{"id":37001,"date":"2022-12-01T08:27:56","date_gmt":"2022-12-01T02:57:56","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=37001"},"modified":"2022-12-01T08:27:56","modified_gmt":"2022-12-01T02:57:56","slug":"ts-inter-1st-year-physics-study-material-chapter-2","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/","title":{"rendered":"TS Inter 1st Year Physics Study Material Chapter 2 Units and Measurements"},"content":{"rendered":"

Telangana TSBIE\u00a0TS Inter 1st Year Physics Study Material<\/a> 2nd Lesson Units and Measurements Textbook Questions and Answers.<\/p>\n

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

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

Question 1.
\nDistinguish between accuracy and precision. [AP Mar. ’15, ’16, May 16, 13, June 15; TS May ’18, Mar. ’15]
\nAnswer:
\nAccuracy :
\nIt indicates the closeness of a measurement to the true value of given quantity.
\n\u2192 If the measurement is nearer to true value then accuracy is more.<\/p>\n

Precision :
\nPrecision of a measuring instrument depends on the limit (or) resolution of the quantity measured with that instrument.
\n\u2192 If the least measurable value is less, then precision is more for that instrument.<\/p>\n

Question 2.
\nWhat are the different types of errors that can occur in a measurement?
\nAnswer:
\nTypes of errors 1) Systematic errors and 2) Random errors.<\/p>\n

Systematic errors are again divided into 1) Imperfectional errors 2) Environmental errors and 3) Personal errors.<\/p>\n

Question 3.
\nHow can systematic errors be minimised or eliminated? [AP May ’17; TS Mar. ’17; AP May ’17; TS Mar. ’17]
\nAnswer:
\nSystematic errors can be minimised<\/p>\n

    \n
  1. by improving experimental techniques,<\/li>\n
  2. by selecting better instruments,<\/li>\n
  3. by taking mean value of number of readings and<\/li>\n
  4. by removing personal errors as far as possible.<\/li>\n<\/ol>\n

    Question 4.
    \nIllustrate how the result of a measurement is to be reported indicating the error involved.
    \nAnswer:
    \nSuppose length of an object is measured with a metre rod with least count equal to 0.1 cm. If the measured length is 62.5 cm, it has to be recorded as (62.5 \u00b1 0.1) cm, stating the limits of error. Similarly, suppose time period of a pendulum is measured to be 2.0 sec, using a stopwatch of least count 0.1 sec, it has to be recorded as (2.0 \u00b1 0.1) sec. It indicates that time period is in the range of 1.9 sec and 2.1 sec.<\/p>\n

    \"TS<\/p>\n

    Question 5.
    \nWhat are significant figures and what do they represent when reporting the result of a measurement? [TS Mar. ’18]
    \nAnswer:
    \nSignificant figures represents all practically measured digits plus one uncertain digit at the end.<\/p>\n

    When a result is reported in this way we can know up to what extent the value is reliable and also the amount of uncertainty in that reported value.<\/p>\n

    Question 6.
    \nDistinguish between fundamental units and derived units. [TS Mar. ’16 ; AP May ’14]
    \nAnswer:
    \n1) Fundamental units are used to measure fundamental quantities. Derived units are used to measure derived quantities.<\/p>\n

    2) Fundamental units are independent. Derived units are obtained by the combination of Fundamental units.
    \nEx: Metre is fundamental unit of length L’. metre\/sec is derived unit of velocity which is a combination of fundamental unit metre and second.<\/p>\n

    Question 7.
    \nWhy do we have different units for the same physical quantity? [TS May ’16. June ’15]
    \nAnswer:
    \nTo measure the same physical quantity we have different units by keeping magnitude of the quantity to be measured.
    \nExample:<\/p>\n

      \n
    1. The measure astronomical distances we will use light year.
      \n1 light year = 9.468 \u00d7 1015<\/sup>m.<\/li>\n
    2. To measure atomic distances we will use Angstrom A (or) Fermi.<\/li>\n<\/ol>\n

      Question 8.
      \nWhat is dimensional analysis?
      \nAnswer:
      \nDimensional analysis is a tool to check the relations among physical quantities by using their dimensions.
      \nDimensional analysis is generally used to check the correctness of derived equations.<\/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 atom = 10-10<\/sup> m,
      \nSize of atomic nucleus = 10-14<\/sup>m.
      \nSize of atom \u00f7 size of nucleus is \\(\\frac{10^{-10}}{10^{-14}}\\) = 104<\/sup>
      \n\u2234 Size of atom is 104<\/sup> times greater than size of nucleus.<\/p>\n

      Question 10.
      \nExpress unified atomic mass unit in kg. [TS Mar. ’19]
      \nAnswer:
      \nBy definition,
      \n1 a.m.u. = \\(\\frac{1}{12}\\) \u00d7 mass of an atom of 12<\/sup>6<\/sub>C
      \n\"TS<\/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:
      \nLeast count of Vernier Callipers = 1 MSD – 1 VSD
      \n\u2234 L.C. = 1 MSD – \\(\\frac{49}{50}\\)MSD = \\(\\frac{1}{50}\\)MSD
      \n= \\(\\frac{1}{50}\\) \u00d7 0.5 = 0.01 m.m<\/p>\n

      Question 2.
      \nIn a system of units, the unit of force is 100N, unit of length is 10m and the unit of time is 100s. What is the unit of mass in this system?
      \nAnswer:
      \nHere, F = MLT-2<\/sup> = 100 N \u2192 (1) ;
      \nL = 10 m ; T = 100s
      \n\u2234 From equation (1)
      \nM \u00d7 (10) \u00d7 (100)-2<\/sup> = 100 \u21d2 M \u00d7 10-3<\/sup> = 100
      \n\u21d2 M = \\(\\frac{100}{10^{-3}}\\) \u21d2 M = 105<\/sup> kg.<\/p>\n

      \"TS<\/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:
      \nSize of galaxy = 1025<\/sup>m,
      \nVelocity of light, c = 3 \u00d7 108<\/sup> ms-1<\/sup>
      \nTime taken by light to reach earth,
      \n\"TS<\/p>\n

      While calculating the order of magnitude we will consider the powers of Ten only.
      \nSo order of magnitude of time taken by light to reach earth from galaxy is 1017 seconds.<\/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:
      \nEarth moon distance, D = 60r.
      \nDiameter of earth, b = 2r
      \n\"TS<\/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:
      \nPrecision is the least measurable value with that instrument in our case precision is \u00b10.1 sec.<\/p>\n

      Calculation of accuracy :
      \nAverage value of measurements
      \n\"TS<\/p>\n

      Error in each measurement =
      \n\"TS<\/p>\n

      Precision \u00b1 1 sec. In these measurements only two significant figures are believable. 3rd one is uncertain.<\/p>\n

      Adjustment of \u2206amean<\/sub> upto given significant figure = 0.156 adjusted to 0.2.<\/p>\n

      So our value is accurate upto \u00b10.2<\/p>\n

      So our result is 39.67 \u00b1 0.2, when significant figures taken into account it is 39.7 \u00b1 0.2 sec.<\/p>\n

      Question 6.
      \n1 calorie = 4.2 J where 1J = 1 kg m\u00b2s-2<\/sup>. Suppose we employ a system of units in which the unit of mass is \u03b1 kg, the unit of length is \u03b2 m and the unit of time \u03b3 s, show that a calorie has a magnitude 4.2 \u03b1-1<\/sup> \u03b2-2<\/sup> \u03b3\u00b2 in the new system.
      \nAnswer:
      \nHere, 1 calorie = 4.2 J = 4.2 kg m\u00b2 \/ s\u00b2 \u2192 (1)
      \nAs new unit of mass = \u03b1 kg
      \n\u2234 1 kg = \\(\\frac{1}{\\alpha}\\) new unit of mass
      \n\u21d2 \u03b1-1<\/sup> new unit of mass<\/p>\n

      Similarly lm = \u03b2-1<\/sup> new unit of length and 1s = \u03b3-1<\/sup> new unit of time<\/p>\n

      Putting these values in (1) we get
      \n1 calorie = 4.2 (\u03b1-1<\/sup> new unit of mass) (\u03b2-1<\/sup> new unit of length)\u00b2 (\u03b3-1<\/sup> new unit of time)-2<\/sup>
      \n= 4.2 \u03b1-1<\/sup> \u03b2-2<\/sup> \u03b3\u00b2 new unit of energy, which was proved.<\/p>\n

      Question 7.
      \nA new unit of length is chosen so that the speed of light in vacuum is 1 ms-2<\/sup>. If light takes 8 min and 20s to cover this distance, what is the distance between the Sun and Earth in terms of the new unit?
      \nAnswer:
      \nGiven that velocity of light in vacuum,
      \nc = 1 new unit of length s-1<\/sup>
      \nTime taken by light of Sun to reach Earth, t = 8 min 20s = 8 \u00d7 60 + 20 = 500 s
      \n\u2234 Distance between the Sun and Earth, x = c \u00d7 t
      \n= 1 new unit of length s-1<\/sup> \u00d7 500s
      \n= 500 new units of length.<\/p>\n

      \"TS<\/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:
      \n\"TS
      \n\u2234 Thickness of hair is 0.035 mm<\/p>\n

      Question 9.
      \nA physical quantity X is related to four mea-surable quantities a, b, c and d as follows:
      \nX = a\u00b2b\u00b3C5\/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:
      \nHere, X = a\u00b2b\u00b3C5\/2<\/sup>d-2<\/sup>
      \n\"TS
      \nThe percentage error in X is \u00b1 23.5 %<\/p>\n

      Question 10.
      \nThe velocity of a body is given by v = At\u00b2 + Bt + C. If v and t are expressed in SI, what are the units of A, B and C?
      \nAnswer:
      \nFrom principle of Homogeneity the terms At\u00b2, Bt and C must have same dimensional formula of velocity ‘v’.
      \nv = Velocity = LT-1<\/sup> \u21d2 CT-1<\/sup> = A [T\u00b2]
      \n\u2234 A = \\(\\frac{LT^{-1}}{T^2}\\) = LT-3<\/sup> . So unit of A is m\/sec\u00b3
      \nLT-1<\/sup> = BT \u21d2 B = LT-2<\/sup> So unit of B is m\/sec\u00b2
      \nLT-1<\/sup> = C So unit of C is m\/sec.<\/p>\n

      Dimensional formulae of physical quantities
      \n\"TS \"TS
      \n\"TS<\/p>\n

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

      Question 1.
      \nIn the expression P = El\u00b2 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 dimensionless, quantity.
      \nSolution:
      \nHere, P = El\u00b2 m-5<\/sup> G-2<\/sup>
      \nHere,
      \nI = energy,
      \nl = angular momentum
      \nm = mass
      \nG = gravitational constant
      \n= [M L\u00b2T-2<\/sup>][ML\u00b2T-1<\/sup>]\u00b2 [M]-5<\/sup> [M-1<\/sup>L\u00b3T-2<\/sup>]-2<\/sup>
      \n= M1+2+5+2<\/sup> L2+4-6<\/sup> T-2-2+4<\/sup>
      \nP = [M\u00b0 L\u00b0 T\u00b0]
      \nHence, P is a dimensionless quantity.<\/p>\n

      \"TS<\/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:
      \nHere, c = [L T-1<\/sup>] ; h = [ML\u00b2T-1<\/sup>]
      \nG = [M-1<\/sup>L\u00b3T-2<\/sup>]
      \n\"TS
      \nApplying the principle of homogeneity of dimensions, we get<\/p>\n

      y – z = 1 \u2192 (2) ; x + 2y + 3z = 0 \u2192 (3) ; – x – y – 2z = 0 \u2192 (4)
      \nAdding eq. (2), eq. (3) and eq. (4),
      \n2y – 1 \u21d2 y = \\(\\frac{1}{2}\\)
      \n\u2234 From eq. (2) z = y – 1 = \\(\\frac{1}{2}\\) – 1 = \\(\\frac{-1}{2}\\)
      \nFrom eq. (4) x = -y – 2z = \\(\\frac{-1}{2}\\) + 1 = \\(\\frac{1}{2}\\)
      \nSubstituting the values of x, y & z in eq. (1), we get
      \n\"TS<\/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:
      \nGiven that
      \nT\u00b2 \u221d r\u00b3 or T \u221d r3\/2<\/sup> Also T is a function of g and R
      \nLet T \u221d r3\/2<\/sup> ga<\/sup> Rb<\/sup> where a, b are the dimen\u00acsions of g and R.
      \n(or) T = k r3\/2<\/sup> ga<\/sup> Rb<\/sup> \u2192 (1)
      \nwhere k is dimensionless constant of proportionality
      \nFrom equation (1)
      \n\"TS
      \nApplying the principle of homogeneity of dimensions, we get<\/p>\n

      a + b + \\(\\frac{3}{2}\\) = 0 \u2192 (2) \u2234 -2a = 1 \u21d2 a = \\(\\frac{-1}{2}\\)
      \nFrom eq (1), \\(\\frac{-1}{2}\\) + b + \\(\\frac{3}{2}\\) = 0 \u21d2 b = -1
      \nSubstituting the values of a’ and b’ in eq. (1), we get
      \n\"TS
      \nThis is the required relation.<\/p>\n

      Question 4.
      \nState the number of significant figures in the following
      \na) 6729 b) 0.024 c) 0,08240 d) 6.032 e) 4.57 x 108
      \nSolution:
      \na) In 6729 all are significant figures.
      \n\u2234 Number of significant figures Four.<\/p>\n

      b) In 0.024 the zeroes to the left of 1st non-zero digit of a number less than one are not significant.
      \n\u2234 Number of significant figures Two.<\/p>\n

      c) 0.08240 – Significant figures Four.<\/p>\n

      d) In 6.032 the zero between two non-zero digits is significant.
      \nSo, number of significant figures in 6.023 are 4.<\/p>\n

      e) 4.57 \u00d7 108<\/sup> – Significant figures Three. [In the representation of powers of Ten our rule is only significant figures must be given].<\/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 and to what is the total length? If the two sticks are placed side by side, what is the difference in their lengths?
      \nSolution:
      \na) When placed end to end total length is l = l1<\/sub> + l2<\/sub>
      \nl1<\/sub> = 12.132 cm and l2<\/sub> = 12.4 cm.
      \n\u2234 l1<\/sub> + l2<\/sub> = 12.132 + 12.4 = 24.532 cm.
      \n\"TS<\/p>\n

      In addition final answer must have least number of significant numbers in that addition, i.e., one after decimal point.
      \nSo our answer is 24.5 cm. b) For difference use l1<\/sub> – l2<\/sub>
      \ni.e., 12.4- 12.132 = 0.268 cm.
      \nIn subtraction final answer must be adjusted to least number of significant figures in that operation.<\/p>\n

      Here least number is one digit after decimal. By applying round off procedure our answer is 0.3 cm.<\/p>\n

      \"TS<\/p>\n

      Question 6.
      \nEach side of a cube is measured to be 7.203 m. What is (i) the total surface area and (ii) the volume of the cube, to appropriate significant figures?
      \nSolution:
      \nSide of cube, a = 7.203 m.
      \nSo number of significant figures are Four.
      \ni) Surface area of cube = 6a\u00b2
      \n= 6x 7.203 \u00d7 7.203 = 311.299<\/p>\n

      But our final answer must be rounded to least number of significant figures is four digits.
      \nSo surface area of cube = 311.3 m\u00b2
      \nii) Volume of cube, V = a\u00b3 = (7.203)\u00b3
      \n= 373.147
      \nBut the answer must be limited to Four significant figures.
      \n\u2234 Volume of sphere, V = 373.1 m\u00b3.<\/p>\n

      Question 7.
      \nThe measured mass and volume of a body are 2.42 g and 4.7 cm\u00b3 respectively with possible errors 0.01 g and 0.1 cm\u00b3. Find the maximum error in density.
      \nSolution:
      \nMass, m = 2.42 g ; Error, \u2206m = 0.01 g.
      \nVolume, V = 4.7 cm\u00b3, Error, \u2206V = 0.1 cc.
      \n\"TS
      \nMaximum % error in density = % error in mass + % error in volume Maximum percentage error in density
      \n= \\(\\frac{1}{2.42}+\\frac{10}{4.7}\\) = 0.413 + 2.127 = 2.54%<\/p>\n

      Question 8.
      \nThe error in measurement of radius of a sphere is 1%. What is the error in the measurement of volume? [AP Mar. ’19]
      \nSolution:
      \n\"\"<\/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:
      \nPercentage change in mass = \\(\\frac{\\Delta \\mathrm{m}}{\\mathrm{m}}\\) \u00d7 100 = 2%
      \n\"TS<\/p>\n

      \"TS<\/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 1\u00c5, what is the ratio of molar volume to the atomic volume of a mole of hydrogen?
      \nSolution:
      \nSize of Hydrogen atom \u2248 1\u00c5 = 10-10<\/sup> m = 10-8<\/sup>cm
      \nV1<\/sub> = Atomic volume = number of atoms \u00d7 volume of atom.
      \nOne mole gas contains n’ molecules.
      \nAvogadro Number, n = 6.022 \u00d7 1023<\/sup>
      \n\"TS
      \nV2<\/sub> = Molar volume of 1 mol. gas = 22.4 lit
      \n= 2.24 \u00d7 104<\/sup>C.C
      \n\u2235 1 lit = 1000 c.c.
      \n\u2234 Ratio of molar volume to atomic volume = V2<\/sub> : V1<\/sub>
      \n= 2.24 \u00d7 104<\/sup> : 2.523 \u2245 104<\/sup> m.<\/p>\n","protected":false},"excerpt":{"rendered":"

      Telangana TSBIE\u00a0TS Inter 1st Year Physics Study Material 2nd Lesson Units and Measurements Textbook Questions and Answers. TS Inter 1st Year Physics Study Material 2nd Lesson Units and Measurements Very Short Answer Type Questions Question 1. Distinguish between accuracy and precision. [AP Mar. ’15, ’16, May 16, 13, June 15; TS May ’18, Mar. ’15] … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[27],"tags":[],"yoast_head":"\nTS Inter 1st Year Physics Study Material Chapter 2 Units and Measurements - TS Board Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"TS Inter 1st Year Physics Study Material Chapter 2 Units and Measurements - TS Board Solutions\" \/>\n<meta property=\"og:description\" content=\"Telangana TSBIE\u00a0TS Inter 1st Year Physics Study Material 2nd Lesson Units and Measurements Textbook Questions and Answers. TS Inter 1st Year Physics Study Material 2nd Lesson Units and Measurements Very Short Answer Type Questions Question 1. Distinguish between accuracy and precision. [AP Mar. ’15, ’16, May 16, 13, June 15; TS May ’18, Mar. ’15] ... Read more\" \/>\n<meta property=\"og:url\" content=\"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/\" \/>\n<meta property=\"og:site_name\" content=\"TS Board Solutions\" \/>\n<meta property=\"article:published_time\" content=\"2022-12-01T02:57:56+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/tsboardsolutions.com\/wp-content\/uploads\/2022\/10\/TS-Board-Solutions.png\" \/>\n<meta name=\"author\" content=\"Srinivas\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Srinivas\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"11 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/\"},\"author\":{\"name\":\"Srinivas\",\"@id\":\"https:\/\/tsboardsolutions.com\/#\/schema\/person\/1d1aa14a8abcb28fa5e85127be78dcea\"},\"headline\":\"TS Inter 1st Year Physics Study Material Chapter 2 Units and Measurements\",\"datePublished\":\"2022-12-01T02:57:56+00:00\",\"dateModified\":\"2022-12-01T02:57:56+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/\"},\"wordCount\":2230,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/tsboardsolutions.com\/#organization\"},\"articleSection\":[\"TS Inter 1st Year\"],\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/\",\"url\":\"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/\",\"name\":\"TS Inter 1st Year Physics Study Material Chapter 2 Units and Measurements - TS Board Solutions\",\"isPartOf\":{\"@id\":\"https:\/\/tsboardsolutions.com\/#website\"},\"datePublished\":\"2022-12-01T02:57:56+00:00\",\"dateModified\":\"2022-12-01T02:57:56+00:00\",\"breadcrumb\":{\"@id\":\"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/tsboardsolutions.com\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"TS Inter 1st Year Physics Study Material Chapter 2 Units and Measurements\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/tsboardsolutions.com\/#website\",\"url\":\"https:\/\/tsboardsolutions.com\/\",\"name\":\"TS Board Solutions\",\"description\":\"Telangana TS Board Textbook Solutions for Class 6th, 7th, 8th, 9th, 10th, Inter 1st & 2nd Year\",\"publisher\":{\"@id\":\"https:\/\/tsboardsolutions.com\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/tsboardsolutions.com\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"Organization\",\"@id\":\"https:\/\/tsboardsolutions.com\/#organization\",\"name\":\"TS Board Solutions\",\"url\":\"https:\/\/tsboardsolutions.com\/\",\"logo\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/tsboardsolutions.com\/#\/schema\/logo\/image\/\",\"url\":\"https:\/\/tsboardsolutions.com\/wp-content\/uploads\/2022\/10\/cropped-TS-Board-Solutions-1.png\",\"contentUrl\":\"https:\/\/tsboardsolutions.com\/wp-content\/uploads\/2022\/10\/cropped-TS-Board-Solutions-1.png\",\"width\":488,\"height\":40,\"caption\":\"TS Board Solutions\"},\"image\":{\"@id\":\"https:\/\/tsboardsolutions.com\/#\/schema\/logo\/image\/\"}},{\"@type\":\"Person\",\"@id\":\"https:\/\/tsboardsolutions.com\/#\/schema\/person\/1d1aa14a8abcb28fa5e85127be78dcea\",\"name\":\"Srinivas\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/tsboardsolutions.com\/#\/schema\/person\/image\/\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/10a27cfafdf21564c686b80411336ece?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/10a27cfafdf21564c686b80411336ece?s=96&d=mm&r=g\",\"caption\":\"Srinivas\"},\"url\":\"https:\/\/tsboardsolutions.com\/author\/srinivas\/\"}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"TS Inter 1st Year Physics Study Material Chapter 2 Units and Measurements - TS Board Solutions","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/","og_locale":"en_US","og_type":"article","og_title":"TS Inter 1st Year Physics Study Material Chapter 2 Units and Measurements - TS Board Solutions","og_description":"Telangana TSBIE\u00a0TS Inter 1st Year Physics Study Material 2nd Lesson Units and Measurements Textbook Questions and Answers. TS Inter 1st Year Physics Study Material 2nd Lesson Units and Measurements Very Short Answer Type Questions Question 1. Distinguish between accuracy and precision. [AP Mar. ’15, ’16, May 16, 13, June 15; TS May ’18, Mar. ’15] ... Read more","og_url":"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/","og_site_name":"TS Board Solutions","article_published_time":"2022-12-01T02:57:56+00:00","og_image":[{"url":"https:\/\/tsboardsolutions.com\/wp-content\/uploads\/2022\/10\/TS-Board-Solutions.png"}],"author":"Srinivas","twitter_card":"summary_large_image","twitter_misc":{"Written by":"Srinivas","Est. reading time":"11 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/#article","isPartOf":{"@id":"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/"},"author":{"name":"Srinivas","@id":"https:\/\/tsboardsolutions.com\/#\/schema\/person\/1d1aa14a8abcb28fa5e85127be78dcea"},"headline":"TS Inter 1st Year Physics Study Material Chapter 2 Units and Measurements","datePublished":"2022-12-01T02:57:56+00:00","dateModified":"2022-12-01T02:57:56+00:00","mainEntityOfPage":{"@id":"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/"},"wordCount":2230,"commentCount":0,"publisher":{"@id":"https:\/\/tsboardsolutions.com\/#organization"},"articleSection":["TS Inter 1st Year"],"inLanguage":"en-US","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/#respond"]}]},{"@type":"WebPage","@id":"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/","url":"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/","name":"TS Inter 1st Year Physics Study Material Chapter 2 Units and Measurements - TS Board Solutions","isPartOf":{"@id":"https:\/\/tsboardsolutions.com\/#website"},"datePublished":"2022-12-01T02:57:56+00:00","dateModified":"2022-12-01T02:57:56+00:00","breadcrumb":{"@id":"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-2\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/tsboardsolutions.com\/"},{"@type":"ListItem","position":2,"name":"TS Inter 1st Year Physics Study Material Chapter 2 Units and Measurements"}]},{"@type":"WebSite","@id":"https:\/\/tsboardsolutions.com\/#website","url":"https:\/\/tsboardsolutions.com\/","name":"TS Board Solutions","description":"Telangana TS Board Textbook Solutions for Class 6th, 7th, 8th, 9th, 10th, Inter 1st & 2nd Year","publisher":{"@id":"https:\/\/tsboardsolutions.com\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/tsboardsolutions.com\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Organization","@id":"https:\/\/tsboardsolutions.com\/#organization","name":"TS Board Solutions","url":"https:\/\/tsboardsolutions.com\/","logo":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/tsboardsolutions.com\/#\/schema\/logo\/image\/","url":"https:\/\/tsboardsolutions.com\/wp-content\/uploads\/2022\/10\/cropped-TS-Board-Solutions-1.png","contentUrl":"https:\/\/tsboardsolutions.com\/wp-content\/uploads\/2022\/10\/cropped-TS-Board-Solutions-1.png","width":488,"height":40,"caption":"TS Board Solutions"},"image":{"@id":"https:\/\/tsboardsolutions.com\/#\/schema\/logo\/image\/"}},{"@type":"Person","@id":"https:\/\/tsboardsolutions.com\/#\/schema\/person\/1d1aa14a8abcb28fa5e85127be78dcea","name":"Srinivas","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/tsboardsolutions.com\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/10a27cfafdf21564c686b80411336ece?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/10a27cfafdf21564c686b80411336ece?s=96&d=mm&r=g","caption":"Srinivas"},"url":"https:\/\/tsboardsolutions.com\/author\/srinivas\/"}]}},"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/posts\/37001"}],"collection":[{"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/comments?post=37001"}],"version-history":[{"count":1,"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/posts\/37001\/revisions"}],"predecessor-version":[{"id":37021,"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/posts\/37001\/revisions\/37021"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/media?parent=37001"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/categories?post=37001"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/tags?post=37001"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}