Chapter 28 Mathematics in Physics free study material by TEACHING CARE online tuition and coaching classes
Chapter 28 Mathematics in Physics free study material by TEACHING CARE online tuition and coaching classes
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.
This law can be expressed by a single mathematical relationship
F µ m1m2
r 2
or F = Gm1m2
r 2
The techniques of mathematics such as algebra, trigonometry, calculus, graph and logarithm can be used to make predictions from the basic equation.
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.
In this introductory chapter we will learn some fundamental mathematics.
Algebra.
 Quadratic equation : An equation of second degree is called a quadratic Standard quadratic equation ax ^{2} + bx + c = 0
Here a is called the coefficient of x^{2}, 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 = 2a
This formula can be written as
x = – Coefficient of x ±
(Coefficient of x)^{2} – 4(Coefficient of x ^{2} ) ´ (Constant term) 2(Coefficient of x ^{2} )
Note : @ If a and b be the roots of the quadratic equation then
Sum of roots a + b = – b and product of roots = c
a a
Problem 1. Solve the equation 10x ^{2} – 27x + 5 = 0
Solution : By comparing the given equation with standard equation a = 10, b = – 27, and c = 5
x = 2a
= – (27) ±
(27)^{2} – 4 ´ 10 ´ 5
2 ´ 10
= 27 ± 23
20
\ x1 = 27 + 23 = 5 and x 2 = 27 – 23 = 1
20 2 20 5
\ Roots of the equation are
5 and 1 .
2 5
 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) x 2 + n(n – 1)(n – 2) x 3 + …..
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 second and the higher order terms being very very small, can be neglected. So the expression can be written as
(1 + x)^{n} = 1 + nx
(1 + x)^{–}^{n} = 1 – nx
(1 – x)^{n} = 1 – nx
(1 – x)^{–}^{n} = 1 + nx Problem 2. Evaluate (1001)^{1/3} upto six places of decimal. Solution : (1001)^{1/3} = (1000 + 1)^{1/3} = 10(1 + 0.001)^{1/3}
By comparing the given equation with standard equation (1 + x)^{n} = 1 + nx + n(n – 1) x ^{2} + ……
2!
x = 0.001 and n = 1/3
é
1 æ 1 – 1ö ´ (.001)^{2} ù
ê ç ÷ ú
\ 10(1 + 0.001)^{1/} ^{3} = 10ê1 + 1 (0.001) + 3 è 3 ø
+ ….ú = 10 é1 + 0.00033 – 1 (0.000001) + ….ù
ê 3 2!
ê
ë
ú êë 9 úû
ú
û
= 10[1.0003301] = 10.003301(Approx.)
Problem 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
(a)
g‘ = gæ1 – h ö
(b)
g‘ = gæ1 – 2h ö
(c)
g‘ = gæ1 + h ö
(d)
g‘ = gæ1 + 2h ö




ç ÷ ç ÷ ç ÷ ç ÷
è ø è ø è ø è ø
æ R ö^{2}
æ 1 ö^{2} æ
h ö2 é
h (2)(3) æ h ö^{2} ù
Solution : (b)
g‘ = gç R + h ÷
= gç 1 + h / R ÷
=ç1 + R ÷
= g ê1 + (2) +
R
2! ç R ÷
+ …….ú
è ø è
ø è ø êë
è ø úû
g‘ = gæ1 – 2h ö ( if h << R then by neglecting higher power of h .)
ç ÷


è ø
 Arithmetic progression : It is a sequence of numbers which are arranged in increasing order and having a constant difference between
Example : 1, 3, 5, 7, 9, 11, 13, …… or 2, 4, 6, 8, 10, 12, …..
In general arithmetic progression can be written as a_{0}, a_{1}, a_{2}, a_{3}, a_{4}, a_{5} …….
 n^{th} term of arithmetic progression
an = a_{0} + (n – 1)d
a_{0} = First term, n = Number of terms, d = Common difference = (a_{1} – a_{0}) or (a_{2} – a_{1}) or (a_{3} – a_{2})
 Sum of arithmetic progression Sn
= n [2a
2
+ (n – 1)d] = n [a
0 2 0
 an ]
Problem 4. Find the sum of series 7 + 10 + 13 + 16 + 19 + 22 + 25
Solution :
Sn = n [a0
2
 an
] = 7 [7 + 25] = 112 [As n = 7; a
2 0
= 7; a_{n}
= a_{7}
= 25]
 Geometric progression : It is a sequence of numbers in which every term is obtained by multiplying the previous term by a constant This constant quantity is called the common ratio.
Example : 4, 8, 16, 32, 64, 128 …… or 5, 10, 20, 40, 80, …….
In general geometric progression can be written as a, ar, ar^{2}, ar^{3}, ar^{4}, …. Here a = first term, r = common ratio
 Sum of ‘n’ terms of P.
 Sum of infinite terms of P.
S = a(1 – r ^{n} )
^{n} 1 – r
S = a (r ^{n} – 1)
^{n} r – 1
S = a
if r < 1
if r > 1 if r < 1
^{¥} 1 – r
S = a
^{¥} r – 1
if r > 1
Problem 5. Find the sum of series Q = 2q + q + q + q
+ ……
3 9 27
Solution : Above equation can be written as Q = q + éq + q + q + q +……….. ù
ëê 3 9 27 ûú
é ù
By using the formula of sum of infinite terms of G.P. Q = q + ê q ú = q + 3 q = 5 q
(5) Some common formulae of algebra
(i) (a + b)^{2} = a^{2} + b^{2} + 2ab
 (a – b)^{2} = a^{2} + b^{2} – 2ab
 (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca
 (a + b) (a – b) = a^{2} – b^{2}
(v) (a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)
(vi) (a – b)^{3} = a^{3} – b^{3} – 3ab(a – b)
(vii) (a + b)^{2} – (a – b)^{2} = 4ab
ê

ê1 –
ë
ú

ú 2 2
3 úû
(viii) (a + b)^{2} + (a – b)^{2} = 2(a^{2} + b^{2})
(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 = c
then
a + b = c + d
Trigonometry.
b d a – b
c – d
 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 q , then side opposite to q, AB, will be termed as perpendicular and BC is called the base of the
sinq = Perpendicular = AB
Hypotenuse AC
cosecq
= Hypotenuse = AC
Perpendicular AB
cosq = Base = BC sec q = Hypotenuse = AC
Hypotenuse AC Base BC
tanq
= Perpendicular = AB
Base BC
cotq =
Base Perpendicular
= BC
AB
(2) Value of trigonometric ratio of standard angles
Angle  0o  30^{o}  45^{o}  60^{o}  90^{o}  120^{o}  135^{o}  150^{o}  180^{o}  270^{o}  360^{o} 
sinq  0  1/2  1/Ö2  Ö3/2  1  Ö3/2  1/ Ö2  1/2  0  – 1  0 
cosq  1  Ö3/2  1/Ö2  1/2  0  – 1/2  – 1/Ö2  – Ö3/2  – 1  0  1 
tanq  0  1/Ö3  1  Ö3  ¥  – Ö3  – 1  – 1/Ö3  0  – ¥  0 
 Important points :
 Value of sinq or cosq lies between – 1 and +1, however tanq and cotq can have any real
 Value of secq and cosecq can’t be numerically less than
 (90^{o} – q) will lie in first quadrant (90^{o} + q) will lie in second quadrant (180^{o} – q) will lie in second quadrant (180^{o} + q) will lie in third quadrant
(270^{o} + q) and (0^{o} – q) will lie in fourth quadrant.
(4) Fundamental trigonometrical relation
(i)
tanq
= sinq
cosq
(ii) cosecq
= 1 sinq
(iii)
sec q
= 1 cosq
 cot q
= 1 tanq
 sin ^{2} q + cos ^{2} q = 1
sec ^{2} q – tan ^{2} q = 1
cosec ^{2}q – cot ^{2} q = 1
 TRatios of allied angles : The angles whose sum or difference with angle q is zero or a multiple of 90° are called angle allied to q.
(i) sin(q ) = – sinq  cos(q ) = cosq  tan(q ) = – tanq 
(ii) sin(90 ^{o} – q ) = cosq
(iii) sin(90 ^{o} + q ) = cosq (iv) sin(180^{o} – q ) = sinq (v) sin(180^{o} + q ) = – sinq (vi) sin(270^{o} – q ) = – cosq (vii) sin(270^{o} + q ) = – cosq (viii) sin(360^{o} – q ) = – sinq (ix) sin(360^{o} + q ) = sinq 
cos(90 ^{o} – q ) = sinq
cos(90 ^{o} + q ) = – sinq cos(180^{o} – q ) = – cosq cos(180^{o} + q ) = – cosq cos(270^{o} – q ) = – sinq cos(270^{o} + q ) = sinq cos(360^{o} – q ) = cosq cos(360^{o} + q ) = cosq 
tan(90^{o} – q ) = cot q
tan(90 ^{o} + q ) = – cot q tan(180^{o} – q ) = – tanq tan(180^{o} + q ) = tanq tan(270^{o} – q ) = cot q tan(270^{o} + q ) = – cot q tan(360^{o} – q ) = – tanq tan(360^{o} + q ) = tanq 
Note : @Angle ( 2np + q )
lies in first quadrant, if q in an acute angle. Similarly
(2np – q )
will lie in fourth
quadrant. Where n = 0, 1, 2, 3, 4
@ Angle (q ) is presumed always lie in fourth quadrant, whatever the value of q .
@ If parent angle is 90° or 270° then to cos ecq .
sinq change to
cosq ,
tanq
change to
cot q
and
sec q
change
@ If parent angle is 180° or 360° then no change in trigonometric function
Problem 6. Find the values of (i) cos(60^{o} )
Solution : (i) cos(60 ^{o} ) = cos 60 ^{o} = 1
2
 tan 210^{o}
 sin 300^{o}
 cos 120^{o}
(v) sin(1485^{o} )
(ii) tan(210 ^{o} ) = tan(180 ^{o} + 30 ^{o} ) = tan 30 ^{o} =
(iii) sin(300 ^{o} ) = sin(360 ^{o} – 60 ^{o} ) = – sin 60 ^{o} =
(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} = –
(6) Addition formulae
 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
Putting B = A in these formulae, we get
 sin 2A = 2 sin A cos A
 cos 2A = cos ^{2} A – sin ^{2} A = 1 – 2 sin ^{2} A = 2 cos ^{2} A – 1
tan 2A =
2 tan A
1 – tan ^{2} A
Problem 7. If
A = 60 ^{o}
then value of
sin 2A will be
(a)
3 (b)
2
1 (c)
2
(d)
Solution : (a) sin 2A = 2 sin A cos A = 2 sin 60 cos 60 = 2 ´
3 ´ 1 = 3
(7) Difference formulae
2 2 2
 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
(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
2
and
B = C – D
2
Putting these values in the above equation we get
 sin C + sin D = 2 sin C +D cos C – D
2 2
 cos C + cos D = 2 cos C +D cos C – D
2 2
 sin C – sin D = 2 cos C +D sin C – D
2 2
 cos C – cos D = 2 sin C +D sin C – D
2 2
 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)
 =
sin A
 =
sin B
c
sin C
 a ^{2} = b ^{2} + c ^{2} – 2bc cos A
 b ^{2} = c ^{2} + a ^{2} – 2ca cos B
 c ^{2} = a ^{2} + b ^{2} – 2ab cos C
 Area of a triangle ABC = ; where, S = (a + b + c) / 3
Logarithm.
Logarithm of a number with respect to a given base is the power to which the base must be raised to represent that number.
If a ^{x} = N then log a N = x
Here x is called the logarithm of N to the base a.
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.
Conversion of natural log into common log :
Important formulae of logarithm :
 log a (mn) = log am + log a n (Product formula)
log e x = 2.3026 log_{10} x
 log
æ m ö = log

n a
m – log a n
(Quotient formula)
è ø
 log am^{n} = n log a m
(Power formula)
 log am = 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
Graphs.
antilogarithm of x. If log n = x
then n = antilog of x
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.
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 xaxis and dependent variable (i.e. effect) is represented along yaxis.
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 xaxis and current on yaxis.
Some important graphs for various equations
Y Y
q
q C
y = mx + c
O X O X
y = mx
m = tanq = slope of line with xaxis
c = Positive intercept on yaxis and positive slope
Y Y
C O q
y = mx – c
X
C q y = – mx + c
O X
Negative intercept and positive slope
Y
Positive intercept and Negative slope
Y
X y^{2} = kx
O
y^{2} = – kx
O X
Symmetric parabola about positive Xaxis Symmetric parabola about negative Xaxis
Y Y
X x^{2} = ky
O
O x^{2} = – ky
X
Symmetric parabola about positive Yaxis
Y
Symmetric parabola about negative Yaxis
Y
X y = ax + bx^{2}
O
O
y = ax – bx^{2}
X
Asymmetric parabola Asymmetric parabola
Differential Calculus.
The differential coefficient or derivative of variable y with respect to variable x is defined as the instantaneous rate of change of y w.r.t. x. It is denoted by dy
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
i.e.
dy = tanq .
dx
Note : @ Actually
dy is a rate measurer.
dx
@If
dy is positive, it means y is increasing with increasing of x and viceversa.
dx
@For small change Dx
we use Dy = dy . Dx
dx
Example: (1) Instantaneous speed v = ds
dt
 Instantaneous acceleration
a = dv =
dt
d ^{2} x
dt ^{2}
 Force
F = dp
dt
 Angular velocity w = dq
dt
 Angular acceleration a = dw
dt
 Power P = dW
dt
 Torque t = dL
dt
(1)
Fundamental formulae of differentiation :
Problem 8. Differentiate the following w.r.t x
 x ^{3}
 ax ^{2} + bx + c
 2x ^{3} – e ^{x}
 6 log e ^{x} – – 7
Solution : (i) d (x ^{3} ) = 3x ^{2}
dx
(ii)
d (x)^{1} ^{/} ^{2} =
dx
1 1 1
2 (x)2 =
1 (x)^{–}^{1} ^{/} ^{2} = 1
2
(iii) d (ax ^{2} + bx + c) = a d (x^{2}) + b d (x) + d (c) = 2ax + b
dx dx dx dx
(iv)
d (2x ^{3} – e ^{x} ) = 2 d (x ^{3} ) – d (e ^{x} ) = 6 x^{2} – e^{x}
dx dx dx
(v) d (6 loge x – – 7) = 6 d (loge x) – d (x^{1} ^{/} ^{2}) – d (7) = 6 – 1
dx dx
dx dx x 2
Problem 9. Differentiate the following w.r.t. x
 sin x + cos x
 sin x + e ^{x}
Solution : (i)
d (sin x + cos x) =
dx
d (sin x) +
dx
d (cos x) = cos x – sin x dx
(ii)
d (sin x + e ^{x} ) =
dx
d (sin x) +
dx
d (e ^{x} ) = cos x + e ^{x}
dx
Problem 10. Differentiate the following w.r.t. t
 sin t ^{2}
 e ^{sin} ^{t}
 sin(wt + q )
Solution : (i)
d (sin t ^{2} ) = cos t ^{2} d (t ^{2} ) = 2t cos t ^{2}
(ii)
dt
d (e sin t ) = e sin t
dt
dt
d (sin t) = e ^{sin} ^{t} .cos t dt
(iii) d [sin(wt + q )] = cos(wt + q ). d (wt + q ) = cos(wt + q ).w
dt
Problem 11. Differentiate
x ^{2} + e ^{x}
log x + 20
dt
w.r.t. x
Solution : Let y =
x ^{2} + e ^{x} . log x + 20
dy d æ x ^{2} + e ^{x} ö

Then
ç ÷
dx dx ç log x + 20 ÷
è ø
(log x + 20) d (x ^{2} + e ^{x} ) – (x ^{2} + e ^{x} ) d (log x + 20)
= dx dx
(log x + 20)^{2}
(log x + 20)(2x + e ^{x} ) – (x ^{2} + e ^{x} )æ 1 + 0 ö
ç x ÷
= è ø
(log x + 20)^{2}
 Maxima and minima : If a quantity y depends on another quantity x in a manner shown in It becomes maximum at x_{1} and minimum at x_{2}.
At these points the tangent to the curve is parallel to Xaxis and hence its
slope is tanq = 0. But the slope of the curve equals the rate of change
dy
dy . Thus,
dx
at a maximum or minimum
dx = 0
Just before the maximum the slope is positive, at the maximum it is zero and just after the maximum it is
negative. Thus
dy decreases at a maximum and hence the rate of change of
dx
dy is negative at a maximum. i.e.,
dx
d æ dy ö < 0 at a maximum.
ç ÷
dx è dx ø
Hence the condition of maxima :
dy = 0
dx
and
d ^{2}y < 0
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 æ dy ö > 0
ç ÷
dx è dx ø
Hence the condition of minima :
dy = 0
dx
and
d2y > 0 . (Second derivative test)
dx^{2}
Problem 12. The height reached in time t by a particle thrown upward with a speed u is given by h = ut – 1 gt ^{2} .
2
Find the
time taken in reaching the maximum height.
Solution : For maximum height
dh = 0
d [ut – 1 gt ^{2} ] = u – 2gt = 0
\t = u
dt dt 2 2 g
Problem 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 m^{2}.
3
What will be the rate of increase of area at t = 10 sec .
Solution : Rate of increase of area
dA = d (3t ^{2} + t + 2) = 6t + 1
dt dt 3 3
æ dA ö
= 6 ´ 10 + 1 = 181 m2 .
ç dt ÷
3 3 sec
è øt =10 sec
Problem 14. The radius of an air bubble is increasing at the rate of when the radius is 1 cm.
Solution : Volume of the spherical bubble V = 4 pR^{3}
3
1 cm / sec . Determine the rate of increase in its volume
2
Differentiating both sides w.r.t. time
dV = d æ 4 pR^{3} ö = 4 p .3R ^{2}. dR = 4pR ^{2} dR



dt dt ç 3 ÷ dt dt
at R = 1cm ,
dV = 4p ´ (1)^{2} ´ 1 = 2p
cm^{3} / sec .
[Given
dR = 1 cm / sec ]
dt 2 dt 2
Problem 15. Find the angle of tangent drawn to the curve y = 3x ^{2} – 7x + 5 at the point (1, 1) with the x– axis.
Solution :
y = 3x ^{2} – 7x + 5
Slope of tangent = dy = 6x – 7
dx
at (1, 1)
Integral Calculus.
dy = 1
dx
\ tanq = 1 Þ q = 135^{o}.
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.
(1) Fundamental formulae of integration :
ò x^{n}dx = x , provided n ¹ – 1 ò sec 2 x dx = tan x
n+1 n + 1 

0+1
ò dx = ò x ^{0} dx = x = x 0 + 1 
ò cos ec ^{2} x dx = – cot x  
ò(u + v) dx = òu dx + òv dx ò sec x tan x dx = sec x
ò cu dx = còu dx ò cosec x cot x dx = cosec x where c is a constant and u is a function of x. n+1 n+1 n+1 ò cx^{n}dx = c x ò(ax + b)^{n} dx = (ax + b) = (ax + b) n + 1 (n + 1) d (ax + b) a(n + 1) dx a log (ax + b) ò x ^{–}^{1}dx = ò dx = log e x ò a dx = ^{e} = log e (ax + b) x (ax + b) d (ax + b) dx ò e x dx = e x ò e ax +b dx = e ax +b = e ax +b d (ax + b) a dx x cx +d cx +d ò a ^{x} dx = a ò a^{cx}^{+}^{d} dx = a = a log e a log e a d (cx + d) c log e a dx ò sin x dx = – cos x ò sec 2 (ax + b) dx = tan(ax + b) = tan(ax + b) d (ax + b) a dx ò sin nx dx = – cos nx ò cosec 2 (ax + b) dx = – cot (ax + b) = – cot (ax + v) n d (ax + b) a dx ò cos x dx = sin x ò sec (ax + b) tan(ax + b) dx = sec (ax + b) = sec (ax + b) d (ax + b) a dx ò cos nx dx = sin nx ò cosec (ax + b) cot (ax + b) dx = –cosec (ax + b) = –cosec (ax + b) n d (ax + b) a dx 
 Method of integration : Sometimes, we come across some functions which cannot be integrated directly by using the standard In such cases, the integral of a function can be obtained by using one or more of the following methods.
 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
 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
Problem 16. Integrate the following w.r.t. x
òuvdx = uò
vdx –
é du ´

ë dx
vdxù dx


û
 x^{1/2} (ii) cot ^{2} x
(iii)
1
 – sin x
1 / 2+1
Solution : (i) ò x^{1/} ^{2}dx = x = 2 (x ^{3} ^{/} ^{2} )
1 + 1 3
2
(ii) ò
cot ^{2} x dx = ò (cosec ^{2} x – 1)dx = ò
cosec ^{2} x dx – ò
dx = – cot x – x
1 dx =
æ 1 ´ 1 + sin x ö. dx =
1 + sin x
dx =
1 + sin x dx
ò 1 – sin x
ò ç 1 – sin x 1 + sin x ÷
ò 1 – sin^{2} x
ò cos ^{2} x cos ^{2} x


= ò(sec ^{2} x + tan x sec x)dx = tan x + sec x.
 Definite integrals : When a function is integrated between definite limits, the integral is called definite For example,
b b
f(x)dx is definite integral of f(x) between the limits a and b and is written as f(x)dx = F(x)^{b} = F(b) – F(a)
òa òa a
Here a is called the lower limit and b is called the upper limit of integration.
b
Geometrically òa f(x) dx equals to area of curve F(x) between the limits a and b.

Problem 17. Evaluate 6 (2x ^{2} + 3x + 5)dx
0


6 6 6
6 é 2x 3 ù ^{6}
é 3x 2 ù ^{6}

Solution :
(2x ^{2} + 3x + 5)dx =
0 0
2x ^{2}dx +
ò0 3x dx +
5 dx = ê
0 êë
ú + ê
3 úû 0 êë
ú
2 úû 0
+ [5x]6 = 144 + 54 + 30 = 228.

Problem 18. Integrate the following
(i)
^{2} 1 dx
(ii)
p / 2
cos x dx
(iii)
r2 Kq1q2 .dr

^{p} ^{/} ^{4} tan ^{2} x dx
ò0
2 1 2
ò0
é x 1 / 2 ù 2 2
òr r 2 ò0
Solution : (i) ò
dx = ò
x 1/ 2 dx = ê
ú = [2x^{1}^{/} ^{2} ]0 = 2
0 0 êë 1 / 2 úû 0
p / 2
 òcos x dx
0
= [sin x]p / 2

= sin p = 1 2
r2 q q

2 1 æ
1 ö^{r}2
é 1 1 ù
é 1 1 ù
 òk
1 2 2 dx = k q1q2 ò
2 dx = kq_{1}q_{2} ç ÷
= –kq1q2 ê
– ú = kq1q2 ê – ú

r1 r
p / 4
r
1
p / 4
è r ø_{r}1
p / 4
p / 4
ë r2
r1 û
p
ë r1
r2 û
ò tan ^{2} x dx = ò(sec ^{2} x – 1)dx = [tan x] 0
– [x] 0
= 1 – 4
0 o
General Formulae for Area and Volume.
 Area of square = (side)^{2}
 Area of rectangle = length ´ breadth
 Area of triangle = 1 ´ base ´ height
2
 Area enclosed by a circle = p r ^{2} ; where r is radius
 Surface area of sphere = 4p r ^{2}
 Surface area of cube = 6 L^{2} ; where L is a side of cube
 Surface area of cuboid = 2[L ´ b + b ´ h + h ´ L]; where L= length, b = breadth, h = height
 Area of curved surface of cylinder = 2p rl ; where r = radius, l = length of cylinder
 Volume of cube = L^{3}
 Volume of cuboid = L ´ b ´ h
 Volume of sphere = 4 p r ^{3}
3
 Volume of cylinder = p r ^{2}l
 Volume of cone = 1 p r ^{2}h
2
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.
Types of Vector.
 Equal vectors : Two vectors A and B are said to be equal when they have equal magnitudes and same
 Parallel vector : Two vectors A and B are said to be parallel when
 Both have same
 One vector is scalar (positive) nonzero multiple of another
 Antiparallel vectors : Two vectors A and B are said to be antiparallel when
 Both have opposite
 One vector is scalar nonzero negative multiple of another
 Collinear vectors : When the vectors under consideration can share the same support or have a common support then the considered vectors are
 Zero vector (0) : A vector having zero magnitude and arbitrary direction (not known to us) is a zero
 Unit vector : A vector divided by its magnitude is a unit vector. Unit vector for A is Aˆ (read as A cap / A hat).
Since,
Aˆ = A
A
Þ A = A Aˆ .
Thus, we can say that unit vector gives us the direction.
 Orthogonal unit vectors : ˆi , ˆj and kˆare 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
ˆi =
x , ˆj =
x
y , kˆ = z
y z
\ x = xˆi ,
y = yˆj ,
z = zkˆ
 Polar vectors : These have starting point or point of application . Example displacement and force etc.
 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
 Coplanar vector : Three (or more) vectors are called coplanar vector if they lie in the same plane. Two (free) vectors are always
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
Q OB = OA + AB
(1) Magnitude of resultant vector
In D ABN cosq
= AN \ AN = B cosq
B
sinq
= BN
B
\ BN = B sinq
In DOBN, we have OB^{2} = ON ^{2} + BN ^{2}
Þ R ^{2} = (A + B cosq )^{2} + (B sinq )^{2}
Þ R ^{2} = A^{2} + B^{2} cos ^{2} q + 2AB cosq + B^{2} sin ^{2} q
Þ R ^{2} = A^{2} + B^{2} (cos ^{2} q + sin ^{2} q ) + 2AB cosq
Þ R ^{2} = A^{2} + B^{2} + 2AB cosq
Þ R =
 Direction of resultant vectors : If q is angle between A and
 A + B=
B, then
If R makes an angle a with A, then in DOBN, then
tana = BN
ON
= BN OA + AN
tana = B sinq
A + B cosq
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
Since, R ^{2} = ON ^{2} + CN ^{2}
Þ R ^{2} = (OA + AN)^{2} + CN ^{2}
Þ R ^{2}
= A^{2}
 B^{2}
 2AB cosq
\ R = R= A + B=
Special cases : R = A + B when q = 0^{o}
R = A – B when q = 180^{o}
R = when q = 90^{o}
 Direction
tan b = CN
ON
= B sinq
A + B cosq
Polygon Law of Vector Addition.
If a number of non zero vectors are represented by the (n – 1) sides of an nsided polygon then the resultant is given by the closing side or the n^{th} side of the polygon taken in opposite order. So,
R = A + B + C + D + E
OA + AB + BC + CD + DE = OE
Note : @ Resultant of two unequal vectors can not be zero.
@Resultant of three coplanar vectors may or may not be zero
@Resultant of three non co planar vectors can not be zero.
Subtraction of Vectors.
Since,
A – B = A + (B)
and  A + B=
Þ  A – B= A^{2} + B^{2} + 2AB cos (180^{o} – q )
Since, cos (180 – q ) = – cosq
Þ  A – B= A ^{2} + B ^{2} – 2AB cosq
tana_{1}
= B sinq
A + B cosq
and
tana _{2} =
B sin (180 – q )
A + B cos (180 – q )
But
sin(180 – q ) = sinq
and cos(180 – q ) = – cosq
Þ tana _{2}
= B sinq
A – B cosq
Problem 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)
Solution : (c) Net movement along xdirection S_{x}
= (6 – 4) cos 45° ˆi
= 2 ´ 1 =
 km
Net movement along ydirection S_{y}
= (6 + 4) sin 45° ˆj
= 10 ´ 1 = 5
2 km
Net movement from starting point  s = = = km
Angle which makes with the east direction tanq = Y – component = 5
X – component
\ q = tan^{–}^{1}(5)
Problem 20. 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
(a) 0°, 180° and 90° (b) 0°, 90° and 180° (c) 0°, 90° and 90° (d) 180°, 0° and 90°
Solution : (a) For 17 N both the vector should be parallel i.e. angle between them should be zero.
For 7 N both the vectors should be antiparallel i.e. angle between them should be 180°
For 13 N both the vectors should be perpendicular to each other i.e. angle between them should be 90°
Problem 21. Given that
A + B + C = 0 out of three vectors two are equal in magnitude and the magnitude of third vector is
times that of either of the two having equal magnitude. Then the angles between vectors are given by (a) 30°, 60°, 90° (b) 45°, 45°, 90° (c) 45°, 60°, 90° (d) 90°, 135°, 135°
Solution : (d) From polygon law, three vectors having summation zero should form a closed polygon. (Triangle) since the
two vectors are having same magnitude and the third vector is 2 times that of either of two having equal magnitude. i.e. the triangle should be right angled triangle
Angle between A and B, a = 90º Angle between B and C, b = 135º Angle between A and C, g = 135º
Problem 22. If
A = 4ˆi – 3ˆj
and B = 6ˆi + 8ˆj
then magnitude and direction of
A + B will be
(a)
5, tan^{–}^{1}(3 / 4)
(b)
5 5, tan^{–}^{1}(1 / 2)
(c)
10, tan^{–}^{1}(5)
(d)
25, tan^{–}^{1}(3 / 4)
Solution : (b)
A + B = 4ˆi – 3ˆj + 6ˆi + 8ˆj = 10ˆi + 5ˆj
 A + B =
tanq = 5 = 1
= 5
Þ q = tan^{–}^{1} æ 1 ö
10 2
ç 2 ÷
è ø
Problem 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 (b)
Solution : (d) From fig.
20 m/s N–W (c) 40 m/s S–W (d)
20 m/s S–W
v_{1} = 20ˆj and v _{2} = 20ˆi
Dv = v _{2} – v_{1} = 20(ˆi + ˆj)
Dv = 20
and direction q = tan^{–}^{1}(1) = 45° i.e. S–W
Problem 24. If the sum of two unit vectors is a unit vector, then magnitude of difference is [CPMT 1995; CBSE PMT 1989]
 (b)
(c) 1 /
(d)
Solution : (b) Let nˆ1 and nˆ2 are the two unit vectors, then the sum is



ns = nˆ1 + nˆ2 or n^{2} = n^{2} + n^{2} + 2n_{1}n_{2} cosq
= 1 + 1 + 2 cosq
Since it is given that n_{s} is also a unit vector, therefore 1 = 1 + 1 + 2 cosq
or cosq = – 1
2
or q = 120°
Now the difference vector is nd = n1 – n2 or n^{2} = n^{2} + n^{2} – 2n1n2 cos q
= 1 + 1 – 2 cos(120°)
d 1 2

\ n^{2} = 2 – 2(1 / 2) = 2 + 1 = 3 Þ nd =
Problem 25. 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 (b) 14, 4 (c) 5, 13 (d) 10, 8
Solution : (c) Let P be the smaller force and Q be the greater force then according to problem –
P + Q = 18………………………………………………………………………………………….. (i)
R = = 12……………………………………………………………….. (ii)
tan f = Q sinq = tan 90 = ¥
P + Q cosq
By solving (i), (ii) and (iii) we will get
\ P + Q cosq = 0
P = 5, and Q = 13
…….(iii)
Problem 26. Two forces
F_{1} = 1 N and
F_{2} = 2 N act along the lines x = 0 and y = 0 respectively. Then the resultant of
forces would be
(a)
ˆi + 2ˆj
(b)
ˆi + ˆj
(c)
3ˆi + 2ˆj
(d)
2ˆi + ˆj
Solution : (d) x = 0 means yaxis Þ
F 1 = ˆj
y = 0 means xaxis
Þ F 2 = 2ˆi
so resultant
F = F 1 + F 2 = 2ˆi + ˆj
Problem 27. Let
A = 2ˆi + ˆj, B = 3ˆj – kˆ and C = 6ˆi – 2kˆ value of
A – 2B + 3C would be
(a)
20ˆi + 5ˆj + 4kˆ
(b)
20ˆi – 5ˆj – 4kˆ
(c)
4ˆi + 5ˆj + 20kˆ
(d)
5ˆi + 4ˆj + 10kˆ
Solution : (b)
A – 2B + 3C = (2ˆi + ˆj) – 2(3ˆj – kˆ) + 3(6ˆi – 2kˆ)
= 2ˆi + ˆj – 6ˆj + 2kˆ + 18ˆi – 6kˆ
= 20ˆi – 5ˆj – 4kˆ
Problem 28. A vector a
respectively
is turned without a change in its length through a small angle dq . The value of Da  and Da
are
(a)
0, a dq
(b)
a dq , 0
(c) 0, 0 (d) None of these
Solution : (b) From the figure OA= a and OB= a
Also from triangle rule OB – OA = AB = Da
Þ Da = AB
Using angle =
arc radius
Þ AB = a . dq
So Da = a dq
Da means change in magnitude of vector i.e.  OB   OA  Þ
a – a = 0
So Da = 0
Problem 29. An object of m kg with speed of v m/s strikes a wall at an angle q and rebounds at the same speed and same angle. The magnitude of the change in momentum of the object will be
(a)
2mv cosq
(b)
2mv sinq
 0 (d)
2 mv
Solution : (a)
P1 = mv sinq ˆi – mv cosq ˆj
and P 2 = mv sinq ˆi + mv cosq ˆj
So change in momentum DP = P 2 – P1 = 2 mv cosq ˆj
DP = 2mv cosq
Resolution of Vector Into Components.
Consider a vector R in x–y plane as shown in fig. If we draw orthogonal vectors R x
and
Ry along x and
y axes respectively, by law of vector addition,
R = Rx + R _{y}
Now as for any vector
A = A nˆ
so, Rx
= ˆi Rx
and Ry
= ˆjRy
so R = ˆi Rx + ˆjRy
…..(i)
But from fig
Rx = R cosq
…..(ii) and
Ry = R sinq
…..(iii)
Since R and q are usually known, Equation (ii) and (iii) give the magnitude of the components of R along x and yaxes respectively.
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 –
 The magnitude of the vector R is obtained by squaring and adding equation (ii) and (iii), e.
R =
 The direction of the vector R is obtained by dividing equation (iii) by (ii), e.
tanq
= (Ry / Rx )
or q = tan ^{1}(Ry / Rx )
Rectangular Components of 3D Vector.
Þ cos g
= Rz = Rz = n R
where l, m, n are called Direction Cosines of the vector R
R ^{2} + R ^{2} + R ^{2}
l ^{2} + m^{2} + n^{2} = cos ^{2} a + cos ^{2} b + cos ^{2} g = ^{x} ^{y} ^{z} = 1
R ^{2} + R ^{2} + R ^{2}
x y z
Note : @When a point P have coordinate (x, y, z) then its position vector OP = xˆi + yˆj + zkˆ
@ When a particle moves from point (x_{1}, y_{1}, z_{1}) to (x_{2}, y_{2}, z_{2}) then its displacement vector
^{®} ˆ ˆ ˆ
r = (x 2 – x1 )i + (y2 – y1 ) j + (z 2 – z1 )k
Problem 30. If a particle moves 5 m in +x– direction. The displacement of the particle will be
(a) 5 j (b) 5 i (c) – 5 j (d) 5 k Solution : (b) Magnitude of vector = 5
Unit vector in +x direction is ˆi So displacement = 5 ˆi
Problem 31. Position of a particle in a rectangularcoordinate system is (3, 2, 5). Then its position vector will be
(a)
3ˆi + 5ˆj + 2kˆ
(b)
3ˆi + 2ˆj + 5kˆ
(c)
5ˆi + 3ˆj + 2kˆ
 None of these
Solution : (b) If a point have coordinate (x, y, z) then its position vector OP = xˆi + yˆj + zkˆ.
Problem 32. If a particle moves from point P (2,3,5) to point Q (3,4,5). Its displacement vector be
(a)
ˆi + ˆj + 10kˆ
(b)
ˆi + ˆj + 5kˆ
(c)
ˆi + ˆj
(d)
2ˆi + 4ˆj + 6kˆ
Solution : (c) Displacement vector r = Dxˆi + Dyˆj + Dzkˆ = (3 – 2)ˆi + (4 – 3)ˆj + (5 – 5)kˆ = ˆi + ˆj
Problem 33. A force of 5 N acts on a particle along a direction making an angle of 60° with vertical. Its vertical component be (a) 10 N (b) 3 N (c) 4 N (d) 5.2 N
Solution : (d) The component of force in vertical direction will be F cosq = F cos 60°
= 5 ´ 1 = 2.5 N
2
Problem 34. If
A = 3ˆi + 4ˆj
and B = 7ˆi + 24ˆj, the vector having the same magnitude as B and parallel to A is
(a)
5ˆi + 20ˆj
(b)
15ˆi + 10ˆj
(c)
20ˆi + 15ˆj
(d)
15ˆi + 20ˆj
Solution : (d)
 B=
= = 25
Unit vector in the direction of A will be
Aˆ = 3ˆi + 4ˆj
5
æ 3ˆi + 4ˆj ö ˆ ˆ
So required vector = 25ç ÷ = 15i + 20 j

ç ÷
è ø
Problem 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 (b) A (c) A
 3
A
Solution : (a) Let the components of A makes angles a, b and g with x, y and z axis respectively then a = b = g
cos ^{2} a + cos ^{2} b + cos ^{2} g = 1
Þ 3 cos ^{2} a = 1 Þ cosa = 1
\ Ax = Ay = Az = A cosa = A
Problem 36. If A = 2iˆ+ 4ˆj – 5kˆ the direction of cosines of the vector A are

 2 , 4 and
Solution : (a)
 A =
\ cosa = 2 ,
=
cos b = 4 ,
cos g = –5
Problem 37. The vector that must be added to the vector ˆi – 3ˆj + 2kˆ and 3ˆi + 6ˆj – 7kˆ so that the resultant vector is a unit vector along the yaxis is
(a)
4ˆi + 2ˆj + 5kˆ
– 4ˆi – 2ˆj + 5kˆ
3ˆi + 4ˆj + 5kˆ
 Null vector
Solution : (b) Unit vector along y axis = ˆj
Scalar Product of Two Vectors.
so the required vector = ˆj – [(ˆi – 3ˆj + 2kˆ) + (3ˆi + 6ˆj – 7kˆ)] = – 4ˆi – 2ˆj + 5kˆ
 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
Thus if there are two vectors A and B having angle q between them, then their scalar product written as
 A. B is defined as
 A. B
= AB cosq
 Properties : (i) It is always a scalar which is positive if angle between the vectors is acute (e., < 90°) and negative if angle between them is obtuse (i.e. 90°<q < 180°).
 It is commutative, e. A. B = B. A
 It is distributive, e. A.(B + C) = A. B + A. C
 As by definition A. B = AB cosq
The angle between the vectors q = cos^{1} é A. B ù

AB
ëê úû
 Scalar product of two vectors will be maximum when cosq = max = 1, e. q = 0^{o} , i.e., vectors are parallel
(A. B)max
= AB
 Scalar product of two vectors will be minimum when cosq = min = 0, e. q = 90^{o}
(A . B)min = 0
i.e., if the scalar product of two nonzero vectors vanishes the vectors are orthogonal.
 The scalar product of a vector by itself is termed as self dot product and is given by
(A)^{2} = A. A = AA cosq = A^{2}
i.e., A =
 In case of unit vector nˆ
nˆ.nˆ = 1 ´ 1 ´ cos 0 = 1
so nˆ.nˆ= ˆi .ˆi = ˆj . ˆj = kˆ.kˆ= 1
 In case of orthogonal unit vectors ˆi, ˆj
and kˆ,
ˆi . ˆj = ˆj .kˆ= kˆ.ˆi = 1 ´ 1cos 90 = 0
 In terms of components
 B =(iAx + jAy + kAz ).(iBx + jBy + kBz ) = [Ax Bx + Ay By + AZ Bz ]
 Example : (i) Work W : In physics for constant force work is defined as, W = Fs cosq………………….. (i)
But by definition of scalar product of two vectors,
 F. s = Fs cosq
…….(ii)
So from eq^{n} (i) and (ii) W = F.s
 Power P :
i.e. work is the scalar product of force with displacement.
As W = F . s or
dW = F . ds
[As F is constant]
dt dt
or P = F .v
i.e., power is the scalar product of force with velocity.
éAs dW = P and ds = vù
 Magnetic Flux f :
Magnetic flux through an area is given by df = Bds cosq
ê
êë dt
……(i)
ú
dt úû
But by definition of scalar product So from eq^{n} (i) and (ii) we have
B . d s = Bds cos q
……(ii)
df = B. d s
or f = ò
 ds
 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 :
UE = – p . E and UB = –M . B
Problem 38.
A = 2ˆi + 4ˆj + 4kˆ and B = 4ˆi + 2ˆj – 4kˆ are two vectors. The angle between them will be (a) 0° (b) 45° (c) 60° (d) 90°
Solution : (d)
cosq =
 A. B
A.B
= a1b1 + a2b2 + a3b3
A. B
= 2 ´ 4 + 4 ´ 2 – 4 ´ 4 = 0
 A. B
\ q = cos ^{–}^{1}(0°) Þ q = 90°
Problem 39. If two vectors 2ˆi + 3ˆj – kˆ and – 4ˆi – 6ˆj – lkˆ are parallel to each other then value of l be (a) 0 (b) 2 (c) 3 (d) 4
Solution : (c) Let
A = 2ˆi + 3ˆj – kˆ and B = 4ˆi – 6ˆj + lkˆ
A and B are parallel to each other
a1 = a2 = a3
i.e. 2 = 3 = –1 Þ l = 2 .
b1 b2 b3 – 4 – 6 l
Problem 40. In above example if vectors are perpendicular to each other then value of l be
(a) 25 (b) 26 (c) – 26 (d) – 25
Solution : (c) If A and B are perpendicular to each other then
 A. B = 0
Þ a1b1 + a2b2 + a3b3 = 0
So, 2(4) + 3(6) + (1)(l) = 0 Þ l = 26
Problem 41. If A = 2ˆi + 3ˆj – kˆ and B = ˆi + 3ˆj + 4kˆ then projection of A on B will be
(a)
3 (b)
3 (c)
(d)
Solution : (b)
A= = =
 B = = =
A . B = 2(1) + 3 ´ 3 + (1)(4) = 3
The projection of
A on B
= A . B =
 B 
Problem 42. 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
Solution : (d)
W = F . S = FS cosq
= 50 ´ 10 ´ cos 60° = 50 ´ 10 ´ 1 = 250 J.
2
Problem 43. A particle moves from position
3ˆi + 2ˆj – 6kˆ to 14ˆi + 13ˆj + 9kˆ due to a uniform force of
4ˆi + ˆj + 3kˆ N.
If the
Solution : (a)
displacement in meters then work done will be
(a) 100 J (b) 200 J (c) 300 J (d) 250 J
S = r2 – r1
W = F . S
= (4ˆi + ˆj + 3kˆ).(11ˆi + 11ˆj + 15kˆ) = (4 ´ 11 + 1 ´ 11 + 3 ´ 15) = 100 J.
Problem 44. If for two vector A and B , sum (A + B) is perpendicular to the difference (A – B). The ratio of their magnitude is
 1 (b) 2 (c) 3 (d) None of these
Solution : (a) (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
Problem 45. A force F = –K(yˆi + xˆj) (where K is a positive constant) acts on a particle moving in the x–y plane. Starting
from the origin, the particle is taken along the positive x– axis to the point (a, 0) and then parallel to the yaxis to the point (a, a). The total work done by the forces F on the particle is [IITJEE 1998]
(a)
 2 Ka ^{2}
2 Ka ^{2}
 Ka ^{2}
Ka ^{2}
Solution : (c) For motion of the particle form (0, 0) to (a, 0)
F = –K(0ˆi + a ˆj) Þ F = –Kaˆj
Displacement
r = (aˆi + 0 ˆj) – (0ˆi + 0 ˆj) = aˆi
So work done from (0, 0) to (a, 0) is given by W = F . r
For motion (a, 0) to (a, a)
= –Kaˆj .aˆi = 0
F = –K(aˆi + aˆj) and displacement r = (aˆi + aˆj) – (aˆi + 0ˆj) = aˆj
So work done from (a, 0) to (a, a) W = F .r
So total work done = –Ka ^{2}
Vector Product of Two Vector.
= –K(aˆi + aˆj).aˆj = –Ka ^{2}
 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
C = A ´ B
Thus, if A and B are two vectors, then their vector product written as
C = A ´ B = AB sinq nˆ
A ´ B is a vector C defined by
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 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
(2) Properties :
A ´ B i.e. C
 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.
 Vector product of two vectors is not commutative, e., Here it is worthy to note that
 A ´ B= B ´ A= AB sinq
A ´ B ¹ B ´ A [but
= –B ´ A]
i.e., in case of vector
A ´ B and
B ´ A magnitudes are equal but directions are opposite.
 The vector product is distributive when the order of the vectors is strictly maintained, e.
A ´ (B + C) = A ´ B + A ´ C
 As by definition of vector product of two vectors A ´ B = AB sinq nˆ
So  A ´ B= AB sinq
i.e.,
q = sin1 é A ´ Bù
ê ú
êë A Búû
 The vector product of two vectors will be maximum when
[A ´ B]max = AB nˆ
i.e., vector product is maximum if the vectors are orthogonal.
sinq = max = 1, i.e., q = 90^{o}
 The vector product of two non zero vectors will be minimum when sinq = minimum = 0, e., q = 0^{o} or 180^{o}
[A ´ B]min = 0
i.e. if the vector product of two nonzero vectors vanishes, the vectors are collinear.
 The self cross product, e., product of a vector by itself vanishes, i.e., is null vector
A ´ A = AA sin 0^{o} nˆ = 0
 In case of unit vector nˆ´ nˆ= 0 so that ˆi ´ ˆi = ˆj ´ ˆj = kˆ´ kˆ = 0
 In case of orthogonal unit vectors, ˆi, ˆj,kˆin accordance with right hand screw rule :
ˆi ´ ˆj = kˆ,
ˆj ´ kˆ = ˆi
and
kˆ´ ˆi = ˆj
And as cross product is not commutative,
ˆj ´ ˆi = –kˆ
kˆ´ ˆj = ˆi
and
ˆi ˆj
ˆi ´ kˆ = ˆj
kˆ
 In terms of components
A ´ B = Ax
Bx
Ay Az
By Bz
= ˆi(Ay Bz – Az By ) + ˆj(Az Bx – Ax Bz ) + kˆ(Ax By – Ay Bx )
 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 It is well – established in physics that :
 Torque t = r ´ F
 Angular momentum
 Velocity v = w ´ r
L = r ´ p
 Force on a charged particle q moving with velocity v in a magnetic field B is given by F = q(v ´ B)
 Torque on a dipole in a field t _{E} = p ´ E and t _{B} = M ´ B
Problem 46. If A = 3iˆ + ˆj + 2kˆ and B = 2iˆ – 2ˆj + 4kˆ then value of  A ´ B  will be
(a)
8 (b)
8 (c)
8 (d) 5
Solution : (b)
ˆi ˆj
A ´ B = 3 1
2 – 2
kˆ
2 = (1 ´ 4 – 2 ´ 2)ˆi + (2 ´ 2 – 4 ´ 3)ˆj + (3 ´ 2 – 1´ 2)kˆ
4
= 8ˆi – 8ˆj – 8kˆ
\ Magnitude of A ´ B = A ´ B= = 8
Problem 47. In above example a unit vector perpendicular to both
A and B will be
(a)
+ 1 (ˆi – ˆj – kˆ)
(b)
– 1 (ˆi – ˆj – kˆ)
(c) Both (a) and (b) (d) None of these
Solution : (c)
nˆ = A´ B =
 A´ B
= 1 (ˆi – ˆj – kˆ)
There are two unit vectors perpendicular to both A and B they are nˆ = ± 1 (ˆi – ˆj – kˆ)
Problem 48. The vectors from origin to the points A and B are of the triangle OAB be
A = 3ˆi – 6ˆj + 2kˆ and
B = 2ˆi + ˆj – 2kˆ respectively. The area
(a)
5 17
2
sq.unit (b)
2 17
5
sq.unit (c)
3 17
5
sq.unit (d)
5 17
3
sq.unit
Solution : (a) Given OA = a = 3ˆi – 6ˆj + 2kˆ and OB = b = 2ˆi + ˆj – 2kˆ
ˆi ˆj
\ (a ´ b) = 3 – 6
2 1
kˆ
2 = (12 – 2)ˆi + (4 + 6)ˆj + (3 + 12)kˆ
– 2
= 10ˆi + 10ˆj + 15kˆ Þ  a ´ b =
= = 5
Area of DOAB = 1  a ´ b = 5 17
sq.unit.
2
Problem 49. The angle between the vectors
2
A and B is q . The value of the triple product
A.(B ´ A ) is
(a)
Solution : (b) Let
A^{2} B
A.(B´ A) = A.C
(b) Zero (c)
A^{2} B sinq
(d)
A^{2} B cosq
Here C = B ´ A Which is perpendicular to both vector A and B \ A . C = 0
Problem 50. The torque of the force F = (2ˆi – 3ˆj + 4kˆ)N
acting at the point r
= (3ˆi + 2ˆj + 3kˆ) m about the origin be
[CBSE PMT 1995]
(a)
6ˆi – 6ˆj + 12kˆ
(b)
17ˆi – 6ˆj – 13kˆ
(c)
– 6ˆi + 6ˆj – 12kˆ
(d)
– 17ˆi + 6ˆj + 13kˆ
Solution : (b)
t = r ´ F
ˆi ˆj
= 3 2
2 – 3
kˆ
3 = [(2 ´ 4) – (3 ´ 3)] ˆi + [(2 ´ 3) – (3 ´ 4)]ˆj + [(3 ´ 3) – (2 ´ 2)]kˆ = 17ˆi – 6 ˆj – 13 kˆ
4
Problem 51. If
A ´ B = C, then which of the following statements is wrong
(a)
C ^ A
(b)
C ^ B
(c)
C ^(A + B)
(d)
C ^(A ´ B)
Solution : (d) 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.
Problem 52. If a particle of mass m is moving with constant velocity v parallel to xaxis in x–y plane as shown in fig. Its angular momentum with respect to origin at any time t will be
(a)
mvb kˆ
(b)
 mvb kˆ
(c)
mvbˆi
(d)
mvˆi
Solution : (b) We know that, Angular momentum
ˆi ˆj kˆ
L = r ´ p
in terms of component becomes L = x
px
y z
py pz
As motion is in x–y plane (z = 0 and Pz = 0 ), so L = k (xpy – ypx )
Here x = vt, y = b,
px = mv
and py = 0
\ L
Lami’s Theorem.
= k [vt ´ 0 – b mv] = –mvb kˆ
In any
D A BC
with sides a, b, c
sina
a
= sin b
b
= sin g
c
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.
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
a + b + c = 0
Þ a + b = –c
[All three sides are taken in order]…………………………………. (i)
…..(ii)
Premultiplying both sides by a
a ´ (a + b) = –a ´ c Þ
0 + a ´ b = –a ´ c Þ
a ´ b = c ´ a
…..(iii)
Premultiplying both sides of (ii) by b
b ´ (a + b) = – b ´ c
Þ b ´ a + b ´ b = –b ´ c Þ
– a ´ b = –b ´ c Þ
a ´ b = b ´ c
…..(iv)
From (iii) and (iv), we get Taking magnitude, we get
a ´ b = b ´ c = c ´ a
a ´ b =b ´ c =c ´ a 
Þ ab sin(180 – g ) = bc sin(180 – a) = ca sin(180 – b )
Þ ab sin g = bc sina = ca sin b
Dividing through out by abc, we have Þ
Relative Velocity.
sina
a
= sin b
b
= sin g
c
 Introduction : When we consider the motion of a particle, we assume a fixed point relative to which the given particle is in motion. For example, if we say that water is flowing or wind is blowing or a person is running with a speed v, we mean that these all are relative to the earth (which we have assumed to be fixed).
Now to find the velocity of a moving object relative to another moving object, consider a particle P whose
®
position relative to frame S is rPS
®
®
while relative to S¢ is rPS¢ . If the position of frames S¢ relative to S at any time is
r S¢S
then from fig.
® ® ®
rPS = rPS¢ + rS¢S
Differentiating this equation with respect to time
®
drPS
dt
®
= drPS¢
dt
®
 drS¢S
dt
® ® ® ® ®
or vPS = vPS¢ + vS¢S
[as
v = dr /dt ]
® ® ®
or vPS¢ = vPS – vS¢S
®
 General Formula : The relative velocity of a particle P_{1} moving with velocity v_{1} with respect to another
® ® ® ®
particle P_{2} moving with velocity v_{2} is given by, v r12 = v_{1} – v_{2}
 If both the particles are moving in the same direction then :


u = u – u
12
 If the two particles are moving in the opposite direction, then :


u = u + u
12
 If the two particles are moving in the mutually perpendicular directions, then:

u =
12
® ®
 If the angle between and
be q, then u
= [u ^{2} + u ^{2} – 2u u
cosq ]1 / 2 .
u_{1} u 2
r12 1 2 1 2
®
 Relative velocity of satellite : If a satellite is moving in equatorial plane with velocity vs
®
and a point
on the surface of earth with
®
ve relative to the centre of earth, the velocity of satellite relative to the surface of earth
® ®
vse
= v_{s} – v _{e}
So if the satellite moves form west to east (in the direction of rotation of earth on its axis) its velocity relative to
earth’s surface will be vse = vs – ve
And if the satellite moves from east to west, i.e., opposite to the motion of earth, vse = vs – (ve ) = vs + ve
®
 Relative velocity of rain : If rain is falling vertically with a velocity vR
®
and an observer is moving
horizontally with speed vM the velocity of rain relative to observer
® ® ®
will be
vRM
= vR – vM
which by law of vector addition has magnitude
vRM =
direction q = tan^{1}(vM / vR ) with the vertical as shown in fig.
 Relative velocity of swimmer : If a man can swim relative to water with velocity
®
v and water is flowing
®
relative to ground with velocity vR
®
velocity of man relative to ground vM
will be given by:
® ® ® ® ® ®
v = vM – vR , i.e., v_{M} = v + v_{R}
So if the swimming is in the direction of flow of water,
vM = v + vR
And if the swimming is opposite to the flow of water,
vM = v – vR
 Crossing the river : Suppose, the river is flowing with
velocity u . A man can swim in still water with velocity u . He is
r m
standing on one bank of the river and wants to cross the river Two cases arise.
 To cross the river over shortest distance : That is to cross the river straight, the man should swim making angle q with the upstream as
® ® ® ® ® ®
Here OAB is the triangle of vectors, in which
OA = v_{m} , AB = u _{r} . Their resultant is given by OB = u . The
direction of swimming makes angle q with upstream. From the triangle OBA, we find,
cos q = ur
u_{m}
Also
sina = ur
u _{m}
where a is the angle made by the direction of swimming with the shortest distance (OB) across the river. Time taken to cross the river : If w be the width of the river, then time taken to cross the river will be given by
t = w = w
1 u
 To cross the river in shortest possible time : The man should swim perpendicular to the bank. The time taken to cross the river will be:
t = w
2 u
m
In this case, the man will touch the opposite bank at a distance AB
down stream. This distance will be given by:
AB = u_{r} t 2
=u w


m
or AB = ur w
u_{m}
Problem 53. Two trains along the same straight rails moving with constant speed 60 km/hr respectively toward each other.
If at time t = 0 , the distance between them is 90 km, the time when they collide is
(a) 1 hr (b) 2 hr (c) 3 hr (d) 4 hr Solution : (a) The relative velocity vrel. = 60 – (30) = 90 km / hr.
Distance between the train s
rel.
= 90 km, \ Time when they collide = srel.
vrel.
= 90 = 1 hr.
90
Problem 54. Two cars are moving in the same direction with the same speed 30 km/hr. They are separated by a distance of 5 km, the speed of a car moving in the opposite direction if it meets these two cars at an interval of 4 minutes, will be
(a) 40 km/hr (b) 45 km/hr (c) 30 km/hr (d) 15 km/hr
Solution : (b) The two car (say A and B) are moving with same velocity, the relative velocity of one (say B) with respect to
the other
A, v_{BA}
= v B – v_{A}
= v – v = 0
So the relative separation between them (= 5 km) always remains the same.
Now if the velocity of car (say C) moving in opposite direction to A and B, is v C
velocity of car C relative to A and B will be v_{rel}_{.} = v_{C} – v
relative to ground then the
But as v is opposite to v_{C} so
vrel
= vc – (30) = (vC + 30)km / hr.
So, the time taken by it to cross the cars A and B
t = d
Þ 4 = 5
Þ vC = 45 km / hr.
vrel
60 vC + 30
Problem 55. A steam boat goes across a lake and comes back (a) On a quite day when the water is still and (b) On a rough day when there is uniform current so as to help the journey onward and to impede the journey back. If the speed of the launch on both days was same, in which case it will complete the journey in lesser time
(a) Case (a) (b) Case (b)
(c) Same in both (d) Nothing can be predicted
Solution : (b) If the breadth of the lake is l and velocity of boat is v_{b}. Time in going and coming back on a quite day
tQ = l + l = 2l
…..(i)
vb vb vb
Now if v_{a} is the velocity of air current then time taken in going across the lake,
t1 = l
vb + va
[as current helps the motion]
and time taken in coming back t 2 = l
vb – va
So tR = t1 + t2 = 2l
vb [1 – (va / vb )^{2} ]
[as current opposes the motion]
…..(ii)
From equation (i) and (ii)
t R = 1 > 1
tQ [1 – (va / vb )^{2} ]
[as 1
2





a 1]
b
i.e.
tR > tQ
i.e. time taken to complete the journey on quite day is lesser than that on rough day.
Problem 56. A man standing on a road hold his umbrella at 30° with the vertical to keep the rain away. He throws the umbrella and starts running at 10 km/hr. He finds that raindrops are hitting his head vertically, the speed of raindrops with respect to the road will be
 10 km/hr (b) 20 km/hr (c) 30 km/hr (d) 40 km/hr
Solution : (b) When the man is at rest w.r.t. the ground, the rain comes to him at an angle 30° with the vertical. This is the direction of the velocity of raindrops with respect to the ground.
Here v rg = velocity of rain with respect to the ground
vmg =
velocity of the man with respect to the ground.
and
vrm = velocity of the rain with respect to the man,
We have
vr g
= vr m + vmg
……(i)
Taking horizontal components equation (i) gives vr g sin 30° = vm g
= 10 km / hr
or vr g
= 10 = 20 km / hr
sin 30°
Problem 57. In the above problem, the speed of raindrops w.r.t. the moving man, will be
(a)
10 /
2 km / h
 5 km/h (c)
10 3 km / h
(d) 5 /
3 km / h
Solution : (c) Taking vertical components equation (i) gives vrg cos 30° = vrm = 20 3
2
= 10
3 km / hr
Problem 58. Two cars are moving in the same direction with a speed of 30 km/h. They are separated from each other by 5 km. Third car moving in the opposite direction meets the two cars after an interval of 4 minutes. What is the speed of the third car
(a) 30 km/h (b) 35 km/h (c) 40 km/h (d) 45 km/h Solution : (d) Let v be the velocity of third car, then relative velocity of third car w.r.t. the either car is
v – (– 30) = (v+30) km/h.
Now (u + 30) ´ (4/60) = 5 Þ v = 45 km/h
Problem 59. To a person, going eastward in a car with a velocity of 25 km/hr, a train appears to move towards north with a
velocity of 25 km/hr. The actual velocity of the train will be
 25 km/hr (b) 50 km/hr (c) 5 km/hr (d) 5 km/hr
Solution : (a)
vT = =
= = = 25 km/hr
Problem 60. A boat is moving with a velocity 3i + 4j with respect to ground. The water in the river is moving with a velocity – 3i – 4j with respect to ground. The relative velocity of the boat with respect to water is [CPMT 1998]
(a) 8j (b) – 6i – 8j (c) 6i +8j (d) 5
Solution : (c) Relative velocity = (3i + 4j) – (– 3i – 4j) = 6i + 8j