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lly expressed as the product of a number and ple of the quantity concerned whic UNITS AND MEASUREMENTS | TOPIC 1| System of Units In order to define units for all fundamental quantities or base quantities, we use the term fundamental units or base units and then to define units for all other quantities as products of powers of the base units that we call derived units. Hence, a complete set of these units, i.e. both the base units and derived units, is known as the system of units. PHYSICAL QUANTITIES All those quantities which can be measured directly or indirectly and in terms of which the laws of physics can be expressed are called physical quantities, Physical quantities can be further divided into two types: (i) Fundamental Quantities Those physical quantities which are independent of other physical quantities and are not defined in terms of other physical quantities, are called fundamental quantities or base quantities. ¢.g.Mass, length, time, temperature, luminous intensity, electric current and the amount of substance, ete. (ii) Derived Quantities Those quantities which can be derived from the fundamental quantities are called derived quantities. e.g. Velocity, acceleration and linear momentum, ete. CHAPTER CHECKLIST ‘al Quantities Physical Unit Significant Figures and Rounding off nal Formulae and Dimensional Equations MEASUREMENT OF PHYSICAL QUANTITIES The measurement of physical quantity is the process of comparing this quantity with a standard amount of the physical quantity of the same kind, called its unit. Hence, to express the measurement of a physical quantity, we need to know two things (i) The unit in which the quantity is measured. (ii) The numerical value or the magnitude of the quai i.e. the number of times that unit (1) is contained in the given physical quantity. or ‘# Numerical Value Inversely Proportional. to the Size of Unit The numerical value (n} is inversely proportional to the size (u) of the unit. 1 Ne— = nu=constant. u e.g. The magnitude of a quantity remains the same, whatever may be the unit of measurement lke = 1000 g Q=n, uy =Mpuy where n,n, are the numerical values and u,v, are the two units of measurement of the same quantity Hence, We may write as PHYSICAL UNIT The standard amount of a physical quantity chosen to measure the physical quantity of the same kind is called a physical unit. The essential requirements of physical unit are given as below: (i) Ir should be of suitable size. (ii) Ir should be easily accessible. (ii) It should not vary with time. (iv) It should be easily reproducible. (v) Ir should not depend on physical conditions like pressure, volume, temperature, etc. The physical unit can be classified into two ways that can be given as below : Fundamental Units The physical units which can neither be derived from one another, nor they can be further resolved into more simpler units are called fundamental units. The units of fundamental quantities, ic. length, mass and time are called fundamental units or base units, Derived Units The units of measurement of all other physical quantities which can be obtained from fundamental units are called derived units. _ Speed = Distance Time Unit of Unit of time istance Unit of speed = Unit of speed (i.c.ms™') is a derived unit. THE INTERNATIONAL SYSTEM OF UNITS A system of units is the complete set of units, both fundamental and derived for all kinds of physical quantities. The common systems of units used in mechanics are given below (i) FPS System It is the British engineering system of units, which uses foor as the unit of length, pound as the unit of mass and second as the unit of time. (ii) CGS System It is based on Gaussian system of units, which uses centimetre, gram and second for length, mass and time, respectively. (iii) MKS System It uses metre, kilogram and second as the fundamental units of length, mass and time, respectively. (iv) International System of Units (SI Units) The system of units, which is accepted internationally for measurement is the System Internationaled’ Units (French for International System of Units) abbreviated as SI. The SI, with standard scheme of symbols, units and abbreviations, was developed and recommended by General Conference on Weights and Measures in 1971 for international usage in scientific, technical, industrial and commercial work. This system of units makes revolutionary changes in the MKS system and is known as rationalised MKS system, It is helpful to obtain all the physical quantities in Physics. SI derived units with special names SI unit =e Name Symbot___Etrason nme esmegsin mere Frequency Hertz Hz - s Force Newton N - kg ms” or kg ms Pressure, stress Pascal Pa Nir? or N mr? kgmrt's™ or kgs*m Energy, work, quantity of heat Joule J Nm kg mis” or kg mis? Power, radiant flux Watt Ww disor Js"! kgm’ s™ or kgms* Quantity of electricity, electric charge [Coulomb = C = As Electric potential, potential difference, | Volt v WIA or WA" kg m?s"? A“ or kgm? 5? A electromotive force Capacitance Farad F cv A? st kin? Electric resistance Ohm Q WA kgm? sta Conductance Siemens S AV mn? kgr's? A? Magnetic flux Weber Wb Vs or JfA kgmis™ At Magnetic field, magnetic flux density, | Tesla T Wo J? kgs Avt magnetic induction Inductance Henry H WhfA kg m’s@a~ Luminous flux, luminous power Lumen im - ed/sr Muminance Lux bi inj? mm? dsr! Activity (of a radio nuclide/ radioactive |Becquerel Bq - = source) Absorbed dose, absorbed dose index | Gray Gy Jikg wie s Some SI derived units expressed by means of SI units Physical quantity Magnetic moment Dipole moment Dynamic viscosity Torque, couple, moment of force Surface tension Power density, irradiance, heat flux density Heat capacity, entropy Specific heat capacity, specific entropy Specific energy, latent heat Radiant intensity Thermal conductivity Energy density Electric field strength Electric charge density Electricity flux density Permittivity Permeability Molar energy Angular momentum, Planck constant Molar entropy, molar heat capacity Exposure (Grays and rays) Absorbed dose rate Compressibility Elastic moduli Pressure gradient Surface potential Pressure energy Impulse Angular impulse Specific resistance Surface energy ‘Sl unit Name Symbol Expression in terms of S| base units Joule per tesla wm mA Coulomb metre cm sAm Poiseuille or pascal second PlorPas or mt kgs"! or newton second per square = Ng mf? metre Newton metre Nm mm kgs* Newton per metre Nim kg s* Watt per square metre Wir? kgs? Joule per kelvin atk mf kgs Kk Joule per kilogram kelvin dikg K ms? Kk" Joule per kilogram dik m s* Walt per steradian wsrt kg ms Sr* Watt per metre kelvin Wm'k™ mkgs@ kK" Joule per cubic metre dim? kg ms Volt per metre vm mkgs@ a7 Coulomb per cubic metre Cit mAs Coulomb per square metre = Cin? nas Farad per metre Fim m3 kg" s* A? Henry per metre Him mkgs™ A~ Joule per mole Jimal mf kgs” mol" Joule second Js kgm’ s™ Joule per mole kelvin imal K mm kgs™ K7 mol" Coulomb per kilogram Cikg kgs A Gray per second Gyls ms Per pascal Pam" mkg™'s? Newton per square metre Nir? or Nr? kg mnt! s* Pascal per metre PalmorNm? = kgm? s™? Joule per kilogram Jikg or Nimvkg mi s® Pascal cubic metre PamPorNm — kg mi?s® Newton second Ns kg mss" Newton metre second Nms kgm? s7* ‘Ohm metre am kgm? s™ A@ Joule per square metre Jim? or Nim kgs? ABBREVIATIONS IN POWERS OF TEN When the magnitudes of the physical quantities are very large or very small, it is convenient to express them in the multiples or submultiples of the SI units. The various prefixes used for powers of 10 are listed below in table Prefixes for Powers of Ten Multiple Prefix Symbol Sub-multiple Prefix Symbol 10 deca da 10" deci d 10° hecto oh 10% centi oc 10° kilo k 107 milli m 10° mega M 10% micro 10° gga «GG 10% nano on 107 tera T io"? pica Pp 10° peta P ion femtof 10° exa E 190° alto a eg. 1 megaohm, MQ. = 10° 1 milliampere or 1] mA =107 A 1 kilometre, | km = 10° m 1 microvolt or IV = 107 V 1 decagram, 1 da g=10 g 1 nanosecond or 1 ns= 107" s 1 centimetre, 1 cm = 107 m 1 picofarad or 1 pF = 107 F y CGS, MKS and SI are Decimal Systems of Units As we know that, CGS, MKS and SI are metric or decimal systems of units. This is because the multiplies and sub-multiplies of their basic units are related to the practical units by powers of 10. SOME IMPORTANT PRACTICAL UNITS For Length/Distance (i) Astronomical Unit It is the mean distance of the earth from the sun, | AU= 1.496 10"! m, (ii) Light year It is the distance travelled by light in vacuum in one year. 1 ly= 9.46% 10" m (iii) Parallactic second Iris the distance at which an arc of length 1 astronomical unit subtends an angle of 1 second of arc. 1 parsec= 3.084% 10! m = 3.26 ly (iv) Micron or micrometer, | jim =10~ m (v) Nanometer, | nm = 107 m (vi) Angstrom unit, 1 A=107'° m (vii) Fermi This unit is used for measuring nuclear sizes 1Fm=107% m EXAMPLE |2| Length Conversion How many parsec are there in one metre? Sol. Given, 1 parsec = 3.084 10" m or 3.084% 10'° m = 1 parsec me 1 3.084% 10° = 325x107" parsec parsee EXAMPLE |3| Relation between different unit of length Deduce relations between astronomical unit, light year and parsec. Arrange them in decreasing order of their magnitudes. Sol Weknow that 1AU=15x 10" m lly= 9.46% 10% m 1 par sec = 3.08 10 m lly _ 9.46% 10° TAU 15x10" 1ly= 63% 10' AU wali Conversion of light year into parsec Lparsec 3.08% 10" tly 9.46% 10" '. 1 parsec = 3.26 ly wii) Comparing results from Eqs. (i) and (ii), we get 1 parsec > 1 ly> 1 AU = 63x 10" iLe., = 326 For Mass (i) Pound, | lb = 0.4536 kg (ii) Slug, 1 Slug = 14.59 kg (iii) Quintal, 1 q = 100 kg (iv) Tonne or Metric ton, 1 ¢ = 1000 kg (v) Atomic mass unit (it is defined as the 1/12th of the mass of one 7C atom) | amu = 1.66 x 1077 kg For Time (i) Solar day It is the time taken by the earth to complete one rotation about its own axis wrt. the sun. (ii) Sedrial year It is the time taken by the earth to complete one rotation about its own axis wrt. a distant star. (iii) Solar year It is the time taken by the earth to complete one revolution around the sun in its orbit. 1 solar year= 365.25 average solar days = 366.25 sedrial days (iv) Tropical year The year in which there is toral solar eclipse is called tropical year. (v) Leap year The year which is divisible by 4 and in which the month of February has 29 days is called a leap year. (vi) Lunar month Ir is the time taken by the moon to complete one revolution around the earth in its orbit. 1 lunar month= 27.3 days (wit) Shake It is the smallest practical unit of time. 1 shake =10~ s EXAMPLE |4| A Clock Which type of phenomena can be used as a measure of time? Give three examples. Sol. A phenomena which repeats itself at regular intervals can be used as a measure of time. Some examples are given below (i) oscillation of a pendulum (ii) rotation of earth around its axis (iii) revolution of earth around the sun For Areas (i) Barn, 1 barn= 1078 m? (ii) Acre, 1 acre= 4047 m? (iii) Hectare, 1 hectare= 104 m For Other Quantities (i) Litre (for volume), 1 L= 10° cc=107™m? Where, cc represents cubic centimetre, i.e. cm?. 2 (ii) Gallon (for volume), In USA, 1 gallon = 3.7854 L In UK, 1 gallon = 4.546 L (iii) Pascal (for pressure), 1 Pa = 1 Nm™ Pressure exerted by earth's, atmosphere. 1 atm = 1.01 10° Pa (iv) Bar (for pressure), 1 bar = 1 atm= 1.01 10° Pa= 760 mm of Hg (v) Torr (for pressure), 1 torr= 1 mm of Hg column = 133.3 Pa (vi) Electron volt, LeV=1.6x 107" J i) Erg (for energy/work), lerg= 1077 J Kilowart hour (for energy), 1 kWh= 3.6x 10 J (ix) Horse power (for power), 1 HP = 746 W (x) Dioprre (for power of a lens), 1D=1m™ ua i) D for angle), P' =—— rad (xi) Degree (For angle), I? = ra TOPIC PRACTICE 1 | OBJECTIVE Type Questions 1. Ais the fundamental quantity. Here, A refers to (a) mass (b) velocity (c) acceleration (d) linear momentum Sol. (a) Mass is the fundamental quantity as it does not depend upon other physical quantities. However, other three quantities, ie. velocity, acceleration and linear momentum are not fundamental quantities as these show their dependency on fundamental quantities. 2. How many wavelength of Kr* are there in 1 m? (a) 1553164.13 (b) 1650763.73 (c) 2348123.73 (d) 652189.63 Sol. (b) The number of wavelengths of Kr" in 1 m is 1650763.73. 3. The solid angle subtended by the periphery of an area lcm? at a point situated symmetrically at a distance of 5 cm from the area is (a) 2x10™ sr (b) 4x10 sr (c) 6x10™ sr (d) 8107 sr dA_ itcm* Sol. (b) Solid angle, dQ = = am? = 004 sr =4x10™ sr 4. Which of the following is not a physical quantity? (a) Time (b) Impulse (c) Mass (d) Kilogram If the velocity of light is taken as the unit of velocity and year as the unit of time, what must be the unit of length? What is it called? Unit of length = Unit of velocity x Unit of time =3x10°ms™ x1 year =3x10°ms™ x 365 x 24 X60 X60 Sol. A =945x10"ms™ =1ly How many metric tons are there in teragram? Sol. In 1 teragram =10" g In 1 metric ton = 10°kg =10° x 10° =10° g :. Number of metric tons are in teragram 2 = 1B oot 10° g¢ 24. Sol. What is common between bar and torr? Both bar and torr are the units of pressure. 1 bar = 1 atmospheric pressure = 760 mm of Hg column =10°N/m* Ltorr = 1 mm of Hg column 1 bar = 760 torr 25. The mass of a proton is L67 x 10” kg. How many protons would make 1 g? Sol Number of protons = Vt! 4ss_ Mass of each proton 10 =, = 599 x 10 1.67% 1 LONG ANSWER Type I Questions 26. Why length, mass and time are chosen as base quantities in mechanics? Sol. In mechanics, length, mass and time are chosen as the base quantities because (i) there is nothing simpler to length, mass and time. (ii) all other quantities in mechanics can be expressed in terms of length, mass and time. (iii) length, mass and time cannot be derived from one another. 27. Express the average distance of earth from the sun in (i) light year (ii) parsec. Sol. Average distance of earth from the sun is (r) =1AU =1496 x10" m _ 1496 x10" 946 x10" Le. ly =158x107 ly Sol. 8 Sol. 30. Sol. _ 1496 x10" Also, p = ————_ 308 x 10° = 4.86 X10 parsec The radius of atom is of the order of 2 A and radius of a nucleus is of the order of fermi. How many magnitudes higher is the volume of atom as compared to the volume of nucleus? Ry, Le. radius of atom is 2 A=2x 107" m Ry, ie. radius of nucleus is 1 fermi 4 4 4 Vv, 3 BRA -|% Vy 4ops LR, w sae by aN ; _fax10] a rca | The unit of length convenient on the atomic scale is known as an angstrom and is denoted. byA 1A=107° m. The size of the hydrogen atom is about 0.5 A. What is the total atomic volume in m* of a mole of hydrogen atoms? [NCERT] Radius of a hydrogen atom (r)=05 A =05 x10" m =8x10"* 4 Volume of each hydrogen atom (V) = qr = < 3.14 x (05 x 107" 8 =5.234 x 10m? Number of atoms in 1 mole of hydrogen = Avogadro's number (N) = 6023 x 10 :. Atomic volume of 1 mole of hydrogen atoms (V") Volume of a hydrogen atom x Number of atoms VisVxN = 5.236 x 10°! «x 6023 x 10%m* = 3152x107 m Why has second been defined in terms of periods of radiations from cesium-133?7 Second has been defined in terms of periods of radiation, because (i) this period is accurately defined. (ii) this period is not affected by change of physical conditions like temperature, pressure and volume ete. (iii) the unit is easily reproducible in any good laboratory. LS) YOUR TOPICAL UNDERSTANDING OBJECTIVE Type Questions Which one of the following is not a unit of British system of units? (a) Foot (b) Metre (c) Pound (d) Second . Which of the following statement is incorrect regarding mass? (a) I is a basic property of matter (b) The SI unit of mass is candela (c) The mass of an atom is expressed in u (d) None of the above . Pascal is the unit of (a) force (b) stress (c) work —(d) energy . The surface area of a solid cylinder of radius 2.0 cm and height A cm is equal to1.5x 10* (mm)’. Here, A refers to (a) 0.9 cm (c) 30 cm (b) 10cm (d) 15cm . If the value of force is 100 N and value of acceleration is 0.001 ms~*, what is the value of mass in this system of units? (a) 10° kg (b)10" kg (c)10* kg (d)10° kg . Young’s modulus of steel is 1.9 10'' N/m?. When expressed in CGS units of dyne/cm’, it will be equal to (1N= 10° dyne, 1m? = 10° cm’) [NCERT Exemplar] (a)1.9x 10 (b) 9x 10" (c)1.9x10" (d)1.9x 10" . If the size of bacteria is 1p, then the number of bacteria in 1 m length will be (a) one hundred (b) one crore (c) one thousand (d) one million . Among the given following units which one is not unit of length? (a) Angstrom (c) Barn (b) Fermi (a) Parsec . Age of the universe is about 10% yr, whereas the mankind has existed for 10° yr. For how many seconds would the man have existed if age of universe were 1 day? (a) 92s (c) 86s {b) 10.28 (a) 10.58 Answer 1. 2 | 3. (b) | 4. (b) | 5. 6. (c) 7 (dd) 8. (c) 9% (c) VERY SHORT ANSWER Type Questions 10. it. 12. 13. Explain the concept of mass, length and time which is related to basic or fundamental quantities in mechanics. How do we make the choice of a standard/unit of measurement? Why MKS system had to be rationalised to obtain SI? What is a coherent system of units? SHORT ANSWER Type Questions 14. 15. 16. 17. 18. 19. Briefly, explain the measuring process of any physical quantity. Why ‘metre’ has been defined in terms of wavelength and ‘second’ in terms of periods of radiation or rotation? Calculate the surface area of a solid cylinder of diameter 4 cm and height 20 cm in mm?. [Ans. 27657.1 mm?] The density of air is 1.293 kg/m, Express it in CGS units. [Ans. 0.001293 g/cc] What is the average distance of earth from the sun? [Ans. 1.496 «10"' m) An Astronomical Unit (AU) is the average distance between earth and the sun, approximately 1.50 x 10® km. The speed of light is about 3 x 10° m/s. Express the speed of light in astronomical unit per minute. [Ams. 0.12 AU/min]} LONG ANSWER Type I Questions 20. 21. 22. 23. The radius of gold nucleus is 41.3 fermi. Express its volume in m?. [Ans. 2.95 x10™ m?] Which unit can be used for measuring of very small masses? “Five litres of benzene will weigh more in summer or winter’. Comment. Is the measure of an angle dependent upon the unit of length? EXAMPLE |1| Uncertainty in Measurement Write down the number of significant figure in the following. (i) 0.072 (ii) 12.000 (iii) 0.060 (iv) 3.08 x 10"? (v) 1.2340 (vi) 0.04 Sol (i) Two (ii) Five (iii) Two (iv) Three (v) Five (vi) One ROUNDING OFF The result of computation with approximate numbers, which contain more than one uncertain digit, should be rounded off. While rounding off measurements, we use the following rules by convention: Rule \ If the digit to be dropped is less than 5, then the preceding digit is left unchanged. e.g. x=7.82 is rounded off to 7.8. Rule 2 \f the digit to be dropped is more than 5, then the preceding digit is raised by one. e.g. x =6.87 is rounded off to 6.9. Rule 3 If the digit to be dropped is 5 followed by digits other than zero, then the preceding digit is raised by one. e.g. x =16.351 is rounded off to 16.4. Rule 4 If the digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is left unchanged, if it is even. e.g. x =3.250 becomes 3.2 on rounding off. Rule 5 If the digit to be dropped is 5 or 5 followed by zeros, then the preceding digits is raised by one, if it is odd. €.g. x =3.750 is rounded off to 3.8. EXAMPLE |2| Precise Value Round off the following number as indicated (i) 18.35 upto 3 digits = 18.4 (ii) 143.45 upto 4 digits = 143.4 (iii) 189.67 upto 3 digits = 190 (iv) 321.1355 upto 5 digits = 321.14 (v) 31.325 x 107 upto 4 digits = 31.32x 10° (vi) 101.55 x 108 upto 4 digits = 101.6 10° RULES FOR ARITHMETICAL OPERATIONS WITH SIGNIFICANT FIGURES Any result is calculated by compounding (ic. adding/ subtracting/multiplying/dividing) two or more variables, which might have been measured with different degrees of accuracy. Inaccuracy in the measurement of any one variable affects the accuracy of the final result. Therefore, the result of arithmetical operations performed with measurements cannot be more accurate than the original measurement themselves. The following rules for arithmetic operations with significant figures that make final result more consistent with the precision of the measured values. Addition and Subtraction In both, addition and subtraction, the final result should retain as many decimal places as are there in the number with the least decimal places. eg. The sum of three measurement of length 2.1 m, 1.78 m and 2.046 m is 5.926 m, which is rounded off to 5.9 m (upto smallest number of decimal places in the three measurements). Similarly, if x = 12.587 m, and y = 12.5 m, then (x = y) is 12587 = 125 =0.087 m, which is rounded off to 0.1 m, upto smallest number of decimal places in y. Multiplication and Division In multiplication or division, the final result should retain as many significant figures, as are there in the original number with the least significant figures. e.g. The speed of light is 3.00 x 10° m/s and one year has 3.1557 x 107s, then light year is given by = 3.00 x 10° x 3.155 x 107 =94685 x 10" m =95 x10" m [Rounded off upto 2 significant figures] EXAMPLE |3| A Rolling Cube Each side of a cube is measured to be 7.203 m. What are the total surface area and the volume of the cube to appropriate significant figures? {NCERT] Sol. Given, Side of the cube = 7.203 m Total surface area = 6x (side)* = 6x (7.203)* = 311.299254 m* = 311.3 m* [Rounded off to 4 significant figures} Volume = (side)? = (7.203)* = 373.714754 m* = 373.7 m*. [Rounded off to 4 significant figures} EXAMPLE |4| Density of a Substance 5.74 g of a substance occupies 1.2 cm? . Express its density keeping significant figures in view. [NCERT] Sol. Density = = 4.783 gem™ =48gcem ~ [Rounded off upto 2 significant figures) RULES FOR DETERMINING UNCERTAINTY IN THE RESULTS OF ARITHMETIC CALCULATIONS Rule 1 Rule 2 Rule 3 Suppose, we use a metre scale to measure length and breadth of a thin rectangular sheets as 15.4 cm and 10,2 cm, respectively. Each measurement has three significant figures and a precision upto first place of decimal. Therefore, we can write length (/)= (15.4 £01) em =1544( 2 x100)=154 cm 20.6% 15.4 Similarly, breadth (6) = (10.2 0.1) em 0.1x 100 = 10.24] ——— ]% = 10.2 cemt+% 10.2 Thus, the error of the product i.e. area of thin sheet is 15.4% 10,24 (0.6+ 1.0) cm? = 157.08 cm? + 1.6% 2,{16 2 =157.08cm~ +] — X157.08 Jem 100 =(15708+251)cm’. As per rule, the final value of area can contain only three significant figures and error can contain only one significant figures, we can write the final result as A=(157 £3)cm? If a set of experimental data is specified to » significant figures, a result obtained by combining the dara will also be valid to m significant figures. eg. «= 13.7 m and y=8.08 m, both have three significant figures. Now, x-y= 13.7—8.08=5.62 m So, the final result should retain as many decimal places as, there is the number with least decimal places. Therefore, rounding off to one place of a decimal, we get x — y= 5.6 cm The relative error of a value of number specified to m significant figures depends not only on m, but also on the number itself. eg. m, = (1.04+ 0.01) kg and my = (9.244 0.01) kg Relative error in 1.04 kg is 001 (2) xl00=+ 1% 1.04 Similarly, the relative error in 9.24 kg is + (= 924 ‘Thus, the relative error depends on the number itself. }>100= 0.1% Rule 4 In a multi-step computation, the intermediate results should be calculated to one more significant figure in every measurement, then the number of digits in the least precise measurement. eg, x=958 has three significant digits. Now, reciprocal of x is +=—L_ =0104, rounded off to 958 three significant digits. When we take reciprocal of 0.104, we get 9.62, rounded off to three significant digits 1 — =01044, rounded x 958 off to four significant figures, then However, if we calculate 1 — £958, rounded off to three significant digits aoa 8 BN Thus, retaining one more extra digits in intermediate steps of complex calculations would avoid additional errors in the process of rounding off the numbers. EXAMPLE |[5| Appropriate Numbers of Significant Number Solve the following and express the result to an appropriate number of significant figures. (i) Add 6.2g, 4.33g and 17.456g (ii) Subtract 63.54 kg from 187.2 kg (iii) 75.5 x 125.2 x 0.51 2.13 x 24.78 458.2 Sol. (i)6.2 g + 4.33 g+ 17.456 g = 27.986= 28.0 ¢ [rounded off to first decimal place] (ii) 187.2kg — 63.54kg = 123.66kg= 123.7kg [rounded off to first decimal place] (ili) 755 x 125.2 * 051 = 4820826 = 4800 [rounded off upto two significant figures] 213 x 2478 (iv) = 0115193 = 0115 4! [rounded off to three significant figures] Sol. As, V =(6040.1)V, 1 =(40+40.2)A Power, P = VI =6.0x4.0= 24 W and maximum error in power measurement AP_AV_Al_ 0.102 a eae aa aaa P 4 IT 60 40 = 0.017 + 0,050 = 0.067 AP = 0.067 x P = 0.0 67 x 24=16W Power consumed by the electric lamp within error limit is (24 +1.6) W. LONG ANSWER Type I Questions il. Sol. Sol. The mass of a box measured by a grocer’s balance is 2.3 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (i) the total mass of the box (ii) the difference in the mass of the pieces to correct significant figures? (NCERT] Given, mass of the box (m) = 23 kg Mass of first gold piece (m, ) = 2015 g = 002015 kg Mass of second gold piece (m,) = 20.17 g = 002017 kg (i) Total mass of the box (M) =m +m, +m, = 23 + 002015 + 0.02017 = 234032 kg As the mass of the box has least decimal place ie. one decimal place, therefore, total mass of the box can have only one decimal place. Rounding off the total mass of the box up to one decimal place, we get Total mass of the box (M) = 23 kg (ii) Difference in masses of gold pieces (Am) = m, — m, = 2017 — 2015 = 002g (The masses of two gold pieces has two decimal places, therefore, it is correct up to two places of decimal. A physical quantity P is related to four observables a, b,c andd are as follows P=a'b?/Jed The percentage errors of measurement in a, b, ¢ and d are 1%,3%, 4% and 2%, respectively. What is the percentage error in the quantity P? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result? [NCERT] Pate? fled Maximum relative error in physical quantity Pis given by Fh) 2)(F] . Maximum percentage error in P is given by Given, AP (= ) (= ) — x 100 = +/ 3] — x 100] + 2} — x100 P a b 2(** ) (+ ) +—|— x 100]+ | — x 100 2he d Given 94 x 100=1%, 92 x100=3% a * x100=4%, “4 100= 2% c & Srimes[sxo+ 2x()+ Lx +a} =+(3+64+2+2)%=+13% As the result (13%) has two significant figures, therefore, the value of P = 3763 should have only two significant figures. Rounding off the value of P up to two significant figures, we get P =3.8 13. State the number of significant figures in the following (i) 0.007 m? (iii) 0.2370 g/cm (v) 6.032 N/m? INCERT] (ii) 2.64 x 10" kg (iv) 6.320 J (vi) 0.0006032 m? Sol. The number of significant figures in the given quantities are given below. (i) In 0.007, the number of significant figures is 1 because in a number less than 1, the zero’s on the right of the decimal point but to the let of the first non-zero digit are not significant. (ii) In 264 x10", the number of significant figures is 3 because all non-zero digits are significant, power of 10 are not taken in significant figure. (iii) In 0.2370, the number of significant figures is 4, as all non-zero digits left to decimal and trailing zero are significant. (iv) In 6.320, the number of significant figures is 4 (reason is same as in part ‘iii’). (v) In 6.032, the number of significant figures is 4 (reason is same as in part ‘iii’). (vi) In 00006032, the number of significant figures is 4 (reason is same as in part ‘i’). 14. Write down the number of significant figures in the following (i) 5238 N (iii) 34.000 m (ii) 4200 kg (iv) 0.02340 N/m Sal. (i) 5238 N has four significant digits. (ii) 4200 kg = 4.200 x 10° kg has four significant figures (iii) 34,000 m has five significant digits (iv) 0.02340 N/m has four significant digits. 15. Compute the following with regards to significant figures. () 46 x 0128 (i) SPRRRR LES 876 + 0.4382 Sol (i) 46x 0128 = 05888 = 059 Mass of the Sun (M) Density of the Sun = The result has been rounded off to have two Volume of the Sun (V) significant digits (as in 4.6) a . _.y 0.9995 x 153 [: Denity= | (ii) ———— = 096057 = 0961 lume tee M__3M 3x 20x10” igni = - - Ee teas sey rounded off to three significant p Tie FRR” axal4 x (70x10) 53). = (iii) 876 + 04382 = 876.4382 = 876 3x10” As, there is no decimal point in 876, therefore, result = ————§_,, =1392x 10° of addition has been rounded off to no decimal point. 628 XS45X10 _ 3 s 16. The length, breadth and thickness of a PwlA x10" kgim rectangular sheet of metal are 4.234 m, 1.005 m This density is of the order of density of solids and and 2.01 cm, respectively. Give the area and liquids and not of gases. volume of the sheet to correct significant The temperature of inner core of the sun is 10” K while figures. (NCERT] the temperature of the outer layers is nearly 6000 K. At ; so high temperature, no matter can exist in its solid or °C): Mt different values of a same quantity are given in liquid state. Every matter is highly ionised and present iferent units, then first of all convert them in : ¥ ates é as a mixture of nucleus, free electrons and ions which perdied nee Shenging ine number. Of is called plasma. The density of plasma is so high due to nee : inward gravitational attraction on outer layers due to Sol. Given, length (/) = 4.234 m, Breadth (b) = 1,005 m inner layers of the sun. Thickness (t) = 201 cm = 00201m 18. Estimate the average mass density of sodium Area oftsheet (A) ='2(/Xb#6%t +61) atom assuming its size to be about 2.5 A (Use m= 3{(4:294 201.005) (1005 00201) the known values of Avogadro's number and + (00201 i 4.234)] the atomic mass of sodium). Compare it with = 2 x 43604739 = 87209478 m’ the density of sodium in its crystalline phase As, thickness has least number of significant figures 3, 970 kg/m‘. Are the two densities of the same therefore, rounding off area up to three significant figures, order of magnitude? If so, why? (NCERT] we get Area of sheet (A) = 872m? Sol. Given, radius of sodium atom, Volume of sheet (V)=/x bxt r=25A=25x10"m (-1A=10-° m] = 4.234 x 1005 x 00201 : i" =00855289 Volume of sodium atom = = tr Rounding off up to three significant figures, we get = £ x 314 x (25 x 107)* = 6542 x 107m? Volume of the sheet = 00855 m* Number of atom in one mole of sodium LONG ANSWER Type II Questions = Avpendy ¢ number (N) . . a N = 6023 x10” 17. The sun is a hot plasma (ionised matter) with its inner core at a temperature exceeding 10’ K and its outer surface at a temperature of about .. Atomic volume of sodium = Volume of one atom of sodium 6000 K. At these high temperatures, no ae pr ienber oaens substance remains in a solid or liquid phase. = 6542 X10" x 6023 x10" = 394 x 107 m’ In what range do you expect the mass density Mass of a mole of sodium = 23 g = 23 x 10°kg. of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess +. Average mass density of sodium = Volume is correct from the following data. (NCERT] ag Mass of the Sun = 2.0x 10% kg = BN = 5.84 x 107 kg/m? = 584 kg/m? d radius of the Sun = 7.0x 10% ae and radius of the Sun = 7.0x 10°m. Density of sodium in crystalline phase Sol. Given, mass of the sun (M) = 20x 10%kg = 970 kg/m? = 97 x 10* kg/m? Radius of the Sun (R) = 7.0 x10* m The two densities are of the same order of magnitude because in solid state, atoms are tightly packed. Problem Solving Strategy (Finding Dimensional Formulae) 1. First read the problem carefully and then find out whether we have given with the formulae or any law to describe in it. 2. Write the formulae of a physical quantity for which the dimensions to be known. 3. Convert the formulae of derived physical EXAMPLE |1| Dimension of Gravitational Constant Find out the dimensions of universal gravitational constant used in Newton's law of gravitation. Sol. According to Newton's law of gravitation, the force F, between two masses m, and m, separated by distance r.can my mt, begivenas F=G——2 Where, G = universal gravitational constant Fr? _ Newton x(metre)* quantity into fundamental quantities, e.g. Ge - acceleration= Length/Time m, mz (kg) 4. Write the corresponding symbols _—_ for gx (mass Xacceleration) x (metze)* fundamental quantities, e.g. mass = [M], length (mass)” = 1_(Cha lo » _ nge in velocity 2 time = [T], ete. ~Tmass ( me ) no dimensions — No units 4 or velocty distance or displacement C 2MeuTy met time T Linear momentum mass x velocity MxLT =[M' LT] kg ms" Acceleration shange in elooty L1T Ar? = [Me Tey ms? ime taken T Acceleration due to change in veloclly CAT Lor? =[Me LT) ms? gravity (9) time taken T Force mass * acceleration Mx LT? =[MUT?) (pester) Impulse force x time MLT*xT=[M' LT") Ns 2 Pressure force/area wr eine u't?) Nev? From Newton's law of gravitation. G F Universal constant of F = or G= nm o= ME Let) Nm? kg? gravitation (G) NM where F is force between masses mm, at a distance r Work force « distance MLT? xL =([M!' L? T?] J (joule) Energy (All types) work (Mm! ? T) J (joule) iength Physical quantity Relation with other quantities Dimensional formula ‘SI unit Moment of force force x distance MLT? xl =(M' LU? TT?) Nem work 2 Power = MUT* Line ery W (watt) me T force Surface tension ion MULT? Li oT?) Nov! Surface energy Energy ol free surface [M2 Tt) J force MLT? Force constant @eplacement =[M oT?) Nm" N Thrust force MUTT?) (newton) , N Tension force IML tT) (newton) force MULT liye MLT™ = (ML T? -2 Stress, Pressure crea al ] Nov change in dimension _ 0 ; Strain oagbal dimeneion =[M Lo 1] No units stress. MELT? ici ici MUN (Mtr? 2 Coetficient of elasticity srain j it J Nev Radius of gyration (K) distance L=[v? Ty m Moment of inertia (i) mass (radius of gyration)? Me =(M' L? T°) kg nf Angle (0) or Angular length () by eye Lom - displacement (6) radius(r) rot ] redian angle (8) Toa oy “1 Angular velocity (a) ime) qr awe ery rad s ; change in angular velocity UT pe ye o Angular acceleration (x) eae r ( J rads Angular momentum lw (ML?) [}=(M' 2 T) kg m? 57? Torque Ja (ML?) [T2] =(M' 2 Ty Nem Wavelength (2) length of one wave, Le., distance L=(Mo LT) m 1 1 ort 7 tenant 2=T eM eT sor Hz Frequency (v) number of vibrations/sec T l i} (hertz) Angular frequency (0) 2 nxtrequency Tei eT) redianise Velocity of light in distance travelled Lowery mst vacuum (c) Time taken Tt loci Velocity gradient ae =e ry st Rate of flow volume ET = Me 2 TY ws" time ‘ energy (E) 2p Planck's constant (h) Tequency () =[ML TT") Js mass M = Linear mass density (m) aed a ee kg mt