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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Systems of Units, , The International System of Units, in 1971 the General Conference on Weights and Measures developed an, internationally accepted system of units for measurement with standard, scheme of symbols, units and abbreviations., This is the Système Internationale d’ Unites (French for International, System of Units), abbreviated as SI system., Now SI system is used in scientific, technical , industrial and commercial, work., In SI system there are seven base units and two supplementary units.
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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Estimation of Very Small Distance, Optical microscope can resolve particles with sizes comparable with, wavelength of light., Electron microscopes can almost resolve atoms and molecules in a material., Tunnelling microscopes can estimate the sizes of molecules.
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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Since dimensions of all terms are the same for Equations (b) and (d) , these, equations can be considered as the equation for kinetic energy., 𝒂, , 9.The Van der waals equation of 'n' moles of a real gas is (P+ 𝟐 )(V−b)=nRT., 𝑽, , Where P is the pressure, V is the volume, T is absolute temperature, R is, molar gas constant and a, b, c are Van der waal constants. Find, the dimensional formula for a and b., a, , (P+ 2 )(V−b)=nRT., V, , By principle of homegeneity, the quantities with same dimensions can be, added or subtracted., a, [P] =[ 2 ], V, , [a] =[PV 2 ], =ML−1 T −2 x L6, [a] = M𝐋𝟓 𝐓 −𝟐, [b] = [V], [b] =𝐋𝟑
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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Problem, We 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, 2.71s and, 2.80 s. Calculate the absolute errors, relative error and percentage error., True value 𝑎𝑚𝑒𝑎𝑛 =, 𝑎𝑚𝑒𝑎𝑛 =, =, , 𝑎1 +𝑎2 +⋯+𝑎𝑛, 𝑛, 2.63+2.56+2.42+2.71+2.80, 5, 13.12, 5, , = 2.624 s, , 𝒂𝒎𝒆𝒂𝒏 = 2.62 s, (As 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.), , The absolute errors in the measurements are, 𝛥𝑎1 = |2.63 s – 2.62 s| = 0.01 s, 𝛥𝑎2 = |2.56 s – 2.62 s | = 0.06 s, 𝛥𝑎3 = |2.42 s – 2.62 s | = 0.20 s, 𝛥𝑎4 = |2.71 s – 2.62 s | = 0.09 s, 𝛥𝑎5 = |2.80 s – 2.62 s | = 0.18 s, , Mean absolute error , 𝛥𝑎𝑚𝑒𝑎𝑛 =, , 𝛥𝑎1 +𝛥𝑎2 +⋯+𝛥𝑎𝑛, , 𝛥𝑎𝑚𝑒𝑎𝑛 =, =, , 𝑛, (0.01+ 0.06+0.20+0.09+0.18)s, 5, 0.54 𝑠, 5, , 𝜟𝒂𝒎𝒆𝒂𝒏 = 0.11 s, Relative Error,, , δa =, =, , 𝛥𝑎𝑚𝑒𝑎𝑛, 𝑎𝑚𝑒𝑎𝑛, 0.11 𝑠, 2.6, , δa = 0.04
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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , (As the arithmetic mean of all the absolute errors is 0.11 s, there is already an error in the tenth of a, second.So second decimal digit 6 can be avoided.), , Percentage Error= 0.04 𝑥 100% = 4%
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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Rule 6:The power of 10, in scientific notation is irrelevant to the determination of, significant figures., , 4.700 m, = 4.700 × 𝟏𝟎𝟐 cm, = 4.700 × 𝟏𝟎𝟑 mm, = 4.700 × 𝟏𝟎−𝟑 km, All these numbers have 4 significant namaste figures.
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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Rounding off the Uncertain Digits, 1) If the insignificant digit to be dropped is more than 5, the preceding, digit is raised by 1, , A number 2.746 rounded off to three significant figures is 2.75, Here the insignificant digit , 6 > 5 and hence 1 is added to the, preceeding digit 4 .(4+1=5), 2) If the insignificant digit to be dropped less than 5, the preceding digit, is left unchanged ., , A number 2.743 rounded off to three significant figures is be 2.74., Here the insignificant digit , 3< 5 and hence the preceeding, number 4 does not change., 3) If the insignificant digit to be dropped is 5,, , Case i) If the preceding digit is even, the insignificant digit, is simply dropped., A number 2.745 rounded off to three significant figures is 2.74., Here the preceding digit 4 is even and hence 5 is simply, dropped., Case ii- ) If the preceding digit is odd, the preceding digit, is raised by 1., A number 2.735 rounded off to three significant figures is 2.74, Here the preceding digit 3 ,is odd and hence 1 is added to, It. (3+1=5)
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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Rules for Arithmetic Operations with Significant Figures, (1)In multiplication or division, the final result should retain as many, significant figures as are there in the original number with the least, significant figures., Eg:If mass of an object is measured to be, 4.237 g (four significant figures), and its volume is measured to be 2.51cm3(3 significant figures), then find, its density in appropriate significant figures., Density =, , 𝑚𝑎𝑠𝑠, 𝑣𝑜𝑙𝑢𝑚𝑒, , =, , 4.237 𝑔, 2.51𝑐𝑚3, , = 1.688047, , As per rule the final result should be rounded to 3 significant figures ., So the answer is 1.69 g/ 𝒄𝒎𝟑, , (2) In addition or subtraction, the final result should retain as many decimal, places as are there in the number with the least decimal places., , Eg:Find the sum of the numbers 436.32 g, 227.2 g and 0.301 g to appropriate, significant figures., 436.32 g +, 227.2 g +, 0.301 g, _______________, 663.821 g, , (2 decimal places), (1 decimal place), (3 decimal places), , As per rule ,the final result should be rounded to 1 decimal place., So the answer 663.8 g, , Seema Elizabeth, MARM Govt HSS Santhipuram
