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Class XIName of Chapter: Units and MeasurementsWeek 1: 6th to 12th April, 2020 Day 1: Pg.No.16 to 18Step – IStudy the following topic from textbook:Chapter introductionSection 2.2Step – IIStudy the same topic in the Extramark app:Step – IIIIf you have any doubts, clear them with your subject teacher. Step – IVRevise using the following bullet points:Physical Quantities: All the quantities in terms of which laws of physics are described, and whose measurement is necessary are called physical quantities.The comparison of any physical quantity with its standard unit is called?measurement.Units: A definite amount of a physical quantity is taken as its standard unit. The standard unit should be easily reproducible, internationally accepted.Fundamental Units: Those physical quantities which are independent of each other are called fundamental quantities and their units are called fundamental units. Eg. Kilogram, metre and second are the fundamental units of the quantities mass, length and time respectively.Derived Units: Those physical quantities which are derived from fundamental quantities are called derived quantities and their units are called derived units.e.g., velocity, acceleration, force, work etc. are derived quantities and their units are respectively m/s, m/s2, kg m/s2 and kg m2/s2 Systems of Units: A system of units is the complete set of units, both fundamental and derived, for all kinds of physical quantities. The common system of units which is used in mechanics are given below:CGS System?In this system, the unit of length is centimetre, the unit of mass is gram and the unit of time is second.FPS System?In this system, the unit of length is foot, the unit of mass is pound and the unit of time is second.MKS System?In this system, the unit of length is metre, the unit of mass is kilogram and the unit of time is second.SI System?This system contains seven fundamental units and two supplementary fundamental units.No.Fundamental QuantitiesFundamental Units?/SI unitsSymbol1.Lengthmetre (m)M2.Masskilogram (kg0Kg3.Timesecond (s)S4.Temperaturekelvin (K)Kg5Electric currentampere (A)A6Luminous intensitycandela (Cd)Cd7Amount of substancemole (mol)MolSupplementary Fundamental UnitsS.No.Supplementary Fundamental QuantitiesSupplementary UnitSymbol1Plane angleradian (rad)Rad2Solid anglesteradian (Sr)SrRelationship between Some?SI Unit and Commonly Used UnitsS.No.Physical QuantityUnit1Length(a)1 micrometre = 10-6?m(b)1 angstrom =10-10?m2Mass(a)1 metric ton = 103?kg(b)1 pound = 0.4537 kg(c)1 amu = 1.66 x10-23?kg3Volume1 litre = 10-32?m34.Force(a)1 dyne = 10-5?N(b)1 kgf = 9.81 N5.Pressure(a)1 kgfm2?= 9.81Nm-2(b)1 mm of Hg = 133 Nm-2(c)1 pascal = 1 Nm-2(d)1 atmosphere pressure = 76 cm of Hg = 1.01 x 105?pascal6.Work and energy(a)1 erg =10-7?J(b)1 kgf-m = 9.81 J(c)1 kWh = 3.6 x 106?J(d)1 eV = 1.6 x 10-19?J7.Power(d)1 kgf- ms-1?= 9.81W1 horse power = 746 WSome Practical Units 1 fermi =10-15?m 1 X-ray unit = 10-13?m 1 astronomical unit = 1.49 x 1011?m (average distance between sun and earth) 1 light year = 9.46 x 1015?m 1 parsec = 3.08 x 1016?m = 3.26 light yearStep – VSolve the following questions in your C/W copy: Which of the following sets cannot enter into the list of fundamental quantities in any system of units?length, time and velocity b) length, mass and velocityc) mass, time and velocity d) length, mass and timewhich of the following systems of units is not based on units of mass, length and time alone?MKS b) SI c) FPS d) CGSWhy length, mass and time is chosen as base quantities in mechanics?Which of the following derived units represent the same physical quantity?Kg m/s, kg m2 /s2, g cm-1s-2, g cm2s-2End of Day – 1Day 2: Pg: 31-32Step – IStudy the following topic from textbook:Sections 2.8 and 2.9Appendix A9, pg no. 214Step – IIStudy the same topic in the Extramark app:Step – IIIIf you have any doubts, clear them with your subject teacher. Step – IVRevise using the following bullet points:Dimensions: The powers to which the fundamental quantities are raised to represent any physical quantity are called dimensions of that quantity.The expression which shows how and which of the fundamental quantities represent the dimensions of a physical quantity, is called the dimensional formula.A dimensional formula is written using [ ].Eg i) area = length x breadth[A] = [L] x [L] = [L2]Area is 2 dimensions in length. ii) speed = distance / time[v] = [L] / [T] = [ LT-1] Speed is 1 dimension in length and -1 dimension in time.Dimensional Constants:? Constants which possess dimensions are called dimensional constants. E.g. Universal Gravitational constant.Dimensional variables: Those physical quantities which possess dimensions but do not have a fixed value are called dimensional variables. E.g. Displacement, Force, velocity etc.Dimensionless quantities: Physical quantities which do not possess dimensions are called dimensionless quantities. E.g. Angle, specific gravity, strain. In general, physical quantity which is a ratio of?two quantities of same dimension will be dimensionless.Dimensional Formula of Some Physical QuantitiesS.No.PhysicalQuantityDimensional FormulaMKSUnit?1Area[L2]metre2?2Volume[L3]metre3?3Velocity[LT-1]ms-1?4Acceleration[LT-2]ms-2?5Force[MLT-2]newton (N)?6Work or energy[ML2T-2]joule (J)?7Power[ML2T-3]J s-1?or watt?8Pressure or stress[ML-1T-2]Nm-2?9Linear momentum or Impulse[MLT-1]kg ms-1?10Density[ML-3]kg m-3Step – VSolve the following questions in your C/W copy:Write the dimensional formula of the following physical quantities in a tabular form-Area, volume, speed, velocity, acceleration, momentum, force, pressure, density, relative density, work, potential energy, kinetic energy, power, refractive index, angle, linear magnification, universal gravitational constant, wavelength, frequency, power of a lens.End of Day – 2Day 3: Pg.No.32 – 33Step – IStudy the following topic from textbook:Section 2.10;, 2.10.1 Step – IIStudy the same topic in the Extramark app:Step – IIIIf you have any doubts, clear them with your subject teacher. Step – IVRevise using the following bullet points:Homogeneity Principle: If the dimensions of left hand side of an equation are equal to the dimensions of right hand side of the equation, then the equation is dimensionally correct. This is known as?homogeneity principle. Mathematically [LHS] = [RHS] This principle indicates that only like quantities can be added or subtracted. Eg, length can be added to length, mass can be subtracted from mass, etc.Applications of Dimensions: i) To check the accuracy or dimensional consistency of physical equations. ii) To deduce a relation between different physical quantities. iii) To change a physical quantity from one system of units to another system of units Checking the dimensional consistency of equations:Dimensions are used for checking the correctness of any physical equation. According to the principle of homogeneity, if the dimensions of all the terms on both the sides of an equation are same, then the equation is dimensionally consistent. However, dimensional consistency does not guarantee correct equations. The arguments of trigonometric, logarithmic and exponential functions, a pure number, ratio of similar quantities has no dimensions. Eg, sin x, log x, 2, ? , π, θ, refractive index, etc are dimensionless. Let us check the correctness of the following equations-v = u + at[L.H.S] = [v] = [LT-1][R.H.S] = [ u + at] = [LT-1] + [LT-2] [T] = [LT-1] + [LT-1] = [LT-1]* velocity added to velocity results in velocity.*[L.H.S] = [R.H.S] Hence the above equation is dimensionally correct.S = ut + ? at2[L.H.S] = [s] [R.H.S] = [ ut + ? at2] = [LT-1] [T] + [LT-2] [T2] = [L] + [L] = [L][L.H.S] = [R.H.S] Hence the above equation is dimensionally correct.S = ut + at2[L.H.S] = [s] [R.H.S] = [ ut + at2] = [LT-1] [T] + [LT-2] [T2] = [L] + [L] = [L][L.H.S] = [R.H.S] Hence the above equation is dimensionally correct.* This equation is numerically incorrect but dimensionally correct. Hence dimensional consistency does not guarantee correct equations. However, if an equation is dimensionally incorrect, that can never be a correct equation.*Step – VDo the following in your C/W copy:Using dimensions, check the correctness of the following equations-v2 = u2 + 2as ii) Ek = ? mv2 iii) T = 2π √ lg iv) Ek = ? mv2 + ma v) Ek = 3/16 mv2( Symbols have their usual meanings)End of Day – 3Day – 4: Pg. No. 33 - 34Step – IStudy the following topic from textbook:Section 2.10.2.Step – IIStudy the same topic in the Extramark app:Step – IIIIf you have any doubts, clear them with your subject teacher. Step – IVRevise using the following bullet points:The method of dimensions can be used to derive a relation among the physical quantities, if the dependence of a quantity on the other quantities are known.Let us consider the following examples-The centripetal force?F?acting on a particle moving uniformly in a circle may depend upon mass?(m), velocity?(v)?and radius?r?of the circle. Derive the formula for?F?using the method of dimensions. Solution: Given, F ∞ ma vb rc, Where, a, b and c are the powers to which the given quantities may be raised.F = k ma vb rc, --------- (1) where k is a dimensionless constant. Taking dimensions of the quantities on both sides, we get [F] = [M]a [LT-1]b [L]c [MLT-2] = [ Ma Lb+c T-b] Comparing the dimensions (powers) on both sides, we get a = 1, b+c = 1, -b = -2 solving these equations, we get a = 1, b = 2, c = 1 – b = 1 – 2 = -1 substituting these values in equation (1), we get F = k m1 v2 r-1 = k mv2/r, which is the required equation for centripetal force.Step – VSolve the above example in your C/W copy.End of Day – 4Day – 5: pg. No. 34; NumericalsStep – IStudy the following topic from textbook:Section 2.10.2Step – IIStudy the same topic in the Extramark app:Step – IIIIf you have any doubts, clear them with your subject teacher. Step – IVRevise using the following bullet points:Limitations of dimensional analysis:Dimensionally correct equation is sometimes incorrect because it doesn’t take into account dimensionless constants like numbers.For e.g :?v2+2as?and?v2+5as?have same dimensions but they are physically incorrect.It does not test whether a physical quantity is a scalar or a vector.It cannot derive relation or formula if a physical quantity depends upon more than three factors having dimensions.It cannot derive a formula containing trigonometric function, exponential function, and logarithmic function.It cannot derive a relation having more than one part in an equation.Step – VSolve the following in your C/W copy:The time period of a simple pendulum depends on the mass (m) of the bob, length (l) of the pendulum and acceleration due to gravity (g). Deduce a relation for the time period of the simple pendulum.Derive by the method of dimensions, an expression for the energy of a body executing simple harmonic motion, assuming that the energy of the body depends upon the mass m, the frequency f and the amplitude of vibration a. A gas bubble from an explosion under water oscillates with a period T proportional to?pa db Ec, where p is the static pressure, d is the density of water, and E is the total energy of the explosion. Find the value of a, b, c.The escape velocity 'v' of a body depends upon (1) the acceleration due to gravity (g) of the planet and (2) the radius (R) of the planet . Use the method of dimensions to obtain a relation between v,g and R.The velocity of water waves (v) may depend on their wavelength?′λ?′, density of water 'd'?and acceleration due to gravity 'g'. Find the relation between these quantities by the method of dimensions.End of Day – 5 ................
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