MATHEMATICAL PHYSICS
ALGEBRA
COMMON IDENTITIES
(i) (a + b)2 = a2 + 2ab + b2 = (a – b)2 + 4ab
(ii) (a – b)2 = a2 – 2ab + b2 = (a + b)2 – 4ab
(iii) a2 – b2 = (a + b)(a – b)
(iv) (a + b)3 = a3 + 3a2b + 3ab2 + b3 = a3 + b3 + 3ab(a + b)
(v) (a – b)3 = a3 – 3a2b + 3ab2 – b3 = a3 – b3 – 3ab(a – b)
(vi) a3 + b3 = (a + b)(a2 – ab + b2) = (a + b)3 – 3ab(a + b)
(vii) a3 – b3 = (a – b)(a2 + ab + b3) = (a – b)3 + 3ab(a – b)
(viii) (a + b)2 + (a – b)2 = 2(a2 + b2)
(ix) (a + b)2 – (a – b)2 = 4ab
(x) (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
QUADRATIC EQUATION
An algebraic equation of second degree having the form ax2 + bx + c = 0 is called
quadratic equation. Here ‘a’ is called the coefficient of x2, ‘b’ is called the coefficient of
x and c is a constant term.
General solution of the quadratic equation is given by
x=
−b
b 2 − 4ac
2a
Clearly, there are two roots of the equation:
−b + b 2 − 4ac
x1 =
2a
b
x1 + x2 = −
a
Coefficient of x
In general, sum of roots = −
Coefficient of x 2
c
x1 x2 =
Again,
a
Constant
In general, product of roots =
Coefficient of x 2
and
−b − b 2 − 4ac
x2 =
2a
BINOMIAL THEOREM
If | x | < 1 i.e., x lies between – 1 and + 1, then
(1 + x) n = 1 + nx +
n(n − 1) 2 n(n − 1)(n − 2) 3
x +
x + ......
2!
3!
where n is any number which may be positive, negative, integer or a fraction.
Here, 2 ! = 2 1, 3 ! = 3 2 1, …...., n ! = n(n – 1)(n – 2)…. 3 2 1
If n is a positive integer, then the expansion will have (n + 1) terms.
If n is a negative integer or a fraction, then the number of terms in the expansion will
be infinite.
When | x | << 1, then only the first two terms of the expansion are significant. The
second and higher order terms can be neglected. In this case, the expansions shall reduce
to the following simplified forms:
(1 + x)n = 1 + nx
(1 + x)–n = 1 – nx
(1 – x)n = 1 – nx
(1 – x)–n = 1 + nx
DEFINITION OF 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.
For example : 1000 = 103
log10 1000 = 3
In general, if
N = ax
then
loga N = x
It follows that
loga 1 = 0
( aº = 1)
and
loga a = 1
( a1 = a)
FOUR STANDARD FORMULAE OF LOGARITHM
These four formulae are commonly used. We are mentioning them without proof.
(i)
loga mn = loga m + loga n …..(product formula)
(ii)
log a
m
= loga m – loga n …..(quotient formula)
n
(iii) loga mn = n loga m ……(power formula)
(iv)
loga m = logb m loga b ….(base change formula)
Note that the product formula, quotient formulae & power formula in log can be
extended to any number of quantities.
TWO SYSTEMS OF LOGARITHM
The two systems of Logarithm in common use are:
(i)
Neperian log or Natural log. Here the base is e, where
e = 1+
(ii)
1 1 1
+ + + ....
1! 2! 3!
e = 2.718(approx).
In all theoretical calculations, we use Natural logarithm
Common Logarithm. Here, the base of the log is 10. In all numerical
calculations, we use common log.
We can convert Natural log into common log, as follows:
loge m = log10 m loge 10
As
loge 10 = 2.3026 log10 10
= 2.3026 1 = 2.3026
[ log1010 = 1]
loge m = 2.3026 log10 m
TRIGONOMETRY
SYSTEMS OF MEASUREMENT OF AN ANGLE
(i)
Sexagesimal system. In this system,
1 right angle = 90º (degree)
1o = 60' (minute)
1' = 60'' (second)
(ii)
Centesimal system. In this system
1 right angle = 100g (grade)
1g = 100' (minute)
1' = 100'' (second)
(iii)
Circular system. In this system, the unit of angle is radian
One radian is the angle subtended at the centre of a circle by an arc whose
length is equal to the radius of the circle.
If l is the length of an arc and is the angle subtended at the centre of the
circle as shown in figure, then
=
Arc
l
= radian
Radius r
Angle subtended at the centre of the circle is
=
Circumference 2 r
=
=2 radian
Radius
r
radian = 180º = 200g
1 radian = 57o16'22'' = 63g 63'64''
TRIGONOMETRICAL RATIOS
In right angled OMP, of figure, OMP = 90º and
trigonometric ratios as follows:
perpendicular PM
=
= sin
hypotenuse
OP
base
OM
cosine =
=
= cos
hypotenuse OP
perpendicular
PM
tangent =
=
= tan
base
OM
hypotenuse
OP
cosecant =
=
= cosec
perpendicular PM
hypotenuse
OP
secant =
=
= sec
base
OM
base
OM
cotangent =
=
= cot
perpendicular PM
sine =
POM = . We can define the
FUNDAMENTAL TRIGONOMETRIC RELATIONS
1
1
1
,
sec =
,
cot =
sin
cos
tan
sin
cos
,
cot =
(ii) tan =
cos
sin
2
2
2
(iii) sin + cos = 1, 1 + tan = sec 2 , 1 + cot 2 = cosec 2
(i) cosec =
SIGNS OF T-RATIOS AND LIMITS OF THE VALUE OF T-RATIOS
In order to understand the signs of T-ratios, let us first understand the rule for signs of
the sides of the triangle OMP.
Keeping in mind this sign convention and the definitions of T-ratios, we shall find
that:
(a) In the first quadrant, all the T-ratios are positive.
(b) In the second quadrant, only the sine and (its reciprocal) cosecant are +ve.
(c) In the third quadrant, only the tangent and (its reciprocal) cotangent are +ve.
(d) In the fourth quadrant, only the cosine and (its reciprocal) secant are +ve.
Let us now mention a few facts about the limits to the values of T-ratios.
(i)
The maximum, i.e., greatest value of sin
least value of sin or cos is – 1.
(ii)
sec
and cosec
(iii)
tan
and cot
or cos
is 1. The minimum, i.e.,
cannot be numerically less than one.
can have any numerical values.
T-RATIOS OF ALLIED ANGLES
sin(− )=− sin
cosec(− ) =−cosec
cos(− )=cos
sec(− ) =sec
tan(− )=− tan
cot(− )=− tan
sin(90o − ) = cos
cosec(90o − ) = sec
cos(90o − ) = sin
sec(90o − ) = cosec
tan(90o − ) = cot
cot(90o − ) = tan
sin(90o + ) = cos
cosec(90o + ) = sec
cos(90o + ) = − sin
sec(90o + ) = −cosec
tan(90o + ) = − cot
cot(90o + ) = − tan
sin(180o − ) = sin
cosec(180o − ) = cosec
cos(180o − ) = − cos
sec(180o − ) = − sec
tan(180o − ) = − tan
cot(180o − ) = − cot
sin(180o + ) = − sin
cosec(180o + ) = −cosec
cos(180o + ) = − cos
sec(180o + ) = − sec
tan(180o + ) = tan
cot(180o + ) = cot
sin(270o − ) = − cos
cosec(270o − ) = −sec
cos(270o − ) = − sin
sec(270o − ) = −cosec
tan(270o − ) = cot
cot(270o − ) = tan
sin(270o + ) = − cos
cosec(270o + ) = −sec
cos(270o + ) = sin
sec(270o + ) = cosec
tan(270o + ) = − cot
cot(270o + ) = − tan
sin(360o − ) = − sin
cosec(360o − ) = −cosec
cos(360o − ) = cos
sec(360o − ) = sec
tan(360o − ) = − tan
cot(360o − ) = − cot
VALUES OF TRIGONOMETRICAL RATIOS OF SOME STANDARD ANGLES
angle
0º
30º
45º
60º
90º
120º
135º
150º
180º
270º
360º
sin
0
cos
1
tan
0
1
2
1
2
1
2
3
2
1
3
1
3
2
1
2
1
0
3
3
2
1
−
2
− 3
1
1
2
2
1
3
−
−
2
2
1
–1 −
3
0
–1
0
–1
0
1
0
–
0
SOME IMPORTANT TRIGONOMETRICAL FORMULAE
(i) sin( A + B) = sin A cos B + cos Asin B
(ii) sin( A − B) = sin A cos B − cos Asin B
(iii) cos( A + B) = cos A cos B − sin Asin B
(iv) cos( A − B) = cos A cos B + sin Asin B
tan A + tan B
1 − tan A tan B
2 tan A
(vii) sin 2 A = 2sin A cos A =
1 + tan 2 A
(v) tan( A + B ) =
(vi) tan( A − B ) =
tan A − tan B
1 + tan A tan B
1 − tan 2 A
(viii) cos 2 A = cos A − sin A = 1 − 2sin A = 2cos A − 1 =
1 + tan 2 A
2 tan A
(ix) tan 2 A =
1 − tan 2 A
(x) sin( A + B) + sin( A − B) = 2sin A cos B
(xi) sin( A + B) − sin( A − B) = 2cos Asin B
(xii) cos( A + B) + cos( A − B) = 2cos A cos B (xiii) cos( A + B) − cos( A − B) = −2sin Asin B
C+D
C−D
cos
(xiv) sin C + sin D = 2sin
2
2
C+D
C−D
sin
(xv) sin C − sin D = 2cos
2
2
C+D
C−D
cos
(xvi) cos C + cos D = 2cos
2
2
C+D
C−D
sin
(xvii) cos C − cos D = −2sin
2
2
2
2
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 hole
good.
a
b
c
=
=
sin A sin B sin C
(iii) b2 = c2 + a2 − 2ca cos B
(i)
(ii) a2 = b2 + c2 − 2bc cos A
(iv) c2 = a2 + b2 − 2ab cos c
(v) Area of a triangle ABC = S ( S − a)( S − b)( S − c) where, S = (a + b + c) / 3
GRAPHS
SOME GRAPHS FOR VARIOUS EQUATIONS
(i) For equation, y = mx, the graph between x and y is
a straight line OA passing through origin (figure).
Here slope of the straight line with
x-axis = m (= tan ).
(ii) For equation, y = mx + c, the graph between x and
y is a straight line BC with positive intercept on yaxis (figure). Here OB = c and slope of straight
line BC = m (= tan ).
(iii)For equation, y = mx – c, the graph between x and y
is a straight line DE, with negative intercept on
Y-axis (figure), where OD, = – c and slope of
straight line DE is m (= tan ).
(iv) For equation y = – mx + c, the graph between x and y
is a straight line FG (figure). Here OF = c and slope
of straight line
= – m (= –tan )
(v) For equations (a) y2 = kx (b) y2 = – kx and (c) x2 = ky, the graphs between x
and y will be symmetric parabolic curves as shown in figures, (a), (b), (c)
respectively.
(vi) For equation, y = ax + bx2, the
graph between x and y will be
asymmetric parabola. If both a and
b are positive constants, the graph
is as shown in figure (a). If a is
positive and b is negative, the
graph is as shown in figure(b).
(vii)For equation xy = c, where c is a constant, the graph
between x and y is a rectangular hyperbola (figure).
(viii) For equations (a) x2 + y2 = a2 and (b)
x2 y 2
= 1 the graph between x and
+
a 2 b2
y is (a) a circle (b) an ellipse as shown in figures, (a) and (b) respectively.
(ix) For equation, y = e–kx, the graph between x and y is as
shown in figure, which is an exponential curve.
GRAPHS OF IMPORTANT T-RATIOS
(i)
(ii)
(iii)
CALCULUS
FUNCTION.
If the value of a variable quantity y depends upon the value of some other variable
quantity x, through some relation, such that there exists only one finite value of y for
each value of x, then y is called a function of x. Mathematically, it is represented as:
y = f (x)
Here, f (x) is a symbol, representing a function of x. It is not of multiplied by x.
In a relation, y = 2x2 + 3x + 4; if x = 0 then y = 4; x = 2, y = 18; x = 3, y = 31 and so on. It
means, for every value of x, there is a definite value of y. Hence, y = 2x2 + 3x + 4
represents a function of x.
LIMIT OF A FUNCTION.
A function f (x) is said to tend to a limit l as its variable x approaches a constant a,
(x → a) such that the absolute difference between f(x) and l, as x → a is negligible
f ( x) = l
small but not zero. It is represented as Lt
x →a
x3 − 8
. This function can be defined at
x−2
23 − 8 0
all points except at x = 2. If we use x = 2, then f(x) =
= which is indeterminate
2−2 0
To understand it, consider a function, y = f(x) =
i.e., cannot be defined. However, for all values of x very close to 2 more or less than 2,
the function tends to 12 as shown below :
(1.98)3 − 8
When, x = 1.98, f(x) =
= 11.88
1.98 − 2
(1.99)3 − 8
When x = 1.99, f (x) =
= 11.94
1.99 − 2
(2.01)3 − 8
When x = 2.01, f (x) =
= 12.06
2.01 − 2
(2.02)3 − 8
When x = 2.02, f(x) =
= 12.12
2.02 − 2
From all values of x very close to 2 but not exactly 2,
x3 − 8
12
x−2
3
So, we write Lt x − 8 = 12.
x →2 x − 2
SOME LMPORTANT LIMITS USED IN PHYSICS ARE GIVEN BELOW:
sin
(i) Lt sin = (in radians)
(ii) Lt
=1
→0
→0
(iii) Lt cos = 1
→0
(v) Lt
→0
tan
=1
(iv) Lt tan =
(in radians)
→0
(vi) Lt x + 1
x →0
x
x
1
y
= Lt (1 + y ) = e = 2.718
DIFFERENTIATION
DIFFERENTIAL COEFFICIENT
Let y be a function of x i.e.,
y = f (x)
Suppose the value of x increases by a small amount
increases by a small amount, say y.
x. Then the value of y also
y
is called the average rate of change of y with respect to x.
x
y
When x approaches zero, the limiting value of
is called differential coefficient or
x
dy
.
derivative of y w.r.t. x and is denoted by
dx
dy
y
= lim
Hence
dx x→0 x
dy
Physically, the derivative
given the instantaneous rate of change of function y with
dx
The ratio
respect to variable x.
SOME IMPORTANT RESULTS ON DIFFERENTIATION
(i) Let c be a constant. Then
d
(c ) = 0
dx
d
dy
(cy) = c.
dx
dx
d n
( x ) = nx n −1
(iii)
dx
(ii)
(iv) Let y = u
, where u and
(v) Product Rule. Let
are functions of x. Then
dy du
=
dx dx
d
dx
y = u , then
dy
d
d
d
du
= ( F .F ). ( S .F .) + ( S .F ). ( F .F .) = u
+
dx
dx
dx
dx
dx
(vi) Quotient rule. Let
dy
=
dx
( Den)
y=
u
, then
d
d
( Num) − ( Num) ( Den)
dx
dx
=
2
( Den)
du
d
−u
dx
dx
2
(vii) Chain rule. Let y be a function of u and u be a function of x. Then
dy dy du
=
.
dx du dx
Let
(viii)
(x)
y = u n .Then
d n
du
(u ) = nu n −1.
dx
dx
d
1
(log e x) =
dx
x
d x
(e ) = e x
dx
(ix)
d
1
(log a x) = log e a
dx
x
(xi)
d x
(a ) = a x log e a
dx
(xii)
d
(sin x) = cos x
dx
(xiii)
d
(cos x) = − sin x
dx
(xiv)
d
(tan x) = sec 2 x
dx
(xv)
d
(cot x) = −cosec 2 x
dx
(xvi)
d
(sec x) = sec x tan x
dx
(xvii)
d
(cosec x) = −cosec x cot x
dx
CONDITIONS FOR MAXIMA OR MINIMA
If y = f (x), then for maximum or minimum value of y for a value of x the first differential
coefficient of y w.r.t. x should be zero. i.e.
dy
= 0.
dx
Find the value of x from here.
d2y
Now, find the second derivative of y w.r.t.x i.e.
and put the value of x.
dx 2
d2y
(i)
If the value of
is negative, then y is maximum for a given value of x.
dx 2
d2y
(ii)
If the value of
is positive, then y is minimum for a given value of x.
dx 2
INTEGRATION
Integration is the reverse process of differentiation. It is the process of finding a function
whose derivative is given. If derivative of function f(x) w.r.t. x is f ' (x), then integration
of f ' (x) w.r.t. x is f (x). Symbolically, we can say
d
[ f ( x)] = f '( x),
dx
if
then
f '( x)dx = f ( x).
FUNDAMENTAL FORMULAE OF INTEGRATION
x n+1
provided n −1.
n + 1'
x0+1
0
(ii) dx = x dx =
= x.
0 +1
dx.
(iii) (u + )dx = u dx +
(i) x n dx =
(iv) cu dx = c u dx where c is a constant and u is a function of x.
xn + 1
(v) cx dx = c
.
n +1
dx
= log e x.
(vi) x −1dx =
x
(vii) e x dx = e x .
n
ax
(viii) a dx =
log e a
x
(ix)
(a) sin x dx = − cos x
− cos nx
.
n
(a) cos x dx = sin x
(b) sin nx dx =
(x)
(xi)
sin nx
.
n
sec2 x dx = tan x.
(xii)
cosec2 x dx = − cot x.
(xiii)
sec x tan x dx = sec x.
(xiv)
cosec x cot x dx = −cosec x.
(b)
cos nx dx =
(xv)
(ax + b) n dx =
(ax + b) n +1
a (n + 1)
a
dx = log e (ax + b).
(ax + b)
eax+b
(xvii) eax +b dx =
.
a
a cx+ d
(xviii) a cx+ d dx =
.
c loge a
tan(ax + b)
.
(xix) sec 2 ( ax + b)dx =
a
cot( ax + b)
.
(xx) cos ec 2 ( ax + b)dx = −
a
sec(ax + b)
.
(xxi) sec(ax + b) tan(ax + b)dx =
a
− cos ec(ax + b)
.
(xxii) cos ec(ax + b) cot(ax + b)dx =
a
(xvi)
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 are the functions of x, then
u dx = u
dx −
du
dx
dx dx
DEFINITE INTEGRAL
When an integral is defined between two definite limits a and b, it is said to be a definite
integral. It is given by
b
f ( x)dx = [ ( x)]ba = (b) − (a)
a
where (x) is the integral of f(x). Here a and b are the lower and upper limits of
integration.
UNITS AND DIMENSIONS
PHYSICAL QUANTITIES.
All those quantities in physics which are capable of being measured are called physical
quantities.
PHYSICAL UNIT.
The standard amount of a physical quantity chosen to measure the physical quantity of
the same kind is called a physical unit.
Measure of a physical quantity = Numerical value of the quantity size of the unit = nu
FUNDAMENTAL AND DERIVED UNITS.
The independent units of length, mass, time, electric current, temperature, light intensity
and amount of substance are called basic or fundamental units and these seven physical
quantities are called fundamental quantities. The units of all other physical quantities can
be expressed in terms of fundamental units and are called derived units.
SYSTEM OF UNITS.
A complete set of units which is used for measuring all kinds of fundamental and derived
quantities is called a system of units.
(i)
The CGS or the metric system. In this system the fundamental units of
length, mass and time are centimetre, gram and second respectively.
(ii)
The FPS or the British system. In this system the fundamental units of
length, mass and time are foot, pound and second respectively.
(iii) The MKS system. In this system the fundamental units of length, mass and
time are metre, kilogram and second respectively.
(iv)
The SI. SI is the abbreviation for Systeme Internationale d’Unites, which is
the French equivalent for international system of units. In this system the
fundamental units of length, mass, time electric current, temperature,
luminous intensity and amount of substance are metre, kilogram, second,
ampere, Kelvin, candela and mole, respectively.
SUPPLEMENTARY SI UNITS.
(i)
Radian (rad). It is defined as the plane angle subtended at the centre of a
circle by an arc equal in length to the radius of the circle.
(in radians) =
(ii)
Arc
l
=
Radius r
Steradian (sr). It is the solid angle subtended at the centre of a sphere by a
surface of a sphere equal in area to that of a square, having each side equal to
the radius of the sphere.
(in steradian) =
Surface area
Radius 2
DIMENSIONS OF THE DERIVED QUANTITY.
These are the powers to which the fundamental units of mass. length and time must be
raised in order to represent a derived quantity completely.
DIMENSIONAL FORMULA.
It is an expression which shows how and which of the fundamental units of mass, length
and time occur in the derived unit of a physical quantity.
DIMENSIONAL EQUATION.
The equation which expresses a physical quantity in terms of the fundamental units of
mass, length and time, is called dimensional equation.
PRINCIPLE OF HOMOGENEITY OF DIMENSIONS.
This principle states that a physical equation will be dimensionally correct if the
dimensions of all the terms occurring on both sides of the equation are the same.
USES OF DIMENSIONAL EQUATIONS:
(i)
To convert a physical quantity from one system of units to another.
Suppose a physical quantity has dimensional formula [MaLbTc]. Let n1 and n2
be its numerical values when the units are u1 and u2. Then
n1 u1 = n2 u2
n1[ M 1a L1b T1c ] = n2 [ M 2a Lb2 T2c ]
M1
n2 = n1
M2
(ii)
(iii)
a
L1
L2
b
T1
T2
c
To check the correctness of a given physical relation.
To derive a relationship between different physical quantities.
DIMENSIONAL FORMULAE AND SI UNITS OF SOME PHYSICAL
QUANTITIES
S. No Physical
Relation with other
Dimensional
SI unit
Quantity
Quantities
formula
A. Mechanical Quantities
(i)
Area
m2
Length breadth
L L = L2 = M0L2T0
(ii)
Volume
Length breadth
L L L = L3 = M0L3T0 m3
height
(iii)
Density
kgm–3
Mass
M
−3 0
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
S. No
Volume
Speed
orDistance
velocity
Time
Acceleration
Changein velocity
Time
= ML T
L3
ms–1
L
0
−1
= M LT
T
ms–2
LT −1
= LT −2 = M 0 LT −2
T
Momentum
Force
Work
Energy
Physical
M LT–1 = MLT–1
M LT–2 = MLT–2
MLT–2 L = ML2T–2
ML2T–2
Dimensional
Mass velocity
Mass acceleration
Force distance
Amount of work
Relation with other
kg ms–1
N
J
J
SI unit
(x)
Quantity
Power
(xi)
Pressure
(xii)
Moment
ofForce
force or torque
(xiii)
Gravitational
constant ‘G’
(xiv)
Impulse of a aForce time
force
Stress
Force
(xv)
Quantities
formula
Work
Time
Force
Area
ML T −2
= ML2T −3
T
MLT −2
= ML−1T −2
2
L
W
MLT2
Nm
2
⊥ distance
Force×(distance)2
Mass×mass
Area
(xvi) Strain
Changein dimension
Originaldimension
(xvii) Coefficient
ofStress
elasticity
Strain
(xviii) Surface tension
Force
Length
(xix) Surface energy
Work
Area
(xx)
Coefficient
ofForce×distance
viscosity
Area×velocity
(xxi) Angle
Arc
Radius
L = ML2T–2
MLT −2 L2
= M −1L3T −2
M M
Nm2 kg–2
MLT–2
Ns
T = MLT–1
MLT −2
= ML−1T −2
2
L
Angular velocity Angle
(xxiii) Angular
acceleration
(xxiv) Moment
inertia
(xxv) Radius
gyration
(xxvi) Angular
momentum
Time
Angular velocity
Time
ofMass
(distance)2
ofDistance
Mass
velocity
radius
Nm–2
M0L0T0 (dimensionless)
—
ML−1T −2
= ML−1T −2
1
MLT −2
= MT −2 = ML0T −2
L
ML2T −2
= MT −2 = ML0T −2
2
L
MLT −2 L
= ML−1T −1
2
−1
L LT
L
= 1 = M 0 L0T 0
L
Nm–2
(dimensionless)
(xxii)
Pa or Nm–2
Nm–1
Jm–2
Nm–2 s
Pa s
rad
1
= T −1 = M 0 L0T −1
T
T −1
= T −2 = M 0 L0T −2
T
rad s–1
ML2 = ML2T0
kg m2
L = M0LT0
m
M
LT–1
L = ML2T–1
rad s–2
kg m2 s–1
or
S. No
Physical
Quantity
T-ratios
(sin , cos ,
tan )
Relation with other
Quantities
Length
Length
L
= 1 = M 0 L0T 0
L
(xxviii)
(xxix)
Time period
Frequency
Time
(dimensionless)
T = M0L0T
1
Time period
1
= T −1 = M 0 L0T −1
T
(xxx)
Planck’s
constant ‘h’
(xxxi)
Relative
density
E
Energy
=
v Frequcncy
Density of substance
Density of water at 4oC
ML2T −2
= ML2T −1
−1
T
ML−3
= 1 = M 0 L0T 0
−3
ML
(xxvii)
Dimensional
formula
(dimensionless)
(xxxii)
Velocity
gradient
(xxxiii)
Pressure
gradient
(xxxiv)
Force constant
Latent heat
(xxxviii) Thermal
conductivity
(xxxix)
Entropy
(xxxx)
Universal
gas constant
(xxxxi)
Boltzmann’s
constant
(xxxxii)
Stefan’s
constant
—
s
s–1 or Hz
Js
—
s–1
Velocity
Distance
Pressure
Distance
LT −1
= T −1 = M 0 L0T −1
L
ML−1T −2
= ML−2T −2
L
Force
Displacement
Nm–1
MLT −2
−2
0 −2
= MT = ML T
L
B. Thermal Quantities
(xxxv)
Heat
orEnergy
enthalpy
(xxxvi)
Specific heat
Heat
(xxxvii)
SI unit
Mass×Temperature
Heat
Mass
Heat×distance
Area×temp.×time
Heat
Temperature
PV
nT
Energy
Temperature
Energy
Area×time×(temp.) 4
Pa m–1
ML2T–2
J
ML2T −2
= M 0 L2T −2 K −1
M .K
ML2T −2
= M 0 L2T −2
M
ML2T −2 .L
= MLT −3 K −1
2
L .K .T
ML2T −2
= ML2T–2K–1
K
−1 −2 3
ML T L
= ML2T −2 K −1mol −1
mol.K
ML2T -2
= ML2T -2 K -1
K
ML2T −2
= ML0T −3 K −4
2
4
L .T .K
J kg–1 K–1
J kg–1
Js–1m–1K–1
JK–1
J mol–1 K–1
JK–1
Js–1 m–2K–4
MEASUREMENTS
SIGNIFICANT FIGURES.
The significant figures are normally those digits in a measured quantity which are known
reliably plus one additional digit that is uncertain.
RULES FOR COUNTING THE NUMBER OF SIGNIFICANT FIGURES IN A
MEASURED QUANTITY:
(i) All non-zero digits are significant. So 13.75 m has four significant figures.
(ii) All zeros between two non-zero digits are significant. Thus 100.05 km has five
significant figures.
(iii)All zeros to the right of a non-zero digit but to left of an understood decimal point are
not significant. For example, 86400 has three significant figures. But such zeros are
significant if they come from a measurement. For example, 86400s has five
significant figures.
(iv) All zeros to the right of a non-zero digit but to the left of a decimal point are
significant. For example, 648700 has six significant figures.
(v) All zeros to the right of a decimal point are significant. So 161 cm, 161.0 cm,
161.00 cm, have three, four and five significant figures respectively.
(vi) All zeros to the right of a decimal point but to the left of a non-zero digit are not
significant. So 0.161 cm and 0.0161 cm, both have three significant figures. Single
zero conventionally placed to the left of the decimal point is not significant.
(vii)The number of significant figure does not depend on the system of units. So
16.4 cm, 0.164 m and 0.000164 km, all have three significant figures.
SIGNIFICANT FIGURES IN THE SUM OR DIFFERENCE OF TWO
NUMBERS.
In addition or subtraction the same number of decimal places as that of the number with
minimum number of decimal places.
RULES FOR ROUNDING OFF A MEASUREMENT:
(i) If the digit to be dropped is less than 5, then the preceding digit is left unchanged.
(ii) If the digit to be dropped is greater than 5, then the preceding digit is increased by 1.
(iii) If the digit to be dropped is 5 followed by non-zero digits, then the preceding digit is
increased by 1.
(iv) If the digit to be dropped is 5, then the preceding digit is left unchanged if it is even.
(v) If the digit to be dropped is 5, then the preceding digit is increased by 1 if it is odd.
ERROR IN A MEASUREMENT.
It is the difference between the measured value and the true value of a physical quantity.
It gives an indication of the limits within which the true value may lie.
ELIMINATION OF RANDOM ERROR.
The errors which occur irregularly and at random, in magnitude and direction, are called
random errors. To eliminate random error, a large number of readings are taken and their
arithmetic mean is taken as the true value.
a=
a1 + a2 + a3 + ..... + an 1 n
=
ai
n
n i =1
ABSOLUTE ERROR.
The magnitude of the difference between the true value and the measured value is called
absolute error. Such errors are given by
a1 = a − a1
a2 = a − a 2
a3 = a − a3
an = a − an
MEAN ABSOLUTE ERROR.
The arithmetic mean of the positive magnitude of all the absolute errors is called mean
absolute error
| a | + | a2 | +.....+ | an | 1 n
a=
=
| ai |
n
n i =1
RELATIVE ERROR.
It is the ratio of the mean absolute error to the true value.
a=
a
a
PERCENTAGE ERROR.
The relative error expressed in percent is called percentage error.
Percentage error =
a
100%
a
ERROR COMBINATION IN A SUM OR A DIFFERENCE.
When two quantities are added or subtracted, the absolute error in the final result is the
sum of the absolute errors associated with the individual quantities.
| Z |=| A | + | B |
ERROR COMBINATION IN A PRODUCT OR A QUOTIENT.
When two quantities are multiplied or divided, the fractional error in the final result is
the sum of the fractional errors of the two quantities.
Z
A
B
=
+
Z
A
B
ERROR DUE TO THE POWER OF A MEASURED QUANTITY.
The fractional error in the nth power of a quantity is equal to n times the fractional error in
the quantity itself.
Z = An ,
If
then
Z
A
=n
Z
A
GENERAL RULE.
A p Bq
, then the maximum fractional or relative error in Z will be
Cr
Z
A
B
C
=p
+q
+r
Z
A
B
C
Z
A
B
C
100 = p
100 + q
100 + r
100
% Error in Z = =
Z
A
B
C
If Z =
THE VERNIER CALLIPERS
It is an instrument used to measure accurately upto
1
th of the millimeter. It is so called
20
after the name of its inventor P. Vernier a French mathematician.
It consists of a rectangular steel bar graduated in centimeter as shown in figure. This is
known as the main scale. Over this scale slides a small scale called the vernier scale. The
instrument has got two jaws, A and B. The jaw A is fixed at the end of the rectangular bar on
the zero side, while the other jaw B is movable and can slide along the main scale. Each jaw
is at right angles to the main scale and the movable jaw can be fixed at any position by the
screw S. A vernier scale is attached to the movable jaw as shown. When the two jaws are
touching each other, the zero of the vernier scale coincides with zero of the main scale.
There also exists two jaws projected in the upper part, shown as P and Q in the figure.
These jaws are used to measure the internal diameters of the tubes. Furthermore, a movable
jaw also carries a thin rectangular rod RS which is used to measure the depth of vessels.
HOW TO DETERMINE THE VALUE OF VERNIER CONSTANT?
If N vernier divisions are equal to (N –1) scale divisions, then
1V =
N −1
S
N
Vernier constant, C = 1S − 1V = 1 −
or
C=
N −1
S
N
1
S
N
POSITIVE ZERO ERROR
The zero error is positive when the zero mark of the vernier scale lies toward the right side
of the zero of the main scale as shown in figure.
It is so called positive because the measured length is always more than the actual length.
In general, if the Nth vernier division coincides with a scale division of the instrument with
vernier constant, C, then the value of positive zero error is
e = + NC
The zero correction for the positive zero error is
negative, i.e. c = –e = –NC
The true reading is Rt = R0 + c = R0 – e = R0 – NC
where R0 is the observed reading.
Note that
The positive zero error is always subtracted from the
observed reading to determine the true reading.
NEGATIVE ZERO ERROR
The zero error is negative when the zero mark of the vernier scale
lies toward the left side of the zero of the main scale as shown in
figure. It is so called negative because the measured length is
always less than the actual length.
In general, if the Nth vernier division coincides with a scale
division of the instrument with vernier constant C, then the value of negative zero error is
e = – NC
The zero correction for the negative error is positive, i.e., c = –e = + NC
The true reading is Rt = R0 + c
or
Rt = R0 + NC
Note that The negative zero error is always added to the observed reading to determine the
true reading.
ILLUSTRATION
The main scale of a vernier calipers reads in millimetre and its vernier is divided into10
divisions which coincide with 9 divisions of the main scale. When the two jaws of the
instrument touch each other the seventh division of the vernier scale coincide with a scale
division and the zero of the vernier lies to the right of the zero of main scale. Furthermore,
when a cylinder is tightly placed along its length between the two jaws, the zero of the
vernier scale lies slightly to the left of 3.2 cm; and the fourth vernier division coincides with
a scale division. Calculate the measured length of the cylinder.
Solution
1S = 1.0mm; N = 10
1
1.0
Vernier constant 1C =
S=
= 0.1 mm
N
10
The instrument has a positive error, e = NC = 7(0.1) = 0.7mm = 0.07 cm
The main scale reading is 3.1cm because the zero of the vernier scale lies to the left of
3.2 cm
Vernier scale reading is x = 4C = 4(0.01) = 0.04 cm
The observed reading is L = 3.1 + x = 3.1 + 0.04 = 3.14 cm
The true reading of the instrument is
Lt = L – e = 3.14 – 0.07 = 3.07 cm
THE SCREW GAUGE
It is an instrument based on the principle of a screw. It is used to measure small thicknesses
or diameters of wires. It consists of a U-shaped frame F having a fixed end at A. Through
the other end B passes an accurately cut screw of uniform pitch. A cap fits on to the screw
and carries on its inner edge H, 50 or 100 equal division marks. This is called the head scale
or circular scale and is used to measure the fraction of a revolution. The number of
complete revolutions can be read on the pitch scale S which is graduated on the nut parallel
to the axis of the screw. The reading of the circular scale is taken against a line known as
the reference line. The screw head is provided with a ratchet arrangement R. When the studs
A and C are in contact either with each other or with some other object placed in between,
the ratchet slips over the screw cap without moving the screw forward. It avoids undue
pressure between the studs or on the object.
PITCH
It is defined as the linear distance moved by the screw forward or backward when one
complete rotation is given to the circular cap. It is also defined as the distance between two
consecutive threads measured along its axis.
HOW TO FIND THE PITCH OF SCREW GAUGE?
Rotate the circular scale so that its zero mark coincides with the reference line and note
down the reading on the pitch scale. Then, give N complete revolutions (say four) to the
circular scale and again note that reading on the pitch scale. Finally, calculate the pitch by
using the formula
pitch =
distance travelled on the pitch scale
number of rotations
LEAST COUNT
For a screw gauge, the least count is defined as the distance through which the screw moves
(forward or backward) when the cap is rotated through one division on the circular scale.
Mathematically,
Least Count =
Pitch of the screw
Number of divisions on the circular scale
POSITIVE ZERO ERROR
If the zero of the circular scale lies below the reference line as shown in figure, then there is
a positive zero error.
If the zero of the circular scale lies N division below the reference line, then the magnitude
of positive error is
e = + NC
The zero correction for the positive zero error is negative, i.e. c = –e = –NC
The true reading is Rt = R0 + c = R0 − e = R0 − NC where R0 is the observed reading.
Note that incase of positive zero error the instrument always reads more than the actual
value, therefore, the true reading is always obtained by subtracting the magnitude of
positive zero error.
NEGATIVE ZERO ERROR
If the zero of the circular scale lies above the reference line,
as shown in figure, then there is a negative zero error.
If the zero of the circular scale lies N divisions above the
reference line, then the magnitude of negative error is
e = – NC
The zero correction for the negative zero error is positive, i.e.
The true reading is Rt = R0 + c = R0 – e = R0 + NC
where R0 is the observed reading.
Note that in case of negative zero error the instrument always reads less than the
actual
value, therefore, the true reading is always obtained by adding the magnitude of negative
zero error.
ILLUSTRATION
The pitch of a screw gauge is 0.5 mm and there are 50 divisions on the circular scale. In
measuring the thickness of a metal plate, there are five divisions on the pitch scale
(or main scale) and thirty fourth division coincides with the reference line. Calculate the
thickness of the metal plate.
Solution
pitch, p = 0.5 mm
No. of circular scale divisions = 50
Least count, C =
0.5
= 0.01mm
50
Main scale reading = 5 (pitch) = 5(0.5) = 2.5 mm
Circular scale reading = NC = 34(0.01) = 0.34 mm
Total Reading = 2.5 + 0.34 = 2.84 mm
VECTORS
SCALARS.
The physical quantities which have only magnitude and no direction are called scalars.
VECTORS.
The physical quantities which have both magnitude and direction are called vectors.
EQUAL VECTORS.
Two vectors are said to be equal if they have the same magnitude and direction.
NEGATIVE VECTOR.
The negative of a vector is defined as another vector having the same magnitude but
having an opposite direction.
ZERO VECTOR.
A vector having zero magnitude and an arbitrary direction is called a zero or null vector.
COLLINEAR VECTORS.
The vectors which either act along the same line or along parallel lines are called
collinear vectors.
COPLANAR VECTORS.
The vectors which act in the same plane are called coplanar vectors.
MODULUS OF A VECTOR.
The magnitude or length of a vector is called its modulus.
→
→
Modulus of vector A =| A |= A
UNIT VECTOR.
A unit vector is a vector of unit magnitude drawn in the direction of a given vector. A
→
unit vector in the direction of A is given by
→
A=
A
→
| A|
RESULTANT VECTOR.
The resultant of two or more vectors is that single vector which produces the same effect
as the individual vectors together would produce.
VECTOR ADDITION
TRIANGLE LAW OF VECTOR ADDITION.
If two vectors can be represented both in magnitude and
direction by the two sides of a triangle taken in the same order,
then their resultant is represented completely both in magnitude
and direction by the third side of the triangle taken in the opposite order. In figure,
→
→
→
OA+ AB = OB
→
→
→
P+ Q = R
or
PARALLELOGRAM LAW OF VECTOR ADDITION.
If two vectors acting simultaneously at a point can be represented both in magnitude and
direction by the two adjacent sides of a parallelogram, then their resultant is represented
completely both in magnitude and direction by the diagonal of the parallelogram passing
through that point. In figure,
→
→
→
OA+ OB = OC or
→
→
→
P+ Q = R
→
The magnitude of the resultant R is given by
R = P2 + Q2 + 2PQ cos
→
where
→
→
is the angle between P and Q . If R makes
→
angle
tan
with P , then
=
Q sin
P + Q cos
POLYGON LAW OF VECTOR ADDITION.
If a number of vectors are represented both in magnitude and direction by the sides of an
open polygon taken in the same order, then their resultant is represented both in
magnitude and direction by the closing side of the polygon taken in opposite order.
IMPORTANT POINTS OF VECTOR ADDITION.
(i) Vector of same nature alone can be added i.e. a force vector can not be added to
velocity vector but can be added to force vector only.
→
→
→
→
(ii) Vector addition is commutative i.e.
A+ B = B + A
(iii)Vector addition is associative i.e.
( A+ B ) + C = A+ ( B + C )
→
→
→
→
→
→
SUBTRACTION OF VECTORS.
→
→
→
Subtraction of a vector B from a vector A is defined as the addition of vector – B
→
(negative of vector B ) to A.
→
Thus
→
→
→
A − B = A + (− B ).
→
If
→
→
→
→
is the angle between A and B , and R = ( A − B ), then
and
R = A2 + B2 + 2 AB cos(180o − )
B sin(180o − )
tan =
A + B cos(180o − )
The vector subtraction does not obey, commutative law and associative law.
LAMI’S THEOREM.
It states that if three forces acting at a point are in equilibrium,
then each force is proportional to the sine of the angle between
the other two forces. i.e.,
A
B
C
=
=
sin
sin
sin
→ → →
where A, B , C are the three forces and , ,
→
→
are the angles
→
→
→
→
between forces B and C , C and A , and A and B
respectively (figure).
RESOLUTION OF A VECTOR.
The process of splitting a vector into two or more vectors is known as resolution of the
vector. The vectors into which the given vector is splitted are called component vectors.
RECTANGULAR COMPONENTS OF A VECTOR INAPLANE. When a vector is
resolved along two mutually perpendicular directions, components so obtained are called
rectangular components of the given vector.
→
→
→
with X-axis and A x and A y are the
As shown in figure, if A makes angle
→
rectangular components
respectively, then
→
of
→
A
along
X-and
Y-axis
→
A = A x + A y = Ax i + Ay j
Also, Ax = A cos , Ay = A sin
A = Ax2 + Ay2 and tan =
Ay
Ax
RECTANGULAR COMPONENTS OF A VECTOR IN THREE DIMENSIONS
→
⎯⎯
→
Consider a vector A represented by OR , as shown in figure. Taking O as origin
construct a rectangular parallelopiped with its three edges along the three rectangular
→
axes i.e. X, Y and Z axes. Here, we note that A represents
the diagonal of the parallelopiped whose intercepts along
→
→
→
X, Y and Z axis are A x , A y and A z respectively: which are three rectangular
→
components of A .
Using triangle law of vectors;
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
⎯⎯
→
(OR) = (OT) + (TR)
Using parallelogram law of vectors,
(OT) = (OS) + (OP)
⎯⎯
→
⎯⎯
→
(OR) = (OS) + (OP) + (OQ)
→
→
→
→
→
→
→
A = Az + Ax + A y = Ax + A y + Az
or
Let
→
⎯⎯
→
[ TR = OQ]
be the unit vectors along X, Y and Z axes respectively, then
i , j, k
→
→
→
A x = A x i ;A y = A y j and A z = A z k
→
Therefore, A = A x i + A y j + A z k
Also,
(OR)2 = (OT)2 + (TR)2
= (OP)2 + (OS)2 + (TR)2
= (OP)2 + (OS)2 + (OQ)2
or
A2 =A2x +A2y +A2z
or
If ,
A = A 2x +A 2y +A 2z
→
and are the angles which the A makes with X, Y and Z axes respectively, then
A
cos = x
or
Ax = A cos
A
A
or
Ay = A cos
cos = y
A
A
cos = z
or
Az = A cos
A
→
Here cos , cos and cos are called the direction cosines of the vector A .
Putting the values of Ax, Ay and Az in we get
A2 = A2 cos2 + A2 cos2 + A2 cos2
or
A2 = A2 (cos2 + cos2 + cos2 )
or
cos2 + cos2 + cos2 = 1
It means sum of the squares of the direction cosines of a vector is always unity.
SCALAR OR DOT PRODUCT
SCALAR OR DOT PRODUCT.
→
→
The scalar or dot product of two vectors A and B is defined as the product of the
→
→
magnitudes of A and B and cosine of the angle
→
→
→
between them. Thus
→
A . B =| A | | B | cos = AB cos
PROPERTIES OF DOT PRODUCT OF TWO VECTORS.
→
→
= 0o , cos = 1, A . B = AB
(i)
For parallel vectors,
(ii)
For antiparallel vectors,
→ →
= 180º, cos = −1, A . B = − AB
→ →
(iii)
→
(iv)
= 90º, cos = 0, A . B = 0
For perpendicular vectors,
→
→
→
A. B = B . A
→
→
→
(Commutative law)
→ →
→ →
(v)
A .( B + C ) = A . B + A . C
(vi)
A . A = A2
(vii)
i . i = j . j = k .k =
(viii)
i . j = j . k = k . i = 0.
→
(Distributive law)
→
DOT PRODUCT IN CARTESIAN CO-ORDINATES.
→
For A = Ax i + Ay j + Az k and B = Bx i + By j + Bz k ,
→
→
A . B = Ax Bx + Ay By + Az Bz
→
Angle
→
between A and B is given by
→
cos =
→
A, B
→
→
| A| | B|
=
Ax Bx + Ay By + Az Bz
Ax2 + Ay2 + Az2 Bx2 + By2 + Bz2
VECTOR OR CROSS PRODUCT
VECTOR OR CROSS PRODUCT.
→
→
For two vectors A and B inclined at an angle , the vector or cross product is defined
as
→
→
A B = AB sin n
→
→
where n is a unit vector perpendicular to the plane of A and B and its direction is that
→
→
in which a right handed screw advances when rotated from A to B .
PROPERTIES OF CROSS PRODUCT OF TWO VECTORS.
→
(i)
For parallel or antiparallel vectors,
→
(ii)
A
→
→
→
→
→
→
B=0
→
B =−B A
→
= 0º or 180º, A
(anti commutation low)
→
→
→
→
(iii)
A (B+ C) = A B+ A C
(iv)
i i = j j=k k = 0
(v)
i j = k, j k = i, k i = j
(vi)
Unit vector perpendicular to the plane of A and B is given by
(Distributive low)
→
→
n=
→
→
→
→
A B
| A B|
→
(vii)
Angle
→
between A and B is given by
→
sin =
→
| A B|
→
→
| A| | B |
→
CROSS PRODUCT IN CARTESIAN CO-ORDINATES.
→
→
A B = ( Ax i + Ay j + Az k ) ( Bx i + By j + Bz k )
i
j
k
= Ax Ay Az = ( Ay Bz − Az By ) i − ( Az Bx − Ax Bz ) j + ( Ax By − Ay Bx ) k .
Bx By Bz
GEOMETRICAL SIGNIFICANCE OF CROSS PRODUCT.
The magnitude of the cross product of two vectors is equal to the area of the
parallelogram having the two vectors as its adjacent sides.
SCALAR TRIPLE PRODUCT
The scalar triple product of vectors can be given by
Ax
Ay
Az
A.( B C ) = Bx
By
Bx
Cx
Cy
Cz
→
→
→
VOLUME OF THE PARALLELLELOPIPED
→
→
→
→
→
→
→
→
→
A.( B C ) = ( A B). C = (C A). B = volume of the parallelopiped whose sides are
represented by the three vectors.
COPLANARITY
→
→ →
→
→
→
If any the given three vectors A, B and C are parallel, then A . ( B C ) = 0.
VECTOR TRIPLE PRODUCT
→
→
→
→ → →
→ → →
Vector triple product of vectors gives A ( B C ) = B ( A .C ) − C ( A . B )
(ASSIGNMENT)
MATHEMATICAL PHYSICS
1. Evaluate 105
106 without multiplying directly.
Ans. 11130
2. Evaluate each of the following using suitable identities:
(i) (104)3
(ii) (999)3
Ans. (i) 1124864 (ii) 997002999
3. Evaluate the following products without multiplying directly:
(i) 103 107
(ii) 95 96
(iii) 104 96
Ans. (i) 11021 (ii) 9120 (iii) 9984
4. Evaluate the following using suitable identities:
(i) (99)3
(ii) (102)3
(iii) (998)3
Ans. (i) 970299 (ii) 1061208 (iii) 994011992
5. Without actually calculating the cubes, find the value of each of the following :
(i) (–12)3 + (7)3 + (5)3
(ii) (28)3 + (–15)3 + (–13)3
Ans. (i) –1260 (ii) –16380
6. Find the remainder when x4 + x3 – 2x2 + x + 1 is divided by x – 1.
Ans. 2
7. Give possible expressions for the length and breadth of each of the following
rectangles, in which their areas are given:
(i) Area : 25a2 – 35a + 12
(ii) Area : 35y2 + 13y –12
Ans. (i) (5a – 3), (5a – 4) (ii) (7y –3), (5y + 4)
8. What are the possible expression for the dimensions of the cuboids whose volumes are
given below?
(i) Volume : 3x2 – 12x
(ii) Volume : 12ky2 + 8ky – 20k
Ans. (i) 3, x, x – 4 (ii) 4k, 3y + 5, y – 1
9. The altitude of right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find
the other two sides.
Ans. 5 cm, 12 cm
10. The diagonal of rectangular field is 60 metres more than the shorter side. If the longer
side is 30 metres more than the shorter side, find the sides of the field.
Ans. 120 m, 90m
11. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would
have taken 1 hour less for the same journey. Find the speed of the train.
Ans. 40 km/h
12. An express train takes 1 hour less than a passenger train to travel 132 km between
Mysore and Bangalore (without taking into consideration the time they stop at
intermediate stations). If the average speed of the express train is 11 km/h more than
that of the passenger train, find the average speed of the two trains.
Ans. 33 km/h, 44 km/h
13. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m,
find the sides of the two squares.
Ans. 18m, 12m
14. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2 ? If so,
find its length and breadth.
Ans. yes, 20m, 20m
3
hours. The tap of larger diameter takes 10
8
hours less than the smaller one to fill the tank separately. Find the time in which each
tap can separately fill the tank.
Ans. 15 hr, 25 hr
15. Two water taps together can fill a tank in 9
16. Compute (98)5
Ans. 9039207968
17. Which is large (1.01)1000000 or 10,000?
Ans. 10000 is smaller
18. Find an approximation of (0.99)5 using the first three terms of its expansion.
Ans. 0.9510
x
19. Expand 1 −
2
−
1
2
, when | x | < 2.
Ans. 1 +
x 3x 2
+
+ .....
4 32
20. Find the sum to infinity of the G.P. ;
−5 5 −5
, , ,....
4 16 64
Ans. –1
21. Find the 5th root of .003, having given
log 3 = .4771213 and log 312936 = 5.4954243
Ans. 0.312936
1
7
2
5
1
5
22. Find the values of (1) 7 , (2) (84) , and (3) (.021) , having given
log 2 = 0.30103,
log 3 = 0.4771213,
log 7 = 0.8450980, log 132057 = 5.1207283,
log 588453 = 5.7697117, and log 461791 = 5.6644438.
Ans. (1) 1.32057 (2) 5.88453 (3) 0.461791
23. Given log 2 = 0.30103, log 3 = 0.4771213, and log 7 = 0.8450980, solve the equations
(1) 2x. 3x + 4 = 7x,
(2) 22x + 1. 33x + 2 = 74x,
(3) 72x 22x – 4 = 33x –7.
24. Given sec
=
13
, calculate all other trigonometric ratios.
12
5
12
5
12
13
Ans. sin = , cos = , tan = , cot = , cosce =
13
13
12
5
5
25. Evaluate the following:
(i) sin 60º cos 30º + sin 30º cos 60º
cos 45o
(iii)
sec30o + cosec30o
(v)
(ii) 2 tan2 45º + cos2 30º – sin2 60º
sin 30o + tan 45o − cosec 60o
(iv)
sec30o + cos 60o + cot 45o
5cos 2 60o + 4sec2 30o − tan 2 45o
sin 2 30o + cos 2 30o
Ans. (i) 1 (ii) 2 (iii)
26. If tan (A + B) =
3 and tan (A – B) =
3 2− 6
43 − 24 3
67
(iv)
(v)
8
11
12
1
; 0º < A + B
3
90º ; A > B, find A and B.
Ans. 45º, 15º
27. Evaluate :
sin18o
(i)
cos 72o
(iii) cos 48º – sin 42º
tan 26o
cot 64o
(iv) cosec 31º – sec 59º
Ans. (i) 1 (ii) 1 (iii) 0 (iv) 0
(ii)
28. If tan 2A = cot (A – 18º), where 2A is an acute angle, find the value of A.
Ans. 36º
29. If sec 4A = cosec (A – 20º), where 4A is an acute angle, find the value of A.
Ans. 22º
30. Evaluate:
sin 2 63o + sin 2 27o
(i)
cos 2 17o + cos 2 73o
(ii) sin 25º cos 65º + cos 25º sin 65º
Ans. (i) 1 (ii) 1
31. Prove the following identities, where the angles involved are acute angles for which the
expressions are defined.
1 − cos
(i) (cosec − cot ) 2 =
1 + cos
cos A 1 + sin A
+
= 2sec A
(ii)
1 + sin A
cos A
tan
cot
+
= 1 + sec cosec
(iii)
1 − cot
1 − tan
1 + sec A
sin 2 A
(iv)
=
sec A
1 − cos A
cos A − sin A + 1
= cosec A + cot A
(v)
cos A + sin A − 1
1 + sin A
(vi)
= sec A + tan A
1 − sin A
sin − 2sin 3
(vii)
= tan
2cos3 − cos
(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan 2 A + cot 2 A
1
(ix) (cosec A − sin A)(sec A − cos A) =
tan A + cot A
1 + tan 2 A
1 − tan A
=
(x)
2
1 + cot A
1 − cot A
2
= tan 2 A
32. From a point P on the ground the angle of elevation of the top of a 10 m tall building is
30º. A flag is hoisted at the top of the building and the angle of elevation of the top of
the flagstaff from P is 45º. Find the length of the flagstaff and the distance of the
building from the point P. (You may take 3 = 1.732 )
Ans. 7.32 m
33. The shadow of a tower standing on a level ground is found to be 40 m longer when the
Sun’s altitude is 30º than when it is 60º. Find the height of the tower.
Ans. 20 3 m
34. The angles of depression of the top and the bottom of an 8 m tall building from the top
of a multi-storeyed building are 30º and 45º, respectively. Find the height of the multistoreyed building and the distance between the two buildings.
Ans. 4(3 +
3 )m, 4(3 +
3 )m
35. From a point on a bridge across a river, the angles of depression of the banks on
opposite sides of the river are 30º and 45º, respectively. If the bridge is at a height of
3m from the banks, find the width of the river.
Ans. 3( 3 + 1)m
36. From the top of a 7m high building, the angle of elevation of the top of cable tower is
60º and the angle of depression of its foot is 45º. Determine the height of the tower.
Ans. 7( 3 + 1)m
37. A straight highway leads to the foot of a tower. A man standing at the top of the tower
observes a car at an angle of depression of 30º, which is approaching the foot of the
tower with a uniform speed. Six seconds later, the angle of depression of the car is
found to be 60º. Find the time taken by the car to reach the foot of the tower from this
point.
Ans. 3 seconds
38. The angles of elevation of the top of a tower from two points at a distance of 4 m and
9 m from the base of the tower and in the same straight line with it are complementary.
Prove that the height of the tower is 6 m.
39. A wheel makes 360 revolutions in one minute. Through how many radians does it turn
in one second?
Ans. 12
40. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor
arc of the chord.
20
Ans.
3
41. Find the degree measure of the angle subtended at the centre of a circle of radius
22
100 cm by an arc of length 22 cm (Use =
).
7
Ans. 12º36 '
3
42. If cos x = − , x lies in the third quadrant, find the values of other five trigonometric
5
functions.
5
4
5
4
3
Ans. sec x = − , sin x = − , cosec = − , tan x = , cot x =
3
5
4
3
4
31
43. Find the value of sin
.
3
3
2
Ans.
44. Find the value of cos (–1710º).
Ans. 0
45. Find the value of
(i) sin 75º
(ii) tan 15º
Ans. (i)
46. Find the principal solutions of the equation sin x =
3 +1
2 2
(ii) 2 –
3
2
Ans. x =
47. Find the value of tan
8
2
3 3
,
.
Ans.
48. Find the principal value of sin–1
2 −1
1
.
2
Ans.
49. Find the principal value of cot–1
4
−1
3
Ans.
50. Show that tan–1
3
2
3
1
2
3
+ tan −1 = tan −1
2
11
4
3
8
84
51. Show that sin −1 − sin −1 = cos −1
5
17
85
52. The sides of a triangle are 119, 111, and 92 metres : prove that its area is 4830 sq. m.
53. If B = 45º C = 60º, and a = 2 ( 3 + 1)cm., prove that the area of the triangle is
6 + 2 3 sq. cm.
54. Given a = 18, b = 24, and c = 30, find sin A, sin B, and sin C.
Ans.
55. Given
a=
3,b=
2 , and c =
3 4
, ,1
5 5
6+ 2
, find the angles.
2
Ans. 60º, 45º, 70º
56. Draw the graph of each of the following linear equations in two variables:
(i) x + y = 4
(ii) x – y = 2
(iii) y = 3x
(iv) 3 = 2x + y
57. Draw the graphs of the equations x – y + 1 = 0 and 3x +2y –12 = 0. Determine the
coordinates of the vertices of the triangle formed by these lines and the x-axis, and
shade the triangular region.
58. Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – k = 0.
Ans. k2
59. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the
reflected ray passes through the point (5, 3). Find the coordinates of A.
13
,0
Ans.
5
60. Find the equation of parabola that satisfies the conditions: vertex (0, 0); focus (3, 0)
Ans. y2 = 12x
61. Find the equation of parabola that satisfies the conditions: vertex (0,0); focus (–2,0)
Ans. y2 = –8x
62. Find the equation of parabola that satisfies the conditions: vertex (0,0) passing through
(2,3) and axis is along x-axis.
Ans. 2y2 = 9x
63. Find the equation of parabola that satisfies the conditions: vertex (0,0), passing through
(5.2) and symmetric with respect to y-axis.
Ans. 2x2 = 25y
64. Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?
Ans. inside
65. Find the equation of the circle with centre (–3, 2) and radius 4.
Ans. (x + 3)2 + (y – 2)2 = 16
66. Find the equation of a circle with centre (2,2) and passes through the point (4,5).
Ans. (x – 2)2 + (y – 2)2 = 13
67. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes
through the point (2, 3).
Ans. (x – 6)2 + y2 = 25 and (x + 2)2 + y2 = 25
68. Find the equation for the ellipse that satisfies the conditions: ends of major axis ( 3, 0),
ends of minor axis (0, 2)
x2 y 2
Ans.
+
=1
9
4
69. Find the equation for the ellipse that satisfies the conditions: ends of major axis
5 ), ends of minor axis ( 1, 0)
(0,
y2
2
Ans. x +
=1
5
70. Find the equation for the ellipse that satisfies the conditions: centre at (0,0), major axis
on the y-axis and passes through the points (3, 2) and (1,6).
x2 y 2
Ans.
+
=1
10 40
71. Find the equation for the ellipse that satisfies the conditions: major axis on the x-axis
and passes through the points (4, 3) and (6,2).
x2 y 2
+
=1
Ans.
52 13
72. Find the limits:
x2 + 1
(i) lim
x →1 x + 100
x2 − 4
x3 − 4 x 2 + 4 x
(iii) lim
x →2
(v) lim
x →1
x3 − 4 x 2 + 4 x
(ii) lim
x →2
x2 − 4
(iv) lim
x →2
x3 − 2 x 2
x2 − 5x + 6
x−2
1
− 3
2
x − x x − 3x 2 + 2 x
Ans. (i)
73. Evaluate:
(i) lim
x →0
sin 4 x
sin 2 x
(ii) lim
x →0
2
(ii) 0 (iii)
101
(iv) –4 (v) 2
tan x
x
Ans. (i) 2 (ii) 1
74. Differentiate the following w.r.t. x
(i) (3x + 5)
(ii) x–2
(iii) x3/2
(iv)
Ans. (i )3 (ii )
x−
1
x
−2
3
1 x +1
(iii ) x1/ 2 (iv)
3
x
2
x
2 x
75. Differentiate the following w.r.t. x
(i) 3x5/2
(ii) (4x)1/3
(iii)
6
x
Ans. (i)
15 3
41/ 3 −2 / 3
x / 2 (ii)
x
(iii) − 3x −3 / 2
2
3
dy
for the following functions:
dx
2
( x − 1)( x − 2)
(i ) y = x3 − 3x 2 + 3x −
(ii ) y =
5
x
76. Find
(iii ) y =
1
x+
x
2
(iv) y = ax 2 + bx + c
Ans. (i )3x 2 − 6 x + 3(ii)
77. Differentiate the following w.r.t. x
(i) (5x2 + 6) (2x3 + 4)
(ii)
3
1
1
1
x−
+ 3 / 2 (iii)1 − 2 (iv) 2ax + b
2
x
2 x x
x ( x2 + 7)
Ans. (i) 50x4 + 36x2 + 40x (ii)
78. Differentiate the following w.r.t. x
(i) x–2 – x3/x4
(ii) (2x2 + 9)3
Ans. (i ) −
79. Differentiate the following functions:
(iii) (1 +
5 3 / 2 7 −1/ 2
x + x
2
2
x )1/2
2 1
1
+ 2 (ii )12 x(2 x 2 + 9) 2 (iii ) 1/ 2
3
x
x
4 x (1 + x )1/ 2
(i)( x 2 − 4 x + 5)( x3 − 2)
(iii)
(ii)
sin x + cos x
sin x − cos x
2x + 3
x2 − 5
(iv)(4 x3 − 5 x 2 + 1) 4
Ans. (i) 5x4 – 16x3 + 15x2 – 4x + 8 (ii) −
(iii)
−2
(sin x − cos x) 2
2 x 2 + 6 x + 10
( x 2 − 5) 2
(iv) 8(4x3 – 5x2 + 1)3 (6x2 – 5x)
80. Differentiate the following w.r.t. x
(i)
sin x
(ii) x sin 2x
(iii)
sin x
x + cos x
cos x
(iv) sin(1 + 2x)3
(ii)sin 2 x + 2 x cos 2 x
4 x sin x
x cos x − sin x + 1
(iii)
(iv)6(1 + 2 x) 2 cos(1 + 2 x)3
2
( x + cos x)
Ans. (i )
81. Differentiate the following w.r.t. x and find the value when x = 9.
(i) sin 5x
(ii) tan 5x
(iii) cos 5x
(iv) sin x .
1
6
Ans. (i )5 / 2 (ii )10 (iii ) − 5 / 2 (iv) cos3
82. Differentiate the following w.r.t. x and find the value when x = 3
(i) (6x2 + 3) (3x + 7)
(ii) (x –1)/(x2 + 1)
Ans. (i) 747 (ii) –1/50
83. Differentiate
x+3 + x−3
w.r.t. x.
x+3 − x−3
Ans.
84. If y = cos (ax2 + b), calculate
dy
.
dx
1
x
1+
3
x2 − 9
Ans. –2ax sin (ax2 + b)
85. Differentiate etan w.r.t. .
Ans. sec2
etan
86. Differentiate log e
2
w.r.t. x.
x
Ans. −
87. Differentiate ex cos x w.r.t. x.
1
x
Ans. ex (cos x – sin x)
88. Differentiate the following w.r.t. x
(i) ex log x
(ii) log (4x2 + 3)
(iii) (loge 2x2)
Ans. (i )e x
1
8x
2
+ log x (ii ) 2
(iii ) log e 2 x
x
4x + 3
x
89. Differentiate the following w.r.t. x
(i) sin 3 x
(ii)
sin x
.
log x + e x
(log x + e x ) cos x − sin x
Ans. (i )3sin 2 x cos x (ii )
90. If x = at3 and y = bt2, calculate the value of
91. Differentiate
92. If x = a cos
1 x
+e
x
(log x + e x )2
dy
.
dx
Ans.
2b
3at
Ans.
1
(1 − t 3 ) 2
t3
w.r.t. t3.
3
1− t
and y = b sin , find
dy
.
dx
b
a
Ans. − cot
93. If x = a ( + sin ) and y = a (1 – cos ), find
dy
.
dx
Ans. tan /2
d2y
, if y = x3 + tan x.
94. Find
dx 2
Ans. (6x + 2 sec2 x tan x)
95. I f y = A sin x + B cos x, then prove that
d2y
+ y = 0.
dx 2
dS
96. Given : S = 3t + 9. Calculate (i)
dt
d 2S
(ii)
.
dt 2
2
Ans. (i) 6t (ii) 6
97. If the displacement x of a particle (in metre) is related with time (in second) according to
related
x = 2t3 – 3t2 + 2t + 2
find the position, velocity and acceleration of a particle at the end of 2 seconds.
Ans. 10 m ; 14 m/s ; 18 m/s2
98. A metallic disc is being heated. Its area A (in m2) at any time t (in sec) is given by
A = 5t2 + 4t + 8
Calculate the rate of increase of increase of area at t = 3s.
Ans. 34 m2/s
99. Find the rate of change of the area of a circle with respect to its radius r when r = 5 cm.
Ans. 10
100. The volume of a cube is increasing at a rate of 9 cubic centimetres per second. How
fast is the surface area increasing when the length of an edge is 10 centimetres?
Ans. 3.6 cm2/s
101. A stone is dropped into a quiet lake and waves move in circles at a speed of 4cm
per second. At the instant, when the radius of the circular wave is 10 cm, how fast is the
enclosed area increasing?
Ans. 80 cm2/s
102. The length x of a rectangle is decreasing at the rate of 3 cm/minute and the width y
is increasing at the rate of 2cm/minute. When x = 10cm and y = 6cm, find the rates of
change of (a) the perimeter and (b) the area of the rectangle.
Ans. (a) –2 cm/min (b) 2 cm2/min
103. Find the intervals in which the function f given by f(x) = x2 – 4x + 6 is
(a) strictly increasing
(b) strictly decreasing
Ans. (a) x < 2 (b) x > 2
104.
Find the slope of the tangent to the curve y = x3 – x at x = 2.
Ans. 11
105.
2
.
3
Ans. (3, 2)
Find the point at which the tangent to the curve y = 4 x − 3 − 1 has its slope
x2 y 2
106. Find points on the curve
+
= 1 at which the tangents are (i) parallel to x-axis
4 25
(ii) parallel to y-axis.
Ans. (i) (0,5), (0 – 5) (ii) (2,0), (–2,0)
107.
Use differential to approximate
36.6.
Ans. 6.05
1
3
108.
Use differential to approximate (25) .
109.
Find the approximate value of f (3.02), where f (x) = 3x2 + 5x + 3.
Ans. 2.926
Ans. 45.46
110.
Find the absolute maximum and minimum values of a function f given by
f (x) = 2x3 – 15x2 + 36x + 1 on the interval [1, 5].
Ans. 56, 24
111. Let AP and BQ be two vertical poles at points A and B, respectively. If AP = 16 m,
BQ = 22 m and AB = 20 m, then find the distance of a point R on AB from the point A
such that RP2 + RQ2 is minimum.
Ans. 10 m
112. If length of three sides of a trapezium other than base are equal to 10 cm, then find
the area of the trapezium when it is maximum.
Ans. 75 3 cm2
113. Prove that the radius of the right circular cylinder of greatest curved surface area
which can be inscribed in a given cone is half of that of the cone.
114. Show that the right circular cone of least curved surface and given volume has an
altitude equal to 2 time the radius of the base.
115.
Show that semi-vertical angle of right circular cone of given surface area and
1
maximum volume is sin–1
.
3
116. Show that the right circular cylinder of given surface and maximum volume is such
that its height is equal to the diameter of the base.
117. A water tank has the shape of an inverted right circular cone with its axis vertical
and vertex lowermost. Its semi-vertical angel is tan–1 (0.5). Water is poured into it at a
constant rate of 5 cubic metre per hour. Find the rate at which the level of the water is
rising at the instant when the depth of water in the tank is 4 m.
35
Ans.
m/h
88
118. A man of height 2 metres walks at a uniform speed of 5 km/h away from a lamp
post which is 6 metres high. Find the rate at which the length of his shadow increases.
5
Ans. km/h
2
119. A circular disc of radius 3 cm is being heated. Due to expansion, its radius increases
at the rate of 0.05 cm/s. Find the rate at which its area is increasing when radius is 3.2
cm.
Ans. 0.32 cm2/s
120. An open topped box is to be constructed by removing equal squares from each
corner of a 3 metre by 8 metre rectangular sheet of aluminium and folding up the sides.
Find the volume of the largest such box.
200 3
m
Ans.
27
121.
Integrate the following with respect to x
(ii) x −
(i) 3 x5
(iv)
1
x+
x
2
1
(v ) x +
x
1
x
(iii) (3x − 4)3
3
(vi) 3x 2 +
1
− 2x
x2
3
2
2
x2
Ans. (i ) x8 / 3 + C (ii ) x3 / 2 − 2 x1/ 2 + C (iii ) (3x − 4)5 / 2 + C (iv) + log e x + 2 x + C
8
3
15
2
4
2
x 3x
1
1
(v ) +
+ 3log e x − 2 + C (vi ) x3 − − x 2 + C
4
2
2x
x
122.
Integrate the following with respect to x.
1
x −1
1
(iv)
1 − cos x
(i)
123.
1
(iii) 1 + cos 2x
1 + sin x
1
(v)
(vi) cos2 x
1 − sin x
Ans. (i) 2 x − 1 + C (ii) tan x – sec x + C (iii) 2 sin x + C
x sin 2 x
+C
(iv) –(cot x + cosec x) + C (v) tan x + sec x + C (vi) +
2
4
(ii)
Integrate the following:
(i) 6 x + 5 x 2 − 2 x 3 +
Ans. (i) 3 x 2 +
1
x2
(ii) 3cos ec2 x − 5x + sin x (iii) 3cos ec2 x + 2sin3x
5 3 1 4 1
5
3
x − x − + c (ii) −3cot x − x 2 − cos x + c (iii) −3cot x − cos 3 x + c
3
2
x
2
2
124. Evaluate the following integrals
(i) cos3 x sin x dx
(ii)
x3 log x dx
(iv)
(v)
x 2 sin x dx
x e2 x dx
(iii)
x cos x dx
(vi) log e x dx
cos 4 x
e2 x
(i) −
+ C (ii)
(2 x − 1) + C (iii) (2 − x 2 cos x) + 2 x sin x + C
x
4
4
x
1
log e x −
+ C (v) x sin x + cos x + C (vi) x(loge x − 1) + C
(iv)
4
4
Ans.
125. Integrate the following functions w.r.t. x and find within the limits x = 2 to 4.
(i) x3
(ii) x .
Ans. (i) 60 (ii) 3.448
126.
Evaluate the following integrals:
4
4
x dx
(i)
(t + 1)dt
(ii)
/4
sec x tan x dx
(v)
0
− /2
1
dx
x
2
Ans. (i )
127.
14
1
3
(ii )25 (iii )2 (iv) 2 − 1 (v) log
3
3
2
Evaluate the following integrals
30
30
cos(4 x − 3) dx
(i)
sin 7x dx
(ii)
15
0
30
10
cos5x dx
(iii)
0
sec 2 (3 x + 6) x dx
(iv)
0
Ans. (i)
128.
cos xdx
(iii)
0
3
1
(iv)
/2
2
Evaluate
1
1
1
(sin117 o − sin 57 o ) (ii) 0.26665 (iii)
(iv) (tan 36o − tan 6o )
4
3
10
/4
/2
(1 + cos x)1/ 2 dx
sin x cos x dx (ii)
(i)
0
0
/2
/4
(1 + sin x)1/ 2 dx
(iii)
(1 − cos 2 x)1/ 2 dx
(iv)
0
0
1
4
Ans. (i ) (ii )2(iii )2(iv)1
129.
Evaluate the following:
/4
1
( 1 − sin 2 x )dx
(i)
0
(ii)
0
x5
dx.
x6 + 1
Ans. (i)
130.
2 − 1 (ii)
1
log 2
6
Find the area of the region bounded by the curve y = x2 and the line y = 4.
32
Ans.
3
131. Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4
and the x-axis.
14
Ans.
3
132.
133.
Find the area of the region bounded by the two parabolas y = x2 and y2 = x.
1
Ans.
3
x2 y 2
Find the area enclosed by the ellipse 2 + 2 = 1
a
b
Ans. ab
134.
Find the area of the parabola y2 = 4ax bounded by its latus rectum.
Ans.
135.
8 2
a
3
Find the area bounded by the curve y = cos x between x = 0 and x = 2 .
Ans. 4
136. In a bank, principal increases continuously at the rate of 5% per year. In how many
years Rs 1000 double itself?
Ans. 20 loge2
137. In a bank, principal increases continuously at the rate of r% per year. Find the value
of r if Rs 100 double itself in 10 years (loge2 = 0.6931).
Ans. 6.93%
138. In a bank, principal increases continuously at the rate of 5% per year. An amount of
Rs 1000 is deposited with this bank, how much will it worth after 10 years
(e0.5 = 1.648).
Ans. 1648
139. In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2
hours. In how many hours will the count reach 2,00,000, if the rate of growth of
bacteria is proportional to the number present?
2 log 2
Ans.
11
log
10
140.
Find the equation of a curve passing through the point (–2, 3), given that the slope
y3
2x
of the tangent to the curve at any point (x, y) is 2 .
Ans.
= x2 + 5
3
y
141. The population of a village increases continuously at the rate proportional to the
number of its inhabitants present at any time. If the population of the village was
20, 000 in 1999 amd 25000 in the year 2004, what will be the population of the village
in 2009?
Ans. 31250
MEASUREMENTS
1. Deduce dimensional formulae for (i) angle (ii) angular velocity (iii) angular acceleration
(iv) torque (v) angular momentum and (vi) moment of inertia.
Ans. (i) Dimensionless (ii) T–1 (iii) T–2 (iv) ML2T–2 (v) ML2T–1 (vi) ML2
2.
Obtain dimension of (i) impulse (ii) power (iii) surface energy (iv) coefficient of
viscosity (v) bulk modulus (vi) force constant.
Ans. (i) MLT–1 (ii) ML2T–3 (iii) ML0T–2 (iv) ML–1T–1 (v) ML–1T–2 (vi) ML0T–2
3.
If force (F), length (L) and time (T) are chosen as the fundamental quantities, then
what would be the dimensional formula for density?
Ans. FL–4T–1
4. Find the dimension of linear momentum and surface tension in terms of velocity
density and frequency v as fundamental quantities.
Ans. 4 v–3, 3 v–1
,
5. If E, M, J and G respectively denote energy, mass, angular momentum and gravitational
constant, calculate the dimension of EJ2/M5 G2.
Ans. A dimensionless quantity
6. The value of acceleration due to gravity at a place is 9.8 ms–2. Find its value in km h–2.
Ans. 127008 km h–2
7. Find the value of 100 J on a system which has 20 cm, 250 g and half minute as
fundament al units of length, mass and time.
Ans. 9 106 new units
8. If the units of force, energy and velocity are 20 N, 200 J and 5 ms–1, find the units of
length, mass and time.
Ans. 10m, 8 kg, 2s
9. If the velocity of light is taken as the unit of velocity and an year as the unit of time,
what is the unit of length? What is it called?
Ans. 9.46 1015 m, light year
10. The viscous force ‘F’ acting on a small sphere of radius ‘r’ moving with velocity
through a liquid is given by F = 6 r . Calculate the dimensions of , the coefficient of
viscosity.
Ans. ML–1 T–1
11. Test if the following equation is dimensionally correct:
h=
where
2 S cos
r g
h = height, S = surface tension, = density,
r = radius and g = acceleration due to gravity.
Ans. Correct
12. The time period of a compound pendulum is given by
T =2
I
mgl
where
I = moment of inertia about the centre of suspension,
g = acceleration due to gravity, m = mass of the pendulum,
l = distance of the centre of gravity from the centre of suspension.
Check the validity of this formula.
Ans. Correct
13. Find the dimensions of (a
b) in the equation : E =
distance and t is time.
b − x2
; where E is energy, x is
at
Ans. M–1 L2 T
14. Find the dimensions of (a/b) in the equation:
a − t2
P=
bx
where P is pressure, x is distance and t is time.
Ans. MT–2
15. Obtain an expression for the centripetal force F acting on a particle of mass m moving
with velocity in a circle of radius r. Take dimensionless constant K = 1.
Ans. F =
m 2
r
16. The orbital velocity of a satellite may depend on its mass m, the distance r from the
centre of the earth and acceleration due to gravity g. Obtain an expression for its orbital
velocity.
Ans. = K rg
17. The velocity of a freely falling body is function of the distance fallen through (h) and
acceleration due to gravity g. Show by the method of dimension that = K gh .
18. A body of mass m hung at one end of the spring executes SHM. Prove that the relation
T = 2 m/k is incorrect, where k is the force constant of the spring. Also derive the correct
relation.
Ans. T = K m / k
19. The depth x to which a bullet penetrates a human body depends upon (i) coefficient of
elasticity and (ii) kinetic energy Ek. By the method of dimensions, show that
x
Ek
1/ 3
.
20. Write the order of magnitude of the following:
(i) 8
(ii) 52
(iii) 999
(iv) 753000
(v) 0.05
Ans. (i) 1 (ii) 2 (iii) 3 (iv) 6 (v) –2
21. What is the order of magnitude of seconds in a day?
Ans. 5
22. Round off the following numbers as indicated:
(i) 15.654 upto 3 digits
(ii) 15.75 upto 3 digits
(iii) 5.996 105 upto 3 digits
(iv) 2.5946 10–4 upto 2 digits.
Ans. (i) 15.7 (ii) 15.8 (iii) 6.00 105 (iv) 2.6
10–4
23. State the number of significant in the following:
(i) 2.653 104
(ii) 0.00368
Ans. (i) 4 (ii) 3
24. State the number of significant figures in the following measurements:
(i) 0.009 m2
(ii) 0.1890 g cm–3
(iii) 0.020800 m
Ans. (i) 1 (ii) 4 (iii) 5
25. Solve the following to the appropriate number of significant figures:
(i)
324 0.08666
5.006
(ii)
2.03 10−5 3.5 10−7
0.6423
Ans. (i) 0.0560 (ii) 11.1
102
26. The mass of a box measured by a grocer’s balance is 4.2 kg. Two additional masses
10.20 g and 15.25 g are added to the box. What is the total mass of the box?
Ans. 4.2 kg
27. The length, breadth and thickness of a metal block are 4.327 m, 2.825 m and 4.32 cm
respectively. Calculate its (i) surface area and (ii) volume and express the results to an
appropriate number of significant figures.
Ans. (i) 25.1 m2 (ii) 0.528 m3
28. Each side of a cube is measured to be 7.203 m. What is (i) the total surface area and
(ii) the volume of the cube to appropriate significant figures?
Ans. (i) 311.3 m2 (ii) 373.7 m3
29. 5.74 g of a substance occupies 1.2 cm3. Express its density by keeping the significant
figures in view.
Ans. 4.8 gm-cm–3
30. The diameter of a wire as measured by a screw gauge was found to be 0.026 cm,
0.028 cm, 0.029 cm, 0.027 cm, 0.024 cm and 0.027 cm. Calculate (i) mean value of the
diameter (ii) mean absolute error (iii) relative error (iv) percentage error. Also express
the result in terms of absolute error and percentage error.
Ans. (i) 0.027 cm (ii) 0.001 cm (iii) 0.037 (iv) 3.7%; (0.027 0.001)cm, (0.027 + 3.7%)cm
31. In an experiment to measure focal length of a concave mirror, the value of focal length in
successive observations turns out to be 17.3 cm, 17.8 cm, 18.3 cm, 18.2 cm, 17.9 cm and
18.0 cm. Calculate the mean absolute error and percentage error. Express the result in a
proper way.
Ans. 0.25 cm, 1.4%, (17.9 0.25) cm
32. The lengths of two rods are recorded as l1 = (25.2
Find their combined length.
0.1) cm and l2 = (16.8
Ans. (42.0
33. The initial and final temperatures of water were recorded as (56.3
(27.5 + 0.3)ºC. Determine the fall in the temperature of water.
Ans. (28.8
0.1)cm.
0.2) cm
0.4)ºC and
0.7)ºC
34. While measuring the volume of a sphere, an error of 1.2% is committed in the
measurement of radius. What percent error is introduced in the measurement of its
volume.
Ans. 3.6%
35. The radius of a sphere is 5.3
0.1 cm. Calculate the percentage error in its volume.
Ans. 5.7 %
36. The measure of the diameter of a cylinder is (1.60
0.01) cm and its length is
(5.0 0.1) cm. Calculate the percentage error in its volume.
Ans. 3.25%
37. The measured mass and volume of a body are 2.00 g and 5.0 cm3 respectively. With
possible errors of 0.01 g and 0.1 cm3, what would be the percent error in density?
Ans. 2.5%
38. A body travels uniformly a distance of (13.8 0.2)m in a time (4.0
velocity with error limits. What is percentage error in velocity?
0.3)s. Calculate its
Ans. 9%
39. The centripetal force acting on a body of mass m moving with speed
path of radius r is given by
along a circular
F=
m 2
r
If the values of m, and r are measured as 0.5 kg, 10 ms–1 and 0.4 m respectively to the
accuracies of 0.005 kg, 0.01 ms–1 and 0.01 m respectively, calculate the percentage error
in the force acting on the body.
Ans. 3.6%
40. The time period of oscillation of a simple pendulum is T = 2
L
. L is about 10 cm and
g
is known to 1 mm accuracy. The time period of oscillation is about 0.5 s. The time of
100 oscillations is measured with a wrist watch to an accuracy of 1 s resolution. What is
the accuracy in determination of g?
Ans. 5%
41. A physical quantity X is related to three observables a, b, c as X = a b2 / c 2 . The
errors of measurement in a, b and c are 2%, 1% and 3% respectively. What is the
percentage error in the quantity X?
Ans. 10%
42. The Young’s modulus Y is determined by stretching a wire by using the formula,
Y=
4FL
d 2l
where F is the stretching force, L is length of wire, l is extension in length and d is its
diameter. If F is of the order of 500 N and known to 1 part in 1000, L is of the order of
3 m and measured with an accuracy of 1 mm, l is of the order of 5 mm measured to
0.1 mm and d is of the order of 1 mm measured correct upto 0.01 mm, estimate the
percentage error in the measurement of Y.
Ans. 4.13%
43. The smallest division on the main scale of a vernier calipers is 1 mm, and 10 vernier
divisions coincide with 9 scale divisions. While measuring the diameter of a sphere, the
zero mark of the vernier scale lies between 2.0 and 2.1 cm and the fifth division of the
vernier scale coincide with a scale division.
(i)
Determine the vernier constant.
(ii)
Find the diameter of the sphere.
Ans. (i) 0.01 cm (ii) 2.05 cm
44. The length of a cylinder is measured with the help of a vernier calipers whose smallest
division on the main scale is 0.5 mm, and nine divisions of the main scale are equal to
ten division of the vernier scale. It is observed that 78th divisions of the main scale
coincides with the sixth division of the vernier scale. Calculate the length of the cylinder.
Ans. 3.63 cm
45. The main scale of a vernier calipers reads in millimeter and 10 vernier divisions coincide
with 9 main scale divisions. While measuring the diameter of a sphere, the vernier scale
lies between 3.0 and 3.1 cm and the fifth division of the vernier scale is in line with a
scale division.
(a) Calculate the vernier constant.
(b) Determine the diameter of the sphere.
Ans. (a) 0.01 cm (b) 3.05 cm
46. In a certain barometer, the least reading on the main scale is 0.5 mm and 50 divisions on
the vernier coincide with 49 main scale divisions. What is the vernier constant?
Ans. 0.001 cm
47. A circular scale is marked in degree and each degree is divided into 3 parts. There are
20 divisions on the vernier scale which coincide with 19 scale divisions. The reading of
an angle is 2.5 degree and 2 small divisons and 17 division of the vernier scale coincides.
(a) Determine the vernier constant.
(b) Find the value of the angle.
Ans. (a) 1 ' (b) 25º57 '
48. The pitch of a screw gauge is 0.5 mm and there are 100 divisions on its circular scale.
The instrument reads +2 divisions when nothing is put in-between its jaws. In measuring
the diameter of a wire, there are 8 divisions on the main scale and 83rd division coincides
with the reference line.
(a) Calculate the least count of the instrument
(b) Find the diameter of the wire
49. The distance advanced by the screw of a screw gauge is 2 mm in four rotations. Its cap is
divided into 50 divisions. There is no zero error.
(a)
Find the pitch of the screw gauge
(b)
Find the least count of the instrument
(c)
Find the diameter of a wire, if the screw reads 3 divisions on the main
scale and 32 divisions on the cap.
Ans. (a) 0.5 mm (b) 0.01 mm
50. The pitch of a screw gauge is 1 mm and there are 100 divisions on its circular scale.
When nothing is put in between its jaws, the zero of the circular scale lies 4 divisions
below the reference line. When a steel wire is placed between the jaws, two main scale
divisions are clearly visible and 67 divisions on the circular scale are observed.
Determine the diameter of the wire.
Ans. 2.63 mm
51. The pitch of a screw gauge is 1 mm and there are 50 divisions on its cap. When nothing
is put in between the studs, 44th division of the circular scale coincides with the
reference line. When a glass plate is placed between the studs, the main scale reads three
divisions and the circular scale reads 26 divisions. Calculate the thickness of the plate.
Ans. 3.64 mm
52. The main scale of a screw gauge reads in millimeter. The cap of the instrument is
divided into 100 equal parts. Find the diameter of a wire if no division on the main scale
is completed and the cap has been moved through 37 divisions, the zero error being –3
division.
Ans. 0.40 mm
VECTORS
→
→
→
→
→
→
1. ABCDE is a pentagon. Prove that AB + BC + CD+ DE + EA = 0.
→
→
2. ABCD is a parallelogram and AC and BD are its diagonals.
→
→
→
→
→
→
Prove that (i) AC + BD = 2 BC and (ii) AC − BD = 2 AB .
3. In figure, ABCDEF is a regular hexagon. Prove that
→
→
→
→
→
→
AB + AC + AD+ AE + AF = 6 AO .
4. Two forces of 5 N and 7 N act on a particle with an angle of 60º between them. Find the
resultant force.
Ans. 10.44 N, 35º29´ with 5 N force
5. Find the resultant of two forces, one 6 N-due east and other 8N due north.
Ans. 10N, 53º8 ' with 6N
6. Two vectors, both equal in magnitude, have their resultant equal in magnitude of the
either. Find the angle between the two vectors.
Ans. 120º
7. Calculate the angle between a 2N force and a 3N force so that their resultant is 4N.
Ans.
= cos −1
1
= 75o31 '
4
8. Two forces whose magnitudes are in the ratio of 3 : 5 give a resultant of 35 N. If the
angle of inclination be 60º, calculate the magnitude of each force.
Ans. 15 N, 25 N.
9. Two equal forces have the square of their resultant equal to three times their product.
Find the angle between them.
Ans. 60º
10. When the angle between two vectors of equal magnitude is 2 /3, prove that the
magnitude of the resultant is equal to either.
11. The resultant of two equal forces acting at right angles to each other is 1414 dyne. Find
the magnitude of either force.
Ans. 1000 dyne
12. A particle is acted upon by four forces simultaneously: (i) 30 N due east (ii) 20 N due
north (iii) 50 N due west and (iv) 40 N due south. Find the resultant force on the particle.
Ans. 20 2 N , 45o south of west
13. Two boys raising a load pull at an angle to each other. If they exert forces of 30 N and
60 N respectively and their effective pull is at right angles to the direction of the pull of
the first boy, what is the angle between their arms? What is the effective pull?
Ans. 120º. 30 3 N
14. Two forces equal to P and 2P newton act on a particle. If the first be doubled and the
second be increased by 20 newton, the direction of the resultant is unaltered. Find the
value of P.
Ans. 10N
15. The sum of the magnitudes of two forces acting at a point is 18 N and the magnitude of
their resultant is 12 N. If the resultant makes an angle of 90º with the force of smaller
magnitude, what are the magnitudes of the two forces?
Ans. 5 N, 13 N
16. At what angle do the two forces (P + Q) and (P – Q) act so that the resultant is 3P 2 + Q2 .
Ans. 60º
→
→
→
→
17. The resultant vector of P and Q is R . On reversing the direction of Q , the resultant
→
vector becomes S . Show that : R2 + S2 = 2 (P2 + Q2)
18. Establish the following vector inequalities:
→
→
→
→
(i )| a + b | | a | + | b |
→
→
→
→
(ii )| a + b | | a | − | b | .
When does the equality sign apply?
19. Establish the following vector inequalities:
→
→
→
→
(i )| a − b | | a | + | b |
→
→
→
→
(ii )| a − b | | a | − | b | .
20. A force is inclined at 30º to the horizontal. If rectangular component in the horizontal
direction is 50 N, find the magnitude of the force and its vertical component.
Ans. 57.74 N, 28.87 N
21. A child pulls a rope attached to a stone with a force of 80 N. The rope makes an angle of
40º to the ground. (i) Calculate the effective value of the pull tending to move the stone
along the ground. (ii) Calculate the force tending to lift the stone.
Ans. (i) 45.96 N (ii) 38.57 N
22. Calculate the resultant of the following forces acting at a point, making use of resolution
process.
(i)
100 2 dyne along north-east
980 2 dyne along north-west
1960 dyne along south.
(ii)
(iii)
2 dyne
Ans. 880
→
23. The x- and y-components of A are 4 and 6 m respectively. The x- and y-components of
→
→
→
vector ( A + B ) are 10 and 9 m respectively. Calculate for the vector B (i) its x-and ycomponents (ii) its length and (iii) the angle it makes with the x-axis.
Ans. (i) 6 m and 3 m (ii)
= tan −1
45 (iii)
1
= 26.6o
2
→
24. Find the vector AB and its magnitude if it has initial point A(1, 2, – 1) and final point B
Ans. 2 i + 3 k
(3, 2, 2).
→
→
25. Find the values of x, y and z so that the vectors a = x i + 2 j + z k and b = 2 i + y j + k are
equal.
Ans. x = 2, y = 2, z = 1
26. Find the direction cosines of the vector joining the points A(1, 2, –3) and
B(–1, –2, 1), directed from A to B.
1 −2 2
Ans. − , ,
3 3 3
27. Find the direction cosines of the vector i + 2 j + 3 k .
→
→
Ans.
→
→
1
2
3
,
,
14 14 14
→
→
28. If A = 3 i + 2 j and B = i − 2 j + 3 k , find the magnitudes of A + B and A − B .
Ans. 5,
29
→
29. Find the unit vector parallel to the resultant of the vectors A = 2 i − 6 j − 3 k and
→
B = 4 i + 3 j− k .
Ans. 1/ 61(6 i − 3 j − 4 k )
→
30. Determine the vector which when added to the resultant of A = 2 i − 4 j − 6 k and
→
B = 4 i + 3 j + 3 k given the unit vector along z-axis.
Ans. −6 i + j + 4 k
→
→
→
31. Given three coplanar vectors a = 4 i − j , b = −3 i + 2 j and c = −3 j . Find the
magnitude of the sum of the three vectors.
Ans. 5
→
→
32. Two forces F1 = 3 i + 4 j and F2 = 3 j + 4 k are acting simultaneously at a point. What
is the magnitude of the resultant force?
Ans. 174 units of force
→
→
→
33. If A = 3 i + 4 j and B = 7 i + 24 j , find a vector having the same magnitude as B and
→
parallel to A .
Ans. 15 i + 20 j
34. Find the position vector of the mid point of the vector joining the points P(2, 3, 4) and
Ans. (3 i + 4 j + k )
Q(4, 1, –2).
35. Calculate the values of (i) j . (2 i − 3 j + k ) and (2 i − j ).(3 i + k ) .
Ans. (i) –3 (ii) 6
→
→
36. Find the angle between the vectors A = 2 i − 4 j + 6 k and B = 3 i + j + 2 k .
Ans. 60º
→
→
37. Show that the vector A = i + j + k is perpendicular to the vector B = − i − j + 2 k .
38. Find the value of m so that the vector 3 i − 2 j + k is perpendicular to the vector
2 i + 6 j+ m k .
Ans. 6
39. Find the angles between the following pairs of vectors:
→
→
(i) A = i + j + k and B = −2 i − j − 2 k .
→
→
(ii) A = −2 i + 2 j − k and B = 3 i + 6 j + 2 k .
→
→
(iii) A = 4 i + 6 j − 3 k and B = −2 i − 5 j + 7 k
Ans. (i) 180º (ii) 79º (iii) 148.9º
→
→
→
→
→
→
40. The sum and difference of two vectors A and B are A + B = 2 i + 6 j + k and A − B =
→
→
4 i + 2 j − 11k . Find the magnitude of each vector and their scalar product A . B .
50, 41, −25
Ans.
41. If the resultant of the vectors 3 i + 4 j + 5 k and 5 i + 3 j + 4 k makes an angle with
x-axis, then find cos .
Ans. 0.5744
→
→
→
→
→
→
42. If a = 5 i − j − 3 k and b = i + 3 j − 5 k , then show that the vectors a + b and a − b are
perpendicular.
→
→
43. Find the projection of the vector a = 2 i + 3 j + 2 k on the vector b = i + 2 j + k .
Ans.
→
→
5 6
3
→
44. Show that the vectors A = 3 i − 2 j + k , B = i − 3 j + 5 k and C = 2 i + j − 4 k form a
right angled triangle.
→
→
→
→
→
→
45. If vectors A , B and C have magnitudes 8, 15 and 17 units and A + B = C , find the
→
→
angle between A and B .
Ans. 90º
→
→
→
→
→
→
46. Three vectors A , B and C are such that A = B + C and their magnitudes are 5, 4 and
→
→
3 respectively. Find the angle between A and C .
= cos–1 (0.6) = 53º
Ans.
→
→
→
→
→
→
47. If | A+ B | = | A− B | , find the angle between A and B .
Ans. 90º
→
→
→
→
→
48. If A = B – C , then determine the angle between A and B .
A2 + B 2 − C 2
= cos
2 AB
−1
Ans.
→
→
→
→
→
→
→
49. For two vectors A and B if A + B = C and A + B = C, then prove that A and B
parallel to each other.
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→
→
→
50. Prove that ( A + 2 B ).(2 A − 3 B ) = 2 A2 + AB cos − 6B 2 .
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→
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51. If A + B = C and A2 + B2 = C2, then prove that A and B perpendicular to each other.
52. If unit vectors a and b are inclined at angle , then prove that
| a − b |= 2sin
2
53. Find the components of a = 2 i + 3 j along the directions of vectors i + j and i − j .
Ans.
→ →
→
→
→
5
1
( i + j ), − ( i − j )
2
2
→
54. Let a , b and c be three vectors such that | a | = 3,| b | = 4,| c | = 5 and each one of them
→
→
→
being perpendicular to the sum of the other two, find | a + b + c | .
→
Ans. 5 2
→
55. Prove that the vectors A = 4 i + 3 j + k and B = 12 i + 9 j + 3 k are parallel to each
other.
56. Find the value of a for which the vectors 3 i + 3 j + 9 k and i + a j + 3 k are parallel.
Ans. a = 2/3
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→
→
→
57. If A = i + 3 j + 2 k and B = 4 i + 3 j + k , then find the vector product A B .
Ans. 4 i + 4 j − 8 k
→
→
→
→
→
→
58. If A = 2 i + 3 j + k and B = 3 i + 2 j + 4 k , then find the value of ( A + B ) ( A − B )
Ans. −20 i + 10 j + 10 k
→
→
59. Find a unit vector perpendicular the vector A = 4 i − j + 3 k and B = −2 i + j − 2 k .
Ans.
→
1
(− i + 2 j + 2 k )
2
→
60. Find the sine of the angle between the vectors A = 3 i − 4 j + 5 k and B = i − j + k .
Ans. 1/5
61. Find a vector of magnitude 18 which is perpendicular to both the vectors 4 i − j + 3 k
and −2 i + j − 2 k .
Ans. −6 i + 12 j + 12 k
62. Determine the area of the parallelogram whose adjacent sides are formed by the vectors
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→
A = i − 3 j + k and B = i + j + k .
Ans. 4 2 square units
63. Find the area of the triangle formed by points O, A and B such that
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OA = i + 2 j + 3 k and OB = −3 i − 2 j + k .
Ans. 3 5 square units
64. Find with the help of vectors, the area of the triangle with vertices A(3, –1, 2),
B(1, –1, –3) and C(4, –3, 1).
Ans. 165 square units
→
65. Find the area of the triangle formed by the tips of the vectors a = i − j − 3 k ,
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→
b = 4 i − 3 j + k and c = 3 i − j + 2 k .
Ans. 6.4 square units
66. The diagonals of a parallelogram are given by the vectors 3 i + j + 2 k and i − 3 j + 4 k .
Find the area of the parallelogram.
Ans. 8.66 square units
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67. Prove that ( a + b ) ( a − b ) = 2( b
→
→
a)
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→
68. For any three vectors A , B and C , prove that
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→
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→
→
→
→
→
→
A ( B + C ) + B (C + A) + C ( A + B ) = O .
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69. Prove that | a
→
→
→
→
b |= a 2b2 − ( a . b )2
→
→
→
→
→
70. Find | A B | if | A | = 10, | B | = 2 and A : B = 12.
Ans. 16
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→
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→
71. Find A , B if | A | = 2, | B | = 5 and | A B | = 8.
Ans.
→ → →
72. If a , b , c are three vectors such that
→ →
→ → →
→
→
→ →
a.b = a.c, a b = a c, a
→
→
o
→
then prove that b = c .
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→
73. If a = i − 2 j − 3 k , b = 2 i + j − k and c = i + 3 j − 2 k , then find a ( b
→
c ).
Ans. − i − 8 j + 5 k
→
→
→
→
74. If a = i − 2 j − 3 k , b = 2 i − j − k and c = i + 3 j − 2 k , find ( a
→
→
b) c .
Ans. −5 i + 15 j + 20 k
75. In any ABC, prove that
a
b
c
=
=
.
sin A sin B sin C
6