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CHAPTER, , 1, , Dimensions and, Error Analysis, in Measurement, Chapter Outline, , ■, ■, , Systems of Units, Length, Mass and Time Measurements, , SYSTEM OF UNIT, 1. The magnitude of a physical quantity is expressed by, giving its numerical value and a unit. The numerical, value tells us how many times the basic unit is contained in the measured value., 2. The dimensions of a physical quantity are the powers to which the fundamental unit of length, mass and, time are to be raised so that the derived quantity can be, completely represented., 3. Dimensions tell us about the nature of the physical, quantity and do not give any idea about the magnitude., 4. The concept of dimension is more generalised, compared to the idea of unit., , ■, ■, , Dimensional Analysis, Error Analysis, , A physical quantity remains the same irrespective of, the system of measurement, i.e.,, a b c, a b c, Q = N U = N U ⇒ N1 M1 L1 T1 = N 2 M 2 L2T2, 1, , 1, , 2, , 2, , a, ,  M1 , ⇒ N 2 = N1 , ,  M2 , So, knowing the quantities on the right, value of N2 can be obtained., Units, , Fundamental units, , Derived units, , Applications of Dimensional Analysis, 1. To find the unit of a given physical quantity in a given, system of units., 2. To convert a physical quantity from one system to the, other., Example: Suppose a physical quantity has the dimensional formula MaLbTc., Let N1 and N2 be the numerical values of a quantity in, the two systems of units, respectively., In first system, physical quantity,, a b c, =, Q N=, N1U1, 1 M1 L1 T1, , In second system, same quantity,, a b c, =, Q N=, N 2U 2, 2 M 2 L2T2, , b, , Independent of each, other and not, , Supplementary, units, , Derived from, fundamental units, , interconvertible, , Units of velocity,, Unit of mass, length,, time, temperature,, , c, ,  L1   T1 ,  L  T ,  2  2, hand side the, , acceleration,, force, work-done etc., , electric current,, amount of substance,, , Units of, , luminous intensity., , 1. Plane angle, 2. Solid angle

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1.2, , ■, , Chapter 1, , TABLE 1.1 Physical Quantities with Symbol and Difinitions, Sr. No., 1., , Basic Physical, Quantity, Length, , Name, metre, , Symbol, , Definition, , m, , One metre is the length of the path travelled by light in vacuum during a, time interval of, , 2., , Mass, , kilogram, , 3., , Time, , 4., , 1, of a second., 29, 97, 29, 458, , kg, , One kilogram is equal to the mass (a platinum-iridium alloy cylinder) kept at, Internation Bureau of Weights and Measures, at Sevres, near Pairs, France., , second, , s, , One second is the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground, state of the cesium 133 atom., , Electric current, , ampere, , A, , One ampere is that constant current, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and, placed 1 metre apart in vacuum, would produce between these conductors, a force equal to 2 × 10–7 newton per metre of length., , 5., , Thermodynamic, , kelvin, , K, , One degree kelvin, is the temperature fraction 1/273.16 of the thermodynamic temperature of the triple point of water., , 6., , Amount of, substance, , mole, , mol, , One mole is the amount of substance of a system, which contains as many, elementary entities as there are atoms in 0.012 kilogram of carbon-12., , 7., , Luminous intensity, , candela, , cd, , One candela is the luminous intensity, in a given direction, of a source that, emits monochromatic radiation of frequency 540 × 1012 hertz and that has, a radiant intensity in that direction of 1/683 watt per steradian., , Supplementary Quantities, 8., , Plane angle dθ = ds/r, , radian, , rad, , Plane angle dθ is defined as the ratio of length of arc ds to the radius r., , 9., , Solid angle, dΩ = dA/r2, , steradian, , sr, , Solid angle dΩ is defined as the ratio of the intercepted area dA of the, spherical surface, described about the apex O as the centre, to the square, of its radius r., , 3. To check the dimensional correctness of given physical, relation. It is based on the principle of homogeneity., According to it, the dimensions of each term on both, sides of the equation are the same. It can be also said as, the physical quantities of same nature can be added or, subtracted., 4. To derive the correct relationship between different, physical quantities., , Limitations of Dimensional Analysis, 1. This method gives no information about the dimensionless constants., 2. Many physical quantities have same dimensions, i.e., it, is not unique., 3. We cannot derive the dimensional formula if a physical quantity depends on more than three unknown, variables., , 4. We cannot derive the relation if the physical quantity, contains more than one term (say sum or difference of, two terms.), e.g., v2 = u2 + 2ax, 5. This method cannot be applied if a quantity depends on, trigonometric functions or exponential functions., 6. This method cannot be applied to derive equation conmm, taining dimensional constants, i.e., F ∝ 1 2 , but we, r2, do not get any idea about the constant G., 7. If an equation contains two or more variables with the, same dimension, then this method cannot be used., , ERRORS AND ACCURACY, 1. If the measured value is other than the true value then, we say that there is an error. One basic thing on which, physical science depends, is measurement., 2. There are always a lot of factors which influence the, measurement. These factors always introduce error,

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Dimensions and Error Analysis in Measurement, , Precision of Measurement: The precision of a measurement, depends upon the least count of the measuring instrument. The, smaller the least count, the more precise the measurement., Accuracy of Measurement: The accuracy of measurement, (if there exists an error) depends upon the number of significant figures in it. The larger the number of significant figures,, the higher the accuracy. If there is no error in a measurement,, then that measurement is most accurate., Examples: If the true value of length is 5.764 m, then:, (i) If LC = 0.1 cm, the instrument gives measured value 5.6 cm., (ii) If LC = 0.01 cm, the instrument gives measured value 5.45 cm., First measurement has more accuracy but less precision and, second measurement is less accurate but more precise., , may be small, no matter whatever be the level of, accuracy. So, no measurement is perfect., 3. We can only minimize the errors using best methods, and techniques, but we cannot eliminate these, permanently., , Accuracy, 1. Accuracy means the extent to which a measured value, agrees with the standard or true value for the measurement., 2. But precision means the extent to which a given set of, measurements of the same quantity agree with their, mean value. This mean value need not be the true value., 3. Precise measurement need not be accurate., 4. As the precision increases, the number of significant figures also increases. Accuracy depends on the systematic, errors where as precision depends on random errors., 5. With the increase in accuracy the error decreases. The, accuracy depends on:, a. The range of the instruement used,, b. Sensitivity of the instruement,, c. The least count and the zero error of the instruement,, d. Effect of environment on the instruement,, e. The size and cost of the instuement., 6. No measurement of any physical quantity is absolutely, correct. The numerical value obtained after measurement is just an approximation., 7. As such it becomes quite important to indicate the degree of accuracy (or precision) in the measurement, done in the experiment., 8. The concept of significant figures helps in achieving, this objective., Significant figures of a measured quatity are all, those digits about which we are absolutely sure plus, one digit that has a little doubt., 9. Significant figures give the number of meaningful digits in a number., , ■, , 1.3, , RULES TO DETERMINE THE SIGNIFICANT, FIGURES, 1. All the digits which are not zero are significant., Example: In number 1987, significant figures are 4., 2. If there are zeros between two non-zero digits, then all, those zeros are significant., Example: In 1708.05, significant figures are 6., 3. If the zeros occur to the right of a decimal point and, to the left of a non-zero digit, those zeros are not, significant., Example: In 0.0001987, significant digits are 4., 4. All the zeros to the right of a decimal point and to the, left of a non-zero digit are significant., Example: The number of significant figures in, 1987.00 is 6. In the number 0.0019870, significant, figures are 5., 5. In the number 0.0019870, the zeros between 1 and the, decimal is not significant. Also, the zero on the left of, decimal is not significant. But the last zero, i.e., to the, right of 7 (i.e., a non-zero digit coming after a decimal), is significant., 6. All the zeros to the right of last non-zero digit are not, significant., Example: The number of significant figures in, 198700 is 4. But all the zero to the right of the last, non-zero digit are significant if they are the result of a, measurment., 7. All the digits in a measured value of physical quantity, are significant., Example: Let the distance between two places measured to the nearest poles is 1090 m. In 1090 significant, digits are 4., 8. Even if we express the measured quantity in different units, then also there will not be any change in the, number of significant figures., 9. If the decimal point in a particular measurment in not, shown, the zeros at the right of the number may or may, not be significant., 10. When we add, subtract, multiply or divide two or more, numbers, the accuracy of the result is taken to be equal to, the least accurate among them. The number of significant, figures in the result will be equal to the number of significant digits in the least accurate number among them., , Rounding Off, Correcting or re-shaping a physical quantity with least deviation from its original value after dropping the last digits, which are not required is called rounding off.

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1.4, , ■, , Chapter 1, , Rules regarding rounding off are as follows:, 1. If digit to be dropped is less than 5, then the preceding, digit remains unchanged., Examples:, (a) 7.32 after rounding off becomes 7.3., (b) 4.934 after rounding off beocmes 4.93., 2. If digit to be dropped is more than 5, then the preceding, digit is increased by one., Examples:, (a) 7.86 after rounding off becomes 7.9., (b) 6.937 after rounding off becomes 6.94., 3. If digit to be dropped is 5:, (i) If it is only 5 or 5 followed by zero, then the preceding digit is raised by one if it is odd and left, unchanged if it is even., Examples:, (a) 5.750 after rounding off becomes 5.8, (b) 5.75 after rounding off becomes 5.8, (c) 5.650 after rounding off becomes 5.6, (d) 5.65 after rounding off becomes 5.6, (ii) If 5 is further followed by a non-zero digit, then the, preceding digit is raised by one., Examples:, (a) 15.352 after rounding off becomes 15.4., (b) 9.853 after rounding off beocmes 9.9., 4. During multi-step calculations one digit more than the, significant figures should be retained and at the end, of the calculation, final result should be round off to, proper significant figures., , Types of Errors, 1. Constant Errors:, a. An error which is continuously and constnatly repeated during all the observations made, is called, constant error., b. This arises due to the faculty calibrations of the, measuring instruments., 2. Systematic Error:, a. Instrumental Errors: Examples are zero error of, screw gauge, vernier calipers, etc., faulty calibration on thermometer, ammeter, voltmeter, etc., in, equality of balance arms in a physical balance,, back lash error in instruments with nut and screw,, like microscope, etc., b. Environmental error: The external conditions can, change the capability of the mesuring instrument., c. Error due to observation, e.g., parallax error., d. Error due to imperfection, e.g., whatever precautions are taken, heat is always lost from a calorimeter due to radiation, etc., , Order of Magnitude, In scientific notation the numbers are expressed as,, Number = M × 10x,, where M is a number lying between 1 and 10 and x is an, integer. Order of magnitude of quantity is the power of 10, required to represent the quantity. For determining this power,, the value of the quantity has to be rounded off. While rounding off, we ignore the last digit which is less than 5. If the last, digit is 5 or more than five, the preceding digit is increased, by one. For example,, 1. Speed of light in vacuum, = 3 × 108 ms-1 ≈ 108 m/s (ignoring 3 < 5), 2. Mass of electron, = 9.1 × 10 -31 kg ≈ 10-30 kg (as 9.1 > 5), , 3. Random Errors:, a. The errors which occurs irregularly and at random in, magnitude and direction are called random errors., b. These errors are not due to any definite cause and, so they are also called accidental errors., c. Such errors may be avoided by taking the measurments a number of times and then finding the arithmetic mean. That is,, a + a + a + a4 + ... + an, 1 n, a= 1 2 3, ⇒ a =   ∑ ai, n,  n  i =1, This arithmetic mean is supposed to be the accurate observation., 4. Gross Errors:, a. The errors caused due to the carelessness of the, person are called gross errors. So, these type of errors are also known as mistakes., Absolute Error = (True value) – (Measured value), b. Taking the arithmetic mean as the true value, the, absolute error in ith observation is, Dai = ( a − ai ), That is, for the first observation, Da1 = a − a1; for, the second observation, Da2 = a − a2 and so on., , Mean Absolute Error, ∆a =, , |∆a1 | + |∆a2 | + |∆a3 | + ... + |∆an |  1 , =   ∑ |∆ai |, n, n i, , a. Relative error =, and, , Mean absolute error  ∆a , =, , True value,  a , ,  ∆a , b. Percentage error = ,  × 100 %.,  a 

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Dimensions and Error Analysis in Measurement, , Propagation of Errors, , ⇒, , The error in the final result depends on the errors in the, individual measurements and on the nature of mathematical, operations performed to get the final result., 1. Addition: Let z = x + y., Let the absolute errors in the two quantities x and y be, Dx and Dy. Their corresponding observed values are, (x ± Dx) and (y ± Dy). Hence,, (z ± Dz) = (x ± Dx) + (y ± Dy), ⇒ (z ± Dz) = (x ± y) ± Dx ± Dy, ∴ ± Dz = ± Dx ± Dy., Hence, the maximum possible error in z is given by,, Dz = Dx + Dy, 2. Subtraction: Let z = x – y, Let the absolute errors in the two quantities x and y be, Dx and Dy. Their corresponding observed values are, (x ± Dx) and (y ± Dy)., Hence, z ± Dz = (x ± Dx) – (y ± Dy), ⇒ z ± Dz = (x – y) ± Dx ± Dy, ∴ ±Dz = ±Dx + Dy., Hence, the maximum possible error in z is,, Dz = (Dx + Dy), 3. Multiplication: Let z = xy, Let the absolute errors in the two quantities x and y be, Dx and Dy. Their corresponding observed values are, (x ± Dx) and (y ± Dy). Hence,, z ± Dz = (x ± Dx) (y ± Dy) = xy ± x Dy ± y Dx ± Dx . Dy, Neglecting Dx . Dy wrt other terms, then, ± Dz = ± x Dy ± y Dx,  ± ∆z ,  x ∆y   y ∆x ,  x ∆y   y ∆x , ⇒ ,  = ±, ±,  = ±, ±, ,  z ,  z   z ,  xy   xy ,  ∆y   ∆x ,  ∆z , ⇒ ±  = ±  ± ,  z ,  y   x , Hence, maximum relative error in z is,  ∆z   ∆x   ∆y ,   =  + ,  z   x   y , Percentage error is,  ∆x ,  ∆y ,  ∆z ,   × 100 =   × 100 +   × 100.,  z ,  x ,  y , x, 4. Division: Let z =  . Let the absolute errors in the,  y, two quantities x and y be Dx and Dy. Their corresponding, observed values are (x ± Dx) and (y ± Dy). Hence,,  x ± ∆x , −1, z ± ∆z = ,  = ( x ± ∆x ) ( y ± ∆y ), ∆, y, y, ±, , ,  ∆x  −1  ∆y , = x 1 ±,  y 1 ±, x , y , , , , −1, , ⇒, ,  x   ∆x   ∆y , z ± ∆z =   1 ±,  1±, x  , y ,  y ,  x   ∆x   ∆y , z ± ∆z =   1 ±,  1, y , x  ,  y , , ■, , 1.5, , −1, , Dividing both sides by z;,  z ± ∆z   ∆x   ∆y ,  = 1 ±, ,  1±,  z  , x  , y ,  ∆z   ∆x   ∆y , ⇒ 1 ±  = 1 ±,  1±, , z  , x  , y , = 1±, , ∆x ∆y  ∆x   ∆y , +, + , x, y  x   y , , ∆x ∆y,  ∆z , +, ⇒ ±  = ±, .,  z , x, y, Hence, the maximum possible relative error in z is,  ∆z   ∆x   ∆y ,   =   +  .,  z   x   y , 5. Power of Observed Quantities:, (a) z = x m. Taking log on both sides, log z = m log x, Differentiating,, 1, 1,   ∆z = m   ∆x, i.e., Relative error in z = m, z, x, times relative error in x.,  xm yn , (b) Let z =  p ,  w , Taking log on both sides, log z = m log x + n log, y – p log w, ∆z,  ∆x ,  ∆y ,  ∆w , = m  + n  − p, Differentiating,, , z,  x ,  w ,  y , Maximum value of relative error in z is obtained by, adding the relative error in the quantity w., Hence, maximum relative error is,,  ∆x ,  ∆y ,  ∆z ,   = m  + n  +,  z ,  x ,  y , ,  ∆w , p, ,  w , , Least Count, 1. The smallest value of a physical quantity which can be, measured accurately with an instrument is called the, least count (LC) of the measuring instrument., 2. For an instrument where vernier is used, its VC (vernier, constant) is its least count. VC is equal to difference of, one main scale division and one vernier scale division.

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1.6, , ■, , Chapter 1, , Vernier Callipers, , Least Count, , 1. It was designed by a French Mathematician Pierre, Vernier, and hence the instrument is named Vernier, after the name of its inventor., 1, 2. It is a device used to measure accurately up to, th of, 10, a millimetre., , Note the value of the main scale division and count the, number n of vernier scale division. Slide the movable jaw, till the zero of vernier scale coincides with any of the mark, on the main scale and find the number of division (n – 1) on, the main scale coinciding with n division on vernier scale., , B, , A, , S, , 0, , 1, , 2, , 2, , 33, , 4, , Main Scale, 5, , 6, , 14, , cm, 15, , N, , Vernier Scale, , C, , D, ,  n −1 , Then, n VSD = (n – 1) MSD or 1 VSD = ,  MSD,  n , or VC or LC = 1 MSD – 1 VSD, 1,  n −1 , = 1 −,  MSD = MSD, n , n, , For example, 1 MSD = 1 mm and 10 VSD = 9 MSD, 9, ∴ 1 VSD = MSD = 0.9., 10, Vernier constant, VC = 1 MSD –1 VSD = (1 – 0.9) mm, = 0.1 = 0.01 cm., , In the figure, 5th vernier scale division is coinciding, with any main scale division., Hence, n = 5, LC = 0.01 cm, Zero error = n × (LC) = 5 × 0.01 = +0.05 cm, Zero correction = –0.05 cm, Actual length will be 0.05 cm less than the observed, (measured) length., , cm 0, 0, , Zero Error and Zero Correction, If the zero marks of the main scale and vernier scale may, not coincide when the jaws are made to touch each other,, then it gives rise to an error called zero error. Zero error can, be positive or negative., 1. If zero of vernier scale coincides with zero of main, scale, then, zero error and zero correction, both are nil., Actual length = Observed (measured) length., , cm 0, 0, , Main Scale, 0.5, 11, , 1.5 1.5, , 5, 10, Vernier Scale, , (Zero error-zero), 2. If zero of vernier scale lies on the right of main scale,, then zero error is positive and zero correction is negative., , Main Scale, 0.5, 11, , 1.5 1.5, , 5, 10, Vernier Scale, , (Zero error-positive), 3. If zero of vernier scale lies on the left of zero of main, scale, then zero error is negative and zero correction is, positive., In the figure, 6th vernier scale division is coinciding, with any main scale division., Hence, n = 6, LC = 0.01 cm, Zero error = n × (LC) = 6 × 0.01 = –0.06 cm, Zero correction = +0.06 cm, Actual length will be 0.06 cm more than the observed, (measured) length., , cm 0, 0, , Main Scale, 0.5, 11, 5, 10, Vernier Scale, , (Zero error-negative), , 1.5 1.5

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Dimensions and Error Analysis in Measurement, , It is defined as ratio of the pitch of the screw to the total, number of divisions on circular scale., LC of the screw gauge, , Screw Gauge, 1. In general vernier calliper can measure accurately upto, 0.01 cm and for greater accuracy mirometer screw devices, e.g., screw guage, spherometer, etc are used., 2. Screw guage works on the priciple of micrometer screw., , Screw, , 1.7, , Least Count, , 4. Measured value by vernier calliper = MSR + (LC ×, VSR) + zero correction., , Stud, , ■, , =, , Screw Cap, , Sleeve, 2 3, , Pitch of the screw, Total number of divisions on the circulaar scale, , 5, , Thimble, , 0, 95, , Spindle, , Main, Scale, , Circular Scale, , Ratchet, , U-Frame, , 15, M. S., , 5, 0, , This implies that the minimum length that can be measured, accurately with the screw gauge is 0.005 mm., , 95, , Circular Scale, , For example, if the pitch of the screw gauge is 0.5 mm and, the total number of divisions on the circular scale is 100,, 0.5 mm, = 0.005 mm., then the least count will be given by,, 100, , 90, , Pitch, It is defined as the linear distance moved by the screw forward or backward when one complete rotation is given to, the circular cap., , Zero of C.S. is below, the zero of M.S., , Distance moved on linear scale, Number of rotation, M. S., , 15, , Zero Error, , 5, , When the two studs of the screw gauge are brought in contact and if the zero of the circular scale does not coincide, with the reference line then the screw gauge has an error., This error is called zero error., , 0, , 1. Positive Zero Error: Zero error is said to be positive if, the zero of the circular scale lies below the reference, line as shown in the figure., , 95, , Circular Scale, , Pitch of the screw =, , (Zero error-zero), , (Zero error-positive), For example, the 4th division of the head scale is in line, with the line of graduation.

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1.8, , ■, , Chapter 1, , M. S., , 5, 0, 95, 90, , Circular Scale, , Zero of C.S. is below, the zero of M.S., , Then, the zero error = +4 × LC = +4 × 0.01 mm =, 0.04 mm Zero correction = –0.04 mm., 2. Negative Zero Error: Zero error is said to negative if, the zero of the circular scale lies above the reference, line as shown in the figure., For example, 97th division of the head scale is in line, with the line of graduation., Thus, zero error = (97 – 100) × LC = –3 × 0.01 mm, = –0.03 mm, Zero correction = +0.03 mm, , Measured Value by Screw Gauge, (Zero error-negative), , = MSR + (LC × CSR) + Zero correction., , Dimensions of Physical Quantities, TABLE 1.2 Continued, Dimensions, , Dimensional, Formula, , length × breadth, , [L2], , [M 0L2T 0], , Volume, , length × breath × height, , [L3], , [M 0L3T 0], , 3., , Density, , mass/volume, , [M ]/[L3] or [ML–3], , [M 0L–3T 0], , 4., , Frequency, , 1/time period, , 1/[T ], , [M 0L0T –1], , 5., , Velocity, , displacement/time, , [L]/[T], , [M 0L0T –1], , 6., , Acceleration, , velocity/time, , [LT –1]/[T ], , [M 0LT –2], , 7, , Force, , mass × acceleration, , [M] [LT –2], , [MTL–2], , 8., , Impulse, , force × time, , [MLT –2][T], , [MLT –1], , 9., , Work, , force × distance, , [MLT –2][L], , [ML2T –2], , 10., , Power, , work/time, , [ML2T –2][L], , [ML2T –3], , 11., , Momentum, , mass × velocity, , [M] [LT –1], , [MLT –1], , 12., , Pressure stress, , force/area, , [MLT –2]/[L2], , [ML–1T –2], , 13., , Strain, , change in dimension, original dimension, , [L]/[L] or [L3]/[L3], , [M 0L0T 0], , 14., , Modulus of elasticity, , stress/strain, , [ML−1 T −2 ], [ M 0 L0T 0 ], , [ML–1T –2], , 15., , Surface tension, , force/length, , [MLT–.2]/[L], , [ML0T –2], , 16., , Surface energy, , energy/area, , [ML2T –2]/[L2], , [ML0T 2], , 17., , Velocity gradient, , velocity/distance, , [LT –1]/[L], , [M 0L0T –1], , 18., , Pressure gradient, , pressure/distance, , [ML–1T –2]/[L], , [ML–2T –2], , 19., , Pressure energy, , pressure × volume, , [ML–1T –2][L3], , [ML2T –2], , Sr. No., , Physical Quantity, , Formula, , 1., , Area, , 2., , (Continued)

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Dimensions and Error Analysis in Measurement, , ■, , 1.9, , TABLE 1.2 Continued, Dimensional, Formula, , Sr. No., , Physical Quantity, , Formula, , Dimensions, , 20., , Coefficient of viscosity, , force, area × velocity gradient, , [MLT −2 ], [ L ][ LT −1 / L], , [ML–1T –1], , 21., , Angle, , arc/radius, , [L]/[L], , [M 0L0T 0], , 22., , Trigonometric ratio, (sinθ, cosθ, tanθ, etc)., , length/length, , [L]/[L], , [M 0L0T 0], , 23., , Angular velocity, , angle/time, , [M 0L0T 0]/[T], , [M 0L0T –1], , 24., , Angular acceleration, , angular velocity/time, , [T –1]/[T], , [M 0L0T –2], , 25., , Radius of gyration, , [L], , [ML2T 0], , 26., , Moment of inertia, , mass × (radius of gyration)2, , [M][L2], , [ML2T 0], , 27., , Angular momentum, , moment of inertia × angular velocity, , [ML2][T –1], , [ML2T –1], , 28., , Moment of force,, moment of couple, , force × distance, , [MLT –2][L], , [ML2T –2], , 29., , Torque, , force × distance, , [MLT –2][L], , [ML2T –2], , 30., , Angular frequency, , 2π × frequency, , [M 0L0T 0][T –1], , [M 0L0T –1], , 31., , Wavelength, , [L], , [M 0LT 0], , 32., , Hubble constant, , recession speed/distance, , [LT –1]/[L], , [M 0L0T –1], , 33., , Intensity of wave, , energy, time × are, , [ML2T −2 ], [T ][ L2 ], , [ML0T –3], , 34., , Radition pressure, , intensity of wave, speed of light, , [ML–3]/[LT –1], , [ML–1T –2], , 35., , Energy density, , energy/volume, , [ML2T –2]/ [L3], , [ML–1T –2], , 36., , Critical velocity, , Reynold’s number ×, , 37., , Escape velocity, , (2 × acceleration due to gravity × earth’s, radius)1/2, , 38., , Heat energy, internal, energy, , 39., , Kinetic energy, , (1/2) mass × (velocity)2, , 40., , Potential energy, , 41., , Rotational kinetic energy, , 42., , Efficiency, , output work or energy, input work or energy, , 43., , Angular impulse, , 44., 45., , 2, , coefficient of viscosity [M 0 L0T 0 ][ML−1T −1 ], [ML−3 ][L], density × radius, , [M 0LT –1], , [LT –2]1/2 × [L]1/2, , [M 0LT –1], , [ML2T –2], , [ML2T –2], , [M][LT –1]2, , [ML2T –2], , mass × acceleration due to gravity × height, , [M][LT –2][L], , [ML2T –2], , (1/2 × moment of inertia) × (angular, velocity)2, , [ML2] × [T –1]2, , [ML2T –2], , [ML2T −2 ], [ML2T −2 ], , [M 0L0T 0], , torque × time, , [ML2T –2][T], , [ML2T –1], , Gravitational constant, , force × (distance) 2, mass × mass, , [MLT −2 ][ L2 ], [ M ][ M ], , [M–1L3T –2], , Planck’s constant, , energy/frequency, , [ML2T –2]/ [T –1], , [ML2T –1], , (Continued)

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Dimensions and Error Analysis in Measurement, , ■, , 1.11, , TABLE 1.2 Continued, Sr. No., , Physical Quantity, , Formula, , Dimensions, , 68., , Inductance, , 69., , Magnetic dipole moment, , 70., , Magnetisation, , magnetic moment, volume, , 71., , Premittivity constant (of, free space) e0, , charge × charge, 4π × electric force × (distance) 2, , 72., , Premeability constant (of, free space) µ0, , 73., , Refractive index, , 74., , Faraday constant, , avogadro’s constant ×, elementary charge, , 75., , Wave number, , 76., 77., , Dimensional, Formula, , magneticflux, current, , [ ML2T −2 A−1 ], [ A], , [ML2T –2A–2], , current × area, , [A][L2], , [M 0L2T 0A], , [ L2 A], [L3 ], , [M 0L–1T 0A], , [ AT ][ AT ], [ MLT −2 ][ L2 ], , [M–1L–3T4A2], , 2π × force × distance, current × current × length, , [M 0 L0T 0 ][MLT −2 ][L], [ A][ A][ L], , [MLT –2A–2], , speed of light in vacuum, speed of light inmedium, , [LT –1/[LT –1], , [M 0L0T 0], , [AT]/[mol], , [M 0L0TA mol–1], , 2π/wavelength, , [M 0L0T 0]/[L], , [M 0L–1T 0], , Radiant flux, radiant power, , energy emitted/time, , [ML2T –2]/[T], , [ML2T –3], , Luminosity of radiant flux, or radiant intensity, , [ML2T –3]/[M 0L0T 0], , [ML2T –3], , radiant power or, , 78., , Luminous power or luminous flux of source, , luminous energy emitted, time, , [ML2T –2]/[T], , [ML2T –3], , 79., , Luminous intensity or illuminating power of source, , luminous flux, solid angle, , [ML2T −3 ], [M 0 L0T 0 ], , [ML2T –3], , 80., , Intensity of illumination of, luminance, , luminous intensity, (distance ) 2, , [ML2T −3 ], [ L2 ], , [ML0T –3], , 81., , Relative luminosity, , (luminous flux of a source of given wavelength)/(luminous flux of peak sensitivity wavelength (555 nm) source of same, power), , [ML2T −3 ], [ML2T −3 ], , [M 0L0T 0], , 82., , Luminous efficiency, , total luminous flux, total radiant flux, , [ML2T −3 ], [ML2T −3 ], , [M 0L0T 0], , 83., , Illuminance or illumination, , luminous flux incident, area, , [ML2T −3 ], [L2 ], , [ML0T –3], , 84., , Mass defect, , (sum of masses of nucleons) –, (mass of the nucleus), , [M], , [ML0T 0], , 85., , Biniding energy of nucleus, , mass defect × (speed of light in vacuum)2, , [M][LT –1]2, , [ML2T –2], , 86., , Decay constant, , 0.693/half life, , [T –1], , [M 0L0T –1], , radiant flux of source, solid angle, , (Continued)

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1.12, , ■, , Chapter 1, , TABLE 1.2 Continued, Sr. No., , Physical Quantity, , Formula, , 87., , Resonant frequency, , (inductance × capacitance)–1/2, , 88., , Quality factor or Q-factor, of coil, , 89., , Power of lens, , 90., , Dimensions, , Dimensional, Formula, , [ML–2T –2A–2]–1/2 ×, [M–1L–2 T 4A2]–1/2, , [M 0L0A0T –1], , [T −1 ]ML2T −2 A−2 ], [ML2T −3 A−2 ], , [M 0L0T 0], , (focal length)–1, , [L–1], , [M 0L–1T 0], , Magnification, , image distance, object distance, , [ L], [ L], , [M 0L0T 0], , 91., , Fluid flow rate, , (π /8) (pressure) (radius) 4, ( viscosity coefficient) × (length), , [ML−1T −2 L4 ], [ML−1T −1 ]L ], , [M 0L3T –1], , 92., , Capacitive reactance, , (angular frequency × (capacitance)–1, , [T –1]–1[M–1L–2T 4A2]–1, , [ML2T –3A–2], , 93., , Inductive reactance, , (angular frequency × inductance), , [T –1][ML2T –2A–2], , [ML2T –3A–2], , resonant frequency × inductance, resistance, , TABLE 1.3 Physical Quantities Having Same Dimensional Formula, Sr., No., , Physical Quantities, , 1., , Frequency, angular frequency, angular velocity, velocity gradient, , [M 0L0T –1], , 2., , Work, internal energy, potential energy, kinetic energy, torque, moment of force, , [ML2T –2], , 3., , Pressure, stress, Young’s modulus, bulk modulus, modulus of rigidity, energy density, , [ML–1T –2], , 4., , Momentum and impulse, , [MLT –1], , 5., , Acceleration due to gravity, gravitational field intensity, , [M 0LT –2], , 6., , Thrust, force, weight, energy gradient, , [MLT –2], , 7., , Angular momentum and Planck’s constant (h), , [ML2T –1], , 8., , Surface tension, surface energy, force gradient, spring constant, , [ML0T –2], , 9., , If l is length, g is acceleration due to gravity, m is mass, k is force constant, R is radius of earth, then, , [M 0L0T], , 1/ 2, , 1/ 2, , Dimensional, Formula, , 1/ 2, , l, m, R, g ,  k  , g,  ,  ,  , , all have the dimensions of time, LC all have the dimensions of time, , [M 0L0T], , 10., , If L is inductance, R is resistance C is capacitance then L /R, CR and, , 11., , Thermal capacity, entropy, Boltzmann constant, , 12., , If p is pressure, V is volume, T is temperature, R is gas constant, m is mass, s is specific heat, L is latent heat,, DT is rise in temperature then pV, RT, mL, (msDT) all have dimensions of energy, , [ML2T –2], , 13., , Work, energy, heat, torque, couple, moment of force have same dimensions, , [ML2T –2], , 1, 1, , 1, , Potential energy (mgh), kinetic energy  mv 2or I ω 2  , energy contained in an inductance  LI 2  and, 2, 2, , 2, , , [ML2T –2], , 14., , 1, Q2 , 1, electrostatic energy of condenser  QV , CV 2 ,, ., 2, 2C , 2, , [ML2T –2K –1]

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Dimensions and Error Analysis in Measurement, , Important Conversions, 1 inch = 2.54 cm, 1 foot = 12 inches = 30.48 cm = 0.3048 m, 1 mile = 5280 ft = 1.609 km, 1 yard = 0.9144 m, 1 slug = 14.59 kg, 1 barn = 10–28 m2, , ■, , 1.13, , 1 litre = 103 cm3 = 10–3 m3, 5, 1 km/h =, m/s, 18, 1 m/s = 3.6 km/h, 1 g/cm3 = 1000 kg/m3, 1 atm = 76 of Hg = 1.013 ×105 N/m2, 1 N/m2 = Pa (Pascal), , CHAPTER-END EXERCISES, BASED ON NCERT EXAMPLES, 1. Fill in the blanks:, (a) The volume of a cube of side 1 cm is equal, to, m3 ., (b) The surface area of a solid cylinder of radius 2.0 cm, and height 10.0 cm is equal to, (mm)2., (c) A vehicle moving with a speed of 18 km/h, covers, m in 1 second., (d) The relative density of lead is 11.3. Its density, is, g/cm3 or, kg m–3., Solution:, (a) 10–6 m, (b) 1.5 × 104 (mm)2, (c) 5 m in 1 second, (d) 11.3 g/cm3; 1.13 × 104 kg m–3, , 3. A 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 of light, takes 8 min and 20 s to cover this distance?, Solution: The new unit of distance = Speed of light in, vacuum = distance travelled by light in 1 second, or new unit of distance = 3 × 108 m., Time taken by light to cover the distance between, the Sun and Earth = 8 minutes and 20 seconds =, 500 seconds., ∴ Distance between the Sun and the Earth, = 3 × 108 × 500 m., Distance between the Sun and the Earth in terms of new unit, , 2. A calorie is a unit of heat or energy and it equals about, 4.2 J where 1 J = 1 kg m2s–2. Suppose we employ a system of units in which the unit of mass equals α kg, the, unit of length equals β m, the unit of time is γ seconds., Show that a calorie has a magnitude 4.2 α –1 β –2γ 2 in, terms of the new units., Solution: Let the unit of mass of the first system be M1, L1 T1 (kg m s) and that of second system be M2 L2 T2 i.e.,, (α kg, β m, γ seconds)., Let n1 and n2 be the numerical values of the first and, second system, then, , 3 × 108 × 500, = 500 new units., 3 × 108, 4. Which of the following is the most precise device for, measuring length?, (a) Vernier callipers with 20 divisions on sliding scale., (b) A screw gauge of pitch 1 mm and 100 divisions on, the circular scale., (c) An optical instrument that can measure length to, within a wavelength of light., Solution: Optical instrument that can measure length, within a wavelength of light, is the most precise device, for measuring length., Hence, the correct answer is option (c)., 5. A student measures the thickness of 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 view of the microscope is 3.5 mm. What is the estimate on the thickness, of hair?, 3.5, Solution: The estimated thickness of hair, 100, = 0.035 mm., , n1 [ M1a L1bT1c ] = n2 [ M 2a Lb2T2c ], We know that 1 J = 1 kg m2 s–2., Hence, the dimensional formula for joule = ML2T–2., Here, a = 1, b = 2 and c = –2, I calorie = 4.2 J = 4.2 kg m2 s–2, 1, , 2, ,  M   L  T , n2 = 4.2  1   1   1 ,  M 2   L2   T2 , 1, , or, , 2, , −2, ,  kg   m   s , n2 = 4.2 ,  ,   ,  α kg   β m   γ s , n2 = 4.2 α –1 β –2γ 2, , −2, , =

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1.14, , ■, , Chapter 1, , 6. Answer the following:, (a) A screw gauge has 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?, (b) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield more, reliable estimate than a set of 5 measurements only?, Solution:, (a) Yes, it is possible to increase the accuracy of the, gauge by increasing the number of divisions on the, circular scale as the least count of the gauge would, become less., ∴ Least count =, Pitch, Number of divisions on circular scale, (b) Due to random errors, a large number of observations (say 100) will give more reliable result than, smaller number of observations (say 5). This is, because the chance (probability) of making a positive random error of given magnitude is equal to, that of making negative random error of the same, magnitude. Thus in a large number of observations,, positive and negative errors are likely to cancel, each other and hence more reliable results can be, observed., 7. State the number of significant figures on the following:, (i) 0.007 m2, (ii) 2.64 × 1024 kg, –3, (iii) 0.2370 g cm, (iv) 6.320 J, (v) 6.032 N m–2, (vi) 0.0006032, Solution:, (i) 0.007 m2 = 7.00 × 10–3 m2 = 7 ×10–3 m2, has only, one significant figure., (ii) 2.64 × 1024 kg. There are three significant figures, viz., 2, 6 and 4., (iii) 0.2370 g cm–3. The number of significant figures is, four, i.e., 2, 3, 7 and 0., (iv) 6.320 J has four significant figure, i.e., 6, 3, 2 and 0., (v) 6.032 N m–2 has four significant figure, i.e., 6, 0, 3, and 2., (vi) 0.0006032 = 6.032 × 10–4, has four significant figure, i.e., 6, 0, 3 and 2., 8. The 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 figure., , Solution:, Area of rectangular sheet, = 2(lb + bh + lh), = 2(4.234 × 1.005 + 1.005 × 0.201 + 0.0201 × 4.234), = 8.7209468 m2, Since the least accuracy in the given data is up to three, places of decimal, hence area of rectangular sheet to, correct significant figure is 8.720 m2, Volume of rectangular sheet, = length × breadth × height, = 4.234 × 1.005 × 0.201 m3 = 0.085 m2, Since the least accuracy in the given data is up to three, places of decimal, hence volume of rectangular sheet to, correct significant figure is 0.085 m2., 9. The mass of a box measured by a grocer’s balance is 2.3, kg. Two gold pieces of masses 20.15 g and 20.17 g are, added to the box. What is (a) the total mass of the box,, (b) the difference in the masses of the pieces to correct, significant figures?, Solution:, (a) Given, mass of box = 2.3 kg (two significant figures), Mass of two gold pieces = 0.02015 kg and 0.02017 kg, Total mass of the box = 2.3 + 0.02015 + 0.02017, = 2.34032 kg, = 2.3 kg, rounded off to two significant figures as mass of box, has two significant figures only., (b) The difference in the mass of the pieces, = 20.17 g – 20.15 g, = 0.02 g., 10. A physical quantity P is related to four observables a, b,, c and d as follows:, P = a3 b 2 /( cd ), 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?, Solution: Given P = a3 b 2 /( cd ),, The percentage error in P is given by,  ∆a, ∆P, ∆b 1  ∆c ∆d  , × 100 = 3, +2, +  ×, , p, a, b 2 c, d  , , 1, , , = 3 × 1% × 2 × 3% + × ( 4 × 2)  = 13%, 2, , , Percentage error in P = 13% or 0.13, Further, P = 3.763 = 3.8 because significant figures in, each quantity is two.

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Dimensions and Error Analysis in Measurement, 11. A book with many printing errors contains four different formulae for the displacement y of a particle undergoing a certain periodic motion:, (a) y = a sin 2π t/T, (b) y = a sin vt, (c) y = (a/T) sin t/a, (d) y = ( a / 2 ) (sin 2π t/T + cos 2π t/T), (a = maximum displacement of the particle, v = speed, of the particle, T = time period of motion). Rule out the, wrong formulae on dimensional grounds., Solution: Dimension of LHS, i.e., displacement is L., Dimension of a = L., Dimensions of RHS in different equations are as, follows:, T, 2π t, , , = L  as angle, is dimensionless , T, T, , , Dimensions of LHS = Dimension of RHS, Hence, the equation is correct., (b) L sin (LT–1 × T) = L sin L, The arguement of a trigonometrical function must, always be dimensionless., Hence, the equation is correct., (c) (L/T) sin (T/L) = LT–1 sin (L–1T), Here the arguements of sin is not dimensionless., Hence, the equation is not correct., T,  T, (d) L  sin + cos  = L,  T, T, i.e., Dimensions of RHS = Dimensions of LHS., , (a) L sin, , ■, , 1.15, , 12. Explain 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’s 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)., Solution: There is a relative motion between the, observer in the train and outside objects which are, stationary. But in case of objects like nearby trees,, houses, etc., the angle subtended by them at the eye, is large and hence changes rapidly with time. On the, other hand, for far objects like hill-top, etc., since the, angle subtended by them is small, the change of angle, with time is also small. Hence far-off objects appear, to be stationary., 13. When the planet Jupiter is at a distance of 824.7 million kilometres from the Earth, its angular diameter is, measured to be 35.72″ of arc. Calculate the diameter of, Jupiter., Solution: Given, D = 824.7 million km = 824.7 × 106 km, a = 35.72″ = 35.72 × 4.85 × 10–6 rad, d=?, Using relationship d = Da, we have, Diameter of Jupiter,, d = 824.7 × 106 × 35.72 × 4.85 × 10–6 km, = 142872.6774 km ≈ 142880 km, , PRACTICE EXERCISES (MCQS), 1. Which of the following is a derived unit?, (a) Unit of mass, (b) Unit of length, (c) Unit of time, (d) Unit of volume, 2. Select the pair whose dimensions are same:, (a) Pressure and stress, (b) Pressure and force, (c) Stress and strain, (d) Power and force, 3. Which of the following is dimensionally correct?, (a) Pressure = Force per unit volume, (b) Pressure = Momentum per unit volume per unit time, (c) Pressure = Energy per unit volume, (d) Pressure = Energy per unit volume per unit time, 4. Which of the following is a derived quantity?, (a) Temperature, (b) Ampere, (c) Candela, (d) Lumen, , 5. Wave number has the dimensions of, (a) Length, (b) Length–1, (c) A dimensionless physical quantity, (d) None of these, 6. Ampere-hour is a unit of, (a) quantity of electricity, (b) strength of electric current, (c) power, (d) energy, 7. Which of the following does not represent the unit of, pressure?, (a) Millibar, (b) mm of mercury column, (c) Water column in meter, (d) Newton/m2

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1.16, , ■, , Chapter 1, , 8. Dimensional analysis cannot be used for, (a) deriving relations, (b) deriving with fourth unknown variable, (c) converting system of units, (d) checking correctness of relations, 9. Identify the pair whose dimensions are equal, (a) Torque and work, (b) Stress and energy, (c) Force and stress, (d) Force and work, 10. Which of the following sets of quantities have same, dimensional formula?, (a) Frequency, angular frequency and angular momentum, (b) Acceleration, momentum and retardation, (c) Thermal capacity, specific heat and entropy, (d) Work, energy and torque, 11. Which of the following, expressed in proper unit?, (a) Torque, :, (b) Stress, :, (c) Modulus of elasticity :, (d) Power, :, (e) Surface tension, :, , quantities has not been, Newton metre, Newton metre–2, Newton metre–2, Newton metre second–1, Newton metre–2, , 12. Which of the following quantities measured from different inertial reference frames are same?, (a) Force, (b) Velocity, (c) Displacement, (d) Kinetic energy, 13. If the units of mass, length and time are doubled, unit of, angular momentum will be, (a) doubled, (b) tripled, (c) quadrupled, (d) 8 times of the original value, 14. The dimensions of specific resistance in terms of charge, Q is, (a) [ML2T –2Q 2], (b) [MLT –2Q], 2 –1 –2, (c) [ML T Q ], (d) [ML3T –1Q –2], 15. The dimensions of specific gravity is, (a) [M 1L–3T 0], (b) [M0L1T –2], (c) [M0L0T –1], (d) [M0L0T 0], 16. The dimensions of the ratio of angular momentum to, linear momentum, (a) [M 1L–3T 0], (b) [M 0L1T –2], 0 0 –1, (c) [M L T ], (d) [M 0L1T 0], 17. The unit of surface tension in SI system is, (a) Dyne/cm2, (b) Newton/m, (c) Dyne/cm, (d) Newton/m2, 18. One poise is equal to, (a) 0.01 N-s/m2, (b) 0.1 N-s/m2, 2, (c) 10 N-s/m, (d) 1 N-s/m2, , 19. Parallactic second is the unit of, (a) time, (b) velocity, (c) distance, (d) angle, 20. The unit of nuclear dose given to a patient is, (a) fermi, (b) curic, (c) rutherford, (d) roentgen, 21. Water equivalent of a body is expressed in, (a) calorie, (b) gram, (c) degree kelvin, (d) erg, 22. The ‘rad’ is the correct unit used to report the measurement of, (a) the ability of a beam of gamma ray photons to produce ions in a target, (b) the energy delivered by radiation to a target, (c) the biological effect of radiation, (d) the rate of decay of a radioactive source, 23. SI unit of Bohr magneton is, (a) ampere, (b) amp-m, (c) amp-m2, (d) kg m2s–1, 24. Select the pair whose dimensions are same, (a) Pressure and stress, (b) Stress and strain, (c) Pressure and force, (d) Power and force, 25. The force F is given by Stoke’s equation:, F = 6πhrv, Then the dimension of viscosity coefficient h are:, (a) [ML–1T –1], (b) [MLT –1], (c) [M –1L–1T –1], (d) [ML2T –1], 26. The force F is given in terms of time t and displacement, x by the equation:, F = a cos αx + b sin βt, where a and b are the amplitudes. The dimensions of, β/α are:, (a) [M 0L0T 0], (b) [M 0L0T–1], 0 –1 0, (c) [M L T ], (d) [M 0L1T –1], 27. µ0 and e0 denote the magnetic permeability and electrical permittivity of free space, then the dimensions of, 1, are similar to, µ0 ε 0, (a) distance, (b) velocity, (c) acceleration, (d) None of these, 28. If e, e0, h and c respectively represent electronic charge,, permittivity of free space, Planck’s constant and the, e2, speed of light, then, has the dimensions of, ε 0 hc, (a) pressure, (b) angle, (c) current, (d) angular momentum

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Dimensions and Error Analysis in Measurement, 29. The unit of absolute permitivity is, (a) Fm (farad-metre), (b) Fm–1 (farad/metre), –2, 2, (c) Fm (farad/metre ), (d) F (farad), 30. Universal time is based on, (a) rotation of earth on its axis, (b) Earth’s orbital motion around the sun, (c) vibrations of cesium atom, (d) oscillations of quartz crystal, 31. The velocity v (in cms–1) of a particle is given in terms, b, of time t (in seconds) by the relation, v = at +, ; the, t +c, dimensions of a, b and c are, (a) a = L2, b = T, c = LT2, (b) a = TL2, b = LT, c = L, (c) a = LT–2, b = L, c = T, (d) a = L, b = LT, c = T2, 32. If x = at + bt2, where x is the distance travelled by the, body in kilometre while t is the time in second, then the, unit of b are, (a) km/s, (b) km-s, (c) km/s2, (d) km-s2, 33. If the velocity of light (c), gravitational constant (G), and Planck’s constant (h) are chosen as fundamental, units, then which of the following represents the dimensions of the mass?, (a) [c1/2 G1/2 h1/2], (b) [c1/2 G–1/2 h–1/2], 1/2, –3/2 1/2, (c) [c G h ], (d) [c–1/2 G1/2 h1/2], ∆V, 34. The quantity X is given by ε 0 L, where e0 is the per∆t, mittivity of free space, L is a length, DV is a potential, difference and Dt is a time interval. The dimensional, formula for X is same as that of, (a) resistance, (b) charge, (c) voltage, (d) current, 35. In the plane progressive wave propagating with velocity, v, the displacement of a wave particle at a position x in, time t is represented by the equation:, y = a sin k(vt – x), where, a is the amplitude. The dimension of k will be, (a) [LT –1], (b) [LT 0], –1 –1, (c) [L T ], (d) [L–1T 0], a , , 36. In the gas equation  P + 2  (V − b) = RT, the dimenV , , sions of constant a is, (a) [L3], (b) [ML3T–2], 5 –2, (c) [ML T ], (d) [ML2T 0], 37. In the relation y = a cos(wt – kx), the dimensional formula for k is, , ■, , 1.17, , (a) [M 0L–1T –1], (b) [M0LT –1], 0 –1 0, (c) [M L T ], (d) [M 0LT], 38. If E, M, L and G denotes energy, mass, angular momentum and universal gravitational constant, respectively,, then EL2/M 5G 2 represents the unit of, (a) length, (b) mass, (c) time, (d) angle, 39. If the energy (E), velocity (v) and force (F) be taken as, the fundamental quantity, then the dimensions of mass, will be, (a) Fv–2, (b) Fv–1, –2, (c) Ev, (d) Ev2, 40. Position of a body with acceleration a is given by, x = Ka mt n, here t is time. Find dimensions of m and n., (a) m = 1, n = 1, (b) m = 1, n = 2, (c) m = 2, n = 1, (d) m = 2, n = 2, 41. If the dimensions of length are expressed as G x c y h z,, where G, c and h are the universal gravitational constant, speed of light and the Planck’s constant, respectively, then, 1, 1, 1, 1, (b) x = , z = −, (a)=, x =, ,y, 2, 2, 2, 2, 3, 1, 1, 3, (c)=, (d) y = − , z =, y =, ,z, 2, 2, 2, 2, 42. To determine Young’s modulus of a wire, the formula, F L, is Y = ⋅, , where F/A is the stress and L/∆L is the, A ∆L, strain. The conversion factor to change Y from CGS to, MKS system is, (a) 1, (b) 10, (c) 0.1, (d) 0.01, 43. If E = energy, G = gravitational constant, I = impulse, and M = mass, the dimensions of GIM2/E2 are same as, that of, (a) time, (b) mass, (c) length, (d) force, 44. Frequency is the function of density (ρ), length (a) and, surface tension (T ). Then its value is, (a) k ρ 1/ 2 a3/ 2 / T, (b) k ρ 3/ 2 a3/ 2 / T, (c) kρ1/2a3/2/T 3/4, , (d) None of these, 45. The velocity of a freely falling body changes as g phq, where g is the acceleration due to gravity and h is the, height. The values of p and q are, 1 1, 1, (b) 1,, (a) ,, 2 2, 2, 1, (c) , 1, (d) 1, 1, 2

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1.18, , ■, , Chapter 1, , 46. A small steel ball of radius r is allowed to fall under, gravity through a column of a viscous liquid of coefficient of viscosity h. After some time the velocity of the, ball attains a constant value known as terminal velocity, vT . The terminal velocity depends on (i) the mass of the, ball m, (ii) h, (iii) r and (iv) acceleration due to gravity g.Which of the following relations is dimensionally, correct?, ηr, mg, (a) vT ∝, (b) vT ∝, ηr, mg, (c) vT ∝ hrmg, , (d) vT ∝, , mgr, η, , 47. In a system of units if force (F), acceleration (A) and, time (T) and taken as fundamental units then the dimensional formula of energy is, (a) FA2T, (b) FAT2, 2, (c) F AT, (d) FAT, 48. The number of significant figure in 6.25 × 105 is, (a) 11, (b) 6, (c) 4, (d) 3, 49. The current flowing through a resistor 10.932 ohm is, 4.25 amp. The potential difference is 46.461 volt. The, potential in significant figures is, (a) 46.461 V, (b) 46.46 V, (c) 46.4 V, (d) 46.0 V, 50. Accuracy in the measurement of a physical quantity can, be increased by using, (a) less trials, (b) more trials, (c) significant digits, (d) order of magnitude, 51. In a slide calliper, n divisions of vernier scale coincides, with (n – 1) divisions of main scale. The least count of, the instrument is, 1, 1, (a) MSD, (b), MSD, n, n −1, n, n −1, (c), (d), MSD, MSD, n −1, n, 52. In a vernier callipers ten smallest divisions of the vernier scale are equal to nine smallest divisions of the, main scale. If the smallest division of the main scale is, half millimeter then vernier constant is, (a) 0.005 mm, (b) 0.05 mm, (c) 0.5 mm, (d) 0.1 mm, 53. A vernier calliper is used to measure the length of a cylinder., 10 divisions of vernier scale coincides with 9 divisions of, the main scale. The best suited length measured is, (a) 9.01 mm, (b) 9.01 cm, (c) 9.628 cm, (d) 9.99 cm, , 54. In a vernier callipers, one main scale division is x cm, and n division of the vernier scale coincide with (n –1), divisions of the main scale. The least count (in cm) of, the callipers is,  n −1 , (a) , x,  n , , (b), , nx, ( n −1), , x, x, (d), ( x −1), n, 55. A spherometer has a least count of 0.005 mm and its, head scale is divided into 200 equal divisions. The distance between the consecutive threads on the spherometer screw is, (a) 1 mm, (b) 0.1 mm, (c) 0.05 mm, (d) 0.005 mm, (c), , 56. A spherical body of mass m and radius r is allowed, to fall in a medium of viscosity h. The time in, which the velocity of the body increases from zero, to 0.63 times the terminal velocity (v) is called, time constant (τ). Dimensionally τ can be represented by, (a), , (c), , mr 2, 6πη, m, 6πη rv, , (b), , 6π mr η, g2, , (d) None of these, , 57. While finding specific heat capacity using calorimeter,, error might occur due to, (a) absence of heat loss reducing covers, (b) absence of water equivalent, (c) both (a) and (b), (d) Neither (a) nor (b), 58. The values of measurement of a physical quantity in 5, trials were found to be 1.22, 1.23, 1.23, 1.24 and 1.25., Then pick up the incorrect answer:, (a) average absolute error is 0.01, (b) relative error is 0.01, (c) percentage error is 1%, (d) percentage error is 0.1%, 59. Choose the incorrect statement out of the following:, (a) Every measuremnt by any measuring instrument, has some error., (b) Every calculated physical quantity that is based on, measured values has some error., (c) A measurement can have more accuracy but less, precision and vice versa., (d) The percentage error is different from relative error.

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Dimensions and Error Analysis in Measurement, 60. The length of a simple pendulum executing simple, harmonic motion is increased by 21%. The percentage, increase in the time period of pendulum of increased, length is, (a) 11%, (b) 21%, (c) 42%, (d) 10.5%, 61. The length of a given cylindrical wire is increased by, 100%. Due to consequent decrease in diameter the, change in the resistance of the wire will be, (a) 200%, (b) 100%, (c) 50%, (d) 300%, 62. If physical quantity x is represented by x = [M aLbT –c], and the maximum percentage errors in M, L and T are, α%, β% and γ %, respectively then the total maximum, error in x is, (a) (αa + βb – γc) × 100%, (b) (αa + βb + γc) × 100%, (c) (αa – βb – γc) × 100%, αa + βb, (d), ×100%, γc, 63. While measuring acceleration due to gravity by a simple pendulum, a student makes a positive error of 2% in, length of the pendulum and a positive error of 1% in the, value of time period. The actual percentage error in the, measurement of the value of g will be, (a) 3%, (b) 4%, (c) 5%, (d) 0%, 64. The percentage errors in the measurement of mass and, speed are 2% and 3%, respectively. How much will be, the maximum error in kinetic energy?, (a) 1%, (b) 5%, (c) 8%, (d) 12%, 65. A physical parameter a can be determined by measuring the parameters b, c, d and e using the relation, a = bαcβ/d γe δ. If the maximum errors in the measurement of b, c, d and e are b1%, c1%, d1% and e1%, then, the maximum error in the value of a determined by the, experiment is, (a) (b1 + c1 + d1 + e1)%, (b) (b1 + c1 – d1 – e1)%, (c) (αb1 + βc1 – γd1 – δe1)%, (d) (αb1 + βc1 + γd1 + δe1)%, 66. Heat is evolved in a resistance on passing current up to, definite time. Measurements for current time and resistance suffer practical errors of magnitudes 1%, 2% and, 2%, respectively. The maximum percentage error in the, heat evolved will be, , (a) 3%, (c) 6%, , ■, , 1.19, , (b) 3/4%, (d) 4%, , 67. The random error in the arithmetic mean of 100 observations is x, then random error in the arithmetic mean, of 400 observations would be, 1, x, 4, (c) 4x, (a), , 1, x, 2, (d) 2x, (b), , 68. If the error in the measurement of momentum of a particle is 100% then the error in the measurement of kinetic, energy would be, (a) 400%, (b) 300%, (c) 200%, (d) 100%, 69. The measured mass and volume of a body are 22.42 g, and 4.7 cm3, respectively. The maximum possible error, in density is approximately, (a) 2%, (b) 0.2%, (c) 1%, (d) 10%, V, where V = 100 ± 5 volts and, i, i = 10 ± 0.2 amperes. What is the total error in R?, (a) 5%, (b) 7%, , 70. The resistance R =, , (c) 5.2%, , (d), , 5, %, 2, , 71. The period of oscillation of a simple pendulum in, the experiment is recorded as 2.63 s, 2.56 s, 2.42 s,, 2.71 s and 2.80 s respectively. The average absolute, error is, (a) 0.1 s, (b) 0.11 s, (c) 0.01 s, (d) 1.0 s, 72. If separation between screen and point source is, increased by 2% what would be the effect on the, intensity?, (a) Increases by 4%, (b) Increases by 2%, (c) Decreases by 2%, (d) Decreases by 4%, 73. The heat generated in a circuit is dependent upon the, resistance, current and time for which the current, is flown. If the errors in measuring the above are 1%,, 2% and 1%, respectively, then the maximum error in, measuring heat is, (a) 8%, (b) 6%, (c) 18%, (d) 12%, A2 B, ,, C 1/ 3 D, the percentage error introduced in the measurements, , 74. In the measurement of physical quantity X =

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1.20, , ■, , Chapter 1, , of the quantities A, B, C and D are 2%, 2%, 4% and, 5%, respectively. Then the minimum amount of percentage error in the measurement of X is contributed, by, (a) A, (b) B, (c) C, (d) D, , 75. The velocity of water waves v may depend upon, their wave length λ, the density of water ρ and, the acceleration due to gravity g. The method of, dimensions gives the relation between these quantities as, (a) v2 ∝ λg–1ρ–1, (b) v2 ∝ gλρ, 2, (c) v ∝ gλ, (d) v2 ∝ g–1λ–3, , ANSWER KEYS, 1. (d), , 2. (a), , 3. (c), , 4. (d), , 5. (b), , 6. (a), , 7. (c), , 8. (b), , 9. (a), , 10. (d), , 11. (e), , 12. (d), , 13. (c), , 14. (d), , 15. (d), , 16. (d), , 17. (b), , 18. (b), , 19. (c), , 20. (d), , 21. (b), , 22. (c), , 23. (c), , 24. (a), , 25. (a), , 26. (d), , 27. (b), , 28. (b), , 29. (b), , 30. (c), , 31. (c), , 32. (c), , 33. (c), , 34. (d), , 35. (d), , 36. (c), , 37. (c), , 38. (d), , 39. (c), , 40. (b), , 41. (d), , 42. (c), , 43. (a), , 44. (a), , 45. (a), , 46. (a), , 47. (b), , 48. (d), , 49. (d), , 50. (b), , 51. (a), , 52. (b), , 53. (b), , 54. (c), , 55. (a), , 56. (d), , 57. (c), , 58. (c), , 59. (d), , 60. (d), , 61. (d), , 62. (b), , 63. (d), , 64. (c), , 65. (d), , 66. (c), , 67. (a), , 68. (b), , 69. (a), , 70. (b), , 71. (b), , 72. (d), , 73. (b), , 74. (c), , 75. (c), , HINTS AND EXPLANATIONS FOR SELECTED QUESTIONS, 2. Both the pressure and stress are measured in the units of, N/m2, hence they have same dimensions., Hence, the correct answer is option (a)., , 8. Fourth variable in the relations cannot be derived with, the help of dimensional analysis., Hence, the correct answer is option (b)., , Force Force × displacement, =, Area, Area × displacement, Energy, =, Volume, Hence, the correct answer is option (c)., 5. Wave number is the reciprocal of wavelength. Hence, dimension is (Length)–1., Hence, the correct answer is option (b)., 6. Charge = Current × Time., Hence, the correct answer is option (a)., 7. Water column in metre can never be used as the unit, pressure., Hence, the correct answer is option (c)., , 9. The dimension of torque are [ML2T –2] which is also the, dimensions of work., Hence, the correct answer is option (a)., , 3. Pressure =, , 10. Work = Force × Displacement, = [MLT –2] [L] = [ML2T –2], Energy = Work, Torque = moment of force, = r⊥F = [L] [MLT –2], = [ML2T –2], Hence, the correct answer is option (d)., 11. Surface tension is defined as the force per unit length, and hence its unit is newton metre–1 or Nm–1., Hence, the correct answer is option (e).

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Dimensions and Error Analysis in Measurement, 12. Kinetic energy is a scalar quantity and is same in all the, inertial reference frames., Hence, the correct answer is option (d)., 13. Angular momentum [L] = [M 1L2T –1]. When M, L and, T are doubled the unit of angular momentum becomes, quadrupled., Hence, the correct answer is option (c)., Area, × Resistance, 14. Specific resistance =, Length, Voltage W / Q W ⋅ t, =, = 2, Current, Q /t, Q, Dimensions of Resistance = [ML2T –2 . TQ–2], ∴ Dimensions of specific resistance, = [Resistance] × [Length], = [ML3T –1Q –2], Hence, the correct answer is option (d)., 15. Specific gravity is the ratio, hence dimensionless., Hence, the correct answer is option (d)., where, Resistance =, , 16., , 17., , 18., , 23., , 24., , 25., , Angular momentum, =r, Linear momentum, Dimensions are [M 0L1T 0], Hence, the correct answer is option (d)., , ■, , 1.21, , MLT −2, = [ ML−1T −1 ], L ⋅ LT −1, Hence, the correct answer is option (a)., 26. Given F = a cos αx + b sin βt, Here, αx and βt are both dimensionless, i.e., [M 0L0T 0], Therefore, [α] = [L–1] and [β] = [T –1], β T −1, =, Then,, or [LT –1] or [M 0L1T –1], α L−1, Hence, the correct answer is option (d)., 27. µ0 ε 0 = (MLT –2A2)1/2 . (M –1L–3T 4A–2)1/2, = (L–2T 2)1/2 = L–1T 1, 1, ∴, = [LT –1] which is the unit of velocity, µ0 ε 0, ∴ [η ] =, , Hence, the correct answer is option (b)., 28., , e2, Q2, =, ε0 hc ( M −1 L−3T 2 Q 2 )(ML2T −1 )(LT −1 ), = [M 0L0T 0] hence, dimensionless., Hence, the correct choice is ‘angle’., Hence, the correct answer is option (b)., , 31. From the principle of homogeneity, [at] = [v], [at] = [LT –1], Force, [a] = [LT –2], = newton/metre., Surface tension =, Length, [b ], Also,, [v] = [LT –1], Hence, the correct answer is option (b)., [t + c], [b] = [LT –1][T ] = [L], Poise is the CGS unit of viscosity, i.e.,, –1, –1, and, [t + c] = [T ], gm × cm × sec, [c] = [T ], Now, 1 Poise = 1 gm × cm–1 × sec–1, Hence, the correct answer is option (c)., = 10–3 kg × 10+2 m–1 × sec–1, = 10–1 kg × m–1 × sec–1, 32. [x] = [bt2] ⇒ [b] = [x/t2] = km/s2., –1, –1, = 0.1 kg × m × sec, Hence, the correct answer is option (c)., or, = 0.1 N-s/m2, 33. Velocity, [c] = [M 0LT –1], (1), Hence, the correct answer is option (b)., Gravitational constant,, [G] = [M –1L3T –2], (2), eh, = nτ A (dimensionally), Bohr magneton µ B =, 2 –1, Planck’s, constant,, [h], =, [ML, T, ], (3), 4π mc, 2, From, Eqs., (1),, (2), and, (3),, we, can, solve,, units of µB is therefore amp-m ., [M] = [c3/2 G–3/2 h1/2], Hence, the correct answer is option (c)., Hence, the correct answer is option (c)., Force, = ML−1T −2, ε A, Pressure =, 34. Capacity, C = 0, Area, d, Hence, the dimensions of e0L are same as that of, Restoring force, −1 −2, Stress =, = ML T, capacity., Area, ∆V dimensions of C × dimension of V, Hence, the correct answer is option (a)., Now ε 0 L, =, ∆T, dimension of t, Given, Force F = 6πhrv, dimensions of charge Q, = = dimensions of current, F, dimensions of t, or η =, 6π rv, Hence, the correct answer is option (d).

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1.22, , ■, , Chapter 1, , 35. k(vt – x) must be an angle, hence dimensionless. But, (vt – x) has the dimensions of length, therefore, k has, the dimensions of (Length)–1, i.e., [L–1T 0]., Hence, the correct answer is option (d)., a, 36. 2 must have the dimensions of pressure which are, V, [ML–1T –2]. Since V is the volume having dimensions, [L3], therefore,, [a] = Pressure × (Volume)2, ∴ [a] = [ML–1T –2] × [L3]2, = [ML5T –2], Hence, the correct answer is option (c)., 37. [kx] = Dimension of wt = (dimensionless), 1 1, = = [ L−1 ] ∴ [k ] = [ L−1 ], X L, Hence, the correct answer is option (c)., , 42., , hence k =, , 43., , E⋅L, M 5G 2, 2, , 38., , Energy × ( Angular momentum) 2, (Mass)5 × (Gravitational constant) 2, Here dimensions of E = [ML2T –2], and dimensions of L = mvr = [ML2T –1], The dimensions of G = [M –1L3T –2], ∴, , 44., , EL2, [ML2T −2 ] × [ML2T −1 ]2, =, [ M ]5 × [ M −1 L3T −2 ]2, M 5G 2, , [M 3 L6T −4 ], = [M 0 L0T 0 ], [M 3 L6T −4 ], These are the dimensions of angle., Hence, the correct answer is option (d)., =, , 39. Let m ∝ EavbFc, Dimensionally, we write it as, [M] = K [ML2T –2]a[LT –1]b[MLT –2]c, For dimensional balance, a + c = 1; 2a + b + c = 0; 2a + b + 2c = 0, solving these equations, we get, a = 1, b = –2, c = 0, ∴ m ∝ Ev–2, Hence, the correct answer is option (c)., 40. As x = Ka m × t m, [M0LT0] = [LT –2]m[T ]n = [LmT –2m+n], ∴ m = 1 and – 2m + n = 0 ⇒ n = 2, Hence, the correct answer is option (b)., 41. Given the dimensions of length L,, L = Gxcyhz, (1), Here the dimensions of G(N-m2/kg2) are [M–1L3T –2], The dimensions of C (m/s) are [LT –1], And the dimensions of h(J-s) are [ML2T –1], , 45., , 46., , Dimensionally Eq. (1) can be written as, [L1] = [M –1L3T –2]x[LT –1]y [ML2T –1]z, For dimensional balance –x + z = 0; 3x + y + 2z = 1, and –2x – y – z = 0, Solving these we get, 1, 3, 1, x = , y = − and z =, 2, 2, 2, Hence, the correct answer is option (d)., Stress, Young’s modulus Y =, ( N/ m 2 ), Strain, Its dimensions are [ML–1T –2], CGS units are gm × cm–1 × sec–2, 10 −3 kg, 1 gm, kg, Now,, = −2, = 10 −1, 2, 10 m-sec 2, cm-sec, m-sec 2, Hence, the correct answer is option (c)., [G] = [M –1L3T –2]; [E] = [ML2T–2], [I] = [MLT –1], GIM 2 [M −1 L3T −2 ][MLT −1 ][M 2 ], ∴, =, = [T ], E2, [M 2 L4T −4 ], Hence, the correct answer is option (a)., Let n = kρaabT c where [ρ] = [ML–3], [a] = [L] and [T] = MT–2], Comparing dimensions both side, we get, 1, −1, −3, a = ,b =, and c =, 2, 2, 2, K T, ∴ h = kρ–1/2a–3/2 T–1/2 = 1/ 2 3/ 2, ρ a, Hence, the correct answer is option (a)., Given velocity v = g phq, Taking dimensions of physical quantities on both the, sides, we get, LT –1 = [LT –2]p[L]q, LT –1 = Lp + q.T –2p, ∴ p+q=1, 1, and –2p = –1 or p =, 2, 1, q, =, ., and also, 2, Hence, the correct answer is option (a)., By substituting dimension of each quantity in RHS of, option (a), we get, ,  mg   M × LT −2 , −1,  η r  =  ML−1T −1 × L  = [ LT ], , , This option gives the dimension of velocity., Hence, the correct answer is option (a)., 47. E = KFaAbT c, [ML2T –2] = [MLT –2]a[LT –2]b[T ]c, [ML2T–2] = [MaLa+bT–2a–2b +c], ∴ a = 1, a + b = 2 ⇒ b = 1

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Dimensions and Error Analysis in Measurement, and –2a – 2b + c = –2 ⇒ c = 2, ∴ E = KFAT2, Hence, the correct answer is option (b)., 49. V = IR = 4.25 × 10.932 V, = 46.461 V = 46.0 V, Hence, the correct answer is option (d)., 1,  n −1 , 51. Least count = 1 −,  MSD = MSD, n , n, , Hence, the correct answer is option (a)., 52. Here, half millimetre of vernier scale is divided into, 10 parts. Therefore, vernier constant is, 0.5, = = 0.05 mm, 10, Hence, the correct answer is option (b)., n, , 53. Least count = 1 −  MSD,  m, 9, , = 1 −  = 0.1 mm = 0.01 cm,  10 , The best suited length measured by the vernier, = 9.01 cm, Hence, the correct answer is option (b)., 54. One main scale division, 1 MSD = x cm, ( n −1) x, One vernier scale division, 1 VSD =, n, nx − nx + x x, = cm., LC = 1 MSD – 1 VSD =, n, n, Hence, the correct answer is option (c)., 55. The distance between the consecutive heads of the, spectrometer is the smallest division of the main scale, of spherometer., Smallest division of main scale,  Least count =, Total divisioon of the head scale, ∴ Distance between the consecutive threads, = Least count × Total division of the head scale, = 0.005 mm × 200 = 1 mm, Hence, the correct answer is option (a)., 56. Time constant is the time of free fall of a body under, gravity in the viscous medium during which the, velocity of the body increases to 63% of the terminal, velocity. Dimensionally, none of the alternatives (a),, (b) or (c) has dimensions of time. Hence, the option (d), is correct., Hence, the correct answer is option (d)., 57. While obtaining heat capacity using calorimeter, error, occurs if we ignore water equivalent and reduce the, covers surrounding the calorimeter., Hence, the correct answer is option (c)., , ■, , 1.23, , 58. Average and relative errors are 0.01. Percentage error is, 0.01 × 100 = 1%, Hence, the correct answer is option (c)., 60. Period of simple pendulum is, l, g, , T = 2π, , ∆T 1 ∆l 1, =, = × 21% = 10.5%, T, 2 l, 2, Hence, the correct answer is option (d)., 61. New length = L′ = L + 100% of L, = L + L = 2L., Volume of the cylindrical wire material remains unchanged, therefore, πR2L = πR ′2L, ∴, , R ′2 =, , L 2, L, R2, R =, ⋅ R2 =, L′, 2L, 2, , Since resistance r = Resistivity (ρ ) ×, , length (l ), area (A), , L, L, =ρ, A, π R2, Now resistance after extending the length of wire,, 2L ⋅ 2, L′, r′ = ρ, =ρ, = 4r, 2, π R′, π R2, DR = r′ – r = 3r, 3r, %∆R = × 100% = 300%, r, Hence, the correct answer is option (d)., 62. Given x = M aLbT –c, ∆L, ∆T, ∆x, ∆M, ∴, =a, +b, −c, x, M, L, T, r′ = ρ, , ∆L, ∆T,  ∆x ,  ∆M, and   × 100% =  a, +b, −c, L, T,  M,  x  max, = (aα + bβ + cγ) × 100%, Hence, the correct answer is option (b)., 63. ∴ T = 2π, , l, g, , or, , g=, , 4π 2 l, T2, , ∆g ∆l, ∆T, =, −2, = 2% − 2 × 1% = zero%, g, l, T, Hence, the correct answer is option (d)., ∴, , 64. Kinetic energy K =, , 1 2, mv, 2, , ∆m, ∆v,  ∆K , =, +2, , ,  K  max, m, v, , ,  × 100%, 

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1.24, , ■, , Chapter 1, , = 2% + 2 × 3% = 8%, Hence, the correct answer is option (c)., 65. a = bαcβ / d γ eδ, So maximum error in a is given by,  ∆a, , × 100 , ,  a,  max, , ∆V, ∆I, ,  ∆R, × 100 +, × 100, × 100  =, 70. ∴ ,  max,  R, V, I, , ∆b, ∆c, ∆d, ∆e, =α, × 100 + β, × 100 + γ, × 100 + δ, × 100, b, c, d, e, = (αb1 + β c1 + γd1 + δe1)%, Hence, the correct answer is option (d)., 66. Heat evolved H = I2Rt, Maximum percentage error in H is, , 71. Average value =, , 5, 02, × 100 + × 100 = (5 + 2)% = 7%, 100, 10, Hence, the correct answer is option (b)., =, , 2.63 + 2.56 + 2.42 + 2.71 + 2.80, 5, = 2.62 sec, Now, ∆T1 = 2.63 – 2.62 = 0.01, ⋅, , ∆T2 = 2.62 – 2.56 = 0.06, ∆T3 = 2.62 – 2.42 = 0.20, , ∴K′ =, , p ′2, 2m, , (1), , (2), 2, , 2, , 1, K  p   100 , =  =,  =, 4, K ′  p′   100 + 100 , Now, percentage error in K is, K′− K,  K′ , × 100% = , =, − 1 × 100%, , K, K, = (4 – 1) × 100% or 300%, Hence, the correct answer is option (b)., M, 69. Density, d =, V, The maximum possible error is,,  ∆d, ,  ∆M ∆V , × 100  = , +, ,  × 100%, V ,  d,  max  M, ∴, ,  0.01 0.1 , =, +,  × 100%,  22.42 4.7 , = (0.000446 + 0.0212765) × 100%, = 0.0217225 × 100%, = 2.17% or ≈ 2%, Hence, the correct answer is option (a)., , ∆T4 = 2.71 – 2.62 = 0.09, ∆T5 = 2.80 – 2.62 = 0.18, Mean absolute error, , ⋅, ,  ∆I ∆R ∆t , ,  ∆H, = 2, +, + %, × 100 , ,  max  I,  H, R, t , = (2 × 1 + 2 + 2) = 6%, Hence, the correct answer is option (c)., 67. Random error in the arithmetic mean reduces as more, number of observations are repeated. Here the number, of observations are increased 4 times, hence the error, 1, x, reduces to, times, i.e., ., 4, 4, Hence, the correct answer is option (a)., p2, 68. K =, 2m, If p changes to p′, then K becomes K′, , ⋅, , ⋅, , ⋅, , ∆T1 + ∆T2 + ∆T3 + ∆T4 + ∆T5, 5, 0.54, =, = 0.108 = 0.11 sec., 5, Hence, the correct answer is option (b)., ∆T =, , 72. Intensity ∝, , 1, 1, ,I ∝ 2, 2, (distance ), d, , ∆I, ∆d, ∝ −2, ∝ −2 × 2% = 4%, I, d, Hence, the correct answer is option (d)., 73. Heat generated is given by, H = I2Rt, ∆H, ∆I ∆R ∆t, ∴, =2, +, +, H, I, R, t, = (2 × 2 + 1 + 1)% = 6%, Hence, the correct answer is option (b)., 74. X =, , A2 B, C 1/ 3 D, ∆X, ∆A ∆B 1 ∆C ∆D, −, −, =2, +, X, A, B 3 C, D, 1, = 2 × 2% + 2% − × 4% − 5%, 3, 4, = 4 + 2% − % − 5%, 2, , 4, The percentage error contributed by C is, which is, 3, minimum among A, B, C and D., Hence, the correct answer is option (c)., 75. Let vx = kgy λzρδ. Now by substituting the dimensions of, each quantities and equating the powers of M, L and T, we get δ = 0 and x = 2, y = 1, z = 1., Hence, the correct answer is option (c).

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Dimensions and Error Analysis in Measurement, , ■, , 1.25, , PREVIOUS YEARS’ QUESTIONS, 1. Which pair do not have equal dimensions?, (a) Energy and torque, (b) Force and impulse, (c) Angular momentum and planck constant, (d) Elastic modulus and pressure, [AIPMT 2000], 2. A force F is given by F = at + bt2, where t is the time., What are dimensions of a and b?, (a) [MLT –4] and [MLT 2], (b) [MLT –2] and [MLT 0], (c) [MLT –3] and [MLT –4], (d) [ML2T –3] and [ML3T –4], [AFMC 2000], 3. The number of significant figures in 3400 is, (a) 7, (b) 6, (c) 12, (d) 2, [AFMC 2000], 4. The dimensions of Planck constant equals to that of, (a) energy, (b) momentum, (c) angular momentum, (d) power, [AIPMT 2001], 5. The dimensions of Planck’s constant are, (a) [ML–3T –2], (b) [ML–2T –1], 2 –3, (c) [ML T ], (d) [ML2T –1], [AFMC 2002], 6. The idea of calculus was given by, (a) Newton, (b) Einstein, (c) Marconi, (d) Planck, [AFMC 2003], 7. The unit of permittivity of free space, e0, is, (a) coulomb/newton-metre, (b) newton-metre2/coulomb2, (c) coulomb2/newton-metre2, (d) coulomb2/(newton-metre)2, [AIPMT 2004], 8. The dimensions of universal gravitational constant are, (a) [M –1L32T –2], (b) [ML2T –1], –2 3 –2, (c) [M L T ], (d) [M –2L2T –1], [AIPMT 2004], 9. Pressure gradient has the same dimensions as that of, (a) velocity gradient, (b) potential gradient, (c) energy gradient, (d) None of these, [AFMC 2004], , 10. The volume of a cube in m3 is equal to the surface area, of the cube in m2. The volume of the cube is, (a) 64 m3, (b) 216 m3, (c) 512 m3, (d) 196 m3, [AFMC 2005], 11. Pascal-second has the dimensions of, (a) force, (b) energy, (c) pressure, (d) coefficient of viscosity, [AFMC 2005], 12. Consider the following equation of Bernoulli’s theorem, 1, P = ρ v 2 + ρ gh = K (constant), 2, The dimensions of K/P are same as that of which of the, following?, (a) Thrust, (b) Pressure, (c) Angle, (d) Coefficient of viscosity, [AFMC 2005], 13. If error in measurement of radius of sphere is 1%, what, will be the error in measurement of volume?, (a) 1%, (b) 1/3%, (c) 3%, (d) 10%, [AFMC 2005], 14. The velocity v of a particle at time t is given by, b, v = at +, , where a, b and c are constants. The, t +c, dimensions of a, b and c are, (a) [L], [LT ] and [LT –2], (b) [LT –2], [L] and [T ], (c) [L2], [T ] and [LT –2], (d) [LT –2], [LT ] and [L], [AIPMT 2006],  2π, , (ct − x )  ,, 15. Given that y = A sin , λ, , , where y and x are measured in metres. Which of the following statements is ture?, (a) The unit of λ is same as that of x and A., (b) The unit of λ is same as that of x but not of A., (c) The unit of c is same as that of 2π/λ., (d) The unit of (ct – x) is same as that of 2π/λ., [AFMC 2006]

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1.26, , ■, , Chapter 1, , 16. Dimensions of resistance in an electrical circuit, in, terms of dimension of mass M, of length L, of time T, and of current I, would be, , (a) [ML2T –2], (c) [ML2T –3I –2], , (b) [ML2T –1I –1], (d) [ML2T –3I –1], [AIPMT 2007], , ANSWER KEYS, 1. (b), 11. (d), , 2. (c), 12. (c), , 3. (d), 13. (c), , 4. (c), 14. (b), , 5. (d), 15. (a), , 6. (a), 16. (c), , 7. (c), , 8. (a), , 9. (d), , 10. (b), , HINTS AND EXPLANATIONS FOR SELECTED QUESTIONS, 1. Dimensions of force = [MLT –2], Dimensions of impulse = [MLT –1]., Hence, the correct answer is option (b)., Energy, 4. Dimensions of Planck constant h =, Frequency, [ ML2T −2 ], =, = [ ML2T −1 ], [T −1 ], Dimensions of angular momentum L, = Momentum of inertia I × Angular velocity w, = [ML2] [T –1] = [ML2T –1], Hence, the correct answer is option (c)., 7. Force between two charges, 1 q2, 1 q2, F=, ε, ⇒, =, = C2 / N-m 2, 0, 4πε 0 r 2, 4π Fr 2, Hence, the correct answer is option (c)., 8. Gravitational constant G, force × (distance) 2, =, mass × mass, , b, t +c, As c is added to t,, , 14. ν = at +, , ∴ c = [T ], −1, , at = [LT –1], , or,, , a=, , [ LT ], [ LT −2 ], [T ], , [b ], = [ LT −1 ] ∴ [b] = [L]., [T ], Hence, the correct answer is option (b)., , 16. According to Ohm’s law,, V, V = RI or R =, I, Dimensions of V =, , W [ ML2T −2 ], =, q, [ IT ], , [ ML2T −2 / IT ], = [ ML2T −3 I −2 ], [I ], Hence, the correct answer is option (c)., , ∴ R=, , [ MLT −2 ][ L2 ], = [ M −1 L3T −2 ], [ M ][ M ], Hence, the correct answer is option (a)., ∴ Dimensions of G =, , QUESTIONS FROM NCERT EXEMPLAR, 1. The number of significant figures in 0.06900 is, (a) 5, (b) 4, (c) 2, (d) 3, 2. The sum of the numbers 436.32, 227.2 and 0.301 in, appropriate significant figures is, (a) 663.821, (b) 664, (c) 663.8, (d) 663.82, , 3. The mass and volume of a body are 4.237 g and 2.5 cm3,, respectively. The density of the material of the body in, correct significant figures is, (a) 1.6048 g/cm3, (b) 1.69 g/cm3, (c) 1.7 g/cm3, (d) 1.695 g/cm3

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Dimensions and Error Analysis in Measurement, 4. The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give, (a) 2.75 and 2.74, (b) 2.74 and 2.73, (c) 2.75 and 2.73, (d) 2.74 and 2.74, 5. The length and breadth of a rectangular sheet are 16.2, cm and 10.1 cm, respectively. The area of the sheet in, appropriate significant figures and error is, (a) (164 ± 3) cm2, (b) (163.62 ± 2.6) cm2, 2, (c) (163.6 ± 2.6) cm, (d) (163.62 ± 3) cm2, 6. Which of the following pairs of physical quantities does, not have same dimensional formula?, (a) Work and torque., (b) Angular momentum and Planck’s constant., (c) Tension and surface tension., (d) Impulse and linear momentum., 7. Measure of two quantities along with the precision of, respective measuring instrument is, A = 2.5 m/s ± 0.5 m/s, B = 0.10 s ± 0.01 s, The value of A B will be, (a) (0.25 ± 0.08) m, (b) (0.25 ± 0.5) m, (c) (0.25 ± 0.05) m, (d) (0.25 ± 0.135) m, , ■, , 1.27, , 8. You measure two quantities as A = 1.0 m ± 0.2 m,, B = 2.0 m ± 0.2 m. We should report correct value for as, (a) 1.4 m ± 0.4 m, (b) 1.41 m ± 0.15 m, (c) 1.4 m ± 0.3 m, (d) 1.4 m ± 0.2 m, 9. Which of the following measurements is most precise?, (a) 5.00 mm, (b) 5.00 cm, (c) 5.00 m, (d) 5.00 km, 10. The mean length of an object is 5 cm. Which of the following measurements is most accurate?, (a) 4.9 cm, (b) 4.805 cm, (c) 5.25 cm, (d) 5.4 cm, 11. Young’s modulus of steel is 1.9 × 1011 N/m2. When, expressed in CGS units of dynes/cm2, it will be equal to, (1 N = 105 dyne, 1 m2 = 104 cm2), (a) 1.9 × 1010, (b) 1.9 × 1011, 12, (c) 1.9 × 10, (d) 1.9 × 1013, 12. If momentum (P), area (A) and time (T) are taken to, be fundamental quantities, then energy has the dimensional formula, (a) (P1 A–1 T 1), (b) (P2 A1 T 1), (c) (P1 A–1/2 T 1), (d) (P1 A1/2 T –1), , ANSWER KEYS, 1. (b), 11. (c), , 2. (b), 12. (d), , 3. (c), , 4. (d), , 5. (a), , 6. (c), , 7. (a), , 8. (d), , 9. (a), , 10. (a), , HINTS AND EXPLANATIONS FOR SELECTED QUESTIONS, 1. In decimals zeroes on left of decimal and before first, non-digit are not significant so number of significant, figures are four (6900)., Hence, the correct answer is option (b)., 2. On calculating sum of digits arithmetically we, obtain 663.821, since the number with least decimal, place is 227.2, so rounding off to one decimal place, therefore 664., Hence, the correct answer is option (b)., 3. Density is 4.237/2.5 = 1.6948, rounding off the number,, we get 1.7., Hence, the correct answer is option (c)., , 4. 2.745 on rounding off = 2.74, 2.735 on rounding off = 2.74., Hence, the correct answer is option (d)., 5. A = l × b, 16.2 × 10.1 = 163.62 cm2, Rounding off to three significant digits, area A = 164 cm2, DA = A × (2.63/163.62) = 3 cm2, Therefore Area A = A ± DA = (164 ± 3) cm2., Hence, the correct answer is option (a).

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1.28, , ■, , Chapter 1, , 7. x = AB = (2.5)(0.10) = 0.25 m, Dx/x = DA/A + DB/B, = 0.075/0.25., Hence, the correct answer is option (a)., 8. Y =, , AB = 1.414 m, 1, DY/Y = (DA/A + DB/B) = 0.212, 2, Rounding off DY = 0.2 m, Thus, 1.4 ± 0.2 m., Hence, the correct answer is option (d)., , 9. Here 5.00 mm is least measurement upto two decimal, places, so it is more precise., Hence, the correct answer is option (a)., 10. l = 5 cm, Dl1 = 5 - 4.9 = 0.1 cm, Dl2 = 5 - 4.805 = 0.195 cm, Dl3 = 5.25 -5 = 0.25 cm, Dl4 = 5 - 5 = 0.4 cm, Error Dl1 is least., Hence, the correct answer is option (a).

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Dimensions and Error Analysis in Measurement, , ■, , 1.29, , AIIMS-ESSENTIALS, ASSERTION AND REASON, In the following questions, a statement of assertion is followed by a statement of reason. You are required to choose, the correct one out of the given five responses and mark it as, (a) If both assertion and reason are true and reason is the, correct explanation of the assertion., (b) If both assertion and reason are true but reason is not, correct explanation of the assertion., (c) If assertion is true, but reason is false., (d) If both assertion and reason are false., (e) If reason is true but assertion is false., 1. Assertion: The measure of a physical quantity is independent of the system of units., Reason: The smaller is unit, bigger is the measure of, the physical quantity and vice-versa., 2. Assertion: The unit of force is written as newton and, not as Newton., Reason: The unit named to commemorate a scientist is, not written with capital initial letter., 3. Assertion: It is wrong to write newton metre (the unit, of moment of force) as Nm., Reason: No full stop is put between the symbols for, units., 4. Assertion: The unit of moment of force Nm, i.e.,, newton metre cannot be written as mN., Reason: Since ‘m’ is symbol for both metre and milli,, the symbol for metre, i.e., m should never be written, before the symbol of another unit., 5. Assertion: Nm and mN are the units of different physical quantities., Reason: Nm (newton metre) is the unit of moment of, force, while mN (millinewton) is the unit of force., 6. Assertion: nm and mN are the units of different physical quantities., Reason: nm (nano metre) is the unit of length, while, mN (millinewton) is the unit of force., 7. Assertion: The parallax method cannot be used to find, the distance of very distant star., Reason: The parallax angle of the star becomes too, small to be measured accurately., 8. Assertion: Mass of a body and the velocity, with which, it is moving, cannot be multiplied., Reason: It is because, mass is a scalar and velocity is, a vector quantity. In addition to this, the dimensions of, mass and velocity are different., , 9. Assertion: The force acting on a body can be divided, by the time, for which it acts; although force and time, have different dimensions., Reason: Any vector quantity can be divided by a scalar, irrespective of their dimensions., 10. Assertion: Force on a body and the velocity, with which, it is moving, cannot be added., Reason: It is because, the dimensions of force and, velocity are different., 11. Assertion: The physical quantities different in nature, but having the same dimensions must possess the same, units., Reason: If the dimensions of two physical quantities, are same, their units must also be same., 12. Assertion: Density (ratio of mass to volume) is different from linear mass density., Reason: Linear mass density is mass per unit length, and is different from density, which usually means volume density., 13. Assertion: Frequency has the dimensions of velocity, gradient., Reason: The dimensional formula of both frequency, and velocity gradient is [M 0L0T –1]., 14. Assertion: A given physical relation may not be correct, even when the dimensions of each and every term, on its either side are the same., Reason: Physical quantities, entirely different in nature,, may possess the same dimensional formula., 15. Assertion: Two students measure the length of a stick, as 1.3 m and 1.30 m. Both the measurements are equally, accurate., Reason: The zero at the end of a number is always, meaningless., 16. Assertion: The number of significant figures in 0.001, is 1., Reason: All zeros to the right of a decimal point and to, the left of a non-zero digit are never significant., 17. Assertion: The number of significant figures in 0.100, is 1., Reason: The zeros at the end of a number are always, meaningless.

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1.30, , ■, , Chapter 1, , ANSWER KEYS, 1. (a), , 2. (a), , 3. (a), , 4. (a), , 5. (a), , 6. (a), , 7. (a), , 11. (d), , 12. (a), , 13. (a), , 14. (a), , 15. (d), , 16. (a), , 17. (d), , 8. (d), , 9. (a), , 10. (a), , HINTS AND EXPLANATIONS FOR SELECTED QUESTIONS, 1. The measure of the physical quantity is given by, X = nu,, where u is the size of the unit and n is the numerical, value of the physical quantity X for the selected unit. It, follows that if the size of the chosen unit is small, then, the numerical value of the quantity will be large and, vice-versa., Thus, both Assertion and Reason are true., Hence, the correct answer is option (a)., 2. Both are true., Hence, the correct answer is option (a)., 3. Both are true., Hence, the correct answer is option (a)., 4. Both are true., Hence, the correct answer is option (a)., 5. Both are true., Hence, the correct answer is option (a)., 6. Both are true., Hence, the correct answer is option (a)., b, 7. The parallax angle is given by θ = ,, S, where b is the base (distance between two points on the, earth) and S is the distance of the star from the earth., For a very distant star, S is very large and hence the, parallax angle θ becomes too small to be measured, accurately., Thus, both Assertion and Reason are true., Hence, the correct answer is option (a)., 8. Any physical quantity can be multiplied with any other, physical quantity irrespective of their dimensions., Thus, both Assertion and Reason are false., Hence, the correct answer is option (d)., 9. Any vector quantity can be divided with a scalar quantity irrespective of their dimensions., Both Assertion and Reason are true., Hence, the correct answer is option (a)., , 10. Force and the velocity have different dimensions. Only, the physical quantities having the same dimensions can, be added (or subtracted)., Thus, both Assertion and Reason are true., Hence, the correct answer is option (a)., 11. The physical quantities different in nature but having the same dimensions possess different units. For, example, the dimensions of force and energy density, are same, i.e., [MLT –2] but their units are N and J m–3, respectively. It is a separate thing that these units turn, out to be equivalent to each other., Thus, both Assertion and Reason are false., Hence, the correct answer is option (d)., 12. Both are true., Hence, the correct answer is option (a)., 13. Both are true., Hence, the correct answer is option (a)., 14. Both are true., Hence, the correct answer is option (a)., 15. A measurement made to the second decimal place is, more accurate., All zeros to the right of the last non-zero digit after the, decimal point are significant., Thus, both Assertion and Reason are false., Hence, the correct answer is option (d)., 16. Zeros at the beginning of a number are not significant., They merely locate the decimal point. Therefore, the, number of significant figures in 0.001 is 1., Thus, both Assertion and Reason are true., Hence, the correct answer is option (a)., 17. All zeros to the right of the last non-zero digit after the, decimal point are significant. Therefore, the number of, significant figures in 0.100 is 3., Thus, both Assertion and Reason are false., Hence, the correct answer is option (d).

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Dimensions and Error Analysis in Measurement, , ■, , 1.31, , PREVIOUS YEARS’ QUESTIONS, SECTION - I (ASSERTION-REASON TYPE), In the following questions, a statement of assertion is followed by a statement of reason. You are required to choose, the correct one out of the given five responses and mark it as, (a) If both assertion and reason are true and reason is the, correct explanation of the assertion., (b) If both assertion and reason are true but reason is not, correct explanation of the assertion., (c) If assertion is true, but reason is false., (d) If both assertion and reason are false., (e) If reason is true but assertion is false., 1. Assertion: SI units are logical and coherent., Reason: SI system of units is a rationalised system., [AIIMS, 2002], 1 T, , where symbols, 2. Assertion: In the relation f =, 2l µ, have standard meaning, m represent linear mass density., Reason: The frequency has the dimensions of inverse, of time., [2008], 3. Assertion: The dimensional formula for product of, resistance and conductance is same as for dielectric, constant., Reason: Both have dimensions of time constant., [2009], , SECTION - II (MULTIPLE CHOICE, QUESTIONS TYPE), 1. The dimension of the modulus of rigidity, is, (a) [ML –2T –2], (b) [MLT–2], –1 –1, (c) [ML T ], (d) [ML–1T –2], [AIIMS, 1994], 2. One nanometre is equal to, (a) 10–7 cm, (b) 109 cm, –9, (c) 10 cm, (d) 10–6 cm, [1994], 3. What is the dimensional formula for the gravitational, constant?, (a) [M –1L3T –2], (b) [M –1L3T –1], (c) [M –2L3T –2], (d) [M –2L–1T –3], [1995], 4. Light year is the unit of, (a) velocity, (b) time, (c) intensity of light, (d) distance, [1996], , 5. Electron-volt (eV) is unit of, (a) energy, (b) charge, (c) current, (d) potential, [1997], 6. The dimension of Plank’s constant is, (a) [ML2T –1], (b) [ML3T –1], –2 –1, (c) [ML T ], (d) [M 0L–1T –3], 7. The dimensions of angular velocity, is, (a) [M 0L0T –1], (b) [M 2L0T –1], –2, (c) [MLT ], (d) [ML2T –2], [1998], 8. How many significant figures are there in 30.00?, (a) 2, (b) 4, (c) 3, (d) 1, [1999], 9. Dimensions [ML–1T –1] are related to, (a) torque, (b) work, (c) energy, (d) coefficient of viscosity, [1999], 10. What is the dimensional formula of gravitational constant G?, (a) [M –1L3T –2], (b) [M –2L3T –2], (c) [M –1L2T –2], (d) [M –1L3T –1], [2000], 11. A body of mass 20.00 g has volume 5.0 cm3. The, maximum possible error in the measurement of mass, and volume respectively are 0.01 and 0.1 cm3. The percentage error in the density will be nearest to, (a) 1%, (b) 2%, (c) 11%, (d) 25%, [2000], 12. Speed in kilometre per hour in SI unit is represented by, (a) KMPH, (b) Kmhr–1, –1, (c) Kmh, (d) kilometre/hour, [2001], 13. Dimension of resistivity is, (a) [ML2T –2I –1], (b) [ML3T –3I –2], 3 –2 –1, (c) [ML T I ], (d) [ML2T –2I –2], [2001], 14. SI unit of velocity is, (a) m/s, (c) mhr–1, , (b) m sec–1, (d) m/hr, [2001], , 15. The heat produced in a long wire is charactrised by, resistance, current and time through which the current, passes. If the errors in measuring these quatities are

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1.32, , ■, , Chapter 1, , respectively 1%, 2% and 1%, then total error in calculating the energy produced is, (a) 4%, (b) 6%, (c) 4/3%, (d) 8%, [2001], 16. What is the dimensional formula for the gravitational, constant?, (a) [M –1L3T –2], (b) [M –1L3T –1], –2 3 –2, (c) [M L T ], (d) [M –2L–1T 3], [2001], 17. Length cannot be measure by, (a) fermi, (b) micron, (c) debye, (d) light year, [2002], 18. The dimension of torque is, (a) [MT –2], (b) [ML–1T –1], 3 –2, (c) [ML T ], (d) [ML3T –3], [2002], 19. Using mass (M), length (L), time (T) and current (A) as, fundamental quantities, the dimension of permeability is, (a) [M –1LT –2A], (b) [M–L2T –2A–1], –2 –2, (c) [MLT A ], (d) [M–LT –1A–1], [2003], 20. Using mass (M), length (L), time (T) and current (A) as, fundamental quantities, the dimension of permittivity is, (a) [ML–2T 2A], (b) [M –1L–3T 4A2], –2, (c) [MLT A], (d) [ML2T –1A2], [2004], 21. “Parsec” is the unit of, (a) time, (b) distance, (c) frequency, (d) angular acceleration, [2005], 22. Dimension of electrical resistance is, (a) [ML2T –3A–1], (b) [ML2T –3A–2], 3 –3 –2, (c) [ML T A ], (d) [ML–1L3T 3A2], [2005], 23. The magnetic moment has dimensions of, (a) [LA], (b) [L2A], –1, (c) [LT A], (d) [L2T –2A], [2006], 24. Which of the following physical quantities do not have, same dimensions?, (a) Pressure and stress, (b) Tension and surface tension, (c) Strain and angle, (d) Energy and work, [2007], 25. In an electrical circuit containing L, C and R which of the, following does not denote the dimensions of frequency?, , (a) LC, , (b), , 1, (c), RC, , R, (d), L, , 26. Lumen is the unit of, (a) luminous flux, (c) illumination, , 1, LC, [2008], , (b) luminosity, (d) quantity of light, [2008], , 27. Which of the following is matched wrongly?, (a) Oil drop experiment → Millikan, (b) Dual nature of light → de Brogile, (c) Uncertainty principle → Heisenberg, (d) None of these, [2008], 28. The dimensions of specific resistance is, (a) [ML2T –2A–1], (b) [ML3T –3A–2], 3 –2 –1, (c) [ML T A ], (d) [ML2T –2A–2], [2009], 29. The dimensional formula of Planck’s constant is, (a) [ML2T –1], (b) [ML2T –2], 0 –2, (c) [ML T ], (d) [MLT 2], [2009], 30. If the energy, E = Gp hq cr, where G is the universal gravitational constant, h is the Planck’s constant and c is the velocity of light, then the values of p, q and r are, respectively, (a) –1/2, 1/2 and 5/2, (b) 1/2, –1/2 and –5/2, (c) –1/2, 1/2 and 3/2, (d) 1/2, –1/2 and –3/2, [2010], 31. Which of the following pairs does not have same, dimensions?, (a) Impulse and momentum, (b) Moment of inertia and moment of force, (c) Angular momentum and Planck’s constant, (d) Work and torque, [2010], 32. What is the dimensions of magnetic field B in terms of, C (= coulomb), M, L, T ?, (a) [M 1L1T –2C], (b) [M 1L0T –1C–1], 1 0 –2, (c) [M L T C], (d) [M 1L0T –1C], [2011], 33. Dimensional formula of DQ, heat supplied to the system is given by, (a) [M 1L2T –2], (b) [M 1L1T –2], 1 2 –1, (c) [M L T ], (d) [ML1T –1], [2012], 34. Dimensional formula of angular momentum is, (a) [ML2T –1], (b) [M 2L2T –2], 2 –3, (c) [ML T ], (d) [MLT –1], [2013], 35. The pressure on a square plate is measaured by measuring the force on the plate and the length of the sides of

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Dimensions and Error Analysis in Measurement, F, . If the maximum, l2, errors in the measurement of force and length are 4%, and 2% respectively, then the maximum error in the, measurement of pressure is, (a) 1%, (b) 2%, (c) 8%, (d) 10%, [2014], the plate by using the formula p =, , ■, , 1.33, , 36. In terms of basic units of mass (M), length (L), time (T ), and charge (Q), the dimensions of magnetic permeability of vacuum (µ0) would be, (a) [MLQ –2], (b) [LT –1Q –1], 2 –1 –2, (c) [ML T Q ], (d) [LTQ –1], [2015], , ANSWER KEYS, Section - I (Assertion-Reason Type), 1. (b), , 2. (b), , 3. (c), , Section - II (Multiple Choice Questions Type), 1. (d), , 2. (c), , 3. (a), , 4. (d), , 5. (a), , 6. (a), , 7. (a), , 8. (b), , 9. (d), , 10. (a), , 11. (b), , 12. (b), , 13. (c), , 14. (b), , 15. (b), , 16. (a), , 17. (a), , 18. (a), , 19. (c), , 20. (b), , 21. (b), , 22. (b), , 23. (b), , 24. (b), , 25. (d), , 26. (a), , 27. (d), , 28. (b), , 29. (a), , 30. (a), , 31. (b), , 32. (b), , 33. (a), , 34. (a), , 35. (c), , 36. (a), , HINTS AND EXPLANATIONS FOR SELECTED QUESTIONS, Section - I (Assertion-Reason Type), 1. A coherent system means a system based on a certain set of basic units from which all derived units are, obtained by multiplication or division without introducing numerical factors. In SI system of units, heat, energy, electric energy and machanical energy are, measured in joule (J). But in CGS system, they have, different units., Hence, the correct answer is option (b)., 2. From f =, or, , µ=, , T, 1 T 2, ,f = 2, 2l µ, 4l µ, T, 2, , 4l f, , 2, , =, , [ MLT −2 ] M, Mass, =, =, [L2T −2 ], L Length, , = Linear mass density., Hence, the correct answer is option (b)., 3. Both the quatities are dimensionless., Resistance × conductance = R × 1/R = (M 0 L0T 0), = 1 and dielectric constant k is dimensionless., Hence, the correct answer is option (c)., , Section - II (Multiple Choice Questions, Type), 1. The modulus of rigidity, , η=, , Shearing stress F / A [ MLT −2 ], =, =, = [ ML−1T −2 ], φ, Shearing strain, [ L2 ], , Hence, the correct answer is option (d)., 3. Gravitational constant (G ) =, , (Force)(Distance) 2, (Mass)2, , [ MLT −2 ][ L2 ], = [M −1 L3T −2 ], [ M ]2, Hence, the correct answer is option (a)., , 6. Planck’s constant (h), =, , Energy in each photon [ E ] [ ML2T −2 ], =, =, Frequency of radiation [ν ], [T −1 ], , = [ML2T –1]., Hence, the correct answer is option (a).

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1.34, , Chapter 1, , ■, ,   , 9. Torque is defined as τ = r × F, ∴ Dimensions of torque [τ] = [r] × [F], = [L] × [MLT –2] = [ML2T –2], Similarly for work and energy the dimensions are same, as that of torque, i.e., [ML2T –2], F, F = 6πhrv ⇒ η =, 6πη rν, F, [ MLT −2 ], =, [ ML−1T −1 ], [r ][ν ] [ L][ LT −1 ], ∴ The given dimensions are related to the coefficient, of viscosity., Hence, the correct answer is option (d)., 10. Gravitational constant comes in the formula, Gm1m2, Fr 2, F=, G, or, =, r2, m1m2, ∴ [η ] =, , metre, ( metre) 2, 2, sec, =, = kg −1 (metre)3 (sec) −2, kg 2, = [M –1L3T –2]., Hence, the correct answer is option (a)., ∆ρ ∆M ∆V, =, +, 11. r = M/V. Hence,, M, V, ρ, kg, , 0.01 0.1, +, = 0.0205 ≅ 2%., 20.00 5.0, Hence, the correct answer is option (b)., 13. By definition, V, A, RA I, l, R=ρ, ⇒ ρ=, =, l, l, A, =, , =, , VA, [V ][ A], ∴ [ρ ] =, [ I ][l ], Il, , ML2T −2 L2, = [ ML2T −2 I −1 ], IL, Hence, the correct answer is option (c)., =, , 15. The heat produced in a wire due to current flow is given by,, ∆H 2∆I ∆R ∆t, ∴, =, +, +, H = I2Rt, H, I, R, t, = 2 × 0.02 + 0.01 + 0.01 = 0.06 = 6%, Hence, the correct answer is option (b)., 17. Debye is the unit of electrie dipole moment. Therefore,, length cannot be measured by debye., Hence, the correct answer is option (a)., 18. Dimension of torque = dimensions of force × dimensions of distance = [MLT -2] [L] = [ML2T -2]., Hence, the correct answer is option (a)., , 19. We know that the force per unit length of a wire carrying current due to another parallel wire carrying current, is given by,, dF µo i1i2, 2π d dF, =, ⇒ µo =, dI, i1i2 dI, 2π d, [ L] [ MLT −2 ], [ A2 ] [ L], or [µo] = [MLT –2] [A–2] = [MLT –2A–2]., Hence, the correct answer is option (c)., 20. Force of attraction between two charges is given by,, 1 q1q2, 1 1 q1q2, F=, ⇒ εo =, 4π F r 2, 4πε 0 r 2, ∴ [ µo ] =, , A2T 2, = M –1 L–3T 4 A2 ., MLT −2 L2, Hence, the correct answer is option (b)., 21. Parsec (pc) is an astronomical unit of length equal to the, distance at which a baseline of one astronomical unit, subtends an angle of one second of arc., 1 parsec = 3.085677 × 1016 m = 3.26 light years., Hence, the correct answer is option (b)., =, , 22. Resistance =, , Potential difference, current, , ML2T −3 A−1, =ML2T –3 A–2, A, Hence, the correct answer is option (b)., 23. The SI unit of magnetic moment is Am2., Therefore its dimensions is [L2A]., Hence, the correct answer is option (b)., 24. Pressure and stress both have the dimensions of force/area., Strain and angle are both dimensionsless. Energy and, work have the same dimensions force × distance., Tension and surface tension refer to two different, physical quantities and their dimensions are different., Tension is a force and surface tension is force per unit, length., Hence, the correct answer is option (b)., 26. Lumen is the SI unit of luminous flux., Hence, the correct answer is option (a)., =, , 28. Specific resistance, ρ =, , RA, l, , = [ML2T –3A–2] [L], = [ML3T –3A–2], Hence, the correct answer is option (b)., 30. E = Gp hq cr, [M1L2T –2] = [M–1L3T –2]p [ML2T –1]q [LT–1]r, = M–p +qL3p + 2q + rT –2p – q – r

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Dimensions and Error Analysis in Measurement, Applying principle of homogeneity of dimensions, we get, –p + q = 1, (1), 3p + 2q + r = 2, (2), –2p – q = –2, (3), Add Eqs. (2) and (3), p + q = 0, (4), 1, Add Eqs. (1) and (4), we get q =, 2, 1, 1, From Eq. (1), we get p = q – 1 = − 1 = − ., 2, 2, 5, 3, Put in Eq. (2), we get − + 1 + r = 2, r = ., 2, 2, Hence, the correct answer is option (a)., m( v2 − v1 ), × t = m(v2 – v1), 31. Impulse = F × t =, t, = Change in momentum, ∴ [Impulse] = [Momentum], Angular momentum, L = mvr, Planck’s constant, [h] = [Energy] × [Time], m(ν 2 −ν 1 ), ×r ×t, ⇒ [F × r × time] =, t, ⇒ m(v2 – v1) × r = (Change of momentum) × r, ∴ [h] = [L],  ,  , Work, W = F ⋅ d ; Torque, τ = r × F, ∴ [W] = [τ], Moment of inertia, I = mr2 = mass × (distance)2,  , Moment of force, τ = r × F = h distance × force, = Distance ×, ∴ [I] ≠ [τ]., , Change of momentum, Time, , ■, , 1.35, , Therefore, moment of inertia and moment of force have, different dimensions., Hence, the correct answer is option (b)., 32. B =, , 33., , 34., , 35., , 36., , [ MLT −2 ], F, =, = [ ML0T −2 A−1 ], qν [ AT ][ LT −1 ], , But [A] = [CT –1] ∴ [B] = [ML0T –1C –1], Hence, the correct answer is option (b)., Heat supplied to a system is in the form of energy., ∴ Dimensional formula is = [ML2T –2]., Hence, the correct answer is option (a)., Angular momentum, = Moment of inertia × Angular velocity, = [ML2] ×[T –1] = [ML2T –1]., Hence, the correct answer is option (a)., Maximum percentage error in p,, p = 4% + 2 × 2% = 8%, Hence, the correct answer is option (c)., The force per unit length experienced due to two wires, in which current is flowing in the same direction is, given by, dF µ0 2 I1 I 2, =, dl 4π d, [ MLT −2 ], [ A2 ], = [ µ0 ], [ L], [ L], −2, [ MLT ], [Q 2 ], ⇒, = [ µ0 ] 2, [ L], [T L], ⇒ µ0 = [MLQ –2], Hence, the correct answer is option (a)., ⇒