UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-1
LESSON 19
UNITS, DIMENSIONS, SIGNIFICANT FIGURES & ERRORS ANALYSIS & EXPERIMENTS
In this lesson, we introduce you to the foundation stone of common man’s science, Physics. As physics is a quantitative science, measurement is its integral part. To present the data quantitavely, understanding of significant figures is very significant. Error is inherent part of measurements. We shall see how these errors are estimated. For the global understanding, the results of an experiment, must be presented in standard units with the mention of confidence level. We shall also inculcate the idea and the importance of dimensions. Measurement constitutes the elementary part of Physics because every physical quantity needs a unit to specify it. We shall understand another related idea called ‘dimensions’. Error in measurement is bound to occur. We shall see how these are calculated.
IITJEE Syllabus: Units and dimensions, least count, significant figures; Methods of measurement and error analysis for physical quantities pertaining to the following experiments: Experiments based on using vernier calipers and screw gauge (micrometer), determination of g using simple pendulum, Young’s modulus by Searle’s method, Specific heat of a liquid using calorimeter, focal length of a concave mirror and a convex lens using u-v method, Speed of sound using resonance column, verification of Ohm’s law using voltmeter and ammeter, and specific resistance of the material of a wire using meter bridge and post office Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-2
Physics is a study of basic laws of nature and natural phenomena. The main thrust of physics is to explain diverse physical phenomena in terms of few concepts and laws. To understand the natural phenomena around us, we need a qualitative as well as quantitative analysis. Qualitative analysis tells us about the various factors on which a physical quantity depends. Quantitative analysis expresses mathematical relationships. To develop the mathematical relationships, a careful measurement of quantities is needed. In this module, we shall deal with measurement methods and the formulation of empirical relations between various physical quantities.
The main objective of physics is to use the limited number of fundamental laws that governs natural phenomena to develop theories that can predict the results of future experiments.
The fundamental laws used in developing theories are expressed in the language of mathematics, the tool that provides a bridge between theory and experiments.
Between 1600 and 1900, three broad areas were developed, which is together called classical physics.
(i) Classical Mechanics deals with the study of the motion of particles and fluids.
(ii) Thermodynamics deals with the study of temperature, heat transfer and properties of aggregations of many particles.
(iii) Electromagnetism deals with electricity, magnetism, electromagnetic wave, and optics.
These three areas explain all the physical phenomena with which we are familiar.
But by 1905 it became apparent that classical ideas failed to explain several phenomena. Then some new theories were developed in what is called Modern Physics.
Three important theories in modern physics are
(i) Special Relativity: A theory of the behavior of particles moving at high speeds. It led to a radical revision of our ideas of space, time and energy.
(ii) Quantum Mechanics: A theory of submicroscopic world of the atom. It also required a profound upheaval in our vision of how nature operates.
(iii) General Relativity: A theory that relates the force of gravity to the geometrical properties of space.
It is also useful in the study of other subjects like biotechnology, geophysics, geology etc.
SCOPES AND EXCITEMENT The scope of physics is very wide. We can understand the scope of physics by looking at its various sub-discipline. It covers a very wide variety of natural phenomena. It deals with the phenomena from microscopic level to macroscopic level. The microscopic domain includes atomic, molecular and nuclear phenomena. The macroscopic domain includes phenomena at laboratory, terrestrial and astronomical scales.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-3
For example forces we encounter in nature are nuclear forces, chemical forces and forces exerted by ropes, springs, fluids, electric charges, magnets, the earth and the Sun. Their ranges and relative strength can be summarized as shown below in the table.
Forces Relative Range Strength
Strong 1 10-15 m
Electromagnetic 10-2 Infinite
Weak 10–6 10–17 m
Gravitational 10-38 Infinite
Similarly the range of distance we study in physics vary from 10-14 m (size of nucleus) to 10+25 m (size of universe) The range of masses includes in study of physics varies from 10-30 kg (mass of electron) to 1055 kg (mass of universe) The range of time interval varies from 10-22 sec (time taken by a light to cross a nuclear distance) to 1018 sec. (life time of sun). So we can say scope of physics is really wide. The study of physics is very exciting in many ways. Excitement of Physics can be seen in every field. Advancement of technology has upgraded the entire scenario of entertainment, starting from a radio to the most advanced cyber park. Communication system has been also deepened its root by bridging distant areas closer. Advances in health science, which has enabled operations without surgery. Telescopes & satellite have broken the limits of knowledge of the undiscovered universe. Exploring the sources of energy from the unexplored sources. Made possible the reach of man beyond the earth, towards the cosmos. Use of Robots in a hazardous places is highly beneficial.
The process of measurement of a physical quantity involves its comparison with a fixed standard. We commonly use these standards. For example, the length of a line drawn on a sheet of paper is measured with help of a scale as, say, 10 cm. Weight of sugar purchased from a shop is expressed as 2 kg or 3 kg. The suffix ‘cm’ or ‘kg’ is the fixed standard.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-4
UNITS A fixed standard, used as a reference for a measurement, is called unit. Like in the above example, length of a line is expressed as 10 cm. Here ‘cm’ is a unit for measurement of length. In general, any measurement can be expressed as X = nu Where, n is the numerical part and u is the unit. For a given value of X, n × u = constant. 1 n u The numerical part of measurement is inversely proportional to the size of unit. The choice of a unit must be both sensible and practical.
For example a length can be written as, either 10 cm or 0.1 m. The length is same in both cases, but as the size of unit changes from 1 cm to 1 m, the numerical part also changes.
PHYSICAL QUANTITY Any quantity, involved in a physical phenomenon, is treated as physical quantity. There are a large number of physical quantities. It is not possible for us to define a separate standard for each physical quantity. As the physical quantities are interdependent, we need not to do such a laborious work. By international agreement, a small number of quantities such as length and time have been chosen, and they have been assigned standards. Remaining quantities have been expressed in terms of the chosen quantities. So, physical quantities have been classified in two categories.
1. Fundamental (or base) quantities: These quantities have been chosen and assigned standards. The units for these quantities are called fundamental units. 2. Derived quantities: These quantities are expressed in terms of fundamental quantities. The units for these quantities are called derived units. A complete set of these units, both the base units and derived unit is known as system of units.
International System of Units (System International d Unit’s)
This system is known as S.I. system of units. In this system, following quantities have been defined as fundamental quantities.
S.No. Name Unit Symbol 1. Length metre m 2. Mass kilogram kg 3. Time second s 4. Electric current ampere A
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-5
5. Temperature kelvin K 6. Amount of substance mole mol 7. Luminous intensity candela cd
In addition to the above, two supplementary quantities are defined. They are plane angle and solid angle. The units for these are radian (rad) and steradian (sr) respectively.
Different system of units were developed in different regions of world. Following are three commonly used system of units.
M.K.S. System: In this system, length is expressed in metre, mass in kilogram and time in second.
C.G.S System: In this system, length is expressed in centimeter, mass in gram and time in second.
F.P.S. System: In this system, unit of length is ‘foot’, unit for mass is ‘pound’ and that for time is ‘second’.
Illustration 1
Write the unit of following derived quantities is SI system.
(i) Velocity (ii) Acceleration (iii) Force (iv) Energy
Solution:
displacement (i) Velocity Its unit is ms 1 time
change in velocity m/s (ii) Acceleration Its unit is ms2 time s
(iii) Force = mass × acceleration Its unit is kg ms2
(iv) Energy = Force × displacement Its unit is kg ms2 m kg m2s2
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-6
Illustration 2
The SI unit of force is newton such that 1N = 1kgms2 . In C.G.S. system, force is expressed in dyne. How many dyne of force is equivalent to a force of 5 N?
Solution:
Let 1 N = n dynes
1kg m g cm n s2 s2
1000g (100)cm g cm or n s2 s2
n 105
5 N = 5 × 105 dyne.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-7
SI BASE QUANTITIES AND UNITS
Base SI Unit
Quantity Name Symbol Definition Length metre m 1 metre is the length of the path traveled by light in vacuum during a time interval of 1/299, 792, 458 of a second. (1983) Mass kilogram kg 1 kilogram is equal to the mass of the international prototype of the kilogram (a platinum-iridium alloy cylinder) kept at International Bureau of Weights and Measures, at Sevres, near Paris, France. (1989) Time second s 1 second is the duration of 9,192,631, 770 periods of the radiation corresponding to the transition between the two-hyperfine levels of the ground state of the cesium- 133 atom. (1967) Electric current ampere A 1 ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to 2107 newton per metre of length. (1948) Thermo kelvin K 1 kelvin, is the fraction 1/273.16 of the thermodynamic dynamic temperature of the triple point of water. (1967) temperature Amount of mole mol 1 mole is the amount of substance of a system, which substance contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12. (1971) Luminous candela cd 1 candela is the luminous intensity, in a given direction, intensity of a source that emits monochromatic radiation of frequency 540 1012 hertz and that has a radiant intensity in that direction of 1/683 watt per steradian. (1979)
Some SI Derived Units expressed in SI Base Units
SI Unit Physical Quantity Name Symbol
Area Square metre m2
Volume cubic metre m3
Speed, velocity metre per second m/s or m1
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-8
Angular velocity radian per second rad/s or rad s1
Acceleration metre per second square m/s 2 or ms 2
Angular acceleration radian per second square rad/s2 or rad s2
Wave number per metre m1
Density, mass density kilogram per cubic metre kg/m3 or kg m3
Current density ampere per square metre A/m2 or Am 2
Magnetic field strength, magnetic intensity, ampere per metre A/m or Am 1 magnetic moment density Concentration (of amount of substance) mole per cubic metre mol/m 3 or mol m3
Specific volume cubic metre per kilogram m3 / kg or m3 kg1
Luminance, intensity of illumination candela per square metre cd/m2 or cd m2
Kinematic viscosity square metre per second m2 / s or m2 s1
Momentum kilogram metre per second kg m s1
Moment of inertia kilogram metre square kg m2
Radius of gyration metre m
Linear/superficial/volume expansivities per kelvin K 1
CONVERSION TABLE
Length Force
1 in = 2.54 cm 1 N = 105 dyn = 0.2248 lbf
1 m = 39.37 in = 3.281 ft 1 lbf = 4.448 N
1 ft = 0.3048 m 1 kgf = 2.2046 lbf
12 in = 1 ft Speed
3 ft = 1 yd 1 mi/h = 1.47 ft/s = 0.447 m/s = 1.61 km/h
1 yd = 0.9114 m 1 m/s = 100 cm/s = 3.28 ft/s
1 km = 0.621 mi 1 mi/min = 60 mi/h = 88 ft/s
1 mi = 1.609 km Acceleration
1 mi = 5280 ft 1m / s2 3.28 ft / s2 100 cm / s2
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-9
1 Å = 1010 m 1ft / s2 0.3048 m / s2 3.48 cm / s2
1 m = 106 m 104 Å Pressure
1 light-year = 9.461 × 1015 m 1bar 105 N/ m2 14.50 lb / in2
Area 1atm 760 mmHg 76.0 cmHg
1m2 104 cm2 10.76 ft2 1atm 14.7 lb / in2 1.013 105 N/ m2
1ft 2 0.0929 m2 144 in2 Time
1in2 6.452 cm2 1year 365 days 3.16 107 s
Volume 1day 24 h 1.44 103 min 8.64 104 s
1m3 106 cm3 6.102 104 in3 Energy
1ft3 1728 in3 2.88 102 m3 1J 0.768 ft. lb 107 ergs
1liter 1000 cm3 1.0576 qt 0.0353 ft3 1 cal = 4.186 J
1ft3 7.481 gal 28.32 liters 2.832 102 m3 1Btu 252 cal 1.054 10 3 J
1gal 3.786 liters 281in3 1 eV = 1.6 × 1019 J
Mass 931.5 MeV is equivalent to 1 u
1000 kg = 1 t (metric ton) 1 kWh = 3.60 × 106J
1 kg = 2 lb Power
1 u = 1.66 × 1027 kg 1 hp = 550 ft. lb/s = 0.746 kW
1 W = 1 J/s = 0.738 ft. lb/s
1 Btu/h = 0.293 W
SI PREFIXES AND SYMBOL
Power of 10 Prefix Symbol 18 exa E 15 peta P 12 tera T 9 giga G 6 mega M
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-10
3 kilo k 2 hecto h 1 deca da – 1 deci d – 2 centi c – 3 milli m – 6 micro – 9 nano n – 12 pico p – 15 femto f – 18 atto a
DIMENSIONS A derived physical quantity is expressed as some combination of seven fundamental quantities. The seven fundamental quantities are termed as dimensions of the physical world. The dimensions are expressed in square brackets ‘* +’.
Following tables gives the dimensions of even physical quantities.
S. No. Fundamental quantity Dimension 1. Length [L] 2. Mass [M] 3. Time [T] 4. Electric current [A] 5. Temperature [K] 6. Amount of substance [mol] 7. Luminous intensity [cd]
Dimensions of a Physical Quantity The dimension of a physical quantity are the powers (or exponents) to be raised on the base quantities to represent the quantity. For example volume occupied by a cuboid is the product of its length, breadth and height. So dimensions of volume are [L] × [L] × [L] = [L]3 . This is expressed by saying the volume has 3 dimensions in length.
DIMENSIONAL FORMULA An expression, which shows how and which of the fundamental quantities represent the dimensions of a physical quantity, is called the dimensional formula of the given physical quantity.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-11
For example the dimensional formula for force can be calculated as follows. displacement Force = mass × acceleration = mass time2
[F] [M][L][T ]2
DIMENSIONAL ANALYSIS Here, we will study the basic tool for qualitative analysis of physical phenomena and a method to find out empirical relationships between various physical quantities. The basis of dimensional analysis is principle of homogeneity of dimensions.
Illustration 3
Find dimensional formula for universal gravitational constant.
Solution:
G m m We know that F 1 2 r 2
F r 2 G m1 m2
[MLT 2][L]2 [G] [M1L3T 2] [M][M]
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-12
HAVING SAME DIMENSIONS
Dimension Quantity
[M 0L0T 1] Frequency, angular frequency, angular velocity, velocity gradient and decay constant
[M1L2T 2 ] Work, internal energy, potential energy, kinetic energy, torque, moment of force
[M1L1T 2] Pressure, stress, Young’s modulus, bulk modulus, modulus of rigidity, energy density
[M 0L1T 2 ] Acceleration due to gravity, gravitational field intensity
[M1L1T 2 ] Thrust, force, weight, energy gradient
[M1L2T 1] Angular momentum and Planck’s constant
[M1L0T 2 ] Surface tension, Surface energy (energy per unit area)
[M0L0T 0 ] Strain, refractive index, relative density, angle, solid angle, distance gradient, relative permittivity (dielectric constant), relative permeability etc.
[M 0L2T 2 ] Latent heat and gravitational potential
[ML2T 2K 1] Thermal capacity, gas constant, Boltzmann constant and entropy
[M 0L0T 1] l m R , , , where l = length g = acceleration due to gravity, m = mass, g k g k = spring constant, R = Radius of earth 0 0 1 L [M L T ] , LC, RC where L = inductance, R = resistance, C = capacitance R
2 2 2 2 [ML T ] 2 V 2 q 2 I Rt, t,VIt, qV, LI , ,CV where I = current, t = time, q = charge, R C L = inductance, C = capacitance, R = resistance
IMPORTANT DIMENSIONS OF COMPLETE PHYSICS
Quantity Unit Dimension Temperature Kelvin [M 0L0T 0K 1]
Heat Joule [ML2T 2]
Specific Heat Joule/kg-K [M 0L2T 2K 1]
Thermal capacity Joule/K [M1L2T 2K 1]
Latent heat Joule/kg [M 0L2T 2 ]
Gas constant Joule/mol-K [M1L2T 2K 1 mol–1]
Boltzmann constant Joule/K [M1L2T 2K 1]
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-13
Coefficient of thermal conductivity Joule/m-s-K [M1L1T 3K 1]
1 0 3 4 Stefan’s constant Watt / m2 K 4 [M L T K ]
Wien’s constant Metre-K [M 0L1TK]
Planck’s constant Joule-s [M1L2T 1]
0 0 0 1 Coefficient of Linear Expansion Kelvin1 [M L T K ]
Mechanical equivalent of Heat Joule/Calorie [M0L0T 0 ]
Vander wall’s constant (a) Joule m3 / mol2 [ML5T 2 mol–2 ]
0 3 0 Vander wall’s constant (b) m3 / mol [M L T mol]
ELECTRICITY
Quantity Unit Dimension
Electric charge Coulomb [M 0L0T 1A1]
Electric current Ampere [M 0L0T 0A1]
Capacitance Coulomb/volt or Farad [M 1L2T 4 A2 ]
Electric potential Joule/coulomb [M1L2T 3A1]
Coulomb2 Permittivity of free space [M 1L3T 4A2] Newton metre2
Dielectric constant Unit-less [M0L0T 0 ]
Resistance Volt/Ampere or ohm [M1L2T 3A2]
Resistivity or Specific resistance Ohm-metre [M1L3T 3A2]
volt second or henry Coefficient of Self-induction ampere [M1L2T 2A2] or ohm-second
Magnetic flux Volt-second or Weber [M1L2T 2A1]
newton
amper metre Joule Magnetic induction [M1L0T 2A1] amper metre2 volt second or Tesla metre2
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-14
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-15
Quantity Unit Dimension
Magnetic Intensity Ampere/meter [M 0L1T 0A1]
Magnetic Dipole Moment Ampere metre 2 [M0L2T 0A1]
Newton
ampere2 Joule or ampere2 metre
Volt second 1 1 2 2 Permeability of Free Space or [M L T A ] ampere metre Ohm second or metre henry or metre 0 2 1 1 Surface charge density Coulomb metre2 [M L T A ]
Electric dipole moment Coulomb – metre [M 0L1T 1A1]
1 2 3 2 Conductance ohm1 [M L T A ]
Conductivity ohm1 metre 1 [M 1L3T 3A2]
2 Current density Ampere / m M 0L2T 0A1
Intensity of electric field Volt/metre, Newton/coulomb M1L1T 3A1
Rydberg constant m1 M0L1T 0
PRINCIPLE OF HOMOGENEITY OF DIMENSIONS The simple fact that the magnitudes of two or more physical quantities can be added or subtracted from one another, only if they have the same dimensions, is called principle of homogeneity of dimensions. It is very simple to understand. We cannot add a velocity of 1 m/s to a force of 10 N. It will be meaningless. Before moving on to dimensional analysis, following points must be remembered.
(a) A + B and A – B are meaningful if A and B have same dimensions.
(b) If A = B is a correct relationship, then A and B have same dimensions.
(c) All trigonometric ratios, logarithmic and exponential functions are dimensionless.
(d) Arguments of the above functions are also dimensionless.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-16
Dimensional analysis can be used for the following points
(a) Check the consistency of a given result or formula.
(i.e., to check the dimensional correctness of a physical relation).
(b) Deduce a relationship among physical quantities.
(c) To convert a physical quantity from one system to another.
We shall discuss them one by one.
CHECK THE CONSISTENCY OF A GIVEN RESULT
A given result is dimensionally consistent if it obeys principle of homogeneity.
1 For example let us consider a well-known formula for position of an object, s ut at 2 . Here u is initial 2 velocity, a is acceleration, t is time and s is the distance travelled. First we shall calculate the dimensions of each terms.
Dimensions of s = [L].
Dimensions of ut [LT 1][T ] [L] .
1 Dimensions of at 2 [LT 2][T 2] [L] 2
So, the above result can be dimensionally expressed as [L] = [L] + [L]
The above equation is dimensionally consistent and therefore the formula is also dimensionally consistent. Note that the dimensional consistency does not assure physical consistency of a formula.
9 For example if we have to verify the dimensional consistency of a formula given by, s 5ut at 2, we 2 will come to same final dimensional equation [L] = [L] + [L]. But the formula is not correct physically. So remember the following two points Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-17
(a) A physically correct relation must be dimensionally correct.
(b) A dimensionally consistent relation may not be physically correct.
DEDUCE A RELATIONSHIP BETWEEN PHYSICAL QUANTITIES
Dimensional analysis is very helpful in determining empirical relationships between physical quantities provided they bear a simple relationship. First we shall discuss a few examples for the deduction of relationship and then we shall study what kind of relations cannot be deduced by dimensional analysis.
For example to determine the time period of a simple pendulum, an experiment is performed to find out the factors on which its time period depends. It is observed that
(i) On changing the length of the pendulum, its time period changes.
(ii) On changing the mass of the bob, there is no effect on the time period.
(iii) On changing the height of bob with respect to ground, time period changes. From these observations, we can guess that time period depends on length and on acceleration due to gravity. (As we known that acceleration due to gravity varies with height from ground)
Let us now assume that time period T of the pendulum depends on some powers of length L and acceleration due to gravity g. Therefore we have,
T La
T g b
or T k La g b [where k is some dimensionless constant]
If the above expression is dimensionally consistent, the dimensions on the two sides must be same.
[M0L0T 1] [M0 L0 T 0 ][L]a [LT 2]b
or M0 L0 T 1 M0 Lab T 2b Thus, we get a + b = 0 and – 2b = 1 1 1 b , a 2 2 So, we have the expression for time period as
1 1 L L T k L2 g 2 k k g g
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-18
L The actual formula is T 2 . g
The value of k (= 2) cannot be determined by dimensional analysis. It has to found experimentally or by other means. In the following examples, we will need to know the dimensional formula for various quantities. We can find them in the table given at the end of this module.
To convert a physical quantity from one system to the other: The measure of a physical quantity is nu = constant
If a physical quantity X has dimensional formula [MaLb T c ] and if (derived) units of that physical a b c a b c quantity in two systems are [M1 L1 T1 ] and [M2 L2 T2 ] respectively n1 and n2 be the numerical values in the two systems respectively, then n1[u1] n2[u2] a b c a b c n1[M1 L1 T1 ] n2 [M2 L2 T2 ]
a b c M1 L1 T1 n2 n1 M2 L2 T2
Where M1, L1 and T1 are fundamental units of mass, length and time in the first (known) system M2, L2 and T2 are fundamental units of mass, length and time in the second (unknown) system. Thus knowing the values of fundamental units in two systems and numerical value in one system, the numerical value in other system may be evaluated.
Example: (i) Conversion of Newton into Dyne. The Newton is the S.I. unit of force and has dimensional formula [MLT 2 ] . So 1 N = 1 Kg-m/sec2 a b c M1 L1 T1 By using n2 n1 M2 L2 T2 1 1 2 kg m sec 1 g cm sec 1 1 3 2 2 10 g 10 cm sec 5 1 10 g cm sec 1 N = 105 Dyne
(ii) Conversion of gravitational constant (G) from C.G.S. to M.K.S. system The value of G in C.G.S. system is 6.67 × 10–8 C.G.S. units while its dimensional formula is [M 1 L3 T 2 ] So G = 6.67 × 10–8 cm3/g s2 a b c M1 L1 T1 By using n2 n1 M2 L2 T2
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-19
1 3 2 8 g cm sec 6.67 10 kg m sec
1 3 2 g cm sec 6.67 10 8 3 2 10 g 10 cm sec 6.67 1011 G = 6.67 × 10–11 M.K.S. units.
Illustration 4
Assuming that critical velocity of flow of a liquid through a narrow tube depends on radius of the tube (r), density of liquid () and viscosity of liquid (), find an expression for critical velocity.
Solution:
Let the expression for critical velocity is v k r a b c [where k is a dimensionless constant]
Now [v] [LT 1], [r ] [L], [] [ML3 ], [] [M L1T 1]
M0 LT 1 [La ][ML3 ]b [M L1T 1]c M bc La3bc T c
Equating powers of M, L and T, we haves
b + c = 0 ..... (i)
a – 3b – c = 1 ..... (ii)
– c = – 1 ..... (iii)
c = 1, b = – 1, a = – 1
The expression for velocity is v k r 1 1 1
k or v . r
Illustration 5
The velocity of sound in a gas depends on its pressure and density. Obtain a relation between velocity, pressure and density.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-20
Solution:
Let the relationship is expressed as v k P a c, where P is pressure, is density and v is velocity of sound.
Now, [v] [LT 1],[P] [M L1T 2], [] [ML3 ]
[M0 LT 1] [M L1T 2]a [M L3 ]c
or [M0 LT 1] [Mac La3c T 2a ]
Equating the powers of M, L and T, we have a + c = 0, – a – 3c = 1, – 2a = – 1 1 1 a , c 2 2
P Expression for velocity of sound is of the form v k .
r P C Actual formula is v where P (adiabatic exponent). CV
Illustration 6
If density [D], acceleration due to gravity (g) and frequency (f) are taken as base quantities, find the dimensions of force.
Solution:
The dimensions of various quantities involved are [D] = [M L3 ], [g] [LT 2], [f ] [T 1]
Let force be expressed in terms of above quantities such that the dimensions of force are given by
[F] [Da g b f c ]
[M LT 2] [M L3 ]a [LT 2]b [T 1]c
or [M LT 2] [Ma L3ab T 2bc ]
Equating powers of M, L and T, we have a = 1, – 3a + b = 1, – 2b – c = – 2 a = 1, b = 4, c = – 6
[F] [D g 4 f 6 ]
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-21
Illustration 7
In a new system of units, unit of mass is taken as 50 kg, unit of length is taken as 100 m and unit of time is 1 minute. What will be the weight of a body in this system, if in SI system, its weight is 10 N.
Solution:
Let the weight of the body in new system is X units
10 N = X units
Let M1, L1,T1 and M2, L2,T2 be the symbols for mass, length and time in the two system respectively, then
2 2 10[M1 L1T1 ] X [M2 L2 T2 ]
M1 1 kg, M2 50 kg,L1 1 m,L2 100 m,T1 1 s,T2 60 s
2 10 1 1 36000 X 7.2units 50 100 60 5000
10 N = 7.2 units.
Illustration 8
bt 3 The distance travelled by an object depends on time according to following equation. x a a , where x is position and t is time. Find dimensions of a and b.
Solution:
b t 3 According to principle of homogeneity x, a and will have same dimensions. a Dimensions of x = [L] Dimensions of a = [L] bt 3 Dimensions of [L] a b[T 3 ] [L] [L]
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-22
Dimension of b = [L2 T 3 ].
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-23
Illustration 9
x 2 Consider the equation F e E , where F is force x is distance E is energy and , are constants. What are the dimensions of and ?
Solution:
As argument of exponential functions is dimensionless, so
x 0 0 0 Dimensions of [M L T ] E
[][L] [M0L0 T 0 ] [M L2 T 2]
[] [M LT 2]
2 Now will have dimensions of force
2 [F]
[2] [M LT 2] [M LT 2]
[] M1 L1T 2
MEASURING INSTRUMENTS
A device used for measurement of a physical quantity is called a measuring instrument. For example, a watch is used for measurement of time a meter scale is used for measurement of length etc.
LEAST COUNT
The smallest measurement that can be taken with the help of a measuring instrument is called its least count. Like, in a simple wristwatch, the second’s hand moves through a smallest time interval of 1 s. So, we can measure a minimum time of 1 s. Thus, the least count of the wristwatch is 1 s similarly the least count of a normal metre scale is 1 mm.
Following two points are worth remembering regarding the least count.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-24
(i) Least count represents the permissible error in the measurement. (ii) It gives the limit of resolution of the instrument. The following illustrations will make the above statements clear.
For example a simple geometry box scale has smallest division of 0.1 cm marked on it. This is the least count. When a measurement is taken from this scale, the result can be 5.1 cm or 7.9 cm, but it cannot be 5.15 cm or 7.93 cm. Reporting the answer as 7.93 cm gives a wrong information, it shows that the least count is 0.01 cm i.e., instrument has greater resolution.
For example a length measured by a geometry box scale of least count 0.1 cm, is reported as 5.5 cm. This actually implies that the length could be any value between 5.4 and 5.6 cm. Thus, there is a maximum possible error of 0.1 cm in the result.
ACCURACY AND PRECISION OF INSTRUMENTS
Accuracy of an instrument represents the closeness of the measured value of actual value. Precision of an instrument represents the resolution of the instrument. It depends on least count.
For example a physical quantity is measured from two instruments A and B. The reading of A is 2.54 cm (say) and that of B is 2.516 cm. The actual value is 2.53 cm. The first reading is closed to actual value, it has more accuracy. The second reading is less accurate, but the instrument B has greater resolution as it can measure upto 3 decimal places.
Sensitivity: the ratio of output signal or response of the instrument to a change of input or measured variable.
Resolution: the smallest change in measured value to which the instrument will respond.
Error: deviation from the true value of the measured variable.
SIGNIFICANT FIGURES
When a certain measurement of length is taken by a scale of least count 0.01 cm, the result is say, 7.62 cm. In this measurement, there exists a maximum possible error of 0.01 cm. Thus, we can say that, in the above measured value, ‘7’ and ‘6’ are certain digits and last digit ‘2’ is uncertain digit. The measured value, in this case, is said to have three significant figures. Following are a few important points regarding significant figures.
(i) Significant figures represent the precision of measurement, which depends on the least count of the measuring instruments.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-25
(ii) A choice of different units does not change the number of significant digits in a measurement.
(iii) Significant figures in a result after calculations are, the number of certain digits plus one uncertain digit.
Rules for obtaining number of significant figures in a number:
(i) All non-zero digits are significant.
(ii) All zeroes between non-zero digits are significant.
(iii) If a number is less than 1, the zeroes to the right of decimal point, but to the left of first non-zero digits are insignificant.
(iv) In a number without decimal, trailing zeroes are not significant.
(v) In a number with decimal, trailing zeroes are significant.
(a) 6 m has 1 significant figure.
(b) 6.0 m has 2 significant figures.
(c) When 6.0 m is converted in cm, we get 600 cm. Now, it has only one significant figure. But a mere change of units should not change the number of significant figures, so the final result should be written as 6.0 102 cm .
(d) 0.1 has 1 significant figure.
(e) 0.010 × 101 has 2 significant figures.
Rounding off The process of rounding off involves the rejection of uncertain digits in a result obtained after calculations. It is done in calculations with approximate numbers, which contain more than one uncertain digit. Following rules are followed while rounding off a result.
Rule 1: If the insignificant digit to be dropped is more than 5, the preceding digit is raised by 1 and if the insignificant digit to be dropped is less than 5, the preceding digit is left as it is.
Rule 2: When the insignificant digit to be dropped is equal to 5 or 50 or 500 or 5000, the preceding digit is left unchanged if it is even and it is raised by 1, if it is odd. (a) 2.746 rounded off to three significant digits is 2.75. (b) 2.743 rounded off to three significant digits is 2.74. (a) 2.745 rounded off of three significant figures is 2.74. (b) 2.735 rounded off to three significant figures is 2.74.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-26
Illustration 10
A cube has side of length 1.2 101m . Calculate its volume.
Solution:
Volume = (Side)3 (1.2 102 )3 1.728 106 m3
As the volume cannot have more than two decimal digits, hence volume should be reported as 1.7 106 m3 .
Illustration 11
Round off the following numbers to three significant digits.
(a) 14.745 (b) 14.750 (c) 14.650 1012
Solution:
(a) 0.045 is less than 0.050 so its dropped out without any change in previous digit so rounded off value will be 14.7.
(b) To remove 0.050, we have to see its previous digit, which is 7 (odd). So rounded off value is 14.8.
(c) In this case the digit before 0.050 is even hence the rounded off value is 14.61012 .
Rules for Arithmetic Operations with Significant Figures
1. Addition/Subtraction: The number of digits after decimal in the final answer should be equal to minimum number of digits after decimal in any of the used values. Write the numbers one below the other with decimal points coinciding add or subtract as required. Now located, from left, first column containing a doubtful digit. All digits to right of this are dropped.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-27
2. Multiplication/Division: The final result contains number of significant digits equal to the least number of significant digits in the values used for calculation.
3. Numeric constants: The constant occurring in calculations like numbers, etc., have infinite significant figures. They can be rounded off as per the most precise quantity involved in the calculations.
ERRORS The measured value of a physical quantity is usually different from its true value. The result of any experimentally measured value contains some uncertainty. This uncertainty is called error.
In general, the errors in measurements can be broadly classified into two categories as
(a) Systematic errors and
(b) Random errors
(c) Accidental errors
Systematic errors: These are the errors occurring due to the instruments used, imperfection in experimental technique or carelessness of the person performing the experiment. These errors tends to move in one direction only.
Random errors: The random errors occur irregularly and therefore random in magnitude and sign. These errors tends to move in both directions. They may arise due to unpredictable fluctuation while performing the experiment. These errors can be minimized by taking a large number of observations for the same physical quantity.
Accidental errors: Some time the measured value is to much small or to much larger then their average value. This is called accidental error. These values are dropped before calculating the average value.
Instrumental errors: These errors depend on the instruments used for measurement. Mainly following two types of errors occur due to instruments.
(i) Zero error: It is a type of systematic error. It occurs due to improper positioning of zero mark of the instrument.
(ii) Least count error: This is the error associated with the resolution of the instrument. Least count has been discussed in details in the earlier section.
For example to measure the length of an object its one end is placed at the zero mark of the scale and other end is found to come at 10th millimeter mark from the start. This implies that the length of the object is 10 mm. But, if one end of the object is placed at 1st mm mark instead of zero mark, the measurement will
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-28
come out be 11 mm. So a zero error of 1 mm is introduced in the measurement.
ABSOLUTE ERROR, RELATIVE ERROR AND PERCENTAGE ERROR
Let a physical quantity is measured in an instrument. Various readings taken are x1, x2, x3 ...... , xn. The true value of the measurement is taken as the mean of above readings.
x x x ..... x x 1 2 3 n n
Absolute error: The magnitude of the difference between the true value of the quantity and the individual measured value is called absolute error of the measurement.
Therefore, x1 x1 x is the absolute error in measurement of x1.
Mean absolute error: The arithmetic mean of all the absolute errors is taken as the final or mean absolute error in the value of physical quantity.
x x x ..... x x 1 2 3 n n
The final answer or result of the experiment is reported as x x. This implies that any measurement of physical quantity x is likely to fall between (x – x) and (x + x).
Relative error: Relative error is the ratio of the mean absolute error to mean value of quantity measured. x Relative error = . It represents the accuracy in measurement i.e., smaller the relative error more is the x accuracy.
Percentage error: When the relative error is expressed as percent, it is called the percentage error. x Percentage error = 100% . x
For example in an experiment, the time period of oscillations of a simple pendulum is measured and the readings are 3.63 s, 3.56 s, 3.42 s, 3.71 s and 3.80 s.
3.63 3.56 3.42 3.71 3.80 The true value is x 3.62 s. 5
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-29
The absolute errors in the measurements are
x1 x1 x 0.01s
x2 x2 x 0.06 s
x3 x3 x 0.20 s
x4 x4 x 0.09 s
x5 x5 x 0.18 s
0.01 0.06 0.20 0.09 0.18 Mean absolute error x = 0.11s . 5
From the above calculations, we can infer following two points: (i) The time period of oscillations of the pendulum is (3.62 0.11) s. (ii) As the absolute error is 0.11 s, there is already an error in the first digit after decimal. So there is no point in reporting to the answer to two digits after decimal. Thus, a better way to write the result is (3.6 0.1) s.
Combination of errors: In many cases, a physical quantity depends on certain other measurable physical quantities. The accuracy or error in the measurement of the physical quantity depends on the errors in the quantities on which it depends. In calculations, errors can combine in different manners.
(a) Error of a sum or difference: Let a quantity Z depends on two quantities A and B as
Z = A + B.
Z Z = (A A) + (B B) Here A and B are absolute errors in the measurements of A and B. The above equation can be written as Z Z = (A + B) (A + B) Z = A + B (i.e., absolute errors are added up) Note: The above result is applicable for Y = A – B also. That is the maximum possible error in Y is Y = A + B.
Illustration 12
The original length of a wire is (153.7 0.6) cm. It is stretched to (155.3 0.2) cm. Calculate the elongation in the wire with error limits. Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-30
Solution: Elongation l'l 155.3 153.7 1.6 cm Error in elongation d = d l'd l = (0.6 + 0.2) = 0.8 cm So elongation = (1.6 0.8) cm.
(b) Error of a product or division: Let Z = A · B
Z Z = (A A) · (B B)
Z Z = A · B A · B B · A A · B
Z = A · B + B · A + A · B
When A and B are small, A · B can be neglected. So, we have,
Z AB BA B A Z = A · B + B · A Z Z Z B A
Z A B or (i.e., relative errors are being added) Z A B
Note: Same result is applicable for division of two quantities.
A Y A B That is if Y , the B Y A B
Alternate Method: Z = A · B or log Z = log A + log B
Differentiating both sides w.r.t. Z, we get
1 1 dA 1 dB Z A dZ B dZ
dZ dA dB Z A B or This is same as Z A B Z A B
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-31
Illustration 13
The mass of a container measured by a beam balance is 2.4 kg. Two small objects of masses 30.12 g and 30.15 g are added to the container. Calculate (a) the total mass of the container and (b) difference in masses of the pieces to correct significant figures? Solution:
30.12 30.15 (a) Total mass of the container = 2.4 1000 1000
= 2.4 + 0.03012 + 0.03015
= 2.46027
= 2.5 kg (final result should contain one digit after decimal).
(b) Difference = 30.15 – 30.12
= 0.03 g (final result contains 2 digits after decimal).
Illustration 14
Area of a rectangle is measured by measuring its sides. The sides are found to be (2.0 0.1) cm and (3.00 0.01) cm. Find its area within proper error limits.
Solution:
Area = side × side
A = 2.0 × 3.00 = 6.0 cm2 (Result must contain 2 significant digits)
As A = L × B
ΔA L B A L B
L B 0.1 0.01 or A A = 6.0 0.32 cm2 L B 2.0 3.00
So, final result is, A = (6.0 0.32)cm2
The result should be reported as A = (6.0 0.3)
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-32
Illustration 15
The length of an object is measured as 1.525 m. Find the error, relative error and percentage error.
Solution:
As nothing is mentioned about the measuring instrument by which reading is taken, we shall assume that last digit of value 1.525 is uncertain. The value can lie between 1.524m – 1.526m. Absolute error = 0.001 m 0.001 Relative error = 0.00065 1.525 Percentage error = 0.065% Note: If significant figures are considered in the above example, relative error would be 0.001 and percentage error would be 0.1%.
Illustration 16
If velocity of light in vacuum (2.998 108 ms 1), acceleration due to gravity (9.81ms 2 ) and density of mercury (13600 kg m3 ) be adopted as the fundamental units, then what will be the corresponding unit of mass.
Solution:
Let the dependence of mass (M) on velocity (V), acceleration due to gravity (g) and density () be as M V x g y z
Substituting the dimensions of V, g and in R.H.S.
M L0 T 0 [M0 LT 1]x [M0 LT 2]y [M L3 ]z
or M1 L0 T 0 M z Lxy 3zT x2y
Comparing the dimensions on both sides, we get x = 6, y = – 3 and z = 1
Therefore, mass [M] = V 6 g 3
(2.998 10 8 )6 Hence, unit of mass = 13600 1.05 10 52 kg . (9.81)3
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-33
Illustration 17
A body moves in a straight line such that half of its journey is covered with velocity
V1 (15.0 0.5) m/s and the next half with velocity V2 (30.0 0.1) m/s . Find average velocity of the body.
Solution: Total displacement It is given that v av Total time 2d v av d d v1 v 2 1 1 2 v1 v2 vav 215.030.0 v av 45.0
vav 20.0 m/s dv dv 2dv Now 1 2 av 2 2 2 v1 v2 vav 0.5 0.1 2dv av (15.0)2 (30.0)2 (20.0)2
1 0.5 0.1 dv (20.0)2 (20.0)2 av 2 2 2 (15.0) (30.0)
dv av 0.5 So average velocity of the body is = (20.0 0.5) m/s.
Maximum Maximum Maximum
Operation Formula Absolute Error Relative Error Percentage Error
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-34
A B A B Sum A + B A + B 100 A B A B
A B A B Difference A – B A + B 100 A B A B
A B A B Multiplication A × B AB + BA 100 A B A B
A BA AB A B Division 100 B B 2 A B
A A Power An nAn-1 A n n 100 A A
EXPERIMENTAL PHYSICS
In this section, we shall study some common experiments based on the measurement of distance (length), acceleration due to gravity, focal length of a lens etc. Our focus will be on the procedure and the possible errors, which occur in the measurement of various physical quantities. We take the experiments one by one.
MEASUREMENT OF LENGTH
Vernier calliper and screw gauge are the instruments used for measuring length with high accuracy. From a normal scale, accuracy of 1 mm is obtained, while with these instruments, accuracies upto 0.01 mm can be achieved.
1. Vernier Calliper
It consists of a fixed main scale marked on a metallic strip. Another metallic strip slides on it with a vernier scale on it. Scales are marked in such a way that when zero’s of both scales are aligned, nth division of vernier scale coincides with (n – 1)th division of main scale. In the figure shown, 10th division of vernier scale coincides with 9th division of main scale.
Main scale (fixed) 0 1 2 3 (cm)
(Figure is not to the scale) 0 10 Vernier scale (sliding)
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-35
Construction:
Internal Jaws Main scale
0 1 2 3 4 5 6 7 Strip for depth 0 10 measurement Vernier scale External Jaws External and internal jaws are used for external and internal measurements respectively. The strip at the end is used for depth measurement.
Working:
n Vernier Scale Divisions (VSD) = (n – 1) Main Scale Divisions (MSD)
n 1 1VSD MSD n
n 1 1 Difference between 1 MSD and 1 VSD is, 1 MSD – 1 VSD = 1 MSD n n
1 The figure MSD is called Vernier Constant (V.C.) or Least Count (L.C.) of the vernier. It is the n smallest distance, which can be accurately measured by the vernier.
For example if main scale division be of 1 mm and 50 divisions of vernier scale coincide with 49 divisions of main scale. MSD = 1 mm 49 mm 1 VSD = , L.C. mm 0.02 mm 50 50
For example if main scale division be of 1 mm and 10 divisions of vernier scale coincide with 9 divisions of main scale.
MSD = 1 mm 9 mm VSD = 0.9 m . So least count = 1 MSD = 1 VSD = 1 mm – 0.9 mm = 0.1 mm 10
READING A VERNIER
Let zero of vernier scale be between two main scale divisions. The division on main scale just before zero of vernier scale is the main scale reading.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-36
Now, count the number of the division of vernier scale that coincides with a division on main scale. Multiply this number to L.C. of the vernier. This is called vernier scale reading. It is to be added to main scale reading.
0 1 2 3 (cm)
0 5 10
The least count of the scale shown is 0.1 mm
The main scale reading = 11 mm
The vernier scale reading = 5 × 0.1 = 0.5 mm
Hence measurement = 11.5 mm.
ZERO ERROR
When jaws of vernier touch each other, the vernier should read 0.0 i.e., Zeros of main scale and vernier scale should coincide. If it is not so, the instrument has a zero error.
If a vernier shows a non-zero reading when jaws are touching each other, it is the zero error, which is to be algebraically subtracted from the reading taken.
ZERO CORRECTION
Zero correction is the correction to be applied to the reading taken to get correct measurement. It is of same magnitude as that of the zero error but is of opposite sign. Zero correction is to be algebraically added to the reading taken.
POSITIVE AND NEGATIVE ZERO ERROR
When the jaws of vernier touch each other.
If zero of vernier scale is to the right of the zero of main scale, then zero error is positive.
If zero of vernier scale is to the left of zero of main scale, then zero error is negative.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-37
For example the figure shows a situation when the jaws of vernier are touching each other.
Main scale 0 1 2 (cm)
0 5 10
Each main scale division is of 1 mm and 5th division of vernier scale coincides with a main scale division.
1 L.C. 0.1mm 10 Zero error = +5 × 0.1 = 0.5 mm
This error is to be subtracted from the reading taken for measurement.
Also, zero correction = – 0.5 mm.
For example figure shows reading when jaws of vernier are touching each other.
0 1 2 (cm)
0 5 10
Each main scale division is 1 mm.
1 L.C. 0.1mm 10
5th division of vernier coincides with a main scale division.
Zero error = – 5 × 0.1 = – 0.5 mm
This error is to be algebraically subtracted from the reading taken.
Zero correction = – (– 0.5) mm
= + 0.5 mm
This is to be added to the reading taken.
Precautions (to be taken) 1. Motion of vernier scale on main scale should be made smooth (by oiling if necessary). 2. Vernier constant and zero error should be carefully found and properly recorded.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-38
3. The body should be gripped between the jaws firmly but gently (without undue pressure on it from the jaws). 4. Observations should be taken at right angles at one place and taken at least at three different places. Sources of Error 1. The vernier scale may be loose on main scale. 2. The jaws may not be at right angles to the main scale. 3. The graduations on scale may not be correct and clear. 4. Parallax may be there in taking observations.
2. Screw Gauge (or Micrometer)
It works on the principle of advancement of a screw when it is rotated inside mating threads.
Metal piece Reference line Stud Screw Nut Circular scale A B 0 Ratchet
Face Main scale Face B Cap (hollow) A Hub
Metal frame
When the screw is rotated inside the nut by rotating the cap (connected internally to the screw), the screw advances or retracts depending on direction of rotation.
A reference line is marked on the Hub and main scale divisions are marked on it (see figure). Circular scale markings (usually 50 or 100 divisions) are made on the cap.
Main scale division value Linear distance advanced in one rotation Least count: L.C. Number of divisions on circular scale Number of divisions on circular scale
If main scale division value is 1 mm and there are 100 divisions in circular scale, then
1 L.C. 0.01mm 100
Reading: Main scale reading is taken on the main scale. To find circular scale reading, the number of the division of circular scale that coincides with reference line is counted. If nth division coincides, then
Final reading = Main scale reading + n × (L.C.)
Illustration 18
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-39
A screw gauge has main scale divisions of 1 mm each and its circular scale has 50 divisions. While measuring the diameter of a ball, cap end is after 19th marking of main scale and 21st division of circular scale coincides with the reference line. What is the measured diameter of the ball?
Solution:
1mm L.C. 0.02 mm 50
Main scale reading = 19 mm Main scale 20 0 1 Circular scale reading = 21 × 0.02 = 0.42 mm Reference line Cap Diameter = 19 + 0.44 = 19.44 mm
ZERO ERROR AND ITS CORRECTION
If the zero mark of circular scale coincides with the reference line when there is no gap between the faces A and B, then there is no zero error. If the zero mark does not coincide, there is a zero error.
If the zero of circular scale is left behind the reference line when the faces meet, zero error is positive. The magnitude of error is to be deducted from the reading taken. If zero of circular scale advances beyond the reference line when the faces meet, the zero error is negative. The magnitude of error is to be added to the reading taken.
For example least count of a screw gauge is 0.01 mm. When the faces touch each other, main scale zero marking is not visible and 97th division of circular scale coincides with the reference line.
As zero of main scale is not visible, the circular cap has advanced by 100 – 97 = 3 divisions.
Hence zero error = – 3 × 0.01 Main scale 0 = – 0.03 mm 95 90 Zero correction = + 0.03 mm
For example a micrometer has L.C. = 0.02 mm. When the faces are th touching each other, only zero marking of main scale is visible and 7 Main scale 10 division of circular scale is coinciding with the reference line. Find zero 0 0 error of the instrument. The circular scale is left behind by 6 divisions
Zero error = + 7 × 0.02 = + 0.14 mm Zero correction = – 0.14 mm Precaution (to be taken)
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-40
1. While taking an observation, the screw must always be turned only in one direction so as to avoid the backlash error. 2. At each place, take readings in pairs i.e. in two directions at right angles to each other. 3. The wire must be straight and free from kinks. 4. Always rotate the screw by the ratchet and stop as soon as it gives one tick sound only. 5. While taking a reading, rotate the screw in only one direction so as to avoid the backlash error. Sources of Error 1. The screw may have friction. 2. The screw gauge may have backlash error. 3. Circular scale divisions may not be of equal size. 4. The wire may not be uniform. 3. Determination of g using a simple pendulum: A simple pendulum is an arrangement consisting of a small steel ball with a string suspended from a fixed point so that it can swing freely.
Arrangement:
Fixed
l
Steel ball
The equation for the periodic motion of a simple pendulum as determined by Galileo is
l T 2 g
Here, l is the length of the pendulum g is acceleration due to gravity and T is time period of periodic motion. To determine g, rearrange the above equation to get,
l g 42 T 2
Procedure:
(a) Determine the time for 20 complete swings for six different lengths. (The length means the distance between point of suspension and centre of the ball).
(b) Repeat the time measurement five times for each length, making sure to get consistent
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-41
readings.
(c) Determine the average time for 20 swings for each length. Then, calculate the time of one swing.
(d) Compute the acceleration due to gravity for each pendulum length.
(e) Calculate the mean g and the percentage error in g for each length.
(f) Plot length versus time graph and length versus time squared, graph.
For example in an experiment to determine acceleration due to gravity, the length of the pendulum is measured as 98 cm by a meter scale of least count 1 cm. The period of swing/oscillations is measured with the help of a stopwatch having a least count of 1 s.
The time period of 50 oscillations is found to be 98 s.
l 0 As T 2 g
Now, time period of 50 oscillations is 98 s. 50 cm
98 Time period of one oscillation is 1.96 s 50 100 cm (1 m)
l As T 2 g
42l 0.98 We have g 4 (3.14)2 10.06 m/s2 . T 2 1.96 1.96
Let us find the permissible error in the measurement.
42l As g T 2
g l 2T We have g l T
1 2 1 g 10 ( Least count of meter scale is 1 cm and least count of stopwatch is 1 s) 98 98
g 0.3 m/s 2
So, final result can be expressed as (10.1 0.3) ms–2.
Illustration 19
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-42
If percentage error in the measurement of length of a simple pendulum be 1% and percentage error in the measurement of acceleration due to gravity by 2%, then find % error in the measurement of time period of the pendulum.
Solution:
l As T 2 g
dT 1 d l 1 d g T 2 l 2 g
dT 1 1 100% (1%) (2%) = 1.5% T 2 2
Illustration 20
The time period of oscillation of a simple pendulum is 0.5 sec and is measured from time of 100 oscillations with a watch of 1-second resolution. What is the accuracy in the determination of g if 10 cm length of the pendulum is known to 1 mm accuracy?
Solution:
l 0.1 1 Here , l 10 100 Time of 100 oscillations = 100 × 0.5 = 50 sec T 1 So T 50 g l 2T So 100% 100% 100% g l T 1 1 = 100% 2 100% 5% 100 50
Precautions:
1. The oscillations amplitude should be kept small (10º or below) as the formula for time period is applicable for small angular displacements. 2. Thread should be strong, weightless and inextensible. 3. Point of suspension should be fixed on rigid support. 4. While measuring the length and time periods, an average of several readings should be taken.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-43
5. The bob should not spin during vibration. 6. Laboratory fan should be switched off. 7. The instruments used should be checked for zero error. 8. Length of pendulum should include the radius of the bob.
Factors Affecting the Time Period: The time period of a simple pendulum is affected by following factors.
(i) Time period is clearly a function of the length of pendulum. From the formula, it is clear that
T l or T 2 l .
The graphical variation is shown here
2 T T T
l l l (a) (b) (c)
(ii) Time period is a function of acceleration due to gravity
1 1 T or T 2 g g
Following graphs represent the above variation
2 2 T T T
g 1/g 1/g (a) (b) (c)
(iii) Time period is independent of mass of ball used in the pendulum? A wooden ball or a steel ball will have same time period if other factors are same.
(iv) The time period of a simple pendulum is independent of amplitude (provided amplitude is small). This type of motions called isochrones motion.
Sources of Error 1. The string may not be weightless and inextensible.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-44
2. Point of suspension may not be rigid. 3. The amplitude may not be small. 4. The bob may spin. 5. The air currents may disturb vibrations. 6. There may be error in counting. 7. The stop clock/watch may be inaccurate. 8. There may be delay in starting and stopping the stop clock/watch.
4. Determination of Young’s Modulus by Searle’s Method: When a deforming force is applied to deform a body, it shows some opposition. This opposition is called stiffness. Young’s modulus is a physical quantity used to describe the stiffness of a body. It is the ratio of stress applied to strain produced, where stress applied is force applied per unit area and strain is the ratio of change in length and original length.
F x F l Y A l A x
Here, F is applied force, A is area of cross-section, l is length and x is increase in length.
Arrangement: The arrangement consists of two wires. One of the wires is a reference wire loaded with a fixed weight. The other wire is the test wire, on which variable load is applied. The reference wire is used to compensate for the thermal expansion of the wire. The extension in the test wire is measured with the help of a spherometer and a spirit level arrangement.
Compensating Experimental wire wire Cross bars Metallic frame
Spirit level Air Spherometer bubble Dead Slotted weight weight
Procedure: The following steps are required to measure the young’s modulus.
(a) Using micrometer determine the radius of the wire. Using the formula for area of circle = r 2, calculate the area.
(b) Measure the length of the wire l.
(c) Note down the load applied F and corresponding increase in length x.
(d) Convert load in kg in weight in newton by the formula 1 kg = 9.81 N.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-45
(e) Plot a graph of extension x (along x-axis) against weight (along y-axis).
F (f) The slope of this graph is the ratio . x
F l (g) Find Y using the formula Y . A x
For example in an experiment for measurement of young’s modulus, following readings are taken. Load = 3.00 kg, length = 2.822 m, diameter = 0.041 cm and extension = 0.87 cm. We shall determine the percentage error in measurement of Y.
M g x Y M x A l As Y Al Y M x A l
Y 0.01 0.01 2 0.001 0.001 0.065 Y 3.00 0.87 0.041 2.822
Y or 100 6.5% . Y
Precautions (to be taken) 1. Both the wires (experimental and auxiliary) should be of same length, material and thickness. 2. Both the wires should be supported from same rigid support. 3. Kink should be removed from experimental wire before starting experiment. 4. Diameter of wire should be measured at different places and along two mutually perpendicular directions at every place. 5. Slotted weights should be added and removed gently. 6. Two minutes wait should be made after adding or removing weight. 7. Load should be increased or decreased in regular steps. Sources of Error 1. Experimental wire may not have uniform cross-section throughout. 2. The slotted weights may not have standard weight.
Illustration 21
In an experiment for the measurement of young’s modulus by Searle’s method if the percentage error in the measurement of weight is 0.5%, length of the wire is 1%, diameter of wire is 0.1% and extension is 2%, then what is the possible error in the measurement of young’s modulus of the wire. Solution:
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-46
F l As y A x
4F l So y D2x
y F l 2D x i.e. 100% = 100% + 100% + 100% + 100% y y l D x
= 0.5% + 1% + 2 × 0.1% + 2% = 3.7%.
5. Measurement of specific heat of a liquid using a calorimeter: Specific heat of a substance is the heat required to raise the temperature of unit mass of a substance by unit degree Celsius. It is Q given by S , where Q is the heat-supplied m is the mass of substance and is the rise in m temperature.
Arrangement: The arrangement consists of a joule calorimeter (JC) with a churner C, thermometer, and variable resistor RH , cell of emf E, an ammeter, a voltmeter and a switch.
+ – V
C T – A + RH
E S
Procedure: The following steps are required.
(a) Weigh the empty calorimeter with churner C. This is mc .
(b) Weight the calorimeter with the liquid in it. The difference in two masses gives the mass of
liquid. This is ml .
(c) Make the set up shown. Keeping the switch open, note the reading of the thermometer 1 .
(d) Close the switch S for a time t and continuously stir the liquid. At the time of opening the switch, note the reading of thermometer 2, voltmeter V and ammeter I.
(e) Calculate heat supplied to the calorimeter using H = VIt.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-47
(f) Calculate rise in temperature using 2 1 d ..... Here d is the correction applied for radiation loss. d is the fall in temperature when the calorimeter and its t contents are left to cool down for time . 2
(g) Let SC be the specific heat of calorimeter and Sl be the specific heat of liquid, then the heat supplied is
Q mlSl mCSC
Q mCSC 1 VIt Sl mCSC ml ml
Precautions: Following precautions must be taken. (a) Correction due to radiation loss d must be taken into account. (b) The stirring of liquid should be slow. (c) While reading voltmeter and ammeter, parallax should be removed.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-48
Source of error
(a) Some heat is lost to atmosphere in the form of radiations.
6. Focal length of a convex lens/concave mirror using u-v method: When an object is placed at a distance u in front of a convex lens/concave mirror, it forms an image at a distance v from the lens/mirror.
The two values u and v are related to each other.
1 1 1 For a convex lens, the relationship is . v u f
1 1 1 For a concave mirror, the relationship is . v u f
Arrangement: The lens/mirror is fixed on an optical bench with a scale marked on it to measure the distance of object and image. There are two other stands in which two pin shaped objects are fixed. One of these is the object pin. This acts as an object. The other one is called image pin. It is used to locate the image position. When there is no parallax between the image pin and image seen in the lens/mirror, the image pin represents the position of the image.
Convex lens
image pin u v
optical bench object pin lens stand u v
Convex mirror v
u u optical bench image pin object pin mirror stand v
Procedure: Following steps are to be followed.
(a) Fix the lens on the lens stand.
(b) Place object pin in front of the lens. Measure the distance between the two. The value of u will be negative of the above distance.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-49
(c) Place the image pin on the other side of the lens at such a distance from the lens, so that there is no parallax between image pin and image seen in the lens. The value of v will be the distance between the lens and image pin.
1 1 1 (d) Compute the focal length of the lens using lens formula . v u f
1 1 (e) Plot a graph between u, v, and . u v
For a convex lens, the shape of graphs obtained are shown.
1 v v
u l
u
(a) (b)
For a concave mirror, the shape of graphs obtained are shown.
1 v v
u l
u
(a) (b)
(f) In the u-v curve, we draw a line at 45º as shown in figure.
v This line intersect the curve at point P. PB and PA are parallel to axes. P B 45º OA Here OA = OB = 2f. So, focal length f . A O u 2
For example in an experiment to measure that focal length of a concave mirror, it was found that for an object distance of 0.30 m, the image distance come out to be 0.60 m. Let us determine the focal length. 1 1 1 By mirror formula, v u f
1 1 1 u = – 0.30 m f 0.30 0.60
3.0 v = – 0.60 m f f = – 0.20 m 0.60
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-50
Precautions (to be taken) Following precautions must be taken. 1. Both u and v should be measured carefully on the bench. 2. While locating the image, parallax should be removed. 3. The uprights should be vertical. 4. Tip to tip parallax should be removed between the needles I and image of the needle O. 5. To locate the position of the image the eyes should be at least 30 cm away from the needle. 6. Tips of the object and image needles should lie at the same height as that of pole of the concave mirror. 7. Index correction for u and v should be applied. 8. The principle axis of the concave mirror must be horizontal and parallel to the scale on the optical bench.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-51
Sources of Errors 1. The upright may not be the vertical. 2. Parallax removal may not be perfect.
7. Determination of speed of sound using resonance column: It is a simple apparatus used to measure speed of sound in air with the help of a tuning fork of known frequency. The resonance column is an air column closed at one end. Its length is variable. It is based on the phenomenon of standing waves. A vibrating tuning fork is held near the open end of the tube, which is partially empty. The air-column vibrates with the frequency of tuning fork. As the length of air column is increased from zero onwards, as stage is reached when very intense sound is observed. At this stage the natural frequency of the air-column matches with the frequency of tuning fork. This state is known as resonance. At resonance, the vibration of air column can be like any of the following figures.
e e l1 l2
Arrangement: The arrangement is shown in the figure. It consists of a metallic tube and a connected glass tube. There is a reservoir containing water. This is connected to the metallic tube by a rubber pipe. Parallel to the glass tube is a scale to measure the length of air-column.
Glass tube Board 100 90 Reservoir 80 70 Water Metallic tube 60 Screw 50 40 Cock 30 20 Rubber tube 10 0
Leveling screw
Procedure: Following steps are used in the experiment.
(a) Hold the vibrating tuning fork at the open end of the column and start increasing the length of the air column by adjusting the height of reservoir.
(b) Determine the first resonating length l1 . This is the length at which an intense sound is observed.
(c) Determine the second resonating length l2 . This the length at which an intense sound is
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-52
observed again.
(d) Compute the wavelength. It can be calculates as shown. From the figure-I, it is clear that
l e 1 4
3 l e 2 4
Here is wavelength of sound and e is end correction (height of the anti-node above the open end)
Subtracting l l 2 1 2
2(l2 l1)
(e) Using the formula v = f , compute the speed of sound.
(f) Compute the end-correction. It can be calculated as shown.
l e 1 4
3 l e 2 4
l2 3l1 l2 3l1 2e 0 e 2
(g) Compute the error in end-correction by comparing it with Reyleigh’s formula e = 0.6 R. Where R is internal radius of resonance tube.
Precautions (to be taken) 1. Resonance tube should be vertical. 2. Pinch cork should be tight. 3. Tuning fork should be vibrated gently by a rubber pad. 4. Prongs should be vibrated in a vertical plane above the mouth (end) of the metallic tube. 5. Prongs should not touch the end of the metallic tube. 6. Reading for water level falling and rising should be noted. 7. Reading should be noted using a set square. 8. While measuring air temperature, thermometer bulb should not touch water or sides of resonance tube.
Sources of Error 1. Resonance tube may not be vertical. 2. Pinch cork may be loose. 3. The edge of open end of metallic tube may not be at zero of metre scale (but this error becomes
eliminated in (l2 – l1). Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-53
For example the internal radius of a 1 m long resonance tube is measured as 3 cm. A tuning fork of frequency 2000 Hz is used. The first resonating length is measured as 4.6 cm and the second resonating length is measured as 14.0 cm. We shall calculate the following
1. Maximum percentage error in measurement of e, as given by Reyleigh’s formula. (Given error in measurement of radius is 0.1 cm).
e = 0.6 R
e = 0.6 R = 0.6 × 0.1 = 0.06 cm
e 0.06 Percentage error is 100 100 3.33% e 0.6 3
2. Speed of sound at the room temperature.
l1 4.6 cm
l2 14.0 cm
2(l 2 l1) 2(14.0 4.6) 18.8 cm
18.8 v f 2000 376 m/s 100
3. End correction obtained in the experiment.
l 3l 14.0 3 4.6 e 2 1 0.1cm 2 2
4. Percentage error in the calculation of e with respect to theoretical value.
0.1 0.6 3 Percentage error = 100 94.44% . 1.8
Illustration 22
In a resonance tube experiment with tuning fork of frequency 500 Hz, first resonance is observed at a length of 16 cm of air column and second resonance is observed at a length of 50 cm of air column. Find speed of sound in air.
Solution
As we known that l l 1 1 2 Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-54
50 cm – 16 cm = 2
So = 68 cm
v = v i.e. v = (500 Hz) (68 cm) = 340 m/s
8. Verification of Ohm’s law using voltmeter and ammeter: According to ohm’s law, the current flowing through a metallic conductor is directly proportional to the potential difference across the ends of the conductor provided the physical conditions like temperature and mechanical strain etc are kept constant.
V = IR, where R is a constant called resistance
Arrangement: The figure shows the arrangement used to verify ohm’s law. it consists of a cell of
emf E, connected to a fixed resistance R and a variable resistance RH (rheostat). An ammeter is connected in the circuit to measure current I and a voltmeter is connected across the fixed resistance R to measure potential difference V.
V
RH (rheostat) R
Ammeter A
E (cell) S
Procedure: Following steps are to be followed.
(a) Close the switch S and note down the readings of voltmeter and ammeter.
(b) Repeat the above process for different values of variable resistance RH .
(c) Plot a graph between V and I by taking V along y-axis and I along x-axis.
V (d) Slope or gradient of this graph is = constant. This shows that V I. 1
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-55
V
I
For example in the circuit shown, voltmeter is ideal and its least count is 0.1 V. The least count of ammeter is 1 mA. Let reading of the voltmeter by 30.0 V and the reading of ammeter is 0.010 A. We shall calculate the value of resistance R within error limits.
V = 30.0
I = 0.010 A
V 30.0 R 3.00 k I 0.010
R
V RH
A
V Error: As R I
R V I V I R R R V I V I
0.1 0.001 = 3.00 10 3 0.310 k 30.0 0.010
So, resistance is (3.00 0.310) k.
Precautions (to be taken) 1. The connections should be neat, clear and tight. 2. Thick copper wires should be used for the connections after removing the insulations near their ends by sand paper.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-56
3. Voltmeter and ammeter should be of proper range. 4. A low resistance rheostat should be used. 5. The key should be inserted only while taking observations to avoid heating of resistance (otherwise its resistance will increase). Source of Error 1. The instrument screws may be loose. 2. Thick connecting wires may not be available. 3. Rheostat may have high resistance.
9. Determination of resistivity of a metal using: (a) Meter bridge and (b) Post office box.
(a) Meter bridge: The resistance of a metal wire depends on its length, area of cross-section and resistivity of the metal. The formula is
l R A
Here, is the resistivity. Its unit is -m (ohm-meter). To measure its resistivity, we use a meter bridge. The working of a meter bridge is based on Wheat stone bridge principle.
P R The circuit shown is called Wheat stone bridge. When , there is no flow of current Q S in the branch BD. At this state, galvanometer shows zero deflection.
B
P Q
I I G
R S
D
Arrangement: The arrangement consists of a 100 cm long wire connected between A and C. It is tapped at point B by a sliding contact called jockey. R is a known resistance. S is the resistance wire
whose resistivity is to be determined. A cell and a variable resistance RH are connected to supply current in the circuit.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-57
R S P
G P Jockey Q A C B l 100 – l 100 cm RH + – ( ) E K
Procedure: Following steps are used in the experiment.
(a) Plug the key and slide the jockey on wire AC to locate point B where the galvanometer does not show deflection. Note down the length l.
P R (b) Compute the resistance S using the formula . Q S
l Here, P wire A
100 l Q wire A
100 l S R . l
(c) Compute the value of S by determining different values of length l. This can be done by using different values of R.
(d) Calculate the percentage error in measurement of S.
(e) Compute the resistivity by measuring length and area of cross-section of resistance wire S l using the formula S . A
For example in an experiment to determine an unknown resistance, a 100 cm long resistance wire is used. The unknown resistance is kept in the left gap and a known resistance is put into the right gap. The scale used to measure length has a least count 1 mm. The null point B is obtained at 40.0 cm from the left gap. We shall determine the percentage error in the computation of unknown resistance.
As shown in the figure
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-58
P(unknown) Q
G
B l 100 – l
P l Q 100 l
l P 100 l
P l (100 l) l l 0.1 0.1 P l 100 l l 100 l 40.0 60.0
P 100 0.42% P
Precaution (To be taken) 1. Connections should be neat, clean and tight. 2. All the plugs in the resistance box should be tight. 3. Move the jockey gently over the bridge wire and do not rub it. 4. The plug in key K should be inserted only when the observations are to be taken. 5. Null point should be brought between 40 cm and 60 cm. 6. Set square should be used to note null point to avoid error of parallax. Source of Error 1. The instrument screws may be loose. 2. The plugs may not be clean. 3. The wire may not have uniform thickness.
(b) POST OFFICE BOX:
This apparatus was initially used in post-offices for measuring the resistance of the telephone or telegraph wires, or for finding the faults in these wires. In post office box, the two arms AB and BC are connected in series. Each of these arms contains resistances 10 , 100 and 1000 . In the third arm AD there are resistances from 1 to 5000 arranged in U shape. In order to insert keys in the arms AC and BD, the point A is connected to a
tapping-key k1 and the point B is connected to another tapping key k2 . The wire whose resistance (S) is to be determined is connected in the arm CD. The galvanometer G is connected
between B and D through the key k2 and the cell is connected between A and C through the
key k1 .
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-59
P Q
A 1000 100 10 B 10 100 1000 C 200 S
5000 2000 2000 1000 500 200 200
100
D 1 2 2 5 10 20 20 50 E G
k2 k1
Equivalent circuit
B
P Q K2
A C G
R S
D K1 E
Working: First of all, from P a 1000 resistor is selected and from Q also 1000 resistor is selected. Now by pulling plugs from R, a balance condition is obtained.
P R As [P = 1000 , Q = 1000 ] Q S
So S = R
Now in order to increase the preciseness, P is selected to be 1000 and Q = 10 . In this case, R S 100
As the least count decreases, hence preciseness increases.
Precautions (To be taken) Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-60
1. The ends of the connecting wires as well as unknown resistance wire should be cleaned by sand paper. 2. All connections should be neat, clean and tight. 3. All the plugs of post-office box should be tight. 4. The wire should not make any loop. 5. While taking an observation, the cell key should always be pressed first and then the galvanometer key. 6. The current should be passed through the bridge only for the minimum time. This will avoid the error due to change in resistance by heating. 7. The length of the wire should be measured from the points where the wire comes out of the terminals of the resistance box.
Illustration 23
B C D For the post office box arrangement to determine the value of unknown resistance, the unknown resistance should be connected between (a) B and C (b) C and D (c) A and D A (d) B1 and C1
B1 C1 Solution: The unknown resistance should be connected between A and D as BC, CD and BA are known resistance. (c)
Illustration 24
What is the maximum resistance that can be measured with a post office box.
Solution:
The total resistance, which can be taken out from the resistance box R of the post office box is R = 1 + 2 + 2 + 5 + 10 + 20 + 20 + 50 + 100 + 200 + 200 + 500 + 1000 + 2000 + 2000 + 5000 R = 16110 Q As X R P By taking = 10 and Q = 1000 1000 X 16110 1611000 10
Illustration 25
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-61
What is the minimum resistance that can be measured with a post office box.
Solution
But taking P = 1000
Q = 10
R = 1
Q 10 X = R 1 0.01 P 1000
Hence the minimum resistance that can be measured from the post office box is 0.01 .
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-62
SOLVED OBJECTIVE EXAMPLES
Example 1:
The dimensional formula of angular velocity is
(a) M°L°T –1 (b) MLT–1 (c) M°L°T1 (d) M1 L1 T–2
Solution:
MLT –1 [Angular velocity] = = M°L°T t T
(a)
Example 2:
The dimensional formula for Planck’s constant (h) is
(a) ML–2 T–3 (b) ML2T–2 (c) ML2T –1 (d) ML–2T–2
Solution:
hc E =
E ML2T 2 L [h] = ML2T–1 c LT 1
(c)
Example 3:
Of the following quantities, which one has dimensions different from the remaining three?
(a) energy per unit volume
(b) force per unit area
(c) product of voltage and charge per unit volume
(d) angular momentum per unit mass
Solution:
ML2T 2 [energy per unit volume] = ML1T 2 L3
MLT 2 [force per unit area] = = ML–1T –2 L2
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-63
[product of voltage and charge per unit volume] = ML–1T–2
[angular momentum per unit mass]= ML2 T–1 L. M–1 = L2T–1
(d)
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-64
Example 4:
E, m, J and G denote energy, mass, angular momentum and gravitational constant respectively. Then the dimensions of EJ 2/m5G2 are
(a) angle (b) length (c) mass (d) time
Solution:
EJ 2 (ML2T 2 )(ML2T 1)2 = M0L0T0 m5G2 (M)5 (M 1L3T 2 )2
(a)
Example 5:
a The equation of state of some gases can be expressed as P (V b) RT . V 2
Here, P is the pressure, V the volume, T the absolute temperature, and a, b, R are constants. The dimensions of a are
(a) ML5T–2 (b) ML–1T–2 (c) M°L3T° (d) M°L6T°
Solution:
a Dimensions of will be same as dimensions of pressure v 2
a –1 –2 = ML T [L3 ]2
[a] = ML5T–2
(a)
Example 6:
When 2.0347 is added to 15.7, the sum is
(a) 17.7347 (b) 17.734 (c) 17.73 (d) 17.7
Solution:
Since the least number of places after decimal is 1, the final result will have same. On rounding off 17.7347 to one place after decimal it becomes 17.7
(d)
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-65
Example 7:
A body travels uniformly a distance of 13.8 0.2 m in a time 4.0 0.3s. The velocity of the body within error limits is (a) 3.45 0.2ms 1 (b) 3.45 0.3ms 1 (c) 3.45 0.4ms 1 (d) 3.45 0.5ms 1
Solution:
Here S 13.8 0.2cm; t 4.0 0.3s
13.8 V 3.45 ms 1 4.0
V S t 0.2 0.3 Also 0.0895 V S t 13.8 4.0
V 0.3 (rounding off to one place of decimal)
V 3.45 0.3ms1
(b)
Example 8:
The length and breadth of a metal sheet are 3.124 m and 3.002 m respectively. The area of this sheet up to four correct significant figures is (a) 9.376 m2 (b) 9.378 m2 (c) 9.379 m2 (d) 9.388 m2
Solution:
Given length (l) = 3.124 m and breadth (b) = 3.002 m.
We know that area of the sheet A l b 3.124 3.002 9.378248m2 .
Since both length and breadth have four significant figures, therefore area of the sheet after rounding off to four significant figures is 9.378 m2
(b)
Example 9:
The value of universal gravitation constant G = 6.67 10-11 N m2 kg-2. The value of G in units of g-1 cm3 s-2 is
(a) 6.67 10-8 (b) 6.67 10-7 (c) 6.67 10-9 (d) 6.67 10-10
Solution:
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-66
G = 6.67 10-11 Nm2 kg-2
= 6.67 10-11 (kg ms-2) (m2) (kg)-2
= 6.67 10-11 [ (1000 g ) (100 cm) s-2] (100 cm)2 (1000 g)-2
= 6.67 10-11 105 104 10-6 g-1 cm3 s-2
= 6.67 10-8 g-1 cm3 s-2
(a)
Example 10:
If energy (E), momentum (P) and force (F) are chosen as fundamental units. The dimensions of mass in new system is
(a) E -1 P3 (b) E -1 P2 (c) E -2 P2 (d) none of these
Solution:
The dimensions of E, P and F in terms of M, L and T are
[E] = ML2T-2
[P] = MLT-1
[F] = MLT-2
Let [M] = Ea Pb FC
or [M] = (ML2T-2)a (MLT-1)b (MLT-2)C
Equating the powers of M, L and T, we have a = -1, b = 2 and c = 0
Hence [M] = E-1 P2
(b)
Example 11:
Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions on circular scale is 50. Further, it is found that screw gauge has a zero error of – 0.03 mm. While measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The diameter of the wire is
(a) 3.67 mm (b) 3.38 mm (c) 3.32 mm (d) 3.73 mm
Solution
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-67
0.5mm Least count = 0.01mm 50 Zero error = – 0.03 mm (Given) Measured diameter = 3 mm + 35 × 0.01 mm = 3.35 mm Correct diameter = 3.35 – (–0.03 mm) = 3.38 mm (b)
Example 12:
A body of mass m = 3.513 kg is moving along the x-axis with a speed of 5.00 ms–1. The magnitude of momentum is recorded as
(a) 17.56 kg ms–1 (b) 17.57 kg ms–1 (c) 17.6 kg ms–1 (d) 17.565 kg ms–1
Solution
p = mv = 3.513 × 5.00 = 17.565 kg ms–1 Since the result cannot have more than three significant figures p = 17.6 kg ms–1 (c)
Example 13:
The dimensions of magnetic field in M L T and C (coulomb) is given as
(a) [MT–1C–1] (b) [M T 2 C 1] (c) [M LT 1C1] (d) [M T 2 C2]
Solution
Using F = q B v [MLT 2] [B] [C][LT 1]
[B] [MT 1C1] (a)
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-68
SOLVED SUBJECTIVE EXAMPLES
Example 1:
Find the dimensions of the following quantities
(i) Acceleration (ii) Angle
(iii) Density (iv) Kinetic energy
(v) Gravitational constant (vi) Permeability
Solution:
Velocity [Velocity] LT 1 (i) Acceleration = [Acceleration] = = LT 2 Time [Time] T
Distance (ii) Angle = angle is dimensionless. Distance
Mass [Mass] M (iii) Density = [Density] = = ML3 Volume [Volume] L3
1 (iv) Kinetic energy = Mass × Velocity2 2
[Kinetic energy] = [Mass] × [Velocity]2 = ML2T–2
(v) Constant of gravitation occurs in Newton’s law of gravitation
m m F = G 1 2 d 2
[F] [d 2 ] MLT 2L2 [G] = M1L3T 2 [m1][m2 ] MM
(vi) Permeability occurs in Ampere’s law of force
(i l )(i l )sin F = 1 1 2 2 r 2
[F][r 2 ] MLT 2 [] = MLT 2 A 2 [i1l1][i 2l2 ] AL.AL
Example 2:
The value of a force on a body is 20 N in SI units. What is the value of this force in CGS units, that is, dynes?
Solution:
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-69
Dimension of force = MLT–2
Unit of force in SI = kg ms–2
Unit of force in CGS = g cm s–2
Let n be the numerical value of the force in CGS system.
20 kg ms–2 = n g cm s–2
2 kg m s 1000 g 100 cm n = 20 20 g cm s g cm
or n = 20 × 1000 × 100 = 2 × 106
20 newtons = 2 × 106 dynes
Example 3:
Check by the method of dimensional analysis whether the following relations are correct.
P (i) v = where v = velocity of sound and D
P = pressure, D = density of medium.
1 F (ii) n = , where n = frequency of vibration, l = length of the string. 2l m
F = stretching force, m = mass per unit length of the string.
Solution:
[P] ML1T 2 (i) [R.H.S.] = = LT–1 [D] ML3
[L.H.S] = [v] = LT–1 Hence the relation is correct.
1 [F] 1 MLT 2 1 (ii) [R.H.S.] = LT 1 T1 [l] [m] L ML1 L
A Number 1 1 [L.H.S.] = = T . Hence the result is correct. Time T
Example 4:
Round off to three significant digits
(i) 0.03927 kg (ii) 4.085 × 105
Solution:
(i) 0.0393 kg ( the last digit 7 > 5)
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-70
(ii) 4.08 × 105 s ( the digit to be rounded off is 5 and preceding digit is even)
Example 5:
c The velocity v of a particle depends on the time t according to the equation v = a + bt + . Write the d t dimensions of a, b, c and d.
Solution:
From principle of homogeneity
[a] = [v]
or [a] = LT–1
[bt] = [v]
[v] LT 1 or [b] = [t] T
or [b] = LT–2
Similarly, [d ] = [t ] = T
[c] Further, [v] [d t]
or [c] = [v] [d + t ]
or [c] = LT -1 .T
or [c] = L
Example 6:
The radius of the earth is 6.37 106 m and its mass is 5.975 1024 kg. Find the earth’s average density to appropriate significant figures.
Solution:
Radius of the earth (R) = 6.37 106 m
4 4 3 Volume of the earth (V) = R m3 3.142 6.37 10 6 m3 3 3
Mass M Average density (D) = = 0.005517 106 kg m-3 = 5.52 103 kg m-3 Volume V
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-71
The density is accurate only up to three significant figures which is the accuracy of the least accurate factor, namely, the radius of the earth.
Example 7:
The moment of inertia of a body rotating about a given axis is 6.0 kg m2 in the SI system. What is the value of the moment of inertia in a system of units in which the unit of length is 5 cm and the unit of mass is 10 g?
Solution:
The dimensions of moment of inertia are (ML2). We have
n1u1 n2 u2
2 2 or n1M1L1 n2 M2L2
2 n M L2 M L n 1 1 1 n 1 1 2 2 1 M2L2 M2 L2
Given n1 = 6.0, M1 = 1 kg, L1 = 1 m, M2 = 10 g and L2 = 5 cm.
2 1kg 1m Therefore, n2 6.0 10 g 5cm
2 1000 g 100 cm 6.0 10 g 5cm
6.0 100 202 = 2.4 105 units
Example 8:
The time period T of the oscillations of a large star, oscillating under its own gravitational attraction, may depend on its mean radius R, its mean density and the gravitation constant G. Using dimensional k analysis, show that T is independent of R and is given by T where k is a dimensionless constant. G
Solution:
Let T kRabGc Substituting the dimensions, we have b c [ T ] LaML3 M1L3T2 Equating powers of M, L and T, we have 0 = b – c 0 = a – 3b + 3c and 1 = -2c which give a = 0, b = -1/2 and c = -1/2 1 1 Hence T kR0 2 G 2
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-72
k or T G Thus T is independent of R
Example 9:
1 The rotational kinetic energy of a body is given by E I2 where I is the moment of inertia of the k 2 body about its axis of rotation and is the angular velocity. Using this relation, obtain the dimensional formula for I.
Solution:
2E Given I k 2
2 2 Now, Ek ML T
T1
ML2T2 Dimensional formula for I is I ML2 T2
Example 10:
A rectangular plate has a length of (21.3 0.2) cm and a width of (9.80 0.10) cm. Find the area of the plate.
Solution:
Area (A) = lb = 21.3 9.8 = 209 cm2
A 0.2 0.1 0.019 A 21.3 9.8
A = 0.019 209= 3.97cm2 4 cm2
A = 209 4cm2
Example 11:
The heat dissipated in a resistance can be obtained by the measurement of resistance, the current and time. If the maximum error in the measurement of these quantities is 1%, 2% and 1% respectively, what is the maximum error in determination of the dissipated heat?
Solution:
Heat produced H is given by
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-73
I 2Rt H = J
H I R t J 2 H I R t J
For maximum percentage error,
H I R t J 100 2 100 100 + 100 100 H I R t J
= 2 ×2% + 1% + 1% + 0% = 6%
Example 12:
In an experiment, values of two resistances are measured to be r1 = (5.0 0.2) ohm and
r2 = (10.0 0.1) ohm. Find the values of total resistance in parallel with limits of percentage error.
Solution:
Combined resistance in parallel (Rp) is given by
1 1 1 r r 2 1 Rp r1 r2 r1r2
r1r2 50 1 1 1 Rp = 3.33 As r1 r2 15 Rp r1 r2
Differentiating we get
dRp dr dr 1 2 R 2 2 p2 r1 r2
dRp dr dr or 1 1 R 2 2 p2 r1 r2
dRp 0.2 0.1 dRp 0.1 (3.33)2 (5.0)2 (10.0)2
So Rp (3.3 0.1)
Example 13:
Calculate focal length of a spherical mirror from the following observations: object distance u = (5.0 0.2) cm and image distance v = (10.0 0.1) cm.
Solution:
1 1 1 v u As f u v uv
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-74
uv 5.0 10.0 f 3.3 cm u v 5.0 10.0
uv As f u v
loge f loge u loge v loge(u v)
Differentiating we get
df du dv d(u v) f u v u v
df 0.2 0.1 0.3 3.3 5.0 10.0 15.0
2 1 3 10 4 1 2 3 df = = 0.1 cm 50 100 150 3 30 30 30 30
So f = (3.3 0.1) cm
Example 14:
A helicopter can hover when the power of its engine is P. A second helicopter is an exact copy of the first one but its linear dimensions are double those of the original. What power output is needed to enable this second helicopter to hover?
Solution
The required power for the hovering helicopter should depend on acceleration due to gravity g,
linear size of helicopter L, average density of its material and density of air 0
a b c d So, let P g L 0
a b c d Or P k g L 0 where k is a dimensionless constant
[M1 L2 T3 ] k [LT 2 ] a[L]b [ML 3 ]c [ML 3 ]d
c + d = 1
a + b – 3c – 3d = 2
3 – 2a = – 3 a = 2
1 Or b 3(c d) 2
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-75
1 b 3 [ c + d = 1] 2
7 b 2
7 So it is clear that P L2
7 P' 22 P 8 2 P so P' 11.31P
So power of the second helicopter should be 11.31 times the power of the first helicopter.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-76
Example 15:
A small bob joins two light un-stretched, identical springs, anchored at their far ends and arranged along a straight line as shown in figure. The bob is displaced in a direction perpendicular to the line of the springs by 1 cm and then released. The period of the ensuring vibrations or the bob is found to be 2 seconds. If the bob were displaced by 2 cm and then released then find its time period of
oscillations. The un-stretched length of the springs is l0 1 cm and gravity is to be ignored.
l0 l0 m
Solution
Let displacement of the bob be x.
2 2 Length of the extended spring = l0 x
x2 = l 1 0 2 l0
x2 l l0 2l0
x2 So extension in the spring is l l0 2l0 l l x2 x Tension in the spring F k 2l0
x2 x Net restoring force on the bob = 2F sin 2k 2l0 l0
k x3 = [ sin tan ] 2 l0
d 2x kx3 d 2x k So the resulting motion is M x3 2 2 2 2 dt l0 dt m l0
k Let C 2 m l0
d 2x So Cx3 d t 2
This equation cannot be solved easily by elementary methods, so let us assume that T Ca Ab
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-77
Where A is amplitude [T ] [C]a [A]b
[T ] [L2 T 2]a [L]b
1 a and b = – 1 2
1 So T A
This means that this time period will reduce to half i.e. will become 1 second.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-78
TIPS & TRICKS
The standard of Weight and Measures Act was passed in India in 1976. It recommended the use of SI in all fields of science, technology, trade and industry.
The dimensions of many physical quantities, especially those in heat, thermodynamics, electricity and magnetism in terms of mass, length and time alone become irrational. Therefore, SI is adopted which uses 7 basic units.
The dimensions of a physical quantity are the powers to which basic units should be raised to represent the derived unit of that physical quantity.
The dimensional formula is very helpful in writing the unit of a physical quantity in terms of the basic units.
The dimensions of a physical quantity do not depend on the system of units.
A physical quantity that does not have any unit must be dimensionless.
The pure numbers are dimensionless.
Generally, the symbols of those basic units, whose dimension (power) in the dimensional formula is zero, are omitted from the dimensional formula.
It is wrong to say that the dimensions of force are MLT 2 . On the other hand we should say that the dimensional formula for force is MLT 2 and that the dimensions of force are 1 in mass, 1 in length and – 2 in time.
The ratio of two similar quantities is dimensionless.
The physical relation involving logarithm, exponential, trigonometric ratios, numerical factors etc. cannot be derived by the method of dimensional analysis.
Physical relations involving addition or subtraction sign cannot be derived by the method of dimensional analysis.
If units or dimensions of two physical quantities are same, these need not represent the same physical characteristics. For example torque and work have the same units and dimensions but their physical characteristics are different.
The standard units must not change with space and time. That is why atomic standard of length and time have been defined. Attempts are being made to define the atomic standard for mass as well.
The unit of time, the second, was initially defined in terms of the rotation of the earth around the sun as well as that about its own axis. This time standard is subjected to variation with time. Therefore, the atomic standard of time has been defined.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-79
Any repetitive phenomenon, such as an oscillating pendulum, spinning of earth about its axis, etc. can be used to measure time.
The product of numerical value of the physical quantity (n) and its unit (U) remains constant.
That is: nU = constant or n1U1 n2U2 .
The product of numerical value (n) and unit (U) of a physical quantity is called magnitude of the physical quantity. Thus: Magnitude = nU .
The order of magnitude of a quantity means its value (in suitable power of 10) nearest to the actual value of the quantity.
Angle is exceptional physical quantity, which though is a ratio of two similar physical quantities (angle = arc/radius) but still requires a unit (degrees or radians) to specify it along with its numerical value.
Solid angle subtended at a point inside the closed surface is 4 steradian.
A measurement of a physical quantity is said to be accurate if the systematic error in its measurement is relatively very low. On the other hand, the measurement of a physical quantity is said to be precise if the random error is small.
A measurement is most accurate if its observed value is very close to the true value.
Errors are always additive in nature.
For greater accuracy, the quantity with higher power should have least error.
The absolute error in each measurement is equal to the least count of the measuring instrument.
Percentage error = relative error × 100.
The unit and dimensions of the absolute error are same as that of quantity itself.
Absolute error is not dimensionless quantity.
Relative error is dimensionless quantity.
value of 1part on main scale (s) Least Count = Number of parts on vernier scale (n)
value of 1 part of value of 1 part of Least count of vernier calipers – main scale (s) vernier scale (v)
Least count of vernier calliper = 1 MSD – 1 VSD
Where MSD = Main Scale Division
VSD = Vernier Scale Division
Pitch(p) Least count of screw gauge = Number of parts on circular scale (n)
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-80
Smaller the least count, higher is the accuracy of measurement.
Larger the number of significant figures after the decimal in a measurement, higher is the accuracy of measurement.
Significant figures do not change if we measure a physical quantity in different units.
Significant figures retained after mathematical operation (like addition, subtraction, multiplication and division) should be equal to the minimum significant figures involved in any physical quantity in the given operation.
Significant figures are the number of digits upto, which we are sure about their accuracy.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-81
If a number is without a decimal and ends in one or more zeros, then all the zeros at the end of the number may not be significant. To make the number of significant figures clear, it is suggested that the number may be written in exponential form. For example 20300 may be expressed as 203.00 × 102, to suggest that all the zeros at the end of 20300 are significant.
1 inch = 2.54 cm
1 foot = 12 inches = 30.48 cm = 0.3048 m
1 mile = 5280 ft = 1.609 km
1 yard = 0.9144 m
1 slug = 14.59 kg
1 barn = 1028 m2
1 liter = 103 cm3 103 m3
5 1 km/h = m/s 18
1 m/s = 3.6 km/h
1 g/cm3 1000 kg/m3
1 atm = 76 cm of Hg = 1.013 × 105 N/m2
1 N/m 2 Pa (Pascal)
When we add or subtract two measured quantities, the absolute error in the final result is equal to the sum of the absolute errors in the measured quantities.
When we multiply or divide two measured quantities, the relative error in the final result is equal to the sum of the relative errors in the measured quantities.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-82
EXERCISE - I
AIEEE SINGLE CHOICE CORRECT
1. If velocity (V), acceleration (A) and force (F) are taken as fundamental quantities instead of mass (M), length (L) and time (T), the dimensions of Young’s modulus of elasticity would be (with M, L and T as fundamental, *Young’s modulus+ = ML–1T–2
(a) FA2V–2 (b) FA2V–3 (c) FA2V–4 (d) FA2V–5
2. The number of particles crossing per unit area perpendicular to X-axis in unit time is
n n N = –D 2 1 x2 x1
Where n1 and n2 are number of particles per unit volume for the value of x1 and x2 respectively. The dimensions of diffusion constant D are
(a) M°LT2 (b) M°L2T–4 (c) M°LT–3 (d) M°L2T–1
3. The ratios L/R and RC (L = inductance, R = resistance and C = capacitance) have the dimensions as those of
(a) velocity (b) acceleration (c) time (d) force
4. The dimensional formula for latent heat is
(a) M°L2T –2 (b) MLT –2 (c) ML2T –2 (d) ML2T –1
5. The result after adding 3.8 × 10–6 with 4.2 × 10–5 with due regard to significant figure is
(a) 4.58 × 10–5 (b) 0.458 × 10–4 (c) 4.6 × 10–5 (d) 45.8 × 10–6
6. A dimensionally consistent relation for the volume V of a liquid of coefficient of viscosity flowing per second through a tube of radius r and length l and having a pressure p across its end is η ML 1T 1
(a) V = pr4/8l (b) V = l/8 pr4 (c) V = 8 pl/r4 (d) V = p /8lr4
7. The volume of a sphere is 1.76 cm3. The volume of 25 such spheres taking into account the significant figure is
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-83
(a) 0.44 × 102 cm3 (b) 44.0 cm3 (c) 44 cm3 (d) 44.00 cm3
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-84
8. Which of the following has not been expressed in proper units? [stress = force /Area, surface tension = force/length]
(a) stress/strain = N/m2 (b) surface tension = N/m
(c) energy = kg × m/s (d) pressure = N/m2
9. When the number 6.03587 is rounded off up to the second place of decimals, it becomes
(a) 6.035 (b) 6.04 (c) 6.03 (d) none of these
10. Suppose the acceleration due to gravity at a place is 10 m/s2. Its value in cm/(minute)2 is
(a) 36 105 (b) 6 104 (c) 36 104 (d) none of these
x 11. If x = an, then fractional error is equal to x
n a a a a (a) (b) n (c) n loge (d) n log a a a a
12. What is the percentage error in the measurement of time period of a pendulum if maximum errors in the measurement of ‘I’ and g’ are 2% and 4% respectively?
(a) 6% (b) 4% (c) 3% (d) 5%
13. A physical quantity is represented by X = MaLbT–c. If percentage error in the measurement of M, L and T are %, % and % respectively, then total percentage error is
(a) (a – b + c)% (b) (a + b + c)% (c) (a – b – c)% (d) none of the above
14. The density of a cube is measured by measuring its mass and the length of its sides. If the maximum errors in the measurement of mass and length are 3% and 2% respectively, then the maximum error in the measurement of density is
(a) 9% (b) 7% (c) 5% (d) 1%
15. What is a vernier constant?
(a) It is the value of one main scale division divided by the total number of division on the main scale
(b) It is the value of one vernier scale division divided by the total number of divisions on the vernier scale Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-85
(c) It is the difference between value of one main scale division and one vernier scale division.
(d) It is not the least count of the vernier scale.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-86
16. Choose the wrong statement for zero error and zero correction.
(a) If the zero of the vernier scale does not coincide with the zero of the main scale then the instrument is said to be having a zero error.
(b) Zero correction has a magnitude equal to zero error but sign is opposite to that of the zero error.
(c) Zero error is positive when the zero of vernier scale lies to the left of the zero of the main scale.
(d) None of these is wrong.
17. How many number of wires are used in Searle’s experiment for determining Young’s modulus of elasticity?
(a) 3 (b) 1 (c) 2 (d) 4
18. The |u|, |v| graph for a concave mirror is as shown in |v| figure. Here |u| > |f|. A line passing through origin of slope 1 cuts the graph at point P. Then coordinates of P point P are 45° (a) (|2f |, |2f |) (b) (|2f |, |f |) |u| (c) (|f |, |2f |) (d) (|f |, |f |)
19. In a vernier callipers N divisions of vernier scale coincides with N – 1 divisions of main scale (in which length of one divisions is 1 mm). The least count of the instrument is ( in mm)
1 1 (a) N (b) N – 1 (c) (d) N N 1
20. Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum. They use different lengths of the pendulum and/or record time for different number of oscillations. The observations are shown in the table.
Least count for length = 0.1 cm. Least count for time = 0.1 sec.
Student Length of the Number of Total time for (n) Time period (s)
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-87
pendulum (cm) oscillations (n) oscillations (s) I 64.0 8 128.0 16.0 II 64.0 4 64.0 16.0 III 20.0 4 36.0 9.0 g If EI, EII, and EIII are the percentage errors in g, i.e., 100 for students I, II and III, g respectively,
(a) EI 0 (b) EI is minimum (c) EI EII (d) EII is maximum
EXERCISE - II
IIT-JEE SINGLE CHOICE CORRECT
50 1. What is the reading of vernier callipers shown in 60 figure mm scale
(a) 54.6 mm (b) 53.1 mm vernier scale Object (c) 52.7 mm (d) 54.7 mm
2. What is the reading of micrometer screw gauge 0 1 2 35 shown in figure. mm 30 25 (a) 2.31 mm (b) 2.29 mm
(c) 2.36 mm (d) 2.41 mm
3. Express which of the following setups can be used to verify Ohm’s law?
A
(a) V (b) A V
A V
(c) (d)
V A
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-88
4. Which of the following statements is not correct?
(a) Resistance box works on the principle of wheat stone bridge.
(b) When post office box is used to measure the internal resistance of a cell, then a resistance box is used in series with the galvanometer.
(c) The wire of the metre bridge is made of a material of high resistivity and high temperature coefficient.
(d) The wheat stone bridge condition does not change even if we inter change the positions of the galvanometer meter and cell.
5. The velocity of sound in air is measured by using a resonance tube method. The observation made during the experiment are as follow.
Frequency of Resonance Position of water level at resonance (cm) tuning fork (Hz)
512 First 16.2 16.0
Second 49.7 49.9
The velocity of sound is
(a) 330 m/s (b) 340 m/s (c) 345 m/s (d) 350 m/s
6. A dimensionally correct equation ______be a correct equation, and a dimensionally incorrect equation ______be incorrect. The words (in order) to be filled in the blank spaces are
(a) must, must (b) must, may (c) may, may (d) may, must
7. Precision in measurement depends on
(a) zero error (b) parallax
(c) least count of instrument (d) calibration of instrument
8. The number of significant figures in the number 0.020740 × 10–3 are
(a) 6 (b) 5 (c) 7 (d) 4
9. 0.205 – 0.2014 can be expressed as
(a) 0.0036 (b) 3.6 × 10–3 (c) 4.0 × 10–3 (d) 4 × 10–3
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-89
10. If speed of light is 3.00 × 108 m/s and 1 year = 365.25 days, then 1 light year is
(a) 9.468 × 1015 m (b) 9.47 fm (c) 9.47 × 1015 m (d) 9.46 × 1015 m
11. The density of wood is 0.5 in CGS system of units. The corresponding value in MKS unit is
(a) 500 (b) 0.5 (c) 5 × 10–2 (d) 5000
12. If n is the numerical value of a physical quantity in the system in which its unit is u, then which of the following relation is correct?
n u (a) constant (b) constant (c) nu = constant (d) n2u constant u n
13. In an experiment to measure the diameter of a wire using a screw gauge of resolution 0.001 cm, which of the following may correctly represents the measurement?
(a) 01.235 cm (b) 01.2351 cm (c) 01.24 cm (d) 01.23500 cm
14. The breadth of a thin rectangular sheet is measured as 10.1 cm. The uncertainty in the measurement is
(a) 1% (b) 0.5 % (c) 0.1 % (d) 5%
1 15. The least count of a stop watch is s . The time of 20 oscillations of a pendulum is measured as 25 5 s. The percentage error in the measurement of time will be
(a) 0.1% (b) 0.8% (c) 1.8% (d) 8%
16. A measured value 0.003749 expressed in two significant digits is
(a) 0.0 (b) 0.0038 (c) 0.0037 (d) 2.8 × 10–3
17. The density of a cube is found by measuring its mass and the length of its side. If the maximum errors in the measurement of mass and length are 0.3% and 0.2% respectively, the maximum error in the measurement of density is
(a) 0.3% (b) 0.5% (c) 0.9% (d) 1.1%
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-90
18. The kinetic energy of a particle depends on the square of speed of the particle. If error in measurement of speed is 40%, the error in the measurement of kinetic energy will be
(a) 40% (b) 80% (c) 96% (d) 20%
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-91
19. The length of a rod is measured by four different instruments and the measurements are reported as (A) 500.0 mm (B) 50.0 cm (C) 0.500 m (D) 5.0 × 10–4 km We can conclude that (a) (A) is most accurate measurement (b) (C) is most accurate measurement (c) (A), (C) and (B) are equally accurate measurements (d) the accuracy of all the measurements is the same
20. The linear momentum p of a particle is given as a function of time t as p At 2 Bt C . The dimensions of constant B are
(a) M L1 T1 (b) M L1 T2 (c) M L T2 (d) M L T1
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-92
EXERCISE - III
ONE OR MORE THAN ONE CHOICE CORRECT
1. A length of 5.0 × 101 cm when converted into meter can be written as
(a) 0.5 m (b) 0.50 m (c) 5.0 × 10–1 m (d) 5.00 × 10–1 m
a bc 2. A dimensionless quantity y is represented by the formula y . Which of following is/are d e correct?
(a) Dimensions of d are e are same (b) abc and de have some dimensions
bc (c) is dimensionless (d) de + bc is not meaningful ae d
3. Let y = A sin (t – kx) represents the variation of distance y of a particle with time t. Which of the following is/are not meaningful?
y y At (a) (b) (c) A – kx (d) A A kx k
4. In a screw gauge, there are 100 divisions in circular scale and each main scale division is of 1 mm. With no gap between the jaws, 98th divisions coincide with the main scale and zero of main scale is not visible. While measuring the diameter of a ball, the circular scale is between 3 mm mark and 4 mm mark. 76th division of circular scale coincides with the reference line. Select the correct alternative (a) the least count of micrometer is 0.01 mm (b) zero error is – 0.02 mm (c) zero error is 0.02 mm (d) diameter of the ball is 3.76 mm
5. Which of the following functions of A and B cannot be performed if A and B possess different dimensions?
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-93
A (a) A + B (b) A – B (c) (d) None of these B
6. The radius of a spherical ball is (10.4 0.4) cm. Select the correct alternative
(a) the percentage error in radius is 4% (b) the percentage error in radius is 0.4%
(c) the percentage error in volume is 12% (d) the percentage error in volume is 1.2 cm3
7. A dimensionless quantity
(a) may have a unit (b) must have a unit
(c) may not have a unit (d) must not have a unit
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-94
2c t 2x 8. The equation of a stationary wave is y 2Asin cos . Which of the following is/are correct?
(a) unit of ct is same as that of (b) unit of x is same as that of
2c 2x c x (c) unit of is same as that of (d) unit of is same as that of c t
9. A voltmeter has a least count of 0.1 V and an ammeter has a least count of 0.1 A. The potential drop V across a resistance is measured as 10.0 V and current through it is measured as 1.0 A. Select the correct alternative
(a) the value of R is (1.0 0.1) × 101
1 (b) the relative error in measurement of current is 10
1 (c) the accuracy in measurement of potential drop is 100
(d) the value of R is (10 0.2)
10. Which of the following unit can possibly be a unit of time?
(a) Leap year (b) Tropical year (c) Siderial year (d) Light year
11. L, C and R represent the physical quantities inductance, capacitance and resistance respectively. The combinations, which have the dimensions of frequency, are
1 R 1 C (a) (b) (c) (d) RC L LC L
12. The dimensions of the quantities in one (or more) of the following pairs are the same. Identify the pair (s)
(a) torque and work (b) angular momentum and work
(c) energy and Young’s modulus (d) light year and wavelength
13. The pairs of physical quantities that have the same dimensions is (are) (a) Reynolds number and coefficient of friction
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-95
(b) Curie and frequency of a light wave (c) Latent heat and gravitational potential (d) Planck’s constant and torque
14. Let [0 ] denote the dimensional formula of the permittivity of the vacuum and [0 ] that of the permeability of the vacuum. If M = mass, L = length, T = time and I = electric current 1 3 2 1 3 4 2 (a) [0 ] [M L T I] (b) [0 ] [M L T I ]
2 2 2 1 (c) [0 ] [M LT I ] (d) [0 ] [M L T I]
15. The SI unit of the inductance, the Henry can by written as (a) weber/ampere (b) volt-second/ampere (c) joule/(ampere)2 (d) ohm-second
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-96
EXERCISE - IV
MATCH THE FOLLOWING
1. In Column – I, some physical quantities are given and in Column – II, dimensions of the physical quantities are given. Match the entries in Column – I to entry in Column – II.
Column – I Column - II
I. Pressure A. M1 L0 T2
II. Potential energy/volume B. M0 L T2
III. Acceleration C. M L1 T2
Kinetic Energy 0 0 0 IV. D. [M L T ] Mass Distance
V. Surface tension
2. Column – I gives parameters/observations on measuring instruments and Column – II gives values of L.C. as readings taken. Match the entries in Column – I to those of Column – II.
Column – I Column - II
I. Main scale divisions of 1 mm and 50 vernier A. Positive zero error scale division
II. Main scale divisions of 0.5 mm and 100 B. Negative zero error circular scale divisions
III. Zero of circular scale has advanced by 2 C. L.C. 0.02 mm divisions from reference line when measuring faces are touching each other
IV. Zero of vernier scale is to the left of zero of D. L.C. 0.005 mm main scale
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-97
3. Some physical quantities are given in Column I and some possible SI units in which these quantities may be expressed are given in Column II. Match the physical quantities in Column I with the units in Column II.
Column -I Column -II
I. GMeMs G – universal gravitational constant, A. (volt) (coulomb) (metre)
Me – mass of the earth,
Ms – mass of the Sun
3RT II. M 3 –2 R – universal gas constant, B. (kilogram) (metre) (second) T – absolute temperature, M – molar mass
2 F III. q 2 B 2 C. (metre)2 (second)–2 F – force, q – charge, B – magnetic field
GM IV. e
Re D. (farad) (volt)2 (kg)–1 G – universal gravitational constant,
Me – mass of the earth,
Re – radius of the earth
REASONING TYPE
Directions: Read the following questions and choose (A) If both the statements are true and statement-2 is the correct explanation of statement-1.
(B) If both the statements are true but statement-2 is not the correct explanation of statement-1.
(C) If statement-1 is True and statement-2 is False.
(D) If statement-1 is False and statement-2 is True.
1. Statement-1: ‘Light year’ and ‘Wavelength’ both measure distance. Statement-2: Both have dimensions of time. (a) A (b) B (c) C (d) D
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-98
2. Statement-1: Force cannot be added to pressure. Statement-2: Because their dimensions are different. (a) A (b) B (c) C (d) D
3. Statement-1: Parallex method cannot be used for measuring distances of stars more than 100 light years away. Statement-2: Because parallex angle reduces so much that it cannot be measured accurately. (a) A (b) B (c) C (d) D
4. Statement-1: Number of significant figures in 0.005 is one and that in 0.500 is three. Statement-2: This is because zeros are not significant. (a) A (b) B (c) C (d) D
5. Statement-1: Mass, length and time are fundamental physical quantities. Statement-2: They are independent of each other. (a) A (b) B (c) C (d) D
6. Statement-1: Density is a derived physical quantity. Statement-2: Density cannot be derived from the fundamental physical quantities. (a) A (b) B (c) C (d) D
7. Statement-1: Radar is used to detect an aeroplane in the sky. Statement-2: Radar works on the principle of reflection of waves. (a) A (b) B (c) C (d) D
8. Statement-1: Surface tension and surface energy have the same dimensions. Statement-2: Both have different S.I. unit. (a) A (b) B (c) C (d) D
9. Statement-1: Radian is the unit of distance. Statement-2: One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius of the circle. (a) A (b) B (c) C (d) D
10. Statement-1: When we change the unit of measurement of a quantity, its numerical value changes. Statement-2: Smaller the unit of measurement smaller is its numerical value. (a) A (b) B (c) C (d) D
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-99
11. Statement-1: Dimensionless constants are the quantities having numerical value. Statement-2: All constants are dimensionless. (a) A (b) B (c) C (d) D
g 12. Statement-1: The time period of a pendulum is given by the formula, T 2 . l Statement-2: According to the principle of homogeneity of dimensions, only that formula may be correct in which the dimensions of L.H.S. is equal to dimensions of R.H.S. (a) A (b) B (c) C (d) D
13. Statement-1: Avogadro number is the number of atoms in one gram mole. Statement-2: Avogadro number is a dimensionless constant. (a) A (b) B (c) C (d) D
14. Statement-1: L/R and CR both have same dimensions. Statement-2: L/R and CR both have dimension of time. (a) A (b) B (c) C (d) D
15. Statement-1: The quantity (1/ 00 ) is dimensionally equal to velocity and numerically equal to velocity of light in vacuum. Statement-2: 0 is permeability of free space and 0 is the permittivity of free space. (a) A (b) B (c) C (d) D
LINKED COMPREHENSION TYPE
Passage – I
The distance between earth and Sun is nearly 1.496 × 1011 m. This is called astronomical unit (AU). The Sun distances of planets from Sun are generally expressed in this unit. To measure the distance of a planet which lies between sun and earth (such planets are called inferior planets) Copernicus method is used. Here, we define the elongation () P AU of a planet, which is the angle formed at the earth between the earth-planet direction and earth-Sun direction. Knowing this value we can estimate the planet’s distance from Sun. Earth
1. As the planet moves around Sun, the elongation changes. Let the path of the planet be circular. When the elongation of the planet is maximum the angle between Sun-planet direction and planet-earth direction is
(a) 0º (b) 90º (c) 45º (d) 180º
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-100
2. Assuming that orbit of planet mercury around Sun to be circular, its orbital radius is found to be 0.38 AU. The angle of maximum elongation for planet mercury is nearly
(a) 30º (b) 38º (c) 22º (d) 68º
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-101
3. The Copernicus method for measuring the distance of a planet from Sun is applicable for
(a) all planets except earth (b) inferior planets
(c) superior planets (d) earth only
Passage – II
Internal micrometer is a measuring instrument used to measure internal diameter (ID) of a large cylinder bore with high accuracy.
Cap Ratchet
Main scale A 5 B P 0 Q 0 10 45 Hub Circular scale Internal diameter (being measured)
Construction is shown in figure. There is one fixed rod B (to the right in figure) and one moving rod A (to the left in figure). It is based on the principle of advancement of a screw when it is rotated in a nut with internal threads. Main scale reading can be directly seen on the hub, which is fixed with respect to rod B. When the cap is rotated, rod A moves in or out depending on direction of rotation. The circular scale reading is seen by checking which division of circular scale coincides with the reference line. This is to be multiplied by L.C. to get circular scale reading.
Least count = value of 1 circular scale division
pitch = number of divisions on circular scale
Length of rod A is chosen to match the ID (PQ) to be measured. Zero error is checked by taking reading between standard blocks fixed at nominal value of ID to be measured.
Zero error checking
5 0 Standard blocks
Zero error is positive if cap end is on the right side of the main scale and negative it is on the left side.
1. In an internal micrometer, main scale division is of 0.5 mm and there are 50 divisions in circular
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-102
scale. The least count of the instrument is
(a) 0.005 mm (b) 0.001 mm (c) 0.05 mm (d) 0.01 mm
2. In the above instrument, while measuring an internal diameter. ID is set of 321 mm with no zero error. If cap end is after 7th division and 17th division of main scale coincides with the reference line, the ID is
(a) 321.717 mm (b) 321.87 mm (c) 328.17 mm (d) 324.67 mm
3. During zero setting of the above instrument, the end of the cap is on left side of the zero of main scale (i.e., zero of main scale is not visible) and 41st division of circular scale coincides with the reference time, the zero error is
(a) – 0.09 mm (b) + 0.41 mm (c) – 0.41 mm (d) + 0.09 mm
EXERCISE - V
SUBJECTIVE PROBLEMS
1. Find the dimensional formula of
(a) linear momentum (b) frequency (c) pressure.
1 2 Show that the expression x = vt + at2 is dimensionally correct, where x is a coordinate and has 2 units of length, v is speed, a is acceleration and t is time.
3. Test with the dimensional analysis whether the following equations are correct or not.
2S cos Pr 4 t (a) h (b) V rg 8l
where h = height, S = surface tension, = density, P = pressure, V = volume, = coefficient of viscosity.
4. The square of the speed of an object undergoing a uniform acceleration a is some function of a and the displacement s, according to the expression v2 = kamsn, where k is a dimensionless constant. Show by dimensional analysis that this expression is satisfied only if m = n = 1.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-103
5. The volume of an object as a function of time is calculated by V = At 3 + B/ t, where t is time measured in seconds and V is in cubic meters. Determine the dimensions of the constants A and B.
Mm 6. Newton’s law of universal gravitation is F G r 2
Here F is the force of gravity, M and m are masses, and r is a length. What is the SI units of the constant G?
7. Write the number of significant digits in (a) 1001, (b) 100.1, (c) 100.10
8. Determine the number of significant figures in the following numbers
(a) 23 cm, (b) 3.589 s, (c) 4.67 × 103 m/s, (d) 0.0032 m.
9. Carry out the following arithmetic operations:
(a) the sum of the numbers 756, 37.2, 0.83, and 2.5;
(b) the product 3.2 × 3.563;
(c) the product 5.6 ×
10. A rectangular building lot is 100.0 ft by 150.0 ft. determine the area of this lot in m 2.
11. A box container of ice cream having volume 27m3 is to be made in the form of a cube. What should be the length of a side in cm?
12. Show that the following are of the dimensions of energy
(i) mc2 where m = mass and c = velocity of light
mP (ii) where P = pressure, = density of liquid and m = mass.
13. The time period (t) of vibration of a liquid drop depends on surface tension (S), radius (R) of the drop and density () of the liquid. Find t.
14. In the formula X = 3YZ2, X and Z have the dimensions of capacitance and magnetic induction, respectively. What are the dimensions of Y?
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-104
15. Assuming that the vibration frequency of atoms in a crystal depends on the atomic mass m, the atomic spacing and compressibility , find an expression for frequency. (Compressibility = 1 ) Bulk modulus
16. In two different systems of units an acceleration is represented by the same number, while a velocity is represented by numbers in the ratio 1: 3, compare the units of length and time.
17. In a certain system of absolute units the acceleration produced by gravity in a body falling freely is denoted by 3, the kinetic energy of a 272.1 kg shot moving with velocity 448 metres per second is denoted by 100, and its momentum by 10; find the units of length, time and mass.
18. If the unit of force be the weight of one kilogram, what must be the unit of mass so that the equation F = ma may still be true?
19. The height h to which a liquid of density and surface tension S rises in a capillary tube of radius r may depend upon , , r and g, where g is the acceleration due to gravity. Experiments show that h is inversely proportional to r. With this experimental information, use the dimensional method to kS show that h . (k is the dimensionless constant) r g
20. The Young’s modulus of steel in the CGS system is 2.0 1012 dyne cm-2 . Express it in the SI system.
21. A cube has a side of length 1.2 102m. Calculate its volume.
22. We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be 2.63 s, 2.56s, 2.42s, 2.71s and 2.80 s. Calculate the (a) absolute errors, (b) relative error (c) percentage error.
23. The mass and density of a solid sphere are measured to be 12.4 0.1kg and 4.6 0.2kg/m3. Calculate the volume of the sphere with error limits.
32 24. A physical quantity is related to four variables , , and as follows . The percentage errors of measurements in , , and are 1%, 3%, 4% and 2 % respectively. Find the percentage errors in .
25. A wire is of mass (0.3 0.003)g. The radius is (0.5 0.005)mm and length is (6.0 0.06)cm then find the % error in density.
26. Speed of light in SI system is 3 × 108 m/s. What is the speed of light in a new system of units in which unit of length is x km and unit of time is y millisecond.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-105
27. In 2 complete revolutions of the circular scale, distance moved on the pitch scale is 1 mm. There are 50 divisions on the circular scale. There is no zero error. The diameter of a wire is measured with the help of this gauge. It reads 6 divisions on the main scale and 32 divisions on the cap. If each main scale division is 0.5 mm, what is the diameter of the wire?
28. A vernier calipers has 50 divisions on its vernier scale. The smallest division on main scale is 1 mm. When the jaws are touching each other, zero of vernier scale is between zero and 1st division of main scale and 8th division of vernier scale coincides with one of the divisions of main scale. While taking a measurement, zero of vernier scale is between 15th and 16th division of main scale and 39th division of vernier scale coincides with one of the divisions of main scale. What is the correct value of measurement?
29. Check the dimensional accuracy of the formula for power transmitted through a cross-section of string. P 22 A2f 2v
Here, A, f, and v are respectively amplitude of displacement, frequency, linear density and speed of wave.
30. The kinetic energy of a particle moving along x-axis varies with the distance x of the particle from A x3 origin as K . Write the dimensional formula for A2B . B (x1/ 4 ) C
42M 31. Spring constant of a spring is calculated using formulae k . Where T is time period of T 2 vertical oscillation when, mass M is lung with help of spring to rigid support, if time of oscillation for 10 oscillations is measured to be 5.0 sec, mass M = 0.20 kg. Find possible error in spring constant k.
32. If (n + 1)th divisions of vernier side coincides with nth division of main scale then find the least count of the vernier. It is given that one main scale division is equal to a units.
33. Explain with a suitable figure how a battery should be Z connected so that the given rheostat can be used as a Y potential divider. Give the points about which we can take the X
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-106
output.
34. The length of the string of a simple pendulum is measured with a metre scale to be 90.0 cm. The radius of the bob plus the length of the hook is calculated to be 2.13 cm using measurements with a slide callipers. What is the effective length of the pendulum? (This effective length is defined as the distance between the point of suspension and the center of the bob).
35. The pitch of a screw quage is 0.5 mm and there are 50 divisions on the circular scale. In measuring the thickness of a metal plate, there are five divisions on the main scale and 34th division concides with the reference line. Find the thickness of the metal plate
36. The length of cylinder is measured with the help of a vernier callipers whose smallest division on the main scale is 0.5 mm and nine divisions of the main scale are equal to ten divisions of the vernier scale. It is observed that 78th divisions of the main scale coincides with the sixth division of the vernier scale. Find the length of the cylinder.
37. A brass ball of mass 100 g is heated to 1000C and then dropped into 200g of turpentine in a calorimeter at 150C. The final temperature is found to be 230 C. Determine the specific heat of turpentine. Take specific heat of brass as 0.092 cal/g0C and water equivalent of calorimeter as 4g.
38. The value of an unknown resistance is obtained by using a post office box. Two consecutive readings of R are observed at which the galvanometer deflects in the opposite directions for three
different values of R1. These two values of R are recorded under the column I and II in the following observation table
X R A 1 B R2 C
1000 100 10 10 100 1000
1 2 5 10 20 50
5000 2000 2000 1000 500 200 200 100 D T T A' K1 K2 B'
G
S. No R1() R2() R lies in between
I() II()
1 10 10 16 17
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-107
2 100 10 163 164
3 1000 10 1638 1639
Determine the value of the unknown resistance (X).
39. The velocity of sound in air is measured by using a resonance tube method. The observations made during the experiment are as follows
Frequency of first tuning fork f1= 480 Hz
Frequency of second tuning fork f1= 512 Hz
Position of upper end of the tube L1= 0.2 cm
Room temperature T = 250 C
Frequency of tuning fork (Hz) Position of water level at resonance L2(cm)
Resonance Water falling Water rising
480 First 17.0 17.2
Second 52.4 52.6
512 First 16.4 16.2
Second 49.9 50.1
Calculate
(i) the velocity of sound at room temperature
(ii) the end-correction for the tube
40. The focal length of a convex lens is measured experimentally by u-v method.
The index error between the object needle and the lens is –0.1 cm.
The index error between the image needle and the lens is –0.1 cm.
The observation made during the experiment are recorded as follows.
Position of
Object Lens Image
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-108
x0 (cm) xl (cm) xi (cm)
10.0 59.1 80.4
10.0 55.3 77.6
10.0 51.5 74.8
10.0 46.7 71.9
10.0 42.3 70.1
10.0 37.4 70.3
10.0 32.1 78.2
1 1 (i) plot a graph between and and determine the focal length of the lens. v u
(ii) plot a graph between +v and –u and determine the focal length of the lens.
41. The Young’s modulus of a steel wire is obtained by using the Searle’s Apparatus.
The observations noted during the experiment are as follows
Length of the wire, L = 30 cm
Diameter of the wire d = 1 mm
Pitch of the micrometer p = 0.5 mm
No. of divisions on the disc N = 100
Least count of the micrometer, C = 0.005 mm
Micrometer reading for no load, R0 = 0.0540 mm
Micrometer reading (mm) S.No Load on the Hanger M(kg)
Load increasing R1(mm) Load decreasing R2(mm)
1 0.5 0.555 0.545
2 1.0 0.565 0.555
3 1.5 0.575 0.565
4 2.0 0.580 0.580
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-109
5 2.5 0.585 0.595
6 3.0 0.595 0.605
7. 3.5 0.610 0.610
8 4.0 0.625 0.615
9 4.5 0.630 0.630
10. 5.0 0.640 0.640
(i) Plot a graph between extension l and load M and show that it is a straight line.
(II) Calculate the value of Young’s Modulus.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-110
ANSWERS
EXERCISE – I
AIEEE SINGLE CHOICE CORRECT
1. (c) 2. (d) 3. (c) 4. (a) 5. (c)
6. (a) 7. (b) 8. (c) 9. (b) 10. (a)
11. (b) 12. (c) 13. (b) 14. (a) 15. (c)
16. (c) 17. (c) 18. (a) 19. (c) 20. (b)
EXERCISE – II
IIT-JEE SINGLE CHOICE CORRECT
1. (d) 2. (c) 3. (b) 4. (d) 5. (c)
6. (a) 7. (c) 8. (a) 9. (a) 10. (b)
11. (c) 12. (c) 13. (c) 14. (a) 15. (c)
16. (b) 17. (a) 18. (a) 19. (c) 20. (c)
EXERCISE – III
ONE OR MORE THAN ONE CHOICE CORRECT
1. (b, c) 2. (a, b, d) 3. (a, c, d,) 4. (a, b) 5. (a, b)
6. (a, c) 7. (a), (c) 8. (a, b) 9. (a, b, c) 10. (a, b, c)
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-111
11.(a, b, c) 12. (a, d) 13. (a, b, c) 14. (b, c) 15. (a, b, c, d)
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-112
EXERCISE – IV
MATRIX MATCH TYPE
1. I – (C), II – (C), III – (B), (IV) – (B), (V) – A)
2. I – (C), II – (D), III – (B), (IV) – (B)
3. I – (A, B), II – (C, D), III – (C, D), (IV) – (C, D)
REASONING TYPE
1. (c) 2. (a) 3. (a) 4. (c) 5. (a)
6. (c) 7. (a) 8. (d) 9. (d) 10. (c)
11. (c) 12. (b) 13. (c) 14. (a) 15. (b)
LINKED COMPREHENSION TYPE
Passage – I
1. (b) 2. (c) 3. (b)
Passage – II
1. (d) 2. (d) 3. (a)
EXERCISE – V
SUBJECTIVE PRBOLEMS
1. (a) [MLT–1] (b) [T–1 ] (c) [ML–1T–2]
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-113
2. Dimensionally correct.
3. Dimensionally correct.
5. [L3 T –3], [L3T]
6. The units of G are m3/(kg.s2). 7. (a) 4, (b) 4, (c) 5
8. (a) 2 (b) 4 (c) 3 (d) 2
9. (a) 795 (b) 11 (c) 2.0 × 101
10. 150.0 × 102 ft2
11. 3.0 102 cm
R 3 13. t = k S
14. M–3L–2 T8A4
15. f = k m
16. 3 : 1 , 9 : 1
17. 153.6 m, 6.857 s, 544.2 kg
18. g kg
20. 2.0 1011 N m-2
21. 1.7 106 m3
22. (a) 0.01s , 0.06 s, 0.20 s, 0.09 s, 0.18 s
(b) .042
(c) 4.2%
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-114
23. 2.7 0.1m3
24. 13%
25. 4%
300y 26. new unit x
27. 3.32 mm
28. 15.60 mm
29. Dimensionally correct.
19 30. [M1 L 4 T 2 ]
31. (31.6 2.8) Nm 1
1 32. units n 1
33. Battery should be connected across XY, with a Z suitable tap it will work like a potential divider. X R Y Output can be taken either across X and Z or across Y and Z.
34. 92.1 cm
35. 2.84 mm
36. 36.8 mm
37. 0.423 cal/g 0C
38. Between 16.38 and 16.39 Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-115
39. (i) 338.4 m/s
(ii) 0.6 cm
40. (i) 15.15
(ii) 15 cm
41. 1.87 1011 N/m2
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-116
SOLUTIONS
EXERCISE – I
F F A2 V 2 1. (c) Dimension of Y = Dimension of pressure = = As L = FA2V 4 . 2 4 L V A
[N][x x ] [L2T 1][L] 2. (d) Dimensions of D = 2 1 = [MºL2T 1] . 3 [n2 n1] [L ]
L 3. (c) = time constant of LR circuit. R
RC = time constant of RC circuit.
4. (a) Q = mL
Q [ML2T 2 ] Latent heat = [Mº L2T 2 ] m [M]
5. (c) 3.8106 4.2105 45.8106 4.6105
6. (a) As volume of fluid flowing through a tube per unit time is pr 4 V 8 l
7. (b) V = 1.76 × 25 = 44.0 cm3
1 1 8. (c) Energy = mv2 mass(velocity)2 2 2 Units of energy = kg(m/s) 2 = kg m2s2
9. (b) When 0.00587 is removed, the previous digit 3 should be increased by one, so the rounded of value should be 6.04
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-117
10 100 cm 10. (a) g 10 m/s2 = 3600000 cm/minute2 = 36 105cm/minute2 2 1 minute 60
dx 11. (b) If x an nan1 dx nan1da x
dx da So n x a
l 12. (c) T 2 g
13. (b) x MaLbT e
dx dM dL dT So = a b c = (a + b + c)% x m L T
M M 14. (a) V L3
d dM dL 100% = 100% 3 100% = 3% + 3 × 2% = 9% M L
15. (c) It is the least count of the vernier calipers which is equal to the difference of one main scale division to one vernier scale division.
16. (c) Zero error is positive when the zero of vernier scale lies to the right of the zero of the main scale.
17. (c) In Searle’s experiment two wires are used. One is called Auxiliary wire (compressed) and other is called experimental wire (extended).
18. (a) As the slope of line is 1 u = v, this possible only when object is placed at 2f.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-118
19. (c) Least count = 1 MSD – 1 VSD
N vernier scale divisions are equal to N – 1 main scale divisions so
N 1 1 VSD = mmand 1 MSD = 1 mm N
N 1 1 So least count = 1 mm – mm = mm N N
g l T 20. (b) 100 2 100 g l T
g 0.1 0.1 5 For student I, EI 100 2 100 % g 64 128 16
g 0.1 0.1 15 For student II, EII 100 2 100 % g 64 64 32
g 0.1 0.1 19 For student III, EIII 100 2 100 % g 20 36 18
Percentage error is minimum for student I.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-119
EXERCISE – II
1. (b) As zero of vernier scale is just above the 53 mm mark of main scale so main scale reading is 53 mm and first line of vernier coincides
So measurement = 53 mm + 1 × 0.1 mm
= 53.1 mm
2. (a) Main scale reading = 2 mm
Circular scale reading = 31 × 0.01 mm
= 0.31 mm
So measurement = 2.31 mm
0.5 mm and least count = 0.01mm 50
3. (a) In first case voltmeter is parallel to the variable resistance and ammeter is in series.
4. (c) The temperature coefficient of the wire of metre bridge should be low.
16.2 16.0 5. (c) l 16.1 cm 1 2
49.7 49.9 l 49.8 cm 2 2
l2 l1 33.7 cm
33.7 cm 2
67.4 cm
67.4 So speed of sound = 512 345 m/s . 100
6. (d) A dimensionally correct equation may be correct equation but a dimensionally incorrect equation must be incorrect.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-120
7. (c) Precision in measurement depends on reliability of the instrument and it is related to the least count of the instrument.
8. (b) As zero before 2 is not significant while zero after 2 and after 4 are significant. So total number of significant figures is five.
9. (d) 0.205 0.2014 0.0036
As there can be only three digits after decimal place in the answer so it should be written as 0.004 or 4103 .
10. (c) 1 ly = 3.00 × 108 × 365.25 × 24 × 60 × 60
= 9.46728 × 1015 m
= 9.47 × 1015 m(Up to three significant figures)
0.5 103 kg 11. (a) 0.5 g/cm3 = 500 kg m3 . 106 m3
12. (c) If system of units is changed, the measurement does not change and hence n1u1 n2u2
So nu = constant
13. (a) If the least count of instrument is 0.001 cm, then there should be three digits (in cm) in the measurement.
14. (a) In 10.1 cm, 0.1 cm is uncertain
0.1100 So % uncertainty = 1% 10.1
1 15. (b) Least count = sec 0.2 sec 5
Measurement = 25 sec
0.2 % Error = 100% 0.8% 25
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-121
16. (c) When 0.000049 is removed for rounding off the previous digit should remain same should remain same.
mass 17. (c) Density = volume
mass = (length)3
d ρ 100% 0.3% 3 0.2% 0.9%
1 1 18. (c) K mv2 ; K' m(1.4v)2 ; K' 1.96 K 2 2
19. (a)
The accuracy of a measurement is related to % error. If the true value is not known, then maximum possible error is equal to the least count of the measurement. So in case of measurement A, maximum possible error is 0.1 mm
0.1 So % error = 100% 0.02% (which is minimum). 500
20. [Bt] = [p]
[p] [M L T1] [B] = [M1 L1 T2] [t] [T ]
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-122
EXERCISE – III
1. (b, c)
5.0 101 cm 0.50 m 5.0 101m
As there should be two significant figures only.
2. (a, b, d)
[a] = [bc] = [d] = [e]
3. (a, c, d)
[t] = [kx] = [M 0 L0 T 0 ]
As [y] = [A]
[yt] = [At]
y t At 1 [t] as kx kx [] [kx]
[y] At So [] kx
4. (a, b)
1mm Least count of vernier = 0.01mm 100
Zero error = – 0.01 × 2 mm = – 0.02 mm
Measurement = 3 + 0.01 × 76 = 3.76 mm
Actual diameter = 3.76 – (– 0.02) = 3.78 mm
5. (a, b)
A and B have different dimensions so A + B and A – B is not possible.
6. (a, c)
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-123
Radius of spherical ball = (10.4 0.4) cm
0.4 So % error in radius is 100% = 3.85% 4%. 10.4
R % Error in volume = 3 × 100% 12% R
7. (a, c)
Dimensionless quantities like refractive index or dielectric constant does not have units while angle, solid angle are dimensionless but have unit’s radian and steradian.
8. (a, b)
[y] = [A]
[ct] = [] = [x]
Dimensions of [ct] is same as that of [] and [x]
9. (a, b, c)
V (10.0 0.1)V
I (1.0 0.1)A
V R 10.0 I
R V I R V I
0.1 0.1 R 10.0 = 1.1 10.0 1.0
So R 10.0 1.1 (1.0 0.1)101
(As answer cannot have more than two significant figures).
10. (a, b, c)
11. (a, b, c)
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-124
12. (a, d)
(a) Torque and work both have the dimensions [M L2 T 2] .
(b) Light year and wavelength both have the dimensions of length i.e., [L]
13. (a, b, c)
Reynold’s number and coefficient of friction are dimensionless quantities.
Curie is the number of atoms decaying per unit time and frequency is the number of oscillations per unit time.
Latent heat and gravitational potential both have the same dimension corresponding to energy per unit mass.
14. (b, c)
1 q q F · 1 2 2 40 r
2 [q1][q2] [IT ] 1 3 4 2 [0 ] = [M L T I ] [F][r 2] [MLT 2][L2]
1 Speed of light, c 00
1 1 [ ] = = [MLT 2 I 2] 0 2 1 3 4 2 1 2 [0 ][c] [M L T I ][LT ]
15. (a, b, c, d)
weber (a) L or henry = i ampere
d i e volt second (b) e L L or henry = d t (d i / d t) ampere
1 2U joule (c) U Li 2 L or henry = 2 i 2 (ampere)2
1 (d) U Li 2 i 2 Rt 2
L = Rt or henry = ohm-second
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-125
EXERCISE – IV
MATCH FHE FOLLOWING
1. Matrix match
Force Pressure = [M1 L1 T2] Area
Potential energy/Volume = [M1 L1 T2]
Acceleration = [Mº L T2 ]
Kinetic Energy [ML2 T2] [Mº L1 T2] Mass Distance [ML]
Surface tension = [M1 Lº T2]
(I – C), (II – C), (III – B), (IV – B), (V) – A)
1 2. (a) Lease count = mm 0.02 mm 50
0.5 (b) Least count = mm 0.005 mm 100
(c) Negative zero error
(d) Negative zero error
(I – C), (II – D), (III – B), (IV – B)
3. (I – A, B), (II – C, D), (III – C, D), (IV – C, D)
ASSERTION AND REASON
1. (c) Light year is the distance travelled by light in one year so it has units of distance similarly wavelength is distance travelled by a wave in a time internal of one time period of particle of medium.
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-126
2. (a) Two quantities of different dimensions cannot be added or subtracted.
3. (a)
4. (c) Zeros before a non zero digit are not significant but zero after a non zero digit in the decimal part of the number are significant.
5. (a)
6. (c) Density of a body is defined as the ratio of mass of the body to its volume.
7. (a) Microwaves are reflected from a moving object like aeroplane is again received by the radar so it works on the principle of reflection.
8. (d) Surface tension is force per unit length and surface energy is energy.
9. (d) Radian is not a unit of distance, it is the unit of plane angle.
10. (a) When we change the system of units, the measurement does not change only numerical value changes.
11. (c) All constants are not dimensionless.
12. (b) If a physical equation is dimensionally correct, it does not mean that it is actually correct.
13. (c) Avagadro number is not dimensionless constant, its dimensions are per mole.
Rt t L RC 14. (a) i i0(1 e ) and i i0 e .
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-127
1 15. (b) Speed of light in vacuum is given of c . 00
PASSAGE BASED PROBLEMS
Passage – I 1. (b) rP orbital radius of planet
re orbital radius of earth
According to Copernicus method elongation will be maximum when = 90º.
2. (c) rP rE sin
0.38 AV = 1 AV sin
sin = 0.38
= 22º as sin 30º = 0.5
3. (b) Copernicus method can be used for inferior plants.
Passage – II
main scale division 1. (d) Least count of screw gauge = number of divisions on circular scale
0.5 mm = = 0.01 mm 50
2. (d) Internal diameter is = 321 mm + 7 × 0.5 mm + 17 × 0.01 mm
= 324.67 mm
3. (a) Zero error = – 0.01 × (50 – 41)
= – 0.09 mm
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-128
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-129
EXERCISE – V
1. (a) Linear momentum = mv [MLT 1]
1 (b) Frequency = [MºLºT 1] T
F [MLT 2] (c) Pressure = [ML1T 2] A [L2]
2. Dimensions of LHS = [L]
Dimension of vt [LT 1T ] [L]
1 Dimensions of at 2 [LT 2T 2] [L] 2
So the above equation is dimensionally correct.
3. (a) Dimensions of LHS = [L]
[MLºT 2 ][Mº LºT º ] Dimensions of RHS = [L] [ML3 ][L][LT 2 ]
(b) Dimension of LHS = [L3 ]
[ML1T 2][L]4[T ] Dimensions of RHS = [L3 ] [ML1T 1][L]
So both the equations are dimensionally correct.
4. v 2 k am sn
[Mº L2 T 2] [LT 2]m [L]n
m n 2
– 2m = – 2
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-130
So m = 1
n = 1
[V ] [L3 ] 5. Dimension of A [Mº L3 T 3 ] [t 3 ] [T 3 ]
Dimensions of B [v][t] [L3 ][T ] [Mº L3 T 1]
Mm 6. F G r 2
Fr 2 G Mm
SI units of G = N m2 kg2
7. 1001 Number of significant digits = 4
100.1 Number of significant digits = 4
100.10 Number of significant digits = 5
8. (a) 23 cm 2
(b) 3.589 s 4
(c) 4.67 103 m/s 3
(d) 0.0032 m 2
9. (a) 756 + 37.2 + 0.83 + 2.5 = 794.53 = 796
(b) 3.2 × 3.563 = 11.4016 = 11
(c) 5.6 × = 5.6 × 3.1 = 17.36 = 1.7 × 101
10. Area = 100.0 × 150.0 ft2 = 150.0 × 102 ft2
11. V 27 m3
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-131
L3 27 m3
L 3 m 3 102 cm
12. (i) mc 2 [ML2T 2]
MP (ii) [ML2T 2 ]
13. t Sa Rb c
[Mº LºT ] [M1LºT 2]a [L]b [ML3 ]c
a + c = 0
b – 3c = 0
– 2a = 1
1 a = 2
1 c = 2
3 b = 2
R 3 So t S
R 3 t k S
Where k is a dimensionless constant.
14. X 3YZ2
X capacitance
Z magnetic induction
X Y 3Z 2 Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-132
(charge)2 energy (charge)4 (velocity)2 (AT )4(LT 1)2 [Y ] = = (force)2 energy (force)2 [ML2T 2][MLT 2]2 (charge)2 (velocity)2
= [M 3L2T 8A4 ]
15. Frequency M a b c
[Mº LºT 1] k[M]a [L]b[M 1LT 2]c
Where k is dimension less constant
a – c = 0
b + c = 0
2c = –1
1 c = 2
1 a = 2
1 b = 2
So frequency = k M
16. Acceleration = velocity × time
Velocity × time = constant
So ratio in units of time = 3 : 1
As Length × time2 = constant
Ration of units of length = 9 : 1
2 2 17. [9.8 ms ] 3 [L2T2 ]
1 (272.1kg)(448 ms1)2 100 [M L2 T 2] 2 2 2 2
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-133
1 1 (272.1kg)(448 ms ) 10[M2 L2 T2 ]
Solving M2 544.2 kg
L2 153.6 m
T2 6.857 s
18. F = Ma
If unit of force be weight of 1 kg i.e., g N
So unit of mass should be g kg.
19. Let h Sar bc g d
So h k Sar bc g d
Where k is a dimensionless constant [L] k [MLºT 2 ]a [L]b [ML3 ]c [LT 2 ]d
a + c = 0
b – 3c + d = 1
–2a – 2d = 0
Experimentally b = – 1
On solving a = 1, c = –1 and d = –1
k S So h r g
dyne 2.0 1012 105 N 20. Y 2.0 1012 = 2.0 1011 Nm 2 cm2 104 m2
21. Volume = (1.2 102 m) 3 1.728 106 m3
So volume = 1.7106 m3
22. T1 2.63 s
T2 2.56 s Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-134
T3 2.42 s
T4 2.71s
T5 2.80 s
2.63 2.56 2.42 2.71 2.80 T 2.62 s av 5
Absolute error = 2.62 2.63 2.56 2.62 2.42 2.62 2.71 2.62 2.80 2.62
0.01 0.06 0.20 0.09 0.18 = = 0.11 s 5
0.11 Relative error = 0.042 2.62
Percentage error = 4.2%
23. Mass = (12.4 0.1) kg
Density = (4.6 0.2) kg m3
12.4 Volume = 2.7 m3 4.6
V M V M
0.1 0.2 V 2.7 0.1 m3 12.4 4.6
So Volume = (2.7 0.1) m3
32 24.
d d d 1 d 1 d So 100% = 3 10% 2 100% 100% 100% 2 2
1 1 = 3(1%) 2(3%) (4%) (2%) = 12% 2 2
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-135
d dm dr dl 25. 100% = 100% 2 100% 100% m r l
0.003 0.005 0.06 2 100% = 4% 0.3 0.5 6.0
26. Speed of sound = 3108m/s
1 3 108 = 1000x 103 y
300y = new unit. x
1 1 103 [ 1 m = km So 1 M = (x Km) and 1s 103 MS (y MS) ] 1000 1000x y
1 mm 27. Main scale reading = 0.5 mm 2
0.5 Least count = 0.01mm 50
Diameter (measurement) = 6 × 0.5 mm + 32 × 0.01 = 3.32 mm
28. Least count of vernier calipers = 1 MSD – 1 VSD
49 = 1 mm mm 50
1 = mm 50
= 0.02 mm
Zero error = 0 + 0.02 × 8 = 0.16 mm
Measurement = 15 mm + 39 × 0.02 = 15.78 mm
Actual measurement = 15.78 mm – 0.18 mm = 15.60 mm
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-136
29. Dimensions of LHS = [M L2 T3 ]
Dimensions of RHS = [L2][T2][ML1] [LT1]
= [M L2 T3 ]
So the above formula is dimensionally correct.
A x3 30. Kinetic Energy K Bx1/ 4 C
Dimensions of A = dimension of x3 [L3 ]
L3 Dimensions of C = [M–1 L T 2 ] M L2 T 2
[C] M–1 LT 2 3 So dimensions of B = [M1L4 T2 ] x1/ 4 1 L4
27 So dimensions of A2B [M–1 L 4 T2 ]
42M 31. K T 2
5.0 T 0.50 s 10
0.1 T 0.01s 10
M = 0.20 kg
M = 0.01 kg
4 (3.14)2 0.20 K = 31.6 (0.50)2
K M 2T K M T
0.01 0.01 K 2 31.6 = 2.8 0.20 0.50
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-137
So spring constant is = (31.6 2.8) Nm–1
32. Least count = 1 MSD – 1 VSD
1 MSD = 1 unit
(n + 1) VSD = n MSD = n units
n So 1 VSD = units n 1
n 1 So least count = 1 unit – units = units. n 1 n 1
33. The battery should be connected between x and y and output can be taken either between x and z or between y and z.
34. Effective length = 90.0 cm + 2.13 cm = 92.13 cm
There can be maximum one digit after decimal point the reported length should be 92.1 cm
(Note that thee is one digit after decimal point in 90.0 cm).
0.5 mm 35. Least count of the screw Gauge = = 0.01 mm 50
Main scale reading = 0.5 × 5 = 2.5 mm
Circular scale reading = 34 × 0.01 mm = 0.34 mm
Thickness of the plate = 2.5 mm + 0.34 mm = 2.84 mm
9 0.5 mm 36. Least count of vernier = 0.5 mm – = 0.05 mm 10
Main scale reading = 73 × 0.5 = 36.5 mm
Vernier scale reading = 6 × 0.05 mm = 0.3 mm
Measurement = 36.5 + 0.3 = 36.8 mm
(If 78th main scale division concide with 6 vernier scale divisions then, before the zero of vernier scale there should be 73rd division of main scale)
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-138
37. Heat given by the ball = heat absorbed by oil and calorimeter
100 (100 – 23) × 0.092 = 200 × (23 – 15) c + 4 × (23 – 15) × 1
9.2 × 77 = 200 × 8 × c + 32
708.4 = 1600 c + 32
676.4 c 0.423 cal g 1 ºC 1 1600
X R 38. 2 R R1
10 10 For third case X 1638 OR X 1639 1000 1000
So X = 16.38 OR X = 16.39
Actually X can be any value between 16.38 and 16.39
39. For the first tuning fork f1 480 Hz
17.0 17.2 l 17 cm 1 2
52.4 52.6 l 52.5 cm 2 2
52.5 cm-17.1cm 2
70.8 cm
70.8 v 480 m/s = 339.8 m/s 1 100
For the second tuning fork f2 512 Hz
16.4 16.2 l ' 16.3 cm 1 2
49.9 50.1 l ' 50.0 cm 2 2
' (50.0 16.3) 2 67.4 cm
67.4 500 v 337 m/s 2 100
339.8 337 So speed of sound is = = 338.4 m/s 2 Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396
UNITS, DIMENSION, SIGNIFICANT FIGURES& ERROR OP-MIII-P-139
l 3l 52.5 51.3 The end correction e 2 1 0.6 cm 1 2 2
l ' 3l ' 50.0 48.9 e 2 1 0.55 cm 2 2 2
0.6 cm 0.55 cm So end correction = 0.6 cm 2
1 v
v
40. (I) (II) u 1
u
l
mgL 41. (I) M (II) Y where is extension. A
Study More with www.puucho.com
Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396