{"id":8200,"date":"2019-11-23T12:31:56","date_gmt":"2019-11-23T07:01:56","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=8200"},"modified":"2021-07-02T15:45:05","modified_gmt":"2021-07-02T10:15:05","slug":"1st-puc-physics-question-bank-chapter-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-physics-question-bank-chapter-2\/","title":{"rendered":"1st PUC Physics Question Bank Chapter 2 Units and Measurements"},"content":{"rendered":"

You can Download Chapter 2 Units and Measurements Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Physics Question Bank Chapter 2 Units and Measurements<\/h2>\n

1st PUC Physics Units and Measurements TextBook Questions and Answers<\/h3>\n

Question 1.
\n(a)
\nThe volume of a cube of side 1 cm is equal to ……….. m3<\/sup>
\nAnswer:
\nVolume of a cube = (side)3<\/sup>
\n= (1 cm)3<\/sup>
\n= (1 \u00d7 10-2<\/sup> m)3<\/sup>
\n= 1 \u00d7 10-6<\/sup> m3<\/sup><\/p>\n

(b)
\nThe surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ………. (mm)2<\/sup>
\nAnswer:
\nSurface area of a solid cylinder is given by S = 2\u03c0r (h+r)
\nWhere r = radius = 2 cm = 20 mm and
\nh = height = 10 cm = 100mm.
\n\u2234Surface area = 2 \u00d7 \u03c0 \u00d7 20 (100 + 20)
\n= 4800 \u00d7 3.14
\n= 1.5 \u00d7 104<\/sup> mm2<\/sup> .
\nNote: This rounding off is because there are only 2 significant figures in the measured value of the radius of the cylinder.<\/p>\n

(c)
\nA vehicle moving with a speed of 18 km h-1<\/sup> covers ……….. m in 1 s
\nAnswer:
\n18 km h-1<\/sup> = \\(\\frac{18 \\times 1000 m}{3600 s}\\) = 5 ms-1<\/sup>
\nNote: km h-1<\/sup> can be converted to ms-1<\/sup> by multiplying by 5\/18.<\/p>\n

(d)
\nThe relative density of lead is 11.3. Its density is ………. g cm-3<\/sup> or …….. kg m-3<\/sup>
\nAnswer:
\nActual density =
\nrelative density \u00d7 density of water.
\nDensity in g cm-3<\/sup> = 11.3 \u00d7 1 g cm-3<\/sup>
\nDensity in kg m-3<\/sup> = 11.3 \u00d7 1000 kgm-3<\/sup>
\n= 11300 kgm-3<\/sup><\/p>\n

Question 2.
\nFill in the blanks by suitable conversion of units
\n(a)
\n1 kg m2<\/sup> s-2<\/sup> = …………. g cm2<\/sup> s-2<\/sup>
\nAnswer:
\n1 Kg m2<\/sup>s-2<\/sup> = \\(\\frac{\\left(1 \\times 10^{3} \\mathrm{g}\\right) \\times\\left(1 \\times 10^{2} \\mathrm{cm}\\right)^{2}}{1 \\mathrm{s}^{2}}\\)
\n=1 \u00d7 107<\/sup>g cm2<\/sup>s-2<\/sup><\/p>\n

(b) 1 m =……….. ly
\nAnswer:
\n1 light year
\n= 3 \u00d7 108<\/sup> \u00d7 3600 \u00d7 24 \u00d7 365
\n= 1 \u00d7 1016<\/sup>m<\/p>\n

(c) 3.0 m s-2<\/sup> =……. km h-2<\/sup>
\nAnswer:
\n3.0 m s-2<\/sup> =\\(\\frac{3 \\times 10^{-3} k m}{(1 \/ 3600 h)^{2}}\\)
\n= 3.888 \u00d7 104<\/sup> km h-2<\/sup>
\n= 3.9 \u00d7 104<\/sup> km h-2<\/sup>
\n( expressed in 2 significant figures)<\/p>\n

(d)
\nG = 6.67 \u00d7 10-11<\/sup> N m2<\/sup> (kg)-2<\/sup> = ……. (cm)3<\/sup> s-2<\/sup> g-1<\/sup>.
\nAnswer:
\nG = 6.67 \u00d7 10-11<\/sup> Nm2<\/sup> (kg) -2<\/sup>
\n1 N = 1 kg m s-2<\/sup>
\n\u2234G = 6.67 \u00d7 10 = 6.67 \u00d7 10-11 \u00d7<\/sup>\u00a0(100 cm)3 <\/sup>s-2<\/sup> (1000g)-1<\/sup>
\n= 6.67 \u00d7 10-8<\/sup> (cm)3<\/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 \u03b1 kg, the unit of length equals \u03b2 m, the unit of time is \u03b3 s. Show that a calorie has a magnitude 4.2 \u03b11<\/sup> \u03b22<\/sup>\u03b32<\/sup> in terms of the new units.
\nAnswer:
\n1 new unit of mass = \u03b1 kg
\n\u21d2 1 kg = \\(\\frac{1}{\\alpha}\\) new units of mass
\n1 new unit of length = \u03b2 m
\n\u21d2 1m = \\(\\frac{1}{\\beta}\\) new units of length
\n1 new unit of time = \u03b3 s
\n\u21d2 1 s = \\(\\frac{1}{\\gamma}\\) new units of time
\n1 calorie = 4.2 kg m2<\/sup> s-2<\/sup>
\n= 4.2 \\(\\left(\\frac{1}{\\alpha}\\right)\\left(\\frac{1}{\\beta}\\right)^{2}\\left(\\frac{1}{\\gamma}\\right)^{-2}\\)
\n= 4.2 \\(\\alpha^{-1} \\beta^{-2} \\gamma^{2}\\) in terms of the new units.<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nExplain this statement clearly:
\n\u201cTo call a dimensional quantity \u2018large\u2019 or \u2018small\u2019 is meaningless without specifying a standard for comparison\u201d. In view of this, reframe the following statements wherever necessary:<\/p>\n

    \n
  1. Atoms are very small objects<\/li>\n
  2. A jet plane moves with great speed<\/li>\n
  3. The mass of Jupiter is very large<\/li>\n
  4. The air inside this room contains a large number of molecules<\/li>\n
  5. A proton is much more massive than an electron<\/li>\n
  6. The speed of sound is much smaller than the speed of light.<\/li>\n<\/ol>\n

    (Note: There is no unique answer for each of these questions.)<\/p>\n

    Answer:<\/p>\n

      \n
    1. Atoms are very small objects when compared to a cricket ball.<\/li>\n
    2. A jet plane moves with great speed when compared to a car.<\/li>\n
    3. The mass of Jupiter is much larger than that of the Earth.<\/li>\n
    4. The air inside this room contains a large number of molecules when compared to the number of objects in the room.<\/li>\n
    5. No change necessary.<\/li>\n
    6. No change necessary.<\/li>\n<\/ol>\n

      Question 5.
      \nA new unit of length is chosen such that the speed of light in vacuum is 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?
      \nAnswer:
      \nNew unit of length = 3 \u00d7 108<\/sup> ms-1<\/sup>
      \nDistance between the Earth and the sun
      \n= \\((8 \\min 20 \\mathrm{s}) \\times 3 \\times 10^{8} \\mathrm{ms}^{-1}\\)
      \n= 500 \u00d7 3 \u00d7 108<\/sup> ms-1<\/sup>
      \n\u2234Distance between the Earth and the Sun in terms of the new units
      \n=\\(\\frac{500 \\times 3 \\times 10^{8}}{3 \\times 10^{8}}\\)
      \n= 500 new units.<\/p>\n

      Question 6.
      \nWhich of the following is the most precise device for measuring length:<\/p>\n

        \n
      1. A vernier caliper with 20 divisions on the sliding scale<\/li>\n
      2. A screw gauge of pitch 1 mm and 100 divisions on the circular scale<\/li>\n
      3. An optical instrument that can measure length to within a wavelength of light?<\/li>\n<\/ol>\n

        Answer:<\/p>\n

          \n
        1. Least measurement possible with vernier callipers =1 \/ 20 mm = 5 \u00d7 10-5<\/sup> m<\/li>\n
        2. Least measurement possible with screw gauge = 1 \/ 100 mm = 1 \u00d7 10-5<\/sup> m<\/li>\n
        3. Least measurement possible with the optical instrument = 300 nm = 3 \u00d7 10-7<\/sup> m<\/li>\n<\/ol>\n

          Clearly (c) is the most precise device.<\/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?
          \nAnswer:
          \nEstimate on the thickness of the hair
          \n= \\(\\frac{\\text { width of the hair }}{\\text { magnification }}\\)
          \n= \\(\\frac{3.5 \\mathrm{mm}}{100}\\)
          \n= 0.035mm<\/p>\n

          Question 8.
          \n(a)
          \nYou are given a thread and a metre scale. How will you estimate the diameter of the thread ?
          \nAnswer:
          \nRepeatedly wind the thread over the breadth of the scale such that there is no space between adjacent windings. Find the number of windings in a given length.
          \nDiameter of the thread
          \n= \\(\\frac{\\text { length on the scale }}{\\text { Number of windings }}\\)<\/p>\n

          (b)
          \nA 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?
          \nAnswer:
          \nNo. When the number of divisions are arbitrarily increased, the accuracy might increase up to a certain extent only.<\/p>\n

          (c)
          \nThe 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?
          \nAnswer:
          \nWhen more measurements are taken, the likelihood of random errors is decreased. Hence a set of 100 measurements is more reliable than a set of 5 measurements.<\/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?
          \nAnswer:
          \nArea of the house = 1.75 cm2<\/sup>
          \nLet\u2019s assume that the house is square in shape.
          \nLength of a side of the house = \\(\\sqrt{1.75}\\) cm
          \nArea of the house on the screen 1.55 m2<\/sup>
          \nLength of a side of the house on the screen = \\(\\sqrt{1.55}\\)m
          \n= \\(\\sqrt{1.55}\\) \u00d7 100 cm
          \nLinear magnification= \\(\\frac{\\text { length on the screen }}{\\text { length on the slide }}\\)
          \n=\\(\\frac{\\sqrt{1.55} \\times 100 \\mathrm{cm}}{\\sqrt{1.75}}\\)
          \n=94.1<\/p>\n

          Question 10.
          \nState the number of significant figures in the following :<\/p>\n

            \n
          1. 0.007 m2<\/sup><\/li>\n
          2. 2.64 \u00d7 1024<\/sup> kg<\/li>\n
          3. 0.2370 g cm-3<\/sup><\/li>\n
          4. 6.320 J<\/li>\n
          5. 6.032 N m-2<\/sup><\/li>\n
          6. 0.0006032 m2<\/sup><\/li>\n<\/ol>\n

            Answer:<\/p>\n

              \n
            1. 0.007 m2<\/sup> = 7×10-3<\/sup>m2<\/sup> \u2192 1 significant figure.<\/li>\n
            2. 2.64 \u00d7 1024<\/sup> Kg \u2192 3 significant figures.<\/li>\n
            3. 0.2370 g cm-3<\/sup> \u2192 4 significant figures.<\/li>\n
            4. 6.320 J \u2192 4 significant figures.<\/li>\n
            5. 6.032 Nm-2<\/sup> \u2192 4 significant figures<\/li>\n
            6. 0.0006032 m2<\/sup> \u2192 6.032 \u00d7 10-4<\/sup> m2<\/sup> \u2192 4 significant figures.<\/li>\n<\/ol>\n

              \"KSEEB<\/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.
              \nAnswer:
              \nLength I = 4.234 m,
              \nBreadth b = 1.005 m,
              \nThickness t = 2.01 cm = 0.0201 m
              \nArea = 2 \u00d7 (lb + bt + It)
              \n= 2 \u00d7 (4.234 \u00d7 1.005 + 1.005 \u00d7 0.0201 + 4.234 \u00d7 0.0201)
              \n= 8.72 m2<\/sup>
              \n(Rounding off to 3 significant figures)
              \nVolume = Ibt
              \n= 4.234 \u00d7 1.005 \u00d7 0.0201
              \n= 8.55 \u00d7 10-2<\/sup> m3<\/sup><\/p>\n

              Question 12.
              \nThe mass of a box measured by a grocer\u2019s balance is 2.300 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is<\/p>\n

                \n
              • the total mass of the box,<\/li>\n
              • the difference in the masses of the pieces to correct significant figures?<\/li>\n<\/ul>\n

                Answer:<\/p>\n

                  \n
                1. Total mass of the box = 2.300 kg + 20.15 g + 20.17 g = 2.34032 kg .
                  \n= 2.34 kg.<\/li>\n
                2. Difference in the masses of the pieces = 20.17 g – 20.15 g.
                  \n= 0.02 g.<\/li>\n<\/ol>\n

                  Question 13.
                  \nA physical quantity \/Ms related to four observables a, b, c and of as follows :
                  \nP = a3<\/sup>b2<\/sup>\\(\/(\\sqrt{\\mathrm{cd}})\\)<\/p>\n

                  The 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?
                  \nAnswer:
                  \nP = a3<\/sup>b2<\/sup>\\(\/(\\sqrt{\\mathrm{c}} \\mathrm{d})\\)
                  \nPercentage error in P
                  \n\"1st<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 14.
                  \nA book with many printing errors contains four different formulas for the displacement y of a particle under going a certain periodic motion :
                  \n(a) y = a sin 2\u03c0t\/T
                  \n(b) y = a sin vt
                  \n(c) y = (a\/T) sin t\/a
                  \n(d) y = \\((a\\sqrt{2})\\) (sin 2\u03c0t\/T + cos 2\u03c0t\/T)
                  \n(a = maximum displacement of the particle, b a speed of the particle. T = time-period of motion). Rule out the wrong formulas on dimensional grounds.
                  \nAnswer:
                  \nHere we need the dimension of the LHS = dimension of RHL = L1<\/sup>
                  \nAlso, the dimension of the argument of the trigonometric function = dimensionless<\/p>\n

                  (a) Dimension of the argument of the sine function
                  \n\\(\\frac{T^{1}}{T^{1}}\\mathrm{T}^{0}\\) = dimensionless.
                  \nDimension of a = L1<\/sup> = dimensions of y.
                  \n\u2234 This formula is dimensionally correct.<\/p>\n

                  (b) Dimension of the argument of the sine function
                  \n= [LT-1<\/sup>] [T1<\/sup>] = L1<\/sup>
                  \n\u2234 This formula is dimensionally incorrect.<\/p>\n

                  (c) Dimension of the argument of the sine function
                  \n= t\/a = \\(\\frac{T}{L}\\) = [L-1<\/sup>] [T1<\/sup>] Also,
                  \ndimension of (\\(\\frac{a}{t}\\)) = \\(\\frac{L}{T}\\) = LT-1<\/sup> Clearly,
                  \n\u2234 This formula is dimensionally incorrect.<\/p>\n

                  (d) Dimension of \\(a\\sqrt{2}\\) = L1<\/sup> = dimension of y Dimension of the argument of the sine function
                  \n= dimension of the argument of the cosine function = 2\u03c0\\(\\frac{t}{T}\\) = \\(\\frac{T^{1}}{T^{1}}\\) = T\u00b0 = dimensionless.
                  \n\u2234 This formula is dimensionally correct.<\/p>\n

                  Question 15.
                  \nA famous relation in physics relates \u2018moving mass’ m to the \u2018rest mass\u2019 m0<\/sub> of a particle In terms of its speed and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant
                  \nc. He writes :m = \\(\\frac{m_{0}}{\\left(1-v^{2}\\right)^{1 \/ 2}}\\) Guess where to put the missing c.
                  \nAnswer:
                  \nWe need the dimension of the RHS to be equal to M1<\/sup>. Since m0<\/sub> has the dimension of M1<\/sup>, the denominator must be dimensionless.<\/p>\n

                  \u21d2 \\(\\left(1-v^{2}\\right)^{\\frac{y}{2}}\\) must be corrected to be dimensionless.
                  \n1 is dimensionless, v\u00b2 has the dimensions [L1<\/sup>T-1<\/sup>]2<\/sup> = L2<\/sup> T-2<\/sup><\/p>\n

                  Using c, we need to make 1 – v\u00b2 dimensionless. Since c has the same dimensions as v, c\/v and v\/c are both dimensionless.
                  \n\u2234 The corrected formula must be
                  \nm = \\(\\frac{m_{0}}{\\left(1-v^{2} \/ c^{2}\\right)^{1 \/ 2}}\\) or m = \\(\\frac{m_{0}}{\\left(1-c^{2} \/ v^{2}\\right)^{1\/2}}\\)<\/p>\n

                  Since m0<\/sub> is the rest mass, m = m0<\/sub> when v = o. (body at rest)
                  \nFor the first formula, when v = 0, m = undefined, which is wrong.
                  \n\u2234 Corrected formula is m = \\(\\frac{m_{0}}{\\left(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 by \u00c5: 1 \u00c5 = 10-10<\/sup> m. The size of a hydrogen atom is about 0.5 \u00c5. What is the total atomic volume in m3<\/sup> of a mole of hydrogen atoms?
                  \nAnswer:
                  \nRadius of a hydrogen atom = r = 0.5 \u00c5 = 0.5 \u00d7 10-10<\/sup>m
                  \nVolume of a mole of hydrogen atoms = Volume of a hydrogen atom \u00d7 Number of hydrogen atoms in 1 mole
                  \n= \\(\\frac{4}{3}\\)\u03c0r3<\/sup> \u00d7 NA<\/sub>
                  \n= \\(\\frac{4}{3}\\) \u00d7 3.14 \u00d7 ( 0.5 \u00d7 10-10<\/sup>)3<\/sup> \u00d7 6.022 \u00d7 1023<\/sup>
                  \n= 3 \u00d7 10-7<\/sup> m3<\/sup> (rounded off to 1 significant figure)<\/p>\n

                  Question 17.
                  \nOne mole of an ideal gas at standard temperature and pressure occupies 22.4 L (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 be about 1<\/p>\n

                  1. Why is this ratio so large?
                  \nAnswer:
                  \nAtomic volume of a mole of hydrogen
                  \n= \\(\\frac{4}{3} \\pi r^{3} \\times N_{A}\\)
                  \n= \\(\\frac{4}{3}\\) \u00d7 3.14 \u00d7 (1 \u00d7 10-10<\/sup>)3<\/sup> \u00d7 6.022 \u00d7 1023<\/sup>
                  \n= 2.52 \u00d7 10-6<\/sup> m3<\/sup>
                  \nRequired ratio
                  \n= \\(\\frac{22.4 L}{2.52 \\times 10^{-6} m^{3}}\\)
                  \n= \\(\\frac{22.4 \\times 10^{-3}}{2.52 \\times 10^{-6}} \\)
                  \n= 9 \u00d7 103<\/sup>
                  \nThis ratio is large because the intermolecular separation in a gas is much larger than the size of a molecule.<\/p>\n

                  Question 18.
                  \nExplain this common observation clearly: If you look out of the window of a fast-moving train, the nearby trees, houses, etc. seem to move rapidly in a direction opposite to the train\u2019s motion, but the distant objects (hilltops, 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).
                  \nAnswer:
                  \nObjects nearer to the eye subtend a greater angle in the eye than the distant objects. When we move, the change in this angle is less for distant objects than for near objects. So the distant objects seem stationary but nearer objects seem to move in the opposite direction.<\/p>\n

                  Question 19.
                  \nThe principle of \u2018parallax\u2019 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth\u2019s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth\u2019s 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 111<\/sup> (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 111<\/sup> (second) of arc from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres?
                  \nAnswer:
                  \nDistance of the earth from the Sun =
                  \n= b = 1\/2 \u00d7 diameter of earth’s orbit
                  \n\u21d2 b= 0.5 \u00d7 3 \u00d7 1011<\/sup> m
                  \nb = 1.5 \u00d7 1011<\/sup> m
                  \n1 parsec = D = b\/\u03b8
                  \n= \\(\\frac{1.5 \\times 10^{11} \\mathrm{m}}{1^{11}}\\) = \\(\\frac{1.5 \\times 10^{11}}{\\left(\\frac{1}{3600}\\right)}\\)
                  \n= \\(\\frac{1.5 \\times 10^{11}}{\\frac{1}{3600} \\times \\frac{\\pi}{180}}\\) = 3 \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 parsecs? How much parallax would this star (named Alpha Centaurl) show when viewed from two locations of the Earth six months apart in Its orbit around the Sun?
                  \nAnswer:
                  \nDistance of Alpha Centauri from the Earth = 4.29 ly
                  \n= 4.29 \u00d7 365 \u00d7 24 v 3600 \u00d7 3 \u00d7 108 <\/sup>= 4.06 \u00d7 1016<\/sup>m
                  \n1 parsec = 3.084 \u00d7 1016<\/sup> m
                  \n\u2234 Distance = \\(\\frac{4.06 \\times 10^{16}}{3.084 \\times 10^{16}}\\) = 1.32parsec.
                  \nD = b\/\u03b8
                  \nb = 3 \u00d7 104<\/sup>, D = 4.06 \u00d7 1016<\/sup> m
                  \nParallax \u03b8 = 2 \u00d7 D = 2.64<\/p>\n

                  Question 21.
                  \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 or 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.
                  \nAnswer:
                  \nActual Density of the Sun = \\(\\frac{\\text { Mass }}{\\text { Volume }}\\)
                  \n= \\(\\frac{2 \\times 10^{30} \\mathrm{kg}}{\\frac{4}{3} \\pi \\times\\left(7 \\times 10^{8}\\right)^{3} \\mathrm{m}^{3}}\\)
                  \n= 1.4 \u00d7 103<\/sup> kg m-3<\/sup><\/p>\n

                  Question 22.
                  \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.
                  \nAnswer:
                  \nD = 824.7 million km
                  \n\u03b8 = 35.7211<\/sup>
                  \nb = D\u03b8
                  \n=> b = 824.7 v 106<\/sup> km \u00d7 35.7211<\/sup>
                  \n824.7 \u00d7 106<\/sup> km \u00d7 \\(\\frac{35.72}{3600}\\) \u00d7 \\(\\frac{\\pi}{180}\\) km
                  \n=1.429 \u00d7 105<\/sup> km<\/p>\n

                  1st PUC Physics Units and Measurements Additional Exercises Questions and Answers<\/h3>\n

                  Question 23.
                  \nA man walking briskly In the rain with speed v must slant his umbrella forward making an angle \u03b8 with the vertical. A student derives the following relation between \u03b8 and v: tan \u03b8 = v, and checks that the relation has a correct limit: as v -> o, \u03b8 -&gt; o, 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.
                  \nAnswer:
                  \nSince tan \u03b8 is dimensionless, we need the RHS of the equation tan \u03b8 = v to be dimensionless.
                  \nLet the speed of the rainfall be v1<\/sup>. Then, by the method of dimensions, we get that tan \u03b8 = v\/v1<\/sup> or v1<\/sup>\/v
                  \nWhen the man is stationary (v = 0), he must hold the umbrella straight.
                  \ni.e \u03b8 = 0\u00b0
                  \ntan \u03b8 = 0 when v = 0
                  \n\u2234 The required relation is tan \u03b8 = v\/v1<\/sup><\/p>\n

                  \"KSEEB<\/p>\n

                  Question 24.
                  \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?
                  \nAnswer:
                  \nSince the maximum error possible for 2 clocks is 0.02 s, the error for 1 clock = 0.01 s.
                  \n\u2234 Accuracy = \\(\\frac{3600 \\times 24 \\times 365 \\times 100}{0.01}\\)
                  \n= 3.15 \u00d7 1011<\/sup>
                  \n= 1 part in to11<\/sup> to 1012<\/sup><\/p>\n

                  Question 25.
                  \nEstimate the average mass density of a sodium atom assuming its size to be about 2.5 \u00c5. (Use the known values of Avogadro\u2019s 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?
                  \nAnswer:
                  \nMass of 1 mole (NA<\/sub>) of sodium atoms = 23 g.
                  \nMass of 1 atom of sodium = \\(\\frac{23 \\mathrm{g}}{\\mathrm{N}_{\\mathrm{A}}}\\)
                  \n= \\(\\frac{23 g \\times 10^{-3} k g}{6.022 \\times 10^{23}}\\)
                  \n= 3.819 \u00d7 10-26<\/sup>
                  \nDensity of sodium atom = \\(\\frac{\\text { Mass }}{\\text { Volume }}\\)
                  \n= \\(\\frac{3.819 \\times 10^{-26}}{\\frac{4}{3} \\pi \\times\\left(2.5 \\times 10^{-10}\\right)^{3}}\\)
                  \n= 583.5 kg m-3<\/sup>
                  \n\"1st
                  \n970 kg m-3<\/sup> and 0.6 \u00d7 103<\/sup> kg m-3<\/sup>
                  \nare of the same order of magnitude because atoms are tightly packed in solid-state, so the atomic mass density is quite close to the density of the solid.<\/p>\n

                  Question 26.
                  \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 fm. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.
                  \nAnswer:
                  \nMass of 1 mole (NA<\/sub>) of atoms of an element = A g
                  \n= A \u00d7 10-3<\/sup> kg
                  \nMass of\u00a0 1 atom = \\(\\frac{A \\times 10^{-3} k g}{N_{A}}\\)
                  \nDensity of the atom = \\(\\frac{\\text { Mass }}{\\text { Volume }}\\)
                  \n\"1st
                  \n\"1st which is a constant.
                  \nMass density of the sodium nucleus
                  \n\"1st
                  \n\"1st
                  \n= 2.30 \u00d7 1017<\/sup> kg m -3<\/sup>
                  \n= 0.2 \u00d7 1018<\/sup> kg m -3<\/sup>
                  \n\\(\\frac{\\text { Nuclear density }}{\\text { Atomic density }}\\) for sodium
                  \n\"1st
                  \n= 0.33 \u00d7 1015<\/sup>
                  \n\u2234 Nuclear density is roughly 1015<\/sup> to 1015<\/sup> times the atomic density.<\/p>\n

                  Question 27.
                  \nA LASER is a source of very intense, monochromatic, and unidirectional beam of light. These properties of 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\u2019s surface. How much Is the radius of the lunar orbit around the Earth?
                  \nAnswer:
                  \nTime for the laser reach the moon
                  \n= 1\/2 \u00d7 2.56 s = 1.28 s
                  \n\u2234 Distance from the Earth to the moon = 1.28 s \u00d7 speed of light in vacuum = 1.28 \u00d7 3 \u00d7 108<\/sup>ms-1<\/sup>
                  \n= 3.84 \u00d7 108<\/sup>m
                  \n\u2234 The radius of the lunar orbit = 3.84 \u00d7 108<\/sup>m.<\/p>\n

                  Question 28.
                  \nA SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects underwater. 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
                  \nan enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of sound in water =
                  \n1450 ms-1<\/sup>).
                  \nAnswer:
                  \nTime for the SONAR to reach the enemy submarine = 1\/2 \u00d7 77.0 s = 38.5s
                  \nDistance of the enemy submarine = 38.5 s \u00d7 speed of Sound in water
                  \n= 38.5 s \u00d7 1450 m s-1<\/sup> = 55825 m = 55.8 km.<\/p>\n

                  question 29.
                  \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 puzzling 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?
                  \nAnswer:
                  \nDistance of the quasar from the earth = Distance travelled by light in 3 billion years
                  \n= 3 \u00d7 105<\/sup> kms-1<\/sup> \u00d7 3 \u00d7 109<\/sup> \u00d7 365 \u00d7 24 \u00d7 3600
                  \n= 2.8 \u00d7 1022<\/sup> km.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 30.
                  \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 .determine the approximate diameter of the moon.
                  \nAnswer:
                  \nAC = AD, AB = AE
                  \n\"1st<\/p>\n

                  From the data, we get the distances of the Sun and Moon from the Earth, as well as the diameter of the Sun.
                  \nUsing the concept of similar triangles
                  \n\u0394ABE ~ \u0394ACD \u21d2 \\(\\frac{A B}{A D}=\\frac{B E}{C D}\\)
                  \n\u21d2 BE = diameter of the moon
                  \n= CD \u00d7 \\(\\frac{\\mathrm{AB}}{\\mathrm{AD}}\\) =<\/p>\n

                  \"1st
                  \n= 3568 km.<\/p>\n

                  Question 31.
                  \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 constant 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?
                  \nAnswer:
                  \n[c] = [LT-1<\/sup>]
                  \n[e] = [IT]
                  \n[me<\/sub>] = [mp] = [M]
                  \n[G] = [M-1<\/sup> L3<\/sup> T2<\/sup>]
                  \n[\u2208o<\/sub>]= [M-1<\/sup> L-3<\/sup> T4<\/sup> I2<\/sup>]
                  \nBy the given condition
                  \n[c]a<\/sup> [e]b<\/sup> [me<\/sub>]c<\/sup>[mp]d<\/sup> [G]e<\/sup> [\u2208o<\/sub>]f<\/sup>\u00a0= [T]
                  \n[LT-1<\/sup>]a<\/sup> [IT]b<\/sup> [M]c<\/sup>[M]d<\/sup> [M-1<\/sup>L3<\/sup>T-2<\/sup>]e<\/sup> \u00d7 [M-1<\/sup> L-3<\/sup> T4<\/sup>I2<\/sup>]f<\/sup> = [T]
                  \n[Mc+d-e-f<\/sup> La+3e-3f<\/sup> T-a+b-2e+4f<\/sup> Ib+2f<\/sup>]
                  \n[M0<\/sup> L0<\/sup> T1<\/sup> I0<\/sup>]<\/p>\n

                  Comparing the powers of M, L, T and I, we get
                  \n\u21d2c + d-e – f\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0= 0\u00a0 \u00a0 \u2192(1)
                  \na + 3e – 3f\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 0\u00a0 \u00a0 \u2192(2)
                  \n-a + b -2e + 4f\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0= 1\u00a0 \u00a0 \u2192(3)
                  \nb + 2f\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0= 0\u00a0 \u00a0 \u2192(4)
                  \nWe have 4 equations but 6 unknowns. Hence we cannot arrive at a unique solution. But we can get the solutions by guessing the values for some of the unknowns.
                  \nLet a = -3, e = -1, f = -2 c = -2, Now we have d = -1 and b = 4
                  \n\u2234 The equation is of the form =
                  \n\"1st
                  \nwhere k is a constant.<\/p>\n

                  1st PUC Physics Units and Measurements One Mark Questions and Answers<\/h3>\n

                  Question 1.
                  \nWhat is a unit?
                  \nAnswer:
                  \nThe standard quantity in terms of which a physical quantity is measured is called the unit of that physical quantity.<\/p>\n

                  Question 2.
                  \nWhat are the fundamental quantities?
                  \nAnswer:
                  \nThe quantities, which are independent of other physical quantities are called fundamental quantities.<\/p>\n

                  Question 3.
                  \nWhat are derived quantities?
                  \nAnswer:
                  \nThe quantities, which are derived from fundamental quantities are called derived quantities.<\/p>\n

                  Question 4.
                  \nWhat is dimensional formula?
                  \nAnswer:
                  \nA formula in which a physical quantity is expressed in terms of fundamental quantities is called the dimensional formula.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 5.
                  \nWhat are the dimensions?
                  \nAnswer:
                  \nDimensions of a physical quantity are the powers to which the fundamental quantities must be raised to obtain the unit of a given physical quantity.<\/p>\n

                  Question 6.
                  \nState the principle of homogeneity of dimensions.
                  \nAnswer:
                  \nThe principle of homogeneity of dimensions states that the dimensional formula of all the terms in an equation must be the same.<\/p>\n

                  Question 7.
                  \nWrite the dimensional formula for velocity.
                  \nAnswer:
                  \nLT-1<\/sup>.<\/p>\n

                  Question 8.
                  \nWrite the dimensional formula for force.
                  \nAnswer:
                  \nMLT2<\/sup>.<\/p>\n

                  Question 9.
                  \nWrite the dimensional formula for power.
                  \nAnswer:
                  \nML2<\/sup>T3<\/sup>.<\/p>\n

                  Question 10.
                  \nName a physical quantity that has a unit but no dimension.
                  \nAnswer:
                  \nAngle. (Bangalore South, North 05)<\/p>\n

                  Question 11.
                  \nWrite the dimensions of the coefficient of viscosity?
                  \nAnswer:
                  \n[\u03b7] = [M1<\/sup>L-1<\/sup>T-1<\/sup>].
                  \n\u2234 Dimensions of co-efficient of viscosity are 1 in mass, -1 in length and -1 in time.<\/p>\n

                  Question 12.
                  \nName the SI unit of weight.
                  \nAnswer:
                  \nSI unit of weight is Newton (N).<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 13.
                  \nWrite the number of significant figures in each of the following measurements<\/p>\n

                    \n
                  1. 1.67 \u00d7 10-27<\/sup> kg<\/li>\n
                  2. 0.270 cm<\/li>\n<\/ol>\n

                    Answer:<\/p>\n

                      \n
                    1. Three significant figures.<\/li>\n
                    2. Three significant figures.<\/li>\n<\/ol>\n

                      Question 14.
                      \nName the S.I unit used to express the amount of substance.
                      \nAnswer:
                      \nMole.<\/p>\n

                      Question 15.
                      \nHow many newtons make 1 kg wt?
                      \nAnswer:
                      \n9.8 N = 1 kg wt.<\/p>\n

                      1st PUC Physics Units and Measurements Two Mark Questions and Answers<\/h3>\n

                      Question 1.
                      \nName the fundamental quantities.
                      \nAnswer:
                      \nMass, length, time, temperature, amount of substance, luminous intensity and electric current.<\/p>\n

                      Question 2.
                      \nWhat are derived quantities? Give examples.
                      \nAnswer:
                      \nThe quantities, which are derived from fundamental quantities, are called derived quantities. E.g: Velocity is derived from length and time, Charge is derived from current and time. Area is derived from length, Volume is derived from the length, etc.<\/p>\n

                      Question 3.
                      \nMention the applications of dimensional equations.
                      \nAnswer:
                      \nDimensional equations are used :<\/p>\n

                        \n
                      1. To check the correctness of an equation.<\/li>\n
                      2. To derive the relation between different physical quantities.<\/li>\n
                      3. To convert one system of units into another system.<\/li>\n<\/ol>\n

                        \"KSEEB<\/p>\n

                        Question 4.
                        \nUsing the method of dimension check the correctness of the equation, v2<\/sup> = u2<\/sup>+2as
                        \nAnswer:
                        \nv2<\/sup> = u2<\/sup>+2as, writing the dimensional formula for each of them
                        \n[LT-1<\/sup>]2<\/sup> = [LT-1<\/sup>]2<\/sup> + [LT-2<\/sup>] [L]
                        \n[IL2<\/sup>T-2<\/sup>] = [L2<\/sup> T-2<\/sup>] + [L2<\/sup> T-2<\/sup>]
                        \nAll the terms have the same dimensional formula & hence the Equation is dimensionally correct.<\/p>\n

                        Question 5.
                        \nFind the area of the circle of radius 3.458 cm up to correct significant figures.
                        \nAnswer:
                        \nA = \u03c0r\u00b2
                        \n= 3.142 \u00d7 3.4582<\/sup> = 37.57 cm2<\/sup><\/p>\n

                        Question 6.
                        \nThe radius of a solid sphere is measured to be 11.24 cm. What is the surface area of the sphere to appropriate significant figures?
                        \nAnswer:
                        \nr = 11.24 cm
                        \nSurface area = 4 \u03c0r\u00b2 = 4 \u00d7 3.142 \u00d7 (11.24)2<\/sup>
                        \n= 1588 cm2<\/sup><\/p>\n

                        Question 7.
                        \nIf A = (12.0 \u00b10.1) cm and
                        \nB = (8.5 \u00b1 0.5) cm, find:
                        \n(i) A + B and (ii) A – B
                        \nAnswer:
                        \n(i) A + B = (12.0 \u00b1 0.1) + (8.5 \u00b1 0.5)
                        \n= (20.5 \u00b10.6) cm
                        \n(ii) A-B = (12.0 \u00b1 0.1) – (8.5 \u00b10.5)
                        \n= (3.5 \u00b1 0.6) cm.<\/p>\n

                        \"KSEEB<\/p>\n

                        Question 8.
                        \nMagnitude of force F experienced by a certain object moving with speed V is given by F = Kv2, where K is a constant. Find the dimensions of K.
                        \nAnswer:
                        \nF =Kv2
                        \nTaking dimensions of each term, we get
                        \n[M L T-2] = [ K ] [L T-1<\/sup>]2<\/sup>
                        \n\u21d2 [K] = \\(\\frac{\\left[M L T^{-2}\\right]}{\\left[L^{2} T^{-2}\\right]}\\)
                        \n[K] = [ML-1<\/sup>]<\/p>\n

                        Question 9.
                        \nFor the determination of \u2018g\u2019 using a simple pendulum, measurenents of I and T are required. Error in the measurement of which of these will have a larger effect on the value of \u2018g\u2019 thus obtained and why? What is done to minimise this error?
                        \nAnswer:
                        \nThe formula for \u2018g\u2019 is
                        \ng = 4 \u03c0\u00b2 \\(\\frac{I}{\\mathrm{T}^{2}}\\)
                        \n\u2234 \\(\\frac{\\Delta g}{g}=\\frac{\\Delta I}{I}+2 \\frac{\\Delta T}{T}\\)
                        \nError in the measurement of T will have a larger effect on the obtained value of \u2018g\u2019. To minimise this error, a large number of measurements of T is taken.<\/p>\n

                        1st PUC Physics Units and Measurements FourFive Mark Questions and Answers<\/h3>\n

                        Question 1.
                        \nThe side (r) of a cube is measured as (11.3 \u00b1 0.1) cm. What is the volume of the cube?
                        \nAnswer:
                        \nVolume V = r3<\/sup>
                        \n\\(\\frac{\\Delta r}{r}\\) = \\(\\frac{0.1}{11.3}\\) = 0.009
                        \n\\(\\frac{\\Delta V}{V}\\)= 3 \u00d7 \\(\\frac{\\Delta r}{r}\\) = 0.027
                        \nV = (11.3)3<\/sup> = 1442.9 cm3<\/sup>
                        \n\u0394V = 1442.9 \u00d7 0.027 = 38.96 cm3<\/sup>
                        \nVolume of the cube = (1442.9 \u00b1 38.96)cm3<\/sup><\/p>\n

                        Question 2.
                        \nA drop of olive oil of radius 1 mm spreads into a circular film of diameter 40 cm on a water surface. Estimate the size of an oil molecule.
                        \nAnswer:
                        \nVolume of the oil is conserved. Radius of oil drop = R = 1 mm
                        \nRadius of the oil film on the water surface is r = 20 cm
                        \nThe thickness of the oil film formed represents the size of the oil molecule.
                        \n\\(\\frac{4}{3} \\pi \\mathrm{R}^{3}\\) = \u03c0r\u00b2 \u00d7 size of a molecule
                        \n\u21d2Size of a molecule = \\(\\frac{4}{3} \\times \\frac{R^{3}}{r^{2}}\\)
                        \n= \\(\\frac{4}{3} \\times \\frac{\\left(1 \\times 10^{-3}\\right)^{3}}{\\left(20 \\times 10^{-2}\\right)^{2}}\\)
                        \n= 0.33 \u00d7 10-7<\/sup> m<\/p>\n

                        Question 3.
                        \nA large fluid star oscillates in shape under the influence of its own gravitational field. Using dimensional analysis, find the expression for period of oscillation (T) in terms of radius of the star (R), mean density of the Fluid (p) and universal gravitational constant (G).
                        \nT \u03b1 Ra<\/sup> \\(\\rho\\)b<\/sup>\u00a0Gc<\/sup>
                        \nConsidering the dimensions of the terms,
                        \n[M0<\/sup> L0<\/sup> T1<\/sup>] = [L1<\/sup>]a<\/sup> [ML-3<\/sup>]b<\/sup> [M-1<\/sup> L3<\/sup> T-2<\/sup>]c<\/sup>
                        \n[M0<\/sup> L0<\/sup> T1<\/sup>] = [Mb-c<\/sup> La-3b+3c<\/sup> T-2c<\/sup>]
                        \nEquating the powers of M, L and T,
                        \nwe get b – c = 0
                        \na – 3b + 3c = 0
                        \n– 2c = 1
                        \n\u21d2b = c, a = 0,c = -1\/2
                        \n\u2234 T = K R\u00b0 \\(\\rho^{-x \/ 2} G^{-y \/ 2}\\) = \\(\\frac{\\mathrm{K}}{\\sqrt{\\rho \\mathrm{G}}}\\)
                        \nwhere K is a constant.<\/p>\n

                        Question 4.
                        \nThe length and breadth of a rectangle are (5.7 \u00b10.1) cm and (3.4 \u00b10.2) cm. Calculate the area of the rectangle with error limits.
                        \nAnswer:
                        \nI = (5.7 \u00b1 0.1) cm
                        \nb = (3.4 \u00b1 0.2) cm
                        \nArea = lb = 5.7 \u00d7 3.4 = 19.38 cm2<\/sup>
                        \n\\(\\frac{\\Delta A}{A}=\\frac{\\Delta I}{I}+\\frac{\\Delta b}{b}\\)
                        \n\u21d2 \u0394A = \\(\\left(\\frac{0.1}{5.7}+\\frac{0.2}{3.4}\\right)\\) \u00d7 19.4
                        \n= 1.5 cm2<\/sup>
                        \n\u2234 A = (19.4 \u00b1 1.5) cm2<\/sup><\/p>\n

                        Question 5.
                        \nFrom the bottom of a rock mountain, If the angle of elevation of its top increases from 30\u00b0 to 45\u00b0 on moving 100 m towards the rock In the horizontal direction through the base of the rock.
                        \nAnswer:
                        \n\u2206 In ABC, \\(\\frac{h}{x}\\) = tan 45\u00b0
                        \n\"1st
                        \nIn ABD, \\(\\frac{\\mathrm{h}}{\\mathrm{x}+100}\\)
                        \n= tan 30\u00b0
                        \n\u21d2 \\(\\frac{h}{h+100}=\\frac{1}{\\sqrt{3}}\\)
                        \n\u21d2 \\(\\mathrm{h}=\\frac{100}{(\\sqrt{3}-1)}\\)
                        \nh = 136.6 m<\/p>\n

                        Question 6.
                        \nIf the units of force, velocity and energy are 100 dyne, 10cm\/sec and 400 ergs respectively, what will be the units of mass, length and time?
                        \nAnswer:
                        \nF = 100 dyne = [M LT-2<\/sup>]
                        \nV = 10 cm s-1<\/sup> = [LT-1<\/sup>]
                        \nE = 400 ergs = [M L2<\/sup>T-2<\/sup>]
                        \n\"1st
                        \nL = 4 cm
                        \n[L T-1<\/sup>] = 10
                        \n\u21d24T-1<\/sup> = 10
                        \n\u21d2T = 0.4 sec
                        \n[M L T-2<\/sup>] = 100
                        \n\u21d2M 4 \u00d7 (0.4)-2<\/sup> = 100
                        \n\u21d2M = 4 gm<\/p>\n

                        Question 7.
                        \nIf the displacement of a body, = (200 \u00b1 5) m and time taken by it is t = (20 \u00b1 0.2) s, find the percentage error In the calculation of velocity.
                        \nAnswer:
                        \n\\(\\frac{\\Delta V}{V} \\times 100 \\%=\\frac{\\Delta S}{S} \\times 100 \\%+\\frac{\\Delta t}{t} \\times 100 \\%\\)
                        \n= \\(\\frac{5}{200}\\) \u00d7 100+\\(\\frac{0.2}{20}\\) \u00d7 100
                        \n= 2.5 + 1
                        \n= 3.5%<\/p>\n

                        Question 8.
                        \nUsing the principle of homogeneity of dimensions, find which of the following is correct.
                        \n\"1st
                        \nWhere T is the time period, G Is the gravitational constant, M is the mass and r is the radius of the orbit.
                        \nAnswer:
                        \nDimensions of LHS in each case =
                        \n\"1st<\/p>\n

                        Question 9.
                        \nA physical quantity Q is given by Q = \\(\\frac{A^{2} \\cdot B^{3 \/ 2}}{C^{4} \\cdot D^{\/ 2}}\\). The percentage error In A, B, C and D are 1%, 2%, 4% and 2% respectively. Find the percentage error in Q.
                        \nAnswer:
                        \n\"1st
                        \n= 22%<\/p>\n

                        1st PUC Physics Units and Measurements Questions from the previous question papers<\/h3>\n

                        Question 1.
                        \nCheck the correctness of the following equation.
                        \n1. v = u + at
                        \n2. s = ut + 1\/2 at2<\/sup>
                        \nAnswer:
                        \n1. Given equation is v = u + at , Writing the dimensional formula of each of them we get,
                        \n[LT-1<\/sup>] = [LT-1<\/sup>] + [LT2<\/sup>][T]
                        \n[LT-1<\/sup>] = [LT-1<\/sup>] + [LT-1<\/sup>]
                        \nAll the terms have the same dimensional formula and hence the equation is dimensionally correct.<\/p>\n

                        2. Given equation is s = ut + 1\/2 at2<\/sup>
                        \nWriting the dimensional formula of each of them we get,
                        \n[L] = [LT-1<\/sup>][T] + [LT-2<\/sup>][T2<\/sup>]
                        \n[L] = [L]+[L]
                        \nAll the terms have the same dimensional formula and hence the equation is dimensionally correct.<\/p>\n

                        \"KSEEB<\/p>\n

                        Question 2.
                        \nThe period of oscillation (T) of a simple pendulum Is found to depend on
                        \ni) mass of the pendulum (m)
                        \nii) length of the pendulum (I)
                        \niii) acceleration due to gravity (g).
                        \nDerive the relation between them.
                        \nAnswer:
                        \nLet T \u221dmx<\/sup> Iy<\/sup> gz<\/sup>
                        \nT = k mx<\/sup> Iy<\/sup> gz<\/sup> ……………….. (1)
                        \nwhere \u2018k\u2019 is constant of proportionality. Writing the dimensional formula of equation (1),
                        \n[T] = [M]x<\/sup>[L]y<\/sup>[LT-2<\/sup>]z
                        \n[M0<\/sup> L0<\/sup> T] = [Mx<\/sup>Ly+z<\/sup>T-2z<\/sup>]
                        \nComparing the dimensions we get,
                        \nx = 0, y + z = 0, -2z = 1
                        \nx = 0, y = – z , z = -1\/2
                        \nx = 0, y = 1\/2 , z = -1\/2
                        \nSubstituting in (1) we get,
                        \n\"1st
                        \nExperimetnally the value of constant ‘k’ is found to be 2\u03c0.
                        \n\"1st
                        \nQuestion 3.
                        \nConvert MKS unit of force Into the CGS unit. Or Relate newton and dyne. OR Show that 1 newton =105<\/sup> dyne by method of dimension.
                        \nAnswer:
                        \nLet MKS unit of force = n \u00d7 CGS unit of force
                        \n\"1st
                        \n\u2234 MKS unit of force = 105<\/sup> CGS unit of force i.e., 1 newton = 105<\/sup><\/p>\n

                        Question 4.
                        \nMention the limitations of dimensional analysis.
                        \nAnswer:
                        \nThe limitations of dimensional equations are:<\/p>\n

                          \n
                        • Even though a given equation is dimensionally correct, it is not possible to conclude that it is truly correct.<\/li>\n
                        • The constant of proportionality cannot be determined.<\/li>\n
                        • This method cannot be used for deriving equations involving trigonometric and logarithm functions.<\/li>\n
                        • This method cannot be used to derive equations where addition or subtraction is involved.<\/li>\n<\/ul>\n

                          Question 5.
                          \nCheck the correctness of the equation sn = u+\\(\\frac{a}{2}\\) (2n-1) by dimensional analysis, where the symbols have the usual meaning.
                          \nAnswer:
                          \nWriting the dimensional formula of each of them, we get
                          \nL1<\/sup> = LT-1<\/sup>+LT-2<\/sup>[T]
                          \nL1<\/sup> = LT-1<\/sup>+LT-1<\/sup>
                          \nSince all the terms do not have the same dimensional formula, the equation is dimensionally not correct.<\/p>\n

                          Question 6.
                          \nCheck the correctness of equation, E = mc2<\/sup> by dimensional analysis, where the symbols have their usual significance.
                          \nAnswer:
                          \nWriting the dimensional formula of each of them,
                          \nwe get M1<\/sup> L2<\/sup>T-2<\/sup> = M1<\/sup> L2<\/sup>T-2<\/sup>
                          \nAll the terms have the same dimensional formula & hence the equation is dimensionally correct.<\/p>\n

                          \"KSEEB<\/p>\n

                          Question 7.
                          \nDefine dimensions of a physical quantity. Write the dimensions of velocity, force, and energy.
                          \nAnswer:
                          \nDimensions of physical quantities are the powers to which the fundamental quantities, must be raised to obtain that physical quantity.
                          \n1) [velocity] = [M0<\/sup>L1<\/sup>T-1<\/sup>]
                          \nDimension of velocity in mass 0, in length 1, & in time -1.
                          \n2) [Force] = [M1<\/sup>L1<\/sup>T-2<\/sup>]
                          \nDimension of force in mass 1, in length 1 & in time -2.
                          \n3) [Energy] = [M1<\/sup>L2<\/sup>T-2<\/sup>]
                          \nDimension of energy in mass 1, in length 2 & in time -2.<\/p>\n

                          Question 8.
                          \nShow that 1 Joule = 107<\/sup>erg by the method of dimension.
                          \nAnswer:
                          \nLet MKS unit of work
                          \n=n \u00d7 CGS unit of work.
                          \n\"1st
                          \n\u2234 MKS unit of work = 107<\/sup> CGS unit of work.
                          \ni.e 1 J = 107<\/sup> erg.<\/p>\n

                          Question 9.
                          \nWrite a note on the SI system.
                          \nAnswer:
                          \nSI system represents the \u201cInternational system of units\u201d. It was adopted in 1971 at the general conference on Weights and Measures :
                          \nS.I. system has seven fundamental units. They are<\/p>\n

                            \n
                          1. meter for length,<\/li>\n
                          2. kilogram for mass<\/li>\n
                          3. second for time<\/li>\n
                          4. ampere for electric current<\/li>\n
                          5. Kelvin for temperature<\/li>\n
                          6. candela for luminous intensity<\/li>\n
                          7. mole for Amount of substance.<\/li>\n<\/ol>\n

                            There are 2 supplementary units, They are<\/p>\n

                              \n
                            1. radian for measurement of angle in a plane.<\/li>\n
                            2. steradian for measurement of angle in solid angle.<\/li>\n<\/ol>\n

                              Question 10.
                              \nWrite any two applications and any two limitations of dimensional analysis.
                              \nAnswer:
                              \nDimensional equations are used :<\/p>\n

                                \n
                              • To check the correctness of an equation.<\/li>\n
                              • To derive the relation between different physical quantities.<\/li>\n
                              • To convert one system of units into another system.<\/li>\n<\/ul>\n

                                The limitations of dimensional equations are:<\/p>\n

                                  \n
                                • Even though a given equation is dimensionally correct, it is not possible to conclude that it is truly correct.<\/li>\n
                                • The constant of proportionality cannot be determined.<\/li>\n
                                • This method cannot be used for deriving equations involving trigonometric and logarithm functions.<\/li>\n
                                • This method cannot be used to derive equations where addition or subtraction is involved.<\/li>\n<\/ul>\n

                                  Question 11.
                                  \nWrite the dimensions for
                                  \ni) force
                                  \nii) pressure
                                  \niii) volume and
                                  \niv) velocity.
                                  \nAnswer:
                                  \ni) [F] = [MLT-2<\/sup>].
                                  \nii) [p] = [ML-1<\/sup>T-2<\/sup>].
                                  \niii) [V] = [M0<\/sup>L3<\/sup>T0<\/sup>].
                                  \niv) [V] = [M0<\/sup>LT-1<\/sup>].<\/p>\n","protected":false},"excerpt":{"rendered":"

                                  You can Download Chapter 2 Units and Measurements Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 1st PUC Physics Question Bank Chapter 2 Units and Measurements 1st PUC Physics Units and Measurements TextBook Questions …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/8200"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=8200"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/8200\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=8200"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=8200"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=8200"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}