, , c = Positive intercept on y-axis and positive slope, Y, , y = mx – c, , C, , , O, , nc, , O, , , , y = – mx + c, , X, , Sc, , ie, , C, , X, , eA, , Y, , Y, , JH, , Y, , O, , Positive intercept and Negative slope, , A, 'S, , Negative intercept and positive slope, , X, , y2 = kx, O, , Symmetric parabola about positive X-axis, , Symmetric parabola about negative X-axis, , Y, , O, , Y, , X, , x2 = ky, , O, , Symmetric parabola about positive Y-axis, y = ax + bx2, , Y, , O, , y2 = – kx, , X, , x2 = – ky, , X, , Symmetric parabola about negative Y-axis, Y, , y = ax – bx2, , X, O, , X

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10 Mathematics In Physics, , Asymmetric parabola, , Asymmetric parabola, Y, , Y, , xy = constant, , O, , a, , X, , O, , Rectangular hyperbola, , Circle of radius 'a', , Y, , Y, 2, , O, , X, , a, , 2, , x, y, + 2 =1, 2, a, b, , y = e–kx, , y, , b, , Exponential curve, , +1, , ca, , de, , Ellipse of semi-major axis a and semi-minor axis b., , 90o, , 270o, , 360o, , 540o, , , , eA, , +1, , 450o, , 180o, , y = sin, , –1, , X, , m, , O, , 0, , x2 + y2 = a2, , X, , 0, , 180o, 90o, , 360o, 270o, , 450o, , , , y = cos, , nc, , –1, , cosine curve, , 0.6 Differential Calculus., , A, 'S, , Sc, , ie, , sine curve, , JH, , The differential coefficient or derivative of variable y with respect to variable x is defined as the instantaneous, dy, Y, y = f (x), rate of change of y w.r.t. x. It is denoted by, dx, Geometrically the differential coefficient of y = f (x ) with respect to x at any, , , , point is equal to the slope of the tangent to the curve representing y = f (x ) at that, point, , X, , i.e., , dy, = tan  ., dx, , Note, , : ❑ Actually, , ❑ If, , dy, is a rate measurer., dx, , dy, is positive, it means y is increasing with increasing of x and vice-versa., dx, , dy, . x, dx, ds, Example: (1) Instantaneous speed v =, dt, , ❑ For small change x we use y =

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12 Mathematics In Physics, dy, d, =, (tan x ) = sec 2 x, dx, dx, , If y = cot x, , dy, d, =, (cot x ) = −cosec 2 x, dx, dx, , If y = sec x, , dy, d, =, (sec x ) = tan x sec x, dx, dx, , If y = cosec x, , dy, d, =, (cosec x ) = − cot x cosec x, dx, dx, , If y = sin u where u is the function of x, , dy, d, d (u), =, (sin u) = cos u, dx, dx, dx, , If y = cos u where u is the function of x, , dy, d, d (u), =, (cos u) = − sin u, dx, dx, dx, , If y = tan u where u is the function of x, , dy, d, d (u), =, (tan u) = sec 2 u, dx, dx, dx, , de, , m, , y, , If y = tan x, , ca, , If y = cot u where u is the function of x, , nc, , Sc, , Differentiate the following w.r.t x, (i) x 3, , Sol:, , (i), , (ii), , JH, , Ex-8., , dy, 1, = log a e, dx, x, , A, 'S, , If y = log a x, , dy, d, d (u), =, (sec u) = sec u tan u, dx, dx, dx, dy, d, d (u), =, (cosec u) = − cosec u cot u, dx, dx, dx, , ie, , If y = cosec u where u is the function of x, , eA, , If y = sec u where u is the function of x, , dy, d, d (u), =, (cot u) = −cosec 2 u, dx, dx, dx, , (iii) ax 2 + bx + c, , x, , (iv) 2 x 3 − e x, , d, (x 3 ) = 3 x 2, dx, 1, , −1, d, 1, 1, 1, (x )1 / 2 = (x ) 2 = (x )−1 / 2 =, dx, 2, 2, 2 x, , (ii), , Ex-9., , (iii), , d, d, d, d, (ax 2 + bx + c) = a, (x 2 ) + b, (x ) +, (c) = 2ax + b, dx, dx, dx, dx, , (iv), , d, d, d x, (2 x 3 − e x ) = 2, (x 3 ) −, (e ) = 6 x 2 − e x, dx, dx, dx, , (v), , d, d 1/2, d, d, 6, 1, (loge x ) −, (x ) −, (7) = −, (6 loge x − x − 7) = 6, dx, dx, dx, dx, x 2 x, , Differentiate the following w.r.t. x, (i) sin x + cos x, , Sol:, , (i), , (ii) sin x + e x, , d, d, d, (sin x + cos x ) =, (sin x ) +, (cos x ) = cos x − sin x, dx, dx, dx, , (v) 6 log e x − x − 7

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Mathematics In Physics 13, , d, d, d x, (sin x + e x ) =, (sin x ) +, (e ) = cos x + e x, dx, dx, dx, , (ii), , Differentiate the following w.r.t. t, , (i), , d, d, (sin t 2 ) = cos t 2 (t 2 ) = 2 t cos t 2, dt, dt, , (ii), , d sin t, d, (e ) = e sin t (sin t) = e sin t . cos t, dt, dt, , d, d, [sin(t +  )] = cos(t +  ). (t +  ) = cos(t +  )., dt, dt, , Ex-11., , Differentiate, , Sol:, , Let y =, , Then, , m, , y, , x2 +ex, ., log x + 20, , dy, d  x 2 + e x , =, dx, dx  log x + 20 , , (log x + 20 ), , d, d, (x 2 + e x ) − (x 2 + e x ) (log x + 20 ), dx, dx, (log x + 20 ) 2, , nc, , =, , x2 +ex, w.r.t. x, log x + 20, , de, , (iii), , eA, , Sol:, , (iii) sin(t +  ), , (ii) e sin t, , (i) sin t 2, , ca, , Ex-10., , A, 'S, , Sc, , ie, , 1, , (log x + 20 )(2 x + e x ) − (x 2 + e x ) + 0 , x, , , =, (log x + 20 ) 2, , JH, , (2) Maxima and minima : If a quantity y depends on another quantity x in a manner shown in figure., It becomes maximum at x1 and minimum at x2., Y, , At these points the tangent to the curve is parallel to X-axis and hence its, slope is tan  = 0. But the slope of the curve equals the rate of change, dy, dy, =0, . Thus, at a maximum or minimum, dx, dx, , O, , x1, , x2, , X, , Just before the maximum the slope is positive, at the maximum it is zero, dy, and just after the maximum it is negative. Thus, decreases at a maximum and hence the rate of change, dx, of, , dy, d  dy , is negative at a maximum. i.e.,,    0 at a maximum., dx, dx  dx , , Hence the condition of maxima :, , dy, d 2y, = 0 and, 0, dx, dx 2, , (Second derivative test), , Similarly, at a minimum the slope changes from negative to positive. The slope increases at such a point, and hence, , d  dy ,  0, dx  dx 

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14 Mathematics In Physics, Hence the condition of minima :, , Ex-12., , dy, d 2y, = 0 and,  0 . (Second derivative test), dx, dx 2, , The height reached in time t by a particle thrown upward with a speed u is given by h = ut −, , 1 2, gt . Find, 2, , the time taken in reaching the maximum height., , dh, d, 1, 2 gt, u, [ut − gt 2 ] = u −, = 0 t =, =0, dt, 2, 2, dt, g, , Sol:, , For maximum height, , Ex-13., , A metal ring is being heated so that at any instant of time t in second, its area is given by, , A = 3t 2 +, , t, + 2 m2., 3, , What will be the rate of increase of area at t = 10 sec ., , dA d, t, 1, = (3 t 2 + + 2) = 6 t +, dt, dt, 3, 3, , ca, eA, , Ex-14., , de, , 1 181 m 2,  dA , = 6  10 + =, ., , , 3, 3 sec,  dt t =10 sec, , The radius of an air bubble is increasing at the rate of, , 4, R 3, 3, , Sc, , ie, , Volume of the spherical bubble V =, , 1, cm / sec . Determine the rate of increase in its, 2, , nc, , volume when the radius is 1 cm., Sol:, , y, , Rate of increase of area, , m, , Sol:, , A, 'S, , Differentiating both sides w.r.t. time, , at R = 1cm ,, , JH, , dR, dR, dV, d 4,  4, = 4R 2, =  R 3  =  . 3 R 2 ., dt, dt, dt, dt  3,  3, , dV, 1, = 4  (1) 2  = 2 cm 3 / sec ., dt, 2, , [Given, , dR 1, = cm / sec ], dt, 2, , Ex-15., , Find the angle of tangent drawn to the curve y = 3 x 2 − 7 x + 5 at the point (1, 1) with the x- axis., , Sol:, , y = 3x 2 − 7x + 5, Slope of tangent =, at (1, 1), , dy, = −1, dx, , dy, = 6x − 7, dx, ,  tan  = −1   = 135 o ., , 0.7 Integral Calculus., The process of integration is just the reverse of differentiation. The symbol ∫ is used to denote integration., If f (x ) is the differential coefficient of function F(x ) with respect to x, then by integrating f (x ) we can get, F(x ) again.

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Mathematics In Physics 15, , (1) Fundamental formulae of integration :, , , , x n dx =, , , , dx =, , , ,  sec, , x n +1, , provided n  – 1, n +1, , x dx = tan x, ,  cos ec, , x 0 +1, =x, 0 +1, , x 0 dx =, , 2, , 2, , x dx = − cot x, ,  (u + v) dx =  u dx +  v dx, ,  sec x tan x dx = sec x, ,  cu dx = c  u dx, ,  cosec x cot x dx = −cosec x, , where c is a constant and u is a function of x., , x n +1, n +1, , , , (ax + b)n +1, (ax + b)n +1, =, d, a(n + 1), (n + 1) (ax + b), dx, , (ax + b)n dx =, , a log e (ax + b ), = log e (ax + b), d, (ax + b), dx, ,  (ax + b) dx =, a, , = log e x, , ca, , dx, , de, , x, , a x dx =, , ax, log e a, ,  sin x dx = − cos x, ,  sin nx dx =, , − cos nx, n, , e ax +b dx =, , nc, , , , , e ax +b, , eA, , dx = e x, , ie, , x, , Sc, , , , dx =, , A, 'S, , e, , −1, , JH, , x, , m, , y, , , , cx n dx = c, , a cx + d dx =, ,  sec, , 2, , d, (ax + b), dx, , 2, , e ax +b, a, , a cx + d, a cx + d, =, d, c log e a, log e a, (cx + d ), dx, , (ax + b) dx =, ,  cosec, , =, , tan( ax + b) tan ( ax + b), =, d, a, (ax + b), dx, , (ax + b ) dx =, , − cot (ax + b) − cot (ax + v), =, d, a, (ax + b), dx, sec (ax + b) sec (ax + b), =, d, a, (ax + b), dx, ,  cos x dx = sin x, ,  sec (ax + b) tan (ax + b) dx =, ,  cos nx dx =, ,  cosec (ax + b) cot (ax + b) dx =, , sin nx, n, , −cosec (ax + b) −cosec (ax + b), =, d, a, (ax + b), dx, , (2) Method of integration : Sometimes, we come across some functions which cannot be integrated, directly by using the standard integrals. In such cases, the integral of a function can be obtained by using, one or more of the following methods.

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16 Mathematics In Physics, (i) Integration by substitution : Those functions which cannot be integrated directly can be reduced to, standard integrand by making a suitable substitution and then can be integrated by using the standard, integrals. To understand the method, we take the few examples., (ii) Integration by parts : This method of integration is based on the following rule :, Integral of a product of two functions = first function  integral of second function – integral of, (differential coefficient of first function  integral of second function)., Thus, if u and v are the functions of x, then, ,  du, , , ,  uv dx = u  v dx −   dx   v dx  dx, , Ex-16. Integrate the following w.r.t. x, (ii) cot 2 x, , (i) x1/2, , 2, ,  (cosec, , x dx =, , m, 2, , , , x − 1)dx =, ,  cosec, , 1 + sin x , , 2, , 1, , 1, , , ,  dx = − cot x − x, , x dx −, ,  1 − sin x dx =   1 − sin x  1 + sin x  . dx = , , (iii), , de, ,  cot, , (ii), , y, , x 1 / 2 +1 2 3 / 2, = (x ), 1, 3, +1, 2, , ca, , , , x 1 / 2 dx =, , eA, , (i), , nc, , Sol:, , 1, 1 − sin x, , (iii), , 1 + sin x, dx =, 1 − sin2 x, ,  cos, , 1, 2, , +, x, , sin x, cos 2 x, , dx, , Sc, , ie, , = (sec 2 x + tan x sec x )dx = tan x + sec x ., , , , b, a, , b, a, , is, , f (x ) dx, , definite, , integral, , of, , f (x ), , between, , the, , limits, , a, , and, , b, , and, , is, , written, , JH, , , , A, 'S, , (3) Definite integrals : When a function is integrated between definite limits, the integral is called definite, integral. For example,, , f (x ) dx =| F(x )| ba = F(b) − F(a), , Here a is called the lower limit and b is called the upper limit of integration., , , , Geometrically, , Ex-17. Evaluate, Sol:, , , , 6, , , , 6, , b, a, , f ( x ) dx equals to area of curve F(x ) between the limits a and b., , (2 x 2 + 3 x + 5)dx, , 0, , (2 x 2 + 3 x + 5)dx =, , 0, , , , 6, 0, , 2 x 2 dx +, , , , 6, , 3 x dx +, , 0, , , , 6, , 6, ,  2x 3 , 3x 2 , 6, 5 dx = ,  +,  + 5 x 0 = 144 + 54 + 30 = 228., 0, 3, 2, ,  0 ,  0, 6, , Ex-18. Integrate the following, (i), , , , 2, 0, , 1, x, , dx, , (ii), , , ,  /2, , cos x dx, 0, , (iii), , , , r2, , r1, , Kq 1 q 2, r, , 2, , .dr, , (iv), , , ,  /4, 0, , tan 2 x dx, , as

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Mathematics In Physics 17, 2, , Sol:, , (i), , , 0, , 1, , , , dx =, , 2, , 2, , x, , −1 / 2, , 0, , x, ,  /2, , (ii), , , ,  x1/2 , 1/2, dx = ,  = 2x,  1 / 2  0, ,  cos x dx, , = sin x 0 / 2 = sin, , 0, , r2, , (iii), , , , k, , r1, , q1 q 2, r2, , r2, , dx = k q 1 q 2, , , 0, ,  (sec, , =2 2, , =1, , , 2, , x − 1)dx = tan x  0 / 4 − [ x ] 0 / 4 = 1 −, ,  /4, , tan 2 x dx =, , 2, , 2, 0, , 1, 1, 1, 1,  1 2, kq, q, =, dx,  −  = −kq 1 q 2  −  = kq 1 q 2  − , 1, 2, 2,  r  r1, r,  r2 r1 ,  r1 r2 , , r1, ,  /4, , (iv), , , , , , r, , 1, , o, , , 4, , 0.8 General Formulae for Area and Volume., , y, , 1. Area of square = (side)2, , de, , 1,  base  height, 2, , eA, , 4. Area enclosed by a circle =  r 2 ; where r is radius, , ca, , 3. Area of triangle =, , m, , 2. Area of rectangle = length  breadth, , nc, , 5. Surface area of sphere = 4 r 2, , ie, , 6. Surface area of cube = 6 L2 ; where L is a side of cube, , Sc, , 7. Surface area of cuboid = 2L  b + b  h + h  L; where L= length, b = breadth, h = height, , A, 'S, , 8. Area of curved surface of cylinder = 2 rl ; where r = radius, l = length of cylinder, , JH, , 9. Volume of cube = L3, , 10. Volume of cuboid = L  b  h, 11. Volume of sphere =, , 4,  r3, 3, , 12. Volume of cylinder =  r 2 l, 13. Volume of cone =, , 1,  r2h, 2, , 0.9 Introduction of Vector., Physical quantities having magnitude, direction and obeying laws of vector algebra are called vectors., Example : Displacement, velocity, acceleration, momentum, force, impulse, weight, thrust, torque,, angular momentum, angular velocity etc., If a physical quantity has magnitude and direction both, then it does not always imply that it is a vector., For it to be a vector the third condition of obeying laws of vector algebra has to be satisfied., Example : The physical quantity current has both magnitude and direction but is still a scalar as it, disobeys the laws of vector algebra.

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18 Mathematics In Physics, , 0.10 Types of Vector., (1) Equal vectors : Two vectors A and B are said to be equal when they have equal magnitudes and same, direction., (2) Parallel vector : Two vectors A and B are said to be parallel when, (i) Both have same direction., (ii) One vector is scalar (positive) non-zero multiple of another vector., (3) Anti-parallel vectors : Two vectors A and B are said to be anti-parallel when, (i) Both have opposite direction., (ii) One vector is scalar non-zero negative multiple of another vector., , m, , y, , (4) Collinear vectors : When the vectors under consideration can share the same support or have a, common support then the considered vectors are collinear., , ca, , de, , (5) Zero vector (0 ) : A vector having zero magnitude and arbitrary direction (not known to us) is a zero, vector., , nc, , ˆ ., ˆ = A  A= AA, A, A, , ie, , Since,, , eA, , (6) Unit vector : A vector divided by its magnitude is a unit vector. Unit vector for A is  (read as A cap / A, hat)., , Sc, , Thus, we can say that unit vector gives us the direction., , A, 'S, , (7) Orthogonal unit vectors : ˆi , ˆj and k̂ are called orthogonal unit vectors. These vectors must form a, Right Handed Triad (It is a coordinate system such that when we Curl the, fingers of right hand from x to y then we must get the direction of z along, thumb). The, , JH, , y, , x, , ˆi = x , ˆj = y , kˆ = z, y, x, z, , , , ˆj, , k̂, , î, , z, , x = xˆi , y = yˆj , z = zkˆ, , (8) Polar vectors : These have starting point or point of application . Example displacement and force, etc., (9) Axial Vectors : These represent rotational effects and are always along the axis of rotation in, accordance with right hand screw rule. Angular velocity, torque and angular momentum, etc., are example, of physical quantities of this type., Axial vector, , Anticlock wise rotation, , Axis of rotation, , Axis of rotation, , Clock wise rotation, , Axial vector

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Mathematics In Physics 19, , (10) Coplanar vector : Three (or more) vectors are called coplanar vector if they lie in the same plane., Two (free) vectors are always coplanar., , 0.11 Triangle Law of Vector Addition of Two Vectors., If two non zero vectors are represented by the two sides of a triangle taken in same order then the, resultant is given by the closing side of triangle in opposite order. i.e. R = A + B,  OB = OA + AB, , B, , (1) Magnitude of resultant vector, , R = A+B, , AN, In  ABN cos  =,  AN = B cos , B, , B, A, , y, , B, , m, , In OBN , we have OB 2 = ON 2 + BN 2, , de, , R, , O, , , A, , A, , eA, ,  R 2 = A 2 + B 2 cos 2  + 2 AB cos  + B 2 sin 2 , , N, B cos, , nc, ,  R 2 = A 2 + B 2 (cos 2  + sin 2  ) + 2 AB cos , , B sin, , B, , , , ca, ,  R 2 = ( A + B cos  )2 + (B sin )2, , ie, ,  R 2 = A 2 + B 2 + 2 AB cos , , A 2 + B 2 + 2 AB cos , , Sc, ,  R=, , A, , O, , BN, sin  =,  BN = B sin, B, , A 2 + B 2 + 2 AB cos , , JH, , | A + B| =, , A, 'S, , (2) Direction of resultant vectors : If  is angle between A and B , then, , If R makes an angle  with A , then in OBN , then, tan  =, , BN, BN, =, ON OA + AN, , tan  =, , B sin , A + B cos , , 0.12 Parallelogram Law of Vector Addition of Two Vectors., If two non zero vector are represented by the two adjacent sides of a parallelogram then the resultant is, given by the diagonal of the parallelogram passing through the point of intersection of the two vectors., (1) Magnitude, , B, , Since, R = ON + CN, 2, , 2, ,  R = (OA + AN ) + CN, 2, , 2, , C, , 2, , R = A+B, ,  R 2 = A 2 + B 2 + 2 AB cos , , B sin, , B, , 2, , , O, , , , , A, , A, , B, N, , B cos

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20 Mathematics In Physics, A 2 + B 2 + 2 AB cos , ,  R =| R | =| A + B | =, , Special cases : R = A + B when  = 0o, , R = A − B when  = 180o, , R = A 2 + B 2 when  = 90o, (2) Direction, CN, B sin , =, ON, A + B cos , , tan  =, , 0.13 Polygon Law of Vector Addition., If a number of non zero vectors are represented by the (n – 1) sides of an n-sided polygon then the, resultant is given by the closing side or the nth side of the polygon taken in opposite order. So,, , R = A+B+C +D+ E, , D, , y, , C, , de, , E, E, , : ❑ Resultant of two unequal vectors can not be zero., , B, , ca, , Note, , m, , OA + AB + BC + CD + DE = OE, , C, , D, , eA, , ❑ Resultant of three co-planar vectors may or may not be zero, , R, , ❑ Resultant of three non co- planar vectors can not be zero., , nc, , O, , A, , A, , ie, , 0.14 Subtraction of Vectors., , B, , | A − B| =, , 2, , Since, cos (180 −  ) = − cos , , B, , , | A − B | = A 2 + B 2 − 2 AB cos , , B sin , A + B cos , B sin (180 −  ), tan  2 =, A + B cos (180 −  ), , tan  1 =, , and, , o, , JH, , , , R sum = A + B, , A + B + 2 AB cos (180 −  ), 2, , A, 'S, , , , Sc, , Since, A − B = A + (−B) and | A + B | = A 2 + B 2 + 2 AB cos , , −B, , 1, 2, , A, , 180 – , , R diff = A + (− B), , But sin(180 −  ) = sin and cos(180 −  ) = − cos , , , tan  2 =, , B sin , A − B cos , , Sample Ex-based on addition and subtraction of vectors, Ex-19., , A car travels 6 km towards north at an angle of 45° to the east and then travels distance of 4 km towards, north at an angle of 135° to the east. How far is the point from the starting point. What angle does the, straight line joining its initial and final position makes with the east, (a), , 50 km and tan −1 (5), , (b) 10 km and tan −1 ( 5 ), , (c), , 52 km and tan −1 (5), , (d), , 52 km and tan −1 ( 5 )

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Mathematics In Physics 21, Sol: (c), , Net movement along x-direction Sx = (6 – 4) cos 45° î = 2 , Net movement along y-direction Sy = (6 + 4) sin 45° ˆj = 10 , , Net movement from starting point | s | =, , s x + sy, 2, , 2, , =, , Angle which makes with the east direction tan  =, , 1, , = 2 km, , 2, 1, , N, 4 km, , = 5 2 km, , 6 km, , 2, , ( 2 ) + (5 2 ), 2, , 2, , Y − component, X − component, , 45o, , W, , E(X, ), , O, , = 52 km, , =, , S (Y), , 5 2, 2, ,   = tan −1 (5), , (a) 0°, 180° and 90°, , (c) 0°, 90° and 90°, , (d) 180°, 0° and 90°, , For 17 N both the vector should be parallel i.e. angle between them should be zero., , de, , Sol: (a), , (b) 0°, 90° and 180°, , y, , There are two force vectors, one of 5 N and other of 12 N at what angle the two vectors be added to get, resultant vector of 17 N, 7 N and 13 N respectively, , m, , Ex-20., , ca, , For 7 N both the vectors should be antiparallel i.e. angle between them should be 180°, , vector is, given by, , nc, , Given that A + B + C = 0 out of three vectors two are equal in magnitude and the magnitude of third, , 2 times that of either of the two having equal magnitude. Then the angles between vectors are, , ie, , Ex-21., , eA, , For 13 N both the vectors should be perpendicular to each other i.e. angle between them should be 90°, , (c) 45°, 60°, 90°, , (d) 90°, 135°, 135°, , From polygon law, three vectors having summation zero should form a, closed polygon. (Triangle) since the two vectors are having same magnitude, , A, 'S, , Sol: (d), , (b) 45°, 45°, 90°, , Sc, , (a) 30°, 60°, 90°, , , , JH, , and the third vector is, 2 times that of either of two having equal, magnitude. i.e. the triangle should be right angled triangle, C, , Angle between A and B,  = 90º, , B, , , , Angle between B and C,  = 135º, , , , A, , Angle between A and C,  = 135º, , Ex-22., , If A = 4ˆi − 3ˆj and B = 6ˆi + 8 ˆj then magnitude and direction of A + B will be, (a) 5, tan −1 (3 / 4 ), , Sol: (b), , (b) 5 5 , tan −1 (1 / 2), , A + B = 4ˆi − 3 ˆj + 6ˆi + 8 ˆj = 10 ˆi + 5 ˆj, , | A + B | = (10 ) 2 + (5 ) 2 = 5 5, , tan  =, , 5, 1, 1, =,   = tan −1  , 10 2, 2, , (c) 10, tan −1 (5), , (d) 25, tan −1 (3 / 4 )

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22 Mathematics In Physics, , Ex-23., , A truck travelling due north at 20 m/s turns west and travels at the same speed. The change in its velocity, be, (a) 40 m/s N–W, , Sol: (d), , (b) 20 2 m/s N–W, , (c) 40 m/s S–W, , (d) 20 2 m/s S–W, , From fig., , v1 =20 m/s, , v 1 = 20 ˆj and v 2 = −20 ˆi, , v2 =20 m/s, , | v | = 20 2 and direction  = tan −1 (1) = 45  i.e. S–W, , If the sum of two unit vectors is a unit vector, then magnitude of difference is, (b), , 2, , (c) 1 / 2, , 3, , (d), , 5, , y, , (a), , Let n̂1 and n̂ 2 are the two unit vectors, then the sum is, , de, , Sol: (b), , – v1, , v, , m, , Ex-24., , O, , , , v = v 2 − v 1 = −20 (ˆi + ˆj), , ca, , n s = nˆ 1 + nˆ 2 or n s2 = n12 + n 22 + 2n1 n 2 cos  = 1 + 1 + 2 cos , , 1, or  = 120 , 2, , nc, , or cos  = −, , eA, , Since it is given that n s is also a unit vector, therefore 1 = 1 + 1 + 2 cos , , The sum of the magnitudes of two forces acting at point is 18 and the magnitude of their resultant is 12. If, the resultant is at 90° with the force of smaller magnitude, what are the, magnitudes of forces, (a) 12, 5, , Sol: (c), , JH, , Ex-25., ,  nd = 3, , A, 'S, ,  n d2 = 2 − 2(−1 / 2) = 2 + 1 = 3, , Sc, , ie, , Now the difference vector is nd = n1 − n 2 or nd2 = n12 + n22 − 2n1n2 cos  = 1 + 1 − 2 cos(120 ), , (b) 14, 4, , (c) 5, 13, , (d) 10, 8, , Let P be the smaller force and Q be the greater force then according to Ex-–, P + Q = 18, , ......(i), , R = P 2 + Q 2 + 2 PQ cos  = 12, , .......(ii), , tan  =, , Q sin , = tan 90 = , P + Q cos , ,  P + Q cos = 0, , .......(iii), , By solving (i), (ii) and (iii) we will get P = 5, and Q = 13, , Ex-26., , Two forces F1 = 1 N and F2 = 2 N act along the lines x = 0 and y = 0 respectively. Then the resultant of, forces would be, (a) ˆi + 2ˆj, , Sol: (d), , (b) ˆi + ˆj, , (c) 3ˆi + 2ˆj, , x = 0 means y-axis  F 1 = ˆj, y = 0 means x-axis  F 2 = 2ˆi so resultant F = F 1 + F 2 = 2ˆi + ˆj, , (d) 2ˆi + ˆj

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Mathematics In Physics 23, , Ex-27., , Let A = 2ˆi + ˆj, B = 3 ˆj − kˆ and C = 6ˆi − 2kˆ value of A − 2 B + 3C would be, (b) 20ˆi − 5 ˆj − 4 kˆ, , (a) 20ˆi + 5 ˆj + 4 kˆ, Sol: (b), , (c) 4ˆi + 5 ˆj + 20 kˆ, , (d) 5ˆi + 4 ˆj + 10 kˆ, , A − 2 B + 3C = (2ˆi + ˆj) − 2(3ˆj − kˆ ) + 3(6ˆi − 2kˆ ), , = 2ˆi + ˆj − 6 ˆj + 2kˆ + 18ˆi − 6kˆ, = 20ˆi − 5 ˆj − 4 kˆ, , Ex-28., , A vector a is turned without a change in its length through a small angle d . The value of |  a | and a, are respectively, (a) 0, a d, , (c) 0, 0, , y, , From the figure | OA | = a and | OB | = a, , de, , Also from triangle rule OB − OA = AB = a  |  a | = AB, , B, , A, d, , a, , O, , nc, , So | a | = a d, , a, , a, , ca, , arc,  AB = a . d, radius, , eA, , Using angle =, , (d) None of these, , m, , Sol: (b), , (b) a d , 0, , ie, , a means change in magnitude of vector i.e. | OB | − | OA |  a − a = 0, , A, 'S, , An object of m kg with speed of v m/s strikes a wall at an angle  and rebounds at the same speed and, same angle. The magnitude of the change in momentum of the object will be, , JH, , Ex-29., , Sc, , So a = 0, , v, , 1, , 0, , (a) 2m v cos , Sol: (a), , , , , , v, , 2, , 0, , (b) 2 m v sin, , (c) 0, , (d) 2 m v, , P 1 = m v sin ˆi − m v cos  ˆj and P 2 = m v sin  ˆi + m v cos  ˆj, So change in momentum P = P 2 − P 1 = 2 m v cos  ˆj, , |  P | = 2 m v cos , , y, p2, , p1, , , , , x, , 

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24 Mathematics In Physics, , 0.15 ReSolof Vector Into Components., Consider a vector R in x-y plane as shown in fig. If we draw orthogonal vectors R x and R y along x and, y axes respectively, by law of vector addition, R = R x + R, , y, , y, , Now as for any vector A = A nˆ so, R x = ˆi R x and R y = ˆjR y, so, , R = ˆi R x + ˆjR y, , But from fig R x = R cos , and, , Ry, , .....(i), , , , .....(ii), , R y = R sin , , R, x, , Rx, , .....(iii), , m, , y, , Since R and  are usually known, Equation (ii) and (iii) give the magnitude of the components of R along, x and y-axes respectively., , ca, , de, , Here it is worthy to note once a vector is resolved into its components, the components themselves can be, used to specify the vector as –, , R x2 + R y2, , nc, , R=, , eA, , (1) The magnitude of the vector R is obtained by squaring and adding equation (ii) and (iii), i.e., ,  = tan −1 (R y / R x ), , A, 'S, , or, , Sc, , tan  = (R y / R x ), , ie, , (2) The direction of the vector R is obtained by dividing equation (iii) by (ii), i.e., , 0.16 Rectangular Components of 3-D Vector., , JH, , R = R x + R y + R z or R = R xˆi + R y ˆj + R z kˆ, If R makes an angle  with x axis,  with y axis and  with z axis, then,  cos  =, , Rx, =, R, ,  cos  =, , Ry, R, , =, , R,  cos  = z =, R, , Rx, R x2, , +, , R y2, , +, , R z2, , +, , R z2, , Ry, R x2, , +, , R y2, , Rz, R x2 + R y2 + R z2, , Y, , =l, , =m, , Ry, , Rz, , =n, , Z, , where l, m, n are called Direction Cosines of the vector R, , l 2 + m 2 + n 2 = cos 2  + cos 2  + cos 2  =, , R x2 + R y2 + R z2, R x2 + R y2 + R z2, , =1, , R, Rx, , X

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Mathematics In Physics 25, , Note, , When a point P have coordinate (x, y, z) then its position vector OP = xˆi + yˆj + zkˆ, , :❑, , ❑ When a particle moves from point (x1, y1, z1) to (x2, y2, z2) then its displacement vector, →, , r = (x 2 − x 1 )ˆi + (y 2 − y 1 )ˆj + (z 2 − z 1 )kˆ, , Sample Ex-based on representation and reSolof vector, Ex-30., , If a particle moves 5 m in +x- direction. The displacement of the particle will be, (a) 5 j, , Sol: (b), , (b) 5 i, , (c) – 5 j, , (d) 5 k, , Magnitude of vector = 5, , y, , Unit vector in +x direction is î, , 5 î, O, , de, , Position of a particle in a rectangular-co-ordinate system is (3, 2, 5). Then its position vector will be, (b) 3ˆi + 2ˆj + 5kˆ, , (a) 3ˆi + 5 ˆj + 2kˆ, , (c) 5ˆi + 3ˆj + 2kˆ, , ca, , Ex-31., , m, , y, , So displacement = 5 î, , x, , (d) None of these, , If a point have coordinate (x, y, z) then its position vector OP = xˆi + yˆj + zkˆ ., , Ex-32., , If a particle moves from point P (2,3,5) to point Q (3,4,5). Its displacement vector be, (b) ˆi + ˆj + 5kˆ, , ie, , (a) ˆi + ˆj + 10 kˆ, , nc, , eA, , Sol: (b), , (c) ˆi + ˆj, , (d) 2ˆi + 4 ˆj + 6kˆ, , Displacement vector r = xˆi + yˆj + zkˆ = (3 − 2)ˆi + (4 − 3)ˆj + (5 − 5)kˆ = ˆi + ˆj, , Ex-33., , A force of 5 N acts on a particle along a direction making an angle of 60° with vertical. Its vertical component, be, (b) 3 N, , (c) 4 N, , The component of force in vertical direction will be F cos  = F cos 60 , , = 5, , (d) 5.2 N, y, , F cos 60o, , Sol: (d), , JH, , (a) 10 N, , A, 'S, , Sc, , Sol: (c), , 1, = 2 .5 N, 2, , F, 60o, F sin 60o, , Ex-34., , If A = 3ˆi + 4 ˆj and B = 7ˆi + 24 ˆj, the vector having the same magnitude as B and parallel to A is, (a) 5ˆi + 20 ˆj, , Sol: (d), , (b) 15ˆi + 10 ˆj, , | B | = 7 2 + (24 )2 = 625 = 25, , ˆ, ˆ, ˆ = 3i + 4 j, Unit vector in the direction of A will be A, 5, ,  3ˆi + 4 ˆj , ˆ, ˆ, , So required vector = 25 ,  5  = 15 i + 20 j, , , , (c) 20ˆi + 15 ˆj, , (d) 15ˆi + 20 ˆj, , x

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26 Mathematics In Physics, , Ex-35., , Vector A makes equal angles with x, y and z axis. Value of its components (in terms of magnitude of A ), will be, (a), , A, , A, , (b), , 3, , (c), , 2, , cos 2  + cos 2  + cos 2  = 1  3 cos 2  = 1  cos  =, , 3, , ,, , 45, , 4, , and, , 45, , −5, , 1, , (b), , 45, , 45, , ,, , 2, , and, , 45, , 3, 45, , , cos  =, , 45, , 4, , , 0 and, , 45, , 4, 45, , (d), , 3, , 2, , ,, , 45, , 45, , and, , 5, 45, , , cos  =, , 45, , −5, , eA, , 2, , 4, , ca, , | A | = (2) 2 + (4 ) 2 + (−5 ) 2 = 45, ,  cos  =, , (c), , m, , 2, , y, , If A = 2ˆi + 4 ˆj − 5kˆ the direction of cosines of the vector A are, (a), , nc, , 45, , ie, , The vector that must be added to the vector ˆi − 3ˆj + 2kˆ and 3ˆi + 6 ˆj − 7kˆ so that the resultant vector is a, , Sc, , unit vector along the y-axis is, (a) 4ˆi + 2ˆj + 5kˆ, , (b) − 4ˆi − 2ˆj + 5kˆ, , (c) 3ˆi + 4 ˆj + 5kˆ, , (d) Null vector, , A, 'S, , Sol: (b), , 3, , de, , Ex-36., , 1, , A, ,  A x = A y = A z = A cos  =, , Ex-37., , 3, A, , Let the components of A makes angles ,  and  with x, y and z axis respectively then  =  = , , Sol: (a), , Sol: (a), , (d), , 3A, , Unit vector along y axis = ˆj so the required vector = ˆj − [(ˆi − 3ˆj + 2kˆ ) + (3ˆi + 6 ˆj − 7kˆ )] = − 4ˆi − 2ˆj + 5kˆ, , JH, , 0.17 Scalar Product of Two Vectors., , (1) Definition : The scalar product (or dot product) of two vectors is defined as the product of the, magnitude of two vectors with cosine of angle between them., Thus if there are two vectors A and B having angle  between them, then their scalar product written, as A . B is defined as A . B = AB cos , (2) Properties : (i) It is always a scalar which is positive if angle between the vectors is acute (i.e., < 90°), and negative if angle between them is obtuse (i.e. 90°< < 180°)., (ii) It is commutative, i.e. A . B = B . A, (iii) It is distributive, i.e. A . (B + C ) = A . B + A . C, (iv) As by definition A . B = AB cos ,  A. B, The angle between the vectors  = cos −1 , ,  AB , , B, , , , A

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Mathematics In Physics 27, , (v) Scalar product of two vectors will be maximum when cos  = max = 1, i.e.  = 0 o , i.e., vectors are, parallel, ( A . B) max = AB, , (vi) Scalar product of two vectors will be minimum when | cos  | = min = 0, i.e.  = 90 o, ( A . B)min = 0, , i.e., if the scalar product of two nonzero vectors vanishes the vectors are orthogonal., (vii) The scalar product of a vector by itself is termed as self dot product and is given by, ( A) 2 = A . A = AA cos  = A 2, , A. A, , y, , i.e., A =, , de, , nˆ . nˆ = 1  1  cos 0 = 1 so nˆ . nˆ = ˆi . ˆi = ˆj . ˆj = kˆ . kˆ = 1, , m, , (viii) In case of unit vector n̂, , eA, , ca, , (ix) In case of orthogonal unit vectors ˆi , ˆj and k̂ , ˆi . ˆj = ˆj . kˆ = kˆ . ˆi = 1  1 cos 90 = 0, , nc, , (x) In terms of components A . B = (iA x + jA y + k A z ). (iB x + jB y + k B z ) = [ A x B x + Ay By + A Z B z ], , ie, , (3) Example : (i) Work W : In physics for constant force work is defined as, W = Fs cos  .......(i), But by definition of scalar product of two vectors, F. s = Fs cos , , Sc, , .......(ii), , A, 'S, , So from eqn (i) and (ii) W = F.s i.e. work is the scalar product of force with displacement., , As, , W = F.s, , or, , JH, , (ii) Power P :, , dW, ds, = F., dt, dt, , [As F is constant], , or P = F . v i.e., power is the scalar product of force with velocity., ,  dW, , ds, = P and, = v,  As, dt, dt, , , , (iii) Magnetic Flux  :, , , ds, , Magnetic flux through an area is given by d = B ds cos , , ......(i), , But by definition of scalar product B . d s = Bds cos , , ......(ii), , So from eqn (i) and (ii) we have, d  = B . d s or  =, , , , , B, , , O, , B.ds, , (iv) Potential energy of a dipole U : If an electric dipole of moment p is situated in an electric field E or a, magnetic dipole of moment M in a field of induction B , the potential energy of the dipole is given by :

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28 Mathematics In Physics, U E = − p . E and U B = − M . B, , Sample Ex-based on dot product, Ex-38., , A = 2ˆi + 4 ˆj + 4 kˆ and B = 4ˆi + 2ˆj − 4 kˆ are two vectors. The angle between them will be, (a) 0°, , Sol: (d), , cos  =, , (b) 45°, , . B, | | .| B |, , =, , (c) 60°, , a1 b1 + a 2 b 2 + a 3 b 3, | | .| B |, , =, , 24 + 4 2 −4 4, , (d) 90°, , =0, , | A | .| B |, ,   = cos −1 (0)   = 90 , , Ex-39., , If two vectors 2ˆi + 3ˆj − kˆ and − 4ˆi − 6 ˆj − kˆ are parallel to each other then value of  be, (a) 0, , (c) 3, , (d) 4, , Let A = 2ˆi + 3 ˆj − kˆ and B = −4ˆi − 6 ˆj + kˆ, , y, , Sol: (c), , (b) 2, , (a) 25, Sol: (c), , m, , ca, , In above example if vectors are perpendicular to each other then value of  be, (b) 26, , (c) – 26, , eA, , Ex-40., , a1 a2 a3, 2, 3, −1, =, =, =, =, i.e.,   = 2., −4 −6, , b1 b2 b3, , de, , A and B are parallel to each other, , (d) – 25, , If A and B are perpendicular to each other then A . B = 0  a1b1 + a2 b 2 + a3 b 3 = 0, , nc, , So, 2(−4) + 3(−6) + (−1)() = 0   = −26, , (a), , 3, 13, , 3, , 26, , (c), , 3, 26, , (d), , 3, 13, , |  | = 2 2 + 3 2 + (−1) 2 = 4 + 9 + 1 = 14, , JH, , Sol: (b), , (b), , Sc, , ie, , If  = 2ˆi + 3ˆj − kˆ and B = −ˆi + 3ˆj + 4 kˆ then projection of A on B will be, , A, 'S, , Ex-41., , | B | = (−1) 2 + 3 2 + 4 2 = 1 + 9 + 16 = 26, A . B = 2 (−1) + 3  3 + (−1) (4 ) = 3, , The projection of A on B =, , A.B, , =, , | B|, , Ex-42., , 3, 26, , A body, acted upon by a force of 50 N is displaced through a distance 10 meter in a direction making an, angle of 60° with the force. The work done by the force be, (a) 200 J, , (b) 100 J, , (c) 300, , (d) 250 J, , 1, = 250 J ., 2, , Sol: (d), , W = F . S = FS cos  = 50  10  cos 60  = 50  10 , , Ex-43., , A particle moves from position 3ˆi + 2ˆj − 6kˆ to 14ˆi + 13 ˆj + 9kˆ due to a uniform force of 4ˆi + ˆj + 3 kˆ N . If the, displacement in meters then work done will be, (a) 100 J, , Sol: (a), , S = r2 − r1, , (b) 200 J, , (c) 300 J, , (d) 250 J

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Mathematics In Physics 29, , W = F . S = (4ˆi + ˆj + 3kˆ ) . (11ˆi + 11ˆj + 15 kˆ ) = (4  11 + 1  11 + 3  15 ) = 100 J., , Ex-44., , If for two vector A and B , sum ( A + B) is perpendicular to the difference ( A − B) . The ratio of their, magnitude is, (a) 1, , Sol: (a), , (b) 2, , (c) 3, , (d) None of these, , ( A + B) is perpendicular to ( A − B) . Thus, , ( A + B) . ( A − B) = 0 or A 2 + B . A − A . B − B 2 = 0, Because of commutative property of dot product A.B = B. A, ,  A 2 − B 2 = 0 or A = B, Thus the ratio of magnitudes A/B = 1, , Ex-45., , A force F = − K(yˆi + xˆj) (where K is a positive constant) acts on a particle moving in the x-y plane. Starting, , y, , from the origin, the particle is taken along the positive x- axis to the point (a, 0) and then parallel to the y-, , F = − K (0 ˆi + a ˆj)  F = − Kaˆj, , nc, , Displacement r = (aˆi + 0 ˆj) − (0 ˆi + 0 ˆj) = aˆi, , de, , For motion of the particle form (0, 0) to (a, 0), , (d) Ka 2, , eA, , Sol: (c), , (c) − Ka 2, , (b) 2 Ka 2, , ca, , (a) − 2 Ka 2, , m, , axis to the point (a, a). The total work done by the forces F on the particle is, , Sc, , For motion (a, 0) to (a, a), , ie, , So work done from (0, 0) to (a, 0) is given by W = F . r = − Kaˆj . aˆi = 0, , A, 'S, , F = − K(aˆi + aˆj) and displacement r = (aˆi + aˆj) − (aˆi + 0 ˆj) = aˆj, , JH, , So work done from (a, 0) to (a, a) W = F . r = − K(aˆi + aˆj). aˆj = − Ka 2, So total work done = − Ka 2, , 0.18 Vector Product of Two Vector., (1) Definition : The vector product or cross product of two vectors is defined as a vector having a, magnitude equal to the product of the magnitudes of two vectors with the sine of angle between them,, and direction perpendicular to the plane containing the two vectors in accordance with right hand screw, rule., , C = AB, Thus, if A and B are two vectors, then their vector product written as A  B is a vector C defined by, C = A  B = AB sin  nˆ, , The direction of A  B, i.e. C is perpendicular to the plane, containing vectors A and B and in the sense of advance of a right, handed screw rotated from A (first vector) to B (second vector), through the smaller angle between them. Thus, if a right handed

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30 Mathematics In Physics, screw whose axis is perpendicular to the plane framed by A and B is rotated from A to B through the, smaller angle between them, then the direction of advancement of the screw gives the direction of A  B, i.e. C, (2) Properties :, (i) Vector product of any two vectors is always a vector perpendicular to the plane containing these two, vectors, i.e., orthogonal to both the vectors A and B , though the vectors A and B may or may not be, orthogonal., (ii) Vector product of two vectors is not commutative, i.e., A  B  B  A [but = −B  A], Here it is worthy to note that, | A  B | =| B  A | = AB sin , , m, , y, , i.e., in case of vector A  B and B  A magnitudes are equal but directions are opposite., , de, , (iii) The vector product is distributive when the order of the vectors is strictly maintained, i.e., , ca, , A  (B + C) = A  B + A  C, , | A  B | = AB sin , , | A  B | , , | A | | B | , ,  = sin −1 , , nc, , i.e.,, , ie, , So, , eA, , (iv) As by definition of vector product of two vectors A  B = AB sin  nˆ, , Sc, , (v) The vector product of two vectors will be maximum when sin = max = 1, i.e.,  = 90 o, , A, 'S, , [ A  B]max = AB nˆ, , JH, , i.e., vector product is maximum if the vectors are orthogonal., (vi) The vector product of two non- zero vectors will be minimum when | sin | = minimum = 0, i.e.,  = 0 o or, , 180 o, , [ A  B] min = 0, , i.e. if the vector product of two non-zero vectors vanishes, the vectors are collinear., (vii) The self cross product, i.e., product of a vector by itself vanishes, i.e., is null vector, A  A = AA sin 0 o nˆ = 0, (viii) In case of unit vector nˆ  nˆ = 0 so that ˆi  ˆi = ˆj  ˆj = kˆ  kˆ = 0, (ix) In case of orthogonal unit vectors, ˆi , ˆj, kˆ in accordance with right hand screw rule :, ˆj, , ˆj, , k̂, , î, , k̂, , î

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Mathematics In Physics 31, , ˆi  ˆj = kˆ ,, , ˆj  kˆ = ˆi, , kˆ  ˆi = ˆj, , and, , And as cross product is not commutative,, kˆ  ˆj = −ˆi, , ˆj  ˆi = −kˆ, , ˆi  kˆ = −ˆj, , and, , ˆi, (x) In terms of components A  B = A x, Bx, , kˆ, A z = ˆi ( A y B z − A z B y ) + ˆj( A z B x − A x B z ) + kˆ ( A x B y − A y B x ), Bz, , ˆj, Ay, By, , y, , (3) Example : Since vector product of two vectors is a vector, vector physical quantities (particularly, representing rotational effects) like torque, angular momentum, velocity and force on a moving charge in a, magnetic field and can be expressed as the vector product of two vectors. It is well – established in physics, that :, , de, , m, , (i) Torque  = r  F, , ca, , (ii) Angular momentum L = r  p, , eA, , (iii) Velocity v =   r, , nc, , (iv) Force on a charged particle q moving with velocity v in a magnetic field B is given by F = q(v  B), , ie, , (v) Torque on a dipole in a field  E = p  E and  B = M  B, , Sc, , Sample Ex-based on vector product, , A, 'S, , If A = 3ˆi + ˆj + 2kˆ and B = 2ˆi − 2ˆj + 4 kˆ then value of | A  B | will be, (a) 8 2, , Sol: (b), , ˆi, AB = 3, 2, , ˆj, 1, −2, , JH, , Ex-46., , (b) 8 3, , (c) 8 5, , kˆ, 2 = (1  4 − 2  −2)ˆi + (2  2 − 4  3)ˆj + (3  −2 − 1  2)kˆ = 8ˆi − 8 ˆj − 8 kˆ, 4, ,  Magnitude of A  B =| A  B | = (8 ) 2 + (−8 ) 2 + (−8 ) 2, , Ex-47., , =8 3, , In above example a unit vector perpendicular to both A and B will be, (a) +, , Sol: (c), , (d) 5 8, , nˆ =, , 1 ˆ ˆ ˆ, (i − j − k ), 3, , A B, | A B|, , =, , 8ˆi − 8 ˆj − 8 kˆ, 8 3, , (b) −, , =, , 1 ˆ ˆ ˆ, (i − j − k ), 3, , (c) Both (a) and (b), , (d) None of these, , 1 ˆ ˆ ˆ, (i − j − k ), 3, , There are two unit vectors perpendicular to both A and B they are nˆ = , , 1 ˆ ˆ ˆ, (i − j − k ), 3

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32 Mathematics In Physics, , Ex-48., , The vectors from origin to the points A and B are A = 3ˆi − 6 ˆj + 2kˆ and B = 2ˆi + ˆj − 2kˆ respectively. The, area of the triangle OAB be, (a), , Sol: (a), , 5, 17 sq.unit, 2, , (b), , 2, 17 sq.unit, 5, , (c), , 3, 17 sq.unit, 5, , (d), , 5, 17 sq.unit, 3, , Given OA = a = 3ˆi − 6 ˆj + 2kˆ and OB = b = 2ˆi + ˆj − 2kˆ, , ˆi,  (a  b) = 3, 2, , kˆ, 2, −2, , ˆj, −6, 1, , = (12 − 2)ˆi + (4 + 6)ˆj + (3 + 12)kˆ, , = 10ˆi + 10 ˆj + 15 kˆ  | a  b | = 10 2 + 10 2 + 15 2 = 425 = 5 17, , m, , The angle between the vectors A and B is  . The value of the triple product A . (B  A ) is, (c), , A 2 B sin, , Let A .(B  A) = A . C, , eA, , Sol: (b), , (b) Zero, ,  A. C = 0, , nc, , Here C = B  A Which is perpendicular to both vector A and B, , The torque of the force F = (2ˆi − 3 ˆj + 4 kˆ )N acting at the point r = (3ˆi + 2ˆj + 3kˆ ) m about the origin be, , Sc, , ie, , Ex-50., , (d) A 2 B cos , , ca, , (a) A 2 B, , de, , Ex-49., , y, , 1, 5 17, | ab| =, sq.unit., 2, 2, , Area of OAB =, , (b) 17ˆi − 6 ˆj − 13 kˆ, , (c) − 6ˆi + 6 ˆj − 12kˆ, , kˆ, 3, 4, , Sol: (b), , ˆi, ˆj,  =r F = 3 2, 2 −3, , Ex-51., , If A  B = C, then which of the following statements is wrong, , = (2  4 ) − (3  −3) ˆi + (2  3) − (3  4 )ˆj + (3  −3) − (2  2)kˆ = 17 ˆi − 6 ˆj − 13 kˆ, , JH, , (a) C ⊥ A, Sol: (d), , (d) − 17ˆi + 6 ˆj + 13 kˆ, , A, 'S, , (a) 6ˆi − 6 ˆj + 12kˆ, , (b) C ⊥ B, , (c) C ⊥ ( A + B), , (d) C ⊥ ( A  B), , From the property of vector product, we notice that C must be perpendicular to the plane formed by, vector A and B . Thus C is perpendicular to both A and B and ( A + B) vector also must lie in the plane, formed by vector A and B . Thus C must be perpendicular to ( A + B) also but the cross product ( A  B), gives a vector C which can not be perpendicular to itself. Thus the last statement is wrong., , Ex-52., , If a particle of mass m is moving with constant velocity v parallel to x-axis in x-y plane as shown in fig. Its, angular momentum with respect to origin at any time t will be, (a) mvb kˆ, , Sol: (b), , (b) − mvb kˆ, , We know that, Angular momentum, , (c) mvb ˆi, , (d) mv ˆi

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Mathematics In Physics 33, , ˆi, L = r  p in terms of component becomes L = x, px, , ˆj, y, py, , kˆ, z, pz, , y, m, , v, b, , As motion is in x-y plane (z = 0 and Pz = 0 ), so L = k (xp y − yp x ), , x, , O, , Here x = vt, y = b, p x = m v and p y = 0,  L = k vt  0 − b mv  = −mvb kˆ, , 0.19 Lami's Theorem., In any  A B C with sides a, b, c, 180 – , , sin  sin  sin , =, =, a, b, c, , y, , , c, , m, , i.e., for any triangle the ratio of the sine of the angle containing the side, to the length of the side is a constant., , de, , , , Pre-multiplying both sides by a, , , , 0 + a  b = −a  c  a  b = c  a, , A, 'S, , a  (a + b ) = −a  c, , , a, , .....(i), .....(ii), , ie, , a + b = −c, , Sc, , , , [All three sides are taken in order], , nc, , a+b +c = 0, , 180 – , , eA, , ca, , For a triangle whose three sides are in the same order we establish the, Lami's theorem in the following manner. For the triangle shown, , b, , .....(iii), , JH, , Pre-multiplying both sides of (ii) by b, , b  (a + b ) = − b  c  b  a + b  b = −b  c  − a  b = −b  c  a  b = b  c, , From (iii) and (iv), we get a  b = b  c = c  a, Taking magnitude, we get | a  b | =| b  c | =| c  a |, ,  ab sin(180 −  ) = bc sin(180 −  ) = ca sin(180 −  ), ,  ab sin  = bc sin = ca sin , Dividing through out by abc, we have , , sin  sin  sin , =, =, a, b, c, , .....(iv), , 180 – 

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34 Mathematics In Physics, , Problems based on fundamentals of vector, 1., , How many minimum number of coplanar vectors having different magnitudes can be added to give zero resultant, (a) 2, , 2., , (b) 3, , (c) 4, , (d) 5, , A hall has the dimensions 10 m  12 m  14 m. A fly starting at one corner ends up at a diametrically opposite corner. What is the, , ca, , eA, , (d) Zero, , nc, , ie, , (c) 4, , (d) 2 10, , Sc, , A, 'S, , (b) 45°, 45°, 45°, , (c) 60°, 60°, 45°, , (c) Vector of magnitude, , (b) tan −1 2 / 3, , (c), , (b) ˆi + ˆj, , (c), , (b) Null vector, , (d) 45°, 45°, 60°, , 2, , (d) Scalar, , Given vector A = 2ˆi + 3ˆj, the angle between A and y-axis is, (a) tan −1 3 / 2, , 9., , (c) 250 N, , (b) 5 2, ,  1 ˆ, 1 ˆ, i+, j  is a, The expression , 2 ,  2, , (a) Unit vector, , 8., , (b) 500 N, , The angles which a vector ˆi + ˆj + 2 kˆ makes with X, Y and Z axes respectively are, (a) 60°, 60°, 60°, , 7., , (d) 0, , 0.8, , The magnitude of a given vector with end points (4, – 4, 0) and (– 2, – 2, 0) must be, (a) 6, , 6., , (c), , 0.2, , 100 coplanar forces each equal to 10 N act on a body. Each force makes angle  / 50 with the preceding force. What is the, resultant of the forces, (a) 1000 N, , 5., , (b), , (d) 21 m, , de, , 0.4ˆi + 0.8 ˆj + ckˆ represents a unit vector when c is, (a) – 0.2, , 4., , (c) 36 m, , JH, , 3., , (b) 26 m, , m, , (a) 17 m, , y, , magnitude of its displacement, , sin −1 2 / 3, , (d) cos −1 2 / 3, , The unit vector along ˆi + ˆj is, (a) kˆ, , ˆi + ˆj, , (d), , ˆi + ˆj, 2, , (d), , 5, , 2, , 10., , A vector is represented by 3 ˆi + ˆj + 2 kˆ . Its length in XY plane is, (a) 2, , 11., , 14, , (c), , 10, , Five equal forces of 10 N each are applied at one point and all are lying in one plane. If the angles between them are equal, the, resultant force will be, (a) Zero, , 12., , (b), , (b) 10 N, , (c) 20 N, , (d) 10 2 N, , (c) 22.5°, , (d) 30°, , The angle made by the vector A = ˆi + ˆj with x- axis is, (a) 90°, , (b) 45°

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Mathematics In Physics 35, 13., , The value of a unit vector in the direction of vector A = 5ˆi − 12 ˆj, is, (a) î, , 14., , (d) (5ˆi − 12 ˆj) / 13, , (c) (ˆi + ˆj) / 13, , (b) ˆj, , Any vector in an arbitrary direction can always be replaced by two (or three), (a) Parallel vectors which have the original vector as their resultant, (b) Mutually perpendicular vectors which have the original vector as their resultant, (c) Arbitrary vectors which have the original vector as their resultant, (d) It is not possible to resolve a vector, , 15., , Angular momentum is, (a) A scalar, , 16., , (b) A polar vector, , (c) An axial vector, , (d) None of these, , If a vector P making angles , , and  respectively with the X, Y and Z axes respectively. Then sin 2  + sin 2  + sin 2  =, (a) 0, , (b) 1, , (c) 2, , (d) 3, , eA, , (b) 60°, , (c) 90°, , A, 'S, , JH, , (c) Lies in the plane containing A + B, , (b) Cannot be zero, (d) Lies in the plane containing A − B, , If the resultant of the two forces has a magnitude smaller than the magnitude of larger force, the two forces must be, (a) Different both in magnitude and direction, , (b) Mutually perpendicular to one another, , (c) Possess extremely small magnitude, , (d) Point in opposite directions, , Forces F1 and F2 act on a point mass in two mutually perpendicular directions. The resultant force on the point mass will be, , (a), , 23., , (d) Between 1 N and 7 N, , Two vectors A and B lie in a plane, another vector C lies outside this plane, then the resultant of these three vectors, , (a) Can be zero, , 22., , (c) 1 N, , ie, , (b) 5 N, , i.e., A + B + C, , 21., , (d) 180°, , A particle is simultaneously acted by two forces equal to 4 N and 3 N. The net force on the particle is, (a) 7 N, , 20., , (d) 60°, , For the resultant of the two vectors to be maximum, what must be the angle between them, (a) 0°, , 19., , (c) 150°, , nc, , 18., , (b) 120°, , ca, , (a) 45°, , de, , Two forces, each of magnitude F have a resultant of the same magnitude F. The angle between the two forces is, , Sc, , 17., , m, , y, , Problems based on addition of vectors, , F1 + F2, , (b) F1 − F2, , (c), , F12 + F22, , (d) F12 + F22, , Find the resultant of three vectors OA, OB and OC shown in the following figure. Radius of the circle is R., C, B, , (a) 2R, (b) R(1 + 2 ), (c), , R 2, , (d) R( 2 − 1), , 45o, 45o, O, , A

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36 Mathematics In Physics, 24., , If | A − B | =| A | =| B |, the angle between A and B is, (a) 60°, , 25., , (c) 120°, ,  2(x 2 − y 2 ) , , (b) cos −1  − 2, , x + y 2 , , , (x 2 + y 2 ), ,  x2 + y2, (c) cos − 1  −,  x2 − y2, , ,  x2 − y2, (d) cos − 1  −,  x2 + y2, , , , , , , , (c), , (b) C must be less than | A − B |, (d) C may be equal to | A − B |, , C must be greater than | A − B |, , Fig. shows ABCDEF as a regular hexagon. What is the value of AB + AC + AD + AE + AF, D, , AO, , m, , (a), , y, , E, , 4 AO, , ca, , (c), , de, , (b) 2 AO, , nc, , (c)  / 2, , (d)  / 4, , Sc, , (b), , A, 'S, , 136, , 13 .2, , (c), , (d), , 202, , 160, , JH, , A particle has displacement of 12 m towards east and 5 m towards north then 6 m vertically upward. The sum of these, displacements is, (b) 10.04 m, , (c) 14.31 m, , (d) None of these, , The three vectors A = 3ˆi − 2ˆj + kˆ , B = ˆi − 3ˆj + 5kˆ and C = 2ˆi + ˆj − 4 kˆ form, (a) An equilateral triangle, , 32., , ie, , (b) , , (a) 12, , 31., , B, , A, , Magnitude of vector which comes on addition of two vectors, 6ˆi + 7 ˆj and 3ˆi + 4 ˆj is, (a), , 30., , O, , The magnitude of vector A, B and C are respectively 12, 5 and 13 units and A + B = C then the angle between A and B is, , (a) 0, , 29., , C, , F, , eA, , (d) 6 AO, , 28., , (b) Isosceles triangle, , (c) A right angled triangle, , (d) No triangle, , For the fig., (a), , A+B =C, C, , B, , (b) B + C = A, (c), , C+A=B, , A, , (d) A + B + C = 0, , 33., , , , , , , Let the angle between two nonzero vectors A and B be 120° and resultant be C, (a) C must be equal to | A − B |, , 27., , (d) 90°, , At what angle must the two forces (x + y) and (x – y) act so that the resultant may be, , x 2 + y 2 , (a) cos −1  −,  2(x 2 − y 2 ) , , , , 26., , (b) 0°, , Let C = A + B then, (a) | C | is always greater then | A |, , (b) It is possible to have | C | | A | and | C | | B |, , (c) C is always equal to A + B, , (d) C is never equal to A + B

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Mathematics In Physics 37, 34., , The value of the sum of two vectors A and B with  as the angle between them is, , A 2 + B 2 + 2 AB cos , , (a), , 35., , (b), , A 2 − B 2 + 2 AB cos , , A 2 + B 2 − 2 AB sin , , (c), , (d), , A 2 + B 2 + 2 AB sin , , Following forces start acting on a particle at rest at the origin of the co-ordinate system simultaneously, , F1 = −4ˆi − 5 ˆj + 5kˆ , F 2 = 5ˆi + 8 ˆj + 6kˆ , F 3 = −3ˆi + 4 ˆj − 7kˆ and F 4 = 2ˆi − 3ˆj − 2kˆ then the particle will move, (a) In x – y plane, (a) 10, 10, 10, , y, , m, , de, , 5, 13, , (c) cos − 1, , 12, 13, , (d) cos − 1, , 7, 13, , eA, , (c) 135°, , (d) None of these, , nc, , (b) 150°, , ie, , (b) − 2ˆi + ˆj − kˆ, , (c), , x-axis, , 2ˆi − ˆj + kˆ, , (d) − 2ˆi − ˆj − kˆ, , (b) tan −1 P / Q, , (c), , tan −1 Q / P, , (d) tan −1 (P − Q) /(P + Q), , (b) cos −1 (−P / Q), , (c), , sin −1 (P / Q), , (d) sin −1 (−P / Q), , Maximum and minimum magnitudes of the resultant of two vectors of magnitudes P and Q are in the ratio 3 : 1. Which of the, following relations is true, (b) P = Q, , (c), , PQ = 1, , (d) None of these, , The resultant of A + B is R 1 . On reversing the vector B, the resultant becomes R 2 . What is the value of R12 + R 22, , A2 + B2, , (b) A 2 − B 2, , (c), , 2( A 2 + B 2 ), , (d) 2( A 2 − B 2 ), , The resultant of two vectors P and Q is R. If Q is doubled, the new resultant is perpendicular to P. Then R equals, (a) P, , 47., , (b) cos − 1, , The resultant of P and Q is perpendicular to P . What is the angle between P and Q, , (a), , 46., , (d) 2 N and 14 N, , What is the angle between P and the resultant of (P + Q) and (P − Q), , (a) P = 2Q, , 45., , (c) 4 N and 12 N, , Sc, , 2ˆi + ˆj − kˆ, , (a) cos −1 (P / Q), , 44., , (b) 8 N and 8 N, , What vector must be added to the two vectors ˆi − 2ˆj + 2kˆ and 2ˆi + ˆj − kˆ , so that the resultant may be a unit vector along, , (a) Zero, , 43., , (d) Moving with an, , The resultant of two vectors A and B is perpendicular to the vector A and its magnitude is equal to half the magnitude of vector B., The angle between A and B is, , (a), , 42., , (b) Moving with a uniform velocity (c) In equilibrium, , ca, , 5, 12, , (a) 120°, , 41., , (d) 10, 20, 40, , If vectors P, Q and R have magnitude 5, 12 and 13 units and P + Q = R, the angle between Q and R is, (a) cos − 1, , 40., , (c) 10, 20, 20, , The sum of two forces acting at a point is 16 N. If the resultant force is 8 N and its direction is perpendicular to minimum force, then the forces are, (a) 6 N and 10 N, , 39., , (d) Along x -axis, , When three forces of 50 N, 30 N and 15 N act on a body, then the body is, (a) At rest, acceleration, , 38., , (b) 10, 10, 20, , A, 'S, , 37., , (c) In x – z plane, , Following sets of three forces act on a body. Whose resultant cannot be zero, , JH, , 36., , (b) In y – z plane, , (b) (P+Q), , (c) Q, , (d) (P–Q), , Two forces, F1 and F2 are acting on a body. One force is double that of the other force and the resultant is equal to the greater, force. Then the angle between the two forces is, (a) cos −1 (1 / 2), , (b) cos −1 (−1 / 2), , (c) cos −1 (−1 / 4 ), , (d) cos −1 (1 / 4 )

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38 Mathematics In Physics, 48., , Given that A + B = C and that C is ⊥ to A . Further if | A | =| C |, then what is the angle between A and B, (a), , , 4, , radian, , (b), , , 2, , (c), , radian, , 3, radian, 4, , (d)  radian, , Problems based on subtraction of vectors, 49., , Figure below shows a body of mass M moving with the uniform speed on a circular path of radius, R. What is the change in, acceleration in going from P1 to P2, P2, , (a) Zero, , v, , (b) v 2 / 2 R, , eA, , nc, , (c) 100 2 km / hr, , (d) 0, , Sc, , ie, , (b) 150 km/hr, , A, 'S, , (b) 32ˆi − 13 ˆj, , (c), , (d) − 25ˆi + 13 ˆj, , − 18ˆi + 6 ˆj, , A body moves due East with velocity 20 km/hour and then due North with velocity 15 km/hour. The resultant velocity, (a) 5 km/hour, , 54., , (d) − 4ˆi − 6 ˆj, , − 4ˆi + 6 ˆj, , What displacement must be added to the displacement 25ˆi − 6 ˆj m to give a displacement of 7.0 m pointing in the x- direction, (a) 18ˆi − 6 ˆj, , 53., , (c), , A plane is revolving around the earth with a speed of 100 km/hr at a constant height from the surface of earth. The change in the, velocity as it travels half circle is, (a) 200 km/hr, , 52., , (b) 4ˆi − 6 ˆj, , ca, , , , A body is at rest under the action of three forces, two of which are F1 = 4ˆi, F2 = 6 ˆj, the third force is, (a) 4ˆi + 6 ˆj, , 51., , y, , v2,  2, R, , m, , (d), , P1, , R, , de, , 2v 2 / R, , JH, , 50., , (c), , (b) 15 km/hour, , (c) 20 km/hour, , (d) 25 km/hour, , A particle is moving on a circular path of radius r with uniform velocity v. The change in velocity when the particle moves from P, to Q is (POQ = 40 ), P, r, , (a) 2v cos 40 , , O, , v, , 40o, , (b) 2v sin 40 , (c), , Q, , 2v sin 20 , , v, , (d) 2v cos 20 , , 55., , The length of second's hand in watch is 1 cm. The change in velocity of its tip in 15 seconds is, (a) Zero, , (b), , , 30 2, , cm / sec, , (c), , , 30, , cm / sec, , (d), ,  2, 30, , cm / sec

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Mathematics In Physics 39, 56., , A particle moves towards east with velocity 5 m/s. After 10 seconds its direction changes towards north with same velocity. The, average acceleration of the particle is, (a) Zero, , (b), , 1, , m / s2 N −W, , 1, , (c), , 2, , m / s2 N − E, , (d), , 1, , m / s2 S −W, , 2, , 2, , Problems based on scalar product of vectors, Consider two vectors F1 = 2ˆi + 5kˆ and F 2 = 3ˆj + 4 kˆ . The magnitude of the scalar product of these vectors is, (a) 20, , m, , (c), , V1 and V2 are mutually perpendicular, , (d) | V 1 | =| V 2 |, , eA, , nc, , A force F = (5ˆi + 3ˆj) Newton is applied over a particle which displaces it from its origin to the point r = (2ˆi − 1ˆj) metres. The work, done on the particle is, (b) +13 joules, , (c) +7 joules, , (d) +11 joules, , (c) 180°, , (d) None of the above, , (c) 60°, , (d) 90°, , (b) 90°, , ie, , The angle between two vectors − 2ˆi + 3ˆj + kˆ and ˆi + 2ˆj − 4 kˆ is, The angle between the vectors (ˆi + ˆj) and (ˆj + kˆ ) is, (b) 45°, , A particle moves with a velocity 6ˆi − 4 ˆj + 3kˆ m / s under the influence of a constant force F = 20ˆi + 15 ˆj − 5kˆ N . The instantaneous, power applied to the particle is, (b) 45 J/s, , (c) 25 J/s, , (d) 195 J/s, , (c) 45°, , (d) 60°, , If P.Q = PQ, then angle between P and Q is, (a) 0°, , 66., , de, , (b) V 1 = V 2, , ca, , V1 is parallel to V2, , (a) 35 J/s, , 65., , (d) A . B = 0, , AB = 0, , (c), , (a), , (a) 30°, , 64., , (b) A − B = 0, , If | V 1 + V 2 | =| V 1 − V 2 | and V2 is finite, then, , (a) 0°, , 63., , (d) 3ˆi − 4 ˆj, , y, , A+B=0, , (a) – 7 joules, , 62., , (c) 7 kˆ, , (b) 6 î, , Sc, , 61., , 4ˆi + 3ˆj, , Two vectors A and B are at right angles to each other, when, (a), , 60., , (d) 26, , Consider a vector F = 4ˆi − 3ˆj. Another vector that is perpendicular to F is, (a), , 59., , (c) 5 33, , A, 'S, , 58., , (b) 23, , JH, , 57., , (b) 30°, , Two constant forces F1 = 2ˆi − 3ˆj + 3kˆ (N) and F2 = ˆi + ˆj − 2kˆ (N) act on a body and displace it from the position, , r1 = ˆi + 2ˆj − 2kˆ (m) to the position r2 = 7ˆi + 10 ˆj + 5kˆ (m). What is the work done, (a) 9 J, , 67., , (b) 18 units, , (c) 11 units, , (d) 5 units, , (b) 45°, , (c) 90°, , (d) 180°, , The vector P = aˆi + aˆj + 3kˆ and Q = aˆi − 2ˆj − kˆ are perpendicular to each other. The positive value of a is, (a) 3, , 70., , (d) None of these, , The angle between the two vector A = 5ˆi + 5 ˆj and B = 5ˆi − 5 ˆj will be, (a) Zero, , 69., , (c) – 3 J, , A force F = 5ˆi + 6 ˆj + 4 kˆ acting on a body, produces a displacement S = 6ˆi − 5kˆ . Work done by the force is, (a) 10 units, , 68., , (b) 41 J, , (b) 4, , (c) 9, , (d) 13, , A body, constrained to move in the Y-direction is subjected to a force given by F = (−2ˆi + 15 ˆj + 6kˆ ) N . What is the work done by, this force in moving the body a distance 10 m along the Y-axis, (a) 20 J, , (b) 150 J, , (c) 160 J, , (d) 190 J

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40 Mathematics In Physics, 71., , A particle moves in the x-y plane under the action of a force F such that the value of its liner momentum (P) at anytime t is, , Px = 2 cos t, p y = 2 sin t. The angle  between F and P at a given time t. will be, (a)  = 0, , (d)  = 180 , , (c)  = 90 , , (b)  = 30 , , Problems based on cross product of vectors, 72., , The area of the parallelogram represented by the vectors A = 2ˆi + 3ˆj and B = ˆi + 4 ˆj is, (a) 14 units, , 73., , (b) 7.5 units, , (c) 10 units, , (d) 5 units, , For any two vectors A and B if A . B = | A  B |, the magnitude of C = A + B is equal to, , A2 + B2, , (a), , (b) A + B, , A2 + B2 +, , (c), , AB, , A 2 + B 2 + 2  AB, , (d), , 2, , A vector F 1 is along the positive X-axis. If its vector product with another vector F 2 is zero then F 2 could be, , (d) Act at an angle of 30°, , ˆ  Bˆ, A, AB sin , , nc, , (d) , , (b) 3, , (c) 1, , (d) 0, , (b), , ˆ  Bˆ, A, AB cos , , (c), , AB, AB sin , , (d), , AB, AB cos , , (b) ˆi B sin  + j B cos , , (c) ˆi B sin  − j B cos , , (d) ˆi B cos  − j B sin , ,  5 , , (b) cos −1 , ,  3, , (c), ,  2 , , sin −1 , ,  3, ,  5, , (d) sin −1 ,  3 , , , , A vector A points vertically upward and B points towards north. The vector product A  B is, (b) Along west, , (c) Along east, , (d) Vertically downward, , (c) 180°, , (d) 60°, , (c) b = 2, , (d) b= – 4, , Angle between the vectors (ˆi + ˆj) and (ˆj − kˆ ) is, (a) 90°, , 84., , , 4, , The angle between two vectors given by 6i + 6 j − 3k and 7i + 4 j + 4 k is, , (a) Zero, , 83., , (c), , Let A = ˆi A cos  + ˆjA sin  be any vector. Another vector B which is normal to A is, ,  1 , , (a) cos −1 , ,  3, , 82., , , 2, , Which of the following is the unit vector perpendicular to A and B, , (a) ˆi B cos  + j B sin , , 81., , (b), , (d)  / 2, , The resultant of the two vectors having magnitude 2 and 3 is 1. What is their cross product, , (a), , 80., , (c)  / 4, , (b) , , What is the angle between ( P + Q) and (P  Q), , (a) 6, , 79., , eA, , The angle between vectors (A  B) and (B  A) is, , (a) 0, , 78., , y, , (c) Act at an angle of 60°, , ie, , 77., , de, , (b) Are parallel to each other, , (a) Zero, , (d) (−4ˆi ), , ca, , (a) Are perpendicular to each other, , Sc, , 76., , If for two vectors A and B, A  B = 0, the vectors, , A, 'S, , 75., , (c) (ˆj + kˆ ), , (b) − (ˆi + ˆj), , m, , (a) 4 ˆj, , JH, , 74., , (b) 0°, , Two vectors P = 2ˆi + bˆj + 2kˆ and Q = ˆi + ˆj + kˆ will be parallel if, (a) b = 0, , (b) b = 1

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Mathematics In Physics 41, 85., , The position vectors of points A, B, C and D are A = 3ˆi + 4 ˆj + 5kˆ , B = 4ˆi + 5 ˆj + 6kˆ , C = 7ˆi + 9 ˆj + 3kˆ and D = 4ˆi + 6 ˆj then the, displacement vectors AB and CD are, (a) Perpendicular, , AB = 0, , 43, , (c) 6  3, , (d) 4  6, , (b) 45°, , (c) 60°, , (d) 90°, , (b) ˆi . ˆi = 0, , (c) ˆj  ˆj = 1, , (d) kˆ . ˆj = 1, , (b) 5  6, , y, , m, , (b), , nc, , ˆi − 10 ˆj + 18 kˆ, , (c) 2 units, , (c), , (d) 4 units, , ˆi − 10 ˆj − 18 kˆ, , (d), , 5 17, , 5 17, , JH, , 61 sq.unit, , ˆi + 10 ˆj + 18 kˆ, 5 17, , (b), , 59 sq.unit, , (c), , 49 sq.unit, , (d), , 52 sq.unit, , (c), , 2 14 sq. unit, , (d), , 14, sq. unit, 2, , The area of the triangle formed by 2ˆi + ˆj − kˆ and ˆi + ˆj + kˆ is, (b) 2 3 sq. unit, , The position of a particle is given by r = (i + 2 j − k ) momentum P = (3i + 4 j − 2k ). The angular momentum is perpendicular to, (a) x-axis, , (b) y-axis, , (c) z-axis, , (d) Line at equal angles to all the three axes, , Two vector A and B have equal magnitudes. Then the vector A + B is perpendicular to, , AB, , (b) A – B, , (c) 3A – 3B, , (d) All of these, , Find the torque of a force F = −3ˆi + ˆj + 5kˆ acting at the point r = 7ˆi + 3ˆj + kˆ, (a) 14 ˆi − 38 ˆj + 16 kˆ, , 99., , ie, , (b) 1 unit, , The area of the parallelogram whose sides are represented by the vectors ˆj + 3kˆ and ˆi + 2ˆj − kˆ is, , (a), , 98., , 41 units, , (d) b  c, , (c) b . c, , Sc, , ˆi + 10 ˆj − 18 kˆ, , (a) 3 sq.unit, , 97., , (d), , What is the unit vector perpendicular to the following vectors 2ˆi + 2ˆj − kˆ and 6ˆi − 3ˆj + 2kˆ, , (a), , 96., , 37 units, , eA, , (b) c, , 5 17, , 95., , (c), , The diagonals of a parallelogram are 2 î and 2ˆj. What is the area of the parallelogram, , (a), , 94., , 31 units, , de, , (b), , Three vectors a, b and c satisfy the relation a . b = 0 and a . c = 0. The vector a is parallel to, , (a) 0.5 units, , 93., , (d) A = 5, , ca, , 29 units, , (a) b, , 92., , A . B = 48, , The linear velocity of a rotating body is given by v =   r, where  is the angular velocity and r is the radius vector. The angular, velocity of a body is  = ˆi − 2ˆj + 2kˆ and the radius vector r = 4 ˆj − 3kˆ , then | v | is, (a), , 91., , (c), , In an clockwise system, (a) ˆj  kˆ = ˆi, , 90., , A 1, =, B 2, , If | A  B | =| A . B |, then angle between A and B will be, (a) 30°, , 89., , (b), , If force (F) = 4ˆi + 5 ˆj and displacement (s) = 3ˆi + 6kˆ then the work done is, (a), , 88., , (d) Inclined at an angle of 60°, , Which of the following is not true ? If A = 3ˆi + 4 ˆj and B = 6ˆi + 8 ˆj where A and B are the magnitudes of A and B, (a), , 87., , (c) Antiparallel, , A, 'S, , 86., , (b) Parallel, , (b) 4ˆi + 4 ˆj + 6kˆ, , (c), , 21ˆi + 4 ˆj + 4 kˆ, , (d) − 14 ˆi + 34 ˆj − 16 kˆ, , (c), , B A, , (d) 2(B  A), , The value of ( A + B) ( A − B) is, (a) 0, , (b) A 2 − B 2

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42 Mathematics In Physics, 100. A particle of mass m = 5 is moving with a uniform speed v = 3 2 in the XOY plane along the line Y = X + 4. The magnitude of, the angular momentum of the particle about the origin is, (a) 60 units, , (b) 40 2 units, , (c) Zero, , (d) 7.5 units, , Problems based on Lami's theorem, 101. P, Q and R are three coplanar forces acting at a point and are in equilibrium. Given P = 1.9318 kg wt, sin 1 = 0.9659, the value, of R is ( in kg wt), (a) 0.9659, , 150o, , P, , (b) 2, , 2, , (c) 1, , 1, R, , y, , 1, 2, , m, , (d), , Q, , de, , 102. A body is in equilibrium under the action the action of three coplanar forces P, Q and R as shown in the figure. Select the correct, , (c), , P, Q, R, =, =, tan  tan  tan , , (d), , P, Q, R, =, =, sin  sin , sin , , eA, , , , Q, , nc, , P, Q, R, =, =, cos  cos  cos , , P, , , , , R, , ie, , (b), , Sc, , P, Q, R, =, =, sin  sin  sin , , A, 'S, , (a), , ca, , statement, , (a) Four, , JH, , 103. If a body is in equilibrium under a set of non-collinear forces, then the minimum number of forces has to be, (c) Two, , (b) 3, , (c) 4, , (b) Three, , (d) Five, , 104. How many minimum number of non-zero vectors in different planes can be added to give zero resultant, (a) 2, , (d) 5, , 105. A metal sphere is hung by a string fixed to a wall. The sphere is pushed away from the wall by a stick. The forces acting on the, sphere are shown in the second diagram. Which of the following statements is wrong, (a), , P = W tan , , , , , (b) T + P + W = 0, (c) T 2 = P 2 + W 2, , W, , (d) T = P + W, , P, , 106. As shown in figure the tension in the horizontal cord is 30 N. The weight W and tension in the string OA in Newton are, (a) 30 3 , 30, (b) 30 3 , 60, , A, 30o, , (c) 60 3 , 30, (d) None of these, , 30 N, O, W

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43 Mathematics In Physics, , Answer Sheet (Practice problems), 2., , 3., , 4., , 5., , 6., , 7., , 8., , 9., , 10., , b, , d, , b, , d, , d, , c, , a, , b, , c, , c, , 11., , 12., , 13., , 14., , 15., , 16., , 17., , 18., , 19., , 20., , a, , b, , d, , c, , c, , c, , b, , a, , d, , b, , 21., , 22., , 23., , 24., , 25., , 26., , 27., , 28., , 29., , 30., , d, , c, , b, , a, , a, , c, , d, , c, , c, , c, , 31., , 32., , 33., , 34., , 35., , 36., , 37., , 38., , 39., , 40., , c, , c, , b, , a, , b, , d, , d, , a, , c, , b, , 41., , 42., , 43., , 44., , 45., , 46., , 47., , 48., , 49., , 50., , b, , a, , b, , a, , c, , c, , c, , c, , d, , d, , 51., , 52., , 53., , 54., , 55., , 56., , 57., , 58., , 59., , 60., , a, , c, , d, , b, , d, , b, , a, , c, , d, , c, , 61., , 62., , 63., , 64., , 65., , 66., , 67., , 68., , 69., , 70., , c, , b, , c, , b, , a, , a, , a, , c, , a, , b, , 71., , 72., , 73., , 74., , 75., , 76., , 77., , 78., , 79., , 80., , c, , b, , d, , d, , b, , b, , d, , c, , c, , 81., , 82., , 83., , 84., , 85., , 86., , 87., , 88., , 89., , 90., , d, , b, , d, , c, , c, , c, , a, , b, , a, , a, , 91., , 92., , 93., , 94., , 95., , 96., , 97., , 98., , 99., , 100., , d, , c, , c, , b, , d, , a, , a, , a, , d, , a, , 101., , 102., , 103., , 104., , 105., , 106., , c, , a, , b, , c, , d, , b, , JH, , m, , de, , ca, , eA, nc, , ie, , A, 'S, , Sc, , b, , y, , 1.