xii physics Chapter 1

ICMS College System for Boys & Girls 1

CHAPTER # 1PHYSICS AND MEAURMETN

INTRODUCTION TO PHYSICS:The branch of & science which deals with the study of Properties of matter, Energy and theirmutual relationship are called Physics.

EXPLANATION: Here the properties of Matter Electrical properties, magnetic properties dc andEnergy means different form of energies like, Thermal energy, Electrical energy, Chemical energyetc all these are mutual relationship of energy and matter i.e. E=mc2 where Erepresents the energy and m is the quantity of matter (mass).Physics explain the Natural Phenomena in terms of Laws and Principles. The use of these laws andPrinciples are visible in everywhere in our daily life. The fastest plane, from the shape of thumb Pinto the shape of missile. The scope Physics is so wide that it includes the motion of bodies from tiniestsub atomic particles to huges galaxies.

BRANCHES OF PHYSICS: -The basic branches of Physics are

i) Mechanicsii) Field Theoryiii) Heat and thermodynamicsiv) Opticsv) Electricityvi) Magnetismvii)Atomic Physicsviii) Nuclear Physics etc.Each branch of Physics has further sub branches. For example:Mechanics has the sub branches: static, Dynamics Ballistics, hydrodynamics and Aerodynamics.Physics has also links to the other fields e.g. Medical Physics, Bio Physics, and Physical Chemistryetc.

MEASURMET:The comparison of unknown quantity with some standard quantity is called comparison.

EXPLANATION:Physics is an experimental and quantities science, which requires measurement, and we use numberto describe the results of a measurement. The number with proper unit that is used to describe aPhysical Phenomena quantitively is called Magnitude.

UNIT:The standard with which the things are compared is called Unit. Measurement has basic importancein our daily life. We cannot buy or sale any thing until and unless we measure it and formeasurement a common standard is needed, called Unit.PHYSICAL QUANTITES:The quantity which can be observed and Measured is called Physical quantity.Physical quantities means, the mass of an object, height of a person, temperature of the classroometc. Physical quantity is usually expressed as the product of a number with proper unit, whichtogether is called the Magnitude of a physical quantity.For Example: Length =5m

Force = 16 N etcPhysical quantities are classified into there (Now these are two) Categories; base, derived andSupplementary


BASE OR FONDAMENTAL QUANTITIES:The minimum numbers of those Physical quantities which are simplest and prominent, in terms ofwhich other quantities are defined and derived are called Base Quantities.The measurement of a Physical quantity depend upon the other fundamental quantities e.g. tomeasure the Area, Length is need so length is simplest as compare to Area and volume.Example: length, mass etc.

DERIVED QUANTITIES:All those quantities whose definition and derivations are based on fundamental quantities are calledDerived Quantities.We know that so Physical quantities are more complex containing two or more than two basequantities, For example, Speed is derived Physical quantity because it can be expressed as the ratioof the length and time,i.e. Speed = Distance / Time = length / Time. Similarly the force; F=ma = Mass. length /time2.

SUPPLEMENTRY QUANTITIES:The two geometrical quantities which are simplest but can be derived from the other quantities arecalled Supplementary Quantities.The general conference on weight, and measurement has not yet classified. The two units either baseor derived units. This class contains only two units of purely geometrical quantities, which are planeangle and solid angle (Now this is included in derived units.)

SYSTEM OF UNITS; SI UNITS:A complete set of units both fundamental and derived units is called system of units

SYSTEM INTERNATIONAL (SI):A complete set of units consists of seven fundamental and large number of derived quantities iscalled system international.We use a quantity, i.e. Length to measure, we run into the difficulty of establishing a standard. Sothose who have the need of comparison one length with another will agree on the measurement butthe other person not necessary to accept it. There for a common standard is needed which is easy toaccessible and not variable to change with the passage of time. This was the need to adopt aninternational system of units.

BASE UNITS:The General conference on weight and measurement, selected seven units i.e. meter (m), kilogram(kg), second (s), ampere (A), Kelvin (K), mole (mol) and candela (cd) for seven base quantities,length, mass, time, current, temperature, amount of substance and Luminous intensity receptivity.These units are defined a:

STANDARD UNIT OF LENGTH; METER (m):The distance traveled by light in vacuum in 1/ 2.99*108 seconds is called one at the beginning themeter was defined as The distance between two finely drawn lines on platinum-iridium rod at 0 C.After this if was redefined as The distance occupied by 1,65076373 wave lengths of orange redlight emitted by Kr-86 atoms.

THE STANDARD OF MASS; KILOGRAM (Kg):The mass of platinum - iridium cylinder having height and diameter 39 Cm is called one Kilogram


1 radian

A

R

B

C

r

r

o

r

THE STANDARD OF TIME; SECOND (S):At the beginning the standard of time was mean solar second defined as 1S(M.S.S)=1/24*1/60*1/60 of mean solar day.This standard depends upon the rotation of the earth, which changes, with the passage of time.Therefore it was redefined as 1 second = the duration of 9,192,631,770 vibration of CS- 133.

THE STANDARD UNIT OF TEMPERATURE; KELVIN (K):One Kelvin is the fraction of 1/273.16 of the thermodynamic temperature of means temperature atwhich water co-exist in there states i.e. solid, Liquid and gaseous in state.

THE STANDARD UNIT OF CURRENT; AMPERE (A):One ampere is that constant current which is maintained in two wires of negligible cross-sectionplaced in vacuum 1m apart, expression a force of attraction or repulsion of 2*10-7 N/m of length.

THE STANDARD UNIT FOR AMOUT OF SUBSTANCE; Mole (mol): The amount of substance which contains as many elementary entities or There are atoms in 0.012kg of 6C12 atoms. Here the entities mean atoms 10ns, electrons etc. One mole of any substancecontains 6.0225*1023 entities.

THE STANDARD OF LUMINOUS TNTENSITY; CANDELA (cd):The Luminous intensity in the perpendicular direction of a surface of 1/600,00 m2 of a black bodyradiator at the temperature of freezing point of platinum under standard pressure (101.325 N/m2).

THE SUPPLEMENTRY UNITS:The units has not yet classified either they are base or derived, called supplementary units. This classcontains only two units of purely geometrical quantities, which are plane angle and solid angle.

RADIAN:The SI unit for plane angle is radian defined as The angle formed by on Arc at the center of thecircle equal to the radius of the circle.In the Fig: the angle AOB is represented by ; measured in radians and it is the length of an Arcdivided by the radius r; = s / r.As is the ratio between two lengths. So it is a it is a number. The number of is the ratio of thecircumference of a circle to the diamet3r of same circle.

So if the Arc = 1 radii --------------- = 1 radArc = 2 radii --------------- = 2 radArc = 2 radii --------------- = 2 rad

So 2 radian = 360 0o = 1 revolution1 radian = 360o/ 2 = 57.3o.

Degree is the most familiar unit but not SI. Degree is the angleformed at the center of a circle by an Arc of length 1/360of the Circumference circle.

A minute is 1/60 a degree and a second is 1/60 of a minute.


r1 sr

r1S

STERDIAN:The SI unit of of solid angle is steradian and is defined as the angle subtended at the center of asphere by an area of its surface equal to the square of its radius of that sphere.Steradian is also are divided by an are so it is a number:The surface are of a closed sphere of radius r is 4r2.So solid angle subtend4ed by a closed sphere at its center is 4r2 divided by r2.

= 4r2 /r2 = 4 SrInfact any closed surface subtends a solid angle of 4 Sr at anyof its interior point.

DERIVED UNITS.Any unit which is obtained from the multiplication or division of base units are called Derivedunits.Derived units are always defined in terms of base units.For example a unit length (1m) multiplied with unit length (1m) gives area i.e.

1m * 1m = 1m2Similarly a unit length divided by a unit time is a unit of speed (1m/s).Densities, acceleration, voltage and charge are derived quantities and their units, are Kg.m-3, ms02volt and coulomb receptivity.Some of the derived quantities are given in the table.

Physically quantity Unit Symbol In terms of base unitsForce Newton N Kg ms-2Work Joule J Nm=kg m2 s -2Power Watt W Js-1=kg m2 s-3Electric Charge Coulomb C AsPotential Difference Volt V J A-1 s -1Pressure Pascal Pa N m2= kgm-1 s-2

PREFIXES:-Any symbol which is put before the unit to represent a large or small value is called prefix.It is convenient to express large number or small number by certain symbol (prefixes) such as centi,milli, kilo, mega etc.

At Are Table 13power Prefix Symbol

10-18

10-15

10-12

10-910-610-310-210-1101101103106

AttoFemoPicoNanoMicroMilliCentiDeciDecaKiloMegaGigaTeraPetaExa

AFPNMCDDaKMGTPE


For example1 centimeter = 1 cm = 10-2

1 kilometer =1km = 103 m1 microfarad = 1 f =10-6 F

The other prefixes are given in the table.

SCIENTIFIC NOTATION:-To reprint a number (large or small) in standard from is called scientific notification or power of ten.In physics, we deal very large quantities as well as very small quantities i.e. I light year = The distancetraveled by light in vacuum in one year.

30 I light year = 946000, 000, 000, 0000 m and the radius of hydrogen nuclei i.e.

R (H1) = 0.000, 000,000,000,0012m

So it is different to memories, writes, multiplied, and divided, therefore the convenient way to expressthese large and small values in power of ten or standard from. Every large or small values N can beexpressed in terms of number M and a power of ten.

N = M X 10n.Where M is a number whose first digit lifts of decimal will non zero digits so

1 light year = 9.46*1015 m .R (H1) = 1.2*10-15 m .

CONNENTION FOR SI UNITS:-The use of S1 units and prefixes need special care. The following stapes should be kept in mind:(1) Full name of unit does not begin with capital letter, if named after a scientist e.g. Newton but not

Newton.(2) The symbol of a unit named after a scientist has initial capital letters such as N for

Newton.(3) The prefix should be writing before the unit without any space.(4) Compound prefixes are not allowed e.g.

1 f =1 PF

(5) A combination of base unit is writing with one space apart.(6) When a multiple of a base unit is raised to a power. The power applies to the whole

multiple. e.g.1 km)2 = (1*103 m)2 = 1*106 m2

CGS AND FPS SYSTEM OF UNITS:The CGS is an abbreviation for the centimeter gram second system which FPS is stand for foot pound second system. These systems gradually disappeared worldwide after the adaptation of MKS(meter, kilogram, second). System which led to the modern standard units.British unit (FPS) is used only in mechanics and thermodynamics. There is no British systemof electrical units. British units are now defined in term of SI unit as:

1 inch =2.54 cm, 1mile = 1.609 km

10910121015


1 hour = 3600 s, 1 pound force = 4.448 N1 m = 3.28 ft.

CONVERSION OF UNITS:-To perform any mathematical operation i.e. addition, subtraction, multiplication etc thephysical qualities must be in the same system of unit.For example A 120 m length is added 60 cm, so one can of them to the unit of another.(1) There are two methods for the conversion between two units.

(1) SIMPLE UNIT METHODS:-(i) First find the conversion factor i.e. 1m = 100 cm, so 100 is the conversion factor between m andcm,

(ii) Then multiply this conversion with the number, which we want to convent in to small or divide thenumber by conversion factor which we want to convert in to bigger number.EXAMPLE: Convert 2m into cm?Here conversion factor is 100, so 2m =2 x 100 cm = 200 cm. and convert 200 cm into m?

200 cm =100200

= 2m.

(2) DIMENSIONAL ANALYSIS:-There are just two simple steps for this method.

(i) Find the conversion factor and write it as an equation in the desired unit which you want forexample convent 2m into cm ?

so 1 m = 100 cm ,m

m

11

=

m

cm

1100

=> 1=m

cm

1100

. Or 1 =m

mm

1001

.

(ii) Now multiply the number which is conversion by 1 (conversion factor) i.e.2m = 2m x 1 = 2 m x 1

m

cm

1100

= 200 cm

Of 200 cm is converted in m ?

200 cm = 200 cm x 1 = 200 cm xcm

m

1001

=2m

ERRORS AND UNCERTATAINTIES:-The difference between the actual value and measured value of a measurement is called error. whereas the rang of error is called uncertainty.

EXPLANATION:-Practically, there is no such measurement which is free from error. Mathematically. Error E is givenby: E = 1x xi / where x is the actual value xi the measured value.The uncertainty in a measurement due to the natural Imperfections of the experimentation or limitationof the apparatus.Suppose we want to meter rod. The last count of meter rod is 1mm = 0.1cm.Of the reading on the meter rod comes as shown in figure. The length is =3.1cm, the extra edge isguessed as = 0.05cm (half of lest cm) ignored if left of the mid point, consider if right from the midpoint so


Incorrect way to read from the ruler

10 20 4030cm

10 20 4030cm

Correct way to read from the ruler

ObjectZero error atthis end

L = 3.1cm I 0.05cmFrom thisL = 3.15 cm ______ L = 3.05

This is called rang of error which is know uncertainty. This means that the true length lies betweenthese two values.

CLASSIFICATION OF ERROR:There are three types of error in the measurement of a physics quality.

(1)SYSTEMATIC ERROR:The error which can predicate and removable is called systematic error. This error occurs in ameasurement due to imperfection or limitation in the equipment being used. Such error can bepredicted and removed before starting the experiment. i.e.In venire caliper, a positive error or negative error can be predicted and removed before takingmeasurement.

(2) Personal Error:The error which is occurs in a measurement due to the natural deficiencies or individual mistake iscalled personal errors. This is also predicated and removable so it is also included in systematic error.This error can be minimizing by placing the object in the line with scale as shown in figure.

(3)RANDOM ERROR:-The error which can not be predicted and removed is called random error.The random error arises due to some unknown causes and gives different values of a measurementunder the same conditions. i.e due to temperature variation or other unknown changes during theexperiment. Repeating the measurement several times can reduce the effect of random error.

CLASSIFICATION OF UNCERTAINTIES:-

(i)ABSOLUTE UNVERTAINTY:-The uncertainty which is equal to the least count of a measuring instrument is called absoluteuncertainty for example the length of this book is measured which meter rod is 13.5cm. The leastcount of meter rod is 1mm = 0.1cm, so the length can be expressed as,

L = 13.5 cm +- 0.01 cm

(ii) RELATIVE UNCERTAINTY OF FRACTIONAL UNCERTAINTY:

1cm 2cm 3cm

0


The ratio of uncertainty X to the measured value is called Fractional uncertainty.If X is the measured value and X is the uncertainty, then Fractional Uncertainty (F.U) = X/X.And percentage Uncertainty = / * 100.Consider the length of this

F.U = 0.1/13.5 = 0.0074And percentage Uncertainty = 0.1/13.5 * 100/100 = 0.74%

PRECISION AND ACCURACY:The measure of similar measurement of a physical quantity is called Precision. Where as Themeasure of a measurement close to the correct value is call Accuracy.

Explanation:In measurement made in physics the precision of a measurement is determined by the instruments ordevice being used while accuracy depends upon the fractional or relative error.For example: if we measure the thickness of an object (block) using meter rod, recorded as 5.4 cm. sowe can write:

Thickness = 5.4cm +- 0.1 cm

On the other hand if the same thickness is recorded with verneir caliper of least count 0.005 cm thenthickness = 5.40 cm +- 0.005 cm. you see the magnitude of error in the first in large and in the 2ndreading is small.Therefore the 2nd one is more precise. If we measure the relative uncertainty (F.U) in both i.e.

F.U = 0.1/5.4 * 100/100 = 1.8%And

F.U = 0.005/5.4 * 100/100 = 0.09 %

These results shows that relative uncertainty in the 2nd result is small so it is more accurate as compareto the 1st reading.Another example:If we measure the value of g 10.6 m/s2, 10.8m/s2 are seems to be similar approximately but awayfrom actual value 9.8 m/s2. we are precise but not accurate.If the results come to be 9.9 m/s2, 9.6 m/s2 etc. these are not close to one another but close to thecorrect so we are accurate but not precise.And if the readings recorded as 9.7 m/s2 etc we are precise as well as accurate. So we can conclude; avalue is precise but not occurate or a value is precise as well as occurate, all are possible.

ASSESSMENT OF UNCERTAINTIES IN FINAL RESULT:The result of an experiment is usually not the final result; they are used to calculate some final resultwhich may be obtained form the addition subtraction multiplication etc. of two or more than towvalues. There fore the uncertainties in the final result must be expressed the following are rules toshow the uncertainties in sum, product etc.

(a) Sum and Difference:If two measured quantities are added or subtracted then the absolute uncertainties are directly added.Suppose two quantities a and b are measured and we need to add or subtracted them to get anotherquantity Q i.e.

Q = a +- b, then the total uncertainty in the Q = uncertainty in a + uncer5tainty in b .For example:


a = 4.5 cm +- 0.1 cmb = 3.2 cm +- 0.1 cm

Q = a + b = 4.5 + 3.2 = 7.7 cm +- 0.2 cmQ = a- b = 4.5 - 3.2 = 1.3 cm +- 0.2 cm

(b) Product and division:If two or more than two quantities are multiplied or divide, then their relative uncertainties are added.For example; we find the resistance of a wire by Ohms Law.

R = V/IAnd

V= 5.2 V +- 0.1 V F.U in V = 0.1/5.2* 100/100 = 2%

I = 0.84 A +- 0.05 A F.U in I = 0.05/0.84 * 100/100 = 6%Now total uncertainty in R (2% + 6%) = 8%

R = 5.2/0.84 = 6.19 +- 8%R = 6.19 +- 8/100 * 6.19R = 6.2 +- 0.5

(c)Power of quantity:The total uncertainty in power quantities is equal the relative uncertainty multiplied with the power ofthat quantity.For example; if the radius of a sphere is measured as 2.25 cm +- 0.01 cm, then the volume of a sphereis;

V = 4/3 r3

Find the Fractional uncertainty is r is

F.U in r = 0.01/2.25 * 100/100 = 0.4 %Total uncertainty is in volume V = 3 * 0.4 % = 1.2%Thus the volume;

V = 4/3 * 4.14 (2.25)3 +- 1.2%V = 47.689 cm3 +- 0.6 cm3

(d) Uncertainties in Average Values:In order to measure the uncertainty in the average value; the following steps are applied.(i) Find the average of measured values.(ii) Find the deviated value from the average.(iii) The mean deviation is the uncertainty in the average.

Fore example:Four reading of screw gauge to measure the diameter of a wire in m as;1.20, 1.22, 1.23, 1.19

The average or mean value = 1.20 + 1.22 + 1.23 + 1.19/4 = 1.21


Mean diameter = 1.21 mm

Deviation of all readings from the mean value; 0.01, 0.01, 0.02,0.02

Mean Deviation = 0.01 + 0.01 + 0.02+ 0.02/4 = 0.015 = 0.02 mm

D = 1.21 mm +- 0.02 mm

(e) Uncertainty in Periodic time Experiment:If the measured quantity is multiplied or divided by a constant then the absolute uncertainty in themeasurement is also divided or multiplied by the same constant.For example;The time recorded for 30 vibrations of a pendulum is 54.6S. Then the time period of the pendulum isobtained as;

T = 54.6/30 = 1.82 S .: t = (54.6+-0.1)SUncertainty = 0.1/30 = 0.003 S

So T = (1.82+- 0.003) S

SIGNIFICANT FIGURES:In any measurement, the accurately known digits and the first doubtful digit are called significantfigures.

Explanation:Physics is an experimental and quantitive science deals with the measured quantities a measurement.i.e. it does not appear to be more accurate then the apparatus, to allow. We must achieve this bycontrolling the significant figures to reports the measurement.For example:Let us we want to measure the length of a pencil with foot ruler, caliberted in cm and mm, with leastcount 1 mm= 0.1 cm.Suppose the length of the pencil when the end point lies between 9.7 cm and 9.8 cm marks. Byconvention, if the end of pencil does not touch or cross the mid point of the mm, the reading isconfined to 9.7 cm, if crossed the mid point the reading is 9.8 cm, so the uncertainly +- 0.05 cm (halfof least count) i.e.

L = 9.7 cm +- 0.05 cmSo

L = 9.75 cm ----------------- 9.69 cm.

According to the lest count the readings should be in one digit after decimal therefore the digit 5 and 9should be dropped then; the length either 9.8 cm or 9.7 cm. here 9 is accurately measured while 8 or 7is the first doubtful.

GENERAL RULES:(1) All digits (non-zero) 1, 2,3,4,5,6,7,8, and 9 are significant digits.For example 45, 112.6 cm are significant.


(2) Zeros digits may or may not significant.(a) Zero digits between two significant digits is significant i.e. 305, 1.032(b) Zeros to the left of a significant digit is not significant i.e. 058, 0.02.(c) Zeros to the right of a significant digit may or many not significant.

(i) In decimal Factors, zero to the right of a significant digit is significant. i.e. 3.570, 7.4000.(ii) in non-decimal factors, zero to the right counting unit (least count) of the measuring device.

For example: the integer 8000 Kg, if

Factor lest count

8000 Kg --------------- 1 Kg ------------------------ 8.000*103 Kg -------------4 digits

8000 Kg --------------- 10 Kg ----------------------- 8.00*103 Kg -------------- 3 digits

8000 Kg --------------- 100 Kg ---------------------- 8.0*103 Kg ----------------2 digits

8000 Kg --------------- 1000 Kg --------------------- 8*103 Kg ------------------1 digit

(3) When a measurement is recorded a scientific notation, the figure other than power of ten aresignificant. For example: 8.70 * 104 Kg has three significant digits

RULES FOR ADDITION OR SUBTRACTION:When measurements are added or subtracted, the result can contain no more decimal places than theleast accurate measurement.For example: tow number 3.210 cm and 10.1 cm are added or subtracted then 3.210+ 10.1 = 13.310 =13.3 cm. this is because 10.1 contain only one decimal place, so the answer can not have more thenone decimal place.

RULE FOR MULTIPLICATION AND DIVISION:When number are multiplied or divided the total answer no more contains significant figures than theleast accurate measurement.For example: 3.6 cm * 1.25 cm = 4.500 cm2But it can be written as = 4.5cm2. Because 3.6 contain two significant figure, therefore answer shouldbe in two significant figure.

ROUNDING OFF:The dropping of non-significant digit is called Rounding off.(1) If the digit dropped (non-sig is greater than the retained digit is increased by 1 e.g. 19.37 = 19.4(2) if the digit dropped is less than 5 the retained digit remains the same e.g. 2.63 = 2.6.(3) If the digit dropped is equal to 5 the retained digit is increased by one if it is odd if even remain assame.For example: 43.75 is rounded off as 43.8, 56.8545 is rounded off as 56.854

DIMENSIONS OF PHYSICAL QUANTITIES:Dimension of a physical quantity is the power to which the fundamental quantities are raised torepresent the physical nature of that quantity is called Dimension


EXPLANATION:To solve the problems in physics, we use equation. To write these equations error may occur. We canprevent it by dimensional check.Quantities can be added subtracted only if they have the same dimensions and quantities on the twosides of an equation must have the same dimension. The dimension of any physically quantity can bewritten in the form of base quantities in square bracket i.e. [Ma, Lb, Tc, Ad, Ce, Kf, Ng]. wherea,b,c,d,e,f,g are integers i.e. -2, -1, 0, 1, 2, 3, etc

Dimension of Speed = S/t = dimension of length / dimension of time.

[V] = [L/T] = [LT-1]

USES:Using the method of dimensions called dimensional analysis we can check;

(i) the correctness of a formula.(ii) The homogeneity of an equation.(iii) To derive a possible formula.

EXAMPLE:Check the correction of the formula V = F*L/m

Dimension of LHS = [V] = [LT-1]Dimension of RHS = ([F] [l] [m-1]) 1/2

= ([MLT-2]) 1/2. [L] 1/2. [M-1]1/2 = [L2 T-2]1/2 = [LT-1]

So L.H.S = R.H.S, so the formula is dimensionally correct.

(ii) Prinple of homogeneity:This mean that all the quantities which are added or subtracted in an equation must have the samedimensions. i.e.

S = vit + 1/2 at2L = L/T * T + L/T2 * T2

L = L + L this is called homogeneity.

(iii) To derive possible formula:Using the dimension derive the possible formula for the time period of simple pendulum. We firstguess that the various quantities on which the time period of the pendulum depends and also gues theirpower. i.e. mass of the bob m, length of the pendulum l angle which the string make with vertical,and acceleration due to gravity g.The relation for the time period T will be of the form;

T ma c gd

Or T = constant ma lb c gd ------------------ (1)Using the dimension

Simple Pendulum

l


[T] = constant [M]a [L]b [LL-1]c [LT-2]d

[T] = constant [M]a [L]b+d [T]-2d

Equating the powers M, L, T on both sides

a = 0, b + d = 0, -2d = 1

a = 0 b = +1/2 d = -1/2

Put the values of a, b, c, d in (1)

T = constant mo L1/2 g-1/2

T = constant l/g

The proportionally is measured by experiment.


SHORT QUESTION:

Q 1. What is a system of units? List the base SI units.Ans. A complete set of units both base and a derived unit is called system of units.

The base units (i) Meter (m) for length (ii) Kilogram (kg) for mass (iii) second (s) for time (iv)Kelvin (K) for temp (v) ampere (A) for current (vi) mole (mol) for substance. (vii) candela (cd) forluminous intensity.

Q 2. Define the number and show that 2 rad = 360o?Ans. The ration fo the circumference of a circle to the diameter of the same circle.

So if theArc = 1 radii --------------- = 1 radArc = 2 radii --------------- = 2 rad

And Arc = 2 radii --------------- = 2 rad2r = 1 revolution = 2 rad 306o = 2 radian

Q 3. How the units of length, mass and time are presently defined?Ans. 1 m = the distance traveled by light in vacuum in 1/ 2.99* 108 S.

1 Kg = the mass of a cylinder having length 39 cm.1 S = the duration time taken by 9.19*109 vibrations of cs 133 atom.

Q 4. Distinguish between the base and derived quantities?Ans. The minimum number of those physical quantities in term of which other physical quantities are

defined and derived. i.e. Temp; Length time etc. derived quantities are those whose definition andderivations are base on fundamental quantities. i.e. Power, Acceleration, Momentum etc.

Q 5. Explain with examples the scientific notation for writing Number?Ans. The representation of a number in standard form (power of ten) is called scientific notation. The

international practice is that there should be non-zero digit left of decimal. i.e.

5392 = 5.392*1038000 = 8.000*1030.00015 = 1.5*10-4

Q 6. Define the terms; Error, uncertainty, precision, and accuracy in Measurement?Ans. Sea the text

Q 7. Explain the principle of dimensional homogeneity of a physical equation?Ans. These principle states that the two terms in an equation can be added or subtracted must in an

equation can be added or subtracted must have the same dimensions.

i.e. Vf = vi + at

L / T = L / T + L /T2 * T=> L / T = L / T + L / T


Q 8. An old saying is that A chain is only as strong as its weakest link. What analogous statement canyou make regarding experimental dates in calculations?

Ans. The similar statement for calculation of experimental data is the number of significant digits can beno greater than the least significant number i.e 4.372 cm 2.1 cm = 6.4 cm.

Q 9. Write the dimension of (i) Pressure (ii) Power (iii) Density (iv) Revolutions?Ans. (i) Pressure:

F/A = so [P] = [F] / [A] = [MLT-2] / [L2]P = [ML-1T-2]

(ii) Power:Work/time = W/t = Fd/t = [MLT-2. L] / [T]

P = [ML2T-3]

(iii) Density:m/v = [M] / [L3] => [D] = MLT-3]

(iv) Revolution:

[Revolution] = [1/T] = [T-1]

Q 10. Time period of a simple pendulum is measured by stopwatch. What types of error are possible intime period?

Ans. The errors which are expected in the measuring of a time period are; systematic errors personal errorand random error.

Q 11 a circle has a diameter of 0.400 m. what is the area?Ans.

Diameter = 0.400 mRadius r = 0.400/2 = 0.200 m

Area; A = r2 = 3.14 *(0.200)2A = 0.1256m2A = 0.126m2

Q 12. Write five units used in Pakistan for measuring mass.Ans. The units of mass used in Pakistan are; (i) Kilogram (ii) gram (iii) Ton i.e. 1 Ton = 907.2 Kg.

(iv) Darri = 4 kg (v) Tola = 12 gm

NUMERICAL PROBLEMS

Problem 1. Express the following quantities using prefixes (a) 3.0 * 10-4 (b) 5.0 *10-5 s (c) 72.0*102 gm

SOLUTION:


(a) 3.0 * 10-4 m = 0.3*10-3 m = 0.3 mm

(b) 5.0 *10-5 s = 5.0*10-6 s = 50 us

(c) 72.0*102 gm = 7.20*103 gm = 7.20 Kg

Problem 2. Estimate your; (a) Age in second (b) mass in second (c) weight in Newtons (d) height inmillimeter?

SOLUTION:

Let us assume that your (a) age is 17 years and 20 days so

So age = 17 year + 20 days= 17*365 + 20

Age = 6225 days

Age = 6225 * 24 * 3600 s = 537840000 sec

Age = 5.3784*108 sec

(b) mass = 65 Kg = 65*103 gm(c) w = mg = 65*9.8 = 637 N(d) your height = 5 ft, 8 inch

H = 5 * 12 inch + 8 inchH = 68 inch = 68*2.54cm = 172.72 cmH = 1727.2 mm

Problem 3. if there are No = 6.02*1023 atoms in 4 gms of helium what is the mass of helium atom?

SOLUTION:6.02 *1023 atoms has mass = 4 gm

1 atom has mass = 4/6.02*1023 gm

1 atom has mass = 6.64*10-2 4 gms

Problem 4. Rest mass of electron in 9.11 * 10 -31 Kg. (a) write it with out use of power of 10. (b) convert itto gms.

SOLUTION:

Rest mass of electron = mo = 9.11*10-31 Kg

(a) mo = 0.000,000,000,000,000,000,000,000,000,000911Kg. in gms (b).

mo = 0.000,000,000,000,000,000,000,000,000,000911 gms


Problem 5: Density of air is 1.3 Kg/m3. Change in to gm/cm3?

SOLUTION:Density of air; S = 1.2 Kg/m3S = 1.2 * 1000 gm/106 cm3S = 1.2 *10-3 gm/cm3

Problem 6. Density of water is 1 gm/cm3 change it into Kg/m3?

SOLUTION:

Density of water; S = 1 gm/cm3

S = 1 * 10-3 Kg/10-6 m3 .: 1 gm = 10-3 Kg

S = 103 Kg/m3 = 1000 Kg/m3 .: 1 cm3 = 10-6 cm3

Problem 7. Express the following in terms of Power of 10? (a) 5 Picofarad (b) 10mega watt (c) 100 million volts.

SOLUTION:

(a) 5 Picofarad = 5*10-12 F .: 1 Pico = 10-12(b) 12 mega watt = 12*106 W .: 1 mega = 106

= 1.7*107 W(c) 100 million volts = 100*106 V = 108 .: 1 million = 106

Problem 8. A light is the distance traveled in vacuum by light in one year? How many meters are there inone light year?

SOLUTION:

Light year is the unit of length i.e.

S = vt=> S = Ct

S = 3*108 * 1 yearS = 3*108 * 365 * 24* 60*60S = 9.46*1015 m.

Problem 9: Compute the following to the correct significant digits? (a) 3.85 m *3.19m (b) 1023 Kg +8.5489 Kg. (c) 22/7 (d) mp/me = 1.67*10-27 Kg /9.11*10-31 Kg

SOLUTION:

(a) 3.85 m *3.19m = 12.2815 m2But according to the rules, final result has significant digits equal to the least significant digitsin the input data. Therefore


3.85*3.19 = 12.3m2

(b) 1023 Kg + 8.5489 Kg = 1031.5489 KgAccording to the rule, the results can no more decimal places than the least accuratemeasurement 1023 Kg + 8.5489 Kg = 1032 Kg.

(c) 22/7 = 3.142857143 These are exact values therefore the number of significant digits is infinite.(d) mp/me = 1.67*10-27/9.11*10-31 = 1.83*102

Problem 10: the length and with of a rectangular plate are measured to be 15.3 cm and 12.80 cm, Find theare of the plate?

SOLUTION:

Area = Length * widthA = 15.3 cm * 12.80 cmA = 195.84 cm2

A = 196cm2

.: As the min : sig: figures are three in the input data.

Problem 11: A rectangular metallic piece is 3.70 +- 0.01 cm long and 2.30 +- 0.01 cm wide.

(a) Find the area of the rectangle and uncertainty.(b) Verify that % age uncertainty in the are is equal to the sum of % age uncertaintities in the

length and width?

SOLUTION:

(a) L = 3.7) cm +- 0.01 cm F.U(L) = 0.01/3.70 * 100/100 = 0.27%W = 2.30 cm +- 0.01 cm F.U(W) = 0.01/2.30 * 1w/1w = 0.44%

A = L*WA = (3.70) * (2.30) Total F.U (Area) = 0.27% + 0.44%A = 8.5101 cm2 +- 0.71% Total F.U = 0.7%

A = 8.5` cm2 +- 0.71/100 * 8.57

A = 8.51 cm2 +- 0.06 cm2

(b) Max length = 3.71 cm Min L = 3.69 cmMax length = 2.31 cm Min W = 2.29 cmMax area = 3.71 * 2.31 Min Area = 3.69*2.29


Amax = 8.5701cm2 A min = 8.4501 cm2

8.5701cm2 < Area < 8.4501 cm2

Taking average:

A = (8.5701cm2 + 8.4501 cm2/ 2) +- (8.5701 - 8.4501/ 2)

A = 8.5701cm2 +- 0.06 cm2

Problem 12: Find the mass of the air in 3.00 m * 8.00 m* 6.00 room? S (air) = 1.29 Kg/m3.

SOLUTION:

V = 3.00 m * 8.00 m * 6.00 mS = 1.29 Kg/m3

S = m/v => m = SV

SV = 1.29* 3.00 *8.00 * 6.00 = 186 Kg

Problem 13: Prove the following equation are homogenous: (a) K.E = 1/2 mv2 (b) a c.p = v2 /r

SOLUTION:

(a) K.E = 1/2 mv2

L.H.S K.E = J = N.m = Kg.m/s2.m = [ML2T-2]

R.H.S 1/2 mv2 = Kg. (m/s) 2 = Kg. m2/s2 = [ML2T-2]

(b) L.H.S a = m/s2 = [LT-2]

R.H.S v2/r = (m/s) 2 / m = (m2/m)/s2 = [LT-2]

So L.H.S [LT-2] = R.H.S [LT-2]

Problem 14: Find the dimension and the unit of the coefficient of viscosity is Stokes Law for drag force Fon a spherical object of radius r moving with velocity V given as F = 6yrv.


SOLUTION:

F = 6yrv = y = F/6rv

So [y] = [MLT-2] .; [F] = [MLT-2]

[L] [LT-1]

[y] = [ML-1T-1]

And the unit of y = Kg/m.s = Kg.m-1. S-1

Problem 15: Find the dimension and hence the unit of universal gravitational constant G in Newton Law forforce between two masses separated by a distance r given as; F = F m1m2/r2

Solution:

F = G m1m2/r2 => G = Fr2/m1m2

[G] = [MLT-2] [L2]/ [M2] = [M-1L3T-2]

And the unit of G = m3 / Kg.S2 = Kg-1. m3. S-2

Problem 16: show that the famous Einsteins equation E = mc2 is dimensionally correct.

Solution:

E = mc2

LHS E = J = N.m = Kg.m/s2.m = [ML2T-2]

RHS mc2 = Kg . m2/S2 = [Ml2T-2]

So

LHS [ML2T-2] = RHS [ML2T-2]

The above equation is correct dimensionally.

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xii physics Chapter 1