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2. unique Notes Physics lsl Year
Measurements
01
Define physics. Name its various branches. What role do the Physics play i
science and technology?
Q1 in the field of
litWBI
Ans. The ranc ot science, which deals with the study of matter, energy and the relationship
between them, is called Physics.
Branches of Physics:
The study ot physics involves investigating, the laws of motion, the structure of space and
time, the type and nature ot torces that hold different materials together, the interaction between
different particles, the interaction of electromagnetic radiation with matter and
By the end ot 19' century many physicists started believing that every thing about Physics
has been discovered. But in the beginning of 20th
century new experimental facts revealed that the
laws formulated by the previous investigators or scientists need modifications.
Further research gave birth to many new branches of Physics such as:
• Nuclear Physics
• Particle Physics
• Relativistic Mechanics
so on.
• Solid State Physics
Conventional Branches of Physics:
Mechanics, Heat and thermodynamics, electromagnetism, optics, Sound, Hydrodynamics,
Quantum Mechanics.
Main Frontiers of Fundamental Science:
At present there are three main frontiers of fundamental science:
The world of the extremely large, the universe itself, radio telescopes now gather information
from the far side of the universe and recently detected as radio waves. The “ firelight” of big
bang which probably started off the expanding universe nearly 20 billion years ago.
The world of the extremely small, that of particles such as, electrons, protons, neutrons,
mesons and others.
The world of complex matter. It is also the world of “ middle-sized” things, from molecules at
one extreme to the earth at the other.
This is all fundamental physics, which is the heart ol science.
Role of Physics in Technology:
Physics plays vital role in development of science and technology that has changed the
outlook of mankind. The information technology has brought all parts of world in close contact with
one another. The computer networks made of chips developed from the basic ideas of Physics
chips are made of silicon.
(i)
(ii)
(iii)
.
. The
01
Jl
3. I
Unique Notes Physics 1 Year
Unique Notes Physics 1° Year
What art* the Physical quantities? Describe international system of units i
base, derived and supplementary quantities.
Ans. Definition:
" The quantities in terms of which laws of Physics are expressed and measured accurately ur ,
called Physical quantities
For example, mass length, time, velocity, force, density etc.
Types of Physical Quantities:
Physical quantities are often divided into two categories:
Fundamental or Base quantities
Derived quantities
Base Quantities:
Q2.
(ii) Supplementary Units:
The general conference on Weights and Mea
^l
under either base units or derived units. These SJ
The unit of plane angle and solid an
supplementary units.
Radian:
11101002
,;is classified certain units of the SI
ule
^supplementary units,
and Stcraijian respectively. These units
its are c:
e
arc
toward 2008)
cut off on the circumference an arc, equal in
The angle between two radiirt
length to the radius is called one radii
a circle
1)
2)
1. r r
Definition: J r f l d
The minimum number of physical quantities in terms of which other physical quantities can
he defined is called base quantifies or fundamental quantities ” ,
Typical examples of base quantities are mass, length and time, electric current,
thermodynamic temperature, amount of substance and luminous intensity.
Derived Quantities:
" Derived quantities are those whose definitions are based on other physical quantities ” ,
For example velocity, acceleration and force etc arc viewed as derived quantities.
Steps to Measure Base Quantity:
The measurement of base quantity involves two steps:
Choice of a standard.
The establishment of a procedure for comparing the quantity to be measured with the
standard so that a number and a unit are determined as the measure of that quantity
Principal Characteristics of Ideal Standard:
An ideal standard has two principal characteristics:
It is accessible.
It is invariable
These two requirements arc often incompatible and compromise has to be made between
r
Stcradian:
^ ( Board 2008)
is the solid angle (three dimensional angle) subtended at the center of a sphere
ccequal to the square of radius of the sphere.
( D
by
2.
( i )
(ii )
(iii) Derived Units:
The units of derived quantities which can be produced by the combination of two or more
base units are called derived units e.g.
( Board 2008)
In terms of
Base Units
( i) Symbol
Units
Physical Quantity
( ii ) 2
kg ms
N
Force newton
N m = kg mV ~
J
joule
Work
them.
International System of Units:
In I 960. an international committee agreed on a el of lefinidons and standard to describe tb
physical quantities. The system that was established is called sysiem International (SI).
Iherc are three kinds of units in SI:
(i) Base Units
Supplementary Units
( iii) Derived Units
(i) Base Units:
Diere are seven base nit for sevei '
•ase quantities namely length, mass, time, temperature
electric current. luminous inter and amo t of substance. These quantities have units, metre!nb
kilogram ( kg), setond(s).kelvinfK mper- candela(cd) and mole( mol), respectively.
J.s 1
= kg mV
W
watt
Power
N m‘
= kgm V
Pa
pascal
Pressure
* A s
C
coulomb
Electric Charge
What do you mean by scientific notation?
standard form called scientific notation, which employs powers of
is that there should be only one non-zero digit left of
11101003
Q3.
Ans.
len- The internationally accepted practice
decimal.
Numbers are expressed in
(ii)
134.7 = 1.347 x l O2
eg-
( Board 2008)
0.0023 =2.3 X 10 3
51767.5 = 5.17675 x 1()4
=5.177 x 10
J
4. Unique Notes Physics 1 Year
Unique Notes Physics l *4
Year
(J4. Whal are the convention*for Indicating units?
Ans.
11101004
Reduction:
Use of SI units require special care, more particularly in writing prelixes.
Following points should be kept in mind while using units.
Full name
Systematic error can be reduced by comparing
to be more accurate. Thus for systematic error a coA
Uncertainty:
of the unit does not begin with a capital letter even if named after a scientist e.g
( I)
IS
ijennkan he applied,
newton.
(ii) The symbol of the unit named after a scientist has initial capital letter such as N for newton,
tiii) The prefixes should be written before the unit without any space, such as
1 x 10 ' m is written as 1 mm.
(iv ) A combination of base unit is written each with one space apart. For example, newton meter
is written as N m.
( v ) Compound prefixes are not allowed. For example, Ifipf may be written as lpF.
A number such as 5.0 x 104 cm may be expressed in scientific notation as 5.0xl02m.
(vii) When a multiple of base unit is raised to a power the power applies the whole multiple and
not the base unit alone,
thus, 1 km
:
= l(km)2
= 1 x 104>
m2
( vili) Measurement in practical work should be recorded immediately in the most convenient unit,
e.g. micrometer screw gauge measurement in mm. The mass of calorimeter in grams (g). But
before calculation for the result all measurements must be converted to the appropriate S I.
base units.
Q5. Define errors and describe its various types.
Ans. All physical measurements are uncertain to some extent. It is very difficult to eliminate
possible errors or uncertainties in a measurement. The error in the measurement may occur due to.
1) Negligence or inexperience of a person.
2) The faulty apparatus.
3) Inappropriate method or technique.
There are two major types of errors:
i) Random error
it) Systematic error
j. Random Error:
Random error n said to occur when repealed measurements
i :i.JC under the iame condition*. It is due to some unkitow n causes
Reduction: N
The uncertainty is also usually descj
Inadequacy or limitation of an imJ
Natural variations of the object hein
Natural imperfections of a$jp
Q6. What do you mean by sigi
and rounding offathMlata.
Ans. It can be defined, as: Jp
rror in a
«
1
?
isurcment. It may occur due to:
1) lent.
2) asured.
3) on's se:
am figures? Describe general rules for delermining ( hem
II101006
ratrty known digits and the first doubtful digit
( vi)
“ In an the a*
asun are called
significan es
General Ri
While calculating a
^esult from the
significant figures and
measurements, it is important to give due attention to
e must know the following rules in deciding how many significant figures
nal result:
in
(i igits 1, 2, 3, 4, 5, 6, 7, 8, 9 are significant. However, zeros may or may not be
leant. In case of zeros, following rules may be adopted,
o between two significant figures is itself significant.
Zeros to the left of significant figure are not significant. For example, none of the zeros in
{0.00467 or 02.59 is significant.
Zeros to the right of a significant figure may or may not significant. In decimal fraction zeros
to the right side of significant figure are significant, e.g. all the zeros in 3.570 or 7.400 are
significant. However, in integers such as 8000 kg the number of significant zeros is determined
by the least count of the measuring instrument. If a measuring scale has a least count of 1 kg
then there are 4 significant figures written in scientific notation as 8.000 x Hr kg.
When a measurement is recorded in scientific notation or standard form, the figures other
than the powers of 10 are significant figures. For example, measurement recorded as 8.70 x
10**
kg has three significant figures.
In multiplying or dividing numbers, keep a number of significant figures in a product or
quotient not more than that contained in the least accurate factor i.e. the factor containing the
least number of significant figures, e.g.. the computation of the following using a calculator
gives;
11101005 gm
different
i rii)
reduce the effect c*
Repea vnt rez& jrcmcnt tbevera, times and raking an average can
random errors
StUematk Error:
S jixematM, >mrA refers Io an
It idjr a Jxwcnt fiercnce m cading
.
•>
effect Out influences all measurement of a particular quan":
5.348 x 101x 3.61 x 104
1.336
A:,LTZ2
Z'.“’.o*.*
. SOT
figures, the answer should be written to three significant ngures only. Theodter fig
insignificant and should be deleted While delettng die figures, the las.srgmfican.figure
rounded remain unchanged
II the first digit dropped is less than the la
Tm% ttv
* » prodded :r, the in-strumer :
i ) /JZ' I zrvx*4 ttnfijwnpA
*
fu> Four tajibracvjn of w J
^«er v or »r ' rrect t zxking%.
are
retained is
(aj 05
04
5. i
0
Unique Notes Physics T Year
Unique Notes Physics lu Year
If the first digit dropped is more than 5. the digit to be retained is increased b> one
If the digit to be dropped is 5. the previous digit which is to be retained is increased by one if
it is odd and retained as such if it is even. For example, the following numbers are rounded
off to 3 significant figures as follows. Following this rule, the correct answer of the
computation given in sec (ii) is 1.46 x 103.
The following numbers are rounded off to three significant figures as follows:
43.75 is rounded off as 43.8
56.8546 is rounded off as 56.9
64.340 is rounded off as 64.3
73.650 is rounded off as 73.6
( iii) In adding or subtracting numbers, the number of decimal places ( a) 77 j
retained in the answer should equal the smallest number of decimal
places in any of the quantities being added or subtracted. In this case,
the number of significant figures is not important. For example,
suppose we wish to add the above quantities measured in metres. The
sum in example (a) and ( b) is rounded off to 75.5 and 8.13 respectively.
Q7. What do you mean by Precision and accuracy?
Ans. Precise measurement is the one which has less fractional or precision or absolute uncertaint
An accurate measurement is the one which has less fractional or percentage uncertainty
The precision of measurement depends upon the instrument being used. The accuracy
measurement depends upon the fractional or percentage uncertainty.
Let the length of an object is recorded as 25.5cm by using meter rod having smallest division
millimetre. The absolute uncertainty is equal to the least count of the measuring instrument.
Precision or absolute uncertainty (least count) = +0.1cm
The length of an object measured = 25.5cm
absolute uncertainty
measured length
= 0.(X)4
(b)
Where as reading 0.45cm taken by Vemii
it is the relative measurement which is immfrtan
instrument should be used. Here the
such as micrometre screw gauge y
conclude that “ A precise measuremer
measurement is the one which tils les
Q8. What are the general rule&fi
following cases? Explain with
Ans. The overall
the factors in&ludei:
lore precise but less accurate. In fact
physical quantity, the more precise
mds that a more precise instrument
t counf
^.OOlcm should have been used. Hence
e which nas less absolute uncertainty and an
J Or percentage uncertainty or error."
essment of total uncertainty in final result in the
11101008
final results can be worked out by calculation of uncertainty in all
The total uncertainty can be worked out as follows:
<c)
The snfalja
I
|
L45cm dei
me em<
we
t is
accurate
s fractioi
(i)
( ii) e
pies.
( iii)
(iv) uncdrtai;
(b) 2.7543 Iculati
in
4.10
3.42 am on:
w 1.273
0.003 Absolute uncei
between separate positi
inties are added. For example, the distance x determined by difference
1 measurements.
8.1273
75.523
Li cm x:= 26.8 ±0.1 cm
X = X2 - X|
11101007 (26.8 ± 0.1) - (10.5 ± 0.1)
16.3 ± 0.2 cm
Multiplication and Division:
The percentage uncertainties are added.
For example in the measurement of resistance R of a conductor of potential difference V and
current I is measured.
V = 5.2 ± 0.1 V
The %age uncertainty for
*
error.
*
Example: I = 0.84 ± 0.05 A
0A jOO
5.2 x 100
0.05 100
0.84 X
100
= About 2%
V
one
= About 6%
I
The %age uncertainty for current
Hence Total uncertainty in resistance R when V is divided by I is 2% + 6% =8%
* Fractional uncertainty =
5.2
0.1 The Resistance R = =6.19 £2
Total uncertainty =8%
8 % o f 6.19 n = — x6.19 =0.5f2
100
R = 6.20 ± 0.5 Q.
4 "
25.5
0.1 100 _ 0-4
“
25.5
x
100"
100
°/c age uncertainty
= 0.4%
Let us lake measurement of length by Vernier Callipers of least count O.OlcnVis 0.45cm.
Absolute uncertainty = +0.01cm
0.01
3. For Power:
Total % age uncertainty is
is calculated by multiplying the power with that percentage
=0.02
Fractional uncertainty = 0.45
uncertainty.
Example:
m 0.01 100
% age uncertainty = x
= 2 %
Thus reading made by meter rod is less precise but more accurate.
=
|
nr3
the percentage uncertainty in
100
, volume of the sphere by using V
Measurement of volume = 3 x % uncertainty
For calculation
in r.
07
-06
n
6. UniqueNotesPhyrdc
* V Yojt
[W hHtm» oi splwiv l
* monied by Vernier UnlUpew with leant count 0,0lent.an 2.25c,,, (jq. What IN uieunt by dimension of physical
Aw* Definition
0 01 0111
Nhsohih nmcH.dnh
Iho rcprofiefltMtlon of a quantity (,i ba
0 01 ion
> ' KM)
0 » * •
'iage MO* eiWlut) in i
HflCli huNe ijuamity In considered iw nsinn iiml denoted l> CCftttJn symbol written with
III mpuur brockets, ll describe
*the qnali
total •Magr niuvitniiitv In V t H 0.4 - iliac ol the
I focal iiilily.
Fxphiiuilion:
•
I »
v t It s ( ?.2^>
» l / t
*M4> on The dimensions ol hnsr
t quonnih
^lcnglb, H and time is denoted by |L|
,|M| and |T)
respectively,
i
.. I7,6H1
) ( ),(» iin
l * *a agi ol l / t»H4> on Hu illlTerenl qiumlijjos m
mo H ten. breadth, diameter and light ycai arc measured in
nu llr have Nimir
^iluj/inioi dimension ol speed is given by:
lleiu » the laud tomtit I
* iccoidcd an
I length
Dimension ol lime
eiimijf)
Ibmeiiftioi
V 17.7 I O d c m NEH
4. Ini Uni t italiih tit the i t c t tige ol itmity mCUNUmilenUi
IVI
-:
I LT |
.
1'iiuliht average value nI mcusuicdI
|
UUIUMICN
tO
i
l md ilu itev mtmn ol emli incasuu*d value hum average value
(lit
itu.
i t
1
=>l«l '
jT| H.T |.
Ih« mean deviation is lh» uiu eOamiy in tin aveiugc value, ha example, the radius ol the
tilt) l
2 2
"in in nun ao I .’O. I * 7, I M, l 0), I T
’, 121,
mu > |F| |M) |LT ) |MI.I |
.
I '0 i I * ’ .t > t i I |0 i I T* •2 2 -2
v i Iage drk l;
S > |Work ) |MI I | |l | |M I I |,
h
2 -2
„ WOrk in ,
1
Inn,
IMLU IMl T I
11*| *|ML I |
2 -.1
I * 1 nun
IIt.ill laiiou ol i at It value sv ilia ail considering the sign utc 0 (II,0.01,0.02, 0.02, O.Ol
leallons:
OOI I 001 I 0 0
’I 002 I 001 » o
Mi m ol deviation* ( lu cking (lie homogeneity of physical equation:
0
In order to check the correctness of an equation, we are to show dial the dimensions ol the
0 01 nun
quantities on both sides ol equation are same, irrespective of the form of the formula. This is called
Hence (hr uncertainty in the mean diamclei I 21 mm is 0 01 mm recorded as 1.21 1 0 01
the principle ol homogeneity of dimension.
I m ci taint) In riming l spei linent ( Time PeriodI:
(ii) Deriving a possible formula:
11tuntaint m tune pciiod is loitnd hy dividing die least coma a timing instrument hy llif
The success of this method for deriving relation for a physical quantity depends on the
IUIIUIHM id N thrillions i c correct guessing ol various factors on which the physical quantity depends.
I east count
lirncunlmy in nine period -No ()fv|hrmiolls
For KMimple:
Ihe tinm period ol 10 vibrations ol simple pendulum ovoid, i by stop walch l)( jeast c0Uni
0 1 s is .S4.65 s
I east count
Uneciminty in tune period No d ihrulions
* 0.1
o.oo.t
to
' S K»s
a 1.82 s.
Time period V U)
I S 2i 0 003 s
Tmic period is written as
7. nUnique Notes Physics 1" Year
Unique Notes Physics V Year
Short Questions 1.7: Does a dimensional analysis give r
information on constant of proportionality
problem
^
Also the amount of substance can be ExplIT
estimated by two methods. Samples: Dimens.onal analysis does
0) For weighing the unit used is, kilogram. give any information about constant
(it ) For counting the number of entities proportionality. Constant of proportion;
contained in a substance the unit used is, mole.. can be determined by experimen
1.4: Three students measured the length of theoretically,
a needle with a scale on which minimum Examples:
division is 1 mm and recorded as (i) 0.2145 (i) Gravitational Com
m (ii) 0.21 m (iii) 0.214m, which record is As the formula of force o
any
as mass (m) per unit
an algebraic
1.1: Name several repetitive phenomenon
occurring in nature which could serve as
reasonable time standards.( Hoard 2<>io,i4)
11101010
Ans.The spin motion of earth about its own
axis, and orbital motion of earth about sun can
be used as time standard because they repeat
themselves in equal intervals of time. The time
period of spin motion of earth is one day and
that of orbital motion about sun is one year.
Time of other phenomenon can be measured
in terms of these periods. Other phenomenon
could be.
( 1 ) Orbital motion of moon about earth
(2) Vibratory motion of an atom or a
molecule and so on.
(3) Radioactive decay of radioactive
elements.
1.2: Give the drawbacks to use the period of a
pendulum as a time standard.
(Board 2009,14) liioioil
Ans.At a particular place, the time period of a
simple pendulum depends upon its length <
and acceleration due to gravity g as.
11101016 ;s
noi ume
={jp}=[ML-3
]
1.9: The wave length X of a wave depends
on the speed v of the wave and its frequency
T knowing that [ X] = [L], [v] = [LT 1
] and
[f] = [T1
]. Decide which of the following is
correct, f =X or f =
^.
Ans.First we try Eq. f = vA.
Writing dimensions:
L.H.S.
R.H.S
As L.H.S * R.H.S.
So above equation is not true.
Now we trv Eq. f = v/X
L.H.S. =[f] = [T1
]
M _ [LT1
]
[ X]"
[L] “
Therefore L.H.S. = R.H.S.
So f =r is true equation.
01
nt:
itati
correct and why?
Ans.A scale can measure up to 1 mm so
significant figure should be up to three digits in this case dimensi
after decimal. Therefore correct answer is (iii) information about C
0.214 m. As the least count of scale is has numerical value and dimei
lmm=0.001m in the above mentioned three (ii) Stoke's Law:
readings, 0.214m is more correct than others. According to stoke’s theorem;
Because in this reading the answer should be F = 6 K r|rv,
up to three decimal places and it is more In this 6n has no dimension and dimension of
accurate than others have less percentage F is equal to rjrv. Therefore, it is not counted
in dimensional analysis.
1.5: An old saying is that “A chain is only (iii) Time period of simple pendulum:
as strong as its weakest link”
analogous statement can u
_
regarding experimental data used in a
11101014
G Mm
11101013 11101018
F = 7
analysis gives some
icause constant
= m= rr']
= [vA] =[LT-'L] = [L2
T']
si
[T 1
R.H.S. =
error.
hal The time period of simple pendulum is:
you make
T = 2TT
Here constant is 2TT which has no dimension.
Ans.The analogous statement ol old saying Therefore, its numerical value cannot be
“ A chain is only as strong as its weakest link , determined from dimensional analysis. It can
in computation is as:
'
^ . be calculated experimentally.
“ A result obtained by an experimental data is 1.8: Write the dimensions of (i) Pressure
only as much accurate as its least accurate (ii) Density?
readings in the experimental data”. . Ans-(i) Pressure
1.6: The period of simple pendulum »s Pressure is defined as normal force F per unit
measured by a stop watch. What types ol area A
errors are possible in the time period?
DQ . r 11101015
of error
i computation?
T = 2rc
g
This could be good time standard at that
particular place. However as value of g
changes from place to place and from planet to
planet etc, so time period of same simple
pendulum will be different at different places.
Due to thermal expansion length of the
pendulum also change. Therefore we can not
use penod of a pendulum as a time standard.
1.3: Why do we find it useful to have two
units for the amount of substance, the
kilogram and the mole?
( Board 2009,14) llioioi?
-5
I
^-IM-'rt '
lA ) [L 1
Ans. Following could be the sources
when time period of a simple pendulum 1
measured by stop watch:
Manufacture’s fault.
* 1 fP| =
11101012
Ans.One kilogram of different substances
contains different number of molecules but
which could result
1.
in incorrect measurement of lime.
2. Error due to calibration of stop watch.
3. Zero error in stop watch.
4. Human error: Delay
starting/stopping of stop watch.
one mole of different substances contains
same number of molecules called A vogadro’s
number. Therefore these two units are
frequently used according to the need of the
* j
8. Unique Notes Physics 1 Year
Solved Examples
The length, breadth and thickness of a sheet are 3.233m, 2.105m and l.05Cll
respectively.Calculate the volume of the sheet correct upto the appropriate significant digits
Example 1: = 2 x 0.8 + 0.2 = 1.8%
v =
3.14x(l.22cm)2
x 5.
11101019 Then
Solution: Given length = l = 3.233m uncertainty
4
Breadth = b =2.105m Thus V =(6.2 ± 0.1) cm3
3 i
Where 6.2 cm is calculated volume and 0.
Thickness = h = 1.05cm = 1.05 x 10 2
m js the uncertainty in it.
Volume = V = / x b x h Mr
= 3.233 x 2.105 x 1.05 x 10"
2
Fx /
m where v is the speed of
Example 4: Check thefecorrec of th lation V =
= 7.14573825 x 10“ 2
m3
As the Factor 1.05cm has minimum number of significant figures equal to three, Therefore transverse wave on a stretched stnn
Solution: Dimensions of L.H.S of rhp
Dimension
;ion F, length / and mass m. 11101022
equation = [ v ] =[ LT-1
]
volume is recorded upto 3 significant figures. Hence
is of R.H.S of
= (iMLT:
] x[ - ' J
—
Example 5: Derive a r
nation = ([F] x [/] x [ m"1
])
,/2
V = 7.15 x 10’2
m3
1
]I/2
= [ L2
rJ]1/2
= [ Lrl ]
Sine •th sides ot the equation are the same, equation is dimensionally
The mass of a metal box measured by a lever balance is 2.2kg. Two silver coin
Example 2: correct.
of masses lO.Olg and 10.02g measured by a beam balance are added to it What is now the toti
mass of the box correct upto the appropriate precision? 11101020
relation for the time period of a simple pendulum (Fig. 1.2) using
Solution: Total mass when silver coins are added to box
dimensional analysis.The various possible factors on which the time period T may depend are:
11101023
= 2.2kg + 0.01001kg + 0.01002 kg
i)
^1 Length of the pendulum (/) •
Since least precise is 2.2 kg, having one decimal place, hence total mass should be to c
^ Mass of the bob (m)
decimal place which is the appropriate precision. Thus the total mass = 2.2 kg. iii) Angle 0 which the thread makes with the vertical
= 2.22003 kg
iv) Acceleration due to gravity (g)
The diameter and length of a metal cylinder measured with the help of veriw Solution:
Example 3:
callipers of least count 0.01 cm are 1.22 cm and 5.35 cm.Calculate the volume V of the c lind The relation for the time period T will be of the form
Tocm° XlbX 0cXgd
11101021
and uncertainty in it.
or T = constant malh 0 gd
Solution: Given data is (1)
Diameter d = 1.22cm with least count 0.01cm Where we have to find the values of powers a, b, c and d.
Length / =5.35cm with least count 0.01cm Writing the dimensions of both sides we get
[ T ] = constant x [ M ]a
[ L ]b
[ LL 1
]c
[ LT“ 2
f
Absolute uncertainty in length =0.01cm
Comparing the dimensions on both sides we have
0.01cm
x 100 = 0.2%
%age uncertainty in length = [ T ] = [ T
5.35cm
I M ]° = [ M ]:
Absolute uncertainty in diameter = 0.01cm
bxl+c-c
[ L )° =[ L ]
0.01cm
%age uncertainty in diameter = x 100 =0.8% Equating powers on both the sides we get F u
1.22cm
1
v =
^ - 2d = 1 or d =
As volume is 2
4
and b + d = 0
a =0
10. Unique Notes Physics 1 ' Year
Solution:
Formula:
Calculation:
=(15.3cm) (12.80cm)
=( 15.3 x 12.80) (cm x cm)
= 195.8cm2
Rounded Area = 196 cm
1.4: Add the following masses given in kg
upto appropriate precision. 2.189,0.089,11.8
( Board 2009)
Solution:
Formula:
-Velocity of so
Density of the r
Modulusof elasticity of the medium =E
To Calculate
Relation between v, p and E.
Solution:
Formula:-
V « En x pm
v = constant Enxpm (A)
Calculation:
[EJ =(F/A) / (Av/v)
= (MLT2
/L2
) / (LVL5
)
= [ML 'T2
]
[P] =[m/V] =[M/L’l = [ML 3
]
v = constani x [ML ’T2
]"[ML'3
]"
[LT1
] =constant x [Mn L"T2
” Mm L'3m
|
[L'r'l =constant x [Mn*"L ” 3m
T2
"]
Comparing powers of L.H.S. to R.H.S.
n + m =0 .
-n -3m = 1
-2n =-1 ...
T = 2K J- 4K
2
? 3ugh medium = v
is = p
Fxr2
=> 8 = G =
T2
9 mim2
Calculation:
Calculation:
2
x- • . ~ 1 40.2
Time period =T =
^= [G] =
= 2.01 sec.
0.001 100
1.0 * 100
[M] [M]
= [M 1
L3
T2
]
and 5.32. 11101028
" % uncertainty in length =
= 0.1 %
Absolute uncertainty in time period
Data:
Units of G will be:
; ,n2 _
m, = 2.189 kg
ra2 =0.089 kg
m3 = 11.8 kg
rru -5.32 kg
Nx
G =
kg2
0.1
1.7: Show that the expressi
dimensionally correct, w
velocity at time t. .
^
Data:
--
^
- = 0.005 Vf = V| + at is
e vr is the
. '
1110103!
Solution: % uncertainty in time period
Addition gives
m = ni| + m2 + m3 + rru
=2.189 + 0.089 + 11.8 + 5.32
= 19.398
= 19.4 kg
1.5: Find the value of ‘g’ and its
0.005 100
"
2.01 x
100 " a25 %
Equation of motion
vf = vJTat
Vj = velocii
vf = velocity at t
a = acceleration
By using formula:
4 x (3.14)2
x 1.0
(2.01r
=9.76 ms 2
t =0
g =
I
( 1 )
% Uncertainty in g =% uncertainty in length Solution*
+ 2 x % uncertainty in time period
L (2)
— from the
uncertainty using T = 2n
(3)
Calculation:
LHS = [Vr] = [LT ']
RHS =[vj + at]
= [L T' + L T2
xT]
=[2LT1
]
=2]LT']
Neglecting constant 2 which is dimensionless,
we find that
9
= 0.1% + 2% x 0.25% = 0.6%
Using formula:
following measurements made during an
experiment. Length of simple pendulum
/ = 100 cm. Time for 20 vibrations = 40.2
sec. Length was measured by a metre scale
of accuracy upto 1 mm and time by stop
watch of accuracy upto 0.1 sec.
Data:
1
=> n = 2
1 9.76 ^ 0.6
0.6 x
= 0.058 ms'2
Put in ( 1 )
=> X =
100"
9.76
x = uncertainty in g = + 0.058 uncertain!
!
rounded off to ± 0.06 ms
'3
100
1
2 + m =0
1
11101029
m = -
-2
g = 9.76 ± 0.06 ms
1.6: What art the dimensions and units of
2
1 LHS = RHS
gravitational constant G in the formula. Hence above equation is dimensionally
correct.
Length of simple pendulum = 100cm From Eq. ( A)
v = constant x E ‘
x 9
100
m = lm = 1.0 m
(f
=:
i,
^
injnu
I = G ~
p—
Data:
100 11101030
v = constant x
1.8: The speed v of sound waves through a
Medium may be assumed to depend on (a )
Least count for length = I mm = 0.001m
Time for 20 vibration = t = 40.2 s
Least count of stop watch = 0.1s
To calculate
Value of g = ?
v = constant x
density p of the medium and (b) its
Modulus of elasticity E which is the ratio of
str«$s to strain. Deduce by the method of
dimensions, the formula for the speed of
sound.
Formula for gravitational constant
F = G
^
To calculate
Dimensions of G = ? 11101032
17
16
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12. Unique Notes Physics 1 Year
Unique Notes Physics 1“ Year
1.9: Shin* that the famous “Einstein
equation” E = me2
is dimensionally
consistent.
Data:
Given equation is E = me
To Determine:
Power of r = n
Power of v = m
To Calculate
Eli
m VECTORS A
11101033
RIUM
n = ?, m = ?
Solution:
Formula:*
a oc r"vm
y
What is the vector quantity
^
Ans. Vectors:
Definition:
Equation is dimensionally consistent.
Solution:
Since
« Ql. 11102001
I
a = constant r" vm
Using dimensional notation
[LT2
] =constant x [Ln x (LT ‘)m
]
[LT2
]
(1)
E = me2
where c is velocity of light
Dimensions of energy are that of work.
Therefore dimensions of L.H.S.
E= [ML'T2
]
Dimensions of RH.S. =[me21= (MILT1
)
2
]
Hence
“ Those physical quantities
description are calledyectors."
For example,
'
velocity; accc
The vectpl
d is ‘d'
(light face
Graphicall
end. While length
scalfy
Q2. What is meant by Rectangular Co-ordinate System or Cartesian Co-ordinate System?
Ans. Rectangular Coordinate System: 11102002
Two lines drawn at right angle to each other are known as
rdinate axes and their point of intersection is called as origin. This
lem of coordinate axes is called Cartesian or rectangular
coordinate system.
In this system one
named as y-axis. The x-axis is taken as horizontal axis with +ve
direction to the right side while y-axis is taken as vertical axis with +ve
direction upward.
uch reqinre both magnitude and direction for their complete
= constant x Ln+m
Tm
Comparing Power on both sides
m + n = 1 (2)
eratiojr' Force etc.
bold face characters such as.A, d, r, v, the magnitude of a vector
;ed
e
and -m= -2 .
From Eq. (3)
m =
~
2
(3) a vector isTepresented by a directed line segment with an arrow' head at its one
‘
thepline segment represents the magnitude of the vector according to the chosen
Dimensions of LHS =dimensions of RHS
U (MLY2
!
Sc above equation is dimensionally correct.
l.lO'
.Suppose, we are told that the
aroi icration of a particle moving in a circle
of radius r with uniform speed v is
proportional to some power of r, say r°, and
vKoe power of v, say v®
determine the
powers of r and v?
Data:
Radius of circle = r
Uniform linear speed = v
Acceleration = a
V
Put the value of m in eq. (2).
2 + n = 1,
n = -1
coo
x x
line is named as x-axis and other line is * O
11101034
Y
P (a.b)
The direction of a vector in a plane is required by the angle
which the representative line of vector makes with positive -axis in the
ami clockwise direction as shown in fig. The point P has coordinates
(a.b). This notation means that if we start at origin, we can reach point
P by moving “ a” unit, along positive x-axis and "b” units along positive
y-axis.
4 b
:
+x
O a
* The direction of a vector in space is denoted by another axis
is. This 3*axis is named as z-
which is at right angle to x-axis and y-ais
axis.
19
—18
