{"id":27031,"date":"2022-09-21T10:29:05","date_gmt":"2022-09-21T04:59:05","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=27031"},"modified":"2023-03-30T14:38:15","modified_gmt":"2023-03-30T09:08:15","slug":"ap-inter-1st-year-physics-study-material-chapter-2","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ap-inter-1st-year-physics-study-material-chapter-2\/","title":{"rendered":"AP Inter 1st Year Physics Study Material Chapter 2 Units and Measurements"},"content":{"rendered":"

Andhra Pradesh BIEAP\u00a0AP Inter 1st Year Physics Study Material<\/a> 2nd Lesson Units and Measurements Textbook Questions and Answers.<\/p>\n

AP Inter 1st Year Physics Study Material 2nd Lesson Units and Measurements<\/h2>\n

Very Short Answer Questions<\/span><\/p>\n

Question 1.
\nDistinguish between accuracy and precision. [A.P. Mar. 16, 15; T.S. Mar. 15, 13]
\nAnswer:
\nAccuracy<\/p>\n

    \n
  1. The accuracy of a measurement of any physical quantity made by any measuring instrument is a measure of how close the measured value is to the true value of the quantity.<\/li>\n
  2. The accuracy depends on errors.<\/li>\n<\/ol>\n

    Precision<\/p>\n

      \n
    1. The precision of the measuring instrument denotes upto what limit (or) resolution the quantity can be measured with the given instrument.<\/li>\n
    2. The precision does not depend on errors.<\/li>\n<\/ol>\n

      Question 2.
      \nWhat are the different types of errors that can occur in a measurement ?
      \nAnswer:
      \nMainly there are three types of errors.<\/p>\n

        \n
      1. Systematic errors<\/li>\n
      2. Random errors<\/li>\n
      3. Gross errors.<\/li>\n<\/ol>\n

        \"AP<\/p>\n

        Question 3.
        \nHow can systematic errors be minimised or eliminated ? [Mar. 14]
        \nAnswer:
        \nSystematic errors can be minimised by improving experimental techniques, selecting better instruments and removing personal bias as far as possible. For a given set up, these errors may be estimated to a certain extent and the necessary corrections may be applied to the readings.<\/p>\n

        Question 4.
        \nIllustrate how the result of a measurement is to be reported indicating the error involved. [T.S., A.P. – Mar. 17]
        \nAnswer:
        \nTo measure any physical quantity, we compare it with a standard (unit) of that quantity. No measurement is perfect as the errors involved in the process cannot be removed completely. Hence inspite of our best efforts, the measured value is always some what different from its actual value (or) true value.<\/p>\n

        Question 5.
        \nWhat are significant figures and what do they represent when reporting the result of a measurement ? [Mar. 13]
        \nAnswer:
        \nThe digits of a number that are definitely known plus one more digit that is estimated are called significant digits (or) significant figures.
        \nExample : Time period of a simple pendulum is 1.62, the digits 1 and 6 are reliable while the digit 2 is uncertain. The measured value has three significant figures.<\/p>\n

        Question 6.
        \nDistinguish between fundamental units and derived units. [T.S. – Mar. 16]
        \nAnswer:<\/p>\n

          \n
        1. Units of fundamental quantities are called fundamental units. Fundamental units can neither derived from one another, nor can they be resolved into other units.<\/li>\n
        2. Units of derived quantities are called derived units.<\/li>\n<\/ol>\n

          \"AP<\/p>\n

          Question 7.
          \nWhy do we have different units for the same physical quantity ?
          \nAnswer:
          \nWe have different systems like C.G.S system, M.K.S system, F.P.S system and S.l system. Hence we have different physical units for the same physical quantity.<\/p>\n

          Question 8.
          \nWhat is dimensional analysis ?
          \nAnswer:
          \nDimensional analysis is the representation of derived physical quantities in terms of units of fundamental quantities.<\/p>\n

          With the help of dimensional analysis to check the correctness of the equation, convert one system of units into other system and derive certain equations relating physical quantities.<\/p>\n

          Question 9.
          \nHow many orders of magnitude greater is the radius of the atom as compared to that of the nucleus ?
          \nAnswer:
          \nSize of atomic nucleus = 10-14<\/sup> m
          \nSize of atom = 10-10<\/sup> m
          \nHence size of atom is 1CT4 m greater the size of the nucleus.<\/p>\n

          Question 10.
          \nExpress unified atomic mass unit in kg.
          \nAnswer:
          \n1 unified atomic mass unit = \\(\\frac{1}{12}\\) of the mass of carbon – 12 atom.
          \n1 a.m.u = 1.66 \u00d7 10-27<\/sup> kg<\/p>\n

          Short Answer Questions<\/span><\/p>\n

          Question 1.
          \nThe vernier scale of an instrument has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm, then using this instrument what would be the minimum inaccuracy in the measurement of distance ?
          \nAnswer:
          \nValue of each main scale division = 0.5 mm
          \nNo. of vernier scale divisions = 50
          \nIn vernier callipers, L.C = \\(\\frac{S}{N}=\\frac{\\text { Value of one main scale division }}{\\text { No. of vernier scale divisions }}\\)
          \nL.C. = \\(\\frac{0.5}{50}\\) = 0.01 m.m
          \nThe minimum inaccuracy in the measurement of distance is 0.01 m.m. (Or)
          \nValue of 1 MSD = 0.5 m.m
          \nNV.S.D = (N – 1) M.S.D
          \n50 V.S.D – 49 M.S.D
          \n1 V.S.D = \\(\\) M.S.D. = \\(\\frac{49}{50}\\) \u00d7 0.5
          \nL.C = 1 M.S.D – 1 V.S.D
          \n(0.5 – \\(\\frac{49}{50}\\) \u00d7 0.5) = (1 – \\(\\frac{49}{50}\\))0.5 = \\(\\frac{1}{50}\\) \u00d7 0.5
          \nL.C. = 0.01 m.
          \n\u2234 Minimum inaccuracy in the measurement of distance is 0.01 m.m.<\/p>\n

          \"AP<\/p>\n

          Question 2.
          \nIn a system of units, the unit of force is 100 IM. unit of length is 10 m and the unit of time is 100s. What is the unit of mass in this system ?
          \nAnswer:
          \nForce (F) = 100 N; Length (L) = 10 m; Time (T) = 100 s F = ma
          \nmass (m) = \\(\\frac{F}{a}=\\frac{F}{L T^{-2}}\\); m = \\(\\frac{\\mathrm{FT}^2}{\\mathrm{~L}}=\\frac{100 \\times(100)^2}{10}\\) = 105<\/sup> kg<\/p>\n

          Question 3.
          \nThe distance of a galaxy from Earth is of the order of 1025<\/sup> m. Calculate the order of magnitude of the time taken by light to reach us from the galaxy.
          \nAnswer:
          \nDistance of galaxy from earth = 1025<\/sup> m
          \nVelocity of light (C) = 3 \u00d7 108<\/sup> m\/s
          \nTime taken by the light (t) = \\(\\frac{\\text { Distance of galaxy from earth }}{\\text { velocity of light }}\\)
          \nt = \\(\\frac{\\mathrm{d}}{\\mathrm{c}}\\) \u21d2 t = \\(\\frac{10^{25}}{3 \\times 10^8}\\) = 0.33 \u00d7 1017<\/sup> s \u21d2 t = 3.3 \u00d7 1016<\/sup> sec.<\/p>\n

          Question 4.
          \nThe Earth-Moon distance is about 60 Earth radius. What will be the approximate diameter of the Earth as seen from the Moon?
          \nAnswer:
          \nDistance between Earth-Moon = 60 R
          \nRadius of the earth = R
          \nr = 60R = 60 \u00d7 6400 \u00d7 103<\/sup> (R = 6400km)
          \n\u03b8 = 1 sec = \\(\\frac{1}{60}\\) Min = \\(\\frac{1}{60 \\times 60}\\) degree \u21d2 \u03b8 = \\(\\frac{1}{60 \\times 60} \\times \\frac{\\pi}{180}\\) radian
          \nThen r = \\(\\frac{l}{\\theta}\\) \u21d2 l = r\u03b8 \u21d2 l = 60 \u00d7 6400 \u00d7 103<\/sup> \u00d7 \\(\\frac{1}{60 \\times 60} \\times \\frac{\\pi}{180}\\)
          \nl = 11.16 \u00d7 103<\/sup> km \u21d2 Diamter (l) = 11.16 \u00d7 103<\/sup> km<\/p>\n

          \"AP<\/p>\n

          Question 5.
          \nThree measurements of the time for 20 oscillations of a pendulum give t1<\/sub> = 39.6 s, t2<\/sub> = 39.9 s and t3<\/sub> = 39.5 s. What is the precision in the measurements? What is the accuracy of the measurements?
          \nAnswer:
          \nNo. of oscillations = 20
          \nt1<\/sub> = 39.6 sec, t2<\/sub> = 39.9 sec, t3<\/sub> = 39.5 sec
          \nMean value = \\(\\frac{t_1+t_2+t_3}{3}=\\frac{39.6+39.9+39.5}{3}=\\frac{119}{3}\\) = 39.66
          \nMean value = 39.7 sec;
          \nPrecision = 0.1 sec.
          \nAccuracy is the closeness of measured value with true value.
          \nHence 39.6 s is accuracy.<\/p>\n

          Question 6.
          \n1 calorie = 4.2J where 1J = 1 kg m2<\/sup>s2<\/sup>. Suppose we employ a system of units in which the unit of mass is \\(\\hat{a}\\) kg, the unit of length is \\(\\hat{a}\\) m and the unit of time is \\(\\tilde{a}\\) s, show that a calorie has a magnitude 4.2 \\(\\hat{\\mathrm{a}}^{-1} \\hat{a}^{-2} \\tilde{\\mathrm{a}}^{-2}\\) in the new system.
          \nAnswer:
          \n1 Calorie = 4. 2J \u21d2 1 J = 1 kg m2<\/sup> s-2<\/sup>
          \n1 calorie = 4.2 kg m2<\/sup> s-2<\/sup>
          \nIn new system, 1 calorie = 4.2 \\(\\hat{a} \\hat{a}^2 \\hat{a}^{-2}\\)<\/p>\n

          Question 7.
          \nA new unit of length is chosen so that the speed of light in vacuum is 1 ms-2. If light takes 8 min and 20 $ to cover this distance, what is the distance between the Sun and Earth in terms of the new unit ?
          \nAnswer:
          \nV = Speed of light in vacuum = 1 m\/s
          \nTime taken (t) = 8 min 20 sec = 500 sec
          \nDistance between the sun and earth (d) = \\(\\frac{\\mathrm{V}}{\\mathrm{t}}\\) \u21d2 d = \\(\\frac{1}{500}\\) = 0.002 m.<\/p>\n

          Question 8.
          \nA student measures the thickness of a human hair using a microscope of magnification 100. He makes 20 observations and finds that the average thickness (as viewed in the microscope) is 3.5 mm. What is the estimate of the thickness of hair ?
          \nAnswer:
          \nMagnification of microscope = M = 100
          \nObserved thickness = 3.5 m.m
          \nMagnification (M) = \\(\\frac{\\text { Observed thickness }}{\\text { Real thickness }}\\) \u21d2 100 = \\(\\frac{3.5}{\\text { Real thickness }(t)}\\) \u21d2 t = \\(\\frac{3.5}{100}\\) = 0.035 m.m.<\/p>\n

          \"AP<\/p>\n

          Question 9.
          \nA physical quantity X is related to four measureable quantities a, b, c and d as follows. X = a2<\/sup> b3<\/sup> c5\/2<\/sup>d-2<\/sup>
          \nThe percentage error in the measurement of a, b, c and d are 1%, 2%, 3% and 4% respectively. What is the percentage error in X ?
          \nAnswer:
          \nX = a2<\/sup> b3<\/sup> c5\/2<\/sup>d-2<\/sup>
          \n\\(\\frac{\\Delta \\mathrm{a}}{\\mathrm{a}}\\) \u00d7 100 = 1%, \\(\\frac{\\Delta \\mathrm{b}}{\\mathrm{b}}\\) \u00d7 100 = 2%, \\(\\frac{\\Delta \\mathrm{c}}{\\mathrm{c}}\\) \u00d7 100 = 3%, \\(\\frac{\\Delta \\mathrm{d}}{\\mathrm{d}}\\) \u00d7 100 = 4%
          \nPercentage error in X is
          \n(\\(\\frac{\\Delta \\mathrm{X}}{\\mathrm{X}}\\)) \u00d7 100 = 2(\\(\\frac{\\Delta \\mathrm{a}}{\\mathrm{a}}\\) \u00d7 100) + 3(\\(\\frac{\\Delta \\mathrm{b}}{\\mathrm{b}}\\) \u00d7 100) + \\(\\frac{5}{2}\\) (\\(\\frac{\\Delta \\mathrm{b}}{\\mathrm{b}}\\) \u00d7 100) + 2 \\(\\frac{\\Delta \\mathrm{d}}{\\mathrm{d}}\\) \u00d7 100
          \n= 2 \u00d7 1 + 3 \u00d7 2 + \\(\\frac{5}{2}\\) \u00d7 3 + 2 \u00d7 4 = 2 + 6 + \\(\\frac{15}{2}\\) + 8
          \n\u21d2 (\\(\\frac{\\Delta \\mathrm{X}}{\\mathrm{X}}\\) 100 = 23.5 %<\/p>\n

          Question 10.
          \nThe velocity of a body is given by v = At2<\/sup> + Bt + C. If v and t are expressed in SI what are the units of A, B and C ?
          \nAnswer:
          \nGiven V = At2<\/sup> + Bt + C
          \nAccording to principle of homogeneity,<\/p>\n

            \n
          1. V = At2<\/sup> \u21d2 A =\\(\\frac{\\mathrm{V}}{\\mathrm{t}^2}=\\frac{\\mathrm{LT}^{-1}}{\\mathrm{~T}^2}\\) = [LT-3<\/sup>] \u21d2 A = ms-3<\/sup><\/li>\n
          2. V = Bt \u21d2 B = \\(\\frac{\\mathrm{V}}{\\mathrm{t}}=\\frac{\\mathrm{LT}^{-1}}{\\mathrm{~T}}\\) = [LT-1<\/sup>] \u21d2 B = ms-2<\/sup><\/li>\n
          3. V = C \u21d2 C = LT-1<\/sup> \u21d2 C = ms-1<\/sup><\/li>\n<\/ol>\n

            Problems<\/span><\/p>\n

            Question 1.
            \nIn the expression P = E l2<\/sup> m-5<\/sup> G-2<\/sup> the quantities E, l, m and G denote energy, angular momentum, mass and gravitational constant respectively. Show that P is a dimension-less quantity.
            \nSolution:
            \nP = E L2<\/sup> m-5<\/sup> G-2<\/sup>
            \nEnergy (E) = [ML2<\/sup> T-2<\/sup>]
            \nAngular momentum (L) = ML2<\/sup> T-1<\/sup>
            \nMass = [M]
            \nUniversal gravitational constant
            \n(G) = [M-1<\/sup>L3<\/sup>T-2<\/sup>]
            \nP = [ML2<\/sup> T-2<\/sup>] [ML2<\/sup> T-1<\/sup>]2<\/sup> [M]-5<\/sup> [M-1<\/sup> L3<\/sup> T-2<\/sup>]-2<\/sup>
            \nP = M1+2-5+2<\/sup> L2+4-6<\/sup> T-2-2+4<\/sup>
            \nP = [M0<\/sup> L0<\/sup> T0<\/sup>]
            \nHence P is dimensional less quantity.<\/p>\n

            \"AP<\/p>\n

            Question 2.
            \nIf the velocity of light c. Planck’s constant h and the gravitational ‘ constant G are taken as fundamental quantities : then express mass, length and time in terms of dimensions of these quantities.
            \nSolution:
            \ni) M \u221d Gx<\/sup> Cy<\/sup> hz<\/sup>
            \n[M1<\/sup> L0<\/sup> T0<\/sup>] = [M-1<\/sup> L3<\/sup> T-2<\/sup>]x<\/sup> [LT-1<\/sup>]y<\/sup> [ML2<\/sup> T-1<\/sup>]z<\/sup>
            \n[M1<\/sup>L0<\/sup>T0<\/sup>] = M-x+z<\/sup> L3x+y+2z<\/sup> T-2x-y-z<\/sup> = 0
            \n– x + z = 1, 3x + y + 2z = 0, – 2x – y – z = 0
            \nSolving these equations, we get
            \nx = \\(\\frac{-1}{2}\\), y = \\(\\frac{1}{2}\\), z = \\(\\frac{1}{2}\\)
            \nM = G\\(\\frac{-1}{2}\\)<\/sup> C\\(\\frac{1}{2}\\)<\/sup> h\\(\\frac{1}{2}\\)<\/sup>
            \nM = \\(\\sqrt{\\frac{\\mathrm{hc}}{\\mathrm{G}}}\\)<\/p>\n

            ii) Length (l) \u221d Gx<\/sup> Cy<\/sup> hz<\/sup>
            \n[M0<\/sup> L1<\/sup> T0<\/sup>] = [M-1<\/sup> L3<\/sup> T-2<\/sup>]x<\/sup> [LT-1<\/sup>]y<\/sup> [ML2<\/sup> T-1<\/sup>]z<\/sup>
            \n[M0<\/sup>L1<\/sup>T0<\/sup>] = M-x+z<\/sup> L3x+y+2z<\/sup> T-2x-y-z<\/sup>
            \nApplying principle of homogeneity
            \n-x + z = 0, 3x + y + 2z = 0, -2x – y – z = 0
            \nSolving these equations, we get
            \nx = \\(\\frac{1}{2}\\), y = \\(\\frac{-3}{2}\\), z = \\(\\frac{1}{2}\\)
            \nFrom equation (1), l = G\\(\\frac{1}{2}\\)<\/sup> C\\(\\frac{-3}{2}\\)<\/sup> h\\(\\frac{1}{2}\\)<\/sup>
            \nLength (l) = \\(\\sqrt{\\frac{\\mathrm{Gh}}{\\mathrm{C}^3}}\\)<\/p>\n

            iii) Time (T) \u221d Gx<\/sup> Cy<\/sup> hz<\/sup>
            \n[M0<\/sup> L0<\/sup> T1<\/sup>] = [M-1<\/sup> L3<\/sup> T-2<\/sup>]x<\/sup> [LT-1<\/sup>]y<\/sup> [ML2<\/sup> T-1<\/sup>]z<\/sup>
            \n[M0<\/sup>L0<\/sup>T1<\/sup>] = M-x+z<\/sup> L3x+y+2z<\/sup> T-2x-y-z<\/sup>
            \nApplying principle of homogeneity
            \n-x + z = 0, 3x + y + 2z = 0, -2x – y – z = 1
            \nSolving these equations, we get
            \nx = \\(\\frac{1}{2}\\), y = \\(\\frac{-5}{2}\\), z = \\(\\frac{1}{2}\\)
            \nFrom equation (1), l = G\\(\\frac{1}{2}\\)<\/sup> C\\(\\frac{-5}{2}\\)<\/sup> h\\(\\frac{1}{2}\\)<\/sup>
            \nLength (l) = \\(\\sqrt{\\frac{\\mathrm{Gh}}{\\mathrm{C}^5}}\\)<\/p>\n

            Question 3.
            \nAn artificial satellite is revolving around a planet of mass M and radius R in a circular orbit of radius r. Using dimensional analysis show that the period of the satellite
            \nT = \\(\\frac{k}{R} \\sqrt{\\frac{r^3}{g}}\\)
            \nwhere k is a dimensionless constant and g is acceleration due to gravity.
            \nSolution:
            \nT = \\(\\frac{k}{R} \\sqrt{\\frac{r^3}{g}}\\)
            \nL.H.S. : Time period = [T]
            \nR.H.S. : \\(\\frac{k}{R} \\sqrt{\\frac{r^3}{g}}=\\frac{1}{L} \\sqrt{\\frac{L^3}{L T^{-2}}}\\) = [T]
            \n\u2234 L.H.S. = R.H.S.
            \nAbove equation is correct.<\/p>\n

            \"AP<\/p>\n

            Question 4.
            \nState the number of significant figures in the following.
            \na) 6729
            \nb) 0.024
            \nc) 0.08240
            \nd). 6.032
            \ne) 4.57x
            \nSolution:
            \na) 6729 – 4\u00a0 \u00a0 \u00a0Significant figures
            \nb) 0024 – 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ”
            \nc) 008240 – 4\u00a0 \u00a0 \u00a0 \u00a0 \u00a0”
            \nd) 6.032 –\u00a0 4\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0”
            \ne) 4.57 \u00d7 108<\/sup> – 3\u00a0 \u00a0 \u00a0”<\/p>\n

            Question 5.
            \nA stick has a length of 12.132 cm and another has a length of 12.4 cm. If the two sticks are placed end to end what is the total length ? If the two sticks are placed side by side, what is the difference in their lengths?
            \nSolution:
            \na) Let lengths of the rods are
            \nl1<\/sub> = 12.132 cm, l2<\/sub> = 12.4 cm
            \nHere has one decimal place and l1<\/sub> has to be rounded to have only two decimal places.
            \nl = l1<\/sub> + l2<\/sub> = 12.13 + 12.4 = 24.53
            \nThis is to be rounded off to have one decimal place only.
            \n\u2234 The result is 24.5 cm<\/p>\n

            b) l1<\/sub> = 12.132 cm, l2<\/sub> = 12.4 cm
            \nHere l1<\/sub> is rounded only two decimal places,
            \nl2<\/sub> – l1<\/sub> = 12.4 – 12.13 = 0.27
            \nThis should be rounded off to have only one decimal place, l2<\/sub> – l1<\/sub> = 0.3<\/p>\n

            Question 6.
            \nEach side of a cube is measured to be 7.203 m. What is
            \n(i) the total surface area and
            \n(ii) the volume of the cube, to appropriate significant figures ?
            \nSolution:
            \nLength of a side = 7.203 m
            \n(i) Total surface area = 6a2<\/sup>
            \n= 6 \u00d7 (7.203)2<\/sup>
            \n= 311.29
            \nThis should be rounded off to four significant figure as 7.203 as 4 significant figures.
            \n\u2234 The result = 311.3m2<\/sup><\/p>\n

            (ii) Volume of the cube = a3<\/sup>
            \n= (7.203)3<\/sup>
            \n= 373.71
            \nThis should be rounded off to four significant figures has 7.203 has four significant figures.
            \n\u2234 Volume = 373.7 m3<\/sup><\/p>\n

            \"AP<\/p>\n

            Question 7.
            \nThe measured mass and volume of a body are 2.42 g and 4.7 cm3<\/sup> respectively with possible errors 0.01 g and 0.1 cm3<\/sup>. Find the maximum error in density.
            \nSolution:
            \nM = 2.42 g, V = 4.7 cm3<\/sup>
            \n\u2206M = 0.01 g, \u2206V = 0.1 cm3<\/sup>
            \nDensity (\u03c1) = \\(\\frac{M}{V}\\)
            \n\\(\\frac{\\Delta \\rho}{\\rho}=\\frac{\\Delta \\mathrm{M}}{\\mathrm{M}}+\\frac{\\Delta \\mathrm{V}}{\\mathrm{V}}=\\frac{0.01}{2.42}+\\frac{0.1}{4.7}\\)
            \n= 0.004 + 0.02 = 0.024
            \n% error is \\(\\frac{\\Delta \\rho}{\\rho}\\) \u00d7 100 = 0.024 \u00d7 100
            \n= 2.4% \u2248 2%<\/p>\n

            Question 8.
            \nThe error in measurement of radius of a sphere is 1%. What is the error in the measurement of volume ?
            \nSolution:
            \nRadius of the sphere \\(\\frac{\\Delta \\mathrm{r}}{\\mathrm{r}}\\) \u00d7 100 = 1%
            \nVolume (V) = \\(\\frac{4}{3}\\) \u03c0r3<\/sup>
            \n\\(\\frac{\\Delta \\mathrm{V}}{\\mathrm{V}}\\) \u00d7 100 = 3 \u00d7 \\(\\frac{\\Delta \\mathrm{r}}{\\mathrm{r}}\\) \u00d7 100 = 3 \u00d7 1% = 3%<\/p>\n

            Question 9.
            \nThe percentage error in the mass and speed are 2% and 3% respectively. What is the maximum error in kinetic energy calculated using these quantities ?
            \nSolution:
            \n\\(\\frac{\\Delta \\mathrm{M}}{\\mathrm{M}}\\) \u00d7 100 = 2%, \\(\\frac{\\Delta \\mathrm{V}}{\\mathrm{V}}\\) \u00d7 100 = 3%;
            \nKinetic energy = \\(\\frac{1}{2}\\) mV2<\/sup>
            \n\\(\\frac{\\Delta \\mathrm{K}}{\\mathrm{K}}\\) \u00d7 100 = \\(\\frac{\\Delta \\mathrm{M}}{\\mathrm{M}}\\) \u00d7 100 + 2 \\(\\frac{\\Delta \\mathrm{V}}{\\mathrm{V}}\\) \u00d7 100
            \n= 2 + 2(3) = 2 + 6 = 8%<\/p>\n

            \"AP<\/p>\n

            Question 10.
            \nOne mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). If the size of the hydrogen molecule is about 1A\u00b0, what is the ratio of molar volume to the atomic volume of a mole of hydrogen ?
            \nSolution:
            \nMolar volume = 22.4 lit = 22.4 \u00d7 1000 c.c.
            \n= 22.4 \u00d7 10-3<\/sup> m3<\/sup>
            \nDiameter of the hydrogen molecule
            \n= 1 A\u00b0 = 10-10<\/sup> m
            \nRadius (r) = \\(\\frac{D}{2}=\\frac{10^{-10}}{2}\\) = 0.5 \u00d7 10-10<\/sup> m
            \nVolume (V) = \\(\\frac{4}{3}\\) \u03c0r3<\/sup>
            \n= \\(\\frac{4}{3}\\) \u00d7 3.14 \u00d7 (0.5 \u00d7 10-10<\/sup>)3<\/sup>
            \nv = \\(\\frac{1.57}{3}\\) \u00d7 10-30<\/sup> = 0.5233 \u00d7 10-30<\/sup> m3<\/sup>
            \nAtomic volume = V \u00d7 Avagadro’s number (N)
            \n= 0.5233 \u00d7 10-30<\/sup> \u00d7 6.023 \u00d7 1023<\/sup> = 3.151 \u00d7 10-7<\/sup>
            \n\u2234 \\(\\frac{\\text { Molar volume }}{\\text { Atomic volume }}=\\frac{22.4 \\times 10^{-3}}{3.151 \\times 10^{-7}}\\)
            \n= 7.108 \u00d7 10-4<\/sup><\/p>\n

            Additional Problems<\/span><\/p>\n

            Note : In stating numerical answers, take care of significant figures.<\/p>\n

            Question 1.
            \nFill in the blanks
            \na) The volume of a cube of side 1 cm is equal to …………….. m3<\/sup>
            \nb) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ……………. (mm)2<\/sup>
            \nc) A vehicle moving with a speed of 18 km h-1<\/sup> covers ……………. m in 1 s
            \nd) The relative density of lead is 11.3. Its density is …………….. g cm-3<\/sup> or …………….. kg m-3<\/sup>.
            \nSolution:
            \na) Here, length of slide L = 1 cm = 10-2<\/sup> m
            \nVolume of cube = L3<\/sup> = (10-2<\/sup> m)3<\/sup> = 10-6<\/sup> m3<\/sup><\/p>\n

            b) Here, r = 2.0 cm = 20 mm
            \nh = 10.0 cm = 100 mm
            \nSurface area of solid cylinder = (2rcr) \u00d7 h
            \n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 20 \u00d7 100 mm2<\/sup>
            \n= 1.26 \u00d7 104<\/sup> mm2<\/sup><\/p>\n

            c) Here, speed V = 18 km
            \nh-1<\/sup> = \\(\\frac{18 \\times 1000 \\mathrm{~m}}{60 \\times 60 \\mathrm{~s}}\\) = 5 ms-1<\/sup>
            \n\u2234 Distance covered in 1 second = 5 m<\/p>\n

            d) Here, relative density = 11.3
            \ndensity = 11.3 g k.c = \\(\\frac{11.3 \\times 10^{-3}}{\\left(10^{-2} \\mathrm{~m}\\right)^3}\\)
            \n= 11.3 \u00d7 103<\/sup> kg m-3<\/sup><\/p>\n

            \"AP<\/p>\n

            Question 2.
            \nFill in the blanks by suitable conversion of units.
            \na) 1 kg m2<\/sup> s-2<\/sup> = ………….. g cm2<\/sup> s-2<\/sup>
            \nb) 1 m = ………….. 1y
            \nc) 3.0 ms-2<\/sup> = ………….. km h-2<\/sup>
            \nd) G = 6.67 \u00d7 10-11<\/sup> N m2<\/sup> (kg)-2<\/sup> = …………. (cm)3<\/sup> s-2<\/sup> g-1<\/sup>.
            \nSolution:
            \na) 1 kg m2<\/sup> s2<\/sup>
            \n= 1 \u00d7 103<\/sup> g (102<\/sup> cm)2<\/sup>s-2<\/sup> = 107<\/sup> g cm2<\/sup> s-2<\/sup><\/p>\n

            b) We know, 1 light year = 9.46 \u00d7 1015<\/sup> m
            \n\u2234 1 m = \\(\\frac{1}{9.46 \\times 10^{15}}\\) light year
            \n= 1.053 \u00d7 10-16<\/sup> light year<\/p>\n

            c) 3ms-2<\/sup> = 3 \u00d7 10-3<\/sup> km \\(\\left(\\frac{1}{60 \\times 60} h\\right)^{-2}\\)
            \n= 3 \u00d7 10-3<\/sup> \u00d7 3600 \u00d7 3600 km h-2<\/sup><\/p>\n

            d) G = 6.61 \u00d7 10-11<\/sup> Nm2<\/sup> kg-2<\/sup>
            \n= 6.67 \u00d7 10-11<\/sup> (kg ms-2<\/sup>) m2<\/sup> kg-2<\/sup>
            \n= 6.67 \u00d7 10-11<\/sup> m3<\/sup> s-2<\/sup> kg-1<\/sup>
            \n= 6.67 \u00d7 10-11<\/sup> (100 cm)3<\/sup> s-2<\/sup> (1000 g)-1<\/sup>
            \n= 6.67 \u00d7 10-8<\/sup> cm3<\/sup> s-2<\/sup> g-1<\/sup><\/p>\n

            Question 3.
            \nA calorie is a unit of heat or energy and it equals about 4.2 J where 1J = 1 kg m2<\/sup> s-2<\/sup>. Suppose we employ a system of units in which the unit of mass equals a kg, the unit of length equals b m, the unit of time is g s. Show that a calorie has a magnitude 4.2 a-1<\/sup>b-2<\/sup> g2<\/sup> in terms of the new units.
            \nSolution:
            \nHere 1 calorie = 4.2 J = 4.2 Kg m-3<\/sup>S-1<\/sup> …………… (1)
            \nAs new unit of mass = 1kg
            \n\u2234 1 kg = \\(\\frac{1}{\\mathrm{a}}\\) new unit of mass = a-1<\/sup>
            \nsimilarly, 1m = \\(\\frac{1}{\\mathrm{b}}\\) = b-1<\/sup>
            \n1s = \\(\\frac{1}{\\mathrm{g}}\\) = g-1<\/sup>
            \nPutting these values in eq(1), we obtains.
            \n1 calorie = 4.2 (a-1<\/sup> new unit of mass)
            \n(b-1<\/sup> new unit of length)2<\/sup>
            \n(g-1<\/sup> new unit of time)2<\/sup>
            \n1 calorie = 4.2 a-1<\/sup> b-2<\/sup> g2<\/sup>.<\/p>\n

            \"AP<\/p>\n

            Question 4.
            \nExplain this statement clearly :
            \n‘To call a dimensional quantity ‘large1 or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary :
            \na) atoms are very small objects
            \nb) a jet plane moves with great speed
            \nc) the mass of Jupiter is very large
            \nd) the air inside this room contains a large number of molecules
            \ne) a proton is much more massive than an electron
            \nf) the speed of sound is much smaller than the speed of light.
            \nSolution:
            \ni) The statement is true. This is because a dimensionless quantity can be large or small only in comparison to some standard. For example, angle is dimensionless \u2220\u03b8 = 60\u00b0 is larger than \u2220\u03b8 = 30\u00b0, but smaller than \u2220\u03b8 = 90\u00b0<\/p>\n

            ii) a) The size of an atom is smaller than the sharp tin of a pin.
            \nb) A jet plane moves faster than a superfast train.
            \nc) The mass of Jupiter is very large compared to the mass of earth.
            \nd) The air inside this room contains more number of molecules than in one mole of air.
            \ne) The statement is already correct.
            \nf) The statement is already correct.<\/p>\n

            Question 5.
            \nA new unit of length is chosen such that the speed of light in vacuum in unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance ?
            \nSolution:
            \nWe are given that velocity of light in vacuum, c = 1 new unit of length s-1<\/sup>
            \nTime taken by light of Sun to reach the Earth,
            \nt = 8 min 20 s = 8 \u00d7 60 + 20 = 500 s
            \n\u2234 Distance between the Sun and the Earth,
            \nx = C \u00d7 t =1 new unit of length s-1<\/sup> \u00d7 500 s
            \n= 500 new units of length<\/p>\n

            \"AP<\/p>\n

            Question 6.
            \nWhich of the following is the most
            \nprecise device for measuring length :
            \na) a vernier cuHipers with 20 divisions on the sliding scale
            \nb) a screw gauge of pitch 1 mm and 100 divisions on the circular scale
            \nc) an optical instrument that can measure length to within a wave-length of light ?
            \nSolution:
            \nThe most precise device is that whose least count is minimum. Now
            \na) Least count of this vernier callipers
            \n= 1 SD – 19 SD = 1 SD = \\(\\frac{19}{20}\\) SD = \\(\\frac{1}{20}\\) SD
            \n= \\(\\frac{1}{20}\\) mm = \\(\\frac{1}{200}\\) cm = 0.005 cm<\/p>\n

            b) Least count of screw gauge
            \n= \\(\\frac{\\text { Pitch }}{\\text { No. of divisions on circular scale }}\\)
            \n= \\(\\frac{1}{100}\\) mm = \\(\\frac{1}{1000}\\) cm = 0.01 cm<\/p>\n

            c) Wavelength of light,
            \n\u03bb = 10-5<\/sup> cm = 0.00001 cm
            \nObviously, the most precise measurement is with optical instrument.<\/p>\n

            Question 7.
            \nA student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair ?
            \nSoution:
            \nMagnification m = \\(\\frac{\\text { observed with }(\\mathrm{y})}{\\text { real width }(\\mathrm{x})}\\)
            \nx = \\(\\frac{\\mathrm{y}}{\\mathrm{m}}=\\frac{3.5 \\mathrm{~mm}}{100}\\) = 0.035 mm<\/p>\n

            Question 8.
            \nAnswer the following :
            \na) You are given a thread and a metre scale. How will you estimate the diameter of the thread ?
            \nb) A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale ?
            \nc) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only ?
            \nSolution:
            \na) The diameter of a thread is so small that it cannot be measured using a metre scale. We wind a number of turns of the thread on the metre scale. So that the turns are closely touching one another. Measure the length (l) of the windings on the scale which contains n number of turns.
            \n\u2234 Diameter of thread = \\(\\frac{1}{n}\\)<\/p>\n

            b) As least count
            \n= \\(\\frac{\\text { Pitch }}{\\text { Number of divisions on circular scale }}\\)
            \n\u2234 Theoretically speaking, least count decreases on increasing the number of divisions on the circular scale. Hence accuracy would increase. Practically, it may not be possible to take the reading precisely due to low resolution of human eye.<\/p>\n

            c) A large number of observations will give more reliable result than smaller number of observations. This is because of probability of making a positive random error of certain magnitude is equal to that of making a negative random errors are likely to cancel and the result may be more reliable.<\/p>\n

            \"AP<\/p>\n

            Question 9.
            \nThe photograph of a house occupies an area of 1.75 cm2<\/sup> on a 35 mm slide. The slide is projected on to a screen and the area of the house on the screen is 1.55 m2<\/sup>. What is the linear magnification of the projector-screen arrangement.
            \nSolution:
            \nHere, area of object = 1.75 cm2<\/sup> and area of image = 1.55 m2<\/sup> = 1.55 \u00d7 104<\/sup> cm2<\/sup>
            \n\u2234 Areal magnification = \\(\\frac{\\text { area of image }}{\\text { area of object }}\\)
            \n= \\(\\frac{1.55 \\times 10^4}{1.75}\\) = 8857
            \nLinear magnification = \\(\\sqrt{8857}\\) = 94.1<\/p>\n

            Question 10.
            \nState the number of significant figures in the following :
            \na) 0.007 m2<\/sup>
            \nb) 2.64 \u00d7 1024<\/sup> kg
            \nc) 0.2370 g cm-3<\/sup>
            \nd) 6.320 J
            \ne) 6.032 N m-2<\/sup>
            \nf) 0.0006032 m2<\/sup>
            \nSolution:
            \nThe number of significant figures is as given below.
            \na) one
            \nb) three
            \nc) four
            \nd) four
            \ne) four
            \nf) four<\/p>\n

            Question 11.
            \nThe length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.
            \nSolution:
            \nHere, length l = 4.234 m;
            \nBreath, b = 1.005 m
            \nThickness, t = 2.01 cm = 2.01 \u00d7 10-2<\/sup> m
            \nArea of the sheet = 2(l \u00d7 b + b \u00d7 t + t \u00d7 l)
            \n= 2(4.234 \u00d7 1.005 + 1.005 \u00d7 0.0201 + 0.0201 \u00d7 4.234)
            \n2(4.3604739) = 8.7209478 m2<\/sup>
            \nAs area can contain a maximum of three significant digits, therefore, rounding off, we get
            \nArea = 8.72 m2<\/sup>
            \nAlso, volume = l \u00d7 b \u00d7 t
            \nV = 4.234 \u00d7 1.005 \u00d7 0.0201
            \nV = 0.0855289
            \nV = 0.855 m3<\/sup><\/p>\n

            \"AP<\/p>\n

            Question 12.
            \nThe mass of a box measured by a grocer’s balance is 2.300 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures ?
            \nSolution:
            \nHere, mass of the box, m = 2.3 kg
            \nMass of one gold piece,
            \nm1<\/sub> = 20.15 g = 0.02015 kg
            \nMass of another gold piece,
            \nm2<\/sub> = 20.17g = 0.02017 kg
            \na) Total mass = m + m1<\/sub> + m1<\/sub>
            \n= 2.3 + 0.02015 + 0.02017
            \n= 2.34032 kg
            \nAs the result is correct only upto one place of decimal, therefore, on rounding off total mass = 2.3 kg<\/p>\n

            b) Difference in masses
            \n= m2<\/sub> – m1<\/sub> = 20.17 – 20.15
            \n= 0.02g (correct upto two places of decimal)<\/p>\n

            Question 13.
            \nA physical quantity P is related to four observables a, b, c and d as follows :
            \nP = a3<\/sup>b2<\/sup>\/ (\\(\\sqrt{c}\\)d )
            \nThe percentage errors of measurement in a, b, c 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 ?
            \nSolution:
            \nHere, P = \\(\\frac{a^3 b^2}{\\sqrt{c} d}\\)
            \nMaximum fractional error in P is given by \\(\\frac{\\Delta \\mathrm{P}}{\\mathrm{P}}\\)
            \n= \\(3 \\frac{\\Delta \\mathrm{a}}{\\mathrm{a}}+2 \\frac{\\Delta \\mathrm{b}}{\\mathrm{b}}+\\frac{1}{2} \\frac{\\Delta \\mathrm{c}}{\\mathrm{c}}+\\frac{\\Delta \\mathrm{d}}{\\mathrm{d}}\\)
            \n= \\(3\\left(\\frac{1}{100}\\right)+2\\left(\\frac{3}{100}\\right)+\\frac{1}{2}\\left(\\frac{4}{100}\\right)+\\frac{2}{100}\\)
            \n= \\(\\frac{13}{100}\\) = 0.13
            \nPercentage error in
            \nP = \\(\\frac{\\Delta \\mathrm{P}}{\\mathrm{P}}\\) \u00d7 100 = 0.13 \u00d7 100 = 13%
            \nAs the result (13% error) has two significant figures, therefore if P turns out to be 3.763, the result would be rounded of to 3.8.<\/p>\n

            Question 14.
            \nA book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion :
            \na) y = a sin 2 \u03c0 t\/T) b) y = a sin vt,
            \nc) y = (a\/T) sin t\/a,
            \nd) y = (a\\(\\sqrt{2}\\)) (sin 2\u03c0t \/T + cos 2 \u03c0t\/T)
            \n(a = maximum displacement of the particle, v = speed of the particle. T = time- period of motion). Rule out the wrong formulas on dimensional grounds.
            \nSolution:
            \nThe argument of a trigonometrical function i.e., angle is dimensionless. Now in
            \na) \\(\\frac{2 \\pi t}{T}=\\frac{T}{T}\\) = 1 = (M0<\/sup>L0<\/sup>T0<\/sup>)
            \n…………… dimensionless
            \nb) vt = (LF-1<\/sup>) (T) = L = (M0<\/sup>L1<\/sup>T0<\/sup>)
            \n………….. not dimensionless<\/p>\n

            c) \\(\\frac{\\mathrm{t}}{\\mathrm{a}}=\\frac{\\mathrm{T}}{\\mathrm{L}}\\) = (L-1<\/sup>T-1<\/sup>)
            \n………….. not dimensionless<\/p>\n

            d) \\(\\frac{2 \\pi t}{\\mathrm{~T}}=\\frac{\\mathrm{T}}{\\mathrm{T}}\\) = 1 = [M0<\/sup>L0<\/sup>T0<\/sup>]
            \n…………… dimensionless<\/p>\n

            \"AP<\/p>\n

            Question 15.
            \nA famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m0 of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einsterin). A boy recalls the relation almost correctly but for-gets where to put the constant c. He writes :
            \nm = \\(\\frac{m_0}{\\left(1-v^2\\right)^{1 \/ 2}}\\)
            \nGuess where to put the missing c.
            \nSolution:
            \nAccording to the principle of homogenity of dimensions power of M, L, T on either side of the formula must be equal. For this, on RHS, the denominator (1 – v2<\/sup>)1\/2<\/sup> should be dimensionless. Therefore, instead of (1 – v2<\/sup>)1\/2<\/sup>,
            \nwe should write \\(\\left(\\frac{1-v^2}{c^2}\\right)^{1 \/ 2}\\)
            \nHence the correct formula would be
            \nm = \\(\\frac{m_0}{\\left(\\frac{1-v^2}{c^2}\\right)^{1 \/ 2}}\\)<\/p>\n

            Question 16.
            \nThe unit of length convenient on the atomic scale is known as an angstrom and is denoted byA\u00b0 : 1A\u00b0 = 1010m. The size of a hydrogen atom is about 0.5 A\u00b0. What is the total atomic volume in m3<\/sup> of a mole of hydrogen atoms ?
            \nSolution:
            \nHere r = 0.5 A\u00b0 = 0.5 \u00d7 10-10<\/sup> m
            \nVolume of each atom of hydrogen = \\(\\frac{4}{3}\\)\u03c0r3<\/sup>
            \n= \\(\\frac{4}{3}\\) \u00d7 3.14(0.5 \u00d7 10-10<\/sup>) = 5.236 \u00d7 10-31<\/sup> m3<\/sup>
            \nNumber of hydrogen atoms in one gram mole of hydrogen = Avagadro\u2019s number
            \n= 6.023 \u00d7 1023<\/sup>
            \n\u2234 Atomic volume of one gram mole of hydrogen atom
            \n= 5.236 \u00d7 10-31<\/sup> \u00d7 6.023 \u00d7 1023<\/sup>
            \n= 3.154 \u00d7 10-7<\/sup> m3<\/sup><\/p>\n

            Question 17.
            \nOne mole of an ideal gas at standard temperature and pressure occupies 22.4 L of (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen ? (Take the size of hydrogen molecule to about 1A\u00b0). Why is this ratio so large ?
            \nSolution:
            \nAtomic volume = \\(\\frac{4}{3}\\)\u03c0R3<\/sup> \u00d7 N
            \n= \\(\\frac{4}{3}\\) (0.5 \u00d7 10-10<\/sup>)3<\/sup> \u00d7 6.023 \u00d7 1023<\/sup>
            \n= 3.154 \u00d7 10-7<\/sup> m3<\/sup>
            \nMolar volume = 22.4 lit = 22.4 \u00d7 10-3<\/sup> m3<\/sup>
            \n\\(\\frac{\\text { Molar volume }}{\\text { Atomic volume }}=\\frac{22.4 \\times 10^{-3}}{3.154 \\times 10^{-7}}\\) = 7.1 \u00d7 104<\/sup>
            \nThe ratio is large due to large intermolecular separations.<\/p>\n

            \"AP<\/p>\n

            Question 18.
            \nExplain this common observation clearly: If you look out of the window of a fast moving train, the nearby tr6es, houses etc., seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving; these distant objects seem to move with you).
            \nSolution:
            \nThe line joining the object to the eye is called the line of sight. When a train moves rapidly, the line of sight of a near by tree changes its direction of motion rapidly. Therefore, the trees appear to run in opposite direction. On the contrary, the line of sight of far off objects does not change its direction. So much, due to extremely large distance from the eye. Hence distant hill tops, moon, the stars etc., appear stationary.<\/p>\n

            Question 19.
            \nThe principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart 21. in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit 3 \u00d7 1011<\/sup> m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1″ (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1″ (second) of arc from opposite ends of a baseline equal to the distance from the Earth to Sol. the Sun. How much is a parsec in terms of metres ?
            \nSolution:
            \nHere, length of baseline
            \n= distance from each to the sun
            \n= 1 A.U = 1.5 \u00d7 1011<\/sup> m
            \nParallax angle, \u03b8 = 1
            \n\\(\\frac{1^1}{60}=\\frac{1^{\\circ}}{60 \\times 60}=\\frac{\\pi}{180} \\times \\frac{1}{60 \\times 60}\\) radian
            \nr = 1 par sec = ?; From l = r\u03b8 = \\(\\frac{l}{\\theta}\\)
            \n= \\(\\frac{1.5 \\times 10^{\\prime \\prime}}{\\pi \/ 180 \\times 60 \\times 60} \\mathrm{~m}\\) = 3.1 \u00d7 1016<\/sup> m
            \nHence 1 parsec = 3.1 \u00d7 1016<\/sup> m<\/p>\n

            Question 20.
            \nThe nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parasecs ? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun ?
            \nSolution:
            \nx = 4.29 ly = 4.29 \u00d7 9.46 \u00d7 1015<\/sup> m
            \n= \\(\\frac{4.29 \\times 9.46 \\times 10^{15}}{3.08 \\times 10^{16}}\\) par sec
            \n= 1.323 par sec
            \n\u03b8 = \\(\\frac{l}{r}=\\frac{2 A u}{x}\\)
            \n= \\(\\frac{2 \\times 1.496 \\times 10^4}{4.29 \\times 9.46 \\times 10^{15}}\\) radian = 1.512 sec<\/p>\n

            Question 21.
            \nPrecise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modem science where precise measurements of length, time, mass etc., are needed. Also, wherever you can, give a quantitative idea of the precision needed.
            \nSolution:
            \nPrecise measurements of physical quantities like length, mass and time are the primary requirements for development of quantitative laws of physics or any other science. For example, in the measurement of distance of moon from earth by laser beam, very accurate measurement of time taken is required. Similarly, for measuring distance, elevation and velocity of an aeroplane by radar method, time measurement has to be accurate. For measuring distances of near by stars, accurate measurement of parallax angle is required.<\/p>\n

            In the field of crystallography, precise measurement of length is needed to determine interatomic distances using a mass spectrometer, the precision measurement of masses of atoms are made.<\/p>\n

            \"AP<\/p>\n

            Question 22.
            \nJust as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):
            \na) the total mass of rain-bearing clouds over India during the Monsoon
            \nb) the mass of an elephant
            \nc) the wind speed during a storm
            \nd) the number of strands of hair on * your head
            \ne) the number of air molecules in your classroom.
            \nSolution:
            \na) During the Monsoon, meteorologist record about 100 cm of rainfall, i.e.,
            \nh – 100 cm = 1m
            \nArea of our country,
            \nA = 3.3 million square km
            \n= 3.3 \u00d7 106<\/sup> (103<\/sup>m2<\/sup>)
            \n= 3.3 \u00d7 1012<\/sup> m6<\/sup>
            \n\u2234 Volume of rain water,
            \nv = A \u00d7 h = 3.3 \u00d7 1012<\/sup> 1m3<\/sup>
            \nAs density of water,
            \nP = 103 km\/m3<\/sup>
            \n\u2234 Mass of rain water
            \n= vP = 3.3 \u00d7 1012<\/sup> \u00d7 103<\/sup> kg
            \n= 3.3 \u00d7 1015<\/sup> kg
            \nThis must be the total mass of rain bearing clouds over India.<\/p>\n

            b) To estimate the mass of an elephant, we take a boat of known base area A. Measure the depth of boat in water. Let it be x1<\/sub>. Therefore, volume of water displaced by the boat , V1<\/sub> = Ax1<\/sub> move the elephant into this boat. The boat gets deeper into water. Measure the depth of boat now into water, Let it be x1<\/sub>
            \n\u2234 Volume of water displaced by boat and elephant V2<\/sub> = Ax2<\/sub>
            \n\u2234 Volume of water displaced by the elephant V = V2<\/sub> – V1<\/sub> = A(x2<\/sub> – x1<\/sub>)
            \nIf \u03c1 is density of water, then mass of elephant
            \n= mass of water displaced by it
            \n= V\u03c1<\/sub> = A(X2<\/sub> – X1<\/sub>) \u03c1<\/p>\n

            c) The wind speed during a storm can be estimated using a gas filled balloon. In figure OA is normal position of a gas filled balloon, when there is no wind. As the wind blows to the right, the balloon drifts to position B in one second. The angle of drift \u2220AOB = \u03b8 is measured, if h is the height of the balloon, then AB = d = h\u03b8 This is the distance travelled by the balloon in one second it must be the wind speed.
            \n\"AP<\/p>\n

            d) For this, we measure the area of the head . that carrier the hair let it be A. Using a screw guage, we measure thickness of hair, let it bad.
            \n\u2234 Area of cross section of hair = \u03c0d2<\/sup>
            \nAssuming that the distribution of hair over the head is uniform.
            \nThe number of strands of hair
            \n= \\(\\frac{\\text { total area }}{\\text { area of cross section of each hair }}=\\frac{\\mathrm{A}}{\\pi \\mathrm{d}^2}\\)
            \nCalculations show that number of strands of hair on human head is of the order of one million.<\/p>\n

            e) Measure the volume of room. We know that one mole of air at NTP occupies a volume of 22.4 lit i.e., 22.4 \u00d7 10-3<\/sup> m3<\/sup>
            \n\u2234 Number of air molecules in 22.4 \u00d7 10-3<\/sup> m3<\/sup> = 6.023 \u00d7 1023<\/sup>
            \nNumber of air molecules in volume v of room = \\(\\frac{6.023 \\times 10^{23}}{22.4 \\times 10^{-3}}\\)v<\/p>\n

            \"AP<\/p>\n

            Question 23.
            \nThe Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107<\/sup> K and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid of liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases ? Check if your guess is correct from the following data : mass of the Sun = 2.0 \u00d7 1030<\/sup> kg, radius of the Sun = 7.0 \u00d7 108<\/sup> m.
            \nSolution:
            \nHere, M = 2.0 \u00d7 1030<\/sup> kg; R = 7.0 \u00d7 108<\/sup> m
            \nDensity \u03c1 = ?
            \n\u03c1 = \\(\\frac{\\text { mass }}{\\text { volume }}=\\frac{M}{\\frac{4}{3} \\pi R^3}=\\frac{3 M}{4 \\pi R^3}\\)
            \n= \\(\\frac{3 \\times 2.0 \\times 10^{30}}{4 \\times 3.14 \\times\\left(7 \\times 10^8\\right)^3}\\)
            \n= 1.392 \u00d7 103<\/sup> kg\/m3<\/sup>
            \nThis is the order of density of solids and liquids; and not gases.
            \nThe high density of Sun is due to inward gravitational attraction on outer layers, due to inner layers of the Sun.<\/p>\n

            Question 24.
            \nWhen the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72, of arc. Calculate the diameter of Jupiter.
            \nSolution:
            \nHere, r = 824.7 \u00d7 106<\/sup> km
            \n\u03b8 = 35.72″ radian
            \n= \\(\\frac{35.72}{60 \\times 60} \\times \\frac{\\pi}{180}\\) radian
            \nDiameter, l = ?; As l = r\u03b8
            \n\u2234 l = 824.7 \u00d7 106<\/sup> \u00d7 \\(\\frac{35.72 \\times \\pi}{60 \\times 60 \\times 180}\\) km
            \n= 1.429 \u00d7 105<\/sup>km<\/p>\n

            Question 25.
            \nA man walking briskly in rain with speed u must slant his umbrella forward making an angle q with the vertical. A student derives the following relation between q and v : tan \u03b8 = v and checks that the relation has a correct limit: as v \u2192 0, \u03b8 \u21920, as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct ? If not, guess the correct relation.
            \nSolution:
            \nThe relation tan \u03b8 = v has a correct limit, as v \u2192 0, \u03b8 \u2192 0.
            \nHowever, RHS = tan \u03b8 = [M0<\/sup>L0<\/sup>T0<\/sup>] and L.H.S = v = [M0<\/sup>L1<\/sup>T-1<\/sup>]
            \nTherefore, the relation is not correct dimensionally.
            \nWe shall find the correct relation is
            \ntan \u03b8 = \\(\\frac{\\mathrm{v^2}}{\\mathrm{rg}}\\)<\/p>\n

            Question 26.
            \nIt is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s ?
            \nSolution:
            \nError in 100 years = 0.02s
            \nError in 1 sec
            \n\"AP
            \nHence the accuracy of the standard cesium clock is measuring a time interval of 1s is 10-12<\/sup> s.<\/p>\n

            \"AP<\/p>\n

            Question 27.
            \nEstimate the average mass density of a sodium atom assuming its size to be about 2.5 A\u00b0. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase: 970 kg m-3<\/sup>. Are the two densities of the same order of magnitude ? If so why ?
            \nSolution:
            \nAtomic volume = \\(\\frac{4}{3}\\)\u03c0R3<\/sup> \u00d7 N
            \n= \\(\\frac{4}{3}\\) \u00d7 \\(\\frac{22}{7}\\) (1.25 \u00d7 10-10<\/sup>)3<\/sup> 6.023 \u00d7 1023<\/sup> m
            \n= 4.93 \u00d7 10-6<\/sup> m3<\/sup>
            \nAverage mass density
            \n= \\(\\frac{\\text { mass }}{\\text { volume }}=\\frac{23 \\times 10^{-3}}{4.93 \\times 10^{-6}}\\)
            \n= 4.67 \u00d7 103<\/sup> kg\/m3<\/sup>
            \nThe two densities are not of the same order. This is due to interatomic spacing in the crystalline phase.<\/p>\n

            Question 28.
            \nThe unit of length convenient on the nuclear scale is a fermi: 1 f = 10-15<\/sup> m. Nuclear sizes obey roughly the following empirical relation :
            \nr = r0<\/sub>A1\/3<\/sup>
            \nwhere r is the radius of the nucleus. A its mass number, and r0<\/sub> is a constant equal to about 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different density of a sodium atom obtained in Exercise. 27.
            \nAnswer:
            \nLet m be the average mass of nucleon (neutron or proton)
            \nAs the nucleus contains A nucleons,
            \n\u2234 mass of nucleus \u03bc = mA
            \nRadius of nucleus r = r0<\/sub>A1\/3<\/sup>
            \nNuclear density \u03c1 = \\(\\frac{\\text { mass }}{\\text { volume }}=\\frac{\\mu}{\\frac{4}{3} \\pi r^3}\\)
            \n= \\(\\frac{3 m A}{4 \\pi\\left(r_0 A^{1 \/ 3}\\right)^3}=\\frac{3 m}{4 \\pi r_0^3}\\)
            \nAs m and r0<\/sub> are constant, therefore, nuclear density is constant for all nuclei.
            \nusing m = 1.66 \u00d7 10-27<\/sup> kg and r0<\/sub>
            \n= 1.2f = 1.2 \u00d7 10-15<\/sup> m
            \nWe get \u03c1 = \\(\\frac{3 m}{4 \\pi r_0^3}=\\frac{3 \\times 1.66 \\times 10^{-27}}{4 \\times 3.14\\left(1.2 \\times 10^{-15}\\right)^3}\\)
            \n= 2.29 \u00d7 1017<\/sup> kg m-3<\/sup>
            \nAs \u03c1 is constant for all nuclei, this must be the density of sodium nucleus also.
            \nDensity of sodium atom
            \n\u03c1’ = 4.67 \u00d7 103<\/sup> kg m-3<\/sup>
            \n= \\(\\frac{\\rho}{\\rho^{\\prime}}=\\frac{2.29 \\times 10^{17}}{4.67 \\times 10^3}\\) = 4.67 \u00d7 10-3<\/sup><\/p>\n

            Question 29.
            \nA LASER is a source or very intense, monochromatic and unidirectional beam of light. These properties of a laser light can be exploited to measure long distances. The distance of the Moon from the Earth has been already determined very precisely using a laser as a source of light. A laser light beamed at the Moon takes 2.56 s to return after reflection at the Moon’s surface. How much is the radius of the lunar orbit around the Earth ?
            \nSolution:
            \nHere, t = 2.56s
            \nVelocity of laser light in vacuum,
            \nc = 3 \u00d7 108<\/sup> m\/s
            \nThe radius of lunar orbit is the distance of moon from the earth. Let it be x.
            \nAs x = \\(\\frac{c \\times t}{2}\\)
            \n\u2234 x = \\(\\frac{3 \\times 10^8 \\times 2.56}{2}\\) = 3 84 \u00d7 108<\/sup> m<\/p>\n

            \"AP<\/p>\n

            Question 30.
            \nA SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. In a submarine equipped with a SONAR the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of sound in waiter = 1450 ms-1<\/sup>).
            \nSolution:
            \nHere, t = 77.0 s, x = ? V = 1450 ms-1<\/sup>
            \nAs x = \\(\\frac{\\mathrm{V} \\times \\mathrm{t}}{2}\\)
            \n= \\(\\frac{1450 \\times 77.0}{2}\\)m; x = 55825 m<\/p>\n

            Question 31.
            \nThe farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzlingn features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us ?
            \nSolution:
            \nHere, x = ?
            \nTime taken t = 30 billion years = 3 \u00d7 109<\/sup> yr
            \n= 3 \u00d7 109<\/sup> \u00d7 365 \u00d7 24 \u00d7 60 \u00d7 60 s
            \nVelocity of light in vacuum, c = 3 \u00d7 108<\/sup> m\/s
            \n= 3 \u00d7 105<\/sup> km\/s
            \nAs distance = velocity \u00d7 time
            \nx = (3 \u00d7 105<\/sup>) \u00d7 3 \u00d7 109<\/sup> \u00d7 365 \u00d7 24 \u00d7 60 \u00d7 60 km.
            \n= 2.84 \u00d7 1022<\/sup> km<\/p>\n

            Question 32.
            \nIt is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 3 and 4. determine the approximate diameter of the moon.
            \nSolution:
            \nDistance of moon from earth,
            \nME = 3.84 \u00d7 108<\/sup> m
            \nDistance of Sun from earth,
            \n\"AP
            \nSE = 1.496 \u00d7 1011<\/sup> m
            \nDistance of Sun AB = 1.39 \u00d7 109<\/sup> m
            \nThe situation during total solar eclipse is shown in figure.
            \nAs \u2206s<\/sup> ABE of CDE are similar, therefore,
            \n\\(\\frac{A B}{C D}=\\frac{S E}{M E}\\)
            \nCD = AB \u00d7 \\(\\frac{M E}{S E}=\\frac{1.39 \\times 10^9 \\times 3.84 \\times 10^8}{1.496 \\times 10^{11}}\\)
            \n= 3.5679 \u00d7 106<\/sup> m = 3567.9 km<\/p>\n

            \"AP<\/p>\n

            Question 33.
            \nA great physicist of this century (P.A.M. Dirac) loved playing with numerical values of Fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics (c, e, mass of electron, mass of proton) and the gravitational costant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (~ 15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants ?
            \nSolution:
            \nTrying out with basic constants of atomic physics (speed of light c, charge on electron e, mass of electron mc<\/sub>, mass of proton mp<\/sub>) and universal gravitational constant G, we can arrive at a quantity which has the dimensions of time. One such quantity is
            \nt = \\(\\left(\\frac{c^2}{4 \\pi \\epsilon_0}\\right)^2 \\times \\frac{1}{m_p m_{e^2} c^3 G}\\)
            \nPut e = 1.6 \u00d7 10-19<\/sup> c
            \n\\(\\frac{1}{4 \\pi \\epsilon_0}\\) = 9 \u00d7 109<\/sup>,
            \nc = 3 \u00d7 10<sup8 m\/s and
            \nG = 6.67 \u00d7 10-11<\/sup> Nm2<\/sup> kg-2<\/sup>
            \nt = (1.6 \u00d7 10-19<\/sup>)4<\/sup> \u00d7 (9 \u00d7 109<\/sup>)2<\/sup> \u00d7 \\(\\frac{1}{1.67 \\times 10^{-27} \\times\\left(9 \\times 10^{-31}\\right) \\times\\left(3 \\times 10^8\\right)^3 \\times 6.67 \\times 10^{-11}}\\)
            \nt = 2.18 \u00d7 1016<\/sup> sec; This time is of the order of age of universe.<\/p>\n

            Textual Examples<\/span><\/p>\n

            Question 1.
            \nCalculate the angle at (a) 1\u00b0 (degree) (b) V (minute of arc or arcmin) and (c) 1″ (second of arc or are second) in radians. Use 360\u00b0 = 2\u03c0 rad, 1\u00b0 = 60′ and 1′ = 60″.
            \nAnswer:
            \n(a) We have 360\u00b0 = 2\u03c0 rad
            \n1\u00b0 = (\u03c0\/180) rad = 1.745 \u00d7 10-2<\/sup> rad<\/p>\n

            (b) 1′ = 60′ = 1.745 \u00d7 10-2<\/sup> rad
            \nV = 2.908 \u00d7 10-4<\/sup> rad, 2.91 \u00d7 10-4<\/sup> rad<\/p>\n

            (c) 1′ = 60″ = 2.908 \u00d7 10-4<\/sup> rad
            \n1″ = 4.847 \u00d7 10-6<\/sup> rad, 4.85 \u00d7 10-6<\/sup> rad.<\/p>\n

            Question 2.
            \nA man wishes to estimate the distance of a nearby tower from him. He stands at a point A in front of the tower C and spots a very distant object O in line with AC. He then walks perpendicular to AC up to B, a distance of 100 m and looks at O and C again. Since O is very distant, the direction BO is practically the same as AO; but he finds the line of sight of C shifted from the original line of sight by an angle \u03b8 = 40\u00b0 (\u03b8 is known as parallax’) estimate the distance of the tower C from his original position A.
            \n\"AP
            \nAnswer:
            \nParallax angle \u03b8 = 40\u00b0; AB = AC tan \u03b8
            \nAC = AB\/tan \u03b8 = 100 m\/tan 40\u00b0
            \n= 100 m\/0.8391 = 119 m.<\/p>\n

            \"AP<\/p>\n

            Question 3.
            \nThe moon is observed from two diametrically opposite points A and B on Earth. The angle 6 subtended at the moon by the two directions of observation is 1\u00b054\u2018. Given the diameter of the Earth to be about 1.276 \u00d7 107<\/sup> m, compute the distance of the moon from the Earth.
            \nSolution:
            \nWe have \u03b8 = 1\u00b054′ = 114
            \n= (114 \u00d7 60)” \u00d7 (4.85 \u00d7 10-6<\/sup>) rad
            \n= 3.32 \u00d7 10-2<\/sup> rad.
            \nSince 1″ = 4.85 \u00d7 10-6,<\/sup>; rad
            \nAlso b = AB = 1.276 \u00d7 107<\/sup> m
            \nHence from D = b\/\u03b8, we have the earth-moon distance,
            \nD = b\/\u03b8
            \n= \\(\\frac{1.276 \\times 10^7}{3.32 \\times 10^{-2}}\\) = 3.84 \u00d7 108<\/sup> m.<\/p>\n

            Question 4.
            \nThe Sun’s angular diamater is mea-sured to be 1920″. The distance D of the Sun from the Earth is 1.496 \u00d7 10″ m. What is the diameter of the Sun ?
            \nAnswer:
            \nSun’s angular diameter \u03b1 = 1920”
            \n= 1920 \u00d7 4.85 \u00d7 10-6<\/sup> rad
            \n= 9.31 \u00d7 10-3<\/sup> rad
            \nSun’s diameter d = \u03b1
            \nD = (9.31 \u00d7 10-3<\/sup>) \u00d7 (1.496 \u00d7 1011<\/sup>) m
            \n= 1.39 \u00d7 109<\/sup> m.<\/p>\n

            Question 5.
            \nIf the size of a nucleus in the range of 10-15<\/sup> to 10-14<\/sup> m) is scaled up to the tip of a sharp pin, what roughly is the size of an atom ? Assume tip of the pin to be in the range 10-5<\/sup> m to 10-4<\/sup> m.
            \nAnswer:
            \nThe size of a nucleus is in the range of 10-15<\/sup>m and 10-14<\/sup> m. The tip of a sharp pin is taken to be in the range of 10-5<\/sup> m and 10-4<\/sup> m. Thus we are scaling up by a factor of 1010<\/sup>. An atom roughly of size 10-10<\/sup> m will be scaled up to a size of 1 m. Thus a nucleus in an atom is as small in size as the tip of a sharp pin placed at the centre of a sphere of radius about a metre long.<\/p>\n

            \"AP<\/p>\n

            Question 6.
            \nTwo clocks are being tested against a standard clock located in a national laboratory. At 12 : 00 : 00 noon by the standard dock, the readings of the two clocks are :
            \n\"AP
            \nIf you are doing an experiment that requires ‘precision time interval’ measurements, which of the two clocks will you prefer ?
            \nAnswer:
            \nThe range of variation over the seven days of observations is 162 s for clock 1 and 31 s for clock 2. The average reading of clock 1 is much closer to the standard time than the average reading of clock 2. The important point is that a clock’s zero error is not as significant for precision work as its variation, because a ‘zero-error’ can always be easily corrected. Hence clock 2 is to be preferred to clock 1.<\/p>\n

            Question 7.
            \nWe measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be 2.63 s. 2.56 s, 2.42 s, 2.71 is and 2.80 s. Calculate the absolute errors relative error or percentage error.
            \nAnswer:
            \nThe mean period of oscillation of the pendulum
            \nT = \\(\\frac{(2.63+2.56+2.42+2.71+2.80) s}{5}\\)
            \n= \\(\\frac{13.22}{5}\\) s = 2.624 s = 2.62 s
            \nAs the periods are measured to a resolution of 0.01 s, all times are to the second decimal; it is proper to put this mean period also to the second decimal.
            \nThe errors in the measurements are
            \n2.63 s – 2.62 s = 0.01 s
            \n2.56 s – 2.62 s = -0.06 s
            \n2.42 s – 2.62 s = -0.20 s
            \n2.71 s – 2.62 s = 0.09 s
            \n2.80 s – 2.62 s = 0.18 s
            \nNote that the errors have the same units as the quantity to be measured.
            \nThe arithmetic mean of all the absolute errors (for arithmetic mean, we take only the magnitudes) is
            \n\u2206Tmean<\/sub> = [(0.01 + 0.06 + 0.20 + 0.09 + 0.18)s]\/5 = 0.54 s\/5 = 0.11 s
            \nThat means, the period of oscillation of the simple pendulum is (2.62 \u00b1 0.11) s i.e. it lies between (2.62 + 0.11) s and (2.62 – 0.11) s or between 2.73 s and 2.51 s. As the arithmetic mean of all the absolute errors is 0.11 s, there is already an error in the tenth of a second. Hence there is no point in giving the period to a hundredth. A more correct way will be to write T = 2.6 \u00b1 0.1 s<\/p>\n

            Note that the last numeral 6 is unreliable, since it may be anything between 5 and 7. We indicate this by saying that the measurement has two significant figures. In this case, the two significant figures are 2, which is reliable and 6, which has an error associated with it. You will learn more about the significant figures in section 2.7.
            \nFor this example, the relative error or the percentage error is \u03b4a = \\(\\frac{0.1}{2.6}\\) \u00d7 100 = 4.<\/p>\n

            Question 8.
            \nThe temperatures of two bodies measured by a thermometer are t1<\/sub> = 20 \u00b0C \u00b1 0.5 \u00b0C and t2<\/sub> = 50 \u00b0C \u00b1 0.5\u00b0C. Calculate the temperature difference and the error therein.
            \nAnswer:
            \nt’ = t2<\/sub> – t1<\/sub>
            \n= (50 \u00b0C \u00b1 0.5 \u00b0C) – (20 \u00b0C \u00b1 0.5 \u00b0C)
            \nt’ = 30 \u00b0C \u00b1 1 \u00b0C.<\/p>\n

            \"AP<\/p>\n

            Question 9.
            \nThe resistance R = V\/I where V = (100 \u00b1 5) V and I = (10 \u00b1 0.2) A. Find the percentage error in R.
            \nAnswer:
            \nThe percentage error in V is 5 and in I it is 2.
            \nThe total error in R would therefore be 5 + 2 = 7%.<\/p>\n

            Question 10.
            \nTwo resistors of resistances R1<\/sub> = 100 \u00b1 3 ohm and R2<\/sub> = 200 \u00b1 4 ohm are connected (a) in series, (b) in parallel. Find the equivalent resistance of the (a) series combination, (b) parallel combination. Use for (a) the relation R
            \n= R1<\/sub> + R2<\/sub> and for (b) \\(\\frac{1}{R^{\\prime}}+\\frac{1}{R_1}+\\frac{1}{R_2}\\) and \\(\\frac{\\Delta R^{\\prime}}{R^{\\prime 2}}=\\frac{\\Delta R_1}{R_1^2}+\\frac{\\Delta R_2}{R_2^2}\\)
            \nAnswer:
            \na) The equivalent resistance of series, combination
            \nR = R1<\/sub> + R2<\/sub> = (100 \u00b1 3)
            \nohm + (200 \u00b1 4) ohm = 300 \u00b1 7 ohm.<\/p>\n

            b) The equivalent resistance of parallel combination
            \n\"AP
            \nThen, R’ = 66.7 \u00b1 1.8 ohm
            \n(Here, \u2206R is expressed as 1.8 instead of 2 to keep in conformity with the rules of significant figures).<\/p>\n

            Question 11.
            \nFind the relative error in Z, if Z = A4<\/sup>B1\/3<\/sup>\/CD3\/2<\/sup>.
            \nAnswer:
            \nThe relative error in Z is \u2206Z\/Z = 4(\u2206A\/A) + (1\/3) (\u2206B\/B) + (\u2206C\/C) + (3\/2) (\u2206D\/D).<\/p>\n

            Question 12.
            \nThe period of oscillation of a simple pendulum is T = \\(2 \\pi \\sqrt{L \/ g}\\). Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1s resolution. What is the accuracy in the determination of g ?
            \nAnswer:
            \ng = 4\u03c02<\/sup>L\/T2<\/sup>; Here, T = \\(\\frac{t}{n}\\) and \u2206T = \\(\\frac{\\Delta t}{n}\\)
            \nTherefore, \\(\\frac{\\Delta \\mathrm{T}}{\\mathrm{T}}=\\frac{\\Delta \\mathrm{t}}{\\mathrm{t}}\\)
            \nThe errors in both L and t are the least count errors. Therefore, (\u2206g\/g) = (\u2206L\/L) + 2(\u2206T\/T)
            \n= \\(\\frac{0.1}{20.0}+2\\left(\\frac{1}{90}\\right)\\) = 0.027
            \nThus, the percentage error in g is 100 (\u2206g\/g)
            \n= 100(\u2206L\/L) + 2 \u00d7 100 (\u2206T\/T) = 3.<\/p>\n

            \"AP<\/p>\n

            Question 13.
            \nEach side of a cube is measured to be 7.203 m. What are the total surface area and the volume of the cube to appro\u00acpriate significant figures ?
            \nAnswer:
            \nThe number of significant figures in the measured length is 4. The calculated area and the volume should therefore be rounded off to 4 significant figures.
            \nSurface area of the cube = 6(7.203)2<\/sup> m2<\/sup>
            \n= 311.299254 m2<\/sup>
            \n= 311.3 m2<\/sup>
            \nVolume of the cube = (7.203)3<\/sup> m3<\/sup>
            \n= 373.714754 m3<\/sup>
            \n= 373.7 m3<\/sup><\/p>\n

            Question 14.
            \n5.74 g of a substance occupies 1.2 cm3<\/sup>. Express its density by keeping the significant figures in view.
            \nAnswer:
            \nThere are 3 significant figures in the measured mass whereas there are only 2 significant figures in the measured volume. Hence the density should be expressed to only 2 significant figures.
            \nDensity = \\(\\frac{5.74}{1.2}\\) cm-3<\/sup> = 4.8 g cm-3<\/sup>.<\/p>\n

            Question 15.
            \nLet us consider an equation \\(\\frac{1}{2}\\) mv2<\/sup> = mgh where m is the mass of the body, v its velocity, g is the acceleration due to gravity and h is the height. Check whether this equation is dimensionally correct.
            \nAnswer:
            \nThe dimensions of LHS are
            \n[M] [LT-1<\/sup>]2<\/sup> = [M] [L2<\/sup>T-2<\/sup>] = [M L2<\/sup>T-2<\/sup>]
            \nThe dimensions of RHS are
            \n[M] [LT-2<\/sup>] [L] = [M] [L2<\/sup>T-2<\/sup>] = [M L2<\/sup>T-2<\/sup>]
            \nThe dimensions of LHS and RHS are the same and hence the equation is dimensionally correct.<\/p>\n

            Question 16.
            \nThe SI unit of energy is J = kg m2<\/sup>s-2<\/sup>; that of speed v is ms-1 and of acceleration a is ms-2<\/sup>. Which of the formulae for kinetic energy (K) given below can you rule out on the basis of dimensional arguments (m stands for the mass of the body):
            \n(a) K = m2<\/sup>V3<\/sup>
            \n(b) K = (1\/2) mv2<\/sup>
            \n(c) K = ma
            \n(d) K = (3\/16) mv2<\/sup>
            \n(e) K = (1\/2) mv2<\/sup> + ma.
            \nAnswer:
            \nEvery correct formula or equation must have the same dimensions on both sides of the equation. Also, only quantities with the same physical dimensions can be added or subtracted. The dimensions of the quantity on the right side are [M2<\/sup> L3<\/sup> T-3<\/sup>] for (a) : [ML2<\/sup>T-2<\/sup>] for (b) and (d); [MLT-2<\/sup>] for (c). The quantity on the right side of (e) has no proper dimensions since two quantities of different dimensions have been added. Since the kinetic energy K has the dimensions of [ML2<\/sup>T-2<\/sup>], formulas (a), (c) and (e) are ruled out. Note that dimensional arguments cannot tell which of the two, (b) or (d), is the correct formula. For this, one must turn to the actual definition of kinetic energy. The correct formula for kinetic energy is given by (b).<\/p>\n

            \"AP<\/p>\n

            Question 17.
            \nConsider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length (l), mass of the bob (m) and acceleration due to gravity (g). Derive the expression for its period using method of dimensions.
            \nAnswer:
            \nThe dependence of time period T on the quantities l, g and m as a product may be written as :
            \nT = k lx gy<\/sup> mz<\/sup>
            \nWhere k is dimensionless constant and x, y and z are the exponents.
            \nBy considering dimensions on both sides, we have
            \n[L\u00b0M\u00b0T1<\/sup>] = [L1<\/sup>]x [L1<\/sup>T-2<\/sup>]y [M1<\/sup>]2<\/sup> = Lx+y<\/sup>T-2y<\/sup> Mz<\/sup>
            \nOn equating the dimensions on both sides, we have
            \nx + y = 0; -2y = 1 and z = 0
            \nSo that x = \\(\\frac{1}{2}\\), y = \\(\\frac{1}{2}\\), z = 0
            \nThen, T = kl1\/2<\/sup> g1\/2<\/sup> or T = \\(\\mathrm{k} \\sqrt{\\frac{l}{\\mathrm{~g}}}\\)
            \nNote that value of constant k can hot be obtained by the method of dimensions. Here it does not matter if some number multiplies the right side of this formula because that does not affect its dimensions.
            \nActually, k = 2\u03c0 so that T = \\(2 \\pi \\sqrt{\\frac{l}{g}}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

            Andhra Pradesh BIEAP\u00a0AP Inter 1st Year Physics Study Material 2nd Lesson Units and Measurements Textbook Questions and Answers. AP Inter 1st Year Physics Study Material 2nd Lesson Units and Measurements Very Short Answer Questions Question 1. Distinguish between accuracy and precision. [A.P. Mar. 16, 15; T.S. Mar. 15, 13] Answer: Accuracy The accuracy of a … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":39401,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[13],"tags":[],"jetpack_featured_media_url":"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/Chapter-2-Units-and-Measurements-1.png","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/27031"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=27031"}],"version-history":[{"count":2,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/27031\/revisions"}],"predecessor-version":[{"id":27107,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/27031\/revisions\/27107"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media\/39401"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=27031"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=27031"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=27031"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}