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S. I. UNITS, STATICS

1.1. S.I. UNITS

1.1.1 IntroductionQuantities which can be measured either directly or indirectly are known as physicalquantities. Every physical quantity is measured in terms of some standard which is calledthe unit of that quantity. When we say that a table is 6 m long, we take metre as the unit oflength and 6 as the magnitude of the physical quantity. Length, mass, electric current, etc.which do not depend on other quantities are known as basic or fundamental quantities.Quantities which are defined in terms of fundamental quantities are called derivedquantities or derived units.

The fundamental units of the CGS system are inconveniently small and hence notsuitable in the fields of physical sciences and engineering. To overcome this difficulty, theMKS system was introduced. It is a coherent system based on the fundamental quantitieslength, mass and time. The units of these quantities in the MKS system are metre, kilogramand second respectively.

1.1.2 System International (S I Units)MKS System of units was found more convenient as compared to the other two (CGS andFPS system). However, in engineering applications large and different conversion factorswere needed to make the units correspond to those used in practice.

Besides, physical parameters, associated with other branches of physics, like optics,heat and thermodynamics, etc. could not be conveniently expressed in terms of thesefundamental units.

In order to overcome these difficulties, it was decided in an International Conferenceon Weights and Measures in 1960 to adopt seven fundamental units and two supplementary

2 Engineering Physics

units. This system is commonly referred to as S I (System International of Units). S I is acoherent system based on seven fundamental quantities and two supplementary quantities.The details are given in Table 1.1

The following definitions in the International Conference on Weights and Measureswere adopted.

Metre:Metre:Metre:Metre:Metre: The metre is the length equal to 1650763.73 wave length in vacuum of theradiation corresponding to the specified transition between the levels 2p10 and 5d5 of theKrypton atom.

Basic Physical Quantities Base Unit Symbol Dimension

Length metre m LMass kilogram kg MTime second s TIntensity of Electric current ampere A AThermodynamic temperature kelvin K θLuminous intensity candella cd IQuantity of a substance mole mol n

Supplementary S I Units

1. Plane angle radian rad –2. Solid angle steradian sr –

Kilogram:Kilogram:Kilogram:Kilogram:Kilogram: Kilogram is the unit of mass which is equal to the mass of the internationalprototype of the kilogram, placed at the International Bureau of Weights and Measures inSevres (near Paris) in France.

Time:Time:Time:Time:Time: Second is the duration of 9192631770 periods of the radiation corresponding to thetransition between the two hyperfine levels of the ground state of the atom caesium.

Ampere: Ampere: Ampere: Ampere: Ampere: The ampere is that constant current, which if maintained in two straight parallelconductors of infinite length of negligible circular cross-section, and placed one metreapart in vacuum, would produce between these conductors a force equal to 2×10–7 permetre of length.

3S I Units, Statics

Kelvin:Kelvin:Kelvin:Kelvin:Kelvin: Kelvin the unit of thermodynamic temperature is 1

th273.16

fraction of

thermodynamic temperature of the triple point of water.

Celsius Kelvin

100 373

0 273

–273 0

Candela. (Luminous intensity): Candela. (Luminous intensity): Candela. (Luminous intensity): Candela. (Luminous intensity): Candela. (Luminous intensity): This is a fundamental physical quantity used inphotometry.

Mole (mol): Mole (mol): Mole (mol): Mole (mol): Mole (mol): A mole of a substance contains Avogadro’s number of particles of thatsubstance. i.e., 6.02217×1023 particles.

Apart from these, the two supplementary units are defined in the following way:

Radian:Radian:Radian:Radian:Radian: This is the unit of plane angle and is given by

360 ' '1 radian 57 14 44 .2

°= = °π

Steradin: Steradin: Steradin: Steradin: Steradin: This is the unit of a solid angle which is defined as the ratio of the interceptedof the spherical surface to the square of the radius. Thus the solid angle made by a sphereat its centre is

2

2

44 sr.

π = πrr

1.1.3 Conventional Standard Quantity (Unit)In order to measure any physical quantity, some definite and convenient quantity of thesame kind is taken as the standard, in terms of which the quantity as a whole is expressed.The conventional standard quantity is termed as unit.

For example, when we say that mass of a body is 100 kg, it implies that a kg (i.e., akilogram) is used as the standard (or Unit) for measurement and the mass of the body is100 times the unit. If we simply say kg not using 100 with it, then it has no meaning.Similarly if we simply say 100 not using kg with it, then also it has no meaning. Thus inorder to express a physical quantity or parameter, its numerical value must be followed bythe unit in which it is measured.

A standard adopted as a unit should possess the following features:

i. It should be universally acceptable and be of reasonable size,ii. Its magnitude should be definite and definable,

iii. Its magnitude should not change with time, temperature and pressure andvi. It should be reproducible.


1.1.4 Some Ponderous PointsWhile using S I Units the following two points have to be kept in mind.

i. The letter s is not added to a symbol to indicate its plurality. For example, 10 kgis not written as 10 kgs. The letter s is the symbol for second and may createconfusion if used for plural.

ii. Units with names of the scientists should not start with a capital letter as is donewhile writing the name of a person i.e., write newton (or N) not Newton; watt(W) not Watt; joule (or J) and not Joule.

1.1.5 Advantages of S I UnitsThe main advantages of S I Units are now summarized:

i. It is comprehenceIt means that its seven base units cover all disciplines of science and technology.

ii. The system is coherentThe unit of any derived physical quantity can be obtained as a product or quotientof two or more fundamental units.

iii. The SI units are internationally used.

1.1.6 Prefixes and UnitsWhen the magnitude of a physical quantity is very large or very small, prefixes are used toexpress them more conveniently. The following table gives commonly used prefixes.

Multiple Prefix Symbol Fraction Prefix Symbol

1012 tetra T 10–1 deci d

10–2 centi c

109 giga G 10–3 milli m

106 mega M 10–6 micro µ

103 kilo k 10–9 nano n

102 hecto h 10–12 pico p

10–15 femto f

101 deca da 10–18 atto a

1.1.7 Some Derived S I UnitsVery important derived S I Units of commonly used physical parameters are given inTable 1.3

5S I Units, Statics

!

Quantity Unit Short form Symbol

Area square metre m2

Volume cubic metre m3

Frequency herts s–1 Hz

Density (mass density) kilogram per cubic metre kg m–3

Velocity metre per socond m s–1

Angular velocity radian per socond rad s–1

Acceleration metre per socond square ms–2

Force newton kg ms–2 NPressure pascal N/m2 PaViscosity pascal second Ns/m2

Work, Energy and joule N m quantity of heat

Power watt J/s

Electric charge coulomb A s

Potential, potential difference, volt W A–1 V electro motive force

Electric field newton/coulomb N/C

Resistance ohm V/A ΩCapacitance farad A s/V FManetic flux weber V s WbInductance henry V s/A HMagnetic flux density tesla Wb/m2 TMagnetic field strength newtom/ampere metre kg/As2

Luminous flux lumen cd sr lmLuminance candela/m2 cd/m2

Illumination lux lm/m2

Wave number per metre m–1

Entropy joule/kelvin J/KSpecific heat joule per kilogram per kelvin J/kg KThermal conductivity watt metre per kelvin W/m KRadiant intensity watt per sterdian W/srActivity (or a radioactive per second s–1

source)

Permeability newton per ampere square N/A2

1.1.8 Dimensions of Physical QuantitiesIt has been pointed out earlier that a physical quantity can be expressed in terms of acombination of fundamental units of mass (M), length (L), time (T), temperature (θ), current(A) and light intensity (I). Thus velocity and acceleration of a body can be written as:


velocity = 1 1displacement°

time− −= = =L

LT M LTT

acceleration =1

2 2velocity°

time

−− −= = =LT

LT M LTT

The equation velocity M° LT–1 is called a dimensional equation, whereas the expressionM° LT–1 is called dimensional formula of velocity. This velocity is said to have zerodimension is mass, 1 dimension length and –1 dimension is time.

1.1.9 Dimension of Some Common Physical ParametersPhysical Quantity Relation with other Quantities Dimesional Formula

Volume length × breadth × height L3

Density mass/volume ML–3

Force mass × acceleration MLT–2

Stress force/area ML–1 T–2

Power work/time ML2 T–3

Surface tension force/length MT–2

Angle lengths of arc/radius No dimension

Coefficient of thermal

× ×heat ×distance

area temp timeMLT–3 θ–1

conductivity

Resistance volt/ampere ML2 T–3 A–2

1.2 STATICS1.2.1 IntroductionPhysical quantities, like mass, time, temperature energy, charge which are specified bytheir magnitudes only are called scalars. Scalars can be added, subtracted and multipliedalgebraically.

Physical quantities like displacement, that are specified by a magnitude and a directionare called vectors. Another commonly known vector is the velocity of a body, which isdescribed by the speed of the body in a definite direction. Force, pressure, momentum,torque, electric and magnetic fields are some other examples of vector quantities.

1.2.2 VectorsVector quantities possess both magnitude and direction. They are added and subtractedaccording to special laws such as parallelogram law of addition e.g., force, velocity, currentdensity, intensity of electric field, etc. It should be noted that all physical quantities havingboth magnitude and direction are not necessarily vectors. All vectors obey the laws ofvector algebra. For example, the electric current and time have both magnitude anddirection; but they are scalars, because they do not obey the laws of vector algebra.

7S I Units, Statics

Representation of a vector: Representation of a vector: Representation of a vector: Representation of a vector: Representation of a vector: A vector quantity is represented by a straight line with anarrow head; the length of the line representing its magnitude and the arrow head indicatingits direction

→

A

1.2.3 Addition of Two Vectors(i) When vectors are acting in the same direction:

The magnitude of the resultant vector is equal to the sum of the magnitudes ofthe two vectors and the direction is the same as that of the two given vectors.

; | | | | | |= + = +

R A B R A B

(ii) When two vectors are acting in opposite directions:The magnitude of resultant vector is the difference between the two vectors andthe direction is that of the bigger vector

( ); | | | | | |= + − = −

R A B R A B

(iii) When two vectors are inclined at an angle:The sum of two vectors can be determined by

(a) the law of triangle of vectors or (b) the law of parallelogram of vectors(a) Triangle Law of Vectors(a) Triangle Law of Vectors(a) Triangle Law of Vectors(a) Triangle Law of Vectors(a) Triangle Law of Vectors

If two sides of a triangle represent two given vectors in the magnitude and directionand in the same order, then the third side of the triangle in the reverse order represents thevector sum of the vectors.

B

O

θ

R = P + Q→ → →

→

→P→ A

Q→

If OA represents vector

P and AB represents vector

Q in the anticlockwise order, then

O B in the clockwise order represents the vector sum of

P and

Q

= +

R P Q

(b) Parallelogram Law of Vectors (Law of Parallelogram of Forces)(b) Parallelogram Law of Vectors (Law of Parallelogram of Forces)(b) Parallelogram Law of Vectors (Law of Parallelogram of Forces)(b) Parallelogram Law of Vectors (Law of Parallelogram of Forces)(b) Parallelogram Law of Vectors (Law of Parallelogram of Forces)If two adjecent sides of a parallelogram represent two vectors in magnitude and

direction, then the diagonal of the parallelogram starting from the point of intersection ofthe two vectors represent the vector sum of the two vectors.


R = P + Q→ → →→

→O

Q→

BC

AP→

→

If OA represents

P and OB represents

O , then the vector sum

R is represented by thediagonal

O C of the parallelogram OACB.

(c) Analytical Method of Vector Addition(c) Analytical Method of Vector Addition(c) Analytical Method of Vector Addition(c) Analytical Method of Vector Addition(c) Analytical Method of Vector AdditionDraw BC perpendicular to OA produced. Let α–be the angle which the resultant makes

with the direction

P .

B

O

θ

R = P + Q→ → →

→

→

P→ A

Q→

C

α

In the triabgle OCB, we haveOB2 = OC2 + CB2 = (OA+AC)2 + CB2

= OA2 + 2OA×AC + AC2+ CB2

= OA2 + 2OA×AC + AB2

R2 = P2 + Q 2 + 2PQ cos θ [since AC = AB cos θ]

R = 2 2 2 cosP Q PQ+ + θ

1.2.4 Equilibrium of a Body Under the Action of Three Concurrent ForcesConsider two forces P and Q acting on a body at a point O. Their resultant can be obtainedby the parallelogram method or by the triangle method which we have studied earlier. Tobalance the resultant, we would require an equal and opposite force. Thus the forces P andQ can be balanced by a force equal in magnitude to the resultant but opposite in direction.The force is called equilibrant. The equilibrant of a set of forces acting on a body is thatforce which along with the set of forces keeps the body in equilibrium.

In Fig. 1.1 the equilibrant of the forces P and Q acting at the point O is R. Its magnitudeis equal to that of the resultant in the opposite direction.

9S I Units, Statics

P Q

→→

R

A

B

C→ →

→

A

B

C→

→

→

"#$%

Let us now apply the triangle law of vectors to find the resultant of P and Q. In thefigure, the forces P and Q are represented by the sides AB and BC of triangle ABC. Then theresultant is obviously given by AC, both in magnitude and direction. To keep the body inequilibrium, the third force to be applied should be equal to that represented by AC but inthe reverse direction. Hence in a vector diagram, the position of the arrow head will tell uswhether a given vector is resultant or equilibrant. AC in figure is the resultant of P and Qbut CA is the equilibrant of the same set. Hence the condition of equilibrium of a bodyunder the action of three concurrent forces can be summarized in the form of a law knownas law of triangle of force.

Law of Triangle of ForcesLaw of Triangle of ForcesLaw of Triangle of ForcesLaw of Triangle of ForcesLaw of Triangle of Forces

If three forces acting at a point be represented in magnitude and direction by the sidesof a triangle taken in order, they will be in equilibrium.

Converse of the law of triangle of forces is also true. If three forces acting at a point bein equilibrium they can be represented in magnitude and direction by the sides of a trianglewhich is drawn so as to have the sides respectively parallel to the directions of the forces.

Lami’s TheoremLami’s TheoremLami’s TheoremLami’s TheoremLami’s Theorem

If three forces acting on a particle keep it in equilibrium, each force is proportional tothe sine of the angle between the other two.

P, Q and R three forces acting at a point keeping it in equilibrium, Fig. 1.2. If α, β and γare the angles apposite to each of them respectively,

sin sin sin= =

α β γP Q R


Q

R

αβ

γP C

→

A

B

C→

→

→z

γ

α

x

βy

This law is a direct consequence of the triangle law. Since the forces are inequilibrium, they can be represented by the sides of the triangle ABC taken in order. Ageneral property of any triangle is that each side is proportional to the sine of theangle opposite to it. Thus in the triangle ABC drawn with the sides parallel to theforces P, Q, and R,

sin sin sin= =AB BC CA

x y z

Here x, y and z are the angles of the triangle ABC. But by the triangle law of forces, thesides of the triangle are proportional to the respective force. From the Fig. 1.2

sin x = sin (180 – α) = sin α

sin y = sin (180 – β) = sin β

sin z = sin (180 – γ) = sin γ

Hencesinα

P=

sin sin=

β γQ Q = constant

Thus, if 3 forces acting on a particle are in equilibrium, each force is proportional to thesine of the angle between the other two.

1.2.5 Experimental Verification of the LawsThe simple arrangement used to verify the above discussed laws is generally calledparallelogram law apparatus as shown in Fig. 1.3.

"

11S I Units, Statics

P

Z

R Q

P Q

X Y

→→→

R

"

Two smooth aluminium pulleys are fixed on a vertical drawing board as shown in thisfigure. A long string is passed over the pulleys. Another string is knotted at a point O asindicated. A sheet of paper is pinned to the board behind the strings. Weights P, Q and Rplaced at the ends of the strings and adjusted such that the knot O rests in a convenientposition against the paper. The position of the point O and the directions of the strings OX,OY and OZ are marked on the paper. The paper is taken out and the experiment is repeatedfor different values P, Q and R.

Verification of the Parallelogram LawVerification of the Parallelogram LawVerification of the Parallelogram LawVerification of the Parallelogram LawVerification of the Parallelogram Law

Taking a suitable scale, lengths OA, OB and OC are selected along the directions OX,OY and OZ. The parallelogram OADB is completed and the diagonal OD is drawn.

A

Q

C

D

x y

B

O

×

Z

r

O 1

C 1

A 1

q

p

"&


Since OC represents the equilibrant of P and Q it should be equal and opposite to theresultant of P and Q. In all cases, it will be seen that OC = OD and angle COD = 180°. Thisverifies the parallelogram law.

Verification of Lami’s TheoremVerification of Lami’s TheoremVerification of Lami’s TheoremVerification of Lami’s TheoremVerification of Lami’s TheoremTo verify Lami’s theorem, the angles α, β and γ, opposite to the forces are measured

from the traces of the string position. The ratios , andsin sin sinα β γ

P Q R are calculated. In all

cases sin sin sin= =

α β γP Q R

which verifies the theorem. The experiment is repeated for

different values of P, Q and R.

Forces Anglesαsin

Pβsin

Qγsin

R

P Q R α β γ

Verification of Triangle LawVerification of Triangle LawVerification of Triangle LawVerification of Triangle LawVerification of Triangle Law

The law of triangle of forces is verified indirectly by verifying the converse of the law.If we have 3 forces in equilibrium, they can be represented by the sides of any trianglewhich is drawn so as to have its sides parallel to the direction of the forces. Using a pair ofset squares draw lines, O1 A1, A1 C1 and C1 O1, parallel to the lines of action of the three

forces P, Q and R respectively to form the triangle O1 A1 C1, 1 1 1 1 1 1

, andP Q R

O A A C C Oare

calculated. It will be seen that these ratios are the same. Hence the sides of the triangle

represent the forces in magnitude. Since the sides are parallel to the lines of action of theforces, they represent magnitudes as well as directions. The experiment is repeated fordifferent values of P,Q and R.


SHORT QUESTIONS1 Mark Questions1 Mark Questions1 Mark Questions1 Mark Questions1 Mark Questions

1.1 Express 25°C in kelvin scale.1.2 What are the two supplementary SI units?1.3 Give the dimensions for work done.1.4 The wave length of sodium light is 589.3 nanometre. Express it in metre.1.5 What is the unit of electric charge?1.6 What is the dimension of the physical quantity stress?1.7 The power of the given electric lamp is 100 Watts. Is it the correct way of

writing this physical quantity?1.8 Strain in elasticity is a physical quantity without any dimension. Is it right

or wrong?1.9 Give two scalar quantities.

1.10 Force acting on unit area is called ...............1.11 Give the value of Avogadro’s number.1.12 Represent a vector.1.13 Write zero degree kelvin in celsius scale.

2 Marks Questions2 Marks Questions2 Marks Questions2 Marks Questions2 Marks Questions

1.1 Explain radian the unit of plane angle. Express one radian in degree.

*1.2 What are the two parameters required to represent a physical quantity?

1.3 Get the unit and dimensions of acceleration.

*1.4 MLT–2 is the dimensions of a physical quantity. Name the physical quantity.From that get the dimension of pressure.

Forces Sides of triangle1 1

PO A 1 1

QA C 1 1

RO C

P Q R O1A1 A1C1 O1C1


1.5 How will you add when two vectors act in the same direction?

*1.6 Define Lami’s theorem.

1.7 Explain the apparatus required to verify parallelogram law of forces

*1.8 Get the unit of thermal conductivity.

1.9 State the converse of triangle law of forces.

1.10 Express 60° in radian measure.

*1.11 Convert 6π

radian into degree.

Answers to Starred QuestiAnswers to Starred QuestiAnswers to Starred QuestiAnswers to Starred QuestiAnswers to Starred Questionsonsonsonsons

1.2 Let us assume the distance between two points is 3 metre. If the magnitude3 is absent, then the representation of the physical quantity is meaningless.Similarly if metre is absent, then also the true meaning is unattainable. Soboth the magnitude and the unit are required t represent a physical quantity.

1.4 The general expression for force isF = ma

The dimension are m → M & a = LT

1

T2−=F MLT Answer

Force for unit area is pressure

i.e. P =2

2

−

=F MLTA L

--1 2Thus the dimension of pressure is −ML T Answer

1.6 If three forces acting on a particle keep it in equilibrium, each force isproportional to the sine of the angle between the other two.

1.8 For solving this one requires the general definition which is explainedbelow:

In the steady state the quantity of heat conducted through a rod ofuniform cross-section is depending on the area of cross-section, time ofconduction and temperature gradient.

i.e. Q A

t

1 2θ − θ d


i.e. 1 2( )θ − θλ Atd

λ 1

1 2( )θ

θ − θAt

dWhen we apply the units, the unit of λ coefficient of thermal conduction is

2

joule metresecm K×

× ×=

1joulesec

−× mK

= W m–1 K–1

This is the unit of λ.

1.11 2π radian corresponds to 360

6π

radian corresponds to 360

306 2

× π = °× π

REVIEW QUESTIONS

10 10 10 10 10 Marks QuestionsMarks QuestionsMarks QuestionsMarks QuestionsMarks Questions1.1 (a) Describe the conventional standards used to represent a physical quantity.

(b) How metre and second in S I units are defined?(c) What are the main advantages of S I units?

1.2 (a) Explain the dimensions of physical quantities with suitable examples.(b) Obtain the unit and dimensions of resistance.(c) Magnitude and unit are essential to represent any physical quantity. Explain.

1.3 (a) Distinguish scalar and vector quantities with suitable examples.(b) Explain addition and subtraction vectors acting in the same direction and

also in the opposite directions.(c) Explain parallelogram law of forces.

1.4 (a) State Lami’s theorem.(b) Explain how Lami’s theorem is verified.(c) If the resultant of two forces 6N and 8N is 12N, find the angle between the

two forces.

1.5 (a) State parallelogram law of forces.


(b) Describe with a neat diagram an experimental arrangement to verify this law.

(c) Discuss how the law of parallelogram of forces is verified.

PROBLEMS AND SOLUTIONS1.11.11.11.11.1 Derive the dimensional formula for the kinetic energy of a body.Derive the dimensional formula for the kinetic energy of a body.Derive the dimensional formula for the kinetic energy of a body.Derive the dimensional formula for the kinetic energy of a body.Derive the dimensional formula for the kinetic energy of a body.

Solution:

212

mv is the expression for kinetic energy. The dimensional formula for velocity is LT–1;

hence that for K-E is ML2T–2.

1.21.21.21.21.2 The value of g in S.I. unit is 9.8 m/sThe value of g in S.I. unit is 9.8 m/sThe value of g in S.I. unit is 9.8 m/sThe value of g in S.I. unit is 9.8 m/sThe value of g in S.I. unit is 9.8 m/s22222. Find its value in ft/s. Find its value in ft/s. Find its value in ft/s. Find its value in ft/s. Find its value in ft/s22222 in FPS system in FPS system in FPS system in FPS system in FPS system

Solution:

The dimensions of g are LT–2. If the units and the numerical magnitudes of a physicalquantity in two systems of unit are U1, ni and U2 n2 respectively, then

n1 U1 = n2 U2

thus n1 (L1 T1–2) = n2 (L2 T2

–2)

n1 = n2

2

2 2

1 1

−

L TL T

= 9.8 × 21 metre 1sec

1 ft 1sec

= 9.8 × 3.281 ft

1 ftn1 = 32.15 ft/s2

21 32.15 ft/s=n Answer

1.31.31.31.31.3 The wave length associated with the green line of mercury spectrum is 546 nm.The wave length associated with the green line of mercury spectrum is 546 nm.The wave length associated with the green line of mercury spectrum is 546 nm.The wave length associated with the green line of mercury spectrum is 546 nm.The wave length associated with the green line of mercury spectrum is 546 nm.Obtain the dimension of Obtain the dimension of Obtain the dimension of Obtain the dimension of Obtain the dimension of PlanckPlanckPlanckPlanckPlanck constant. Calculate also the frequency of the line. constant. Calculate also the frequency of the line. constant. Calculate also the frequency of the line. constant. Calculate also the frequency of the line. constant. Calculate also the frequency of the line.

Solution:

E =c h

ν =λ

h

The dimension of energy F × distance i.e E → MLT–2 × L and ν → T–1.


Thus, the dimension of h is 2 2

1

−

−

ML TT

= ML2T–1

ν =8

9

3 10546 10−

×=λ ×c

145.5 10ν = × Hz Answer

1.41.41.41.41.4 Explain what is meant by dimensions? Write down the dimensional formulaeExplain what is meant by dimensions? Write down the dimensional formulaeExplain what is meant by dimensions? Write down the dimensional formulaeExplain what is meant by dimensions? Write down the dimensional formulaeExplain what is meant by dimensions? Write down the dimensional formulaefor the following physical quantities:for the following physical quantities:for the following physical quantities:for the following physical quantities:for the following physical quantities:(((((aaaaa) momentum () momentum () momentum () momentum () momentum (bbbbb) surface tension () surface tension () surface tension () surface tension () surface tension (ccccc) angular velocity) angular velocity) angular velocity) angular velocity) angular velocity

Solution:

It has been pointed out in many places that a physical quantity can be expressed interms of a combination of fundamental units of mass (M), length (L), time (T), temperature(θ), current (A) and light intensity (I). Thus, velocity and acceleration of a body can bewritten as

--1

-1--2

Displacement LengthVelocity = = =

Time TimeVelocity

Acceleration = = =Time

LT

LTLT

T

Answer

The equatin for veleocity namely LT–1 is called a dimensional equation, whereas theexpresssion M° LT–1 is called dimensional formula. Thus, velocity is said to have zerodimension in mass, 1 in length and –2 in time.

1

22

1

(a) Momentum : mass velocity :

force(b) Surface tension : : :

lengthangle angle

(c) Angular velocity : : :time time

−

−−

−

× MLT

MLTMT

L

TAnswer

1.51.51.51.51.5 Deduce the dimensions of universal constant of gravitation.Deduce the dimensions of universal constant of gravitation.Deduce the dimensions of universal constant of gravitation.Deduce the dimensions of universal constant of gravitation.Deduce the dimensions of universal constant of gravitation.

Solution:

From Newton’s law of universal graviation, we know that force F, between two bodiesof masses m1 and m2, kept at a distance d apart is given by

F = 1 2

2

m mG

d

G =2

1 2

F dm m

×


where G is a constant called gravitational constant

G =2 2

2

− ×MLT LM

1 3 2: − −G M L T Answer

1.61.61.61.61.6 Get the unit of electrical resistance and its dimension. Also that of electricGet the unit of electrical resistance and its dimension. Also that of electricGet the unit of electrical resistance and its dimension. Also that of electricGet the unit of electrical resistance and its dimension. Also that of electricGet the unit of electrical resistance and its dimension. Also that of electricchargechargechargechargecharge

Solution:

Ohm’s law states

V = IR

:volt/ampere= VR

I Answer

Dimension of R

R =VI

But V = El

with F = q E; or E = Fq

E =2 2−

→ →F MLT MLTq I t A T

V =2−MLT L

A T

Thus, R =2−MLT L

A T A = MA–2 L2T–3

i.e. the dimension of resistance

2 –3 –2

Dimension of Electric Charge, = It = ML T A

q AT Answer


1.71.71.71.71.7 A famous relation in physics relates moving mass m to rest mass mA famous relation in physics relates moving mass m to rest mass mA famous relation in physics relates moving mass m to rest mass mA famous relation in physics relates moving mass m to rest mass mA famous relation in physics relates moving mass m to rest mass mooooo in terms in terms in terms in terms in terms

of its speed of its speed of its speed of its speed of its speed vvvvv and the speed of light and the speed of light and the speed of light and the speed of light and the speed of light ccccc. A boy writes wrongly as . A boy writes wrongly as . A boy writes wrongly as . A boy writes wrongly as . A boy writes wrongly as 21

omm=

- v.....

Guess where to putGuess where to putGuess where to putGuess where to putGuess where to put c c c c c?????

Solution:

The dimensions of physical quantities appearing on both sides must be possible. This

is possible only when you put 2

2

vc

instead of v2. The correct equation is.

2 21 ( /om

mv c

=−

Answer

1.81.81.81.81.8 Get the dimension of gas constantGet the dimension of gas constantGet the dimension of gas constantGet the dimension of gas constantGet the dimension of gas constant

Solution:

R =2 3

2

− = θ

PV MLT LT L

R → ML2T–2θ–1

2 2 1− −→ θR ML T Answer

1.91.91.91.91.9 Convert one newton into dynesConvert one newton into dynesConvert one newton into dynesConvert one newton into dynesConvert one newton into dynes

Solution:

Let n1 newton be equal to n2 dynes

n1 (N1) = n2 (N2)

n1 (M1L1T1–2) = n2 (N2) (M2L2T2

–2)

n2 =1 1 2

1 1 11

2 2 2

M L Tn

M L T

−

=1 1 2

1 1 11

2 2 2

M L Tn

M L T

−

= 1×1000×100 = 105

51 newton 10 dyne= Answer


1.101.101.101.101.10 Two forces 15N and 10N act at a point at an angle 30°. Find the resultant.Two forces 15N and 10N act at a point at an angle 30°. Find the resultant.Two forces 15N and 10N act at a point at an angle 30°. Find the resultant.Two forces 15N and 10N act at a point at an angle 30°. Find the resultant.Two forces 15N and 10N act at a point at an angle 30°. Find the resultant.

Solution:P = 15 N Q = 10N Q = 30°

O

Q

C

AP

D

α

D

θθ

R

R2 = P2 + Q2 + 2PQ cos θ= 152 + 102 +2×15×10 cos 30°

or

24.2 newtonR = Answer

1.111.111.111.111.11 The resultant of two unequal forces acting at an angle 150° is perpendicularThe resultant of two unequal forces acting at an angle 150° is perpendicularThe resultant of two unequal forces acting at an angle 150° is perpendicularThe resultant of two unequal forces acting at an angle 150° is perpendicularThe resultant of two unequal forces acting at an angle 150° is perpendicularto the smaller of the forces. If the larger force is 3 newton, find out the smallerto the smaller of the forces. If the larger force is 3 newton, find out the smallerto the smaller of the forces. If the larger force is 3 newton, find out the smallerto the smaller of the forces. If the larger force is 3 newton, find out the smallerto the smaller of the forces. If the larger force is 3 newton, find out the smallerforce and the resultant.force and the resultant.force and the resultant.force and the resultant.force and the resultant.

Solution:

D C

Q

A B3N

60°

90°

R

→

→

Let Q and R represent the smaller force and the resultant respectively. Q is perpendicularto R.

From triangle ACD

sin 60° =3

=AD QCD

or Q = 3 sin 60 = 2.6 newtonAlso, R2 + Q2 = 32


or R2 = 9 – Q2

R2 = 9 – 2.62

R2 = 9 2.62−

1.5 NR = Answer

1.121.121.121.121.12 A body is acted on by two forces 150 N towards north and another of 200A body is acted on by two forces 150 N towards north and another of 200A body is acted on by two forces 150 N towards north and another of 200A body is acted on by two forces 150 N towards north and another of 200A body is acted on by two forces 150 N towards north and another of 200 NNNNNtowards west. Find the resultant in magnitude and direction.towards west. Find the resultant in magnitude and direction.towards west. Find the resultant in magnitude and direction.towards west. Find the resultant in magnitude and direction.towards west. Find the resultant in magnitude and direction.

Solution:

150N

N

R

α200N

R2 = 2002 + 1502

= 62500

250 newtonR = Answer

tan α =150

0.75200

=

36.9α = ° Answer

1.131.131.131.131.13 Find graphically the resultant of two forces 10Find graphically the resultant of two forces 10Find graphically the resultant of two forces 10Find graphically the resultant of two forces 10Find graphically the resultant of two forces 10 N and 5N and 5N and 5N and 5N and 5 N acting at a pointN acting at a pointN acting at a pointN acting at a pointN acting at a pointinclined at an angle 60°. Verify by calculationinclined at an angle 60°. Verify by calculationinclined at an angle 60°. Verify by calculationinclined at an angle 60°. Verify by calculationinclined at an angle 60°. Verify by calculation

Solution:

O A

B R

→

→

→

60°

(2”) OA corresponds to 10 units O R (2.7) corresponds to

02.7

21 × = 13.5

13.5=OR Answer

s. i. units, statics - newagepublishers.comnewagepublishers.com/samplechapter/001507.pdf · very...

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