1.

1 Units and Dimensions

To measure a physical quantity we need some
standard unit of that quantity. The measurement of the
quantity is mentioned in two parts, the first part gives how
many times of the standard unit and the second part gives
the name of the unit. Thus, suppose I say that length of this
wire is 5 m. The numeric part 5 says that it is 5 times of the
unit of length and the second part metre says that unit
chosen here is metre.
Fundamental and Derived Quantities
There are a large number of physical quantities and
every quantity needs a unit.
However, not all the quantities are independent. For
example, if a unit of length is defined, a unit of volume is
automatically obtained. Thus, we can define a set of
fundamental quantities and all other quantities may be
expressed in terms of the fundamental quantities.
Fundamental quantities are only seven in numbers. Unit
of all other quantities can be expressed in terms of the
units of these seven quantities by multiplication or
division.
Many different choices can be made for the
fundamental quantities. For example, if we take length
and time as the fundamental quantities, then speed is a
derived quantity and if we take speed and time as
fundamental quantities then length is a derived quantity.
Several system of units are in use over the world. The
units defined for the fundamental quantities are called
fundamental units and those obtained for derived
quantities are called the derived units.
SI Units
In 1971, General Conference on Weight and
Measures held its meeting and decided a system of units
which is known as the International System of Units. It is
abbreviated as SI from the French name Le Systems
International d Unites. This system is widely used
throughout the world. Table below gives the seven
fundamental quantities and their SI units.
Table 1.1
S. No. Quantity SI Unit Symbol
1. Length metre m
2. Mass kilogram kg
3. Time second s
4. Electric current ampere A
5. Thermodynamic temperature kelvin K
6. Amount of substance mole mol
7. Luminous intensity candela cd
Two supplementary units namely plane angle and
solid angle are also defined. Their units are radian (rad)
and steradian (st) respectively.
(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 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, pound and second. The
unit of force in this system is poundal.
SI Prefixes
The most commonly used prefixes are given below in
tabular form.
Chapter Snapshot
G
Units and Dimensions
G
Significant Figures
G
Error Analysis
1 Units, Dimensions
Error Analysis
and
Table 1.2
Power of 10 Prefix Symbol
6 mega M
3 kilo k
2 centi c
3 mili m
6 micro
9 nano n
12 pico p
Definitions of Some Important SI Units
(i) Metre 1 m = 1,650,763.73 wavelengths in
vacuum, of radiation corresponding to orange red
light of krypton-86.
(ii) Second 1 9 192 631 770 s , , , time periods of a
particular radiation from cesium-133 atom.
(iii) Kilogram 1 kg=mass of 1 Lvolume of water at 4C.
(iv) Ampere It is the current which when flows through
two infinitely long straight conductors of negligible
cross-section placed at a distance of 1 m in vacuum
produces a force of 2 10
7

N/m between them.
(v) Kelvin 1 K = 1 / 273.16 part of the thermodynamic
temperature of triple point of water.
(vi) Mole It is the amount of substance of a system
which contains as many elementary particles
(atoms, molecules, ions etc.) as there are atoms in
12 g of carbon-12.
(vii) Candela It is luminous intensity in a perpendicular
direction of a surface of
1
600000
m
2

_
,

of a black
body at the temperature of freezing platinumunder a
pressure of 1.013 10
5
N/m
2
.
(viii) Radian It is the plane angle between two radii of a
circle which cut-off on the circumference, an arc
equal in length to the radius.
(ix) Steradian The steradian is the solid angle which
having its vertex at the centre of the sphere, cut-off an
area of the surface of sphere equal to that of a square
withsides of lengthequal tothe radius of the sphere.
Dimensions
Dimensions of a physical quantity are the powers to
which the fundamental quantities must be raised to
represent the given physical quantity.
For example, 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.
For convenience, the fundamental quantities are
represented by one letter symbols. Generally mass is denoted
by M, length by L, time by T and electric current by A.
The thermodynamic temperature, the amount of
substance and the luminous intensity are denoted by the
symbols of their units K, mol and cd respectively. The
physical quantity that is expressed in terms of the base
quantities is enclosed in square brackets.
Thus, Eq. (i) can be written as
[density] = [ML ]
3
Such an expression for a physical quantity in terms of
the fundamental quantities is called the dimensional
formula.
Here, it is worthnoting that constants such as 5, or
trigonometrical functions such as sin , cos etc., have no
units and dimensions.
[sin ] [cos ] [tan ] [log ] x
[ ] e
x
[M L T ]
0 0 0
Table 1.3 given below gives the dimensional
formulae and SI units of some physical quantities
frequently used in physics.
2 Objective Physics Volume 1
Table 1.3
S. No. Physical Quantity SI Units Dimensional Formula
1. Velocity displacement/time m/s [ ] M LT
0 1
2. Acceleration velocity/time m/s
2
[ ] M LT
0 2
3. Force mass acceleration kg-m/s
2
newton or N [ ] MLT
2
4. Work force displacement kg-m
2
/s
2
N-m joule or J [ ] ML T
2 2
5. Energy 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
V
V
or
No units
[ ] M L T
0 0 0
12. Stress force/area N/m
2
[ ] ML T
1 2
13. Pressure force/area N/m
2
[ ] ML T
1 2
Key-Terms for Concepts
G Astronomical unit
1 AU mean distance of earth from sun
1.5 10 m
11
G Light year
1 ly distance travelled by light in vacuum in 1 year
9.46 10 m
15
G Parsec
1 parsec 3.07 10 m 3.26 light year
16
G X-ray unit
1 10
13
U m

G 1shake 10 s
8
=

G 1 bar = 10
5
N/m
2
10
5
pascal
G 1 torr 1 mm of Hg 133.3 Pa
G 1 barn

10
28
m
2
G 1 horse power 746 W
G 1 pound 453.6 g 0.4536 kg
Example 1.1 Find the dimensional formulae of
(a) coefficient of viscosity
(b) charge q
(c) potential V
(d) capacitance C, and
(e) resistance R.
Some of the equations containing these quantities are
F A
v
l

_
,
, q It , U VIt ,
q CV and V IR
where Adenotes the area, v the velocity, l is the length, I the
electric current, t the time and U the energy.
Solution (a)

F
A
l
v
[ ]
[ ][ ]
[ ][ ]
[ ][ ]
[ ][ ]

F l
A v
MLT
2
L
L LT
2 1

[ ] ML T
1 1
(b) q It
[ ] [ ][ ] [ ] q I t AT
Chapter 1 Units, Dimensions and Error Analysis 3
S. No. Physical Quantity SI Units Dimensional Formula
14. Modulus of elasticity stress/strain N/m
2
[ ] ML T
1 2
15. Frequency 1/time period per sec 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-m
2
[ML T ]
2 0
18. Surface tension force/length N/m [ML T ]
0 2
19. Gravitational constant
force (distance)
(mass)
2
2
N-m
2
/kg
2
[M L T ]
3 1 2
20. Angular momentum kg-m
2
/s [ML T ]
2 1
21. Coefficient of viscosity N-s/m
2
[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

24. Coefficient of thermal conductivity (K) watt/m-K [MLT ]

3 1

25. Gas constant (R ) J/mol-K [ML T mol ]

2 1 2 1

26. Boltzmann constant (k ) J/K [ML T ]

2 2 1

27. Wien's constant (b) m-K [L ]

28. Stefan's constant () watt/m
2
-K
4
[MLT ]
3 4

29. Electric charge C [AT]

30. Electric intensity N/C [MLT ]
3 1
A
31. Electric potential volt (V) [ML T ]
2 3 1
A
32. Capacitance farad (F) [M L T ]
2 1 4 2
A
33. Permittivity of free space C
2
N
1
m
2
[M L T A ]
1 3 4 2
34. Electric dipole moment C-m [LTA]
35. Resistance Ohm [ML T A ]
2 3 2
36. Magnetic field tesla (T) or weber/m
2
(Wb/m
2
) [MT A ]
2 1
37. Coefficient of self-induction henry (H) [ML T A ]
2 2 2
(c) U VIt
V
U
It

or [ ]
[ ]
[ ][ ]
[ ]
[ ][ ]
V
U
I t

ML T
A T
2 2

[ ] ML T A
2 3 1
(d) q CV
C
q
V

or [ ]
[ ]
[ ]
[ ]
[ ]
C
q
V


AT
ML T A
2 3 1


[ ] M L T A
4 2 2 2
(e) V IR
R
V
I

or [ ]
[ ]
[ ]
[ ]
[ ]
R
V
I


ML T
2 3 1
A
A
=[ ] ML T A
2 3 2
Uses of Dimensions
Theory of dimensions have following main uses
(i) Conversion 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 n
1 1 2 2
[ ] [ ] u u
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 T
1
and in the other system are M L
2 2
, and T
2
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)
Here, n
1
and n
2
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 1.2 The value of gravitational constant is
G 6.67 10
11
N-m
2
/kg
2
in SI units. Convert it into CGS
system of units.
Solution The dimensional formula of G is[ ] M L T
3 1 2
.
Using Eq. (i), i.e.,
n n
1 2
[M L T ] [M L T ]
1
1
1
3
1
2
2
1
2
3
2
2

n n
2 1

1
]
1

1
]
1

1
]
1
M
M
L
L
T
T
1
2
1
2
1
2
1 3 2
Here, n .
1
11
667 10

M
1
1 kg, M
2
1 g = 10 kg
3
L
1
1 m, L
2
1 cm = 10 m,
2

T T
1 2
1 s
Substituting in the above equation, we get
n .
2
11
667 10

1
]
1

1
]
1

1 kg
10 kg
1 m
10 m
3 2
1
3
1 s
1 s

1
]
1
2
or n
2
8
10

6.67
Thus, value of G in CGS system of units is
667 10
8
.

dyne cm /g
2 2
.
(ii) To check the dimensional correctness of a given
physical equation Every physical equation should
be dimensionally balanced. This is called the
Principle of Homogeneity. 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.
Example 1.3 Showthat the expression of the time periodT of
a simple pendulum of length l given by T l g 2 / is
dimensionally correct.
Solution T
l
g
2
Dimensionally, [T]
[L]
[LT ]
[T]
2

As in the above equation, the dimensions of both sides

are same. The given formula is dimensionally correct.
Principle of Homogeneity of Dimensions
This principle states that the dimensions of all the
terms in a physical expression should be same. For
example, 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, have the
same dimensions.
Example 1.4 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.
Solution From principle of homogeneity
[ ] [ ] a v
or [ ] [ ] a

LT
1
[ ] [ ] bt v
or [ ]
[ ]
[ ]
[ ]
[ ]
b
v
t

LT
1
T
or [ ] [ ] b LT
-2
Similarly, [ ] [ ] [ ] d t T
Further,
[ ]
[ ]
[ ]
c
d t
v
+
or [ ] [ ][ ] c v d t +
or [ ] [ ][ ] c LT
-1
T
or [ ] [ ] c L
(iii) To establish the relation among various physical
quantities If we knowthe factors on which a given
physical quantity may depend, we can find a
formula relating the quantity with those factors.
Let us take an example.
Example 1.5 The frequency ( ) f of a stretched string depends
upon the tension F (dimensions of force), length l of the string
4 Objective Physics Volume 1
and the mass per unit length of string. Derive the formula for
frequency.
Solution Suppose, that the frequency f depends on the
tension raised to the power a, length raised to the power band
mass per unit length raised to the power c.
Then, f F l
a b c
( ) ( ) ( )
or f k F l
a b c
( ) ( ) ( ) (i)
Here, k is a dimensionless constant.
Thus, [ ] [ ] [ ] [ ] f l
a b c
F
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

or f
k
l
F

Experimentally, the value of k is found to be

1
2

Hence, f
l
F

1
2
Limitations of Dimensional Analysis
The method of dimensions has the following
limitations
(i) By this method the value of dimensionless constant
cannot be calculated.
(ii) By this method the equation containing trigono-
metrical, exponential and logarithmic terms
cannot be analysed.
(iii) If a physical quantity depends on more than three
factors, then relation among them cannot be
established because we can have only three
equations by equalising the powers of M, L and T.
1.2 Significant Figures
Significant figures in the measured value of a physical
quantity tell the number of digits in which we have
confidence. Larger the number of significant figures
obtained in a measurement, greater is the accuracy of the
measurement.
All accurately known digits in a measurement plus
the first uncertain digit together formsignificant figures.
Significant figures depends on the least count
measuring instrument.
For example, when we measure the length of a
straight line using a metre scale and it lies between 7.4 cm
and 7.5 cm, we may estimate it as l 743 . cm. This
expression has three significant figures out of these 7 and
4 are precisely known but the last digit 3 is only
approximately known.
Rules for Counting Significant Figures
For counting significant figures, we use the following
rules
Rule 1 All non-zero digits are significant. For
example, x 2567 has four significant figures.
Rule 2 The zeros appearing between two non-zero
digits are counted in significant figures, no matter where
the decimal point is, if any. For example, 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.
For example, 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.
For example, 4.600 or 0.002300 have four significant
figures each.
Point of confusion and its remedy
Suppose we change the units,then we will write
2.30 m 23.0 cm
2300 mm
000230 . km
When we are writing 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 rule.
Rule 6 The power of 10 is irrelevant to the
determination of significant figures. For example, in the
measurements
230 . m 2.30 10
2
cm
2.30 10
3
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
significant.
Chapter 1 Units, Dimensions and Error Analysis 5
Rule 7 A choice of change of different units does not
change the number of significant digits or figures in a
measurement.
For example, 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.
This shows that location of decimal point is of no
consequence in determining the number of significant
figures.
Measured value Number of significant figures Rule
12376 5 1
6024.7 5 2
0.071 2 3
410 m 3 4
2.40 3 6
1.6 10
10
2 7
Rounding off a Digit
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.
For example, 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) it 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.
For example, 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.
Algebraic Operations with Significant
Figures
In addition , subtraction, multiplication or division the
final result should not have more significant figures than
the original data from which it was obtained. To
understand this, let us consider a chain of which all links
are strong except the one. The chain will obviously break
at the weakest link. Thus, the strength of the chain cannot
be more than the strength of the weakest link in the chain.
(i) Addition and 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.
Example 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).
Example 12.63 10.2 2.43 2.4
(ii) 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.
Example 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.
Example
1100
10.2
107.8431373 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
1100 m /s
10.2 m /s
= 107.8431373 108
Note In this case answer becomes 108. Think why?
1.3 Error Analysis
No measurement is perfect, as the errors involved in a
measurement cannot be removed completely. Measured
value is always somewhat different from the true value.
The difference is called an error.
Errors can be classified in two ways. First
classification is based on the cause of error. Systematic
errors and random errors fall in this group.
Second classification is based on the magnitude of
error. Absolute error, mean absolute error and relative (or
fractional) error lie on this group. Now let us discuss them
separately.
(i) Systematic errors These are the errors whose
causes are known to us. Such errors can therefore
be minimised. Following are few causes of these
errors.
(a) Instrumental errors may be due to erroneous
instruments. These errors can be reduced by
using more accurate instruments and applying
zero correction, when required.
(b) Sometimes errors arise on account of ignoring
certain facts. For example, 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.
(c) Change in temperature, pressure, humidity
etc., may also sometimes cause errors in the
6 Objective Physics Volume 1
result. Relevant corrections can be made to
minimise their effects.
(ii) 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. For example, 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 errors 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. Thus,
a
a a a
n
n
mean

+ + +
1 2
K
(iii) Absolute errors The difference between the true
value and the measured value of a quantity is called
an absolute error. Usually the mean value a
m
is
taken as the true value. So, if
a
a a a
n
m
n

+ ++
1 2
Then by definition, absolute errors in the
measured values of the quantity are,
a a a
m 1 1

a a a
m 2 2


a a a
n m n

Absolute error may be positive or negative.
(iv) Mean absolute error It is the arithmetic mean of
the magnitudes of absolute errors. Thus,

a
| a| | a| | a|
n
n
mean

+ + +
1 2
K
The final result of measurement can be written as
a a a
m
t
mean
This implies that value of ais likely to lie between
a a
m
+
mean
and a a
m

mean
.
(v) 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
mean

a
a
m
Relative error expressed in percentage is called as
the percentage error, i.e.,
Percentage error
mean

a
a
m
100
Example 1.6 The diameter of a wire as measured by a screw
gauge was found to be 2.620, 2.625, 2.630, 2.628 and 2.626 cm.
Calculate
(a) mean value of diameter,
(b) absolute error in each measurement,
(c) mean absolute error,
(d) fractional error,
(e) percentage error, and
(f) express the result in terms of percentage error.
Solution (a) Mean value of diameter,
a
m

2.620 + 2.625 + 2.630 + 2.628 + 2.626
5
= 2.6258 cm = 2.626 cm
(rounding off to three decimal places)
(b) Taking a
m
as the true value, the absolute errors in
different observations are,
a
1
2.626 2.620 = + 0.006 cm
a
2
= 2.626 2.625 = + 0.001 cm
a
3
= 2.626 2.630 = 0.004 cm
a
4
= 2.626 2.628 = 0.002 cm
a
5
= 2.626 2.626 = 0.000 cm
(c) Mean absolute error,

a
a a a a a
mean

+ + + + | | | | | | | | | |
1 2 3 4 5
5

+ + + + 0006 0001 0004 0002 0000

5
. . . . .
. 00026
. 0003
(rounding off to three decimal places)
(d) Fractional error t
a
a
m
mean
t
0003
2626
.
.
t . 0001
(e) Percentage error = 0.001 100 t
= 0.1% t
(f) Diameter of wire can be written as,
d t 2.626 cm 0.1%
Combination of Errors
(i) Errors in sum or difference Let x a b t .
Further, let a is the absolute error in the
measurement of a, b is the absolute error in the
measurement of b and x is the absolute error in
the measurement of x. Then,
x x a a b b + t t t ( ) ( )
t t t t ( ) ( ) a b a b
t t t x a b ( )
or x a b t t
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 t ( )
i e . ., the maximum absolute error in sum and
difference of two quantities is equal to sum of the
absolute errors in the individual quantities.
Chapter 1 Units, Dimensions and Error Analysis 7
Example 1.7 The volumes of two bodies are measured to be
V cm
1
3
t (10.2 0.02) and V cm
2
3
t (6.4 0.01) . Calculate
sum and difference in volumes with error limits.
Solution V
1
t (10.2 0.02) cm
3
and V
2
t (6.4 0.01) cm
3
V V V t + ( )
1 2
= t (0.02 + 0.01) cm
3
t 0.03 cm
3
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) cm
3
t
and difference of volumes = (3.8 0.03) cm
3
t
(ii) Errors in a product Let x ab
Then, ( ) ( )( ) x x a a b b t t t
or x
x
x
ab
a
a
b
b
1 1 1 t

_
,

_
,

_
,

or 1 1 t t t t
x
x
b
b
a
a
a
a
b
b
(as x ab )
or t t t t
x
x
a
a
b
b
a
a
b
b
Here,
a
a
b
b
is a small quantity, so can be
neglected.
Hence, t t t
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
t

_
,

Therefore, maximum fractional error in product of

two (or more) quantities is equal to sum of
fractional errors in the individual quantities.
(iii) Error in division Let x
a
b

Then, x x
a a
b b
t
t
t

or x
x
x
a
a
a
b
b
b
1
1
1
t

_
,

_
,

_
,

or 1 1 1
1
t

_
,

_
,

_
,

x
x
a
a
b
b
as x
a
b

_
,

As
b
b
<<1, so expanding binomially, we get
1 1 1 t

_
,

_
,

_
,

x
x
a
a
b
b
or 1 1 t t + t
x
x
a
a
b
b
a
a
b
b
Here,
a
a
b
b
is a small quantity, so can be
neglected.
Hence, t t +
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
t

_
,

Or, the maximum value of fractional error in

division of two quantities is equal to the sum of
fractional errors in the individual quantities.
(iv) 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,
t t +
x
x
n
a
a
m
b
b
Therefore, maximum value of
x
x
n
a
a
m
b
b
t +

_
,

Example 1.8 The mass and density of a solid sphere are

measured to be(12.4 0.1) kg t and(4.6 0.2) t kg/m
3
. Calculate
the volume of the sphere with error limits.
Solution Here, m m t t (12.4 0.1) kg
and t t (4.6 0.2) kg/m
3
Volume V
m

12.4
4.6
= 2.69 m = 2.7 m
3 3
(rounding off to one decimal place)
Now,
V
V
m
m
t +

_
,

or

V
m
m
V t +

_
,

=
0.1
12.4
+
0.2
4.6
2.7 t

_
,

= 0.14 t
V V t t (2.7 0.14) m
3
8 Objective Physics Volume 1
Example 1.9 Calculate percentage error in determination of
time period of a pendulum
T
l
g
2
where l and g are measured with t 1% and t 2% errors.
Solution
T
T
l
l
g
g
t +

_
,
100
1
2
100
1
2
100
t +

_
,
t
1
2
1
1
2
2 1.5%
Least Count
The minimum measurement that can be measured
accurately by an instrument is called the least count. The
least count of a metre scale graduated in millimetre mark
is 1 mm. The least count of a watch having seconds hand is
1 s.
Key-Terms for Concepts
G Least count of vernier callipers
=
Value of 1 part of
main scale ( )

s

'

Value of one part on

vernier scale ( ) v

'

Least count of vernier callipers = 1MSD 1VSD

where, MSD Main Scale Division
VSD Vernier Scale Division
G Least count
Value of 1 part on main scale ( )
Number
s
of parts on vernier scale ( ) n
G Least count of screw guage
=
Pitch ( )
Number of parts on circular scale ( )
p
n
Solved Examples
Example 1. Check the correctness of the relation s ut at +
1
2
2
where u is initial velocity, a the acceleration, t the time and s
the displacement.
Solution Writing the dimensions of either side of the given
equation.
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, formula is dimensionally correct.
Example 2. 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.
Solution 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 2
[L
[ML T ][T]
2 3

[M L T ]
1 0 2
Example 3. 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.
Solution Let F k m v r
x y z
( ) ( ) ( ) (i)
Here, k is a dimensionless constant of proportinality.
Writing the dimensions of RHS and LHS in Eq. (i), we
have
[ [ ] [ MLT ] =[M] LT L]
2 x y z 1

[ ] M L T
+ x y z y
Equating the powers of M, L and T of 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)
Example 4. Write down the number of significant figures in
the following :
(a) 6428 (b) 62.00 m
(c) 0.00628 cm (d) 1200 N
Solution (a) 6428 has four significant figures.
(b) 62.00 m has four significant figures.
(c) 0.00628 cm has three significant figures.
(d) 1200 N has four significant figures.
Example 5. Round off to four significant figures :
(a) 45.689 (b) 2.0082
Solution (a) 45.69
(b) 2.008
Example 6. Add6.75 10
3
cm to 4.52 10
2
cm with regard to
significant figures.
Chapter 1 Units, Dimensions and Error Analysis 9
Solution a 6.75 10 cm
3

b 4.52 cm 10
2
0.452 10
3
cm
0.45 cm 10
3
(upto 2 places of decimal)
a b + + ( ) 6.75 0.45 cm 10 10
3 3
7.20 cm 10
3
Example 7. A thin wire has a length of 21.7 cm and radius
0.46 cm. Calculate the volume of the wire to correct significant
figures.
Solution Given l 21.7 cm,
r 0.46 mm 0.046 cm
Volume of wire V r l
2

22
7
2
( ( ) 0.046) 21.7
0.1443 cm
3
0.14 cm
3
Note The result is rounded off to least number of significant figures in the given
measurements i e . ., 2 ( in 0.46 mm ).
Example 8. The refractive index (n) of glass is found to have
the values 1.49, 1.50, 1.52, 1.54 and 1.48. Calculate
(a) the mean value of refractive index,
(b) absolute error in each measurement,
(c) mean absolute error,
(d) fractional error, and
(e) percentage error.
Solution (a) Mean value of refractive index,
n
m

+ + + + 1.49 1.50 1.52 1.54 1.48
5
1.505 1.51
(rounded off to two decimal places)
(b) Taking n
m
as the true value, the asbolute errors in
different observations are,
n
1
+ 1.51 1.49 0.02
n
2
+ 1.51 1.50 0.01
n
3
1.51 1.52 0.01
n
4
1.51 1.54 0.03
n
5
+ 1.51 1.48 0.03
(c) 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
(d) Fractional error
t

t n
n
m
mean
0.02
1.51
t 0.0132
(e) Percentage error t t ( ) 0.0132 1.32% 100
Example 9. The radius of sphere is measured to be
( ) 2.1 0.5 t cm. Calculate its surface area with error limits.
Solution Surface area, S r 4
2

_
,
( ) ( ) 4
22
7
2
2.1
55.44 55.4 cm
2
Further,
S
S
r
r
2
or

S
r
r
S

_
,
2 ( )

2 554 0.5
2.1
.
26.38 26.4 cm
2
S t ( ) 55.4 26.4 cm
2
Example 10. Calculate focal length of a spherical mirror from
the following observations. Object distance u t ( ) 50.1 0.5 cm
and image distance v t ( ) 20.1 0.2 cm.
Solution
1 1 1
f v u
+
or f
uv
u v

+

( ) (
( )
50.1 20.1)
50.1 20.1
14.3 cm
Also,
f
f
u
u
v
v
u v
u v
t + +
+
+

1
]
1
t + +
+
+

1
]
1
0.5
50.1
0.2
20.1
0.5 0.2
50.1 20.1
+ + [0.00998 0.00995 0.00997]
t ( ) 0.0299
f 0.0299 14.3
0.428 0.4 cm
f t ( ) 14.3 0.4 cm
10 Objective Physics Volume 1
NCERT Corner
1. Fill in the blanks
(a) The volume of a cube of side 1 cmis equal to m
3
.
(b) The surface area of a solid cylinder of radius
2.0 cm and height 10.0 cm is equal to mm
2
.
(c) A vehicle moving with a speed of 18 kmh
1
coversm in 1s.
(d) The relative density of lead is 11.3. Its density is
g cm
3
or kg m
3
Solution
(a) The volume of a cube of side 1 cmis given by,
V (10 m) = 10 m
2 3 6 3

(b) The surface area of a solid cylinder of radius r and height
h is given by
A = Area of two circles + curved surface area
2 + 2
2
r rh
+ 2r r h ( )
Here, r h 2 20 cm mm, = 10 cm = 100 mm
A = 2
22
7
20 (20 + 100) (mm)
2

15099 mm
2
1.5099 10 mm
4 2
1.5 10 mm
4 2
(c) Here v = 18 km h 5 ms
1 1

18 1000
3600
m
s
t 1s
x vt = = 5 1 5m
(d) Relative density of lead 11.3
Density of water

1 g cm
3
Relative density of lead
density of lead
density of water
Density of lead
relative density of lead density of water


11.3 1 g cm
3

11.3 g cm
3
In SI system density of water

10 kg m
3 3
Density of lead

11.3 10 kg m
3 3


1.13 10 kg m
4 3
2. Fill in the blanks by suitable conversion of units.
(a) 1 kg m s
2 2
g cm s
2 2
(b) 1 m ly (light year)
(c) 3.0 ms
2
km h
2
(d) G

6.67 10 Nm kg
11 2 2
cm s g
3 2 1
Solution
(a) 1 kg m s 1 (10 g) (10 cm) s
2 2 3 2 2 2

10 10
3 4
g cm s
2 2

10
7
g cm s
2 2
(b) 1 light year (ly) 9.46 10 m
15

1m =
1
9.46 10
ly 1.057 ly
15
16

(c) 3.0 ms = 3 10 km
1
60 60
h
2 3 1
2

_
,

3 10 3600
3 2
( ) kmh
2


3.888 10 kmh
4 2


3.9 10 kmh
4 2
(d) G =6.67 10 Nm kg
11 2 2

6.67 10 (10 dyne) (10 cm) (10 g)
11 5 2 2 2 3


6.67 10 10 10
11 5 4
10
6
dyne cm g
2 2
= 6.67 10 (g cm s ) cm g
8 2 2 2

6.67 10 cm g s
8 3 1 2
3. A calorie is a unit of heat or energy and it equals
about 4.2 J, where 1J = 1kgm s
2 2
. Suppose we
employ a system of units in which the unit of mass
equals kg, the unit of length m , and the unit of
time is s. Show that a calorie has a magnitude
4.2
1 2 2
in terms of new units.
Solution n u n u
1 1 2 2

or n n
u
u
n
a b c
a b c
2 1
=
1
2
1

[M L T ]
[M L T ]
1 1 1
2 2 2

1
]
1

1
]
1

1
]
1
n
a b c
1
1
2
1
2
M
M
1
2
L
L
T
T
1 cal = 4.2 J =4.2 kg m s ,
2 2
a b c = 1, = 2, = 2
n
2
1 2
4.2
1 kg
kg
1 m
m
1s
s

1
]
1

1
]
1

1
]
1

2
n
2
1 2 2
= 4.2

1 cal = 4.2
1 2 2
in new system.
4. A new unit of length is chosen such that the speed of
light in vacuum is unity. What is the distance
between the sun and the earth in terms of the new
unit if light takes 8 min and 20s to cover this distance?
Solution We are given that velocity of light in vacuum, c 1 new
unit of length s
1
.
Time taken by light of sun of reach the earth,
t 8 min 20s.
+ 8 60 20 500s
Distance between the sun and the earth,
x c t =
= 1new unit of length s s

1
500
500 new units of length
5. Which of the following is the most precise device for
measuring length?
(a) A vernier callipers with 20 divisions on the sliding
scale.
(b) A screw gauge of pitch 1 mm and 100 divisions on
the circular scale.
(c) An optical instrument that can measure length to
within a wavelength of light.
Solution The most precise device is that whose least count is
minimum.
Chapter 1 Units, Dimensions and Error Analysis 11
(a) Least count of vernier callipers
1 MSD 1 VSD
1 MSD
19
20
MSD =
1
20
MSD

1
20
mm
=
1
200
cm
0005 . cm
(b) Least count of screw gauge

Pitch
Number of divisions of circular scale

1
100
1
1000
mm cm
=0.001 cm
(c) Wavelength of light,

10
5
cm 0.00001 cm
Least count of optical instrument 0.00001 cm
Thus, clearly the optical instrument is the most precise.
6. State the number of significant figures in the
following
(a) 0.007 m
2
(b) 2.64 10 kg
24

(c) 0.2370 g cm
3
(d) 6.320 J
(e) 6.032 Nm
2
(f) 0.0006032 m
2
Solution The number of significant figures is as given below.
(a) 1 (b) 3
(c) 4 (d) 4
(e) 4 (f) 4
7. The length, breadth and thickness of a rectangular
sheet of metal are 4.234 m, 1.005 m and 2.01 cm
respectively. Give the area and volume of the sheet to
correct significant figures.
Solution Here length, l 4.234 m
Breadth, b = 1.005 m
Thickness, h = 0.0201 m = 2.01 cm
Area of the sheet + 2 ( + ) lb bh hl
+ + 2 4234 1005 1005 00201 00201 4234 ( . . . . . . )
8.7209468 m
2
As the least number of significant figures in thickness is 3,
Area 8.72 m
2

Volume l b h
4234 1005 00201
3
. . . m 00855
3
. m
8. The mass of a box measured by a grocers balance is
2.3 kg. Two gold pieces of masses 20.15 g and20.17 g
are added to the box. What is (a) the total mass of the
box, (b) the difference in the mass of the pieces to
correct significant figures?
Solution (a) Total mass (2.300 + 0.02015 + 0.02017) kg
234032 . kg
As the least number of significant figures in the mass of
box is 2, so maximum number of significant figures in
the result can be 2.
Total mass 2.3 kg
(b) Difference in masses 2017 2015 . . 002 . g
Since there are two significant figures, so the difference
in masses to the correct significant figures is 0.02 g.
9. A physical quantity P is related to four observables
a b c , , and das follows P
a b
cd

3 2
. The percentage errors
of measurement in a b c , , and d are 1%, 3%, 4%and 2%
respectively. What is the percentage error in the
quantity P? If the value of P calculated using the
above relation turns out to be 3.763, to what value
should you round off the result?
Solution P
a b
c d
=
3 2
( )
Percentage error in P is given by
P
P
a
a
b
b

_
,
+

_
,
100 3 100 2 100
+

_
,
+

_
,

1
2
100 100
c
c
d
d
(i)


a
a
c
c
b
b
d
d


1
]
1
1
1
100 1 100 4
100 3 100 2
%, %
%, %
...(ii)
From Eqs. (i) and (ii), we get
P
P
+ + + 100 3 1 2 3
1
2
4 2 % % % %
+ + + 3 6 2 2 13%
The calculation of error clearly shows that the number of
significant figures is 2, so the result of P may be rounded off
to two significant digits i.e. P = 3.763 = 3.8.
10. A book with many printing errors contains four
different formulae for the displacement y of a particle
under going a certain periodic motion :
(a) y a
t
T
= sin
2
(b) y a vt = sin
(c) y
a
T
t a = sin( / )
(d) y
a t
T
t
T
=
2
2 2

_
,

_
,

sin cos

(where a maximum displacement of the particle,
v speed of the particle, T time period of motion).
Rule out the wrong formulae on dimensional grounds.
Solution The argument of a trigonometrical function i e . ., angle is
dimensionless. Now here in each case dimensions of LHS is
[L] and dimensions of RHS in
(a) [L] angle
2
is dimensionless
t
T

_
,

(b)

[L] sin [LT (T)] = [L] sin [L]
1
(angle is not dimensionless here)
(c)

[L]
[T]
sin
[T]
[L]
= [LT ] sin [TL ]
1 1
(angle is not dimensionless here)
(d) +

1
]
1
[L] sin cos [ ]
T
T
T
T
L
Formulae (b) and (c) are wrong.
11. The unit of length convenient on the atomic scale is
known as an angstrom and is denoted by . 1
10 m
10

. The size of the hydrogen atom is about
0.5 . What is the total atomic volume in m
3
of a
mole of hydrogen atoms?
12 Objective Physics Volume 1
Solution r = 0.5

0.5 10 m
10
V
1
Volume of each hydrogen atom
4
3
3
r


4
3
3.14 (0.5 10 )
10 3


5.236 10 m
31 3
According to Avogadros hypothesis, one mole of hydrogen
contains
N = 6.023 10
23
atoms
Atomic volume of 1 mole of hydrogen atoms,
V NV =
1
or V = 6.023 10 5.236 10
23 31




3.154 10 m
7 3


3 10 m
7 3
12. One mole of an ideal gas at standard temperature and
pressure occupies 22.4 L (molar volume). What is the
ratio of molar volume to the atomic volume of a mole
of hydrogen? (Take the size of hydrogen molecule to
be about 1 ). Why is this ratio so large?
Solution d diameter of hydrogen molecule 1
Molar volume of one mole of hydrogen


22.4 L = 22.4 10 m
3 3
r radius of one molecule of hydrogen

2
=0.5
d


0.5 10 m
10
Volume of one molecule of hydrogen


4
3
=
4
3
(0.5 10 )
3 10 3
r
= 5.236 10 m
31 3

1 mole has 6.023 10
23
atoms or molecules of H
2
Atomic volume of one mole of hydrogen


6.023 10 5.236 10 m
23 31 3


3154 10
7 3
. m

Molar volume
Atomic volume
22.4 10 m
3.154
3 3

10 m
7 3
7.1 10
4
7 10
4
The large value of the ratio shows that the inter molecular
separation in a gas in much larger than size of a mulecule.
13. The nearest star to our solar system is 4.29 light years
away. Howmuch is this distance in terms of parsecs?
Solution Distance 4.29 light year
4.29 9.46 10 m
15
( 1 ly 9.46 10 m)
15
Q

4.29 9.46 10
3.08 10
15
16
parsec
(Q1parsec = 3.08 10 m)
16

1.318 parsec 1.32parsec

14. It is claimed that the two cesium clocks, if allowed
of run for 100 yr, free from any disturbance, may
differ by only about 0.02s. What does this imply for
the accuracy of the standard cesium clock in
measuring a time interval of 1s?
Solution Time interval 100 years
100 365 24 60 60s
3.155 10 s
9
Difference in time 0.2s
Fractional error
Difference in time(s)
Time interval(s)

0.2
3.155 10
9


6.34 10
12


10 10 10
12 11
In 1s, the difference is 10
11
to 6.34 10
12

Hence degree of accuracy shown by the atomic clock in 1s is

1 part in
1
10
11
to
1
6.34 10
12

or 10
11
to 10
12
.
15. Estimate the average mass density of sodium atom
assuming, its size to be about 2.5 (Use the known
values of Avogadro's number, and the atomic mass of
sodium). Compare it with the density of sodium in its
crystalline phase970 kg m
3
. Are the two densities of
the same order of magnitude? If so, why?
Solution Average radius of sodium atom,
r = 2.5 = 2.5 10 m
10

Volume of sodium atom
4
3
3
r


4
3
3.14 (2.5 10 )
10 3


65.42 10 m
30 3
Mass of a mole of sodium 23 g

23 10
3
kg.
One mole contains 6.023 10
3

2
atoms, hence the mass of
sodium atom,
M =
23 10
6.023 10
3
23

kg =3.82 10
26

kg
Average mass density of sodium atom,

M
V

3.82 10
65.42 10
kgm
26
30
3
= 0.64 10 kgm
3 3

Density of sodium in crystalline phase 970 kgm
3

0.970 10 kgm
3 3

Average mass density of sodium atom

Density of sodium of crystalline phase
0.64 10
0.970 10
3
3

0.66
Both densities are of the same order i.e. of the order of 10
3
.
This is because in the solid phase atoms are tightly packed,
so the atomic mass density is close to the mass density of the
solid.
16. A SONAR (sound navigation and ranging) uses
ultrasonic waves to detect and locate objects under
water. In a submarine equipped with a SONAR the
time delay between generation of a probe wave and
the reception of its echo after reflection from an
enemy submarine is found to be 77.0s. What is the
distance of the enemy submarine? (Speed of sound in
water 1450 m s
1

).
Chapter 1 Units, Dimensions and Error Analysis 13
Solution Time taken by the wave to go from submarine to enemy
submarine is
t
77
2
385 . s
Speed of sound, v

1450
1
ms
Distance of enemy submarine,
S vt = = 1450 38.50
55825 m 55.825 km
17. The farthest objects in our universe discovered by
modern astronomers are so distant that light emitted
by them takes billions of years to reach the earth.
These objects (known as quasars) have many
puzzling features which have not yet been
satisfactorily explained. What is the distance in km of
a quasar from which light takes 3.0 billion years to
reach us?
Solution Time taken, t = 3 10
9
years
3 10 365 24 60 60
9
s
Velocity of light, c = 3 10 ms
8 1

Distance of quasar from earth ct
3 10 3 10 365 24 3600
8 9
m
2.8 10 m
25
=2.8 10 km
22

Objective Problems (Level 1)

Measurement of Length and Time
1. Which one is not a unit of time?
(a) Leap year (b) Year
(c) Shake (d) Light year
2. Parsec is the unit of
(a) time (b) distance
(c) frequency (d) angular acceleration
3. One light-year distance is equal to
(a) 9.46 10
10
km (b) 9.46 10
12
km
(c) 9.46 10
9
km (d) 9.46 10
15
km
4. Parallactic second is equal to
(a) 9.4605 10
15
m (b) 3.07 10
16
m
(c) 1.496 10
11
m (d) 3 10
8
m
5. A new unit of length is chosen such that the speed of light in
vacuum is unity. What is the distance between the sun and
the earth in terms of the newunit, if light takes 8 min and 20 s
to cover this distance?
(a) 300 (b) 400
(c) 500 (d) 600
6. Astudent measures the thickness of a human hair by looking
at it through a microscope of magnification 100. He makes
20 observations and finds that the average width of the hair
is 3.5 mm. What is the estimate on the thickness of the hair?
(a) 0.0035 mm (b) 0.035 mm
(c) 0.01 m (d) 0.7 mm
7. Which of the following is the most precise device for
measuring length?
(a) a vernier calipers with 20 divisions on the sliding scale
(b) an optical instrument that can measure length to within
a wavelength of light
(c) a screw gauge of pitch 1 mm and 100 divisions on the
circular scale
(d) All the above are equally precise
Units and Dimensions
8. The dimensions of impulse are equal to that of
(a) force (b) linear momentum
(c) pressure (d) angular momentum
9. In the SI system, the unit of temperature is
(a) degree centigrade (b) kelvin
(c) degree celsius (d) degree fahrenheit
10. Which one of the following have same dimensions?
(a) Torque and force
(b) Potential energy and force
(c) Torque and potential energy
(d) Plancks constant and linear momentum
11. Which of the following does not posses the same dimensions
as that of pressure?
(a) Stress (b) Bulk modulus
(c) Thrust (d) Energy density
12. Which of the following is a dimensional constant?
(a) Poissions ratio (b) Refractive index
(c) Relative density (d) Gravitational constant
13. Which one of the following is not the dimensionless
quantity?
(a) Plancks constant (b) Dielectric constant
(c) Solid angle (d) Strain
14. Joule second is the unit of
(a) energy (b) momentum
(c) angular momentum (d) power
15. Which of the following is not equal to watt?
(a) joule/second (b) ampere volt
(c) (ampere)
2
ohm (d) ampere/volt
16. Which of the following is not the units of surface tension?
(a) N m / (b) J m /
2
(c) kg / s
2
(d) None of these
17. Wb/m
2
is equal to
(a) dyne (b) tesla (c) watt (d) henry
18. Dimensional formula for electromotive force is same as that
for
(a) potential (b) current
(c) force (d) energy
19. Which of the following has the dimensions of pressure?
(a) [ML T ]
2 2
(b) [M L ]
1 1
(c) [MLT ]
2
(d) [ML T ]
1 2
14 Objective Physics Volume 1
20. Dimensions of torque are
(a) [M L T ]
2 2 2
(b) [ML T ]
2 2
(c) [ML T ]
0 1
(d) [ML T ]
2 1
21. Dimensions of impulse are
(a) [ML T ]
2 3
(b) [ML ]
2
(c) [MLT ]
1
(d) [MLT ]
2
22. 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
23. Dimensions of surface tension are
(a) [M L T ]
2 2 2
(b) [M LT ]
2 2
(c) [MT ]
2
(d) [MLT ]
2
24. The dimensional formula for Youngs modulus is
(a) [ML T ]
1 2
(b) [M LT ]
0 2
(c) [MLT ]
2
(d) [ML T ]
2 2
25. Which of the following is the dimensions of the coefficient of
friction?
(a) [M L T]
2 2
(b) [M L T ]
0 0 0
(c) [ML T ]
2 2
(d) [M L T ]
2 2 2
26. The dimensional formula for the action will be
(a) [MLT ]
2
(b) [M LT ]
2 2
(c) [ML T ]
2 1
(d) [M L T ]
2 2 2
27. [ML T ]
1 1
stand for dimensions of
(a) work (b) torque
(c) linear momentum (d) coefficient of viscosity
28. Dimensions of relative density is
(a) [ML ]
2
(b) [ML ]
3
(c) dimensionless (d) [M L ]
2 6
29. The dimensions of the ratio of angular to linear momentum
us
(a) [M LT ]
0 0
(b) [MLT ]
1
(c) [ML T ]
2 1
(d) [M L T ]
1 1 1
30. The dimensional formula for thermal resistance is
(a) [ML T K ]
2 3 1
(b) [ML T A ]
2 2 1
(c) [ML T K ]
2 3 2
(d) [M L T K]
1 2 3
31. [ML T A ]
2 3 1
is the dimensional formula for
(a) capacitance (b) resistance
(c) resistivity (d) potential difference
32. Temperature can be expressed as a derived quantity in
terms of any of the following
(a) length and mass (b) mass and time
(c) length, mass and time (d) None of these
33. Given that y a
t
p
qx

_
,

cos , where t represents time in

following statements is 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
34. The dimensional formula[ML T ]
0 3
is more closely associated
with
(a) power (b) energy
(c) intensity (d) velocity gradient
35. Which of the following is dimensionally correct?
(a) Pressure = energy per unit area
(b) Pressure = energy per unit volume
(c) Pressure = force per unit volume
(d) Pressure = momentumper unit volume per unit time
36. 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 and the acceleration due to gravity g.
Then, m is directly proportional to
(a) v
3
(b) v
4
(c) v
5
(d) v
6
37. If p represents radiation pressure, c represent speed of light
and Q represents radiation energy striking a unit area per
second, then non-zero integers x y , and z such that p Q c
x y z
is dimensionless are
(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 , ,
38. 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
39. Which of 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
40. 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
41. Which of the following will have the dimensions of time?
(a) LC (b)
R
L
(c)
L
R
(d)
C
L
42. If C and R denote capacity and resistance, the dimensions of
CR are
(a) [M L T]
0 0
(b) [ML T]
0
(c) [M L T ]
0 0 2
(d) not expressible in terms of M, L and T
43. The force F on a sphere of radius a moving in a mediumwith
velocity v is given by F a v 6 . The dimensions of are
(a) [ML ]
3
(b) [MLT ]
2
(c) [MT ]
1
(d) [ML T ]
1 1
44. The equation of a wave is given by y a
x
v
k

_
,
sin
where is angular velocity and v is the linear velocity. The
dimension of k will be
(a) [T ]
2
(b) [T ]
1
(c) [T] (d) [LT]
45. A force is given by F at bt +
2
, where t is the time. The
dimensions of a and b are
(a) [MLT ]
4
and [MLT] (b) [MLT ]
1
and [MLT ]
0
(c) [MLT ]
3
and [MLT ]
4
(d) [MLT ]
3
and [MLT ]
0
46. The dimensional formula for Plancks 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
Chapter 1 Units, Dimensions and Error Analysis 15
47. The dimension of
1
2
0
2
E (
0
is the permittivity of the space
and E is electric field), is
(a) [ML T ]
2 1
(b) [ML T ]
1 2
(c) [ML T ]
2 2
(d) [MLT ]
1
48. 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
49. Dimension of velocity gradient is
(a) [M L T ]
0 0 1
(b) [ML T ]
1 1
(c) [M LT ]
0 1
(d) [ML T ]
0 1
50. The dimensional formula for emf e is MKS system will be
(a) [ML T Q ]
2 2 1
(b) [ML T ]
2 1
(c) [ML Q ]
2 1
(d) [MLT Q ]
2 2
51. The velocityv of a particle at timet is given byv at
b
t c
+
+
,
where a, b and c are constants. The dimensions of a, b and c
are respectively
(a) [LT ], [L]
2
and [T] (b) [L ], [T]
2
and [LT ]
2
(c) [LT ], [LT]
2
and [L] (d) [L], [LT] and [T
2
]
52. What is the units of k
1
4
0

?
(a) C N m
2 1 2
(b) Nm C
2 2
(c) Nm C
2 2
(d) Unitless
53. Pressure gradient has the same dimensions as that of
(a) velocity gradient (b) potential gradient
(c) energy gradient (d) None of these
54. The unit of permittivity of free space,
0
is
(a) coulomb/newton-metre
(b) newton-metre
2
/coulomb
2
(c) coulomb
2
/newton-metre
2
(d) coulomb
2
/(newton-metre)
2
55. Dimensions of electrical resistance are
(a) [ML T A ]
2 3 1
(b) [ML T A ]
2 3 2
(c) [ML T A ]
3 3 2
(d) [ML L T A ]
1 3 3 3
56. The magnetic moment has dimensions of
(a) [LA] (b) [L A]
2
(c) [LT A]
1
(d) [L T A]
2 1
57. 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
Significant Figures
58. The significant figures of the number 6.0023 is
(a) 2 (b) 5
(c) 4 (d) 1
59. What is the number of significant figures in 0.0310 10
3
?
(a) 2 (b) 3
(c) 4 (d) 6
60. The number of significant figures in 11.118 10 V
6

is
(a) 3 (b) 4
(c) 5 (d) 6
61. In which of the following numerical values, all zeros are
significant?
(a) 0.2020 (b) 20.2
(c) 2020 (d) None of these
62. A student measured the diameter of a wire using a screw
gauge with least count 0.001 cm and listed the
measurements. The correct measurement is
(a) 8.320 cm (b) 5.3 cm
(c) 5.32 cm (d) 5.3200 cm
63. 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 to correct significant figures is
(a) 0.0855 m
3
(b) 0.086 m
3
(c) 0.08556 m
3
(d) 0.08 m
3
64. Three measurements are made as 18.425 cm, 7.21 cm and
5.0 cm. The addition should be written as
(a) 30.635 cm (b) 30.64 cm
(c) 30.63 cm (d) 30.6 cm
65. 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 (d) None of these
66. 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
67. When 97.52 is divided by 2.54, the correct result is
(a) 38.3937 (b) 38.394
(c) 65.81 (d) 38.4
68. The radius of a thin wire is 0.16 mm. The area of
cross-section of the wire in mm
2
with correct number of
significant figures is
(a) 0.08 (b) 0.080
(c) 0.0804 (d) 0.080384
69. What is the number of significant figure in
(3.20 + 4.80) 10
5
?
(a) 5 (b) 4
(c) 3 (d) 2
70. What is the value of [( . ) ( . )] 50 10 50 10
6 8


with due
regards to significant digits?
(a) 25 10
14

(b) 250 10
14
.

(c) 2.50

10
13
(d) 250 10
15

71. The mass of a box is 2.3 kg. Two gold pieces of masses
20.15 g and 20.17 g are added to the box. The total mass of
the box to correct significant figures is
(a) 2.3 kg (b) 2.34 kg (c) 2.3432 kg (d) 2.31 kg
72. Subtract 0.2 kg from 34 kg. The result in terms of proper
significant figure is
(a) 33.8 kg (b) 33.80 kg
(c) 34 kg (d) 34.0 kg
73. The length, breadth and thickness of a block are given by
l b 12 6 cm cm , andt 245 . cm. The volume of the block
according to the idea of significant figures should be
(a) 1 10
2
cm
3
(b) 2 10
2
cm
3
(c) 1.763 10 cm
2 3
(d) None of these
Error Analysis
74. The length of a rod is (11.05 0.2) cm t . What is the length of
the two rods?
(a) (22.1 0.05) cm t (b) (22.1 0.1) cm t
(c) (22.10 0.05) cm t (d) (22.10 0.2) cm t
75. The radius of a ball is(5.2 0.2) cm t . The percentage error in
the volume of the ball is approximately
(a) 11% (b) 4% (c) 7% (d) 9%
16 Objective Physics Volume 1
76. A physical quantity Q is calculated according to the
expression
Q
A B
C D

3 3
If percentage errors in A B , , C D , are 2%, 1%, 3% and 4%
respectively. What is the percentage error in Q?
(a) t 8% (b) t 10%
(c) t 14% (d) t 12%
77. Abody travels uniformly a distance of (13.8 0.2) m t in a time
(4.0 0.3) s t . The velocity of the body within error limit is
(a) (3.45 0.2) ms
1
t

(b) (3.45 0.3) ms
1
t

(c) (3.45 0.4) ms
1
t

(d) (3.45 0.5) ms
1
t

78. If the error in the measurement of momentumof a particle is
( %) + 100 , then the error in the measurement of kinetic
energy is
(a) 100% (b) 200%
(c) 300% (d) 400%
79. 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%
80. The values of two resistors are (5.0 0.2) k t and
(10.0 0.1) k t . What is the percentage error in the
equivalent resistance when they are connected in parallel?
(a) 2% (b) 5%
(c) 7% (d) 10%
81. The heat generated in a wire depends 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%
82. 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%
83. A cuboid has volumeV 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%
Miscellaneous Problems
84. The ratio of the SI unit to the CGS unit of modulus of rigidity
is
(a) 10
2
(b) 10
2
(c) 10
1
(d) 10
85. Imagine a system of unit in which the unit of mass is 10 kg,
length is 1 km and time is 1 min. Then, 1 J in this system is
equal to
(a) 360 (b) 3.6
(c) 36 10
5
(d) 36 10
5

86. The dimensional formula for molar thermal capacity is same
as that of
(a) gas constant (b) specific heat
(c) Boltzmanns constant (d) Stefans constant
87. In measuring electric energy, 1kWh is equal to
(a) 3.6 10 J
4
(b) 3.6 10 J
6

(c) 7.3 10 J
6
(d) None of these
88. Out of the following four dimensional quantities, which one
qualifies to be called a dimensional constant?
(a) Acceleration due to gravity
(b) Surface tension of water
(c) Weight of a standard kilogram mass
(d) The velocity of light in vacuum
89. The square root of the product of inductance and
capacitance has the dimensions of
(a) length (b) time
(c) mass (d) no dimension
90. With usual notation, the following equation, said to give the
distance covered in the nth second. i.e.,
S u a
n
n
+
( ) 2 1
2
is
(a) numerically correct only
(b) dimensionally correct only
(c) both dimensionally and numerically only
(d) neither numerically nor dimensionally correct
Objective Problems (Level 2)
1. A quantity is given by X
V
t

0
l
, where V is the potential
difference and l is the length. Then, X has dimensional
formula same as that of
(a) resistance (b) charge
(c) voltage (d) current
2. The length of a strip measured with a metre rod is 10.0 cm.
Its width measured with a vernier calipers is 1.00 cm. The
least count of the metre rod is 0.1 cm and that of vernier
calipers 0.01 cm. What will be error in its area?
(a) t 13% (b) t 7%
(c) t 4% (d) t 2%
3. The length of cylinder is measured with a metre rod having
least count 0.1 cm. Its diameter is measured with vernier
calipers having least count 0.01 cm. Given that length is 5.0 cm
and radius is 2.0 cm. The percentage error in the calculated
value of the volume will be
(a) 1.5% (b) 2.5% (c) 3.5% (d) 4%
4. The random error in the arithmetic means of
100 observations is x, then random error in the arithmetic
mean of 400 observation would be
Chapter 1 Units, Dimensions and Error Analysis 17
(a) 4x (b)
1
4
x
(c) 2x (d)
1
2
x
5. Dimensions of ohm are same as
(a)
h
e
(b)
h
e
2
(c)
h
e
2
(d)
h
e
2
2
(where h is Plancks constant and e is charge)
6. Given that
dx
ax x
a
x a
a
n
2
2
1

1
]
1

sin
wherea constant. Using dimensional analysis, the value of
n is
(a) 1 (b) zero
(c) 1 (d) None of these
7. If E energy, G gravitational constant, I impulse and
M mass, then dimensions of
GIM
E
2
2
are same as that of
(a) time (b) mass
(c) length (d) force
8. The dimensional formula for magnetic flux is
(a) [ML T A ]
2 2 1
(b) [ML T A ]
3 2 2
(c) [M L T A ]
0 2 2 2
(d) [ML T A ]
2 1 2
9. Using mass ( ) M , length ( ) L , time ( ) T and current ( ) A as
fundamental quantities, the dimension of permeability is
(a) [M LT A]
1 2
(b) [ML T A ]
2 2 1
(c) [MLT A ]
2 2
(d) [MLT A ]
1 1
10. Let g be the acceleration due to gravity at earths surface
and K the rotational kinetic energy of the earth. Suppose the
earths radius decreases by 2%. Keeping mass to be
constant, then
(a) g increases by 2% and K increases by 2%
(b) g increases by 4% and K increases by 4%
(c) g increases by 4% and K increases by 2%
(d) g increases by 2% and K increases by 4%
11. If the energy ( ) E , velocity ( ) v and force ( ) F be taken as
fundamental quantities, then the dimension of mass will be
(a) Fv
2
(b) Fv
1
(c) Ev
2
(d) Ev
2
12. In a system of units, the units of mass, length and time are
1 quintal, 1 km and 1 h respectively. In this system 1 N
force will be equal to
(a) 1 new unit (b) 129.6 new unit
(c) 427.6 new unit (d) 60 new unit
13. If force F, length L and time T are taken as fundamental
units, the dimensional formula for mass will be
(a) [FL T ]
1 2
(b) [FLT ]
2
(c) [FL T ]
1 1
(d) [FL T ]
5 2
14. Given that y A ct x

_
,

1
]
1
sin ( )
2

, where y and x are

measured in metre. Which of the following statements is
true?
(a) The unit of is same as that of x and A
(b) The unit of is same as that of x but not of A
(c) The unit of c is same as that of
2

(d) The unit of ( ) ct x is same as that of

2

15. The frequency of vibration of string is given by

f
p
l
F
m

1
]
1
2
1 2 /
. Here, p is number of segments in the string
and l is the length. The dimensional formula for m will be
(a) [M LT ]
0 1
(b) [ML T ]
0 1
(c) [ML T ]
1 0
(d) [M L T ]
0 0 0
Assertion and Reason
Directions (Q. Nos. 1-17) These questions consists of two
statements each printed as Assertion and Reason. While
answering these questions you are required to choose any one
of the following five responses.
(a) If both Assertion and Reason are correct and Reason in
the correct explanation of Assertion.
(b) If both Assertion and Reason are correct but Reason is
not correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
(e) If both Assertion and Reason are false.
1. Assertion Pressure has the dimensions of energy density.
Reason Energy density

energy
volume
[ML T ]
[L ]
2 2
3


[ML T ]
1 2
pressure .
2. Assertion Method of dimension cannot be used for deriving
formulae containing trigonometrical ratios.
Reason This is because trigonometrical ratios have no
dimensions.
3. Assertion When percentage errors in the measurement of
mass and velocity are 1% and 2% respectively, the
percentage error in KE is 5%.
Reason KE or E mv
E
E
m
m
v
v
+
1
2
2
,
2
4. Assertion The error in the measurement of radius of the
sphere is 0.3%. The permissible error in its surface area is
0.6%.
Reason The permissible error is calculated by the
formula
A
A
r
r
4 .
5. Assertion The light year and wavelength consist of
dimensions of length.
Reason Both light year and wavelength represent time.
6. Assertion Number of significant figures in 0.005 is one and
that in 0.500 are three.
Reason This is because zeros before decimal are non
significant.
7. Assertion Out of two measurements l 07 . mand l 070 . m,
the second one is more accurate.
18 Objective Physics Volume 1
Reason In every measurement, more the last digit is not
accurately known.
8. Assertion When we change the unit of measurement of a
quantity, its numerical value changes.
Reason Smaller the unit of measurement smaller is its
numerical value.
9. Assertion L R / and CR both have same dimensions.
Reason L R / and CR both have dimension of time.
10. Assertion
Magnetic dipole moment moment induction
Moment of

inertia
Dimensional formula [M L T]
0 0
Reason The given dimension is that of frequency.
11. Assertion
Modulus of elasticity
Density
has the unit ms
1
.
Reason Acceleration has the dimensions of
1
0
( )
0
t
.
12. Assertion If x
a
b
n
m
the
x
x
n
a
a
m
b
b

_
,

t

_
,

The change in a or b i.e., a or b may be comparable to a

and b.
Reason The above relation is valid when a a << and
b b << .
13. Assertion Systematic errors and random errors fall in the
same group of errors.
Reason Both systematic and randomerrors are based on the
cause of error.
14. Assertion Absolute error may be negative or positive.
Reason Absolute error is the difference between the real
value and the measured value of a physical quantity.
15. Assertion The watches having hour hand, minute hand and
seconds hand have least count as 1 s.
Reason Least count is the maximum measurement that can
be measured accurately by an instrument.
16. Assertion Pendulum bob is preferred to be spherical.
Reason Sphere has minimum surface area.
17. Assertion A screwgauge having a smaller value of pitch has
greater accuracy.
Reason The least count of screw gauge is directly
proportional to the number of divisions on circular scale.
Match the Columns
1. Match the following columns.
Column I
Column II
(A) R L / (p) Time
(B) C R / (q) Frequency
(C) E B / (r) Speed
(D)
0 0
(s) None
2. Match the following columns.
Column I Column II
(A) Stress (p) Pressure
(B) Strain (q) Energy density
(C) Modulus of elasticity (r) Angle
(D) Torque (s) Energy
3. Suppose force( ) F , area( ) A and time( ) T are the fundamental
units, then the match the following columns.
Column I Column II
(A) Work (p) [A T ]
1/ 2 1
(B) Moment of inertia (q) [FA ]
1/ 2
(C) Velocity (r) [FA T ]
1/ 2 2
4. Match the following columns.
Column I Column II
(A) Electrical resistance (p) [M L T A ]
1 2 4 2
(B) Capacitance (q) [ML T A ]
2 2 2
(C) Magnetic field (r) [ML T A ]
2 3 2
(D) Inductance (s) [MT A ]
2 1
5. Match the following two columns.
Column I Column II
(A)
GM M
e s
(p)
[ ] M L T
2 2 3
(B) 3RT
M
(q)
[ ] ML T
3 2
(C)
F
q B
2
2 2
(r)
[ ] L T
2 2
(D) GM
R
e
e
(s)
None
Entrance Corner
1. The SI unit of activity of a radioactive sample is
[J& K CET 2011]
(a) Curie
(b) Rutherford
(c) Becquerel
(d) Millicurie
2. SI unit of power is [J& K CET 2011]
(a) Joule (b) Erg
(c) Newton (d) Watt
3. The SI unit of thermal conductivity is [J& K CET 2011]
(a) Jsm K
1 1
(b) W m K
1 1 1
(c) Wm K
1 1
(d) Wm K
2 1
Chapter 1 Units, Dimensions and Error Analysis 19
4. The dimensions of (
0 0
1 2
)
/
are
[CBSE AIPMT 2011]
(a) [L T]
1
(b) [LT ]
1
(c) [L T ]
1/ 2 1/ 2
(d) [L T ]
1/ 2 1/ 2
5. Surface tension has the same dimensions as that of
[Kerala CEE 2011]
(a) coefficient of viscosity
(b) impulse
(c) momentum
(d) spring constant
(e) frequency
6. The dimension of impulse is [J&K CET 2011]
(a) [MLT ]
1
(b) [ML T ]
2 1
(c) [ML T ]
1 1
(d) [MT ]
1
7. If C be the capacitance andV be the electric potential, then
the dimensional formula of CV
2
is [KCET 2011]
(a) [ML T A ]
2 2 0
(b) [MLT A ]
2 1
(c) [M LT A ]
0 2 0
(d) [ML TA]
3
8. What is the dimension of surface tension? [WB JEE 2011]
(a) [MLT ]
0
(b) [MLT ]
1
(c) [ML T ]
0 2
(d) [ML T ]
0 2
9. The unit of magnetic moment is [Guj. CET 2010]
(a) TJ
1
(b) JT
1
(c) Am
2
(d) Am
1
10. Unit of electrical conductivity is [UP CPMT 2010]
(a) ohm (b) siemen
(c) m mho / (d) mho m /
11. From the dimensional consideration which of the following
equations is correct? [Haryana PMT 2010]
(a) T
R
GM
2
3
(b) T
GM
R
2
3

(c) T
GM
R
2
2
(d) T
R
GM
2
2

12. If force F, length L and time T be considered fundamental

units, then units of mass will be [VMMC 2010]
(a) [FLT ]
2
(b) [FL T ]
2 1
(c) [FL T ]
1 2
(d) [F LT ]
2 2
13. Dimensions of capacitance is [Manipal 2010]
(a) [M L T A ]
1 2 4 2
(b) [MLT A ]
3 1
(c) [ML T A ]
2 3 1
(d) [M L T A ]
1 2 3 1
14. A uniform wire of length L, diameter D and density is
stretched under a tensionT. The correct relation between its
fundamental frequency f , the length L and the diameter D
is [KCET 2010]
(a) f
LD

1
(b) f
L D

1
(c) f
D

1
2
(d) f
LD

1
2
15. The dimensions of resistance are same as those of ......
where h is the Planck's constant, e is the charge.
[KCET 2010]
(a)
h
e
2
2
(b)
h
e
2
(c)
h
e
2
(d)
h
e
16. The equation of state of some gases can be expressed as
p
a
V
V b RT +

_
,

2
( ) where, p is absolute the pressure, V
is the volume, T is absolute temperature and a and b are
constants. The dimensional formula of a is [JCECE 2010]
(a) [ML T ]
5 2
(b) [M L T ]
1 5 2
(c) [ML T ]
1 2
(d) [ML T ]
5 2
17. The relation p e
Z
k

, where p is pressure, Z is distance,

k is Boltzmann constant and is temperature. The
dimensional formula of will be [AFMC 2010]
(a) [M L T ]
0 2 0
(b) [ML T]
2
(c) [ML T ]
0 1
(d) [M L T ]
0 2 1
18. The dimension of electromotive force in terms of current A
is [BVP 2010]
(a) [ML A ]
2 2
(b) [ML T A ]
2 2 2
(c) [ML T A ]
2 2 2
(d) [ML T A ]
2 3 1
19. The dimensional formula of
1
0 0

is
[Guj. CET 2010]
(a) [M LT ]
0 2
(b) [M L T ]
0 2 2
(c) [M LT ]
0 1
(d) [M L T ]
0 2 2
20. If p
RT
V b
e
V RT

/
, then dimensional formula of is
[UP CPMT 2010]
(a) p (b) R
(c) T (d) V
21. Velocity v is given by v at bt c + +
2
, where t is time. What
are the dimensions of a, b and c respectively?
[UP CPMT 2010]
(a) [LT ], [LT ]
3 2
and [LT ]
1
(b) [LT ], [LT ]
1 2
and [LT ]
3
(c) [LT ], [LT ]
2 3
and [LT ]
1
(d) [LT ], [LT ]
1 3
and [LT ]
2
22. If E M L , , and G denote energy, mass, angular momentum
and gravitation constant respectively, then the quantity
( / ) E L M G
2 2 5 2
has the dimensions of
[AMU 2010]
(a) angle (b) length
(c) mass (d) None of these
23. A capillary tube is attached horizontally to a constant heat
arrangement. If the radius of the capillary tube is increased
by10%, then the rate of flowof liquid will change nearly by
[WB JEE 2010]
(a) + 10% (b) + 46%
(c) 10% (d) 40%
24. If momentum is increased by 20%, then kinetic energy
increases by [WB JEE 2010]
(a) 48% (b) 44%
(c) 40% (d) 36%
25. If increase in linear momentum of a body is 50%, then
change in its kinetic energy is [Manipal 2010]
(a) 25% (b) 125%
(c) 150% (d) 50%
26. At constant temperature, the volume of a gas is to be
decreased by 4%. The pressure must be increased by
[BVP 2010]
(a) 4% (b) 4.16%
(c) 8% (d) 3.86%
27. Choose the incorrect statement out of the following.
[AMU 2010]
(a) Every measurement by any measuring instrument has
some errors
(b)
20 Objective Physics Volume 1
Every calculated physical quantity that is based on
measured values has some error
(c) A measurement can have more accuracy but less
precision and vice versa
(d) The percentage error is different form relative error
28. Which one of the following quantities has not been
expressed in proper units? [Kerala CEE 2009]
(a) Torque Newton metre
(b) Stress Newton metre
2
(c) Modulus of elasticity Newton metre
2
(d) Power Newton metre/second
1
(e) Surface tension Newton metre
2
29. The unit of specific conductivity is [Manipal 2009]
(a) - cm
1
(b) - cm
2
(c)
1
- cm (d)
1
- cm
1
30. An object is moving through the liquid. The viscous damping
force action on it is proportional to the velocity. Then
dimensional formula of constant of proportionality is
[UP CPMT, Punjab PMET 2009]
(a) [ML T ]
1 1
(b) [MLT ]
1
(c) [M LT ]
0 1
(d) [ML T ]
0 1
31. By what percentage should the pressure of a given mass of a
gas be increased, so as to decrease its volume by 10% at a
constant temperature? [AIIMS 2009]
(a) 5% (b) 7.2%
(c) 12.5% (d) 11.1%
32. Percentage error in the measurement of mass and speed are
2% and 3% respectively. The error in the estimation of
kinetic energy obtained by measuring mass and speed will
be [AIIMS 2009]
(a) 12% (b) 10%
(c) 2% (d) 8%
33. If the length of a seconds pendulum is increased by 2% then
in a day the pendulum [Kerala CEE 2009]
(a) loses 764 s (b) loses 924 s
(c) gains 236 s (d) loses 864 s
(e) gains 346 s
Answers
Objective Problems (Level 1)
Objective Problems (Level 2)
Assertion and Reason
Match the Columns
Entrance Corner
Chapter 1 Units, Dimensions and Error Analysis 21
1. (d) 2. (b) 3. (b) 4. (b) 5. (c) 6. (b) 7. (b) 8. (b) 9. (b) 10. (c)
11. (c) 12. (d) 13. (a) 14. (c) 15. (d) 16. (d) 17. (b) 18. (a) 19. (d) 20. (b)
21. (c) 22. (c) 23. (c) 24. (a) 25. (b) 26. (a) 27. (d) 28. (c) 29. (a) 30. (d)
31. (d) 32. (d) 33. (d) 34. (c) 35. (b) 36. (d) 37. (b) 38. (c) 39. (c) 40. (a)
41. (c) 42. (a) 43. (d) 44. (c) 45. (c) 46. (b) 47. (b) 48. (b) 49. (a) 50. (a)
51. (a) 52. (b) 53. (d) 54. (c) 55. (b) 56. (b) 57. (a) 58. (b) 59. (b) 60. (c)
61. (b) 62. (a) 63. (a) 64. (d) 65. (a) 66. (d) 67. (d) 68. (b) 69. (c) 70. (a)
71. (a) 72. (c) 73. (b) 74. (d) 75. (a) 76. (d) 77. (b) 78. (c) 79. (c) 80. (c)
81. (b) 82. (a) 83. (c) 84. (d) 85. (d) 86. (c) 87. (b) 88. (d) 89. (b) 90. (c)
1. (d) 2. (d) 3. (b) 4. (b) 5. (c) 6. (b) 7. (a) 8. (a) 9. (c) 10. (b)
11. (c) 12. (b) 13. (a) 14. (a) 15. (c)
1. (a) 2. (a) 3. (a) 4. (c) 5. (a) 6. (c) 7. (b) 8. (c) 9. (a) 10. (d)
11. (b) 12. (d) 13. (a) 14. (a) 15. (d) 16. (a) 17. (c)
1. (A q, B p, C r, D s) 2. (A r, B p, C s, D q)
3. (A q, B r, C p) 4. (A s, B p, C r, D q)
5. (A q, B r, C r, D s)
1. (c) 2. (d) 3. (c) 4. (b) 5. (d) 6. (a) 7. (a) 8. (c) 9. (b) 10. (d)
11. (a) 12. (c) 13. (a) 14. (a) 15. (c) 16. (a) 17. (a) 18. (d) 19. (d) 20. (a)
21. (a) 22. (d) 23. (b) 24. (b) 25. (b) 26. (b) 27. (d) 28. (c) 29. (d) 30. (c)
31. (d) 32. (d) 33. (d)
Solutions
Objective Problems (Level 1)
1. Leap year, year and shake are the units of time.
3. 1 light year ( ) ( ) ( ) ( ) 3 10 365 24 3600
5
9.416 10
12
km
8. Impulse change in linear momentum.
13. Solid angle, strain and dielectric constant are dimensionless
constant.
14. Since ( ) mvr n
h

2
and E h
So, unit of h joule second angular momentum
17. Wb/m
2
and tesla are the units of magnetic field.
21. Impulse = Force time
24. Young's modulus and pressure have the same dimensions.
26. Action is a force.
28. Relative density

Density of substance
Density of water at 4 C temperature
Dimensionless
36. m v g
a b c
. Writing the dimensions on both sides
[M] = [LT ] [ML ] [LT ]
1 2 2 a b c
[M] = [M L T
b a b c a c + 3 2
]
b 1
a b c + 3 0
a c 2 0
Solving these we get
a 6
Hence, m v
6
37. Since p Q c
x y z
is dimensionless. Therefore,
[ML T ] [MT ] [LT ] [M L T ]
1 2 3 1 0 0 0

x y z
Only option (b) satisfies this expression
So x y z 1 1 1 , ,
38. Since units of length, velocity and force and doubled
Hence, [ ] m
[force] [time]
[velocity]
, [time]
[length]
[ve

locity]
Hence unit of mass, and time remains same.
Momentum is doubled.
40. Since, R
l
A

, where is specific resistance.
[ ] , ,

1
]
1

RA
l
R
V
i
V
W
Q
[ ]

[ML T Q ]
3 1 2
41. i i e
t L R


0
1 { }
/ ( / )
Where
L
R
is time constant and its dimension is same as for
time.
42. CR is time constant.
44. k is dimensionless.
45. [ ] a
F
t

1
]
1
and [ ] b
F
t

1
]
1
2
47.
1
2
0
2
E is energy density or energy per unit volume.
48. p
a t
bx

2
, where p-pressure, t-time
[ ] [ ] [ ] pbx a t
2
Hence, [ ]
[ ]
[ ]
b
t
px

2
Dimensions of
a
b
px

[ ] [ ] MT
2
49. Velocity gradient is change in velocity per unit length.
50. Unit of emf e is volt.
51. [ ] : [ ] [ ] : [ ] [ ] a
v
t
b vt c t

1
]
1

54. F
q q
r

1
4
0
1 2
2

0
1 2
2
1
4

q q
Fr

0

(coulomb)
newton- metre
2
2
55. From definition of time constant t RC , where R is
resistance and C is capacitance.
R
t
C

[ ]
[ ]
T
M L T A
1 2 4 2
R

[ ] ML T A
2 3 2
56. M NIA
57. Since, R
A

l
, where is specific resistance
[ ] , ,

1
]
1

RA
R
V
i
V
W
Q l
[ ]

[ML T Q ]
3 1 2
68. R 0.16 mm
Hence, A R
2

22
7
016
2
( . )
0.080384
Since radius has two significant figure so answer also will
have two significant figures.
A 0.080
73. Minimum number of significant figure should be 1.
75. Radius of ball 5.2 cm
V R
4
3
3

22 Objective Physics Volume 1

V
V
R
R

_
,
3
V
V

_
,

_
,
100 3
02
52
100
.
.
11%
78. Since error in measurement of momentum is + 100%
p p p p
1 2
2 ,
K
P
m
K
p
m
1
2
2
2
2
2
2
,
( )
% in K
K K
K

_
,

2 1
1
100

_
,

4 1
1
100
300%
81. H i R t
2
% error in H 2 (% error in i)
+ (% error in R) + (% error in t)
82. p
F
A
F
L
FL

2
2
% error in pressure (% error in F) + 2 (error in L)
+ ( %) ( %) 4 2 2
8%
89. f
LC

1
2
or LC
f
T

1
2 2
Thus, LC has the dimensions of time.
Objective Problems (Level 2)
3. Volume of cylinder
V r L r
D

_
,

2
2
,

V
V
D
D
L
L

_
,

_
,
+

_
,
100 2 100 100

_
,
+

_
,
2
001
40
100
01
05
100
.
.
.
.
25 . %
4. Since error is measured for 400 observations instead of 100
observations. So error will reduce by 1/4 factor.
Hence,
x
4
5. Dimension of (ohm) R

h
e
2
; (e charge = current time)

[ ]
[ ]
Et
it
2

P
i
R
2
( ) as P
E
t

_
,

8. [ ] [ ] [ ] [ ] [ ]

BS MT A L ML T A
2 1 2 2 2 1
10. g
GM
R
K I
L
I

2
2
2
1
2 2
:
Further, L will remain constant.
K
I

1
or K
MR

1
2
5
2
or K R
2
and g R
2
11. Energy
1
2
2
mv
[ ]
[ ]
[ ]
[ ] m
E
v
Ev

2
2
12. [Froce] [MLT ]
2

1
1
100
1
1000
3600
2
N

_
,

_
,
( )
1296 . units.
13. [ ] [ ] [ ] [ ] [ ] FL T MLT L T M


1 2 2 1 2
14. Here,

( ) ct x is dimensionless.
Hence,
ct

is also dimensionless and unit of ct is same as that

of x.
Therefore, unit of is same as that of x. Also unit of y is
same as that of A, which is also that unit of x.
15. m is mass per unit length.
Match the Columns
3. [ ] [ ] A L
2
[ ] [ ]
/
L A
1 2
[ ] [ ] T T
[ ] [ ] F MLT
2
[ ] [ ] [ ]
/
M FL T FA T
1 2 1 2 2
Now, [ ] [ ] [ ]
/
W FL FA
1 2
[ ] [ ] [ ] [ ]
/ /
I ML FA T A FA T
2 1 2 2 1 2 2
[ ] [ ] [ ]
/
V LT A T
1 1 2 1
Entrance Corner
1. The SI unit of activity of a radioactive sample is Becqueral.
2. The SI unit of power is watt.
3. The SI unit of thermal conductivity is Wm K
1 1
.
4. This expression for speed of light and the dimensions of
speed of light are [LT ]
1
.
5. Surface tension has the same dimension as that of spring
constant.
6. The dimension of impulse [MLT ]
1

.
7. We know, Energy, E CV
2
Dimensions of CV
2
Dimensions of energy, E

[ ] ML T
2 2
Chapter 1 Units, Dimensions and Error Analysis 23
8. We know, T
F
l

Dimensions of T
F
L

Dimensions of
Dimension of
T
[MLT ]
[L]
2

[ML T ]
0 2
9. Magnetic moment is the strength of magnet. Its SI unit is
A m
2
or N-m/T or JT
1
.
10. Unit of electrical conductivity is mho / mor siemem/ m.
11. TakingT
R
GM
2
3

Substituting the dimensions,

LHS T T , [ ]
RHS, 2
3 3
1 3 2

R
GM

[ ]
[ ] [ ]
L
M L T M
[T] T]
2
= [
Thus, LHS RHS forT
R
GM
2
3

12. Let [ ] [ ] M F L T
a b c
So, using dimensions, we have
[ML T ] = [MLT ] [L] [T]
1 0 0 2
K
a b c
a a b 1 0 , b 1
and + 2 0 a c c 2
So unit of mass is [ ] FL T
1 2
13. The capacitance C of a conductor is defined as the ratio of
charge q given to raise the potential V of the conductor.
i e . ., C
q
V

Farad =
coulomb
volt
coulomb
joule / coulomb

coulomb
joule
2

(ampera - sec)
newton- metre
ampere - sec
(kg - m sec
2 2

2
) metre

ampera - sec
kg - metre
2 4
2

kg - metre - sec - amp
1 2 4 2
So, the dimension of capacitance is [M L T A ]
1 2 4 2
.
14. The fundamental frequency is f
L
T

1
2
f
L
T
D LD
T

1
2
4
1
2

f
LD

1
15. Resistance, R
V
i
W
qi

[ML T ]
[A T]
2 2
2
R

[ML T A ]
2 3 2
Now for
h
e
2

1
]
1

[ML T ]
[AT]
[ML T A ]
2 1
2
2 3 2
16. In the equation p, V and T are pressure, volume and
temperature respectively
p
a
V
V b RT +

_
,

2
( )
Dimensions of
a
V
2
will be same as that of pressure
Dimensions of
a
V
2
dimensions of p
Dimension of a dimension of p dimension of V
2
[ML T ] [L ] = [ML T ]
1 2 6 5 2
17. In the given equation,

Z
k
should be dimensionless

k
Z
[ ] =
[ML T K ] [K]
[L]
[MLT ]
2 2
2

1
and p

[ ]

=
p

1
]
1

[MLT ]
[ML T ]
2
1 2
[M L T ]
0 2 0
18. Electromotive force = potential difference
V
W
q

[ML T ]
[AT]
2 2

[ML T A ]
2 2 1
19. Velocity of electromagnetic waves v
1
0

0

1
0 0
2

v
Thus the dimensional formula of
1
0 0

[M LT ]
0 1 2

[M L T ]
0 2 2
20. Given p
RT
V b
e
V RT

/
So,
V
RT
is dimensionless.
Hence, [ ] =
RT
V

1
]
1


[ML T ] [ ]
[L ]
2 2 1
3

[ ] ML T
1 2
This is also the dimension less formula of pressure.
21. Dimensions of velocity is [ ] [ ] [ ] v L T
1
So, dimensions of [ ] [ ] at LT
2 1

[ ] [ ] [ ] a T LT
2 1

[ ] [ ] a LT
3
Dimensions of [ ] [ ] bt LT
1
[ ] [ ] [ ] b T LT
1
[ ] [ ] b LT
2
Dimensions of [ ] [ ] c LT
1
24 Objective Physics Volume 1
22. The dimension of E

[ ] ML T
2 2
Dimensions of M [ ] M
Dimensions of L

[ ] ML T
2 1
Dimensions of G

[ ] M L T
1 3 2
Dimensions of
E L
M G
[ML T ] [ML T ]
[M] [M L T ]
2 2
5 2
2 2 2 2 1 2
5 1 3 2

1
]
1


2


[ML T ]
2 2
23. Volume of liquid coming out of the tube per second
V
p r
l

4
8

V
V
r
r
2
1
2
1
4

1
]
1
V V
2 1
4
110
100

1
]
1
V
1
(1.1) 1.4641
4
volt

V
V
V V
V
V
V

2 1
1.4641
46%
24. The kinetic energy is given by KE
p
m
2
2
So,

KE
2
2
p p
m
p p
m

4 2 KE
KE

p
p
Thus, the final momentum becomes 1.2 p.
So, percentage change in KE
final KE initial KE
initial KE

100

144 2 2
2
44
2 2
2
. ( / ) ( / )
( / )
%
p m p m
p m
25. We know that linear momentum p mK 2
Now we have p p p p
1 2 1
50 + , % of p p
1 1
15 .

K
K
p
p
1
2
1
2
2
2
K
p
p
K K
2
2
2
1
2
1
225 .
So change in KE 225 1 125 125 . . %
26. At constant temperature, p V p V
1 1 2 2

p
p
V
V
1
2
2
1

Hence, tractional change in volume

V V
V
1 2
1
4
100
1
25

1
1
25
2
1

V
V

V
V
2
1
24
25

p
p
V
V
1
2
2
1
24
25

p p
p
2 1
1
25
24
1
1
24

Percentage increase in pressure
100
24
4.16%
27. When the relative error is expressed in percentage, we cal it
percentage error.
28. The correct unit of surface tension is newton/metre.
29. Specific conductivity
1 1
specific resistance -c m

( ) -cm
1
30. We have F v F kv
[ ] k

1
]
1

1
]
1

F
v
[MLT ]
[LT ]
2
1


[ML T ]
0 1
31. When T is constant, pV constant. When volume is
decreased by 10%that is volume becomes
90
100
, the pressure
must become 100/99. Thus percentage increase in pressure

( ) 100 90 100
90
11.1%
32. Kinetic energy K mv
1
2
2
Fractional error in kinetic energy
K
K
m
m
v
v
+
2
Percentage error in kinetic energy is
+
m
m
v
v
100
2
100
As we know,
m
m
100 2% and
2
100 3
v
v
%
So, percentage error in kinetic energy
+ + 2 2 3 2 6 8%
33. Time period, T
l
g
2 or
T
T
l
l

1
2
For 1s,

T
l
l
T T T

_
,

1
2
1
2
002 001 001 . . . s
For a day, T 24 60 60 001 864 . s
Chapter 1 Units, Dimensions and Error Analysis 25