PHYSICAL QUANTITIES

All quantities that can be measured are called physical quantities.

eg. time, length, mass, force, work done, etc. In physics we study

about physical quantities and their inter relationships.

MEASUREMENT

Measurement is the comparison of a quantity with a standard of

the same physical quantity. Different countries followed different

standards.

UNITS

All physical quantities are measured with respect to standard

magnitude of the same physical quantity and these standards are

called UNITS. e.g. second, meter, kilogram, etc.

Four basic properties of units are:

1. They must be well defined.

2. They should be easily available and reproducible.

3. They should be invariable e.g. step as a unit of length is not

invariable.

4. They should be accepted to all.

Set of Fundamental Quantities

A set of physical quantities which are completely independent of

each other but all other physical quantities can be expressed in

terms of these physical quantities is called Set of Fundamental

Quantities.

The Fundamental Quantities that are currently being accepted

by the scientific community are mass, time, length, current,

temperature, luminous intensity and amount of substance.

Derived Physical Quantities

The physical quantities that can be expressed in terms of

fundamental physical quantities are called derived physical

quantities. E.g. speed = distance/time.

System of Units

1. FPS or British Engineering system: In this system length,

mass and time are taken as fundamental quantities and their

base units are foot (ft), pound (lb) and second (s) respectively.

2. CGS or Gaussian system: In this system the fundamental

quantities are length, mass and time and their respective units

are centimetre (cm), gram (g) and second (s).

3. MKS system: In this system also the fundamental quantities

are length, mass and time but their fundamental units are

meter (m), kilogram (kg) and second (s) respectively.

Table: Units of some physical quantities in different systems

Type of physical

Quantity

Physical

Quantity

System

CGS MKS FPS

Fundamental

Length cm m ft

Mass g kg lb

Time s s s

4. International system (SI) of units: This system is

modification over the MKS system. Besides the three base

units of MKS system four other fundamental and two

supplementary units are also included in this system.

Table: SI base quantities and their units

S. No. Physical quantity unit Symbol

1 Length metre m

2 Mass kilogram kg

3 Time second s

4 Temperature kelvin K

5 Electric current ampere A

6 Luminous Intensity candela cd

7 Amount of substance mole mol

Physical Quantity

(SI Unit)

Definition

Length (m) The metre, symbol m, is the SI unit of length.

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

m s–1, where the second is defined in terms

of the caesium frequency ∆νCs.

CHAPTER

1 Units and Measurements

2 JEE (XI) Module-1 P

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Physical Quantity

(SI Unit)

Definition

Mass (kg) The kilogram, symbol kg, is the SI unit

of mass. 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 J s, which is equal to kg m2 s–1,

where the metre and the second are defined

in terms of c and ∆νCs.

Time (s) The second, symbol s, is the SI unit of time.

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 (A)

The ampere, symbol A, is the SI unit of

electric current. 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, which is equal to

A s, where the second is defined in terms

of ∆νCs.

Thermodynamic

Temperature (K)

The kelvin, symbol K, is the SI unit of

thermodynamic temperature. It is defined

by taking the fixed numerical value of the

Boltzmann constant k to be 1.380649×10–23

when expressed in the unit J K–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)

The mole, symbol mol, is the SI unit of

amount of substance. One mole substance

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 (cd)

The candela, symbol cd, is the SI unit of

luminous intensity in given direction. It is

defined by taking the fixed numerical value

of the luminous efficacy 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–2s3, where the kilogram, metre and second

are defined in terms of h, c and ∆νCs.

DIMENSIONS AND

DIMENSIONAL FORMULA

All the physical quantities of interest can be derived from the base

quantities. The power (exponent) of base quantity that enters into

the expression of a physical quantity, is called the dimension of

the quantity in that base. To make it clear, consider the physical

quantity force.

Force = Mass × Acceleration

= Length / Time Mass

Time × = Mass × Length × (Time)–2

So the dimensions of force are 1 in mass, 1 in length and –2 in

time. Thus

[Force] = MLT–2

Similarly, energy has dimensional formula given by

[Energy ] = ML2T–2

i.e. energy has dimensions 1 in mass, 2 in length and –2 in time.

Such an expression for a physical quantity in terms of base

quantities is called dimensional formula

Physical quantity can be further of four types:

1. Dimensionless constant i.e. 1, 2, 3, π etc.

2. Dimensionless variable i.e. angle θ etc.

3. Dimensional constant i.e. G, h etc.

4. Dimensional variable i.e. F, v, etc.

Table: Units and dimensions of some physical quantities

Quantity SI Unit Dimension

Density kg/m3 M/L3

Force newton (N) ML/T2

Work joule (J)(= N m) ML2/T2

Energy joule(J) ML2/T2

Power Watt (W) (= J/s) ML2/T3

Momentum kg m/s ML/T

Gravitational constant N m2/kg2 L3/MT2

Angular velocity radian/s T–1

Angular acceleration radian/s2 T–2

Angular momentum kg m2/s ML2/T

Moment of inertia kg m2 ML2

Torque N m ML2/T2

Angular frequency radian/s T–1

Frequency hertz (Hz) T–1

Period s T

Surface Tension N/m M/T2

Coefficient of viscosity N s/m2 M/LT

Wavelength m L

Intensity of wave W/m2 M/T3

Temperature Kelvin (K) K

Units and Measurements 3 P

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Quantity SI Unit Dimension

Specific heat capacity J/kg K L2/T2K

Stefan’s constant W/m2 K4 M/T3K4

Heat J ML2/T2

Thermal conductivity W/m-K ML/T3K

Current density A/m2 I/L2

Electrical conductivity 1/Ω m(= mho/m) I

2T3/ML3

Electric dipole moment C m LIT

Electric field V/m (=N/C) ML/IT3

Potential (voltage) Volt (V) (=J/C) ML2/IT3

Electric flux V m ML3/IT3

Capacitance Farad (F) I

2T4/ML2

Electromotive force Volt (V) ML2/IT3

Resistance ohm (Ω) ML2/I2T3

Permittivity of space C2/N m2 (=F/m) I2T4/ML3

Permeability of space N/A2 ML/I2T2

Magnetic field Tesla (T) (= Wb/m2) M/IT2

Magnetic flux Weber (Wb) ML2/IT2

Magnetic dipole moment A m2 IL2

Inductance Henry (H) ML2/I2T2

DIMENSIONAL EQUATION

Whenever the dimension of a physical quantity is equated with its

dimensional formula, we get a dimensional equation.

PRINCIPLE OF HOMOGENEITY

The magnitude of a physical quantity may be added or subtracted

from each other only if they have the same dimension. Also the

dimension on both sides of an equation must be same. This is

called as principle of homogeneity.

Train Your Brain

Example 1: The distance covered by a particle in time t is

given by x = a + bt + ct2 + dt3; find the dimensions of a,

b, c and d.

Sol. The equation contains five terms. All of them should

have the same dimensions. Since [x] = length, each of

the remaining four must have the dimension of length.

Thus, [a] = length = L

[bt] = L, or [b] = LT–1

[ct2] = L, or [c] = LT–2

and [dt3] = L or [d] = LT–3

Example 2: Calculate the dimensional formula of energy

from the equation E = 1

2

mv2.

Sol. Dimensionally, E = mass × (velocity)2, since 1

2

is a

number and has no dimension.

or, [E] = M ×

2      

L

T

= ML2T–2.

Example 3: Kinetic energy of a particle moving along

elliptical trajectory is given by K = αs2 where s is the

distance travelled by the particle. Determine dimensions

of α.

Sol. K = αs2

52

k

⇒α=

[α] =

2 2

2

( )

( )

− MLT

L

[α] = M1 L0 T–2

[α] = M T–2

Example 4: The position of a particle at time t, is given by

the equation, x(t) = 0

α

v (1 – e–αt

), where v0 is a constant and

α > 0. The dimensions of v0 and α are respectively.

(a) M0 L1 T0 and T–1 (b) M0 L1 T–1 and T

(c) M0 L1 T–1 and T–1 (d) M1 L1 T–1 and LT–2

Sol. (c) [α] [t]= M0L0T0 and [v0] = [x] [α]

[α] = M0L0T–1 = M0L1T–1

Example 5: If 2 ;

v As F

Bt

+ = find the dimension of A and B

where F = force, v = velocity, s = displacement and t = time

Sol. Using principle of homogeneity; AS v = [ ]

1

2 1 AL LT  

1

2 1 A LT − ∴ =

Also, we can write, B = 2

v AS

Ft



1

10 1

2 2 [ ] [ ] ( )( )

LT B B M LT

MLT T

−

− −

− = ⇒=

Concept Application

1. If v Bu at

A − = , find the dimension of A and B where

u, v = velocity, a = acceleration, t = time.

2. If 2

v As B

t = − find the dimension of A and B.

4 JEE (XI) Module-1 P

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3. If 2 sin B A C

t

  +     is equation of displacement of a

body. Find dimensions of A, B, C.

4. If displacement, y = 3

2

A Bt

t − find dimension of

(A × B).

USES OF DIMENSIONAL ANALYSIS

To Check the Dimensional Correctness of a

Given Physical Relation

It is based on principle of homogeneity, which states that a given

physical relation is dimensionally correct if the dimensions of the

various terms on either side of the relation are the same.

Remark:

™ Powers are dimensionless

™ sinθ, eθ, cosθ, logθ give dimensionless value and in above

expression θ is dimensionless

™ We can add or subtract quantity having same dimensions only.

Train Your Brain

Example 6: Check the accuracy of the relation

= π2 L T

g

for a simple pendulum using dimensional analysis.

Sol. From principle of homogeneity of dimension, the

dimensions of LHS = The dimension of RHS

Now, T = [M0L0T1]

The dimensions of

1/2 dimension of length RHS

dimension of acceleration

  =    

( 2π is a dimensionless constant)

[RHS] [ ] 1/2 1/2 2 001

2

L T T [M L T ]

LT−

  = = = =         =[LHS].

So, equation is correct.

Example 7: Check whether the given relation

2

KE Fv

t =

is dimensionally correct? Where F = force, v = velocity and

t = time?

Sol. [LHS] =

2 Fv

t

2 22 MLT L T 3 5 ML T

T

− − × − = =    

[RHS] = [KE] =

1 2 22 22 M L T ML T

2

mv   − − =× =          

⸫ [LHS] ≠ [RHS], so the given relation in incorrect

dimensionally.

Concept Application

5. Consider the following equation:

2

2 , t a qvbt c

x

  = +    

where a, b, c are constants (not necessarily

dimensionless) and q, v, x and t represent charge,

velocity, distance and time respectively. For the

equation to be dimensionally correct,

(a)

1

2 1 LT a

c

  − =         (b)

1

2 1 1 L T a

c

  − − =        

(c) 3 1 L A b

c

  − − =         (d) [ ] LTA a

b

  =    

6. Consider the statements below.

(i) A dimensionally consistent equation is a

physically correct equation.

(ii) A dimensionally consistent equation may or may

not be correct.

(iii) A dimensionally inconsistent equation is an

incorrect equation.

(iv) A dimensionally inconsistent equation may or

may not be incorrect.

The correct statement(s) is/are

(a) (i), (iii) and (iv) (b) (ii) and (iv)

(c) only (iii) (d) (ii) and (iv)

To Establish a Relation Between Different Physical

Quantities

If we know the various factors on which a physical quantity

depends, then we can find a relation among different factors by

using principle of homogeneity.

Train Your Brain

Example 8: Find an expression for the time period T of

a simple pendulum. The time period T may depend upon

(i) mass m of the bob of the pendulum, (ii) length  of

pendulum, (iii) acceleration due to gravity g at the place

where the pendulum is suspended.

Sol. Let (i) a T m ∝ (ii) b T ∝  (iii) c T g ∝

Combining all the three factors, we get

ab c Tmg ∝  or ab c d T Km g = θ  ...(i)

where K is a dimensionless constant of proportionality.

Writing down the dimensions on either side of equation

(i), we get

[T] = [Ma][Lb][LT–2]c

= [MaLb+c T–2c

]

Units and Measurements 5 P

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Comparing dimensions, a = 0, b + c = 0 , – 2c = 1

∴ a = 0, c = – 1/2, b = 1/2

From equation (i) T = Km01/2g–1/2

or

1/2

TK K

g g

  = =    

 

The value of K, as found by experiment or mathematical

investigation, comes out to be 2π.

∴ T 2

g = π 

Example 9: When a solid sphere moves through a liquid,

the liquid opposes the motion with a force F. The magnitude

of F depends on the coefficient of viscosity η of the liquid,

the radius r of the sphere and the speed v of the sphere.

Assuming that F is proportional to different powers of

these quantities, guess a formula for F using the method of

dimensions.

Sol. Suppose the formula is F = k ηa rb vc

Then, MLT–2 = [ML–1 T–1]

a Lb      

c L

T

= Ma L–a + b + c T –a – c

Equating the exponents of M, L and T from both sides,

a = 1

– a + b + c = 1

– a – c = –2

Solving these, a = 1, b = 1 and c = 1

Thus, the formula for F is F = kηrv.

Example 10: If P is the pressure of a gas and ρ is its density,

then find the dimension of velocity in terms of P and ρ.

(a) P1/2ρ–1/2 (b) P1/2ρ1/2

(c) P–1/2ρ1/2 (d) P–1/2ρ–1/2

Sol. (a) Method - I

[P] = [ML–1T–2] ...(i)

[ρ] = [ML–3] ...(ii)

Dividing eq. (i) by (ii)

[Pρ–1] = [L2T–2]

⇒ [LT–1] = [P1/2ρ–1/2]

⇒ [v] = [P1/2ρ–1/2]

Method - II

v ∝ Pa ρb

v = kPa ρb

[LT–1] = [ML–1T–2]a [ML–3]b

⇒ a = 1

2

, b = – 1

2

(Equating dimensions)

⇒ [v] = [P1/2ρ–1/2]

Example 11: Find relationship between speed of sound in

a medium (v), the elastic constant (E) and the density of the

medium (ρ).

Sol. Let the speed depends upon elastic constant and density

according to the relation

v ∝ Ea ρb

or v = KEaρb ...(i)

Where K is a dimensionless constant of proportionality.

Considering dimensions of the quantities

[v] = M0 L T–1

[E] = 11 2 2 11 2

1 1

[stress] [force] / [area] [M L T ] / [L ] [M L T ] [strain] [ ] / [ ] [L ]/ [L ]

− − − = = = ∆ 

∴ [Ea] = [Ma L–a T–2a]

[ρ] = [mass]/[volume] = [M]/[L3] = [M1L–3T0]

∴ [ρb] = [Mb L–3b T0]

Equating the dimensions of the LHS and RHS quantities

of equation (i), we get

[M0 L1 T–1] = [Ma L–a T–2a] ≠ [Mb L–3b T0]

or [M0 L1 T–1] = [Ma+b L–a–3b T–2a]

Comparing the individual dimensions of M, L and T

a + b = 0, ...(ii)

– a – 3b = 1, and ...(iii)

– 2a = – 1 ...(iv)

Solving we get

1 1 ,

2 2

a b = = −

Therefore the required relation is

E v K= ρ .

Example 12: Pressure (P) acting due to a fluid kept in a

container depends on, weight of liquid (w), Area of crosssection of container (A) and density of fluid (ρ). Establish a

formula of pressure (P).

Sol. According to the question

P ∝ wx

Ay

ρz

P = K[wx Ay ρz

]

Now writing the dimension of each quantity on either

side.

ML–1T–2 = K [MLT–2]x [L2]y [ML–3]z

ML–1T–2 = K [M]x+z [L]x+2y–3z [T]–2x

Now comparing the powers

For M, 1 = x + z ...(i)

L, –1 = x + 2y – 3z ...(ii)

T, –2 = –2x ...(iii)

⇒ Solving we get

⸫ 1 0 P KwA− = ρ

6 JEE (XI) Module-1 P

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Concept Application

7. If c is the velocity of light, h is Planck's constant and

G is Gravitational constant are taken as fundamental

quantities, then the dimensional formula of mass is

8. Taking frequency f, velocity (v) and Density (ρ) to

be the fundamental quantities then the dimensional

formula for momentum will be

(a) (ρv4f –3) (b) (ρv3f

–1)

(c) (ρvf 2 ) (d) (ρ2v2f 2)

9. If momentum (P), mass (M) and time (T) are chosen

as fundamental quantities the dimensional formula for

length is ________.

(a) (P1T1M1) (b) (P1T1M2)

(c) (P1T1M–1) (d) (P2T2M1)

10. For the equation F = Aavbdc

where F is force, A is area,

v is velocity and d is density, with the dimensional

analysis gives the following values for exponents.

(a) a = 1, b = 2, c = 1

(b) a = 2, b = 1, c = 1

(c) a = 1, b = 1, c = 2

(d) a = 0, b = 1, c = 1

To Convert Units of a Physical Quantity from

One System of Units to Another

It is based on the fact that,

Numerical value × unit = constant

So on changing unit, numerical value will also gets changed. If

n1 and n2 are the numerical values of a given physical quantity

and u1 and u2 be the units respectively in two different systems

of units, then

n1u1 = n2u2

1 11

2 1

2 22

a bc

M LT

n n

M LT

    =        

LIMITATIONS OF

DIMENSIONAL ANALYSIS

(i) It supplies no information about dimensionless constants

and the nature (vector and scalar) of physical quantities.

(ii) This method fails to derive the exact form of a physical

relation, if a physical quantity depends upon more than

three other mechanical physical quantities.

(iii) This method is applicable only if relation is of product

type. It fails in the case of exponential and trigonometric

relations.

(iv) It does not predict numerical correctness of formula.

Train Your Brain

Example 13: Convert 1 newton (SI unit of force) into dyne

(CGS unit of force)

Sol. The dimensional equation of force is [F] = [M1 L1T–2]

Therefore if n1, u1 and n2, u2 corresponds to SI and

CGS unit respectively, then

11 2

1 11

2 1

2 22

MLT

MLT

n n

−     =        

= 1 ( ) ( ) ( )

2 kg m s 5 1 1000 100 1 10

g cm s

−     = × =        

Example 14: A calorie is a unit of heat or energy and it

equals about 4.2 J, where 1 J = 1 kg m2/s2. Suppose we

employ a system of units in which the unit of mass equals

α kg, the unit of length equals β metre, the unit of time is γ

second. Show that a calorie has a magnitude 4.2 α–1β–2γ2 in

terms of the new units.

Sol. 1 cal = 4.2 kg m2s–2

SI New system

n1 = 4.2 n2 = ?

M1 = 1 kg M2 = α kg

L1 = 1 m L2 = β metre

T1 = 1 s T2 = γ second

Dimensional formula of energy is [ML2T–2]

Comparing with [MaLbTc

],

We find that a = 1, b = 2, c = –2

Now, 1 11

2 1

2 22

MLT

MLT

a bc

n n

    =        

=

122 1 kg 1 m 1 s 1 22 4.2 4.2

kg m s

−

− −      = αβγ      αβγ     

Example 15: Young's modulus of steel is 19 × 1010 N/m2.

Express it in dyne/cm2. Here dyne is the CGS unit of force.

Sol. The unit of Young's modulus is N/m2.

This suggest that it has dimensions of 2

Force

(Distance) .

Thus, [Y] = 2

[ ] F

L =

-2

2

MLT

L

= ML–1T–2.

N/m2 is in SI units, so, 1 N/m2 = (1 kg)(1m)–1 (1s)–2

and 1 dyne/cm2 = (1g)(1cm)–1 (1s)–2 so,

2

2

1N/m

1dyne / cm

= 1kg

1g

     

1 1m

1cm

−      

–2

1s

1s

     

= 1000 × 1

100

× 1 = 10

or, 1 N/m2 = 10 dyne/cm2

or, 19 × 1010 N/m2 = 19 × 1011 dyne/cm2.

Units and Measurements 7 P

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Example 16: The dimensional formula for viscosity of

fluids is η = M1L–1T–1. Find how many poise (CGS unit

of viscosity) is equal to 1 poiseuille (SI unit of viscosity).

Sol. η = M1 L–1 T–1

1 CGS units = g cm–1 s–1

1 SI units = kg m–1 s–1

= (1000 g)(100 cm)–1 s–1

= 10 g cm–1 s–1

Thus, 1 Poiseuilli = 10 poise

Concept Application

11. If minute is the unit of time, 10 ms–2 is the unit of

acceleration and 100 kg is the unit of mass, the new

unit of work in joule is

(a) 105 (b) 106

(c) 6 × 106 (d) 36 × 106

12. The magnitude of force is 100 N. What will be its

value if the units of mass and time are doubled and

that of length is halved?

(a) 25 (b) 100

(c) 200 (d) 400

13. The value of universal gravitational constant G in

CGS system is 6.67 × 10–8 dyne cm2 g–2. Its value in

SI system is

(a) 6.67 × 10–11Nm2 kg–2

(b) 6.67 × 10–5 Nm2 kg–2

(c) 6.67 × 10–l0 Nm2 kg–2

(d) 6.67 × 10–9 Nm2 kg–2

14. Which equation cannot be derived using dimensional

analysis from the following? (k is a dimensionless

constant and x does not necessarily represent distance.

Remaining variables follow standard meaning)

(a) x = 2at2 (b) x = ksin(wt)

(c)

2 r g x ρ = η (d) All of these

MEASUREMENT OF LENGTH

You are already familiar with some direct methods for the

measurement of length. For example, a metre scale is used

for lengths from 10–3 m to 102 m. A vernier callipers is used

for lengths to an accuracy of 10–4 m. A screw gauge and a

spherometer can be used to measure lengths as less as 10–5 m.

To measure lengths beyond these ranges, we make use of some

special indirect methods.

Range of Lengths

The sizes of the objects we come across in the universe vary over

a very wide range. These may vary from the size of the order of

10–15 m of the proton to the size of the order of 1026 m of the extent

of the observable universe.

We also use certain special length units for short and large

lengths. These are

1 fermi = 1 f = 10–15 m

1 angstrom = 1 Å = 10–10 m (It is used mainly in measuring

wavelength of light)

1 astronomical unit = 1 AU (average distance of the Sun from

the Earth) = 1.496 × 1011 m

1 light year = 1 ly = 9.46 × 1015 m (distance that light travels

with velocity of 3 × 108 m s–1 in 1 year)

1 parsec = 3.08 × 1016 m (Parsec is the distance at which

average radius of earth’s orbit subtends an angle of 1 arc second)

MEASUREMENT OF LARGE DISTANCES

Parallax Method

Large distances such as the distance of a planet or a star from

the earth cannot be measured directly with a metre scale. An

important method in such cases is the parallax method. When you

hold a pencil in front of you against some specific point on the

background (a wall) and look at the pencil first through your left

eye A (closing the right eye) and then look at the pencil through

your right eye B (closing the left eye), you would notice that the

position of the pencil seems to change with respect to the point on

the wall. This is called parallax.

The distance between the two points of observation is called

the basis. In this example, the basis is the distance between the

eyes. To measure the distance D of a far away planet S by the

parallax method, we observe it from two different positions

(observatories) A and B on the Earth, separated by distance AB

= b at the same time as shown in figure. We measure the angle

between the two directions along which the planet is viewed at

these two points. The ∆ASB in figure represented by symbol θ is

called the parallax angle or parallactic angle.

As the planet is very far away, 1 b

D

<< and therefore, θ is

very small. Then we approximately take AB as an arc of length

b of a circle with centre at S and the distance D as the radius

AS = BS so that AB = b = D θ where θ is in radians.

b D = θ ...(i)

d

D

N

A

M

D

a

S

D

b A B

D

q

8 JEE (XI) Module-1 P

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Having determined D, we can employ a similar method to

determine the size or angular diameter of the planet. If d is the

diameter of the planet and α the angular size of the planet (the

angle subtended by d at the earth), we have

α = d/D ....(ii)

The angle α can be measured from the same location on

the earth. It is the angle between the two directions when two

diametrically opposite points of the planet are viewed through the

telescope. Since D is known, the diameter d of the planet can be

determined using equation (ii).

ESTIMATION OF VERY SMALL DISTANCES

Size of a Molecule

To measure a very small size, like that of a molecule (10–8 m

to 10–10 m), we have to adopt special methods. We cannot use a

screw gauge or similar instruments. Even a microscope has certain

limitations. An optical microscope uses visible light to ‘look’ at

the system under investigation. As light has wave like features,

the resolution to which an optical microscope can be used is the

wavelength of light.

For visible light the range of wavelengths is from about 4000

Å to 7000 Å (1 angstrom = 1 Å = 10–10 m). Hence an optical

microscope cannot resolve particles with sizes smaller than this.

Instead of visible light, we can use an electron beam. Electron

beams can be focused by properly designed electric and magnetic

fields. The resolution of such an electron microscope is limited

finally by the fact that electrons can also behave as waves.

The wavelength of an electron can be as small as a fraction of

an angstrom. Such electron microscopes with a resolution of 0.6

Å have been built. They can almost resolve atoms and molecules

in a material. In recent times, tunneling microscopy has been

developed in which again the limit of resolution is better than an

angstrom. It is possible to estimate the sizes of molecules.

ORDER OF MAGNITUDE

If a number P can be expressed as

P = A × 10x

where 0.5 ≤ A < 5, then x is called order of magnitude of the

number.

SI Prefixes: The magnitudes of physical quantities vary over a

wide range. The mass of an electron is 9.1 × 10–31 kg and that of

our earth is about 6 × 1024 kg. Standard prefixes for certain power

of 10. Table shows these prefixes

Power of 10 Prefix Symbol

12 tera T

9 giga G

6 mega M

3 kilo k

2 hecto h

1 deka da

Power of 10 Prefix Symbol

–1 deci d

–2 centi c

–3 milli m

–6 micro µ

–9 nano n

–12 pico p

–15 femto f

Train Your Brain

Example 17: The Sun’s angular diameter is measured to be

1920''. The distance D of the Sun from the Earth is 1.496 ×

1011 m. What is the diameter of the Sun?

Sol. Sun’s angular diameter a.

= 1920\"

= 1920 × 4.85 × 10–6 rad

= 9.31 × 10–3 rad

Sun’s diameter d = αD

= (9.31 × 10–3) × (1.496 × 1011) m

= 1.39 × 109 m

Example 18: The moon is observed from two diametrically

opposite points on the earth with angle subtended = 1°54'

given diameter of earth = 1.276 × 107 m. Find the distance

of moon from the earth.

Sol. From parallax method,

θ = b

D

We have, θ = 1°54' = 60' + 54' = 114'

Also 1' = 2.91 × 10–4 rad

⸫ θ = 114 × 2.91 × 10–4 = 0.033 rad

⸫ D =

7 1.2760 10 8 3.8 10 m

0.033

× = ×

Example 19: If the size of a nucleus (in the range of

10–15 to 10–11m) is scaled up to the tip of a sharp pin, what

roughly is the size of an atom? Assume tip of the pin to be

in the range 10–5 m to 10–4 m.)

Sol. The size of a nucleus is in the range of 10–15 m and

10–14 m. The tip of a sharp pin is taken to be in the

range of 10–5 m and 10–4 m.

Thus we are scaling up by a factor of 1m. An atom

roughly of size 10–10 m will be scaled up to a size of

1 m. Thus a nucleus in an atom is as small in size as

the tip of a sharp pin placed at the centre of a sphere of

radius about a metre long.

Units and Measurements 9 P

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Concept Application

15. Considering the distance between sun and moon to

be 15 × 1010 m, the angular diameter of the sun as

observed from the moon is close to (Take the radius

of sun to be 7 × 108 m.)

(a) 214 rad (b) 107 rad

(c) 0.005 rad (d) 0.009 rad

16. The mass of an object is 75.2 × 104 kg. The order of

magnitude of the mass is

(a) 4 (b) 5

(c) 6 (d) 7

ERROR ANALYSIS IN EXPERIMENTS

Significant Figures or Digits

The significant figures (SF) in a measurement are the figures or

digits that are known with certainty plus one that is uncertain.

Significant figures in a 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 its accuracy and vice versa.

1. Rules to find out the number of significant figures:

I Rule:All the non-zero digits are significant E.g. 1984 has

4 SF.

II Rule: All the zeros between two non-zero digits are

significant. E.g. 10806 has 5 SF.

III Rule: All the zeros to the left of first non-zero digit are

not significant. E.g.00108 has 3 SF.

IV Rule: If the number is less than 1, zeros on the right of the

decimal point but to the left of the first non-zero digit are not

significant. E.g. 0.002308 has 4 SF.

V Rule: The trailing zeros (zeros to the right of the last nonzero digit) in a number with a decimal point are significant.

E.g. 01.080 has 4 SF.

VI Rule: The trailing zeros in a number without a decimal

point are not significant e.g. 010100 has 3 SF. But if the

number comes from some actual measurement then the

trailing zeros become significant. E.g. m = 100 kg has 3 SF.

VII Rule: When the number is expressed in exponential

form, the exponential term does not affect the number

of S.F. For example in x = 12.3 = 1.23 × 101 = .123

× 102 = 0.0123 × 103 = 123 × 10–1, each term has

3 SF only. (Note: It has 3 significant figure in each expression.)

2. Rules for arithmetical operations with significant figures:

I Rule: In addition or subtraction the number of decimal

places in the result should be equal to the number of decimal

places of that term in the operation which contain lesser

number of decimal places. E.g. 12.587 – 12.5 = 0.087 = 0.1

( second term contain lesser i.e. one decimal place)

II Rule: In multiplication or division, the number of SF in the

product or quotient is same as the smallest number of SF in

any of the factors. E.g. 5.0 × 0.125 = 0.625 = 0.62

To avoid the confusion regarding the trailing zeros of the

numbers without the decimal point the best way is to report

every measurement in scientific notation (in the power of

10). In this notation every number is expressed in the form

a × 10b , where a is the base number between 1 and 10 and b

is any positive or negative exponent of 10. The base number

a is written in decimal form with the decimal after the first

digit. While counting the number of SF only base number is

considered (Rule VII).

Note: The change in the unit of measurement of a quantity

does not affect the number of SF. For example in 2.308 cm =

23.08 mm = 0.02308 m = 23080 µm each term has 4 SF.

ROUNDING OFF

To represent the result of any computation containing more than

one uncertain digit, it is rounded off to appropriate number of

significant figures.

Rules for rounding off the numbers:

I Rule : If the digit to be rounded off is more than 5,

then the preceding digit is increased by one.

e.g. 6.87≈ 6.9

II Rule : If the digit to be rounded off is less than 5, than the

preceding digit is unaffected and is left unchanged. e.g.

3.94 ≈ 3.9

III Rule : If the digit to be rounded off is 5 then the preceding digit

is increased by one if it is odd and is left unchanged if

it is even. e.g. 14.35 ≈ 14.4 and 14.45 ≈ 14.4

Train Your Brain

Example 20: Write down the number of significant figures

in the following.

(i) 165 (ii) 2.05

(iii) 34.000 m (iv) 0.005

(v) 0.02340 N m–1 (vi) 26900

(vii) 26900 kg

Sol. (i) 3 SF (following rule I)

(ii) 3 SF (following rules I and II)

(iii) 5 SF (following rules I and V)

(iv) 1 SF (following rules I and IV)

(v) 4 SF (following rules I, IV and V)

(vi) 3 SF (see rule VI)

(vii) 5 SF (see rule VI)

10 JEE (XI) Module-1 P

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Example 21: The length, breadth and thickness of a metal

sheet are 4.234 m, 1.005 m and 2.01 cm respectively. Give

the area and volume of the sheet to correct number of

significant figures.

Sol. Length () = 4.234 m, Breadth (b) = 1.005 m

Thickness (t) = 2.01 cm = 2.01 × 10–2 m

Therefore, area of the sheet = 2( × b + b × t + t × )

= 2(4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)m2

= 2(4.3604739) m2 = 8.720978 m2

Since area can contain a maximum of 3 SF (Rule II of

arithmetic operations) therefore, rounding off, we get

Area = 8.72 m2

Like wise volume =  × b × t

= 4.234 × 1.005 × 0.0201 m3 = 0.0855289 m3

Since volume can contain 3 SF, therefore, rounding off,

Volume = 0.0855 m3

Example 22: Find the number of significant figures in each

(i) 36.72 (ii) 0.003303

Sol. (i) It has four significant figures. All non-zero digits

are significant.

(ii) Here first 3 zeros are insignificant but zeros

between 3 are significant. So it has four

significant figures.

Example 23: Round off following values to four significant

figures.

(i) 36.879 (ii) 1.0084

(iii) 11.115 (iv) 11.1250

(v) 11.1251

Sol. The following values can be rounded off to four

significant figures as follows:

(i) 36.879 ≈36.88 ( 9 > 5 ∴7 is increased by one

i.e. I Rule)

(ii) 1.0084 ≈1.008 ( 4 < 5 ∴8 is left unchanged i.e.

II Rule)

(iii) 11.115 ≈11.12 ( last 1 is odd it is increased by

one i.e. III Rule)

(iv) 11.1250 ≈11.12 ( 2 is even it is left unchanged

i.e. III Rule)

(v) 11.1251 ≈11.13 ( 51 > 50 ∴ 2 is increased by

one i.e. I Rule)

Concept Application

17. The number of significant figures in 0.0006032 is

(a) 7 (b) 4 (c) 5 (d) 2

18. The radius of disc is 1.2 cm, its area according to idea

of significant figures is _______

(a) 4.5216 cm2 (b) 4.521 cm2

(c) 4.52 cm2 (d) 4.5 cm2

19. The number of significant figures in 5.69 × 1015 kg is

(a) 1 (b) 2

(c) 3 (d) 4

20. When 57.986 is rounded off to 4 significant figures,

then it becomes ______.

(a) 58 (b) 57

(c) 57.90 (d) 57.99

ERRORS IN MEASUREMENT

The difference between the true value and the measured value of

a quantity is known as the error of measurement.

Classification of Errors

Errors may arise from different sources and are usually classified

as follows:

Systematic or controllable errors: Systematic errors are

the errors whose causes are known. They can be either positive

or negative. Systematic errors can further be classified into three

categories:

(i) Instrumental errors: These errors are due to imperfect

design or erroneous manufacture or misuse of the measuring

instrument. These can be reduced by using more accurate

instruments.

(ii) Environmental errors: These errors are due to the changes

in external environmental conditions such as temperature,

pressure, humidity, dust, vibrations or magnetic and

electrostatic fields.

(iii) Observational errors: These errors arise due to

improper setting of the apparatus or carelessness in taking

observations. This can be reduced by proper setting of the

instrument before we start using it.

Random errors: These errors are due to unknown causes.

Therefore they occur irregularly and are variable in magnitude

and sign. Since the causes of these errors are not known precisely

they can not be eliminated completely. For example, when the

same person repeats the same observation in the same conditions,

he may get different readings different times.

Random errors can be reduced by repeating the observation

a large number of times and taking the arithmetic mean of all the

observations. This mean value would be very close to the most

accurate reading.

Note: If the number of observations is made n times then the

random error reduces to 1

n

      times. E.g. If the random error in

the arithmetic mean of 100 observations is 'x' then the random

error in the arithmetic mean of 500 observations will be

5

x

Units and Measurements 11 P

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For example:

(i) Reading instrument without proper initial settings.

(ii) Taking the observations wrongly without taking necessary

precautions.

(iii) Exhibiting mistakes in recording the observations.

(iv) Putting improper values of the observations in calculations.

These errors can be minimised by increasing the sincerity

and alertness of the observer.

REPRESENTATION OF ERRORS

Errors can be expressed in the following ways:

1. Mean Absolute Error: The mean of measurements of a

physical quantity (a1, a2, .... an) is expressed as

m

1

...... n n i

i

aa a a

a

n n =

++ + = = ∑ 1 2

am is taken as the true value of a quantity if the same is not

known.

The absolute error in the measurements are expressed as

∆a1 = am – a1

∆a2 = am – a2

.....................

∆an = am – an

Mean absolute error is defined as

1 2

1

... n n i

i

aa a a

a

n n =

∆ +∆ + +∆ ∆ ∆ = = ∑

Final result of measurement may be written as:

a = am ± ∆a

2. Relative Error or Fractional Error: It is given by

Mean absolute Error

Mean value of measurement m

a

a

∆ =

3. Percentage Error 100%

m

a

a

∆ = ×

Train Your Brain

Example 24: The period of oscillation of a simple pendulum

in an experiment is recorded as 2.63 s, 2.56 s, 2.42 s,

2.71 s and 2.80 s respectively. Find (i) mean time period

(ii) absolute error in each observation and percentage error.

Sol. (i) Mean time period is given by

2.63 2.56 2.42 2.71 2.80

5

T ++++ =

13.12 2.62

5 = = s

(ii) The absolute error in each observation is

2.62 – 2.63 = –0.01, 2.62 – 2.56

= 0.06, 2.62 – 2.42

= 0.20, 2.62 – 2.71

= –0.09, 2.62 – 2.80

= –0.18

Mean absolute error, | |

5

T T Σ ∆ ∆ =

0.01 0.06 0.2 0.09 0.18 0.11sec

5

+ ++ + = =

∴ Percentage error is

0.11 100 100 4.2%

2.62

T

T

∆ =×= ×=

Example 25: In an experiment the values of refractive

indices of glass material recorded as 1.56, 1.53, 1.54, 1.45,

1.44 and 1.43 in repeated measurements. Find

(i) Mean value of μ (ii) Mean absolute error

(iii) Relative error (iv) Percentage error

Sol. (i) Mean value of μ

1 2 ..... n

n

µ +µ + +µ µ =

1.56 1.54 1.53 1.45 1.44 1.43 1.491 1.50

6

+++++

µ = = ≈

(ii) Absolute error Ɠ

1 1 ∆µ = µ −µ = 0.06

2 2 ∆µ = µ −µ = 0.04

3 3 ∆µ = µ −µ = 0.03

4 4 ∆µ = µ −µ = 0.05

5 5 ∆µ = µ −µ = 0.06

6 6 ∆µ = µ −µ = 0.07

⸫ mean absolute error

1 2 6 ............

6

∆µ + ∆µ + ∆µ ∆µ =

0.06 0.04 0.03 0.05 0.06 0.07

6

 

= 0.051

⸫ Reading = µ+ ∆µ = 1.50 ± 0.05

(iii) Relative error =

0.05 0.033

1.50

 

(iv) Relative percentage error = Relative error × 100%

= ± 0.033 × 100%

= ± 3.3% (approx)

12 JEE (XI) Module-1 P

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Concept Application

21. The diameter of a wire as measured by a screw

gauge was found to be, 1.004 cm, 1.000 cm. Find

the absolute error in first reading.

22. The diameter of a thick wire measured by a screwgauge.

D1 = 1.006 cm, D2 = 1.004 cm, D3 = 1.002 cm

Find mean absolute error and write the reading.

COMBINATION OF ERRORS

(i) In Sum: If Z = A + B, then ∆Z = ∆A + ∆B.

Maximum fractional error in this case is

ZA B

Z AB AB

∆∆ ∆

= +

+ +

i.e. when two physical quantities are added then the

maximum absolute error in the result is the sum of the

absolute errors of the individual quantities.

(ii) In Difference: If Z = A – B, then maximum absolute error

is ∆Z = ∆A + ∆B and maximum fractional error in this case

ZA B

Z AB AB

∆∆ ∆

= + − −

(iii) In Product: If Z = AB, then the maximum fractional error,

Z AB

Z AB

∆ ∆∆

= +

where ∆Z/Z is known as fractional error.

(iv) In Division: If Z = A/B, then maximum fractional error is

Z AB

Z AB

∆ ∆∆

= +

(v) In Power: If Z = An then Z A

n

Z A

∆ ∆ =

In more general form if

x y

q

A B Z

C =

then the maximum fractional error in Z is

Z ABC

xyq Z ABC

∆ ∆∆∆

=++

Applications:

1. For a simple pendulum, T ∝ l

1/2

⇒

1

2

T l

T l

∆ ∆ =

2. For a sphere, area and volume are given as

4 4 , 3

A rV r =π = π 2 3

So, relative errors in them are given as

2 A r

A r

∆ ∆ = and 3 V r

V r

∆ ∆ =

3. When two resistors R1 and R2 are connected

(i) In series Rs

= R1 + R2 ⇒ ∆Rs

= ∆R1 + ∆R2

⇒ s

s

R R R

R RR

∆ ∆ +∆ = +

1 2

1 2

(ii) In parallel,

1 2

1 11

R RR P

= + ⇒ 222

1 2

p

p

R R R

RRR

∆ ∆ ∆

= + 1 2

Train Your Brain

Example 26: In an experiment of simple pendulum, the

errors in the measurement of length of the pendulum (L) and

time period (T) are 3% and 2% respectively. The maximum

percentage error in the value of L/T2 is

(a) 5% (b) 7% (c) 8% (d) 1%

Sol. (c) Maximum percentage in the value of L/T2

100% 2 100% L T

L T

∆ ∆

=× + ×

= 3 + 2 × 2 = 7%

Example 27: If X =

2 A B

C , then

(a) ∆X = ∆A + ∆B + ∆C

(b) X ABC 2

X A BC

∆ ∆∆∆

= ++

(c) 2

2

X ABC

X A BC

∆ ∆∆∆

= ++

(d) X ABC

X ABC

∆ ∆∆∆

=++

Sol. (c)  X = A2B1/2C

∴

2

2

X ABC

X A BC

∆ ∆∆∆

= ++

Example 28: A body travels uniformly a distance

(13.8 ± 0.2) m in a time (4.0 ± 0.3) s. Calculate its velocity

with error limits. What is the percentage error in velocity?

Sol. Given distance, s = (13.8 ± 0.2) m

and time t = (4.0 ± 0.3) s

Velocity v = 13.8 1 1 3.45m s 3.5m s

4.0

s

t

− − = = =

0.2 0.3

13.8 4.0

v st

v st

∆ ∆∆    =± + =± +      

0.8 4.14

13.8 4.0

  +

= ±    ×

∴ ∆v = ± 0.0895 × v = ± 0.0895 × 3.45 = ± 0.3087

= ± 0.31

Hence v = (3.5 ± 0.31) ms –1

Percentage error in velocity

= 100 v

v

∆ × = ±0.0895 × 100 = ± 8.95% = ±9%

Units and Measurements 13 P

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Example 29: The heat generated in a circuit is given by

Q = I 2 Rt, where I is current, R is resistance and t is time.

If the percentage errors in measuring I, R and t are 2%, 1%

and 1% respectively, then the maximum error in measuring

heat will be

(a) 2 % (b) 3 % (c) 4 % (d) 6 %

Sol. (d) The percentage error in heat is given as

%2 % % % Q IRt

Q I Rt

∆ ∆∆ ∆

= ++

Substituting the values, maximum possible

percentage error is

= 2 × 2% + 1 × 1% + 1 × 1% = 6%

Example 30: Given: Resistance, R1 = (8 ± 0.4) Ω and

Resistance, R2 = (8 ± 0.6) Ω. What is the net resistance

when R1 and R2 are connected in series?

(a) (16 ± 0.4) Ω (b) (16 ± 0.6) Ω

(c) (16 ± 1.0) Ω (d) (16 ± 0.2) Ω

Sol. (c) R1 = (8 ± 0.4)Ω, R2 = (8 ± 0.6) Ω

Rs

= R1 + R2 = (16 ± 1.0)Ω

Example 31: The following observations were taken for

determining surface tension of water by capillary tube

method: Diameter of capillary, D = 1.25 × 10–2 m and rise

of water in capillary, h = 1.45 × 10–2 m.

Taking g = 9.80 ms–2 and using the relation

T = (rgh/2) × 103 Nm–1, what is the possible error in surface

tension T?

(a) 2.4 % (b) 15 %

(c) 1.6 % (d) 0.15 %

Sol.(c) Given T = (rgh/2) × 103 Nm–1,

D = 1.25 × 10–2 m, h = 1.45 × 10–2 m,

g = 9.80 ms–2

T rhg

T rhg

δ δδδ

=++

after applying the above values in this relation

we get δT % = 1.6%

Concept Application

23. If the length of cylinder is measured to be 4.28 cm

with an error of 0.01 cm. The percentage error in the

measured length is approximately.

(a) 0.4% (b) 0.5% (c) 0.2 % (d) 0.1 %

24. The pressure on square plate is measured by measuring

the force on the plate and the length of the sides of

the plate. If the maximum error in measurement of

force and length are respectively 4% and 2% then the

maximum error in measurement of pressure is

(a) 1% (b) 2% (c) 6 % (d) 8%

25. The length and breadth of rectangular object are

25.2 cm and 16.8 cm respectively and have been

measured to an accuracy of 0.1 cm. Relative error and

percentage error in the area of the object are

(a) 0.01 and 1% (b) 0.02 and 2%

(c) 0.03 and 3% (d) 0.04 and 4%

26. The error in the measurement of length of a simple

pendulum is 0.1% and error in the time period is 2%.

The possible maximum error in the quantity having

dimensional formula LT–2 is

(a) 1.1% (b) 2.1%

(c) 4.1% (d) 6.1%

MEASURING INSTRUMENTS

Measurement is an important aspect of physics. Whenever we

want to know about a physical quantity, we take its measurement

first of all. Instruments used in measurement are called measuring

instruments.

Least Count: The least value of a quantity, which the

instrument can measure accurately, is called the least count of the

instrument.

Error: The measured value of the physical quantity is usually

different from its true value. The result of every measurement

by any measuring instrument is an approximate number, which

contains some uncertainty. This uncertainty is called error. Every

calculated quantity, which is based on measured values, has an

error.

Accuracy and Precision: The accuracy of a measurement

is a measure of how close the measured value is to the true value

of the quantity. Precision tells us to what resolution or limit the

quantity is measured.

VERNIER CALLIPER

It is a device used to measure accurately upto 0.1 mm. There are

two scales in the vernier calliper, vernier scale and main scale. The

main scale is fixed whereas the vernier scale is movable along the

main scale.

Its main parts are as follows:

Main scale: It consists of a steel metallic strip M, graduated in cm

and mm at one edge and in inches and tenth of an inch at the other

edge on same side. It carries fixed jaws A and C projected at right

angle to the scale as shown in figure.

3 5 6 7 8 9 10 E 0

C

V

S

P

A B

M

D

1

Main Scale

14 JEE (XI) Module-1 P

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Vernier Scale: A vernier V slides on the strip M. It can be

fixed in any position by screw S. It is graduated on both sides. The

side of the vernier scale which slides over the mm side has ten

divisions over a length of 9 mm, i.e., over 9 main scale divisions

and the side of the vernier scale which slides over the inches side

has 10 divisions over a length of 0.9 inch, i.e., over 9 main scale

divisions.

Movable Jaws: The vernier scale carries jaws B and

D projecting at right angle to the main scale. These are called

movable jaws. When vernier scale is pushed towards A and C,

then as B touches A, straight side of D will touch straight side of

C. In this position, in case of an instrument free from errors, zeros

of vernier scale will coincide with zeros of main scales, on both

the cm and inch scales.

The object whose length or external diameter is to be measured

is held between the jaws A and B, while the straight edges of C and

D are used for measuring the internal diameter of a hollow object.

Metallic Strip: There is a thin metallic strip E attached to the

back side of M and connected with vernier scale. When the jaws

A and B touch each other, the edge of strip E touches the edge of

M. When the jaws A and B are separated, E moves outwards. The

strip E is used for measuring the depth of a vessel.

Determination of Least Count (Vernier Constant)

Note the value of the main scale division and count the number n

of vernier scale divisions. Slide the movable jaw till the zero of

vernier scale coincides with any of the mark of the main scale and

find the number of divisions (n – 1) on the main scale coinciding

with n divisions of vernier scale. Then

n V.S.D. = (n – 1) M.S.D. or 1 V.S.D. = n –1

n

      M.S.D.

1 V.C. = L.C. = 1 M.S.D. – 1 V.S.D. = –1 1– n

n

      M.S.D.

= 1

n

M.S.D.

Determination of Zero Error and Zero Correction

For this purpose, movable jaw B is brought in contact with fixed

jaw A.

One of the following situations will arise.

(i) Zero of Vernier scale coincides with zero of main scale

0.5 M

0 5 V 10

0 1

In this case, zero error and zero correction, both are nil.

Actual length = observed (measured) length.

(ii) Zero of vernier scale lies on the right of zero of main scale

0.5 M

0 5 V 10

0 1

Here 5th vernier scale division is coinciding with any main

sale division and zero of vernier is ahead of Nth main scale

division.

Hence, N = 0, n = 5, L.C. = 0.01 cm.

Zero error = N + n × (L.C.) = 0 + 5 × 0.01 = + 0.05 cm

Zero correction = – 0.05 cm.

Actual length will be 0.05 cm less than the observed

(measured) length.

(iii) Zero of the vernier scale lies left of the main scale.

0.5 M

0 5 V 10

0 1

Here, 5th vernier scale division is coinciding with any main

scale division.

In this case, zero of vernier scale lies on the right of –0.1

cm reading on main scale.

Hence, N = – 0.1 cm, n = 5, L.C. = 0.01 cm

Zero error = N + n × (L.C.)

= – 0.1 + 5 × 0.01 = –0.05 cm.

Zero correction = +0.05 cm.

Actual length will be 0.05 cm more than the observed

(measured) length.

Measurement Using Vernier Calliper

Let us measure the diameter of a small spherical/cylindrical body

using a vernier calliper. Insert the object between the jaws as

shown in the figure below.

3 5 6 7 8 9 10 E 0

C

V

S

P

A B

M

D

1

Main Scale

SPHERE

If with the body between the jaws, the zero of vernier scale

lies ahead of Nth division of main scale, then main scale reading

(M.S.R.) = N.

If nth division of vernier scale coincides with any division of

main scale, then vernier scale reading (V.S.R.)

= n × (L.C.) (L.C. is least count of vernier calliper)

= n × (V.C.) (V.C. is vernier constant of vernier calliper)

Total reading,

T.R. = M.S.R. + V.S.R.

= N + n × (V.C.)

Units and Measurements 15 P

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Train Your Brain

Example 32: The least count of vernier callipers is 0.1 mm.

The main scale reading before the zero of the vernier scale

is 10 and the zeroth division of the vernier scale coincides

with the main scale division. Given that each main scale

division is 1 mm, what is the measured value?

Sol. Length measured with vernier callipers

= reading before the zero of vernier scale + number of

vernier divisions coinciding

with any main scale division × least count

= 10 mm + 0 × 0.1 mm = 10 mm = 1.00 cm

Example 33: A vernier callipers has its main scale of 10 cm

equally divided into 200 equal parts. Its vernier scale of 25

divisions coincides with 12 mm on the main scale. The least

count of the instrument is

(a) 0.020 cm (b) 0.002 cm

(c) 0.010 cm (d) 0.001 cm

Sol. (b) 10 cm divided in 200 divisions, so

1 division = 10

200

= 0.05 cm.

Now, 25V = 24S.

⇒ V = 24

25

S

LC = S – V

= S – 24

25

S = 1

25

S

 1S = 0.05 cm, thus

So, LC = 0.05

25

= 0.002 cm

Example 34: One centimetre on the main scale of vernier

callipers is divided into ten equal parts. If 10 divisions of

vernier scale coincide with 8 small divisions of the main

scale, the least count of the callipers is

(a) 0.005 cm (b) 0.05 cm

(c) 0.02 cm (d) 0.01 cm

Sol. (c) Now, 10V = 8S

⇒ V = 8

10

S.

Now, LC = S – V

= S – 8

10

S = 2

10

S = 1

5

S

But 1S = 0.1 cm, thus

LC = 0.1

5

= 0.02 cm

Concept Application

27. In an experiment the angles are required to be

measured using an instrument. 29 divisions of the

main scale exactly coincide with the 30 divisions of

the vernier scale. If the smallest division of the main

scale is half-a-degree (= 0.5°), find the least count of

the instrument (in min.).

28. The diameter of a cylinder is measured using a vernier

callipers with no zero error. It is found that, the zero

of the vernier scale lies between 5.10 cm and 5.15 cm

of the main scale. The vernier scale has 50 division

equivalent to 2.45 cm. If 24th division of the vernier

scale exactly coincides with one of the main scale

division, the diameter of the cylinder is

(a) 5.112 cm (b) 5.124 cm

(c) 5.136 cm (d) 5.148 cm

29. The main scale of vernier calliper is calibrated in mm

and 19 divisions of main scale are equal in length to 20

divisions of vernier scale. In measuring the diameter

of a cylinder by this instrument, the main scale reads

35 division and 4th division of vernier scale coincides

with a M.S.D. Find LC in (cm).

SCREW GAUGE

This instrument (shown in figure) works on the principle of

micrometer screw. It consists of a U-shaped frame M. At one end

of it is fixed a small metal piece A of gun metal. It is called stud

and it has a plane face. The other end N of M carries a cylindrical

hub H. The hub extends few millimetre beyond the end of the

frame. On the tubular hub along its axis, a line is drawn known as

reference line. On the reference line graduations are in millimetre

and half millimeter depending upon the pitch of the screw. This

scale is called linear scale or pitch scale. A nut is threaded through

the hub and the frame N. Through the nut moves a screw S made

of gun metal. The front face B of the screw, facing the plane face

A, is also plane. A hollow cylindrical cap K is capable of rotating

over the hub when screw is rotated. It is attached to the right hand

end of the screw. As the cap is rotated the screw either moves in

or out. The bevelled surface E of the cap K is divided into 50 or

100 equal parts. It is called the circular scale or head scale. Right

hand end R of K is milled for proper grip.

Ratchet

E

0

Sleeve

N H

K

A

B S

Frame

R

Stud

M

16 JEE (XI) Module-1 P

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In most of the instrument the milled head R is not fixed to the

screw head but turns it by a spring and ratchet arrangement such

that when the body is just held between faces A and B, the spring

yields and milled head R turns without moving in the screw.

In an accurately adjusted instrument when the faces A and B

are just touching each other, the zero marks of circular scale and

pitch scale exactly coincide.

Determination of Least Count of Screw Gauge

Note the value of linear (pitch) scale division. Rotate screw to

bring zero mark on circular (head) scale on reference line. Note

linear scale reading i.e. number of divisions of linear scale

uncovered by the cap.

Now give the screw a few known number of rotations. (one

rotation completed when zero of circular scale again arrives on

the reference line). Again note the linear scale reading. Find

difference of two readings on linear scale to find distance moved

by the screw.

Then, pitch of the screw

= Distance moved in rotation

No. of full rotation ( )

n

n

Now count the total number of divisions on circular (head)

scale. Then, least count is

LC = Pitch

Total number of divisions on thecircular scale

The least count is generally 0.001 cm.

Determination of Zero Error and Zero Correction

For this purpose, the screw is rotated forward till plane face B of

the screw just touches the fixed plane face A of the stud and edge

of cap comes on zero mark of linear scale. Screw gauge is held

keeping the linear scale vertical with its zero downwards.

One of the following three situations will arise.

(i) Zero mark of circular scale comes on the reference line

In this case, zero error and zero correction, both are nil.

Actual thickness = Observed (measured) thickness.

(ii) Zero mark of circular scale remains on right of reference

line and does not cross it.

Here 2nd division on circular scale comes on reference line.

Zero reading is already 0.02 mm. It makes zero error =

+ 0.02 mm and zero correction = – 0.02 mm.

Actual thickness will be 0.02 mm less than the observed

(measured) thickness.

Circular

Scale 0

Reference

line

5

H

N

(iii) Zero mark of circular scale goes to left on reference line

after crossing it. Here zero of circular scale has advanced

from reference line by 3 divisions on circular scale. A

backward rotation by 0.03 mm will make reading zero. It

makes zero error = –0.03 mm & zero correction = + 0.03

mm.

Circular

Scale 5

Reference

line

0

H

N

3

Actual thickness will be 0.03 mm more than the observed

(measured) thickness.

Measurement Using Screw Gauge

To measure diameter of a given wire using a screw gauge:

1. Determine of least count of screw gauge

2. If with the wire between plane faces A and B, the edge of the

cap lies ahead of Nth division of linear scale, then, linear scale

reading (L.S.R.) = N

If nth division of circular scale lies over reference line,

then, circular scale reading (C.S.R.) = n × (L.C.)

(L.C. is least count of screw gauge)

Total reading (T.R.) = L.S.R. + C.S.R. = N + n × (L.C.)

E

0

N H

K

A B S

R

M

wire

Train Your Brain

Example 35: In four complete revolutions of the cap, the

distance traveled on the pitch scale is 2 mm. If there are fifty

divisions on the circular scale, then

(i) Calculate the pitch of the screw gauge.

(ii) Calculate the least count of the screw gauge.

Units and Measurements 17 P

W

Sol. Pitch of screw = Linear distance traveled in one

revolution P = 2mm

4 = 0.5 mm = 0.05 cm

Least count =

Pitch

no.of divisions in circular scale

= 0.05

50

= 0.001 cm

Example 36: The pitch of a screw gauge is 0.5 mm and

there are 50 divisions on the circular scale. In measuring

the thickness of a metal plate, there are five divisions on

the pitch scale (or main scale) and thirty fourth division

coincides with the reference line. Calculate the thickness

of the metal plate.

Sol. Pitch of screw = 0.5 mm., LC = 0.5

50

= 0.01 mm.

Thickness = (5 × 0.5 + 34 × 0.01) mm

= (2.5 + 0.34) = 2.84 mm

Concept Application

30. In a screw gauge, five complete rotations of the screw

causes it to move a linear distance of 0.25 cm. There

are 100 circular divisions. Four main scale divisions

and 30 circular scale divisions is obtained as the

thickness of the wire measured by this instrument.

Assuming negligible zero error, find the thickness of

wire (in cm).

31. The thickness of a marker measured using a screw

gauge whose LC = 0.001 cm, comes out to be 0.802

cm. The percentage error in the measurement would

be

(a) 0.125% (b) 2.43%

(c) 4.12% (d) 2.14%

Short Notes

Fundamental Quantity Derived Quantity

The physical quantities which do

not depend on any other physical

quantities for their measurements.

E.g., Mass, Length, Time

Temperature, current, luminous

Intensity & mole

Those quantities which can

be expressed in terms of

fundamental/base quantities.

E.g., Angle, speed or velocity

Acceleration, force etc.,

System of Units

(a) FPS System: Here length is measured in foot, mass in

pounds and time in second.

(b) CGS System: In this system, L is measured in cm, M is

measured in g and T is measured in sec.

(c) MKS System: In this system, L is measured in metre, M is

measured in kg and T is measured in sec.

Principle of Homogeneity

According to this, the physical quantities having same dimension

can be added or subtracted with each other and for a given equation,

dimensions of both sides must be same.

For eg, in equation = ++

B F Am C

v ,

all the three parts of R.H.S have same dimension as force on L.H.S.

Dimensions

The fundamental or base quantities along with their powers needed

to express a physical quantity is called dimensions

E.g.: [F] = [MLT–2] is dimension of force.

Usage of Dimensional Analysis

(i) To check the correctness of a given formula.

(ii) To establish relation between quantities dimensionally.

(iii) To convert the value of a quantity from one system of

units to other system.

Limitations of Dimensional Analysis

(i) It does not predict the numerical value or number

associated with a physical quantity in a relation

eg, 1

3 5

u

v = + at & v = u + at

Both are dimensionally valid.

(ii) It does not derive any relations involving trigonometric,

logarithmic or exponential functions

E.g. P = P0e–bt2

cannot be derived dimensionally.

(iii) It does not give any information about dimensionally

constants or nature of a quantity (vector/scalar) associated

with a relation.

Significant Figure or Digits

1. Rules to find out the number of significant figures:

I Rule:All the non-zero digits are significant E.g. 1984 has

4 SF.

II Rule: All the zeros between two non-zero digits are

significant. E.g. 10806 has 5 SF.

III Rule: All the zeros to the left of first non-zero digit are

not significant. E.g.00108 has 3 SF.

18 JEE (XI) Module-1 P

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IV Rule: If the number is less than 1, zeros on the right of the

decimal point but to the left of the first non-zero digit are not

significant. E.g. 0.002308 has 4 SF.

V Rule: The trailing zeros (zeros to the right of the last nonzero digit) in a number with a decimal point are significant.

E.g. 01.080 has 4 SF.

VI Rule: The trailing zeros in a number without a decimal

point are not significant e.g. 010100 has 3 SF. But if the

number comes from some actual measurement then the

trailing zeros become significant. E.g. m = 100 kg has 3 SF.

VII Rule: When the number is expressed in exponential

form, the exponential term does not affect the number of S.F.

For example in x = 12.3 = 1.23 × 101 = .123 × 102 = 0.0123

× 103 = 123 × 10–1, each term has 3 SF only.

2. Rules for arithmetical operations with significant figures:

I Rule: In addition or subtraction the number of decimal

places in the result should be equal to the number of decimal

places of that term in the operation which contain lesser

number of decimal places. E.g. 12.587 – 12.5 = 0.087 = 0.1

( second term contain lesser i.e. one decimal place)

II Rule: In multiplication or division, the number of SF in the

product or quotient is same as the smallest number of SF in

any of the factors. E.g. 5.0 × 0.125 = 0.625 = 0.62.

Rounding Off

Rules for rounding off the numbers:

I Rule: If the digit to be rounded off is more than 5, then the

preceding digit is increased by one. e.g. 6.87≈ 6.9

II Rule: If the digit to be rounded off is less than 5, than the

preceding digit is unaffected and is left unchanged. e.g. 3.94 ≈ 3.9

III Rule: If the digit to be rounded off is 5 then the preceding digit

is increased by one if it is odd and is left unchanged if it is even.

e.g. 14.35 ≈ 14.4 and 14.45 ≈ 14.4

Representation of Errors

1. Mean absolute error is defined as

1 2

1

... n n i

i

aa a a

a

n n =

∆ +∆ + +∆ ∆ ∆ = = ∑

Final result of measurement may be written as:

a = am ± ∆a

2. Relative Error or Fractional Error: It is given by

Mean absolute Error

Mean value of measurement m

a

a

∆ =

3. Percentage Error 100%

m

a

a

∆ = ×

Combination of Errors

(i) In Sum: If Z = A + B, then ∆Z = ∆A + ∆B.

Maximum fractional error in this case is

ZA B

Z AB AB

∆∆ ∆

= +

+ +

(ii) In Difference: If Z = A – B, then maximum absolute error

is ∆Z = ∆A + ∆B and maximum fractional error in this case

ZA B

Z AB AB

∆∆ ∆

= + − −

(iii) In Product: If Z = AB, then the maximum fractional error,

Z AB

Z AB

∆ ∆∆

= +

(iv) In Division: If Z = A/B, then maximum fractional error is

Z AB

Z AB

∆ ∆∆

= +

(v) In Power: If Z = An then Z A

n

Z A

∆ ∆ =

In more general form if

x y

q

A B Z

C =

then the maximum fractional error in Z is

Z ABC

xyq Z ABC

∆ ∆∆∆

=++

To Find Smaller Measurements

Vernier Calliper

(i) Least count: Suppose movable Jaw is slided till the zero

of vernier scale coincides with any of the mark of the main

scale.

Let, n V.S.D = (n – 1) MSD

1 1VSD n

n

  − ⇒ =     M.S.D

\\ Vernier constant = 1 M.S.D – 1 V.S.D

1 1 1 MSD MSD n

n n

  − =− =    

(ii) Total reading = MSR + VSR

= MSR + n ×VC

where MSR = Main scale reading

VC = Vernier constant i.e. least count

n = nth division of vernier scale coinciding with main scale.

Screw Gauge

This instrument works on the principle of micro-meter screw. It

is used to measure very small (mm) measurements. It is provided

with linear scale and a circular scale.

(i) Pitch of the screw gauge

Distance moved in -rotation of cir-scale

No.of full-rotation

n =

(ii) L.C = Pitch

Total number of division on thecircular scale

(iii) Total Reading (T.R) = L.S.R + C.S.R

L.S.R = Linear scale Reading = N where

C.S.R = Circular Scale Reading = n × L.C

If nth division of circular scale coincides with the linear

scale line, then

\\ Total reading = N + n × (L.C)

Units and Measurements 19 P

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Solved Examples

1. 1

2 sin 1 .

2

xdx n x

a

ax x a

−   = −   −   ∫ The value of n is

(a) 0 (b) –1

(c) 1 (d) None of these

Use dimensional analysis to solve the problem.

Sol. (c) 1

2 sin 1

2

x dx n x

a

ax x a

−   = −   −   ∫

Denominator 2ax – x2 must have the dimension

of [x]2

( we can add or subtract only if quantities same

diemnsion)

∴ 2 2 [] ax x x   − =    

Also, dx has the dimension of [x]

∴ 2 2

x dx

ax x −

is having dimension L

Equating the dimension of L.H.S. and R.H.S. we have

[an] = M0L1T0

{ 1 sin 1 x

a

−   −     must be dimensionless}

2. In the formula;

a

RTV nRT P e

V b

−

= − find the dimensions

of ‘a’ and ‘b’, where P = pressure, n = no. of moles,

T = temperature, V = volume and R = universal gas constant.

Sol. [b] = [V] = L3

[a] = [RTV] =

2 23 [ ] ML T L [ ] [ ] mol

PV V

n

−

= ( nRT = PV)

= ML5T–2mol–1.

3. A particle is performing SHM along the axis of a fixed ring.

Due to gravitational force, its displacement at time t is given

by x = a sinwt.

m

x = 0

r

In this equation w is found to depend on radius of the ring

(r), mass of the ring (m) and gravitational constant (G).

Using dimensional analysis, find the expression of w in terms

of m, r and G.

Sol. Let w = KMarbGc

where K is a dimensionless constant

Writing the dimension of both the sides and equating them

we have,

T–1 = MaLb(M–1L3T–2)c

= Ma–cLb+3c

T–2c

Equating the exponents,

–2c = –1 or c = 1/2;

b + 3c = 0 or –3c = b = 3

2 − ;

a – c = 0, c = a = 1

2

+

Thus the required equation is 3 . Gm K r ω =

4. Using screw gauge, the observation of the diameter of a

wire are 1.324, 1.326, 1.334, 1.336 cm respectively. Find

the average diameter, the mean error, the relative error and

% error.

Sol. Average diameter:

( ) 1.324 1.326 1.334 1.336 1.330

4

D D

N

Σ +++ = = =

DD1 = 1.324 – 1.330 = –0.006

DD2 = 1.326 – 1.330 = –0.004

DD3 = 1.334 – 1.330 = 0.004

DD4 = 1.336 – 1.330 = 0.006

Means absolute error:

1234 | || || || |

4

DDDD

D

∆ +∆ +∆ +∆ ∆ =

0.006 0.004 0.004 0.006 0.020 0.005cm

4 4

+++ = = =

Relative error =

0.005 0.004

1.330

D

D

∆ = =

% error = 100 0.4% D

D

∆ × =

5. If a tuning fork of frequency (f

0) 340 Hz and tolerance 1%

is used in resonance column method [v = 2f

0 (2 – 1)], the

first and the second resonance are measured at 1 = 24.0 cm

and 2 = 74.0 cm. Find maximum permissible error in speed

of sound.

Sol. v = 2f

0(2 – 1)

⇒ 0 1 2

max 0 2 1

v f

v f

  ∆ ∆ ∆ +∆

= +     −

 

 

1 0.1 0.1 1.4%

100 74 24

+

=+ = −

20 JEE (XI) Module-1 P

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6. Side of a cube is measured with the help of vernier calliper.

Main scale reading is 10 mm and vernier scale reading is 1.

It is known that 9 M.S.D. = 10 V.S.D. Mass of the cube is

2.735 g. Find density of the cube upto appropriate significant

figure

Sol. Least count = 1 M.S.D. – 1 V.S.D. = 1 M.S.D. – 9

10

M.S.D.

= 1

10 M.S.D. =

1

10 × 1 mm

Least count = 0.1 mm

Length of side of cube = M.S.R. + V.S.R. × least count

= 10 + 1 × 0.1

= 10.1 mm

Density = 3

mass 2.735 0.0026546

volume (10.1) = =

Using significant figures the correct answer would be

0.00265 (with 3 significant figures).

7. For a cubical block, error in measurement of sides is + 1%

and error in measurement of mass is + 2%, then maximum

possible error in density is

(a) 1% (b) 5% (c) 3% (d) 7%

Sol. (b)

m



Density, 3

m m

V ρ= = 

Given: 2 2 2% 2 10 , 1% 1 10 m

m

∆ ∆ − − =± =± × =± =± × 



3 m

m

∆ρ ∆ ∆ = +

ρ



 = 2 × 10–2 + 3 × 10–2 = 5 × 10–2 = 5%

8. The length of a rectangular plate is measured by a meter scale

and is found to be 10.0 cm. Its width is measured by vernier

callipers as 1.00 cm. The least count of the meter scale

and vernier callipers are 0.1 cm and 0.01 cm respectively

(Obvious from readings). Maximum permissible error in

area measurement is

(a) +0.2 cm2 (b) +0.1 cm2 (c) +0.3 cm2 (d) Zero

Sol. (a) A = b = 10.0 × 1.00 = 10.00

Now, A b

A b

∆ ∆∆

= +





0.1 0.01

10.00 10.0 1.00

∆A

= +

⇒

11 2 2 10.00 10.00 0.2cm

100 100 100

A    ∆ = + = =±      

9. To estimate ‘g’ (from g = 4p2 2

L

T ), error in measurement of

L is + 2% and error in measurement of T is + 3%. The error

in estimated ‘g’ will be

(a) +8% (b) +6% (c) +3% (d) +5%

Sol. (a) 2 g 4

T

= π



–2 2% 2 10 ∆

= =± × 



–2 3% 3 10 T

T

∆

=+ =± ×

⇒

2 –2 –2 2 10 2 3 10 g T

g T

∆∆ ∆

= + = × +××





= 8 × 10–2 = ±8%

10. The mass of a ball is 1.76 kg. The mass of 25 such balls is

(a) 0.44 × 103 kg (b) 44.0 kg

(c) 44 kg (d) 44.00 kg

Sol. (b) m = 1.76 kg, M = 25 m = 25 × 1.76 = 44.0 kg

Mass of one unit has three significant figures and it is

just multiplied by a pure number (magnified). So result

should also have three significant figures.

11. To measure the diameter of a wire, a screw gauge is used.

In a complete rotation, spindle of the screw gauge advances

by 1/2 mm and its circular scale has 50 division. The main

scale is graduated to 1/2mm. If the wire is put between

the jaws, 4 main scale divisions are clearly visible and 10

divisions of circular scale co–inside with the reference line.

The resistance of the wire is measured to be (10W ± 1%).

Length of the wire is measured to be 10 cm using a scale of

least count 1mm. Maximum permissible error in resistivity

measurement is

Sol. L.C. of screw gauge =

1/ 2 mm 0.01mm

50 =

Diameter of wire as measured by the screw gauge

= 4 × 1

2

+ 10 × 0.01 = 2.1 mm

Since,

2

4

d Rπ

ρ = 

⇒

max

R d 2

R d

  ∆ρ ∆ ∆ ∆   =+ +

  ρ





max

1 2 0.01 1 2.9%

100 2.1 100

  ∆ρ ×   =+ +=

  ρ

12. The number of circular divisions on the shown screw gauge is

50. It moves 0.5 mm on main scale for one complete rotation.

Main scale reading is 2. The diameter of the ball is

5 0

Units and Measurements 21 P

W

25 2

(a) 2.25 mm (b) 2.20 mm

(c) 1.20 mm (d) 1.25 mm

Sol. (c) Least count =

0.5 0.01

50 =

Zero error = 5 × 0.01 = 0.05 mm

Actual measurement = 2 × 0.5 + 25 ×

0.5

50 – 0.05

= 1 mm + 0.25 mm – 0.05 mm = 1.20 mm.

13. A student performs an experiment for determination of

2

2

4 g L, 1m T

  π   = =

 

 and he commits an error of ∆L. For

For the takes the time of n oscillations with the stop watch

of least count ∆T and he commits a human error of 0.1 sec.

For which of the following data, the measurement of g will

be most accurate ?

(a) ∆L = 0.5, ∆T = 0.1, n = 20

(b) ∆L = 0.5, ∆T = 0.1, n = 50

(c) ∆L = 0.5, ∆T = 0.01, n = 50

(d) ∆L = 0.1, ∆T = 0.05, n = 50

Sol. (d) 2 gL T

gL T

∆∆ ∆

= +

In option (d) error in ∆g is minimum and number of

repetition of measurement are maximum. In this case

the error in g is minimum.

14. Student I, II and III perform an experiment for measuring

the acceleration due to gravity (g) using a simple pendulum.

They use different lengths of the pendulum and /or record

time for different number of oscillations. The observations

are shown in the table.

Least count for length = 0.1 cm

Least count for time = 0.1 s

Student Length

of the

pendulum

Number of

oscillations

(n)

Total

time for n

oscillations

(s)

Time

period

(s)

I 64.0 8 128.0 16.0

II 64.0 4 64.0 16.0

III 20.0 4 36.0 9.0

If EI

, EII and EIII are the percentage error in g, i.e. 100 g

g

  ∆

  ×

 

(a) EI

= 0 (b) EI

is minimum

(c) EI

= EII (d) EIII is maximum

Sol. (b) The least count of length ∆ = 0.1 cm

The least count of length ∆t = 0.1 s

% error of g = 100 g

g

∆

×

Now,

2

2

4 1 Tg T 2 where

g n T

π

=π ⇒ = =

 

So, 2

2

2

4

g n

t

π =  t T

n

  =    

⇒ 2 g t

g t

∆∆ ∆

= +





For student I,

0.1 2 0.1 100 100

64.0 128.0

g

g Ι

  ∆ ×     ×= + ×      

1

0.2 20 100

64.0 64

E = ×=

For student II,

0.1 0.1 100 2 100

64.0 64.0

g

g ΙΙ

  ∆     × = +× ×      

II

0.3 30 100

64.0 64

E = ×=

For student III,

0.1 0.1 19 100 100

20.0 18.0 18

g

g ΙΙΙ

  ∆     × = + ×=      

III

0.1 0.1 19 100

20.0 18.0 18

E   = + ×=    

⇒ EI

is least.

15. The density of a solid ball is to be determined in an

experiment. The diameter of the ball is measured with a

screw gauge whose pitch is 0.5 mm and there are 50 divisions

on the circular scale. The reading on the main scale is 2.5

mm and that on the circular scale is 20 divisions. If the

measured mass of the ball has a relative error of 2%, the

relative percentage error in the density is

(a) 0.9% (b) 2.4%

(c) 3.1% (d) 4.2%

Sol. (c) Least count =

0.5 0.01mm

50 =

Diameter of ball D = 2.5 mm + (20)(0.01)

⇒ D = 2.7 mm

3 Vol 4

3 2

M M

D

ρ= =

 

π   

⇒

max

3 M D

M D

  ∆ρ ∆ ∆   = +

  ρ

max

0.01 2% 3 100%

2.7

  ∆ρ     =+ ×     ρ  

max

3.1%   ∆ρ   =

  ρ

22 JEE (XI) Module-1 P

W

UNITS, SYSTEM OF UNITS

1. Which of the following is not the unit of time?

(a) Solar day (b) Parallactic second

(c) Leap year (d) Lunar month

2. A unit less quantity

(a) Never has a non zero dimension

(b) Always has a non zero dimension

(c) May have a non zero dimension

(d) Does not exit

3. Which of the following is not the name of a physical

quantity?

(a) Kilogram (b) Impulse

(c) Energy (d) Density

4. Parsec is a unit of

(a) Time (b) Angle

(c) Distance (d) Velocity

5. Which of the following system of units is not based on the

unit of mass, length and time alone

(a) FPS (b) SI

(c) CGS (d) MKS

6. In the S.I. system the unit of energy is:

(a) Erg (b) Calorie

(c) Joule (d) Electron volt

7. The SI unit of the universal gravitational constant G is

(a) N m kg–2 (b) N m2 kg–2

(c) N m2 kg–1 (d) N m kg–1

8. Surface tension has unit of

(a) Joule m2 (b) Joule m–2

(c) Joule m–1 (d) Joule m3

9. The specific resistance has the unit of:

(a) ohm/m (b) ohm/m2 (c) ohm m2 (d) ohm m

10. The unit of magnetic moment is:

(a) Amp m2 (b) Amp m–2

(c) Amp m (d) Amp m–1

11. The SI unit of the universal gas constant R is

(a) Erg K–1 mol–1 (b) Watt K–1 mol–1

(c) Newton K–1 mol–1 (d) Joule K–1 mol–1

12. The SI unit of Stefan's constant is

(a) Ws–1 m–2 K–4 (b) J s m–1 K–1

(c) J s–1 m–2 K–1 (d) W m–2 K–4

DIMENSION, FINDING DIMENSIONAL

FORMULA

13. In SI unit the angular acceleration has unit of

(a) N m kg–1 (b) m s–2 (c) rad s–2 (d) N kg–1

14. The angular frequency is measured in rad s–1. Its exponent

in length are

(a) – 2 (b) –1 (c) 0 (d) 2

15. [M L T–1] are the dimensions of

(a) Power (b) Momentum

(c) Force (d) Couple

16. What are the dimensions of Boltzmann's constant?

(a) MLT–2K–1 (b) ML2T–2K–1

(c) M0LT–2 (d) M0L2T–2K–1

17. A pair of physical quantities having the same dimensional

formula is

(a) Angular momentum and torque

(b) Torque and energy

(c) Force and power

(d) Power and angular momentum

18. Which one of the following has the dimensions of ML–1T–2?

(a) Torque (b) Surface tension

(c) Viscosity (d) Stress

19. The dimension of work done per unit mass per unit relative

density would be equivalent to dimension of

(a) (Acceleration)2 (b) (Velocity)2

(c) (Force)2 (d) (Torque)2

20. Which of the following is dimension of intensity?

(a) MT–3 (b) M–1L2T–2 (c) ML½T–1 (d) None

21. The dimension of

h

G

 

    where h = Planck’s constant and

G = gravitational constant is

(a) ML–1T2 (b) M–1L3T2 (c) M2L–1T (d) M3L0T–1

22. A dimensionless quantity:

(a) Never has a unit (b) Always has a unit

(c) May have a unit (d) Does not exit

PRINCIPLE OF HOMOGENEITY OF

DIMENSION

23. Force F is given in terms of time t and distance x by F = A

sin C t + B cos D x Then the dimensions of A/B and C/D are

given by

(a) MLT–2, M0L0T–1 (b) MLT–2, M0L–1T0

(c) M0L0T0, M0L1T–1 (d) M0L1T–1, M0L0T0

Exercise-1 (Topicwise)

Units and Measurements 23 P

W

24. The equation for the velocity of sound in a gas states that v =

b

T k

m

γ . Velocity v is measured in m/s. γ is a dimensionless

constant, T is temperature in kelvin (K), and m is mass in

kg. What are the units for the Boltzmann constant, kb?

(a) kg m2 s–2 K–1 (b) kg m2 s2 K

(c) kg m/s K–2 (d) kg m2 s–2 K

25. A wave is represented by y = a sin (At – Bx + C) where A,

B, C are constants and t is in seconds & x is in metre. The

dimensions of A, B, C are

(a) T–1, L, M0L0T0 (b) T–1, L–1, M0L0T0

(c) T, L, M (d) T–1, L–1, M–1

26. If v = γ P

ρ

, then the dimensions of γ are (P is pressure, ρ

is density and v is speed of sound has their usual dimension)

(a) M0L0T0 (b) M0L0T–1

(c) M1L0T0 (d) M0L1T0

27. Consider the equation d F ds A F P

dt

  =     ∫  

     . Then

dimension of A will be (where F =

 force, ds =  small

displacement, t = time and P =

 linear momentum)

(a) MºLºTº (b) M1LºTº

(c) M–1LºTº (d) MºLºT–1

28. If F = A

m

+ B where F = Force, m = Mass.

Then dimension of [A × B] is,

(a) M5/2L2T–4 (b) M2/5L2T–1

(c) M2L2/5T–1 (d) M–1L2/5T–2

29. If v = At3 +

B

m

, where m = mass, v = velocity and t = time.

Then dimension of A in the given equation would be

(a) LT–2 (b) L2T–3

(c) L3T–2 (d) LT–4

APPLICATION OF DIMENSIONAL ANALYSIS

Deriving New Relation

30. The velocity of water waves may depend on their wavelength

λ, the density of water ρ and the acceleration due to gravity

g. The method of dimensions gives the relation between

these quantities as (where k is a dimensionless constant)

(a) v2 = kλ–1 g –1 ρ–1 (b) v2 = k g λ

(c) v2 = k g λ ρ (d) v2 = k λ3 g–1 ρ–1

31. Force applied by water stream depends on density of water

(ρ), velocity of the stream (v) and cross–sectional area of

the stream (A). The expression of the force should be

(a) ρAv (b) ρAv2

(c) ρ2Av (d) ρA2v

32. If velocity (v), frequency (f) and mass (m) are taken as

fundamental quantity. How energy (E) may be described

using above quantity.

(a) Kvf 2m (b) Kv2f 0m

(c) Kvf 2m0 (d) Kv1/2f

–1m2

33. If P = power delivered by a motor is dependent of force

= F, velocity = v and density of material = r. Then power

may be proportional to

(a) F2vr–1 (b) Fv2r

(c) Fvr0 (d) None of these

34. The velocity of a freely falling body changes as gphq where

g = acceleration due to gravity and h is height. The value of

p and q are

(a) 1, 1

2 (b) 1

2 , 1

2

(c) 1

2

, 1 (d) 1, 1

APPLICATION OF DIMENSIONAL ANALYSIS

To Convert from One System of Unit

35. One watt-hour is equivalent to

(a) 6.3 × 103 Joule (b) 6.3 × 10–7 Joule

(c) 3.6 × 103 Joule (d) 3.6 × 10–3 Joule

36. The pressure of 106 dyne/cm2 is equivalent to

(a) 105 N/m2 (b) 106 N/m2

(c) 107 N/m2 (d) 108 N/m2

37. Consider ρ = 2 g/cm3. Convert it into MKS system

(a) 2 × 10–3

3

kg

m (b) 2 × 103

3

kg

m

(c) 4 × 103

3

kg

m (d) 2 × 106

3

kg

m

38. The density of mercury is 13600 kg m–3. Its value of CGS

system will be

(a) 13.6 g cm–3 (b) 1360 g cm–3

(c) 136 g cm–3 (d) 1.36 g cm–3

39. Force in CGS system is 20 N. Its value in SI unit will be

(a) 20 × 105 (b) 20 × 10–5

(c) 200 N (d) 2 × 10–3 N

40. If in a system of unit mass is measured in a kg, length in b

m and time in g sec. Find the value of 100 joule in the above

system.

(a) 100 α–1β–2γ2 (b) 100 α–2β–1γ–2

(c) 100 αβ–2γ (d) 1000 α–2β2γ–1

ERRORS IN MEASUREMENT

41. Which of the following measurements is most accurate?

(a) 9 × 10–2 m (b) 90 × 10–3 m

(c) 900 × 10–4 m (d) 0.090 m

24 JEE (XI) Module-1 P

W

42. A system takes 70.40 second to complete 20 oscillations.

The time period of the system is:

(a) 3.52 s (b) 35.2 × 10 s

(c) 3.520 s (d) 3.5200 s

43. The percentage error in the measurement of mass and speed

are 1% and 2% respectively. What is the percentage error in

kinetic energy?

(a) 5% (b) 2.5% (c) 3% (d) 1.5%

44. Number 15462 when rounded off to numbers to three

significant digits will be

(a) 15500 (b) 155

(c) 1546 (d) 150

45. Value of expression 25.2 1374

33.3

× will be

(All the digits in this expression are significant.)

(a) 1040 (b) 1039 (c) 1038 (d) 1041

46. Value of 24.36 + 0.0623 + 256.2 will be (considering rules

of significant digits)

(a) 280.6 (b) 280.8

(c) 280.7 (d) 280.6224

47. The percentage errors in the measurement of mass and speed

are 2% and 3% respectively. How much will be the maximum

error in the estimation of the kinetic energy obtained by

measuring mass and speed?

(a) 11% (b) 8%

(c) 5% (d) 1%

48. The random error in the arithmetic mean of 100 observations

is x; then random error in the arithmetic mean of 400

observations would be

(a) 4x (b) 1

4

x (c) 2x (d) 1

2

x

49. What is the number of significant figures in 0.310 × 103

(a) 2 (b) 3

(c) 4 (d) 6

50. Error in the measurement of radius of a sphere is 1%. The

error in the calculated value of its volume is

(a) 1% (b) 3%

(c) 5% (d) 7%

51. The mean time period of second’s pendulum is 2.00 s and

mean absolute error in the time period is 0.05 s. To express

maximum estimate of error, the time period should be written

as

(a) (2.00 ± 0.01) s (b) (2.00 +0.025) s

(c) (2.00 ± 0.05) s (d) (2.00 ± 0.10) s

52. The unit of percentage error is

(a) Same as that of physical quantity

(b) Different from that of physical quantity

(c) Percentage error is unit less

(d) Errors have got their own units which are different from

that of physical quantity measured

53. The decimal equivalent of 1/20 upto three significant figures

is

(a) 0.0500 (b) 0.05000

(c) 0.0050 (d) 5.0 × 10–2

54. A thin copper wire of length l metre increases in length by

2% when heated through 10 ºC. What is the percentage

increase in area when a square copper sheet of length

l metre is heated through 10 ºC?

(a) 4% (b) 8%

(c) 16% (d) 32 %

55. The length and breadth of a rectangle are 20 ± 0.2 cm and

10 ± 0.1 cm. Find the percentage error in area would be

(a) 1% (b) 2%

(c) 3% (d) 4%

56. The mass rate of flow of liquid through a pipe is given by

dm

dt = rAV, where r = density, A = area of cross-section

and v → velocity. The readings of area, A = 10 ± 0.1 m2 and

v = 30 ± 0.3 m/sec. Find the percentage error in measurement

of mass flow rate.

(a) 5% (b) 4%

(c) 8% (d) 3.5%

MEASURING INSTRUMENTS

57. In a vernier calliper, ten smallest divisions of the vernier

scale are equal to nine smallest division on the main scale.

If the smallest division on the main scale is half millimeter,

then the vernier constant is:

(a) 0.5 mm (b) 0.1 mm

(c) 0.05 mm (d) 0.005 mm

58. A vernier calliper has 20 divisions on the vernier scale, which

coincide with 19 on the main scale. The least count of the

instrument is 0.1 mm. The main scale divisions are of

(a) 0.5 mm (b) 1 mm (c) 2 mm (d) 1/4 mm

59. A vernier calliper having 1 main scale division = 0.1 cm is

designed to have a least count of 0.02 cm. If n be the number

of divisions on vernier scale and m be the length of vernier

scale, then

(a) n = 10, m = 0.5 cm (b) n = 9, m = 0.4 cm

(c) n = 10, m = 0.8 cm (d) n = 10, m = 0.2 cm

60. The pitch of a screw gauge is 0.05 cm. In how many

revolutions of hollow cylinder the Screw will advance 0.35

cm in the straight line?

(a) 7 (b) 10

(c) 15 (d) 14

61. A student in the laboratory measures thickness of a wire

using screw gauge. The readings are 1.22 mm, 1.23 mm,

1.19 mm, 1.20 mm. The percentage error in measurement is

± m.71%. Find m.

(a) 2.20 (b) 2.71

(c) 2.85 (d) 3.52

Units and Measurements 25 P

W

1. The unit of impulse is the same as that of

(a) Moment force

(b) Linear momentum

(c) Rate of change of linear momentum

(d) Force

2. Which of the following is not the unit of energy?

(a) Watt-hour (b) Electron-volt

(c) N × m (d) kg × m/sec2

3. If a and b are two physical quantities having different

dimensions then which of the following can denote a new

physical quantity?

(a) a + b (b) a – b (c) a/b (d) ea/b

4. The time dependence of a physical quantity

P = P0 exp(–αt

2) where α is a constant and t is time The

constant α

(a) Will be dimensionless

(b) Will have dimensions of T–2

(c) Will have dimensions as that of P

(d) Will have dimensions equal to the dimension of P

multiplied by T–2

5. Which pair of following quantities has dimensions different

from each other?

(a) Impulse and linear momentum

(b) Plank's constant and angular momentum

(c) Moment of inertia and moment of force

(d) Young's modulus and pressure

6. The product of energy and time is called action. The

dimensional formula for action is same as that for

(a) Power (b) Angular energy

(c) Force × velocity (d) Impulse × distance

7. What is the physical quantity whose dimensions are

[M L2 T–2]?

(a) Kinetic energy (b) Pressure

(c) Momentum (d) Power

8. If E, M, J and G denote energy, mass, angular momentum

and gravitational constant respectively, then

2

5 2

EJ

M G

has the

dimensions of

(a) Length (b) Angle (c) Mass (d) Time

9. The position of a particle at time 't' is given by the relation

x(t) = 0 – [1– ] V t e α

α

where V0 is a constant and α > 0. The

dimensions of V0 and α are respectively.

(a) M0L1T0 and T–1 (b) M0L1T0 and T–2

(c) M0L1T–1 and T–1 (d) M0L1T–1 and T–2

10. If force (F) is given by F = Pt–1 + αt, where t is time. The

unit of P is same as that of

(a) Velocity (b) Displacement

(c) Acceleration (d) Momentum

11. When a wave traverses a medium, the displacement of a

particle located at x at time t is given by y = asin(bt – cx)

where a, b and c are constants of the wave. The dimensions

of b are the same as those of

(a) Wave velocity (b) Amplitude

(c) Wavelength (d) Wave frequency

12. In a book, the answer for a particular question is expressed

as

2 1 ma kl b

k ma

 

= +  

 

here m represents mass, a represents

accelerations, l represents length. The unit of b should be

(a) m/s (b) m/s2 (c) meter (d) /sec

13. α = 2 sin( ) F

t

v

β (here v = velocity, F = force, t = time). Find

the dimension of α and β

(a) α = [M1L1T0], β = [T–1]

(b) α = [M1L1T–1], β = [T1]

(c) α = [M1L1T–1], β = [T–1]

(d) α = [M1L–1T0], β = [T–1]

14. If force, acceleration and time are taken as fundamental

quantities, then the dimensions of length will be

(a) FT2 (b) F–1 A2 T–1

(c) FA2T (d) AT2

15. If the unit of length is micrometer and the unit of time is

microsecond, the unit of velocity will be

(a) 100 m/s (b) 10 m/s

(c) micrometers (d) m/s

16. In a certain system of units, 1 unit of time is 5 s, 1 unit of

mass is 20 kg and unit of length is 10m. In this system, one

unit of power will correspond to

(a) 16 watts (b) 1/16 watts

(c) 25 watts (d) None of these

17. If the unit of force is 1 kilonewton, the length is 1 km and

time is 100 second, what will be the unit of mass?

(a) 1000 kg (b) 10 kg (c) 10000 kg (d) 100 kg

18. 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

Exercise-2 (Learning Plus)

26 JEE (XI) Module-1 P

W

19. If the units of force and that of length are doubled, the unit

of energy will be:

(a) 1/4 times (b) 1/2 times

(c) 2 times (d) 4 times

20. If the units of M, L are doubled then the unit of kinetic energy

will become

(a) 2 times (b) 4 times

(c) 8 times (d) 16 times

21. The angle subtended by the moon's diameter at a point on

the earth is about 0.50°. Use this and the fact that the moon

is about 384000 km away to find the approximate diameter

of the moon (D).

(a) 192000 km (b) 3350 km

(c) 1600 km (d) 1920 km

22. The least count of a stop watch is 0.2 second. The time of

20 oscillations of a pendulum is measured to be 25 seconds.

The percentage error in the time period is

(a) 16% (b) 0.8 %

(c) 1.8 % (d) 8 %

23. The dimensions of a cuboidal block measured with a vernier

calliper having least count of 0.1 mm is 5 mm × 10 mm

× 5 mm. The maximum percentage error in measurement of

volume of the block is

(a) 5 % (b) 10 % (c) 15 % (d) 20 %

24. An experiment measures quantities x, y, z and then t is

calculated from the data as t =

2

3

xy

z . If percentage errors

in x, y and z are respectively 1%, 3%, 2%, then percentage

error in t is:

(a) 10 % (b) 4 %

(c) 7 % (d) 13 %

25. The external and internal diameters of a hollow cylinder

are measured to be (4.23 ± 0.01) cm and (3.89 ± 0.01) cm.

The thickness of the wall of the cylinder is

(a) (0.34 ± 0.02) cm

(b) (0.17 ± 0.02) cm

(c) (0.17 ± 0.01) cm

(d) (0.34 ± 0.01) cm

26. The mass of a ball is 1.76 kg. The mass of 25 such balls is

(a) 0.44 × 103 kg (b) 44.0 kg

(c) 44 kg (d) 44.00 kg

27. Two resistors R1 (24 ± 0.5) Ω and R2 (8 ± 0.3) Ω are joined

in series. The equivalent resistance is

(a) 32 ± 0.33 Ω (b) 32 ± 0.8 Ω

(c) 32 ± 0.2 Ω (d) 32 ± 0.5 Ω

28. The pitch of a screw gauge is 0.5 mm and there are 100

divisions on its circular scale. The instrument reads +2

divisions when nothing is put in-between its jaws. In

measuring the diameter of a wire, there are 8 divisions on

the main scale and 83rd division coincides with the reference

line. Then the diameter of the wire is

(a) 4.05 mm (b) 4.405 mm

(c) 3.05 mm (d) 1.25 mm

29. The pitch of a screw gauge having 50 divisions on its circular

scale is 1 mm. When the two jaws of the screw gauge are in

contact with each other, the zero of the circular scale lies 6

division below the line of graduation. When a wire is placed

between the jaws, 3 linear scale divisions are clearly visible

while 31st division on the circular scale coincide with the

reference line. The diameter of the wire is

(a) 3.62 mm (b) 3.50 mm

(c) 3.5 mm (d) 3.74 mm

30. The smallest division on the main scale of a vernier calipers

is 1 mm, and 10 vernier divisions coincide with 9 main scale

divisions. While measuring the diameter of a sphere, the zero

mark of the vernier scale lies between 2.0 and 2.1 cm and

the fifth division of the vernier scale coincide with a scale

division. Then diameter of the sphere is

(a) 2.05 cm (b) 3.05 cm

(c) 2.50 cm (d) None of these

31. A satellite orbiting around the Earth, its orbital velocity (v0)

is found to depend on mass of Earth M, radius of earth R

and universal gravitational constant G. The expression for

orbital velocity is proportional to

(a) G–1M1R–1 (b) G1M1R–1

(c) G1/2M1/2R–1/2 (d) None of these

32. The vernier constant of vernier calliper is 0.1 mm. and it has

zero error of (– 0.05) cm. While measuring the diameter of

a sphere, the main scale reading is 1.7 cm and coinciding

vernier division is 5. The corrected diameter will be n × 10–2

cm. Find n.

(a) 2180 (b) 2220 (c) 2160 (d) 2200

33. In a particular system, the unit of length, mass and time are

chosen to be 10 cm, 10g and 0.1s respectively. The unit of

force in this system will be equivalent to

(a) 1/10 N (b) 1 N

(c) 10 N (d) 100 N

34. In sub-atomic physics, one often associates a characteristic

wavelength l with a particle of mass m. If

2

h = π

 (h being

Planck's constant) and c is the speed of light, which of the

following expression is most likely to be correct one ? (Use

formula E = hf)

(a) hc

m

λ = (b) 2 mc

λ = 

(c) m

c

λ =  (d)

mc

λ = 

Units and Measurements 27 P

W

35. If the mass, time and work are taken as fundamental physical

quantities then dimensional formula of length is

(a) [M1/2 T1 W–1/2] (b) [M–1/2 T1 W1/2]

(c) [M–1 T2 W] (d) None of these

36. Given that ln(a/pb) = az/kBq where p is pressure, z is

distance, kB is Boltzmann constant and q is temperature.

The dimensions of b are (Useful formula: Energy = kB ×

temperature)

(a) L0M0T0 (b) L1M–1T2

(c) L2 M0 T0 (d) L–1M1T–2

37. The dependence of g on geographical latitude at sea level

is given by g = g0(1 + bsin2f) where f is the latitude angle

and b is a dimensionless constant. If Dg is the error in the

measurement of g, then the error in measurement of latitude

angle is

(a) zero (b)

0 sin(2 )

g

g

∆ ∆φ = β φ

(c)

0 cos(2 )

g

g

∆ ∆φ = β φ (d)

0

g

g

∆ ∆φ =

38. Let y = l

2 –

3

l

z where l = 2.0 ± 0.1, z = 1.0 ± 0.1, then the

value of y is given by

(a) –4 ± 2.4 (b) –4 ± 1.6

(c) –4 ± 0.8 (d) None of these

39. If measured time period are T1 = 8.01 s and T2 = 8.41 s by a

student who used stop watch having least count = 0.01 sec,

then find best reported time (in sec) is

(a) 8.2 ± 0.2 (b) 8.41 ± 0.2

(c) 8.21 ± 0.01 (d) 8.41 ± 0.01

40. Assume pressure (P), length (L) and velocity (V) are

fundamental quantities. The dimension of coefficient of

viscosity (h) is

(a) [PL–1V] (b) [PLV–1] (c) [P–2LV–1] (d) [PL–1V–2]

41. A gas bubble from an explosion under water oscillates with a

period T proportional to padbEc

, where p is static pressure, d

is the density of water, E is the total energy of the explosion.

The values of c, b, a respectively will be

(a)

511 , ,

623

− (b)

111 , ,

334

−

(c)

11 5 , ,

32 6 − (d)

511 , ,

632

−

42. The diagram shows part of the vernier scale on a pair of

calipers.

3 cm 4 cm

0 10

Which reading is correct ?

(a) 2.74 cm (b) 3.10 cm (c) 3.26 cm (d) 3.64 cm

Exercise-3 (JEE Advanced Level)

MULTIPLE CORRECT TYPE QUESTIONS

1. Choose the correct statement(s).

(a) All quantities may be represented dimensionally in

terms of the base quantities.

(b) A base quantity cannot be represented dimensionally in

terms of the rest of the base quantities.

(c) The dimension of a base quantity in other base quantities

is always zero.

(d) The dimension of a derived quantity is never zero in

any base quantity.

2. The dimensions ML–1T–2 may correspond to

(a) Work done by a force

(b) Linear momentum

(c) Pressure

(d) Energy per unit volume

3. A student curiously picks up Resnick and Halliday and

tries to understand the answers given at the end of the book

using his new found knowledge of physics. He marks four

answers. In which of them A has the same units as that of

angular momentum ?

Useful formula L = r × p, 2 , T

π

ω = c is velocity of light,

t = r × F, E represent energy, l represent length and f

represents frequency.

(a) 1 2

2

mv Af =

(b) 2

2

2 1 A v

p

v c

ω ∆= −

(c)

2

sin( ) 2

A kl k

mE =

(d) t = Al

4. The quantity/quantities that does/do not have mass in its/

their dimensions (when we take standard 7 quantities as

fundamental) is/are

(a) Specific heat

(b) Latent heat

(c) Luminous intensity

(d) Mole

28 JEE (XI) Module-1 P

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5. The power for the hovering helicopter depends on the

factors, its linear size, the density of air and g × density of

the helicopter as

P ∝ (linear size)x

(density of air)y (g × density of helicopter)z

where g is acceleration due to gravity.

[Given: Power = ML–2T–3, Linear size = L, Density = ML–3,

g × density = ML–2T–2]

(a) The value of y is –1/2.

(b) The ratio of power output of engines of two hovering

having helicopters when linear size of one helicopter is

one fourth of linear size of other and all other parameters

are same is 64.

(c) The ratio of the power output to hover a helicopter

on the earth and on the imaginary planet, where gplanet

= g/4 and density of air on imaginary planet is same

as that of earth is 8.

(d) If helicopter is to hover at higher altitudes (like Siachin

glacier) then we need less powerful engine.

6. A student taking a quiz finds on a reference sheet the two

equations n = 1/T and v T = / µ

(µ = mass/length, torque = r × F and rest of symbols have

usual notations.)

He has forgotten what T represents in each equation. Use

dimensional analysis to determine the units required for T

in each equation.

(a) In first equation T represents tension.

(b) In first equation T represents time.

(c) In second equation T represents torque.

(d) In second equation T represents tension.

7. A chunk of unknown rock masses 38.254 ± 0.003 grams and

has a volume of 15.0 cm3.

(a) The density of the rock is 2.55 g/cm3.

(b) The absolute error in density is 0.02 g/cm3.

(c) The relative error in density is 0.007.

(d) The number of significant figure in density is 3.

8. Using screw gauge the diameter of the wire is found to be

5.00 mm. The length of wire is measured by using a scale

and is found to be 50.0 cm. If mass of wire is measured as

25 g, then mark the correct statement(s) (Take p = 3.14).

(a) The density has to be computed upto 2 significant digits.

(b) The least count of scale used to measure length of wire

is 1 mm.

(c) The density of wire is 2.5 g/cm3

(d) The least count of screw gauge is 0.01 mm

COMPREHENSION BASED QUESTIONS

Comprehension (Q. 9 to 11): Let us consider a particle P which

is moving straight on the X-axis. We also know that the rate of

change of its position is given by dx

dt ; where x is its separation

from the origin and t is time. This term dx

dt is called the velocity

of particle (v). Further the second derivation of x, with respect to

time is called acceleration (a) or rate of change of velocity and

represented by

2

2

d x

dt

or

dv

dt . If the acceleration of this particle is

found to depend upon time as follows a = At + Bt2 + 2

Ct

D t +

then

9. The dimensions of A are

(a) LT–2 (b) LT–3 (c) LT3 (d) L2T3

10. The dimensions of B are

(a) LT–4 (b) L2T–3 (c) LT4 (d) LT–2

11. The dimensions of C are

(a) L2T–2 (b) LT–2 (c) LT–1 (d) T2

Comprehension (Q. 12 to 14): According to coulombs law of

electrostatics there is a force between two charged particles q1 &

q2 separated by a distance r such that F ∝ q1, F ∝ q2 & F ∝ 2

1

r ;

combining all three we get F ∝ 1 2

2

q q

r or F = 1 2

2

kq q

r , where k is a

constant which depends on the medium and is given by 1/4πε0εr

where ε0 is absolute permittivity & εr

is relative permittivity.

But in case of protons of a nucleus there exists another

force called nuclear force; which is much higher in magnitude

in comparison to electrostatic force and is given by F = 2

kr Ce

r

−

.

12. What are the dimensions of C ?

(a) M2L3T–1 (b) ML3T–3 (c) ML3T–2 (d) ML2T–3

13. What are the dimensions of k?

(a) L (b) L2 (c) L–3 (d) L–1

14. What are the SI units of C?

(a) Nm–2 (b) Nm2 (c) Nm–3 (d) Nm

Comprehension (Q. 15 to 17): The Van-der waals equation is

2 () , a P V b nRT

V

  + −=     where P is pressure, V is volume and T

is the temperature of the given sample of gas. R is called molar gas

constant, a and b are called Van-der wall constants

15. The dimensional formula for a is same as that for

(a) V2 (b) P (c) PV2 (d) RT

16. Which of the following does not possess the same dimensional

formula as that for nRT ?

(a) PV (b) Pb (c) a/V2 (d) ab/V2

17. The dimensional formula of nRT is same as that of

(a) Energy (b) Force

(c) Specific heat (d) Latent heat

Comprehension (Q. 18 to 20): Max Planck noted in 1899 the

existence of a system of units based on the three fundamental

constants G, c, and h. These constants are dimensionally

independent in the sense that no combination is dimensionless

and a length, a time, and a mass may be constructed from them.

Specifically, with 1.05

2

h ≡ = π

 × 10–34 SI units in preference to

h, the Planck scale is represented by P, TP, MP.

Units and Measurements 29 P

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18. Which of the following formula can represent length in terms

of  , c and G?

(a) 3

G

c

 (b) c

G



(c)

Gc

 (d) G c

19. Which among the following would be the order of numerical

value of a unit of mass defined in terms of , c and G?

(a) 10–8 kg (b) 10–10 kg

(c) 10–5 kg (d) 10–27 kg

20. The dimensional formula for acceleration in terms of these

quantities would be

(a) 1/2 3/2 1/2 c G−  (b) 3/2 5/2 1/2  c G

(c) 1/2 7/2 1/2 c G − −  (d) 5/2 3/2 1/2  c G

Comprehension (Q. 21 to 23): When numbers having

uncertainties or errors are used to compute other numbers, these

will be uncertain. It is especially important to understand this when

a number obtained from measurements is to be compared with

a value obtained from theoretical prediction. Assume a student

wants to verify the value of p as the ratio of circumference to

diameter of a circle. The correct value of ten digits is 3.141592654.

He draws a circle and measures its diameter and circumference to

its nearest millimeter obtaining the values 135 mm and 424 mm,

respectively. Using a calculator he finds p = 3.140740741.

21. Why does measured value not match with calculated value?

(a) Due to systematic error

(b) Due to error in calculation

(c) Due to random error

(d) Lack of precision in measuring

22. What is the value of p in the passage measured by the

student?

(a) 3.140 (b) 3.141

(c) 3.1407 (d) 3.14

23. If diameter and circumference both have an error of 1%,

what is the error in value of p?

(a) 2% (b) 1%

(c) 0.5% (d) 0%

MATCH THE COLUMN TYPE QUESTIONS

24. Match the following columns

Physical quantity Dimension Unit

A. Gravitational

constant 'G'

p. M1L1T–1 (i) N m

B. Torque q. M–1L3T–2 (ii) N s

C. Momentum r. M1 L–1T–2 (iii) N m2/kg2

D. Pressure s. M1L2T–2 (iv) pascal

(a) A-(p)-(iii); B-(r)-(i); C-(q)-(iv); D-(r)-(ii)

(b) A-(q)-(iii); B-(s)-(i); C-(p)-(ii); D-(r)-(iv)

(c) A-(q)-(iii); B-(r)-(i); C-(r)-(ii); D-(s)-(iv)

(d) A-(p)-(iv); B-(s)-(ii); C-(p)-(i); D-(r)-(iii)

25. Match the following

Physical quantity Dimension Unit

A. Stefan's constant 'σ' p. M1L1T–2A–2 (i) W/m2

B. Wien's constant 'b' q. M1LºT–3K–4 (ii) K.m.

C. Coefficient of

viscosity 'η'

r. M1LºT–3 (iii) tesla .m/A

D. Emissive power of

radiation (Intensity

emitted)

s. MºL1TºK1 (iv) W/m2.K4

E. Mutual inductance 'M' t. M1L2T–2A–2 (v) Poise

F. Magnetic

permeability 'µ0'

u. M1L–1T–1 (vi) Henry

(a) A-(p)-(iv); B-(s)-(iii); C-(q)-(v); D-(r)-(i), E-(u)-(vi),

F-(t)-(ii)

(b) A-(q)-(iii); B-(s)-(ii); C-(r)-(i); D-(u)-(v), E-(p)-(vi),

F-(t)-(iv)

(c) A-(p)-(iv); B-(s)-(ii); C-(r)-(i); D-(u)-(v), E-(t)-(vi),

F-(q)-(iii)

(d) A-(q)-(iv); B-(s)-(ii); C-(u)-(v); D-(r)-(i), E-(t)-(vi),

F-(p)-(iii)

26. In Column-I, some physical quantities are given and some

possible SI units are given in Column-II. Match the physical

quantities in Column-I with the units in Column-II. Some

useful formulas

4 , ,, , F v P E hcR T b E AT F A

A L = = λ = =σ =η

(P-Pressure, F-force, v-velocity, A-Area, l-wavelength

h-Planck's constant, g-Gravitational acceleration, R-Rydberg

constant, b-Wien's constant, h-Coefficient of viscosity,

L-length, c-speed of light, t-time, E-energy, s-Stefan's

constant and T-temperature)

Column-I Column-II

A. PAvt p.

3

watt second

meter

B. hgR q. Joule

C. 4 b

A

σ

r.

2

Newton

metre

D.

t

η s. Newton metre

second

t. Newton metre

(a) A-(q,t); B-(s,t); C-(p,t); D-(p,r)

(b) A-(p,t); B-(s); C-(q,t); D-(p)

(c) A-(q,t); B-(s); C-(q,t); D-(p,r)

(d) A-(t); B-(s); C-(q); D-(p,r)

30 JEE (XI) Module-1 P

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27. Suppose two students are trying to make a new measurement

system so that they can use it like a code measurement

system and others do not understand it. Instead of taking

1 kg, 1 m and 1 sec. as basic unit they took unit of mass as

a kg, the unit of length as b m and unit of time as g second.

They called power in new system as SHAKTI then match

the two columns.

Column-I Column-II

A. 1N in new system p. a–1 b–2 g2

B. 1J in new system q. a–1 b–1 g2

C. 1 Pascal (SI unit of pressure) in new system

r. a–1 b g2

D. a SHAKTI in watt s. a2 b2 g–3

(a) A-(q); B-(p); C-(r); D-(s)

(b) A-(p); B-(q); C-(r); D-(s)

(c) A-(q); B-(p); C-(s); D-(r)

(d) A-(p); B-(r); C-(q); D-(s)

28. Match the following

Column-I Column-II

A. Latent heat constant p. M0 L0 T0

B. Reynold number q. M L2

C. Coefficient of friction r. M L0 T–3

D. Avogadro constant s. L2 T–2

E. Intensity of wave r. M0 L0 T0

F. Moment of inertia s. mol–1

(a) A-(p); B-(s); C-(t); D-(u); E-(r); F-(q)

(b) A-(s); B-(t); C-(p); D-(r); E-(u); F-(q)

(c) A-(s); B-(p); C-(t); D-(u); E-(q); F-(r)

(d) A-(s); B-(p); C-(t); D-(u); E-(r); F-(q)

NUMERICAL TYPE QUESTIONS

29. Number of significant figures in 0.007 m2 .

30. Number of significant figures in 2.64 × 1024 kg

31. Number of significant figures in 6.032 N m–2

32. The velocity of sound in a gas depends on its pressure and

density. The relation between velocity, pressure and density

is given by v = Kpa Db, then (a + b) is

33. A gas bubble, from an explosion under water, oscillates with a

period proportional to PadbEc

. Where P is the static pressure,

d is the density and E is the total energy of the explosion.

Find the values of a + b + c

34. The pitch of a screw gauge is 1 mm and there are 100

divisions on the circular scale. While measuring the diameter

of a wire, the linear scale reads 1 mm and 47th division on the

circular scale coincides with the reference line. The length

of the wire is 5.6 cm. Find the curved surface area (in cm2)

of the wire in two number of significant figures.

35. The density of a cube is measured by measuring its mass

and the length of its sides. If the maximum errors in the

measurement of mass and length are 3% and 2% respectively,

then the maximum error in the measurement of density is.

36. The length of the string of a simple pendulum is measured

with a metre scale to be 90.0 cm. The radius of the bob plus

the length of the hook is calculated to be 2.13 cm using

measurements with a slide callipers. What is the effective

length of the pendulum? (This effective length is defined as

the distance between the point of suspension and the center

of the bob).

37. Using the approximation (1 + x) n = 1 + nx, |x| << 1

Find the value of 99 .

38. The time period of oscillation of a body is given by

T = 2 mgA

K

π

K represents the kinetic energy, m mass, g acceleration due

to gravity and A is unknown. If [A] = Mx

Ly

Tz

, then what is

the value of x + y + z?

39. The radius of a sphere is measured to be 5.3 ± 0.1 cm.

Calculate percentage error in volume. Round off to nearest

integer.

40. The main scale of a Vernier calliper reads in millimeter and

its vernier is divided into 10 divisions which coincides with

9 divisions of the main scale. The length of the object for

situation is found to be 12

10

x mm. Find the value of x.

0 1 cm 2 3

0 10

0 1 2 3 cm

0 10

When not in use

When in use

41. In an experiment of simple pendulum, time period measured

was 50 s for 25 oscillations when the length of the simple

pendulum was taken 100 cm. If the least count of stop watch

is 0.1 s and that of meter scale is 0.01 cm, calculate the

maximum possible percentage error (p) in the measurement

of value of g. Quote100p.

Units and Measurements 31 P

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JEE MAIN

1. The density of a material in SI units is 128 kg m– 3. In certain

units in which the unit of length is 25 cm and the unit of

mass 50 g, the numerical value of density of the material is

(2019)

(a) 40 (b) 16 (c) 640 (d) 410

2. Expression for time in terms of G (universal gravitational

constant), h (Planck constant) and c (speed of light) is

proportional to (2019)

(a)

5 hc

G (b)

3

c

Gh (c) 5

Gh

c

(d) 3

Gh

c

3. Let L, R, C and V represent inductance, resistance,

capacitance and voltage, respectively. The dimension of

L

RCV

in SI units will be (2019)

(a) [LA–2] (b) [A–1] (c) [LTA] (d) [LT2]

4. In the formula X = 5YZ2, X and Z have dimensions of

capacitance and magnetic field, respectively. What are the

dimensions of Y in SI units? (2019)

(a) [M–2 L–2 T6 A3] (b) [M–1 L–2 T4 A2]

(c) [M–3 L–2 T8 A4] (d) [M–2 L0 T–4 A–2]

5. If surface tension (S), Moment of inertia (I) and Planck’s

constant (h), were to be taken as the fundamental units, the

dimensional formula for linear momentum would be (2019)

(a) S3/2I1/2h0 (b) S1./2I1/2h0

(c) S1/2I1/2h–1 (d) S1/2I3/2h–1

6. The pitch and the number of divisions, on the circular scale,

for a given screw gauge are 0.5 mm and 100 respectively.

When the screw gauge is fully tightened without any object,

the zero of its circular scale lies 3 divisions below the mean

line. The readings of the main scale and the circular scale for

a thin sheet are 5.5 mm and 48 respectively, the thickness of

this sheet is (2019)

(a) 5.755 mm (b) 5.950 mm

(c) 5.725 mm (d) 5.740 mm

7. The diameter and height of a cylinder are measured by

a meter scale to be 12.6 ± 0.1 cm and 34.2 ± 0.1 cm

respectively. What will be the value of its volume in

appropriate significant figures? (2019)

(a) 4264 ± 81 cm3 (b) 4264 ± 81.0 cm3

(c) 4260 ± 80 cm3 (d) 4300 ± 80 cm3

8. The least count of the main scale of a screw gauge is 1 mm.

The minimum number of divisions on its circular scale

required to measure 5 µm diameter of a wire is (2019)

(a) 50 (b) 200 (c) 100 (d) 500

9. The area of a square is 5.29 cm2. The area of 7 such squares

taking into account the significant figures is (2019)

(a) 37 cm2 (b) 37.0 cm2

(c) 37.03 cm2 (d) 37.030 cm2

10. If momentum (P), area (A) and time (T) are taken to be the

fundamental quantities then the dimensional formula for

energy is (2020)

(a)

1

2 1 P AT −       (b) [P2AT–2]

(c)

1

2 1 PA T −       (d) [PA–1T–2]

11. If speed V, area A and force F are chosen as fundamental

units, then the dimension of Young’s modulus will be (2020)

(a) FA–1V0 (c) FA2V–1

(b) FA2V–2 (d) FA2V–3

12. Amount of solar energy received on the earth’s surface per

unit area per unit time is defined a solar constant. Dimension

of solar constant is (2020)

(a) ML2T–2 (b) MLT–2

(c) M2L0T–1 (d) ML0T–3

13. A quantity x is given by (IFv2/WL4) in terms of moment

of inertia I, force F, velocity v, work W and length L. The

dimensional formula for x is same as that of (2020)

(a) Coefficient of viscosity

(b) Force constant

(c) Energy density

(d) Planck’s constant

14. The quantities x =

0 0

1

µ ε , y = E

B

and z = I

CR

are defined

where C-capacitance, R-resistance, l-length, E-electric

field, B-magnetic field and ε0, µ0-free space permitivity and

permeability respectively. Then (2020)

(a) Only x and y have the same dimension.

(b) Only x and z have the same dimension.

(c) x, y and z have the same dimension.

(d) Only y and z have the same dimension.

15. A simple pendulum is being used to determine the value of

gravitational acceleration g at a certain place. The length of

the pendulum is 25.0 cm and a stop watch with 1s resolution

measures the time taken for 40 oscillations to be 50 s. The

accuracy in g is (2020)

(a) 4.40% (b) 3.40%

(c) 2.40% (d) 5.40%

Exercise-4 (Past Year Questions)

32 JEE (XI) Module-1 P

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16. If the screw on a screw–gauge is given six rotations, it moves

by 3mm on the main scale. If there are 50 divisions on the

circular scale, the least count of the screw gauge is (2020)

(a) 0.01 cm (b) 0.02 mm

(c) 0.001 mm (d) 0.001 cm

17. The least count of the main scale of a vernier calipers is 1

mm. Its vernier scale is divided into 10 divisions and coincide

with 9 divisions of the main scale. When jaws are touching

each other, the 7th division of vernier scale coincides with a

division of main scale and the zero of vernier scale is lying

right side of the zero of main scale. When this vernier is

used to measure length of a cylinder the zero of the vernier

scale between 3.1 cm and 3.2 cm and 4 VSD coincides with

a main scale division. The length of the cylinder is (VSD is

vernier scale division) (2020)

(a) 3.21 cm (b) 2.99 cm

(c) 3.07 cm (d) 3.2 cm

18. A physical quantity z depends on four observables a, b, c and

d, as

2

2 3

3

a b

c d .The percentages of error in the measurement of

a, b, c and d are 2%, 1.5%, 4% and 2.5% respectively. The

percentage of error in z is (2020)

(a) 13.5% (b) 14.5%

(c) 16.5% (d) 12.25%

19. A student measuring the diameter of a pencil of circular

cross-section with the help of a vernier scale records the

following four readings 5.50 mm, 5.55 mm, 5.45 mm, and

5.65 mm. The average of these four readings is 5.5375 mm

and the standard deviation of the data is 0.07395 mm. The

average diameter of the pencil should therefore be recorded

as (2020)

(a) (5.5375 ± 0.0739) mm

(b) (5.54 ± 0.07) mm

(c) (5.538 ± 0.074) mm

(d) (5.5375 ± 0.0740) mm

20. The density of a solid metal sphere is determined by

measuring its mass and its diameter. The maximum error in

the density of the sphere is 100

  x     %. If the relative errors

in measuring the mass and the diameter are 6.0% and 1.5%

respectively, the value of x is (2020)

(a) 5010 (b) 5100

(c) 1050 (d) 5101

21. The vernier scale used for measurement has a positive zero

error of 0.2 mm. If while taking a measurement it was noted

that zero on the vernier scale lies between 8.5 cm and 8.6

cm, and vernier coincidence is 6, then the correct value of

measurement is (least count = 0.01 cm) (2021)

(a) 8.58 cm (b) 8.54 cm

(c) 8.56 cm (d) 8.36 cm

22. In order to determine the Young’s modulus of a wire of

radius 0.2 cm (measured using a scale of least count =0.001

cm) and length 1m (measured using a scale of least count

= 1 mm), a weight of mass 1kg (measured using a scale of

least count = 1g) was hanged to get the elongation of 0.5

cm (measured using a scale of least count 0.001 cm). What

will be the fractional error in the value of Young’s modulus

determined by this experiment? (2021)

(a) 0.14% (b) 9% (c) 1.4% (d) 0.9%

23. One main scale division of a vernier calipers is ‘a’ cm and nth

division of the vernier scale coincide with (n – 1)th division

of the main scale. The least count of the calipers (in mm) is

(2021)

(a) 10

1

na

n − (b) 1

10

  −    

n

a

n

(c) 10

−1

a

n

(d)

10 a

n

24. If velocity [V], time [T] and force [F] are chosen as the base

quantities, the dimensions of the mass will be (2021)

(a) [FT–1V–1] (b) [FTV–1]

(c) [FT2V] (d) [FVT–1]

25. Assertion (A): If in five complete rotations of the circular

scale, the distance travelled on main scale of the screw gauge

is 5 mm and there are 50 total divisions on circular scale,

then least count is 0.001 cm.

Reason (R): Least count Pitch

Totaldivisions on circular scale

In the light of the above statements, choose the most

appropriate answer from the option given below (2021)

(a) A is not correct but R is correct.

(b) Both A and R are correct and R is correct explanation

of A

(c) A is correct but R is not correct

(d) Both A and R are correct and R is not the correct

explanation of A.

26. The force is given in terms of time t and displacement x by

the equation F = AcosBx + CsinDt. The dimensional formula

of AD/B is (2021)

(a) [M0LT–1] (b) [ML2T–3]

(c) [M1L1T–2] (d) [M2L2T–3]

27. If the length of the pendulum in pendulum clock increases

by 0.1%, then the error in time per day is (2021)

(a) 86.4 s (b) 4.32 s

(c) 43.2 s (d) 8.64 s

28. The acceleration due to gravity is found upto an accuracy of

4% on a planet. The energy supplied to a simple pendulum

to known mass ‘m’ to undertake oscillations of time period T

is being estimated. If time period is measured to an accuracy

of 3%, the accuracy to which E is known as _____. (2021)

Units and Measurements 33 P

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29. An expression for a dimensionless quantity P is given by

log ; e

kt P

x

α   =   β β 

where a and b are constants, x is distance

k is Boltzmann constant and t is the temperature. Then the

dimensions of a will be (2022)

(a) [M0L–1T0] (b) [ML0T–2]

(c) [MLT–2] (d) [ML2T2]

30. The SI unit of a physical quantity is pascal-second. The

dimensional formula of this quantity will be (2022)

(a) [ML–1T–1]

(b) [ML–1T–2]

(c) [ML2T–1]

(d) [M–1L3T0]

31. In Vander Waals equation 2 [] ; a P V b RT

V

  + −=    

P is

pressure. V is volume, R is universal gas constant and T is

temperature. The ratio of constants a/b is dimensionally

equal to (2022)

(a) P/V (b) V/P

(c) PV (d) PV3

32. A torque meter is calibrated to reference standards of mass,

length and time each with 5% accuracy. After calibration,

the measured torque with this torque meter will have net

accuracy of (2022)

(a) 15% (b) 25%

(c) 75% (d) 5%

33. Consider the efficiency of Carnot’s engine is given by

log , sin e

x

kT

αβ β η = θ

where a and b are constants. If T

is temperature, k is Boltzmann constant, q is angular

displacement and x has the dimensions of length. Then,

choose the incorrect option. (2022)

(a) Dimension of b is same as that of force.

(b) Dimension of a–1x is same as that of energy.

(c) Dimension of h–1sin q is same of ab.

(d) Dimension of a is same of b.

34. Match Column-I with Column-II

Column-I Column-II

A. Torque p. Nms–1

B. Stress q. J kg–1

C. Latent heat r. Nm

D. Power s. Nm–2

Choose the correct answer from the options given below

(2022)

(a) A-(r); B-(q); C-(p); D-(s)

(b) A-(r); B-(s); C-(q); D-(p)

(c) A-(s); B-(p); C-(r); D-(q)

(d) A-(q); B-(r); C-(p); D-(s)

35. In a Vernier Calliper, 10 divisions of Vernier scale is equal

to the 9 divisions of main scale. When both jaws of Vernier

calipers touch each other, the zero of the Vernier scale is

shifted to the left of zero of the main scale and 4th Vernier

scale division exactly coincides with the main scale reading.

One main scale division is equal to 1 mm. While measuring

diameter of a spherical body, the body is held between two

jaws. It is now observed that zero of the Vernier scale lies

between 30 and 31 divisions of main scale reading and 6th

Vernier scale division exactly, coincides with the main scale

reading. The diameter of the spherical body will be (2022)

(a) 3.02 cm (b) 3.06 cm

(c) 3.10 cm (d) 3.20 cm

36. In an experiment to find out the diameter of wire using screw

gauge, the following observations were noted? (2022)

45

P Q

A. Screw moves 0.5 mm or main scale in one complete

rotation.

B. Total divisions on circular scale = 50

C. Main scale reading is 2.5 mm

D. 45th division of circular scale is in the pitch line

Then the diameter of wire is

(a) 2.92 mm (b) 2.54 mm

(c) 2.98 mm (d) 3.45 mm

37. Given below are statements: One is labelled as Assertion

(A) and other is labelled as Reason (R) (2022)

Assertion (A): Time period of oscillation of a liquid drop

depends on surface tension (S). If density of the liquid

is ρ and radius of the drop is r, then 3 3/2 TK rS = ρ / is

dimensionally correct, where K is dimensionless.

Reason (R): Using dimensional analysis we get R.H.S

having different dimension than that of time period.

In the light of above statements, choose the correct answer

from the options given below.

(a) Both (A) and (R) are true and (R) is the correct explanation

of (A).

(b) Both (A) and (R) are true but (R) is not the correct

explanation of (A).

(c) (A) is true but (R) is false.

(d) (A) is false but (R) is true.

34 JEE (XI) Module-1 P

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JEE ADVANCED

38. In the determination of Young’s modulus 2

4MLg Y

ld

  =     π

by using Searle’s method, a wire of length L = 2 m and

diameter d = 0.5 mm is used. For a load M = 2.5 kg, an

extension  = 0.25 mm in the length of the wire is observed.

Quantities d and  are measured using a screw gauge and a

micrometer, respectively. They have the same pitch of 0.5

mm. The number of divisions on their circular scale is 100.

The contributions to the maximum probable error of the Y

measurement (2012)

(a) Due to the errors in the measurements of d and  are the

same.

(b) Due to the error in the measurement of d is twice that

due to the error in the measurement of .

(c) Due to the error in the measurement of  is twice that

due to the error in the measurement of d.

(d) Due to the error in the measurement of d is four time

that due to the error in the measurement of .

39. Match Column-Ι with Column-ΙΙ and select the correct

answer using the codes given below the lists: (2013)

Column-I Column-II

A. Boltzmann constant p. [ML2T–1]

B. Coefficient of viscosity q. [ML–1T–1]

C. Planck constant r. [MLT–3K–1]

D. Thermal conductivity s. [ML2T–2K–1]

(a) A-(r); B-(p); C-(q); D-(s)

(b) A-(r); B-(q); C-(p); D-(s)

(c) A-(s); B-(q); C-(p); D-(r)

(d) A-(s); B-(p); C-(q); D-(r)

40. The diameter of a c ylinder is measured using a vernier

callipers with no zero error. It is found that the zero of the

vernier scale lies between 5.10 cm and 5.15 cm of the main

scale. The vernier scale has 50 division equivalent to 2.45

cm. The 24th division of the vernier scale exactly coincides

with one of the main scale divisions. The diameter of the

cylinder is (2013)

(a) 5.112 cm (b) 5.124 cm

(c) 5.136 cm (d) 5.148 cm

41. Using the expression 2d sin θ = λ, one calculates the

values of d by measuring the corresponding angles θ in

the range 0 to 90º. The wavelength λ is exactly knowns

and the error in θ is constant for all values of θ. As

θ increases from 0º (2013)

(a) The absolute error in d remains constant.

(b) The absolute error in d increases.

(c) The fractional error in d remains constant.

(d) The fractional error in d decreases.

42. To find the distance d over which a signal can be seen clearly

in foggy conditions, a railways engineer uses dimensional

analysis and assumes that the distance depends on the mass

density ρ of the fog, intensity (power/area) S of the light

from the signal and its frequency f. The engineer find that d

is proportional to S1/n. The value of n is (2014)

(a) 1

2 (b) 3

2

(c) 3 (d) 2

3

43. Planck's constant h, speed of light c and gravitational constant

G are used to form a unit of length L and a unit of mass M.

The relation between M, c, G, h are as under (2015)

(a) M c ∝ (b) M G ∝

(c) L h ∝ (d) L G ∝

44. Consider a Vernier calipers in which each 1 cm on the main

scale is divided into 8 equal divisions and a screw gauge with

100 divisions on its circular scale. In the Vernier calipers, 5

divisions of the vernier scale coincide with 4 divisions on

the main scale and in the screw gauge, one complete rotation

of the circular scale moves it by two divisions on the linear

scale. Then (2015)

(a) If the pitch of the screw gauge is twice the least count of

the Vernier calipers, the least count of the screw gauge

is 0.01 mm.

(b) If the pitch of the screw gauge is twice the least count of

the Vernier calipers, the least count of the screw gauge

is 0.005 mm.

(c) If the least count of the linear scale of the screw gauge

is twice the least count of the Vernier calipers, the least

count of the screw gauge is 0.01 mm.

(d) If the least count of linear scale of the screw gauge is

twice the least count of the Vernier calipers, the least

count of the screw gauge is 0.005 mm.

45. In terms of potential difference V, electric current I,

permittivity ε0, permeability μ0 and speed of light c, the

dimensionally correct equation(s) is (are) (2015)

(a) μ0I2 = ε0V2 (b) μ0I = μ0V

(c) I = ε0cV (d) μ0cI = ε0V

46. In an experiment to determine the acceleration due to gravity

g, the formula used for the time period of a periodic motion

is T = 2π 7( )

5

R r

g

− . The values of R and r are measured

to be (60 ± 1) mm and (10 ± 1) mm, respectively. In five

successive measurements, the time period is found to be

0.52 s, 0.56s, 0.57s, 0.54s and 0.59 s. The least count of the

watch used for the measurement of time period is 0.01 s.

Which of the following statement(s) is (are) true? (2016)

(a) The error in the measurement of r is 10%

(b) The error in the measurement of T is 3.57%

(c) The error in the measurement of T is 2%

(d) The error in the determined value of g is 11%

Units and Measurements 35 P

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47. A length scale (l) depends on the permittivity (ε) of a

dielectric material, Boltzmann constant (kB), the absolute

temperature (T), the number per unit volume (n) of certain

charged particles, and the charge (q) carried by each of the

particles. Which of the following expression(s) for l is (are)

dimensionally correct? (2016)

(a)

2

B

nq l

k T

  =  

ε  

(b) 2

Bk T l

nq

  ε =    

(c)

2

2/3

B

q l

n kT

  =  

ε  

(d)

2

1/3

B

q l

n kT

  =  

ε  

48. There are two Vernier calipers both of which have 1 cm

divided into 10 equal divisions on the main scale. The Vernier

scale of one of the calipers (C1) has 10 equal divisions that

correspond to 9 main scale divisions. The Vernier scale of

the other caliper (C2) has 10 equal divisions that correspond

to 11 main scale division. The readings of the two calipers

are shown in the figure. The measured values (in cm) by

calipers C1 and C2, respectively, are (2016)

2 3 4

0 5 10

C1

2 3 4

0 5 10

C2

(a) 2.87 and 2.86 (b) 2.85 and 2.82

(c) 2.87 and 2.87 (d) 2.87 and 2.83

Comprehension (Q. 49 to 50): In electromagnetic theory,

the electric and magnetic phenomena are related to each other.

Therefore, the dimensions of electric and magnetic quantities must

also be related to each other. In the questions below, [E] and [B]

stand for dimensions of electric and magnetic fields respectively,

while [e0] and [µ] stand for dimensions of the permittivity and

permeability of free space respectively. [L] and [T] are dimensions

of length and time respectively. All the quantities are given in SI

units. (2018)

49. The relation between [E] and [B] is

(a) [E] = [B] [L] [T]

(b) [E] = [B] [L]

–1 [T]

(c) [E] = [B] [L] [T]–1

(d) [E] = [B] [L]

–1 [T]–1

50. The relation between [e0] and [µ0] is:

(a) [µ0] = [e0] [L]2 [T]–2

(b) [µ0] = [e0] [L]–2 [T]2

(c) [µ0] = [e0]–1 [L]2 [T]–2

(d) [µ0] = [e0]–1 [L]–2 [T]2

51. Let us consider a system of units in which mass and angular

momentum are dimensionless. If length has dimension of L,

which of the following statement(s) is/are correct? (2019)

(a) The dimension of force is L–3.

(b) The dimension of energy of L–2.

(c) The dimension of power is L–5.

(d) The dimension of linear momentum is L–1.

52. Sometimes it is convenient to construct a system of units

so that all quantities can be expressed in terms of only

one physical quantity. In one such system, dimensions of

different quantities are given in terms of a quantity x as

follows: [position] = [xα]; [speed] = [xβ]; [acceleration]

= [xp]; [linear momentum] = [xq]; [force] = [xt

]. Then (2020)

(a) α + p = 2β

(b) p + q – r = β

(c) p – q + r = α

(d) p + q + r = β

53. The smallest division on the main scale of a Vernier callipers

is 0.1 cm. Ten divisions of the Vernier scale correspond to

nine divisions of the main scale. The figure below on the

left shows the reading of this calliper with no gap between

its two jaws. The figure on the right shows the reading with

a solid sphere held between the jaws. The correct diameter

of the sphere is (2021)

0 1

Vernier scale 0 10

main scale 3 4

0 1 Vernier scale 0

main scale

(a) 3.07 cm (b) 3.11 cm

(c) 3.15 cm (d) 3.17 cm

54. A physical quantity S



is defined as 0 S EB =× µ ( )/ ,   

where E

 is electric field, B

 is magnetic field and m0 is the

permeability of free space. The dimensions of S



are the

same as the dimensions of which of the following quantity

(ies)? (2021)

(a) Energy

Charge Current × (b) Force

Length Time ×

(c) Energy

Volume (d) Power

Area

55. In a particular system of units, a physical quantity can

be expressed in terms of the electric charge e, electron

mass me, Planck's constant h, and coulomb's constant

k =

0

1 , 4πε

where e0 is the permittivity of vacuum. In terms

of these physical constants, the dimension of the magnetic

field is [B] = [el

a [me

]b [h]

g

[k]

d. The value of a + b + g + d

is _______________. (2022)

36 JEE (XI) Module-1 P

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CONCEPT APPLICATION

1. 00 0 A = M LT , B = °°° MLT 2. 1/2 3 A (L T ) − = , |B| = [LT–3] 3. |A| = |L|, |B| = M°L°T2, |C| = M°L°T° 4. L2 T–1

5. (c) 6. (d) 7.

11 1

m Kc h G 22 2 −

= 8. (a) 9. (c) 10. (a) 11. (d) 12. (a) 13. (a)

14. (d) 15. (d) 16. (c) 17. (b) 18. (d) 19. (c) 20. (d) 21. [0.001] 22. [1.004 ± 0.001]

23. (c) 24. (d) 25. (a) 26. (c) 27. [1] 28. (b) 29. [0.005] 30. [0.215] 31. (a)

EXERCISE-1 (TOPICWISE)

1. (b) 2. (a) 3. (a) 4. (c) 5. (b) 6. (c) 7. (b) 8. (b) 9. (d) 10. (a)

11. (d) 12. (d) 13. (c) 14. (c) 15. (b) 16. (b) 17. (b) 18. (d) 19. (b) 20. (a)

21. (c) 22. (c) 23. (c) 24. (a) 25. (b) 26. (a) 27. (c) 28. (a) 29. (d) 30. (b)

31. (b) 32. (b) 33. (c) 34. (b) 35. (c) 36. (a) 37. (b) 38. (a) 39. (b) 40. (a)

41. (c) 42. (c) 43. (a) 44. (a) 45. (a) 46. (a) 47. (b) 48. (b) 49. (b) 50. (b)

51. (c) 52. (c) 53. (a) 54. (a) 55. (b) 56. (b) 57. (c) 58. (c) 59. (c) 60. (a)

61. (b)

EXERCISE-2 (LEARNING PLUS)

1. (b) 2. (d) 3. (c) 4. (b) 5. (c) 6. (d) 7. (a) 8. (a) 9. (c) 10. (d)

11. (d) 12. (c) 13. (d) 14. (d) 15. (d) 16. (a) 17. (d) 18. (c) 19. (d) 20. (c)

21. (b) 22. (b) 23. (a) 24. (d) 25. (a) 26. (b) 27. (b) 28. (b) 29. (d) 30. (a)

31. (c) 32. (a) 33. (a) 34. (b) 35. (b) 36. (c) 37. (b) 38. (a) 39. (a) 40. (b)

41. (c) 42. (a)

EXERCISE-3 (JEE ADVANCED LEVEL)

1. (a,b,c) 2. (c,d) 3. (a,b,c) 4. (a,b,c,d) 5. (a,c) 6. (b,d) 7. (a,b,c,d) 8. (a,b,c,d) 9. (b) 10. (a)

11. (c) 12. (c) 13. (d) 14. (b) 15. (c) 16. (c) 17. (a) 18. (a) 19. (a) 20. (c)

21. (d) 22. (d) 23. (a) 24. (b) 25. (d) 26. (c) 27. (a) 28. (d) 29. [1] 30. [3]

31. [4] 32. [0] 33. [0] 34. [2.6] 35. [9] 36. [92.1] 37. [9.95] 38. [3] 39. [6] 40. [6]

41. [41]

EXERCISE-4 (PAST YEAR QUESTIONS)

JEE Main

1. (a) 2. (c) 3. (b) 4. (c) 5. (b) 6. (b) 7. (c) 8. (b) 9. (b) 10. (c)

11. (a) 12. (d) 13. (c) 14. (a) 15. (a) 16. (d) 17. (c) 18. (b) 19. (b) 20. (c)

21. (b) 22. (c) 23. (d) 24. (b) 25. (a) 26. (b) 27. (c) 28. [14] 29. (c) 30. (c)

31. (c) 32. (b) 33. (d) 34. (c) 35. (c) 36. (a) 37. (d)

JEE Advanced

38. (a) 39. (c) 40. (b) 41. (d) 42. (c) 43. (a,c,d) 44. (b,c) 45. (a,c) 46. (a,b) 47. (b,d)

48. (b) 49. (c) 50. (d) 51. (a,b,d) 52. (a,b) 53. (d) 54. (b,d) 55. [4]

ANSWER KEY