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11. TRIGONOMETRY
It is one of the important branch of mathematics which deal with relations of
sides and angles of triangle and also with the relevant functions of any angle.
Consider a ray OA. If this ray revolves about its end point O in anti-clockwise
direction and takes position OB, then we say that the angle ∠AOB has been
generated as shown in the following figure.
Or simply say that angle is a measure of an amount of revolution of a given ray
about its initial point.
The angle is positive (or negative), if the initial line revolves in anti-clockwise
(or clockwise) direction to get the terminal line.
System of measurement of angles
(i) Sexagesimal system In this system, a right angle is divided into 90 equal
parts, called degree.
Thus, 1 right angle = °
90 (degrees)
1 60
° = ′ (minutes)
1 60
′ = ′′ (seconds)
0
CHAPTER
Basic
Mathematics
Inside
1
2
3
Trigonometry
Graphs
Calculus (Differentiation)
Integration
4
O A
B
Initial line
Terminal line O A
B
Initial line
Terminal line
(i) Positive angle (ii) Negative angle
Fig. 0.1

12. (ii) Circular system In this system, the unit of
measurement is radian.
θ =
l
r
radian or rad
1 right angle =
π
2
rad
1 straight angle = π rad
and 1 complete angle = 2π rad
One radian (denoted by 1c
) is the measure of an
angle subtended at the centre of a circle by an arc of
length equal to the radius of the circle.
1 rad (1c
) =
°
≈ ° ′ ′′ ≈ °
180
57 17 45
π
57.3
Note (i) π = =
22
7
314
. (ii) 360 2
° = π radian,
π
2
radian = °
90
Example 0.1 Find the radian measures corresponding to the
following degree measures.
(i) 75° (ii) − ° ′
37 30 (iii) 5 37 30
° ′ ′′
Sol. We have,1
180
c
=
°
π
⇒ 1
180
° =






π
c
(i) 75 75
180
5
12
° = ×





 =






π π
c c
(ii) − ° ′ = −






°
= −






°
= − ×





 =
37 30 37
1
2
75
2
75
2 180
π
c
−






5
24
π
c
(iii) 5 37 30 5
5
8
45
8
45
8 180
° ′ ′′ =






°
=






°
= ×






π
c
=






π
32
c
Example 0.2 Find the degree measures corresponding to the
following radian measures.
(i)
2
15
π






c
(ii)
π
8






c
(iii) ( )
− 2 c
Sol. We have,
(i)
2
15
2
15
180
24
π π
π





 = ×






°
= °
c
(ii)
π π
π
8 8
180 45
2





 = ×






°
=






°
c
=






°
= ° ×






′
22
1
2
22
1
2
60
= ° ′
22 30
(iii) ( )
− = − ×






°
= − × ×






°
2 2
180
2
180
22
7
c
π
= −






°
114
6
11
= − ° ×





 = − °






′ ′
114
6
11
60 114 32
8
11
= − ° ′ ×






″
− ° ′ ′′
114 32
8
11
60 114 32 44
• [ ]
Trigonometrical ratios (or T-ratios)
Consider the two fixed lines X OX
′ andYOY′ intersecting
each other at right angle at point O as shown in the
following figure
Then,
(i) Intersection point O is called origin.
(ii) X OX
′ andYOY′ are called X-axis andY-axis,
respectively.
(iii) The portion XOY,YOX′, X OY
′ ′ andY OX
′ are
known as I, II, III and IV quadrants, respectively.
Now, consider that the revolving line OA has traced out an
angle θ in anti-clockwise direction (in I quadrant).
From point A, draw AB OX
⊥ which results a right angled
∆ABO, where AB = perpendicular, OA = hypotenuse and
OB = base.
The three sides of right angled triangle are related to each
other through side having different ratios, called
trigonometrical ratios or T-ratios, which are given as
(i) sinθ = =
Perpendicular
Hypotenuse
AB
OA
(From Fig. 0.3)
(ii) cos θ = =
Base
Hypotenuse
OB
OA
(iii) tan θ = =
Perpendicular
Base
AB
OB
(iv) cot θ = =
Base
Perpendicular
OB
AB
(v) sec θ = =
Hypotenuse
Base
OA
OB
(vi) cosec
Hypotenuse
Perpendicular
θ = =
OA
AB
Fundamental of T-ratios or
trigonometric functions
For any acute angle say θ ( )
< °
90 , the functions are given as
(i) cosec θ
θ
=
1
sin
(ii) sec
cos
θ
θ
=
1
2 OBJECTIVE Physics Vol. 1
X′ X
Y′
Y
θ
A
B
O
Fig. 0.3 Trigonometrical ratios
θ
O r
l
Fig. 0.2 Circular system

13. (iii) cot
tan
θ
θ
=
1
(iv) sin cos
2 2
1
θ θ
+ =
(v) 1 2 2
+ =
tan sec
θ θ
(vi) 1 2
+ =
cot θ θ
cosec2
Signs of trigonometric ratios or T-ratios in
various quadrants
(i) In I quadrant, all T-ratios are positive.
(ii) In II quadrant, sin θ is + ve, cos θ and tan θ are − ve .
(iii) In III quadrant, tan θ is + ve, sin θ and cos θ are − ve .
(iv) In IV quadrant, cos θ is + ve, sin θ and tan θ are − ve.
Example 0.3 If sin θ =
4
5
, where θ lies in the first quadrant,
then find all the other T-ratios.
Sol. Let ∆PQR be right angled triangle, right angled at Q.
sin θ =
4
5
(Given)
∴ PR = 5 and PQ = 4
On applying Pythagoras theorem in ∆PQR, we have
( ) ( ) ( )
PR PQ QR
2 2 2
= +
⇒ ( ) ( ) ( )
5 4
2 2 2
= + QR
⇒ QR = − = =
25 16 9 3
[Taking positive value of square root]
Now, cos θ = =
QR
PR
3
5
,
tan θ = =
PQ
QR
4
3
,
cot θ = =
QR
PQ
3
4
,
sec θ = =
PR
QR
5
3
and cosec θ = =
PR
PQ
5
4
T-ratios of allied angles
In trigonometry two angles are said to be allied angles
when their sum or difference is a multiple of 90°.
The T-ratios of the following allied angles are as
(i) When angle (say) θ is negative, then
(a) sin( ) sin
− = −
θ θ (b) cos( ) cos
− =
θ θ
(c) tan ( ) tan
− = −
θ θ
(ii) When angle θ is less than 90°
(i.e., lies in I quadrant), then
(a) sin( ) cos
90° − =
θ θ (b) cos( ) sin
90° − =
θ θ
(c) tan( ) cot
90° − =
θ θ
(iii) When angle θ lies between 90° and 180°
(i.e., lies in II quadrant), then
(a) ● sin( ) cos
90° + =
θ θ
● cos(90 ) sin
° + = −
θ θ
● tan( ) cot
90° + = −
θ θ
(b) ● sin( ) sin
180° − =
θ θ
● cos(180 cos
° − = −
θ) θ
● tan( ) tan
180° − = −
θ θ
(iv) When angle θ lies between 180° and 270°
(i.e., lies in III quadrant), then
(a) ● sin( ) sin
180° + = −
θ θ
● cos( ) cos
180° + = −
θ θ
● tan( ) tan
180° + =
θ θ
(b) ● sin( ) cos
270° − = −
θ θ
● cos( ) sin
270° − = −
θ θ
● tan( ) cot
270° − =
θ θ
(v) When angle θ lies between 270° and 360°
(i.e., lies in IV quadrant), then
(a) ● sin( ) cos
270° + = −
θ θ
● cos( ) sin
270° + =
θ θ
● tan( ) cot
270° + = −
θ θ
(b) ● sin( ) sin
360° − = −
θ θ
● cos( ) cos
360° − =
θ θ
● tan( tan
360° − = −
θ) θ
Values of T-ratios of some standard angles
Angle (θ)
0° 30°
=






π
6
45°
=






π
4
60°
=






π
3
90°
=






π
2
120°
=






2
3
π
135°
=






3
4
π
150°
=






5
6
π
180°
(= π)
sin θ
0 1
2
1
2
3
2
1 3
2
1
2
1
2
0
cos θ
1 3
2
1
2
1
2
0
−
1
2
−
1
2
−
3
2
− 1
tan θ
0 1
3
1 3 ∞ − 3 − 1
−
1
3
0
Basic Mathematics 3
I
All + ve
IV
only cos and
sec are + ve
II
only sin and cosec
are + ve
III
only tan and
cot are + ve
Fig. 0.4 Sign of T-ratios
Q
P
R
θ
4
5

14. Example 0.4 Find the value of
(i) sin( )
− °
45 (ii) tan 225°
(iii) cos 300° (iv) sec 120°
Sol. (i) sin( ) sin
− ° = − °
45 45 [Qsin( sin
− = −
θ) θ]
= −
1
2
[Qsin 45
1
2
° = ]
(ii) tan tan( ) cot
225 270 45 45
° = ° − ° = °
[Qtan( cot
270° − =
θ) θ]
= 1 [Qcot 45 1
° = ]
(iii) cos cos( ) sin
300 270 30 30
° = ° + ° = °
[Qcos( sin
270° + =
θ) θ]
=
1
2
[Qsin 30
1
2
° = ]
(iv) sec sec( ) sec
120 180 60 60
° = ° − ° = − °
[Qsec( sec
180° − = −
θ) θ]
= − 2
Some important formulae of
trigonometry
● sin( ) sin cos cos sin
A B A B A B
+ = +
● sin( ) sin cos cos sin
A B A B A B
− = −
● cos( ) cos cos sin sin
A B A B A B
+ = −
● cos( ) cos cos sin sin
A B A B A B
− = +
● tan( )
tan tan
tan tan
A B
A B
A B
+ =
+
−
1
● tan( )
tan tan
tan tan
A B
A B
A B
− =
−
+
1
● sin sin cos
tan
tan
2 2
2
1 2
A A A
A
A
= =
+
● cos cos sin cos
2 2 1
2 2 2
A A A A
= − = −
= − =
−
+
1 2
1
1
2
2
2
sin
tan
tan
A
A
A
● tan
tan
tan
2
2
1 2
A
A
A
=
−
● sin sin sin
3 3 4 3
A A A
= −
● cos cos cos
3 4 3
3
A A A
= −
● tan
tan tan
tan
3
3
1 3
3
2
A
A A
A
=
−
−
● sin( ) sin( ) sin cos
A B A B A B
+ + − = 2
● sin( ) sin( ) cos sin
A B A B A B
+ − − = 2
● cos( ) cos( ) cos cos
A B A B A B
+ + − = 2
● cos( ) cos( ) sin sin
A B A B A B
+ − − = − 2
● sin sin sin cos
C D
C D C D
+ =
+





 ⋅
−






2
2 2
● sin sin cos sin
C D
C D C D
− =
+





 ⋅
−






2
2 2
● cos cos cos cos
C D
C D C D
+ =
+





 ⋅
−






2
2 2
● cos cos sin sin
C D
C D C D
− = −
+





 ⋅
−






2
2 2
Example 0.5 Find the value of (i) sin15° (ii) tan 75°
Sol. (i) We have,
sin sin( )
15 45 30
° = ° − °
= ° ° − ° °
sin cos cos sin
45 30 45 30
[Qsin( ) sin cos
A B A B
− = − cos sin
A B]
= ⋅ − ⋅
1
2
3
2
1
2
1
2
(Qsin cos
45 45
1
2
° = ° = , cos 30
3
2
° =
and sin 30
1
2
° = )
=
−
3 1
2 2
(ii) We have, tan tan( )
75 45 30
° = ° + °
=
° + °
− °⋅ °
tan tan
tan tan
45 30
1 45 30
Q tan ( )
tan tan
tan tan
A B
A B
A B
+ =
+
− ⋅






1
=
+
− ⋅
1
1
3
1 1
1
3
Q tan tan
45 1 30
1
3
° = ° =






and
=
+
−
3 1
3 1
4 OBJECTIVE Physics Vol. 1

15. 1. Find the radian measures corresponding to the following
degree measures.
(i) 25° (ii) − ° ′
47 30
(iii) 39 22 30
° ′ ′′
Ans. (i)
5
36
π
(ii) −
19
72
π
(iii)
7
32
π
2. Find the degree measures corresponding to the following
radian measures.
(i)
18
5
π


 


c
(ii) ( )
− 3 c
(iii) −


 


5
6
π
c
(iv)
9
5
π


 


c
Ans. (i) 648° (ii) − ° ′ ′′
171 49 5 (iii) − °
150
(iv) 324°
3. Find sin θ and tan θ, if cos θ = −
12
13
and θ lies in the third
quadrant.
Ans. sin θ = −
5
13
and tan θ =
5
12
4. Find the values of other five T-ratios, if tan θ = −
3
4
and θ
lies in II quadrant.
Ans. sin θ =
3
5
, cos θ = −
4
5
, cosec θ =
5
3
, sec θ = −
5
4
and cot θ = −
4
3
5. Find the values of the following T-ratios
(i) cosec 315° (ii) cos 210°
(iii) sin( )
− °
330
Ans. (i) − 2 (ii) −
3
2
(iii)
1
2
6. Find the value of
(i) sec165° (ii) cot 105°
Ans. (i) ( )
2 6
− (ii)
1 3
1 3
−
+
CHECK POINT 0.1
CALCULUS (DIFFERENTIATION)
Differentiation in calculus, is the process of finding
the derivative. The derivative is the instantaneous
rate of change of a function with respect to one of its
variable. This is equivalent to finding the slope of the
tangent line to the function at a point.
Variable
A quantity, which can take different values, is called a
variable quantity. A variable is usually represented by
x, y, z etc.
Constant
A quantity, whose value remains unchanged during
mathematical operations, is called a constant quantity.
The integers, fractions such as π, e, etc are all
constant.
Function
A quantity y is called a function of a variable x, if
corresponding to any given value of x, there exists a
single definite value of y. The phrase ‘y is function of
x’ is represented as y f x
= ( ).
e.g., Consider that y is a function of the variable x
which is given by y x x
= + +
4 3 7
2
and
y x ex
= +
sin .
Here, we will treat x as independent variable and y as
dependent variable, i.e., the value y depends on x. If
we change the value of x, then y will change.
Physical meaning of
dy
dx
(i) The ratio of small change in the function y and the
variable x is called the average rate of change of y w.r.t.
x.
e.g. If a body covers a small distance ∆s in small time ∆t,
then average velocity of the body, v
s
t
av =
∆
∆
(ii) The limiting value of
∆
∆
y
x
, when ∆ x → 0, i.e.,
lim
∆
∆
∆
x
y
x
dy
dx
→
=
0
, is called the instantaneous rate of
change of y w.r.t. x.
Thus, the differentiation of a function w.r.t. a variable
implies the instantaneous rate of change of the function
w.r.t. that variable.
e.g., Instantaneous velocity of the body,
lim
∆
∆
∆
t
s
t
ds
dt
→
=
0
Theorems of differentiation
(i) If c is constant, then
d
dx
c
( ) = 0
(ii) If y cv
= , where c is a constant and v is a function of x,
then
dy
dx
d
dx
c v c
dv
dx
= ⋅ =
( )

16. (iii) If y xn
= , where n is a real number, then
dy
dx
d
dx
x nx
n n
= = −
( ) 1
(iv) If y u v w
= ± ± , where u, v and w are functions of
x, then
dy
dx
d
dx
u v w
du
dx
dv
dx
dw
dx
= ± ± = ± ±
( )
(v) Product rule If y u v
= ⋅ , where u and v are
functions of x, then
dy
dx
d
dx
u v u
dv
dx
v
du
dx
= ⋅ = +
( )
(vi) Division rule If y
u
v
= , where u and v are functions
of x, then
dy
dx
d
dx
u
v
v
du
dx
u
dv
dx
v
=





 =
−
2
Example 0.6 Differentiate the following functions
(i) y x
= − 3
(ii) 6 4 3 2 7
5 3 2
x x x x
+ − + −
(iii) y x x
= + +
( ) ( )
2 1
2
(iv) y
x
x
=
+
+
( )
( )
3
4
1
Sol. (i) We have, y x
= − 3
On differentiating both sides w.r.t. x, we get
dy
dx
x
= − − −
3 3 1
Q
d
dx
x nx
n n
=






−1
= − −
3 4
x =
− 3
4
x
(ii) Let y x x x x
= + − + −
6 4 3 2 7
5 3 2
On differentiating both sides w.r.t. x, we get
dy
dx
d
dx
x
d
dx
x
d
dx
x
d
dx
x
= + − +
6 4 3 2
5 3 2
−
d
dx
7
Q
d
dx
c v c
dv
dx
( )
⋅ = ⋅






⇒
dy
dx
x x x
= ⋅ + ⋅ − ⋅
− − −
6 5 4 3 3 2
5 1 3 1 2 1
+ ⋅ −
−
2 0
1 1
x
Q
d
dx
x nx
d
dx
c
n n
= =






−1
0
and
⇒
dy
dx
x x x
= + − + ×
30 12 6 2 1
4 2
( )
Q x x
1 1 0
1
−
= =
⇒
dy
dx
x x x
= + − +
30 12 6 2
4 2
(iii) We have, y x x
= + +
( ) ( )
2 1
2
On differentiating both sides w.r.t. x, we get
dy
dx
x
d
dx
x x
d
dx
= + + + +
( ) ( ) ( )
2 1 1
2 2
( )
x + 2
(By product rule)
= + + + + +
( ) ( ) ( ) ( )
x x x
2 2 0 1 1 0
2
= + + +
2 2 1
2
x x x
( ) = + + +
2 4 1
2 2
x x x = + +
3 4 1
2
x x
(iv) We have, y
x
x
=
+
+
3
4
1
On differentiating both sides w.r.t. x, we get
dy
dx
x
d
dx
x x
d
dx
x
x
=
+ + − + +
+
( ) ( ) ( ) ( )
( )
1 4 4 1
1
3 3
2
(By division rule)
dy
dx
x x x
x
=
+ + − + +
+
( ) ( ) ( ) ( )
( )
1 3 0 4 1 0
1
2 3
2
=
+ − −
+
3 1 4
1
2 3
2
x x x
x
( )
( )
=
+ − −
+
3 3 4
1
3 2 3
2
x x x
x
( )
=
+ −
+
2 3 4
1
3 2
2
x x
x
( )
Formulae for differential coefficient of
trigonometric, logarithmic and exponential function
●
d
dx
x x
(sin ) cos
= ●
d
dx
x x
(cos ) sin
= −
●
d
dx
x x
(tan ) sec
= 2 ●
d
dx
x x
(cot ) = − cosec2
●
d
dx
x x x
(sec ) sec tan
= ⋅
●
d
dx
x x x
( ) cot
cosec cosec
= − ⋅
●
d
dx
x
x
(log ) =
1 ●
d
dx
e e
x x
( ) =
Example 0.7 Differentiate the following functions
(i) y x ex
= +
sin (ii) y x x ex
= + + +
3 4 5
2
log
(iii) y e x
x
= ⋅ tan
Sol. (i) We have, y x ex
= +
sin
On differentiating both sides w.r.t. x, we get
dy
dx
d
dx
x ex
= +
(sin ) = +
cos x ex
(ii) We have, y x x ex
= + + +
3 4 5
2
log
On differentiating both sides w.r.t. x, we get
dy
dx
d
dx
x x ex
= + + +
( log )
3 4 5
2
= + + +
3 4 5
2
d
dx
x
d
dx
x
d
dx
e
d
dx
x
(log )
= ⋅ + + +
−
3 2x
x
ex
2 1 1
4 0 = + +
6
1
4
x
x
ex
(iii) We have, y e x
x
= ⋅ tan
On differentiating both sides w.r.t. x, we get
dy
dx
e
d
dx
x x
d
dx
e
x x
= ⋅ +
tan tan (By product rule)
= ⋅ + ⋅
e x x e
x x
sec tan
2
= +
2
e x x
x
(sec tan )
6 OBJECTIVE Physics Vol. 1

17. Chain rule
This rule is applied, when the given function is the
function of function, i.e., a function is in the form of
f g x
[ ( )].
∴
d
dx
f g x f g x g x
[ ( )] [ ( )] ( )
= ′ ⋅ ′
Example 0.8 Differentiate sin( )
x2
5
+ w.r.t. x.
Sol. Let y x
= +
sin( )
2
5
On differentiating both sides w.r.t. ‘x’, we get
dy
dx
d
dx
x
= +
sin( )
2
5 = + ⋅ +
cos( ) ( )
x
d
dx
x
2 2
5 5
= + ⋅ +
cos( ) ( )
x x
2
5 2 0 = +
2 5
2
x x
cos( )
Maxima and minima
Let y f x
= ( ), where y is a function of x.
For y to be minimum or maximum, put
dy
dx
= 0, then
find x.
If
d y
dx
2
2
0
< , y is maximum and if
d y
dx
2
2
0
> , y is minimum.
Note Most of time, it is known from physical situation whether the
quantity is a maximum or minimum; therefore, there is no need
to check maximum or minimum with the help of
d y
dx
2
2
.
Example 0.9 Divide a number 1000 in two parts such that
there product is maximum.
Sol. Let the two parts be x and ( )
1000 − x .
∴Their product, P x x
= −
( )
1000 ⇒ P x x
= −
1000 2
On differentiating both sides w.r.t. x, we get
dP
dx
x
= −
1000 2 ...(i)
For P to be maximum or minimum,
dP
dx
= 0 ⇒ 1000 2 0
− =
x ⇒ x = 500
On differentiating both sides of Eq. (i) w.r.t. x, we get
d P
dx
2
2
2 0
= − <
∴ P is maximum at x = 500
On dividing equally, the two parts are ( , )
500 500 .
Remembering points
● All the problems of maxima/minima cannot be solved by the above
methods e.g., y x
= 2
, y is maximum when x is maximum.
● If y x
=sin , by simple observation, y is maximum if sin x is
maximum, i.e., sin x = 1. The value of sine or cosine functions lies
between − 1and + 1.
● The product of the two parts is the maximum when the parts are
equal.
Applications of differentiation in physics
(i) If the displacement is a function of time t, then to
find the velocity, differentiate s w.r.t. t.
s f t
= ( ), v
ds
dt
=
(ii) If the velocity is a function of time t, then to find
acceleration, differentiate v w.r.t. t
v f t
= ( ), a
dv
dt
=
=





 =
d
dt
ds
dt
d s
dt
2
2
(iii) If the velocity is a function of displacement s, then
to find the acceleration, differentiate v w.r.t. t and
use the expression, a v
dv
ds
=
(iv) Consider the motion along the X-axis
● If v > 0, s is increasing, then the particle is moving
along the positive X-axis.
● If v < 0, s is decreasing, then the particle is moving
along the negative X-axis.
● If a > 0, v > 0, then speed is increasing along the
positive X-axis.
● If a > 0, v < 0, then speed is decreasing along the
negative X-axis.
● If a < 0, v > 0, then the speed is decreasing along the
positive X-axis.
● If a < 0, v < 0, then the speed is increasing along the
negative X-axis.
Note
(i) If v and a have same sign, then the speed is increasing.
(ii) If v and a have opposite sign, then the speed is decreasing.
Example 0.10 The displacement of a particle as a function of
time t is given by s t t t
= + + + 4
α β γ δ
2
, where α, β, γ and
δ are constants. Find the ratio of the initial velocity to the
initial acceleration.
Sol. First find the velocity and acceleration in terms of time t,
then use t = 0 to find the initial values.
s t t t
= + + +
α β γ δ
2 4
(Given)
On differentiating both sides w.r.t. t, we get
ds
dt
t t
= + +
β γ δ
2 4 3
⇒ v t t
= + +
β γ δ
2 4 3
Q
ds
dt
v
=





 ...(i)
On differentiating both sides of Eq. (i), w.r.t. t, we get
dv
dt
t
= + ⋅
2 2
γ δ
4 3
Basic Mathematics 7

18. ⇒ a
dv
dt
t
= = + 2
2 12
γ δ Q
dv
dt
a
=






At t = 0, v = β and a = 2γ
∴
Initial velocity
Initial acceleration
=
β
γ
2
Example 0.11 The position of a particle moving along the
X-axis varies with time t as x t t
= − +
6 4
2
. Find the time
interval during which the particle is moving along the positive
x-direction.
Sol. Given, x t t
= − +
6 4
2
v
dx
dt
d
dt
t t t t
= = − + = − + = −
( ) ( )
6 4 6 2 0 2 3
2
At t < 3, v > 0, then the particle is moving along the positive
x-direction.
At t > 3, v < 0, then particle is moving along the negative
x-direction.
At t = 3, v = 0
For time-interval t = 0 to t = 3, the particle is moving along
the positive x-direction.
1. Differentiate the following function
(i) y x
x
x
= + +
3 2
1
4
2
log (ii) y x x
= + +
( ) ( )
2
1 2
(iii) y
x
x
=
+
3
1
2
(iv) y x
=sin
(v) y x x
= + +
tan( )
2
3 1
(vi) y x e x
x
= + + +
3 7 1
2
cos log( )
Ans. (i)12 4
1 1
3
3
x
x x
− + (ii) 3 4 1
2
x x
+ +
(iii)
3 6
1
2
2
x x
x
+
+
( )
(iv) cos x
(v) ( ) sec ( )
2 3 3 1
2 2
x x x
+ + + (vi) − + +
+
3 7
2
1
2
sin x e
x
x
x
2. A particle is moving with velocity v t t
= − +
3 2
6 4, where v is
in m/s and t is in seconds. At what time will the velocity be
maximum/minimum and what is it equal to?
Ans. vmax = 4 m/s at t = 0 s and vmin = − 28 m/s at t = 4 s
3. If the time and displacement of particle along the positive
X-axis are related as t x
= −
( )
2
1
2
1 , then find the acceleration
in terms of x.
Ans.
1
3
x
8 OBJECTIVE Physics Vol. 1
CHECK POINT 0.2
INTEGRATION
It means summation. It is the process of finding the function,
whose derivative is given. In other word, integration is the
reverse process of differentiation. It’s symbol is .
∫
Consider a function f x
( ), whose derivative w.r.t. x is equal to
′
f x
( ), then f x C
( ) + is called integration of ′
f x
( ), where C is
called constant of integration. Symbolically, it is written as
′ = +
∫ f x dx f x C
( ) ( )
Here, ′
f x dx
( ) is called element of integration and ∫ is called
indefinite integral.
Some basic formulae of integration
● x dx
x
n
C
n
n
=
+
+
+
∫
1
1
; n ≠ − 1 Q
d
dx
x
n
x
n
n
+
+





 =






1
1
● 1dx x C
= +
∫ Q
d
dx
x =






1
● sin cos
x dx x C
= − +
∫ Q
d
dx
x x
− =






(cos ) sin
● cos sin
x dx x C
= +
∫ Q
d
dx
x x
(sin ) cos
=






●
1
x
dx x C
e
= +
∫ log Q
d
dx
x
x
e
log =






1
● e dx e C
x x
= +
∫ Q
d
dx
e e
x x
=






Example 0.12 Evaluate the following integrals.
(i) ( )
e
x
x dx
x
+ + +
∫
1
2 3
2
(ii) cos /
x x
x x
dx
+ + +






∫ 3
3 4
1 2
2
Sol. (i) Let I e
x
x dx
x
= + + +






∫
1
2 3
2
= + + + ∫
∫
∫
∫e dx
x
dx x dx dx
x 1
2 3 1
2
= + + ⋅
+
+ +
+
e x
x
x C
x
e
log 2
2 1
3
2 1

19. = + + + +
e x x x C
x
e
log
2
3
3
3
(ii) Let I x x
x x
dx
= + + +






∫ cos /
3
3 4
1 2
2
= + + + ∫
∫
∫
∫ cos /
x dx x dx
x
dx
x
dx
3 3
1
4
1
1 2
2
= + ⋅
+
+ +
− +
+
+ − +
sin
( / )
log
( / )
x
x
x
x
C
e
3
1 2 1
3 4
2 1
1 2 1 2 1
= + + − +
sin log
/
x x x
x
C
e
2 3
4
3 2
Other important formula of integration
I f ax b dx
f ax b
d
dx
ax b
= ′ + =
+
+






∫ ( )
( )
( )
Example 0.13 Evaluate the following
(i) ( )
2 1 3
x dx
+
∫ (ii)
1
a x
dx
−






∫
(iii) ( )
x x dx
2 4
3 4
+ +
∫ (iv) sin ( )
2 2
x dx
∫
Sol. (i) Let I x dx
= +
∫ ( )
2 1 3
=
+
+ +






+
+
( )
( ) ( )
2 1
3 1 2 1
3 1
x
d
dx
x
C
=
+
⋅
+ =
+
+
( ) ( )
2 1
4 2
2 1
8
4 4
x
C
x
C
(ii) Let I
a x
dx
=
−






∫
1
=
−
−
+
log ( )
( )
a x
d
dx
a x
C
=
−
−
+
log ( )
a x
C
1
= − − +
log ( )
a x C
(iii) Let I x x dx
= + +
∫ ( )
2 4
3 4
=
+ +
+ + +
+
+
( )
( ) ( )
x x
d
dx
x x
C
2 4 1
2
3 4
4 1 3 4
=
+ +
+
+
( )
( )
x x
x
C
2 5
3 4
5 2 3
(iv) Let I x dx
= ∫ sin ( )
2 2
=
−
+
cos ( )
( )
2
2
2
2
x
d
dx
x
C
=
−
+
cos ( )
2
4
2
x
x
C
Definite integral
When a function is integrated between a lower limit and
an upper limit, it is called a definite integral.
A definite integral has definite value.
If
d
dx
f x f x
( ) ( )
= ′ , then ′
∫ f x dx
a
b
( ) is called definite integral,
where a and b are called lower and upper limit,
respectively of variable x.
After carrying out integration, the result is evaluated
between upper and lower limits as shown below
′ = = −
∫ f x dx f x f b f a
a
b
a
b
( ) | ( )| ( ) ( )
Example 0.14 Evaluate the following
(i) ( )
4 2 2 1
3 2
0
2
x x x dx
+ + +
∫
(ii) (sin cos )
/
x x dx
+
∫
0
4
π
(iii)
dx
x
2
4
∫
Sol. (i) Let I x x x dx
= + + +
∫ ( )
4 2 2 1
3 2
0
2
= + + + ∫
∫
∫
∫
4 2 2 1
3 2
0
2
0
2
0
2
0
2
x dx x dx x dx dx
= + + +
4
4
2
3
2
2
4
0
2
3
0
2
2
0
2
0
2
x x x
x
| |
=
−





 +
−





 +
−






4
2 0
4
2
2 0
3
2
2 0
2
4 4 3 3 2 2
+ −
( )
2 0
= + × + +
16
2
3
8 4 2
I = + =
+
=
22
16
3
66 16
3
82
3
(ii) Let I x x dx
= +
∫ (sin cos )
/
0
4
π
= + ∫
∫ sin cos
/
/
x dx x dx
0
4
0
4 π
π
= − +
| cos | |sin |
/ /
x x
0
4
0
4
π π
= − + + −
cos cos sin sin
π π
4
0
4
0
= − + + −
1
2
1
1
2
0 = 1
(iii) Let I
dx
x
= ∫
2
4
= loge x 2
4
= −
log log
e e
4 2
=





 =
log log
e e
4
2
2 Qlog log log
e e e
a b
a
b
− =












Basic Mathematics 9

20. Example 0.15 Evaluate the following
(i) e dx
x
( )
+
∫
4
1
2
(ii) cos ( )
/
2 2
0
4
x x dx
+
∫
π
(iii)
dx
x
( )
3 4
1
2
+
∫ (iv) ( )
3 4 5
2
0
4
x x dx
+ +
∫
Sol. (i) Let I e dx
x
= +
∫
( )
4
1
2
=
+
+
e
d
dx
x
x 4
1
2
4
( )
=
+
ex 4
1
2
1
= −
+ +
e e
2 4 1 4
= −
e e
6 5
= −
e e
5
1
( )
(ii) Let I x x dx
= +
∫ cos ( )
/
2 2
0
4
π
=
+
+
sin ( )
( )
/
2
2
2
2
0
4
x x
d
dx
x x
π
=
+
+
sin ( )
/
2
4 1
2
0
4
x x
x
π
=
+






⋅ +
−
× +
× +
sin .
sin( )
2
16 4
4
4
1
2 0 0
4 0 1
2
2
π π
π
=
+






+
−
sin
( )
sin
π π
π
2
2
8
1
0
1
=
+






+
−
sin
π π
π
2
2
8
1
0 =
+






+






1
1
2
8
2
π
π π
sin
(iii) Let I
dx
x
=
+
∫ ( )
3 4
1
2
=
+
+
log ( )
( )
3 4
3 4
1
2
x
d
dx
x
=
+
log ( )
3 4
3 1
2
x
=
× + − × +
log ( ) log ( )
3 2 4 3 1 4
3
= −
1
3
10 7
[log log ] =






1
3
10
7
log
(iv) Let I x x dx
= + +
∫ ( )
3 4 5
2
0
4
=
+
+
+
+
+ +
3
2 1
4
1 1
5
2 1 1 1
0
4
x x
x
= + +
3
3
4
2
5
3 2
0
4
x x
x
= + +
| |
x x x
3 2
0
4
2 5
= + × + × −
4 2 4 5 4 0
3 2
= + +
64 32 20 = +
64 52 = 116
Application of integration in physics
We know that, v
ds
dt
a
dv
dt
= =
, or v
dv
ds
If the displacement is given and we have to find the
velocity and acceleration, then we use differentiation.
But if the acceleration is given and we have to find the
velocity and displacement, then we use integration.
Example 0.16 A particle is moving under constant
acceleration a t t
= +
3 4 2
. If the position and velocity of the
particle at start, i.e., t = 0 are x0 and v0, respectively, then
find the displacement and velocity as a function of time t.
Sol.
Given, a t t
= +
3 4 2
dv
dt
t t
= +
3 4 2
Qa
dv
dt
=






⇒ dv t t
t
v
v
= +
∫
∫ ( )
3 4 2
0
0
dt (Integrating both sides)
⇒ | |
v
t t
v
v
t
0
3
2
4
3
2 3
0
= +
⇒ ( ) ( ) ( )
v v t t
− = − + −
0
2 3
3
2
0
4
3
0
⇒ v v t t
= + +
0
2 3
3
2
4
3
…(i)
⇒
dx
dt
v t t
= + +
0
2 3
3
2
4
3
Qv
dx
dt
=






⇒ dx v t t dt
t
x
x
= + +






∫
∫ 0
2 3
0
3
2
4
3
0
⇒ x x v t
t t
t
− = + ⋅ + ⋅
0 0
3 4
0
3
2 3
4
3 4
= + ⋅ + −
v t t t
0
3 4
1
2
1
3
0
⇒ x x v t t t
− = + ⋅ +
0 0
3 4
1
2
1
3
⇒ x x v t t t
= + + +
0 0
3 4
1
2
1
3
Average value
If the velocity is variable and depends on time t, then find
the average value of velocity (v) for time interval t t
= 1 to
t t
= 2.
Let v f t
= ( ); v
v dt
dt
t
t
t
t
=
∫
∫
1
2
1
2
10 OBJECTIVE Physics Vol. 1
A
O
t
x x
v v
= 0
=
=
0
0
t t
x x
v v
=
=
=

21. To find the average value of square of velocity
v
v dt
dt
t
t
t
t
2
2
1
2
1
2
=
∫
∫
⇒ v
v dt
dt
t
t
t
t
4
4
1
2
1
2
=
∫
∫
If velocity is a function of displacement, v f x
= ( ), for
average of v from x x
= 1 to x x
= 2
⇒ v
v dt
dt
x
x
x
x
=
∫
∫
1
2
1
2
The above procedure can be applied to find the average
value of any quantity; like acceleration, force, etc.
Example 0.17 The velocity of a particle is given by
v v
= 0 sin ωt, where v0 is constant and ω π
= 2 /T. Find the
average velocity in time interval t = 0 to t T
= /2.
Sol. We have, v v
= 0 sin ωt
Average velocity, v
v t dt
dt
T
T
=
∫
∫
0
0
2
0
2
sin ( )
/
/
ω
=
−
v t
t
T
T
0 0
2
0
2
cos ( )
| |
/
/
ω
ω
=
−





 +






−






v T
T
0
2
0
2
0
ω
ω
cos cos
=
−





 +






v
T T
T
T
0
2 2
2
1
2
π π
cos .
Qω
π
=






2
T
=
−
2
2
1
0
v T
T
π
π
( cos )
= − −
v0
1 1
π
[ ( )]
( cos cos )
Q π = ° = −
180 1
= +
v0
1 1
π
( ) =
2 0
v
π
1. Evaluate
(i) sin x
x x
x dx
+ + +


 


∫
1
2
1
3
2
3
(ii) ( cos )
3 4 5
2
x e x x dx
x
+ + + +
∫
Ans. (i)− + − + +
cos log
x x
x
x C
2 3
4
4
(ii) 3
4
3 2
5
3
2
sin x e x
x
x C
x
+ + + + +
2. Evaluate the following
(i) ( )
2 4 1
2 2
x x dx
+ +
∫ (ii)
1
2
2
x
dx
+
∫
(iii) cos ( )
x dx
+
∫ 2
Ans. (i)
1
12
2 4 1
1
2 3
( )
( )
x x
x
C
+ +
+
+ (ii)
1
2
2
2
x
x C
log ( )
+ +
(iii) sin ( )
x C
+ +
2
3. Evaluate
(i) 4
1
1
1
3
x
x
dx
+ +


 


∫ (ii) (sin cos )
/
x x dx
−
∫
0
4
π
Ans. (i)18 3
+ log (ii)1 2
−
4. Evaluate the following
(i)
dx
x x
( )
2
0
2
4 1
+ +
∫
(ii) ( )
4 3 2 1
3 2
1
3
x x x dx
+ + +
∫
Ans. (i)
1
8
13
log (ii) 116
5. A particle is moving in a straight line under acceleration
a kt
= , where k is a constant. Find the velocity in term of t, if
the motion starts from rest.
Ans.
kt2
2
6. A particle is moving in a straight line such that its velocity
varies as v v e t
= −
0
λ
, where λ is a constant. Find the average
velocity during the time interval in which the velocity
decrease from v0 to
v0
2
.
Ans.
v
e
0
2
log ( )
Basic Mathematics 11
CHECK POINT 0.3

22. GRAPHS
It is defined as pictorial representation showing the
relation between variable quantities, typically two
variables, each measured along one of a pair of axes at
right angles.
(i) If a graph is concave up (curved upward), the slope
is increasing.
(ii) If a graph is concave downward
(curved downward), the slope is decreasing.
(iii) If the graph is a straight line, the slope is constant.
(iv) The general equation of a straight line is of the form
y mx c
= + where, m is the slope of line, m = tanθ
and c is the intercept on theY-axis.
(a) If m is + ve, i.e., 0 90
° < < °
θ , then lines will be of
the type,
(b) If m is –ve, i.e., 90 180
° < < °
θ , then the lines will
be of the type,
(c) If c is + ve, then the lines will cut theY-axis above
the origin.
(d) If c is –ve, then the lines will cut theY-axis below
the origin.
(e) If c = 0, then the lines will pass through the origin.
(v) Parabola Some standard forms of parabola are as
follows
(a) y kx
2
= , a parabola passing through the origin and
opens rightward.
12 OBJECTIVE Physics Vol. 1
Y
X
O Increasing slope
Fig. 0.5
Y
X
O Decreasing slope
Fig. 0.6
Y
X
O Slope is constant
Fig. 0.7
c
Y
θ
X
y mx c
= +
Fig. 0.8
θ
0° < < 90°
θ
Fig. 0.9
θ
90° < < 180°
θ
Fig. 0.10
O
X
Y
O
X
Y
c c
or
Fig. 0.11
O
X
Y
O
X
Y
c c
or
Fig. 0.12
O
X
Y
O
X
Y
or
Fig. 0.13
Y
X
O
Fig. 0.14

23. (b) y kx
2
= − , a parabola passing through the origin and
opens leftward.
(c) x ky
2
= , a parabola passing through the origin and
opens upward.
(d) x ky
2
= − , a parabola passing through the origin and
opens downward.
(vi) Exponential graph The most popular graph based
on exponential (e) are
(a) y e x
= −
(b) y e x
= − −
1
(vii) Circle If equation of circle is x y a
2 2 2
+ = , where
centre of circle ≡ ( , )
0 0 and radius of circle = a
(viii) Ellipse Equation of ellipse is
x
a
y
b
a b
2
2
2
2
1
+ = >
( ),
where, coefficient of x 2
≠ coefficient of y 2
.
Here, 2a is major axis and 2b is minor axis.
Eccentricity, e
b
a
= −
1
2
2
For ellipse, e < 1
Focus, S ae S ae
≡ ′ ≡ −
( , ), ( , )
0 0
Area of ellipse = π ab
(ix) Hyperbola Equation of hyperbola is
x
a
y
b
2
2
2
2
1
− = .
(x) Rectangular Hyperbola If x
y
∝
1
or xy = constant,
then
Basic Mathematics 13
Y
X
O
Fig. 0.15
Y
X
O
Fig. 0.16
Y
X
O
Fig. 0.17
y
x
O
Fig. 0.18
y
x
O
Fig. 0.19
a
a
X
Y
O (0,0)
Fig. 0.20
X
Y
2b
S′ S
2a
(– ,0)
ae ( ,0)
ae
Fig. 0.21
Y
X
O (0,0)
Fig. 0.22
Y
X
Fig. 0.23

24. Sketches of some standard curves
Equation of curve Diagram/Sketch
1. (a) Straight lines
x a
= and x a
= − ,
where a > 0
(b) Straight lines
y b
= and y b
= – ,
where b > 0
2. Straight lines
y x
= and y x
= –
3. Straight lines
x
a
y
b
+ =1
, a b
≠
x y a a b
+ = =
,
4. Modulus function
y x
= | |
y
x x
x x
=
≥
− <



,
,
for
for
0
0
5. Circle x y a
2 2 2
+ =
Centre = ( , )
0 0
Radius = a
Equation of curve Diagram/Sketch
6. (a) Parabola y ax
2
4
=
and y ax
2
4
= –
Vertex, O = ( , )
0 0
Focus, S a
= ( , )
0
and (– , )
a 0
Length of
latusrectum = 4a
(b) Parabola x ay
2
4
=
and x ay
2
4
= −
Vertex, O = ( , )
0 0
Focus, S a
= ( , )
0
and ( , )
0 − a
Length of
latusrectum = 4 a
7. (a) Ellipse
x
a
y
b
2
2
2
2
1
+ = ,
when a b
>
Vertices = ±
( , )
a 0
Centre, O = ( , )
0 0
(b) Ellipse
x
a
y
b
2
2
2
2
1
+ = ,
when a b
<
Vertices = ±
( , )
0 b
Centre, O = ( , )
0 0
8. (a) Sine function
y x
= sin
(b) Cosine function
y x
= cos
14 OBJECTIVE Physics Vol. 1
Y′
X′ X
Y
O
x = a
( , 0)
a
( , 0)
– a
x a
= –
Y′
X′ X
Y
O
y = b
(0, )
b
(0, – )
b
y = b
–
Y ′
X ′
Y
O
y
=
x
y = x
–
X
Y′
X′ X
Y
O
a
b
A a
( , 0)
x x
B b
(0, )
a b
+ =1
Y¢
X¢ X
Y
O
y
x
=
-
y
x
=
Y¢
X¢ X
Y
O (0, 0)
( , 0)
a
(0, )
a
( , 0)
− a
(0, )
−a
Y¢
X¢ X
Y
( , 0)
a
L¢
L
L
S
( , –2 )
a a
( , 2 )
a a
O
( , 0)
– a
( , 2 )
– a – a
( 2 )
–a, a
L¢ y = ax
2
4
y =– ax
2 4
1
1
Y
X′ X
Y′
(0, )
−a
L′
O
S
( 2 )
− −
a, a (2 )
a, a
−
L(2 )
a, a
L
L′
( 2 )
− a, a
(0, )
a
x = ay
2 4
x = ay
2 − 4
1
1
Y′
X′ X
Y
O (0, 0)
( , 0)
a
(0, )
b
( , 0)
−a
(0, )
−b
Y′
X′ X
Y
O
( , 0)
a
(0, )
b
( , 0)
−a
(0, )
−b
(0, 0)
Y′
X′ X
Y
π/2
π 2π
O
Y′
X′ X
Y
π/2
π
2π
O

25. Example 0.18 Find the area of the region in the first
quadrant enclosed by the X-axis, the line y x
= and the
circle x y
2 2
32
+ = .
Sol. We have, circle
x y
2 2
32
+ = ...(i)
and line y x
= ...(ii)
It is clear from the figure that region for finding area is
OABO.
On putting the value of y from Eq. (ii) in Eq. (i), we get
x x
2 2
32
+ =
⇒ 2 32
2
x = ⇒ x2
16
= ⇒ x = ± 4
From Eq. (ii), we get y = ± 4
Thus, line and circle intersect each other at two points ( , )
4 4
and ( , )
− −
4 4 . So, coordinates of A( , )
4 2 0 , and B( , )
4 4 in
I quadrant.
Now, area of OABO = Area of ODBO + Area of DABD
= +
∫ ∫
y dx y dx
line circle
0
4
4
4 2
= + −
∫ ∫
xdx x dx
0
4
2
4
4 2
32
[From Eq. (i) and Eq. (ii)]
=





 + −
∫
x
x dx
2
0
4
2
4
4 2
2
32
=
−
+ − +












−
4 0
2 2
32
32
2 4 2
2 2
2 1
4
4 2
x
x
x
sin
= +
16
2
4 2
2
32 32
4
2
32 16
32
2
4 2
4 2
32
2
4
4 2
1 1
− − − +
−









− −
sin sin



= + − × + −






− −
8 0 2 4 16 1 16
1
2
1 1
sin ( ) sin
= − +





 −






8 8 16
2
16
4
π π
= −
8 4
π π = 4π sq unit
Hence, the required area of region is 4π sq units.
Example 0.19 Find the area of region bounded by the
curve y x
2
4
= and the line x = 4.
Sol. Given curve is a parabola, y x
2
4
= ...(i)
Which is of the form y aX
2
4
= having vertex ( , )
0 0
and line, x = 4 ...(ii)
Then, the region for which we have to find area is OACBO.
Also, region OACO is symmetrical about X-axis.
On putting the value of x from Eq. (ii) in Eq. (i), we get
y2
4 4 16
= =
( ) ⇒ y = ± 4
∴ Area of bounded region OACBO = 2( )
Area of region OACO
(QParabola is symmetrical about X-axis)
= ∫
2
0
4
y dx
(parabola) = ⋅
∫
2 2 1 2
0
4
x dx
/
[From Eq. (i)]
= ∫
4 1 2
0
4
x dx
/
=






4
3 2
3 2
4
0
x /
/
= ⋅
4
2
3
3 2
0
4
[ ]
/
x
= −
8
3
4 0
3 2
[ ]
/
= ×
8
3
4 4 = × × =
8
3
4 2
64
3
sq units
Hence, the required area is
64
3
sq units.
Example 0.20 Find the area of the region bounded by the
line y x
= +
3 2, the X-axis and the ordinates x = − 1 and
x = 1.
Sol. Given lines are y x
= +
3 2 ...(i)
y = 0 [on X-axis] ...(ii)
x = − 1 ...(iii)
and x = 1 ...(iv)
Now, table for y x
= +
3 2
x
0 −
2
3
y 2 0
The given region bounded by y x
= +
3 2, X-axis and the
ordinates x = − 1 and x = 1 is represented by shaded region.
Basic Mathematics 15
X′ X
Y′
Y
O
D A
B
y x
=
x y
2 2
+ = 32
Y
Y′
X′ X
A(4,4)
O
(0,0) (4, 0)
C
x = 4
B
X′ X
Y′
Y B
A
O
F
C(0,2)
x=1
y
x
=3
+2
E
x = –1
D
2
3
0
,
−





26. ∴ Required area
= Area of region EFDE + Area of region ABDA
= +
−
−
−
∫ ∫
y dx y dx
1
1
2 3
2
2 3
1
/
/
= + + +
−
−
−
∫ ∫
( ) ( )
/
/
3 2 3 2
1
2 3
2 3
1
x dx x dx
= +





 + +






−
−
−
3
2
2
3
2
2
2
1
2 3
2
2 3
1
x
x
x
x
/
/
=





 −






− −












3
2
4
9
4
3
3
2
2 + +





 −





 −












3
2
2
3
2
4
9
4
3
= − +





 + − +






2
3
4
3
1
2
7
2
2
3
4
3
=
− +





 +
− +






4 8 3
6
21 4 8
6
= −





 + = + =
1
6
25
6
1
6
25
6
26
6
=
13
3
sq units
Hence, the required area is
13
3
sq units.
Example 0.21 Find the area of the region bounded by the
ellipse
x
a
y
b
2
2
2
2
1
+ = .
Sol. We have,
x
a
y
b
2
2
2
2
1
+ = ...(i)
Since, power of x and y both are even in the equation of the
curve. So, it is symmetrical about the axes as shown in figure.
Area enclosed by the ellipse = 4 [Area enclosed by the ellipse
and coordinate axes in 1st quadrant]
⇒ A y dx
a
= ∫
4
0
= −
∫
4 2 2
0
b
a
a x dx
a
[From Eq. (i)]
= − +






−
4 1
2
1
2
2 2 2 1
0
b
a
x a x a
x
a
a
sin
= +






= ×






−
4
0
1
2
1
4 1
2 2
2 1 2
b
a
a
b
a
a
sin ( )
π
= πab sq units
1. Find the area of region bounded by the curve y x
= 3
,
y x
= + 6 and x = 0.
Ans. 10 sq units
2. Find the area of the region included between the parabola
y
x
=
3
4
2
and the line 3 2 12 0
x y
− + = .
Ans. 27 sq units
3. For the curve
x y
2 2
4 9
1
+ = , evaluate the area of the region
under the curve and above the X-axis.
Ans. 3π sq units
4. Find the area of the region bounded by y x
= and y x
= .
Ans.
1
6
sq units
16 OBJECTIVE Physics Vol. 1
O
Y′
B b
′(0,– )
dx
B b
(0, )
P x y
( , )
Y
A a
( ,0)
X
A a
′( ,0)
–
X′
CHECK POINT 0.4

27. In this chapter, we will discuss about units and dimensions of different physical
quantities and errors that occur in measurement.
PHYSICAL QUANTITIES AND UNITS
All the quantities which are used to describe the laws of physics are called
physical quantities, e.g. length, mass, volume, etc.
To express the measurement of a physical quantity, we need to know two things
as given below
(i) The unit in which the quantity is measured.
(ii) The numerical value or the magnitude of the quantity.
i.e. The number of times that unit is contained in the given physical quantity
= nu
∴ n
u
∝
1
⇒ nu = constant
Here, n = numerical value of the physical quantity and u = size of unit.
We may also write it as n u n u
1 1 2 2
=
where, n1 and n2 are values of the physical quantity in two different units u1
and u2.
The standard amount of a physical quantity chosen to measure the physical
quantity of same kind is called a physical unit.
The essential requirements of physical unit are given below
(i) It should be of suitable size.
(ii) It should be easily accessible.
(iii) It should not vary with time.
(iv) It should be easily reproducible.
(v) It should not depend on physical conditions like pressure, volume, etc.
01
CHAPTER
Units, Dimensions
Error Analysis
&
Inside
1
2
3
Physical quantities and units
System of units
Dimensions of physical
quantities
Applications of dimensional
analysis
Defects or limitations
of dimensional analysis
Significant figures
Rules to determine significant
figures
Mathematical operations of
significant figures
Rounding off
Order of magnitude
Error in measurement
Expression of errors
Combination of errors
4

28. System of units
A complete set of units which is used to measure all kinds of
fundamental and derived quantities is called a system of
units.
Some of the commonly used systems of units are as follows
(i) CGS system In this system, the units of length, mass
and time are centimetre (cm), gram (g) and second (s),
respectively. The unit of force in this system is dyne
and that of work or energy is erg.
(ii) FPS system In this system, the units of length, mass and
time are foot (ft), pound (lb) and second (s), respectively.
The unit of force in this system is poundal.
(iii) MKS system In this system, the units of length, mass
and time are metre (m), kilogram (kg) and second (s),
respectively. The unit of force in this system is newton
(N) and that of work or energy is joule (J).
(iv) International system (SI) This system of units helps in
revolutionary changes over the MKS system and is
known as rationalised MKS system. It is helpful to
obtain all the physical quantities in physics.
Note
(i) The FPS system is not a metric system. This system is not in much use
these days.
(ii) The drawback of CGS system is that many of the derived units on this
system are inconveniently small.
Fundamental quantities and
fundamental units
Those physical quantities which are independent of each
other and not defined in terms of other physical quantities,
are called fundamental quantities or base quantities. The
units of these quantities are called fundamental or base
units.
Derived quantities and derived units
The quantities which can be expressed in terms of the
fundamental quantities are called derived quantities. The
units of these quantities are called derived units.
e.g. Unit of speed = −
ms 1
can be derived from fundamental
units, i.e. unit of length and time as
Speed
Distance
Time
= = = −
m
s
ms 1
Supplementary quantities and
supplementary units
Other than fundamental and derived quantities, there are
two more quantities called as supplementary quantities.
The units of these quantities are known as supplementary
units.
Fundamental quantities and their SI units
Base
quantity
SI units
Name/
Symbol
Definition
Length Metre
(m)
It is defined by taking the fixed
numerical value of the speed of light in
vacuum c to be 299792458 when
expressed in the unit ms−1
, where the
second is defined in terms of the
caesium frequency ∆νCs.
Mass Kilogram
(kg)
It is defined by taking the fixed
numerical value of the Planck constant h
to be 6.62607015 × −
10 34
when
expressed in the unit Js, which is equal
to kg-m2
s−1
, where the metre and the
second are defined in terms of c and
∆νCs.
Time Second
(s)
It is defined by taking the fixed
numerical value of the caesium
frequency ∆νCs, the unperturbed
ground state hyperfine transition
frequency of the caesium-133 atom, to
be 9192631770 when expressed in the
unit Hz, which is equal to s−1
.
Electric
current
Ampere
(A)
It is defined by taking the fixed
numerical value of the elementary
charge e to be 1.602176634 × −
10 19
when expressed in the unit C.
Thermo-
dynamic
temperature
Kelvin
(K)
It is defined by taking the fixed
numerical value of the Boltzmann
constant k to be1.380649 × −
10 23
when
expressed in the unit JK−1
, which is
equal to kg-m2
s−2
K−1
, where the
kilogram, metre and second are defined
in terms of h c
, and ∆νCs.
Amount of
substance
Mole
(mol)
One mole contains exactly
6 02214076 1023
. × elementary entities.
This number is the fixed numerical
value of the Avogadro constant NA,
when expressed in the unit mol−1
and is
called the Avogadro number. The
amount of substance, symbol n, of a
system is a measure of the number of
specified elementary entities. An
elementary entity may be an atom, a
molecule, an ion, an electron, any other
particle or specified group of particles.
Luminous
intensity
Candela
(cd)
It is defined by taking the fixed
numerical value of the luminous
intensity of monochromatic radiation of
frequency 540 1012
× Hz, Kcd, to be 683
when expressed in the unit lm W−1
,
which is equal to cd sr W−1
, or cd sr kg
−1
m−2
s3
, where the kilogram, metre and
second are defined in terms of h, c and
∆νCs.
18 OBJECTIVE Physics Vol. 1

29. Supplementary quantities and their SI units
Supplementary
quantity
SI units
Name/Symbol Definition
Plane angle Radian
(rad)
One radian is the angle subtended
at the centre by an arc equal in
length to the radius of the circle.
i.e. d
ds
r
θ =
Solid angle Steradian
(sr)
One steradian is the solid angle
subtended at the centre of a
sphere, by that surface of the
sphere, which is equal in area, to
the square of radius of the sphere.
i.e. d
dA
r
Ω = 2
Some other units
(Not contained in SI units)
Length
(i) 1 micron = = −
1 10 6
µm m
(ii) 1 nanometre = = −
1 10 9
nm m
(iii) 1 Angstrom = = = =
− − −
1 10 10 10
10 8 4
Å m cm m
µ
(iv) 1 fermi = 1fm = −
10 15
m
(v) 1 astronomical unit = 1 AU = ×
1.496 1011
m
(vi) 1 light year = 1ly = ×
9467 1015
. m
(vii) 1 parsec = × =
3 08 10 3 26
16
. .
m ly = 206267 AU
Mass
(i) 1 quintal = 100 kg
(ii) 1 tonne or 1metric ton = 1000 kg = 10 quintal
(iii) 1 megagram = 103
kg
(iv) 1 gigagram = 106
kg
(v) 1 teragram = 109
kg
(vi) 1 slug = 1457
. kg
(vii) 1 pound = 1lb = 0 4536
. kg
Time
(i) 1 millisecond = −
10 3
s
(ii) 1 microsecond = −
10 6
s
(iii) 1 shake = −
10 8
s
(iv) 1 nanosecond = −
10 9
s
(v) 1 picosecond = −
10 12
s
(vi) 1 hour = 60 min = 3600 s
(vii) 1 day = 24 hours = 86400 s
(viii) 1 year = 365 days = ×
3.15 107
s
(ix) 1 century = 100 years
Common SI Prefixes and Symbols for
Multiples and Sub-multiples
Multiple Sub-multiple
Factor Prefix Symbol Factor Prefix Symbol
1018
Exa E 10 18
−
atto a
1015
Peta P 10 15
−
femto f
1012
Tera T 10 12
−
pico p
109
Giga G 10 9
−
nano n
106
Mega M 10 6
−
micro µ
103
Kilo k 10 3
−
milli m
102
Hecto h 10 2
−
centi c
101
Deca da 10 1
−
deci d
Example 1.1 The acceleration due to gravity is 9.8 ms−2
.
Give its value in ft s −2
.
Sol. As, 1 m = 3.28 ft
∴ 9.8 ms−2
= ×
9 8 328
. . ft s−2
= 32.14 ft s−2
≈ 32 ft s−2
Example 1.2 The value of gravitational constant G in MKS
system is 6 67 10 11
. × −
N-m2
kg−2
. What will be its value in
CGS system?
Sol. Given, G = × −
6 67 10 11
. N-m2
kg−2
= × −
6 67 10 11
. (kg-ms−2
) m2
kg−2
= × − − −
6 67 10 11 3 2 1
. ( ) ( ) ( )
m s kg
= × − − −
6 67 10 10 10
11 2 3 2 3 1
. ( ) ( ) ( )
cm s g
= × − −
6 67 10 8 3 1
. cm g s−2
= × −
6 67 10 8
. dyne-cm g
2 2
−
Example 1.3 The wavelength of a light is of the order of
6400 Å. Express this in micron and metre.
Sol. As,1Å 10 m
10
= −
∴ Wavelength of light = 6400 Å
= × −
6400 10 m
10
= × −
6.4 10 m
7
Units, Dimensions & Error Analysis 19
r
ds
dθ
O
r
dΩ
dA
O

30. Also, 1 micron = −
10 6
metre
∴ Wavelength of light (in micron) =
× −
−
6 4 10
10
7
6
.
micron
= 0 64
. micron
Example 1.4 How many microns are there in 1 light year?
Sol. 1 ly = ×
9 46 1015
. m
As, 1m = 106
micron
∴ 1 ly = × ×
9 46 10 10
15 6
. micron
= ×
9 46 1021
. micron
≈ 1022
micron (approx.)
Example 1.5 How many microseconds are there in
10 minutes?
Sol. As, 1 second = 106
microseconds
10 minutes = ×
10 60 seconds
= × ×
10 60 106
= ×
6 108
microseconds
Example 1.6 Calculate the angle of
(i) 1° (degree)
(ii) 1′ (minute of arc or arc minute) and
(iii) 1′ ′ (second of arc or arc second)
in radians. (Use 360 2
° = π rad,1 60
° = ′ and1 60
′ = ′ ′)
Sol. (i) 1
2
360
° =
π
rad =
π
180
rad
=
×
22
7 180
rad = × −
1746 10 2
. rad
(ii) 1 arc min = ′ =
°
= ×
1
1
60
1
60 180
π
rad = × −
2 91 10 4
. rad
(iii) 1 arc second = ′ ′ =
′
=
°
×
1
1
60
1
60 60
=
×
×
1
60 60 180
π
rad
= × −
4 85 10 6
. rad
DIMENSIONS OF PHYSICAL
QUANTITIES
The dimensions of a physical quantity are the powers (or
exponents) to which the fundamental quantities must be
raised to represent that quantity completely.
e.g. Density = =
Mass
Volume
Mass
(Length)3
or Density = −
(Mass) (Length) 3
…(i)
Thus, the dimensions of density are 1 in mass and − 3 in
length. The dimensions of all other fundamental quantities
are zero.
Dimensional representation of
physical quantities
For convenience, the fundamental quantities are
represented by one letter symbols. The dependence of all
other physical quantities on these base quantities can be
expressed in terms of their dimensions.
Thus, the seven dimensions of physical quantities are
represented as follows
[M] for mass
[L] for length
[T] for time
[A] for electric current
[K] or [ ]
θ for thermodynamic
temperature
[cd] for luminous intensity
[mol] for amount of substance
The physical quantity that is expressed in terms of the
base quantities is enclosed in square brackets.
Thus, from Eq. (i), dimensions of density can be
represented as[ML ]
3
−
.
Dimensional formula and
dimensional equation
The expression of a physical quantity in terms of its
dimensions is called its dimensional formula. e.g.
Dimensional formula for density is [ ],
ML T
−3 0
the
dimensional formula of force is [ ]
MLT−2
and that for
acceleration is[M LT ]
0 2
−
.
An equation which contains a physical quantity on one side
and its dimensional formula on the other side, is called the
dimensional equation of that quantity.
Dimensional equations for a few physical quantities are
given below
Speed [ ] [ ]
v = −
M LT
0 1
Area [ ] [ ]
A = M L T
0 2 0
Force [ ] [ ]
F = −
MLT 2
, etc.
The physical quantities having same derived units have
same dimensions.
20 OBJECTIVE Physics Vol. 1

31. Dimensional formulae of some physical quantities
The table given below gives the dimensional formulae and SI units of some physical quantities frequently used in physics
S. No. Physical quantity SI units Dimensional formula
1. Velocity = displacement/time m/s [M LT ]
0 1
−
2. Acceleration = velocity/time m/s2
[M LT ]
0 2
−
3. Force = mass × acceleration kg-m/s2
= newton or N [MLT ]
2
−
4. Work = force × displacement kg-m2
/s2
= N-m = joule or J [ML T ]
2 2
−
5. Energy Joule or J [ML T ]
2 2
−
6. Torque = force × perpendicular distance N-m [ML T ]
2 2
−
7. Power = work/time J/s or watt [ML T ]
2 3
−
8. Momentum = mass × velocity kg-m/s [MLT ]
1
−
9. Impulse = force × time N-s [MLT ]
1
−
10. Angle = arc/radius radian or rad [M L T ]
0 0 0
11. Strain =
∆L
L
or
∆V
V
No units [M L T ]
0 0 0
12. Stress = force/area N/m2
[ML T ]
1 2
− −
`
13. Pressure = force/area N/m2
[ML T ]
1 2
− −
14. Modulus of elasticity = stress/strain N/m2
[ML T ]
1 2
− −
15. Frequency = 1/time period per second or hertz (Hz) [M L T ]
0 0 1
−
16. Angular velocity = angle/time rad/s [M L T ]
0 0 1
−
17. Moment of inertia = (mass) × (distance)2
kg-m2
[ML T ]
2 0
18. Surface tension = force/length N/m [ML T ]
0 2
−
19. Gravitational constant =
×
force (distance)
(mass)
2
2 N-m2
/kg2
[M L T ]
1 3 2
− −
20. Angular momentum kg-m2
/s [ML T ]
2 1
−
21. Coefficient of viscosity N-s/m2
[ML T ]
1 1
− −
22. Planck's constant J-s [ML T ]
2 1
−
23. Specific heat (s) J/kg-K [M L T ]
0 2 2 1
− −
θ
24. Coefficient of thermal conductivity (K) watt/m-K [M LT 3
− −
θ 1
]
25. Gas constant (R ) J/mol-K [M L T
2 2 1
− − −
θ mol ]
1
26. Boltzmann constant (k ) J/K [ M L T
2 2
− −
θ 1
]
27. Wien's constant (b) m-K [ M LT
0 0
θ]
28. Stefan's constant (σ) watt/m2
-K4
[ M L T ]
0 3 4
− −
θ
29. Electric charge C [ M L T
0 0
A]
30. Electric intensity N/C [ M LT A ]
3 1
− −
31. Electric potential volt (V) [ M L T A ]
2 3 1
− −
32. Capacitance farad (F) [ M L T A ]
1 2 4 2
− −
33. Permittivity of free space C2
N−1
m−2
[ M L T A ]
–1 –3 4 2
Units, Dimensions & Error Analysis 21

32. S. No. Physical quantity SI units Dimensional formula
34. Electric dipole moment C-m [M0
LTA]
35. Resistance Ohm [ ML T A ]
2 –3 –2
36. Magnetic field tesla (T) or weber/m2
(Wb/m2
) [ ML T A ]
0 –2 –1
37. Coefficient of self-induction henry (H) [ ML T A ]
2 –2 –2
38. Magnetic flux Wb (weber) [ ML T A ]
2 –2 –1
39. Permeability of free space Hm−1
[ MLT A ]
–2 –2
40. Magnetic moment Am2
[ M L T A]
0 2 0
Quantities having same dimensions
22 OBJECTIVE Physics Vol. 1
Example 1.7 Find the dimensional formulae of
(i) coefficient of viscosity, η (ii) charge, q
(iii) potential,V (iv) capacitance, C and
(v) resistance, R
Some of the equations containing above quantities are
F A
v
l
= −
∆
∆






η , q It
= , U VIt
= ,
q CV
= and V IR
=
where, A is the area, v is the velocity, l is the length, I is the
electric current, t is the time andU is the energy.
Sol.
(i) η = −
∆
∆
F
A
l
v
∴ [ ]
[ ][ ]
[ ][ ]
η = =
−
−
F l
A v
[MLT ] [L]
[L ][LT ]
2
2 1
= − −
[ML T ]
1 1
(ii) q It
=
∴ [ ] [ ][ ]
q I t
= = [AT]
S. No. Quantities Dimensions
1. Strain, refractive index, relative density, angle, solid angle, phase, distance gradient, relative
permeability, relative permittivity, angle of contact, Reynolds number, coefficient of friction,
mechanical equivalent of heat and electric susceptibility.
[M L T ]
0 0 0
2. Mass and inertia [M L T ]
1 0 0
3. Momentum and impulse [M L T ]
1 1 1
−
4. Thrust, force, weight, tension and energy gradient. [M L T ]
1 1 2
−
5. Pressure, stress, Young’s modulus, bulk modulus, shear modulus, modulus of rigidity and energy
density.
[M L T ]
1 1 2
− −
6. Angular momentum and Planck’s constant (h). [M L T ]
1 2 1
−
7. Acceleration, acceleration due to gravity and gravitational field intensity. [M LT ]
0 1 2
−
8. Surface tension, free surface energy (energy per unit area), force gradient and spring constant. [M L T ]
1 0 2
−
9. Latent heat and gravitational potential. [M L T ]
0 2 2
−
10. Thermal capacity, Boltzmann constant and entropy. [ML T ]
2 2 1
− −
θ
11. Work, torque, internal energy, potential energy, kinetic energy, moment of force, ( / ),
q C
2
( )
LI2
, ( )
qV ,
( )
V C
2
, ( )
I Rt
2
,
V
R
t
2





, ( )
VIt , ( ), ( ), ( )
pV RT mL and ( )
mc T
∆ .
[M L T ]
1 2 2
−
12. Frequency, angular frequency, angular velocity, velocity gradient, radioactivity,
R
L





 ,
1
RC





 and
1
LC





 . [M L T ]
0 0 1
−
13.
l
g
m
k
R
g
L
R
RC
























1 2 1 2 1 2
/ / /
, , , , ( ), ( )
LC and time. [M L T ]
0 0 1
14. Power ( ), ( )
VI I R
2
and ( )
V R
2
. [ML T ]
2 3
−

33. (iii) U VIt
=
∴ V
U
It
=
or [ ]
[ ]
[ ][ ]
V
U
I t
= =
−
[ML T ]
[A] [T]
2 2
= − −
[ML T A ]
2 3 1
(iv) q CV
=
∴ C
q
V
=
or [ ]
[ ]
[ ]
C
q
V
= = − −
[AT]
[ML T A ]
2 3 1
= − −
[M L T A ]
1 2 4 2
(v) V IR
=
∴ R
V
I
=
or [ ]
[ ]
[ ]
R
V
I
= =
− −
[ML T A ]
[A]
2 3 1
= [ML T A ]
2 3 2
− −
Example 1.8 If C and R denote capacitance and resistance,
then find the dimensions of CR.
Sol. The capacitance of a conductor is defined as the ratio of the
charge given to the rise in the potential of the conductor,
C
q
V
q
W
= =
2
QV
W
q
=






C =
ampere -s
kg-metre /s
2 2
2 2
Hence, dimensions of C are [M L T A ]
1 2 4 2
− −
.
From Ohm’s law, V iR
= , therefore dimensions of resistance,
R
V
i
= =
Volt
Ampere
= − −
kg-metre s ampere
2 3 2
Dimensions of R = − −
[ML T A ]
2 3 2
∴ Dimensions of RC [ML T A ][M L T A ]
2 3 2 1 2 4 2
= − − − −
= [M L T ]
0 0 0
A
Example 1.9 Which amongst the following quantities is (are)
dimensionless?
(i)
Work
Energy
(ii) sin θ (iii)
Momentum
Time
Sol. (i) Since, work and energy both have the same
dimensions [ ML T ],
2 2
−
therefore their ratio is a
dimensionless quantity.
(ii) sin θ, here θ represents an angle. An angle is the
ratio of two lengths, i.e. arc length and radius.
Therefore, θ is dimensionless, hence sin θ is
dimensionless.
(iii)
Momentum
Time
MLT
T





 =






−1
= −
[ ]
MLT 2
Hence, the given ratio is not dimensionless.
Example 1.10 In the formula x yz x
= 3 2
, and y have
dimensions of capacitance and magnetic induction
respectively, then find the dimensions of y.
Sol. Given, x yz
= 3 2
⇒ y
x
z
= =
3 2 2
Capacitance
Magneticinduction
( )
[ ]
y =
− −
− −
[M L T ]
[M T ]
1 2 2
1 2
4
2
A
A
= − −
[M L T ]
3 2 4
8
A
Applications of dimensional
analysis
The method of studying a physical phenomenon on the
basis of dimensions is called dimensional analysis.
The three main uses of a dimensional analysis are
described in detail in the following section
1. Checking the dimensional
consistency of equation
Every physical equation should be dimensionally balanced.
This is called the principle of homogeneity. This
principle states that, the dimensions of each term on both
sides of an equation must be the same. On this basis, we
can judge whether a given equation is correct or not. But a
dimensionally correct equation may or may not be
physically correct.
e.g. In the physical expression s ut at
= +
1
2
2
, the
dimensions of s ut
, and
1
2
2
at all are same.
Note The physical quantities separated by the symbols + − = > <
, , , ,
etc., should have the same dimension.
Example 1.11 Show that the expression of the time period T
of a simple pendulum of length l given by T l g
= 2π / is
dimensionally correct.
Sol. Given, T
l
g
= 2π
Dimensionally, [T]
[L]
[LT ]
[T]
2
= =
−
As in the above equation, the dimensions of both sides are
same. Therefore, the given expression is dimensionally
correct.
Example 1.12 Check the correctness of the relation
s ut at
= +
1
2
2
, where u is initial velocity, a is the
acceleration, t is the time and s is the displacement.
Units, Dimensions & Error Analysis 23

34. Sol. Writing the dimensions of either side of the given relation.
LHS = =
s displacement = [M LT ]
0 0
RHS = =
ut velocity × time = −
[M LT ] [T] = [M LT ]
0 1 0 0
and
1
2
2
at = (acceleration) × (time)2
= −
[M LT ] [T] = [M LT ]
0 2 2 0 0
As LHS = RHS, so the relation is dimensionally correct.
Example 1.13 Write the dimensions of a and b in the relation,
P
b x
at
=
− 2
where, P is power, x the distance and t the time.
Sol. The given equation can be written as Pat b x
= − 2
Now, [ ] [ ] [ ]
Pat b x
= = 2
or [ ] [ ]
b x
= =
2
[M L T ]
0 2 0
and [ ]
[ ]
[ ]
a
x
Pt
= = −
2
[L ]
[ML T ] [T]
2
2 3
= −
[M L T ]
1 0 2
Example 1.14 The velocity v of a particle depends upon the
time t according to the equation v a bt
c
d t
= + +
+
⋅ Write
the dimensions of a b c
, , and d.
Sol. From principle of homogeneity,
[ ] [ ]
a v
= or [ ]
a = −
[LT ]
1
[ ] [ ]
bt v
=
or [ ]
[ ]
[ ]
b
v
t
= =
−
[LT ]
[T]
1
or [ ]
b = −
[LT ]
2
Similarly, [ ] [ ] [ ]
d t
= = T
Further,
[ ]
[ ]
[ ]
c
d t
v
+
= or [ ] [ ][ ]
c v d t
= +
or [ ]
c = −
[LT ] [T]
1
or [ ] [ ]
c = L
∴ Dimensions of a = −
[LT 1
]
Dimensions of b = −
[ ]
LT 2
Dimension of c = [ ]
L
Dimension of d = [T]
Example 1.15 The following equation gives a relation between
the mass m1, kept on a surface of area A and the pressure p
exerted on this area
p
m m x
A
=
+
( )
1 2
What must be the dimensions of the quantities x and m2?
Sol. Since, all the terms of a mathematical equation should have
the same dimensions.
Therefore, [ ]
( )
p
m m x
A
=
+






1 2
...(i)
Only the quantities having same dimensions and nature can
be added to each other.
Here, m2 is added to mass m1.
Hence, [ ] [ ] [ ]
m m
2 1
= = M
Also, the quantity obtained by the addition of m1 and m2
would have the same dimensions as that of mass.
∴ [ ] [ ]
m m
1 2
+ = M
Now, going back to Eq. (i),
[ ]
[ ][ ]
[ ]
p
m m x
A
=
+
1 2
⇒ [ ]
[ ][ ]
[ ]
ML T
− −
=
1 2
2
M
L
x
⇒ [ ] [ ][ ]
ML T
− − −
=
1 2 2
ML x
⇒
[ ]
[ ]
[ ]
ML T
ML
− −
−
=
1 2
2
x ⇒ [ ] [ ]
x = −
LT 2
Hence, the quantity x represents acceleration. In this example,
it is the acceleration due to gravity g. ( )
m m g
1 2
+ represents
the weight exerted by two masses m m
1 2
, on the area A.
2. To convert a physical quantity from one
system of units to other system of units
This is based on the fact that the product of the numerical
value (n) and its corresponding unit (u) is a constant, i.e.
n u
( ) = constant or n u n u
1 1 2 2
[ ] [ ]
=
Suppose the dimensions of a physical quantity are a in
mass, b in length and c in time. If the fundamental units in
one system are M L
1 1
, and T1 and in the other system are
M2, L2 and T2, respectively. Then, we can write
n n
a b c a b c
1 1 1 1 2 2 2 2
[ ] [ ]
M L T M L T
= ...(i)
n n
u
u
n
a b c
2 1
1
2
1
1
2
1
2
1
2
= =


















M
M
L
L
T
T
Here, n1 and n2 are the numerical values in two system of
units, respectively. Using Eq. (i), we can convert the
numerical value of a physical quantity from one system of
units into the other system.
Example 1.16 Find the value of 100 J on a system which has
20 cm, 250 g and half minute as fundamental units of
length, mass and time.
Sol. The dimensional formula of work is = −
[ML T ]
2 2
To convert a physical quantity from one system of units to
other system of units, we use the following formula
n n
a b c
2 1
1
2
1
2
1
2
=


















M
M
L
L
T
T
n2
2
100
250 20 0 5
=















1 kg
g
1 m
cm
1 s
min
1
.



−2
=


















−
100
1000
250 20 30
2 2
g
g
100 cm
cm
1 s
s
1
= × × × ×
100 4 25 30 30 = ×
9 106
new units
24 OBJECTIVE Physics Vol. 1

35. Example 1.17 The value of gravitational constant is
G = ×
6.67 10–11
N-m2
/kg2
in SI units. Convert it into
CGS system of units.
Sol. The dimensional formula of G is [ ]
M L T
3
− −
1 2
.
To convert a physical quantity from one system of units to
other system of units, we use the following formula
n n
1 1
1
1
3
1
2
2 2
1
2
3
2
2
[M L T ] [M L T ]
− − − −
=
n n
2 1
=


















− −
M
M
L
L
T
T
1
2
1
1
2
3
1
2
2
= ×












−
−
6.67 10
1 kg
10 kg
1 m
10 m
11
–3
1
–2
3 2
1 s
1 s






−
or n2 = × −
6.67 10 8
Thus, value of G in CGS system of units is
6 67 10 8
. × −
dyne cm / g
2 2
.
3. Deducing relation between the
physical quantities
If we know the factors on which a given physical
quantity depends, we can find a formula relating to
those factors.
Example 1.18 The frequency ( )
f of a stretched string
depends upon the tension F (dimensions of force), length l
of the string and the mass per unit length µ of string.
Derive the formula for frequency.
Sol. Suppose, the frequency f depends on the tension raised to
the power a, length raised to the power b and mass per unit
length raised to the power c.
Then, f F l
a b c
∝ ( ) ( ) ( )
µ
or f k F l
a b c
= ( ) ( ) ( )
µ …(i)
Here, k is a dimensionless constant of proportionality.
Thus, [ ] [ ] [ ] [ ]
f l
a b c
= F µ
or [M L T ] [MLT ] [L] [ML ]
0 0 1 2 1
− − −
= a b c
or [M L T ] [M L T ]
0 0 1 2
− + + − −
= a c a b c a
For dimensional balance, the dimensions on both sides
should be same.
Thus, a c
+ = 0 …(ii)
a b c
+ − = 0 …(iii)
and − = −
2 1
a …(iv)
Solving these three equations, we get
a =
1
2
, c = −
1
2
and b = − 1
Substituting these values in Eq. (i), we get
f k F l
= − −
( ) ( ) ( )
/ /
1 2 1 1 2
µ or f
k
l
F
=
µ
Experimentally, the value of k is found to be
1
2
⋅
Hence, f
l
F
=
1
2 µ
Example 1.19 The centripetal force F acting on a particle
moving uniformly in a circle may depend upon mass (m),
velocity (v) and radius (r) of the circle. Derive the formula for
F using the method of dimensions.
Sol. Let F k m v r
x y z
= ( ) ( ) ( ) …(i)
Here, k is a dimensionless constant of proportionality.
Writing the dimensions of RHS and LHS in Eq. (i), we have
[MLT ] [M] [LT ] [L]
2 1
− −
= x y z
= −
[M L T ]
+
x y z y
Equating the powers of M, L and T on both sides, we have
x = 1, y = 2 and y z
+ = 1
or z y
= − = −
1 1
Putting the values in Eq. (i), we get
F kmv r k
mv
r
= =
−
2 1
2
F
mv
r
=
2
(where, k = 1)
Defects or limitations of
dimensional analysis
The method of dimensional analysis has the following
limitations
(i) The value of dimensionless constant involved in a
formula cannot be deduced from this method.
(ii) By this method, the equation containing
trigonometrical, exponential and logarithmic terms
cannot be analysed.
(iii) This method does not work when physical quantity
depends on more than three variables because we only
have three equations by equalising the power of M, L
and T.
(iv) If dimensions are given, physical quantity may not be
unique. e.g. Work, energy and torque all have the
same dimensional formula [ML T ].
2 2
−
(v) It gives no information whether a physical quantity is
a scalar or a vector.
Units, Dimensions & Error Analysis 25

36. 1. In the SI system, the unit of temperature is
(a) degree centigrade
(b) kelvin
(c) degree celsius
(d) degree Fahrenheit
2. Which amongst the following is not equal to watt?
(a) joule/second (b) ampere × volt
(c) (ampere)2
× ohm (d) ampere/volt
3. Joule × second is the unit of
(a) energy (b) momentum
(c) angular momentum (d) power
4. Which amongst the following pairs has the same units?
(a) Wavelength and Rydberg constant
(b) Relative velocity and relative density
(c) Thermal capacity and Boltzmann constant
(d) Time period and acceleration gradient
5. Density of liquid in CGS system is 0.625 g cm 3
−
. What is its
magnitude in SI system?
(a) 0.625 (b) 0.0625
(c) 0.00625 (d) 625
6. Dimensions of surface tension are
(a) [M L T ]
2 2 2
−
(b) [M LT ]
2 2
−
(c) [MT ]
2
−
(d) [MLT ]
2
−
7. The dimensions of impulse are equal to that of
(a) force (b) linear momentum
(c) pressure (d) angular momentum
8. Which of the following does not possess the same
dimensions as that of pressure?
(a) Stress (b) Bulk modulus
(c) Thrust (d) Energy density
9. What is the dimensional formula of gravitational constant?
(a) [ML T ]
2 2
−
(b) [ML T ]
1 1
− −
(c) [M L T ]
1 3 2
− −
(d) None of these
10. Which one of the following have same dimensions?
(a) Torque and force
(b) Potential energy and force
(c) Torque and potential energy
(d) Planck’s constant and linear momentum
11. The force F on a sphere of radius a moving in a medium
with velocity v is given by F a v
= 6π η . The dimensions of η
are
(a) [ML ]
3
−
(b) [MLT ]
2
−
(c) [MT ]
1
−
(d) [ML T ]
1 1
− −
12. The dimensional representation of specific resistance in
terms of charge Q is
(a) [ML T Q ]
3 1 2
− −
(b) [ML T Q ]
2 2 2
−
(c) [MLT Q ]
2 1
− −
(d) [ML T Q ]
2 2 1
− −
13. The dimensional formula for Planck’s constant and angular
momentum is
(a) [ML T ]
2 2
−
and [MLT ]
1
−
(b) [ML T ]
2 1
−
and [ML T ]
2 1
−
(c) [ML T ]
3 1
−
and [ML T ]
2 2
−
(d) [MLT ]
1
−
and [MLT ]
2
−
14. The dimensions of
1
2
0
2
ε E (ε0 is the permittivity of the space
and E is electric field), are
(a) [ML T ]
2 1
−
(b) [ML T ]
1 2
− −
(c) [ML T ]
2 2
−
(d) [MLT ]
1
−
15. The units of length, velocity and force are doubled. Which of
the following is the correct change in the other units?
(a) Unit of time is doubled
(b) Unit of mass is doubled
(c) Unit of momentum is doubled
(d) Unit of energy is doubled
16. Given that y a
t
p
qx
= −






cos , where t represents time and x
represents distance; which amongst the following
statements which is(are) true?
(a) The unit of x is same as that of q
(b) The unit of x is same as that of p
(c) The unit of t is same as that of q
(d) The unit of t is same as that of p
17. The dimensions of
a
b
in the equation p
a t
bx
=
− 2
, where p is
pressure, x is distance and t is time, are
(a) [M LT ]
2 3
−
(b) [MT ]
2
−
(c) [LT ]
3
−
(d) [ML T ]
3 1
−
18. The equation of a wave is given by y a
x
v
k
= −






sin ω
where, ω is angular velocity and v is the linear velocity. The
dimensions of k will be
(a) [T ]
2
−
(b) [T ]
1
−
(c) [T] (d) [LT]
19. If ‘muscle times speed equals power’, then what is the ratio
of the SI unit and the CGS unit of muscle?
(a)105
(b)103
(c)107
(d)10 5
−
20. If p represents radiation pressure, c represents speed of light
and Q represents radiation energy striking a unit area per
second, then for what values of non-zero integers x y
, and z,
p Q c
x y z
is dimensionless?
(a) x y z
= = = −
1 1 1
, , (b) x y z
= = − =
1 1 1
, ,
(c) x y z
= − = =
1 1 1
, , (d) x y z
= = =
1 1 1
, ,
21. Assuming that the mass m of the largest stone that can be
moved by a flowing river depends upon the velocity v of the
water, its density ρ and the acceleration due to gravity g.
Then, m is directly proportional to
(a) v3
(b) v4
(c) v5
(d) v6
OBJECTIVE Physics Vol. 1
CHECK POINT 1.1

37. SIGNIFICANT FIGURES
The significant figures are normally those digits in a
measured quantity which are known reliable or about
which we have confidence in our measurement plus one
additional digit that is uncertain.
e.g. If length of some object is 185.2 cm, then it has four
significant figures. The digits 1,8 and 5 are reliable and digit
2 is uncertain.
Note Significant figures indicate the precision of the measurement
which depends on the least count of the measuring instrument.
Rules to determine significant
figures
For determining number of significant figures, we use the
following rules
Rule 1 All non-zero digits are significant, e.g. x = 2567 has
four significant figures.
Rule 2 The zeros appearing between two non-zero digits
are significant, no matter where the decimal point is, if at
all, e.g. 6.028 has 4 significant figures.
Rule 3 If the number is less than 1, the zero(s) on the right
of decimal point but to the left of first non-zero digit are not
significant.
e.g. 0.0042 has two significant digits.
Rule 4 The terminal or trailing zero(s) in a number without
a decimal point are not significant. Thus, 426 m = 42600 cm
= 426000 mm has three significant figures.
Rule 5 In a number with decimal, zeros to the right of last
non-zero digit are significant.
e.g. 4.600 or 0.002300 have four significant figures each.
Point of confusion and its remedy
Suppose we change the units of a physical quantity, then we
will write
2.30 m 230 cm
= = 2300 mm = 0 00230
. km
When we are considering 2300 mm, then from Rule-4, we
would conclude erroneously that the number has two
significant figures, while in fact it has three significant
figures and a mere change of units cannot change the
number of significant figures.
To remove such ambiguities in determining the number of
significant figures, apply following rules
Rule 6 The power of 10 is irrelevant to the determination
of significant figures. e.g. In the measurements,
2.30 m 2.30 102
= × cm = ×
2.30 103
mm
= × −
2.30 10 3
km
The significant figures are three in each measurement,
because all zeros appearing in the base number in the
scientific notation (in the power of 10) are not
significant.
Rule 7 A choice of change of different units does not
change the number of significant digits or figures in a
measurement.
e.g. The length 7.03 cm has three significant figures. But
in different units, the same value can be written as
0.0703 m or 70.3 mm. All these measurements have the
same number of significant figures (digits 7, 0 and 3)
namely three.
Rule 8 The exact numbers appearing in the
mathematical formulae of various physical quantities
have infinite number of significant figures. e.g.
Perimeter of a square is given by 4 × side. Here, 4 is an
exact number and has infinite number of significant
figures.
∴ It can be written as 4.0, 4.00, 4.0000 as per the
requirement.
Some significant figures of measured values given in the
table below
Measured values
Number of significant
figures
Rule
12376 5 1
6024.7 5 2
0.071 2 3
410 2 4
2.40 3 5
1.6 × 1010
2 6
ln 2 ( ),
l b
+ digit 2 Infinite 8
Example 1.20 How many significant figures are there in the
following measured values?
(i) 227.2 g (ii) 3600 g
(iii) 0.00602 g (iv) 2 50 1010
. × g
Sol. (i) 227.2 g has all the non-zero digits. Hence, it has four
significant figures.
(ii) According to rule number 4, trailing zeros are not
significant. Hence, 3600 g has 2 significant figures.
(iii) According to the rule number 3, the zeros on the right
of decimal point but to the left of first non-zero digit
are not significant. Hence, 0.00602 g has 3 significant
figures.
(iv) According to the rule number 6, it has 3 significant
figures.
Units, Dimensions & Error Analysis 27

38. Mathematical operations of
significant figures
The result of a mathematical operation involving measured
values of quantities cannot be more accurate than the
measured value themselves.
So, certain rules have to be followed while doing
mathematical operations with significant figures, so that
precision in final result is consistent with the precision of
the original measured values.
Addition or subtraction
Suppose in the measured values to be added or subtracted,
the least number of significant digits after the decimal is n.
Then, in the sum or difference also, the number of
significant digits after the decimal should be n.
e.g. 1.2 3.45 6.789 11.439 11.4
+ + = ≈
Here, the least number of significant digits after the
decimal is one. Hence, the result will be 11.4 (when
rounded off to smallest number of decimal places).
Similarly, e.g. 1263 10 2 243
. . .
− = ≈ 2.4
Example 1.21 Add 6.75 × 103
cm to 4.52 × 102
cm with
regard to significant figures.
Sol. Let a = ×
6.75 10 cm
3
, b = ×
4.52 10 cm
2
= ×
0.452 103
cm = ×
0.45 10 cm
3
(upto 2 places of decimal)
∴ Addition of significant figures
a b
+ = × + ×
(6.75 10 0.45 10 ) cm
3 3
= ×
7.20 10 cm
3
Example 1.22 Two sticks of lengths 12.132 cm and 10.2 cm
are placed end to end. Find their total length with due
regard to significant figures.
Sol. Length of first stick = 12.132 cm (5 significant figures)
Length of second stick = 10.2 cm (3 significant figures)
∴ Total length of two sticks = + =
12132 102
. . 22.332
The answer should be rounded off with least number of
significant digits after the decimal.
∴ Total length of two sticks will be 22.3 cm.
Multiplication or division
Suppose in the measured values to be multiplied or
divided, the least number of significant digits be n, then in
the product or quotient, the number of significant digits
should also be n.
e.g. 1.2 36.72 44.064 44
× = ≈
The least number of significant digits in the measured
values are two. Hence, the result when rounded off to two
significant digits become 44. Therefore, the answer is 44.
Similarly, e.g.
1100
10 2
1078431373 110
.
.
= ≈
As 1100 has minimum number of significant figures
(i.e. 2), therefore the result should also contain only two
significant digits. Hence, the result when rounded off to
two significant digits becomes 110.
Example 1.23 The voltage across a lamp is 6.32V when the
current passing through it is 3.4 A. Find the power
consumed upto appropriate significant figures.
Sol. Voltage across a lamp,V = 6.32 V (3 significant figures)
Current flowing through lamp, I = 3.4A (2 significant figures)
∴ Power consumed, P VI
= = =
(6.32)(3.4) 21.488 W
Answer should have minimum number of significant figures.
Here, the minimum number of significant figures is 2.
∴ Power consumed = 21W
Example 1.24 A thin wire has a length of 21.7 cm and radius
0.46 mm. Calculate the volume of the wire upto correct
significant figures.
Sol. Given, l = 21.7 cm, r = =
0.46 mm 0.046 cm
Volume of wire,V r l
= π 2
=
22
7
(0.046) (21.7)
2
= 0.1443 cm3 ~
− 0.14 cm3
Example 1.25 The time taken by a pendulum to complete
25 vibrations is 88.0 s. Find the time period of the pendulum
in seconds upto appropriate significant figures.
Sol. Time period of oscillation =
Totaltime taken
Number of oscillations
=
88.0
25
s = 3.52 s
Out of the two quantities given in the data, 25 is exact, hence
has infinite significant figures. Therefore, the answer should
be reported to three significant figures, i.e. 3.52 s.
Example 1.26 5.74 g of substance occupies 1.2 cm 3
. Express
its density by keeping the significant figures in view.
Sol. Here, mass, m = 5 74
. g, volume,V = 1.2 cm3
As density, ρ = = = −
mass
volume
g
1.2 cm
g cm
5 74
4 783
3
3
.
.
As mass has 3 significant digits and volume has 2 significant
digits, therefore as per rule, density will have only two
significant digits, rounding off, we get ρ = −
4 8 3
. gcm .
Rounding off
The process of omitting the non-significant digits and
retaining only the desired number of significant digits,
incorporating the required modifications to the last
significant digit is called rounding off the number.
In physics, calculation is a vital part and during that we
shall reduce the number to the required extent and that is
why there is a need to round off numbers.
Like mathematical operations of significant figures,
rounding off numbers also follow certain rules.
28 OBJECTIVE Physics Vol. 1

39. Rules for rounding off a measurement
Following are the rules for rounding off a measurement
Rule 1 If the number lying to the right of cut-off digit is less
than 5, then the cut-off digit is retained as such. However, if
it is more than 5, then the cut-off digit is increased by 1.
e.g. x = 6.24 is rounded off to 6.2 to two significant digits
and x = 5.328 is rounded off to 5.33 to three significant
digits.
Rule 2 If the insignificant digit to be dropped is 5, then the
rule is
(i) if the preceding digit is even, the insignificant digit
is simply dropped.
(ii) if the preceding digit is odd, the preceding digit is
raised by 1.
e.g. x = 6.265 is rounded off to x = 6.26 to three
significant digits and x = 6.275 is rounded off to x = 6.28
to three significant digits.
Rule 3 The exact numbers like π, 2, 3 and 4, etc., that
appear in formulae and are known to have infinite
significant figures, can be rounded off to a limited number
of significant figures as per the requirement.
Example 1.27 Round off the following numbers upto three
significant figures.
(i) 2.520 (ii) 4.645 (iii) 22.78 (iv) 36.35
Sol. (i) 2.520 : Since, 0 is less than 5, preceding digit is left
unchanged. Hence, 2.52.
(ii) 4.645 : Since, the digit to be dropped is 5 and the
preceding digit 4 is even. Hence, 4.64.
(iii) 22.78 : Since, the digit to be dropped is 8 and is greater
than 5, therefore the preceding digit 7, is raised by 1.
Hence, 22.8.
(iv) 36.35 : Since, the digit to be dropped is 5 and the
preceding digit 3 is odd, we can write the answer as 36.4.
Example 1.28 The length and the radius of a cylinder measured
with slide callipers are found to be 4.54 cm and 1.75 cm,
respectively. Calculate the volume of the cylinder.
Sol. Length of cylinder, h = 4.54 cm (3 significant figures)
Radius of cylinder, r = 1.75 cm (3 significant figures)
∴ Volume of cylinder = = × ×
πr h
2
3.14 (1.75) 4.54 cm
2 3
= 43 657775
. cm3
= 43 6
. cm3
(Rounded off upto 3 significant figures)
Order of magnitude
Any physical quantity can be expressed in the form of
a b
× 10 (in terms of magnitude), where a is a number lying
between 1 and 10; and b is any negative or positive
exponent of 10, then the exponent b is called the order of
magnitude of the physical quantity. And the expression of a
quantity as a b
× 10 is called scientific notation.
e.g. The speed of light is given as 3 00 108
. × m/s. So, the
order of magnitude of the speed of light is 8.
The order of magnitude gives an estimate of the magnitude
of the quantity. The charge on an electron is16 10 19
. × −
C.
Therefore, we can say that the charge possessed by an
electron is of the order10 19
−
or its order of magnitude
is −19.
Units, Dimensions & Error Analysis 29
1. What is the number of significant figures in 0.0310 103
× ?
(a) 2 (b) 3
(c) 4 (d) 6
2. The number of significant figures in 11.118 10 V
6
× −
is
(a) 3 (b) 4
(c) 5 (d) 6
3. In which of the following numerical values, all zeros are
significant?
(a) 0.2020 (b) 20.2
(c) 2020 (d) None of these
4. What is the number of significant figure in
(3.20 4.80) 105
+ × ?
(a) 5 (b) 4
(c) 3 (d) 2
5. Subtract 0.2 J from 7.26 J and express the result with
correct number of significant figures.
(a) 7.1 (b) 7.06
(c) 7.0 (d) None of these
6. Multiply 107.88 by 0.610 and express the result with correct
number of significant figures.
(a) 65.8068 (b) 64.807 (c) 65.81 (d) 65.8
7. The length, breadth and thickness of rectangular sheet of
metal are 4.234 m, 1.005 m and 2.01 cm, respectively. The
volume of the sheet upto correct significant figures is
(a) 0.0855 m3
(b) 0.086 m3
(c) 0.08556 m3
(d) 0.08 m3
8. The radius of a thin wire is 0.16 mm. The area of
cross-section of the wire (in mm2
) with correct number of
significant figures is
(a) 0.08 (b) 0.080
(c) 0.0804 (d) 0.080384
9. When 97.52 is divided by 2.54, the correct result
(considering significant figures) is
(a) 38.3937 (b) 38.394
(c) 65.81 (d) 38.4
10. What is the order of magnitude of [(5.0 10 ) (5.0 10 )]
6 8
× ×
− −
with due regards to significant digits?
(a) − 14 (b) − 15 (c) + 15 (d) + 14
CHECK POINT 1.2

40. ERROR IN MEASUREMENT
We use different kinds of instruments for measuring
various quantities. However, these measurements
always has a degree of uncertainty related to it. This
uncertainty is called as error in the measurement. Thus,
the difference between the measured value and the true
value of a quantity is known as the error of
measurement.
∴ Error = True value − Measured value
Errors may arise from different sources and are usually
classified as follows
1. Systematic errors
These are the errors whose causes are known to us.
They can be either positive or negative.
One of the common source of systematic errors is as
follows
Instrumental errors
These errors are due to imperfect design or erroneous
manufacture or misuse of the measuring instrument.
These are of following types
(i) Zero error If the zero mark of vernier scale does
not coincide with the zero mark of the main scale,
the instrument is said to have zero error. A metre
scale having worn off zero mark also has zero
error.
(ii) Least count or permissible error This error is
due to the limitation imposed by the least count of
the measuring instrument. It is an uncertainty
associated with the resolution of the measuring
instrument.
Note Least Count ( )
LC of
(i) Vernier callipers =
Value of MSD
Number of divisions
1
or LC = 1 MSD − 1VSD on vernier scale
(ii) Screw gauge =
Pitch
Number of divisions on circular scale
(iii) Constant error The errors which affect each
observation by the same amount are called
constant errors. Such errors are due to faulty
calibration of the scale of the measuring
instrument.
(iv) Backlash error Backlash error occurs in screw
gauge, when we try to rotate the screw very fast
to measure a reading. Due to this, there is some
slipping between the different screws instead of
the rotation, which gives an incorrect reading. To
avoid this we should rotate the screw slowly in
only one direction.
Causes of systematic errors
Few causes of systematic errors are as follows
(i) Instrumental errors may be due to erroneous
instruments. These errors can be reduced by using
more accurate instruments and applying zero
correction, when required.
(ii) Sometimes errors arise on account of ignoring certain
facts. e.g. In measuring time period of simple
pendulum, error may creap because no consideration is
taken of air resistance. These errors can be reduced by
applying proper corrections to the formula used.
(iii) Change in temperature, pressure, humidity, etc., may
also sometimes cause errors in the result. Relevant
corrections can be made to minimise their effects.
2. Random errors
The errors which occur irregularly and at random, in
magnitude and direction are called random errors. The causes
of random errors are not known. Hence, it is not possible to
remove them completely. These errors may arise due to a
variety of reasons.
e.g. The reading of a sensitive beam balance may change by
the vibrations caused in the building, due to persons moving
in the laboratory or vehicles running nearby. The random
error can be minimised by repeating the observation a large
number of times and taking the arithmetic mean of all the
observations. The mean value would be very close to the
most accurate reading.
Example 1.29 In a vernier callipers, 1 main scale reading is
1 mm and 9th main scale division coincide with 10th vernier
scale. Find the least count of vernier.
Sol. Given, 1 main scale reading or division (MSD)
= 1 mm
9 MSD = 10 VSD
⇒ 1 VSD =
9
10
MSD = ×
9
10
1 =
9
10
mm
∴ LC = −
1
9
10
=
1
10
mm or 0.1 mm
Expression of errors
Errors can be expressed in following way
(i) Absolute error The difference between the true value
and the measured value of a quantity is called an
absolute error. Usually the mean value am is taken as
the true value. So, if
a
a a a
n n
a
m
n
i
n
i
=
+ + … +
= ∑
=
1 2
1
1
30 OBJECTIVE Physics Vol. 1

41. Then by definition, absolute errors in the measured
values of the quantity are
∆a a am
1 1
= −
∆a a am
2 2
= −
M M M
∆a a a
n n m
= −
Absolute error may be positive or negative.
Mean absolute error It is the arithmetic mean of
the magnitudes of absolute errors. Thus,
∆
∆ ∆ ∆
∆
a
a a a
n n
a
n
i
n
i
mean =
| + | + … + |
= ∑ |
=
1 2
1
1
| | |
|
Thus final result of measurement can be written as
a a a
m
= ± ∆ mean
This implies that value of a is likely to lie between
a a
m + ∆ mean and a a
m − ∆ mean .
(ii) Relative or fractional error The ratio of mean
absolute error to the mean value of the quantity
measured is called relative or fractional error.
Thus, Relative error =
∆a
am
mean
(iii) Percentage error When the relative error is
expressed in percent, it is called percentage error. It
is denoted by δa.
Thus, δa
a
am
= ×
∆ mean
100%
Example 1.30 The length of a rod as measured in an
experiment is found to be 2.48 m, 2.46 m, 2.49 m, 2.49 m
and 2.46 m. Find the average length, the absolute error in
each observation and the percentage error.
Sol. Average length = Arithmetic mean of the measured
values
xmean =
+ + + +
2.48 2.46 2.49 2.49 2.46
5
= =
12.38
5
2.476 m
∴ True value, xmean = 2.48 m
Absolute errors in various measurements,
| | | | | |
∆x x x
1 1
= − = − =
mean 2.48 2.48 0.00 m
| | |
∆x2 = − =
2.46 2.48| 0.02 m
| | |
∆x3 = − =
2.49 2.48| 0.01 m
| | |
∆x4 = − =
2.49 2.48| 0.01 m
| | |
∆x5 = − =
2.46 2.48| 0.02 m
Mean absolute error =
+ + + +
| | | | | | | | | |
∆ ∆ ∆ ∆ ∆
x x x x x
1 2 3 4 5
5
=
+ + + +
=
(0.00 0.02 0.01 0.01 0.02)
5
0.06
5
= 0 012
.
∆xmean = 0.01 m
Thus, x = 2.48 ± 0.01 m
Percentage error, δx
x
x
= ×
∆ mean
100
= × =
0.01
2.48
100 0.40%
Example 1.31 The diameter of a wire as measured by a
screw gauge was found to be 2.620 cm, 2.625 cm,
2.630 cm, 2.628 cm and 2.626 cm. Calculate
(i) mean value of diameter,
(ii) absolute error in each measurement,
(iii) mean absolute error,
(iv) fractional error,
(v) percentage error and
(vi) express the result in terms of percentage error.
Sol. (i) Mean value of diameter,
am =
+ + + +
2.620 2.625 2.630 2.628 2.626
5
= 2.6258 cm = 2.626 cm
(rounding off to three decimal places)
(ii) Taking am as the true value, the absolute errors in
different observations are
∆a1 0 6
= − = −
2.62 2.62 0.006 cm
∆a2 5 6
= − = −
2.62 2.62 0.001 cm
∆a3 = − = +
2.630 2.636 0.004 cm
∆a4 = − = +
2.628 2.626 0.002 cm
∆a5 = − =
2.626 2.626 0.000 cm
(iii) Mean absolute error,
∆
∆ ∆ ∆ ∆ ∆
a
a a a a a
mean =
+ + + +
| | | | | | | | | |
1 2 3 4 5
5
=
+ + + +
0.006 0.001 0.004 0.002 0.000
5
= 0.0026 = 0.003
(rounding off to three decimal places)
(iv) Fractional error = ±
∆a
am
mean
= ±
0.003
2.626
= ± 0.001
(v) Percentage error = ± ×
0 001 100
. = ± 0.1%
(vi) Diameter of wire can be written as
d = 2.626 cm ± 0.1%
Example 1.32 The refractive index (n) of glass is found to
have the values 1.49, 1.50, 1.52, 1.54 and 1.48. Calculate
(i) the mean value of refractive index,
(ii) absolute error in each measurement,
(iii) mean absolute error,
(iv) fractional error and
(v) percentage error.
Sol. (i) Mean value of refractive index,
nm =
+ + + +
1.49 1.50 1.52 1.54 1.48
5
= =
1.50 1.51
6
(rounded off to two decimal places)
Units, Dimensions & Error Analysis 31

42. (ii) Taking nm as the true value, the asbolute errors in
different observations are
∆n1 49 51
= − = −
1. 1. 0.02
∆n2 0 1
= − = −
1.5 1.5 0.01
∆n3 2
= − = +
1.5 1.51 0.01
∆n4 = − = +
1.54 1.51 0.03
∆n5 48 51
= − = −
1. 1. 0.03
(iii) Mean absolute error,
∆
∆ ∆ ∆ ∆ ∆
n
n n n n n
mean =
+ + + +
| | | | | | | | | |
1 2 3 4 5
5
=
0.02 + 0.01 + 0.01 + 0.03 + 0.03
5
= 0.02
(iv) Fractional error =
±
=
±
∆n
nm
mean 0.02
1.51
= ± 0.0132
(v) Percentage error = ± × = ±
( 0.0132 100) 1.32%
Combination of errors
Most of our experiments involves the measurement of various
physical quantities. We then put these measurements in
appropriate formula, to calculate the required quantity.
Therefore, we must know how the errors in all the
measurements combine and appear in the final quantity.
1. Error in sum or difference
Let x a b
= ±
Further, let ∆a be the absolute error in the measurement
of a, ∆b be the absolute error in the measurement of b and
∆x be the absolute error in the measurement of x.
Then, x x a a b b
+ = ± ± ±
∆ ∆ ∆
( ) ( )
= ± ± ± ±
( ) ( )
a b a b
∆ ∆
= ± ± ±
x a b
( )
∆ ∆ or ∆ ∆ ∆
x a b
= ± ±
The four possible values of ∆x are ( ),
∆ ∆
a b
− ( ),
∆ ∆
a b
+
( )
− −
∆ ∆
a b and ( )
− +
∆ ∆
a b . Therefore, the maximum
absolute error in x is
∆ ∆ ∆
x a b
= ± ( )
+
i.e. The maximum absolute error in sum or difference of
two quantities is equal to sum of the absolute errors in the
individual quantities.
Example 1.33 The volumes of two bodies are measured to be
V cm
1
3
= ±
(10.2 0.02) andV cm
2
3
= ±
(6.4 0.01) .
Calculate the sum and difference in volumes with error
limits.
Sol. Given,V1 = ±
(10.2 0.02) cm3
and V2 = ±
(6.4 0.01) cm3
∆ ∆ ∆
V V V
= ± +
( )
1 2 = ± +
(0.02 0.01) cm3
= ± 0.03 cm3
V V
1 2
+ = + =
(10.2 6.4) cm 16.6 cm
3 3
and V V
1 2
− = − =
(10.2 6.4) cm 3.8 cm
3 3
Hence, sum of volumes = ±
(16.6 0.03) cm3
and difference of volumes = ±
(3.8 0.03) cm3
2. Error in product
Let x ab
=
Then, ( ) ( ) ( )
x x a a b b
± = ± ±
∆ ∆ ∆
or x
x
x
ab
a
a
b
b
1 1 1
±





 = ±





 ±






∆ ∆ ∆
or 1 1
± = ± ± ± ⋅
∆ ∆ ∆ ∆ ∆
x
x
b
b
a
a
a
a
b
b
(Qx ab
= )
or ± = ± ± ± ⋅
∆ ∆ ∆ ∆ ∆
x
x
a
a
b
b
a
a
b
b
Here,
∆ ∆
a
a
b
b
⋅ is a very small quantity, so can be
neglected.
Hence, ± = ± ±
∆ ∆ ∆
x
x
a
a
b
b
Possible values of
∆x
x
are
∆ ∆
a
a
b
b
+





,
∆ ∆
a
a
b
b
−





,
− +






∆ ∆
a
a
b
b
and − −






∆ ∆
a
a
b
b
.
Hence, maximum possible value of
∆ ∆ ∆
x
x
a
a
b
b
= ± +






Therefore, maximum fractional error in product of two
(or more) quantities is equal to sum of fractional errors in
the individual quantities.
3. Error in division
Let x
a
b
=
Then, x x
a a
b b
± =
±
±
∆
∆
∆
or x
x
x
a
a
a
b
b
b
1
1
1
±





 =
±






±






∆
∆
∆
or 1 1 1
1
±





 = ±





 ±






−
∆ ∆ ∆
x
x
a
a
b
b
Q x
a
b
=






As
∆b
b
<< 1, so expanding binomially, we get
1 1 1
±





 = ±





 +






∆ ∆ ∆
x
x
a
a
b
b
or 1 1
± = ± + ± ⋅
∆ ∆ ∆ ∆ ∆
x
x
a
a
b
b
a
a
b
b
Here,
∆ ∆
a
a
b
b
⋅ is a very small quantity, so can be neglected.
32 OBJECTIVE Physics Vol. 1

43. Hence, ± = ± +
∆ ∆ ∆
x
x
a
a
b
b
Possible values of
∆x
x
are
∆ ∆
a
a
b
b
−





,
∆ ∆
a
a
b
b
+





,
− −






∆ ∆
a
a
b
b
and − +






∆ ∆
a
a
b
b
. Therefore, the maximum
value of
∆ ∆ ∆
x
x
a
a
b
b
= ± +






Therefore, the maximum value of fractional error in
division of two (or more) quantities is equal to the sum of
fractional errors in the individual quantities.
Example 1.34 Calculate focal length of a spherical mirror
from the following observations. Object distance
u = ±
(50.1 0.5) cm and image distance v = ±
(20.1 0.2) cm.
Sol. Formula for focal length of a spherical mirror,
1 1 1
f v u
= + …(i)
or f
uv
u v
=
+
=
+
=
(50.1) (20.1)
(50.1 20.1)
14.3 cm
On differentiating Eq. (i), we get
∆ ∆ ∆
f
f
u
u
v
v
2 2 2
= +
or ∆ ∆ ∆
f
f
u
u
f
v
v
= × +
2
2
2
2
=





 × +





 ×
143
501
0 5
143
201
02
2 2
.
.
.
.
.
.
= +
0 0407 01012
. .
= ± 01419
. cm ~ .
− ± 01 cm
∴ f = ±
( . . )
143 01 cm
4. Error in quantity raised to some power
Let x
a
b
n
m
=
Then, ln ( ) ln ( ) ln ( )
x n a m b
= −
Differentiating both sides, we get
dx
x
n
da
a
m
db
b
= −
In terms of fractional error, we may write
± = ± +
∆ ∆ ∆
x
x
n
a
a
m
b
b
Therefore, maximum value of
∆ ∆ ∆
x
x
n
a
a
m
b
b
= ± +






Example 1.35 The radius of sphere is measured to be
( 2.1 0.5)
± cm. Calculate its surface area with error limits.
Sol. Surface area, S r
= 4 2
π =






(4)
22
7
(2.1)2
= =
55.44 55.4 cm2
Further,
∆ ∆
S
S
r
r
= 2 or ∆
∆
S
r
r
S
=






2 ( )
=
× ×
2 0.5 55.4
2.1
= =
26.38 26.4 cm2
∴ S = ±
(55.4 26.4) cm2
Example 1.36 The mass and density of a solid sphere are
measured to be (12.4 0.1)
± kg and (4.6 0.2)
± kg m–3
.
Calculate the volume of the sphere with error limits.
Sol. Here, m m
± = ±
∆ (12.4 0.1) kg
and ρ ρ
± = ±
∆ (4.6 0.2) kgm−3
Volume, V
m
= =
ρ
12.4
4.6
= =
2.69 m 2.7 m
3 3
(rounding off to one decimal place)
Now,
∆ ∆ ∆
V
V
m
m
= ± +






ρ
ρ
or ∆
∆ ∆
V
m
m
V
= ± +





 ×
ρ
ρ
= ± +





 ×
0.1
12.4
0.2
4.6
2.7 = ± 0.14
∴ V V
± = ±
∆ (2.7 0.14) m3
Example 1.37 A thin copper wire of length L increase in
length by 2% when heated from T1 to T2. If a copper cube
having side 10 L is heated from T1 to T2, what will be the
percentage change in
(i) area of one face of the cube and
(ii) volume of the cube ?
Sol. (i) Area, A = × =
10 10 100 2
L L L
Percentage change in area
= × = × ×
∆ ∆
A
A
L
L
100 2 100
= × =
2 2 4
% %
(ii) Volume, V = × × =
10 10 10 1000 3
L L L L
Percentage change in volume
= ×
∆V
V
100 = × = × =
3 100 3 2 6
∆L
L
% %
Example 1.38 Calculate percentage error in determination of
time period of a pendulum
T
l
g
= 2π
where, l and g are measured with ± 1% and ± 2% errors.
Units, Dimensions & Error Analysis 33

44. Sol. Percentage error in time period,
∆ ∆ ∆
T
T
l
l
g
g
× = ± × × + × ×






100
1
2
100
1
2
100
= ± × + ×





 = ±
1
2
1
1
2
2
% % 1.5 %
Example 1.39 Find the relative error in Z, if Z
A B
CD
=
4 1 3
3 2
/
/
and
the percentage error in the measurements of A, B, C and D
are 4%, 2%, 3% and 1%, respectively.
Sol. Q
∆ ∆ ∆ ∆ ∆
Z
Z
A
A
B
B
C
C
D
D
=





 +





 + +






4
1
3
3
2
Given,
∆A
A
× =
100 4%
∆B
B
× =
100 2%
∆C
C
× =
100 3%
and
∆D
D
× =
100 1%
∴
∆Z
Z
× = × + ×





 + + ×






100 4 4
1
3
2 3
3
2
1
( %) % % %
= + + +






16
2
3
3
3
2
%
= 21.16%
The percentage error in the measurement of Z is 21.16%.
Therefore, the relative error in Z is 0.2116.
1. A spherometer has 100 equal divisions marked along the
periphery of its disc and one full rotation of the disc
advances on the main scale by 0.01 cm. The least count of
this system is
(a) 10 2
−
cm (b) 10 4
−
cm
(c) 10 5
−
cm (d) 10 1
−
cm
2. Three measurements are made as 18.425 cm, 7.21 cm and
5.0 cm. The mean of measurements should be written as
(a) 10.212 cm (b) 10.21 cm (c) 10.22 cm (d) 10.2 cm
3. If error in measuring diameter of a circle is 4%, the error in
measuring radius of the circle would be
(a) 2% (b) 8%
(c) 4% (d) 1%
4. The length of a rod is (11.05 0.2) cm
± . What is the net
length of the system of rods, when these two rods are
joined side by side?
(a) (22.1 0.05) cm
± (b) (22.1 0.1) cm
±
(c) (22.10 0.05) cm
± (d) (22.10 0. ) cm
± 4
5. A body travels uniformly a distance of (13.8 0.2) m
± in a
time (4.0 0. ) s
± 3 . The velocity of the body within error limit
is
(a) (3.45 0.2) ms 1
± −
(b) (3.45 0.3) ms 1
± −
(c) (3.45 0.4) ms 1
± −
(d) (3.45 0.5) ms 1
± −
6. A cuboid has volume V = × ×
l l l
2 3 , where l is the length of
one side. If the relative percentage error in the
measurement of l is 1%, then the relative percentage error
in measurement of V is
(a) 18% (b) 6%
(c) 3% (d) 1%
7. A force F is applied on a square plate of side L. If the
percentage error in the determination of L is 2% and that in
F is 4%. What is the permissible error in pressure?
(a) 8% (b) 6%
(c) 4% (d) 2%
8. The heat generated in a wire depends directly on the
resistance, current and time. If the error in measuring the
above are 1%, 2% and 1%, respectively. The maximum error
in measuring the heat is
(a) 8% (b) 6% (c) 18% (d) 12%
9. If the error in the measurement of momentum of a particle
is ( %)
+ 100 , then the error in the measurement of kinetic
energy is
(a) 100% (b) 200% (c) 300% (d) 400%
10. The radius of a ball is (5.2 0.2) cm
± . The percentage error in
the volume of the ball is (approximately)
(a) 11% (b) 4% (c) 7% (d) 9%
11. The values of two resistors are (5.0 0.2) k
± Ω and
(10.0 0.1) k
± Ω. What is the percentage error in the
equivalent resistance when they are connected in parallel?
(a) 2% (b) 5% (c) 7% (d) 3%
34 OBJECTIVE Physics Vol. 1
CHECK POINT 1.3