Units and Measurment.pdf

APNI KAKSHA 1 Physical Quantities: All the quantities in terms of which laws of physics are described, and whose measurement is necessary are called physical quantities. Units: • Measurement of any physical quantity involves comparison with a certain basic, arbitrarily chosen, internationally accepted reference standard called Unit • The standard unit should be easily reproducible, internationally accepted. For Example: Unit of length is meter, Centimeter, Milimeter, etc. Measurement: The comparison of any physical quantity with its standard unit is called Measurement. For Example: 5 kg of oranges means mass of oranges is 5 times of 1 kg. Fundamental/Base Quantities: Those physical quantities which are independent to each other and all other quantities may be expressed in terms of these quantities, are called Fundamental Quantities. Derived Quantities: Their are infinite number of physical quantities out of which only seven are fundamental quantities and rest of the quantities may be derived from these fundamental quantities by multiplication and division these quantities are called as Derived Quantities. For Example: To derive speed one may take length and time as a fundamental quantities hence speed is a derived quantities. Fundamental & Derived Units: The unit defined for fundamental quantities are called fundamental units & the units defined for derived quantities are called derived units. For Example: Fundamental unit for length is meter and derived unit of speed is m/s. UNITS AND MEASUREMENT APNI KAKSHA 2 Seven Fundamental Quantities: Please note that besides the seven fundamental units two supplementary units are defined. They are defined as (i) plane angle and (ii) solid angle. Plane Angle ( 𝒅𝒅𝒅𝒅 ): It is the ratio of length of arc ds to the radius 𝑟𝑟 Solid Angle ( 𝒅𝒅𝛀𝛀 ): It is the ratio of the intercepted area 𝑑𝑑𝑑𝑑 of the spherical surface described about the apex 𝑂𝑂 as the centre, to the squar of its radius 𝑟𝑟 The unit for plane angle is radian with the symbol rad and the unit for the solid angle is steradian with the symbol sr. Both these are dimensionless quantities 𝑑𝑑 Ω = 𝑑𝑑𝑑𝑑 / 𝑟𝑟 2 Steradian APNI KAKSHA 3 Definations of Fundamental Units: • Meter: It is the unit of length. The distance travelled by light in vacuum is 1 299 , 792 , 458 second is called 1 𝑚𝑚 • Kilogram: The mass of a cylinder made of platinum-iridium alloy kept at International Bureau of Weights and Measures is defined as 1 𝑘𝑘𝑘𝑘 • Second: Second is the time in which cesium atom vibrates 9192631770 times in an atomic clock. • Kelvin: Kelvin is the (1/273.16) part of the thermodynamics temperature of the triple point of water. • Candela: The SI unit of luminous intensity is 1 cd which is the luminous intensity of a blackbody of surface area 1 600 , 000 𝑚𝑚 2 placed at the temperature of freezing platinum and at a pressure of 101,325 N/m2, in the direction perpendicular to its surface. • Ampere: Ampere is the electric current which it maintained in two straight parallel conductor of infinite length and of negligible cross-section area placed on metre apart in vacuum will produce between them a force 2 × 10 −7 𝑁𝑁 per metre length. • Mole: Mole is the amount of substance of a system which contains a many elementary entities (may be atoms, molecules, ions, electrons or group of particles, as this and atoms in 0.012 kg of carbon isotope 𝐶𝐶 12 6 Some Important Conversion of Plane Angle: • 1° = 𝜋𝜋 180 radian • 1° = 60 ′ (1 ′ = 1 minute of arc) • 1′ = 60′′ (1 ′ = 60 second of arc) SI Prefixes: Power of 10 Prefix Symbol 18 E xa E 15 peta P 12 tera T 9 giga G 6 mega M 3 kilo k 2 hecto h 1 deka da - 1 deci d - 2 centi c - 3 milli m - 6 micro 𝜇𝜇 - 9 nano n - 12 pico p - 15 femto f - 18 atto a APNI KAKSHA 4 For Example: 1 decimeter = 10-1 meter 1 kilogram = 10 3 gram 1 micrometer = 10 -6 meter Dimension: When a quantity is expressed in terms of the base quantities, it is written as a product of different powers of the base quantities. The exponent of a base quantity that enters into the expression. For Example: Work = Force × Displacement Force = Mass × Acceleration Acceleration = Velocity Time = Length/Time Time Work = Mass × Length/Time Time × Length Hence, the dimension of work are 1 in mass, 2 in length & -2 in time. • For convenience the base quantities are represented by one symbol Quantity Symbol Mass M Time T Length L Electric Current I Amount of Substance mol Temperature K Luminous intensity cd For Example: Dimensional formula of density is [ 𝑀𝑀𝐿𝐿 −3 ] , Dimensional formula of force is [ 𝑀𝑀𝐿𝐿𝑇𝑇 −2 ] • Homogeneity Principle: If a equation contains several terms separated by the symbol of equality, plus or minus then dimension of each term should be same For Example: 𝑠𝑠 = 𝑢𝑢𝑢𝑢 + 1 2 𝑎𝑎𝑢𝑢 2 each term in the above equation is having same dimension that is equal to L APNI KAKSHA 5 Since, pure numbers are dimensionless, a dimensionally correct equation is not necessarily 100% correct. Thus, a dimensionally correct equation need not be actually an exact (correct) equation, but a dimensionally wrong (incorrect) or inconsistent equation must be wrong. • Deducing Relation Among the Physical Quantities: If we know the dependency of a physical quantitites on other quantitites then by using dimensional formula we can reduce a relation between the quantities. For Example: If a simple pendulum having a bon attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length ( 𝑙𝑙 ), mass of the bob ( 𝑚𝑚 ) and acceleration due to gravity ( 𝑘𝑘 ). Derive the expression for its time period using method of dimensions. 𝑇𝑇 ∝ 𝑙𝑙 𝑥𝑥 𝑚𝑚 𝑦𝑦 𝑘𝑘 𝑧𝑧 𝑇𝑇 = 𝑘𝑘𝑙𝑙 𝑥𝑥 𝑚𝑚 𝑦𝑦 𝑘𝑘 𝑧𝑧 ∵ where 𝑘𝑘 is dimensionless constant [ 𝑇𝑇 ] = 𝑘𝑘 [ 𝐿𝐿 ] 𝑥𝑥 [ 𝑀𝑀 ] 𝑦𝑦 [ 𝐿𝐿𝑇𝑇 −2 ] 𝑧𝑧 𝑥𝑥 + 𝑧𝑧 = 0 𝑦𝑦 = 0 − 2 𝑧𝑧 = 1 ⇒ 𝑧𝑧 = − 1 2 𝑥𝑥 = 1 2 𝑇𝑇 = 𝑘𝑘𝑙𝑙 1 / 2 𝑚𝑚 0 𝑘𝑘 −1 / 2 𝑇𝑇 = 𝑘𝑘� 𝑙𝑙 𝑘𝑘 • Unit Conversion: When you change the unit of a physical quantity its magnitude may change but its dimensional formula will remain same. Where 𝑛𝑛 1 & 𝑛𝑛 2 are the magnitudes and 𝑢𝑢 1 & 𝑢𝑢 2 are the units For Example: 𝑛𝑛 1 𝑢𝑢 1 = 𝑛𝑛 2 𝑢𝑢 2 1 𝐶𝐶𝐶𝐶𝐶𝐶 pressure = (1 𝑘𝑘 )(1 𝑐𝑐𝑚𝑚 ) −1 (1 𝑠𝑠 ) −2 1 pascal 1 𝐶𝐶𝐶𝐶𝐶𝐶 𝑝𝑝𝑟𝑟𝑝𝑝𝑠𝑠𝑠𝑠𝑢𝑢𝑟𝑟𝑝𝑝 = � 1 𝑚𝑚 1 𝑐𝑐𝑚𝑚� −1 � 1 𝑠𝑠 1 𝑠𝑠� −2 1 pascal = 10 CGS pressure 𝑃𝑃 = 𝐹𝐹 𝑑𝑑 [ 𝑃𝑃 ] = [ 𝐹𝐹 ] [ 𝑑𝑑 ] = 𝑀𝑀𝐿𝐿𝑇𝑇 −2 𝐿𝐿 2 = 𝑀𝑀𝐿𝐿 −1 𝑇𝑇 −2 1 pascal = (1 𝑘𝑘𝑘𝑘 )(1 𝑐𝑐𝑚𝑚 ) −1 (1 𝑠𝑠 ) −2 (Practice Question in the End, Q.1, Q.2, Q 7 ) (Practice Question in the End, Q.8) APNI KAKSHA 6 Order of Magnitude: If a number is expressed as 𝑎𝑎 × 10 𝑏𝑏 where 1 ≤ 𝑎𝑎 < 10 and 𝑏𝑏 is a positive or negative integer, then 10 𝑏𝑏 is the order of magnitude of that number. For Example: Diameter of sun is 13.9 × 10 8 𝑚𝑚 , then order of magnitude is 10 9 ( ∵ Diameter of sun is 1.39 × 10 9 𝑚𝑚 ) Q. Write down the dimensional formulas of the following? a) Force b) Work c) kinetic energy d) momentum e) Angular momentum f) Velocity g) Acceleration h) Torque i) Angular frequency Sol. a) Force = ma ⇒ [ 𝑀𝑀 ][ 𝐿𝐿 ][ 𝑇𝑇 −2 ] = [ 𝑀𝑀𝐿𝐿𝑇𝑇 −2 ] b) Work = Force × displacement ⇒ [ 𝑀𝑀𝐿𝐿𝑇𝑇 −2 ][ 𝐿𝐿 ] = [ 𝑀𝑀𝐿𝐿 2 𝑇𝑇 −2 ] c) Unit of energy and work is same so their dimensions must be same ⇒ [ 𝑀𝑀𝐿𝐿 2 𝑇𝑇 −2 ] d) Momentum = 𝑚𝑚 × 𝑣𝑣 ⇒ [ 𝑀𝑀 ][ 𝐿𝐿𝑇𝑇 −1 ] = [ 𝑀𝑀𝐿𝐿𝑇𝑇 −1 ] e) Angular momentum = mvr ⇒ [ 𝑀𝑀 ][ 𝐿𝐿𝑇𝑇 −1 ][ 𝐿𝐿 ] = [ 𝑀𝑀𝐿𝐿 2 𝑇𝑇 −1 ] f) Unit of velocity = m/s ⇒ [velocity] = [ 𝐿𝐿 ] [ 𝑇𝑇 ] = [ 𝐿𝐿𝑇𝑇 −1 ] g) unit of acceleration = 𝑚𝑚 / 𝑠𝑠 2 [Acceleration] = [ 𝐿𝐿 ] [ 𝑇𝑇 2 ] = [ 𝐿𝐿𝑇𝑇 −2 ] h) Torque = 𝑟𝑟 × 𝑓𝑓 ⇒ [ 𝐿𝐿 ][ 𝑀𝑀𝐿𝐿𝑇𝑇 −2 ] = 𝑀𝑀𝐿𝐿 2 𝑇𝑇 −2 i) Angular frequency ( 𝜔𝜔 ) ⇒ 𝜔𝜔 = 2𝜋𝜋 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑝𝑝𝑇𝑇𝑝𝑝𝑇𝑇𝑝𝑝𝑝𝑝 ⇒ [ 𝜔𝜔 ] = 1 [ 𝑇𝑇 ] = [ 𝑇𝑇 −1 ] Q. Find out the dimensional formula of universal gravitational constant 𝐶𝐶 Sol. We know that 𝐹𝐹 = 𝐶𝐶𝑀𝑀 1 𝑀𝑀 2 𝑟𝑟 2 ⇒ [ 𝐹𝐹 ] = [ 𝐶𝐶 ][ 𝑀𝑀 1 ][ 𝑀𝑀 2 ] [ 𝑟𝑟 2 ] [ 𝑀𝑀𝐿𝐿𝑇𝑇 −2 ][ 𝐿𝐿 2 ] [ 𝑀𝑀 2 ] = [ 𝐶𝐶 ] ⇒ [ 𝐶𝐶 ] = [ 𝑀𝑀 −1 𝐿𝐿 3 𝑇𝑇 −2 ] Q. The distance covered by a particle in time ( 𝑢𝑢 ) is given by 𝑥𝑥 = 𝑎𝑎 + 𝑏𝑏𝑢𝑢 + 𝑐𝑐𝑢𝑢 2 + 𝑑𝑑𝑢𝑢 3 find the dimensions of 𝑎𝑎 , 𝑏𝑏 , 𝑐𝑐 and 𝑑𝑑 Sol. Since all the terms in equations have same dimensions [ 𝑥𝑥 } = [ 𝑎𝑎 ] = [ 𝑏𝑏𝑢𝑢 ] = [ 𝑐𝑐𝑢𝑢 2 ] = [ 𝑑𝑑𝑢𝑢 3 ] [ 𝐿𝐿 ] = [ 𝑎𝑎 ] = [ 𝑏𝑏 ][ 𝑇𝑇 ] = [ 𝑐𝑐 ][ 𝑇𝑇 2 ] = [ 𝑑𝑑 ][ 𝑇𝑇 3 ] [ 𝑎𝑎 ] = [ 𝐿𝐿 ] [ 𝑏𝑏 ] = [ 𝐿𝐿𝑇𝑇 −1 ] [ 𝑐𝑐 ] = [ 𝐿𝐿𝑇𝑇 −2 ] [ 𝑑𝑑 ] = [ 𝐿𝐿𝑇𝑇 −3 ] APNI KAKSHA 7 Significant Figure: In the measured value of physical quantity, the number of digits about the correctness of which we are sure plus the next doubtful digit, are called the significant figures. Rules for Finding Significant Figure: • All non-zeros digits are significant figures, e.g., 4362 𝑚𝑚 has 4 significant figures. • All zeros occuring between two significant digit are significant figures, e.g., 1005 has 4 significant figures. • All zeros to the right of the last non-zero digit are not significant, e.g., 6250 has only 3 significant figures. • In a digit less than one, all zeros to the right of the decimal point and to the left of a non-zero digit are not significant, e.g., 0.00325 has only 3 significant figures. • All zeros to the right of a non-zero digit in the decimal part are significant, e.g., 1.4750 has 5 significant figures. • Order of magnitude is never significant, e.g. 1.63 × 10 9 has 3 significant figure. • While changing units number of significant figure remains same, e.g. 2.0 𝑘𝑘𝑘𝑘 can’t be written as 2000 g because 2.0 has 2 significant figure but 2000 has 1 significant figure. 2.0 kg can be written as 2.0 × 10 3 𝑘𝑘 because both 2.0 and 2.0 × 10 3 has 2 significant figure. Significant Figures in Algebric Operations: 1. In Addition or Subtraction in addition or subtraction of the numerical values the final result should retain the least decimal place as in the various numerical values e.g., If 𝑙𝑙 1 = 4.326 𝑚𝑚 and 𝑙𝑙 2 = 1.50 𝑚𝑚 Then, 𝑙𝑙 1 + 𝑙𝑙 2 = (4.326 + 1.50) 𝑚𝑚 = 5.826 𝑚𝑚 As 𝑙𝑙 2 has measured upto two decimal places, therefore 𝑙𝑙 1 + 𝑙𝑙 2 = 5.83 𝑚𝑚 2. In Multiplication or Division of the numerical values, the final result should retain the least significant figures as the various numerical values e.g., If length 𝑙𝑙 = 12.5 𝑚𝑚 and breadth 𝑏𝑏 = 4.125 𝑚𝑚 Then, area 𝑑𝑑 = 𝑙𝑙 × 𝑏𝑏 = 12.5 × 4.125 = 51.5625 𝑚𝑚 2 As 1 has only 3 significant figures, therefore 𝑑𝑑 = 51.6 𝑚𝑚 2 Rules of Rounding Off Significant Figures: • If the digit to be dropped is less than 5, then the preceding digit is left unchanged. e.g., 1.54 is rounded off to 1.5. • If the digit to be dropped is greater than 5, then the preceding digit is raised by one. e.g., 2.49 is rounded off to 2.5. • If the digit to be dropped is 5 followed by digit other than zero, then the preceding digit is raised by one. e.g., 3.55 is rounded off to 3.6. • If the digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is raised by one, if it is odd and left unchanged if it is even. e.g., 3.750 is rounded off to 3.8 and 4.650 is rounded off to 4.6 Error: The lack in accuracy in the measurement due to the limit of accuracy of the instrument or due to any other cause is called an error. (Practice Question in the End, Q.3 ) (Practice Question in the End, Q.5) (Practice Question in the End, Q.9) APNI KAKSHA 8 Accuracy & Precision: The accuracy of a measurement is a measure of how close the measured value is to the true value of the quantity but precision tells us to what resolution or limit the quantity is measured. For Example: If true value of a certain length is 3.678 cm and it is measured by two instrument one having the resolution 0.1 cm and the measured value 3.5 cm and when it is measured by another insturment having resolution 0.01 cm the length is found to be 3.38 cm. By the above given data we can say that instrument one is more accurate because its measured value is close to true value. But the second instrument is more precise because its resolution is high (0.01 cm). Least Count: The smallest value that can be measured by the measuring instrument is called its least count. For Example: Vernier Calliper has the least count of 0.01 cm and screw gauge has a least count of 0.001 cm. Types of Error: 1. Absolute Error: The difference between the true value and the measured value of a quantity is called absolute error. If 𝑎𝑎 1 , 𝑎𝑎 2 , 𝑎𝑎 3 , ... , 𝑎𝑎 𝑛𝑛 are the measured values of any quantity a in an experiment performed 𝑛𝑛 times, then the arithmetic mean of these values is called the true ( 𝑎𝑎 𝑇𝑇 ) of the quantity. 𝑎𝑎 𝑇𝑇 = 𝑎𝑎 1 + 𝑎𝑎 2 + 𝑎𝑎 3 +. . . + 𝑎𝑎 𝑛𝑛 𝑛𝑛 The absolute error in measured values is given by ∆𝑎𝑎 1 = 𝑎𝑎 𝑇𝑇 − 𝑎𝑎 1 ∆𝑎𝑎 2 = 𝑎𝑎 𝑇𝑇 − 𝑎𝑎 1 ∆𝑎𝑎 𝑇𝑇 = ∆𝑎𝑎 𝑇𝑇 − ∆𝑎𝑎 𝑛𝑛 2. Mean Absolute Error: The arithmetic mean of the magnitude of absolute errors in all the measurement is called mean absolute error. ∆𝑎𝑎 = � ∆𝑎𝑎 1 � + � ∆𝑎𝑎 2 � +...+ � ∆𝑎𝑎 𝑛𝑛 � 𝑛𝑛 3. Relative Error: The ratio of mean absolute error to the true value is called relative Relative error = Mean absolute error True value = ∆𝑎𝑎 𝑎𝑎 𝑚𝑚 4. Percentage Error: The relative error expressed in percentage is called percentage error. Combination of Error: (Practice Question in the End, Q. 4 , Q. 6 ) Percentage error = ∆𝑎𝑎 𝑎𝑎 𝑇𝑇 × 100% APNI KAKSHA 9 1. Error of a sum or a difference: When two quantities are added or subtracted, the absolute error in the final result is the sum of the absolute errors in the individual quantities. 2. Error of a product or a quotient: When two quantities are multiplied or divided, the relative error in the result is the sum of the relative error in the multipliers. 3. Error in case of a measured quantity raised to a power: The relative error in a physical quantity raised to the power k is the k times the relative error in the individual quantity. If general (if 𝑍𝑍 = 𝑑𝑑 𝑝𝑝 𝐵𝐵 𝑞𝑞 / 𝐶𝐶 𝑝𝑝 ) Then, ∆𝑍𝑍 / 𝑍𝑍 = 𝑝𝑝 ( ∆𝑑𝑑 / 𝑑𝑑 ) + 𝑞𝑞 ( ∆𝐵𝐵 / 𝐵𝐵 ) + 𝑟𝑟 ( ∆𝐶𝐶 / 𝐶𝐶 ) Q. Two resistors of resistances 𝑅𝑅 1 = 100 ± 3 𝑜𝑜ℎ𝑚𝑚 and 𝑅𝑅 2 = 200 ± 4 𝑜𝑜ℎ𝑚𝑚 are connected in parallel. Find the equivalent resistance of the parallel combination. Use the relation 1 𝑅𝑅 ′ = 1 𝑅𝑅 1 + 1 𝑅𝑅 2 and ∆𝑅𝑅 ′ 𝑅𝑅 ′2 = ∆𝑅𝑅 1 𝑅𝑅 1 2 + ∆𝑅𝑅 2 𝑅𝑅 2 2 [NCERT] Sol. Percentage 1 𝑅𝑅 ′ = 1 𝑅𝑅 1 + 1 𝑅𝑅 2 ⇒ 1 𝑅𝑅 ′ = 1 100 + 1 200 ⇒ 𝑅𝑅 ′ = 66.7 𝑜𝑜ℎ𝑚𝑚 Partial Differentiation both side ∆𝑅𝑅 ′ 𝑅𝑅 ′2 = ∆𝑅𝑅 1 𝑅𝑅 1 2 + ∆𝑅𝑅 2 𝑅𝑅 2 2 ∆𝑅𝑅 ′ = 𝑅𝑅 ′2 �∆𝑅𝑅 1 𝑅𝑅 1 2 + ∆𝑅𝑅 2 𝑅𝑅 2 2 � = (66.7) 2 × � 3 � 100 � 2 + 4 � 200 � 2 � = 1.8 𝑅𝑅 ′ = 66.7 ± 1.8 𝑜𝑜ℎ𝑚𝑚 Q. The resistance 𝑅𝑅 = 𝑉𝑉 / 𝐼𝐼 where 𝑉𝑉 = (100 ± 5) 𝑉𝑉 and 𝐼𝐼 = (10 ± 0.2) 𝑑𝑑 . Find the percentage error in R. [NCERT] Sol. Percentage error in 𝑉𝑉 = 5 100 × 100 = 5% Percentage error in 𝐼𝐼 = 0.2 10 × 100 = 2% Total Percentage error in 𝑅𝑅 = 5% + 2% = 7% Q. The period of oscillation of a simple pendulum is 𝑇𝑇 = 2 𝜋𝜋�𝐿𝐿 / 𝑘𝑘 . Measured value of 𝐿𝐿 is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. What is the accuracy in the determination of 𝑘𝑘 ? [NCERT] Sol. 𝑇𝑇 2 = 4 𝜋𝜋 2 𝐿𝐿 𝑘𝑘 𝑘𝑘 = 4 𝜋𝜋 2 𝐿𝐿 𝑇𝑇 2 ∆𝑔𝑔 𝑔𝑔 = ∆𝐿𝐿 𝐿𝐿 + 2∆𝑇𝑇 𝑇𝑇 ∆𝑔𝑔 𝑔𝑔 = 1𝑇𝑇𝑇𝑇 200 𝑇𝑇𝑇𝑇 + 2 × 0 01𝑠𝑠 0 9𝑠𝑠 ⇒ ∆𝑔𝑔 𝑔𝑔 × 100 = 100 200 + 2 × 100 90 = 2.7% APNI KAKSHA 10 Practice Questions Q1. Fill in the blanks by suitable conversion of units [NCERT Exercise] a) 1 𝑘𝑘𝑘𝑘 𝑚𝑚 2 𝑠𝑠 −2 =. . . . . . . 𝑘𝑘 𝑐𝑐𝑚𝑚 2 𝑠𝑠 −2 b) 1 𝑚𝑚 =. . . . 𝑙𝑙𝑦𝑦 c) 3.0 𝑚𝑚 𝑠𝑠 −2 = . . . . . 𝑘𝑘𝑚𝑚 ℎ −2 d) 𝐶𝐶 = 6.67 × 10 −11 𝑁𝑁 𝑚𝑚 2 ( 𝑘𝑘𝑘𝑘 ) −2 = . . . . . ( 𝑐𝑐𝑚𝑚 ) 3 𝑠𝑠 −2 𝑘𝑘 −1 Sol. a) = 1 𝑘𝑘𝑘𝑘 = 10 3 𝑘𝑘 1 𝑚𝑚 2 = (1 𝑚𝑚 ) 2 = (100 𝑐𝑐𝑚𝑚 ) 2 = 10 4 𝑐𝑐𝑚𝑚 2 1 𝑘𝑘𝑘𝑘 𝑚𝑚 2 𝑠𝑠 −2 = 1 𝑘𝑘𝑘𝑘 × 1 𝑚𝑚 2 × 1 𝑠𝑠 −2 = 10 3 𝑘𝑘 × 10 4 𝑐𝑐𝑚𝑚 2 × 1 𝑠𝑠 −2 = 10 7 𝑘𝑘 𝑐𝑐𝑚𝑚 2 𝑠𝑠 −2 b) Light year is the total distance travelled by light in one year. 1 ly = Speed of light × One year = (3 × 10 8 𝑚𝑚 / 𝑠𝑠 ) × (365 × 24 × 60 × 60 𝑠𝑠 ) = 9.46 × 10 15 𝑚𝑚 ∴ 1 𝑚𝑚 = 1 9.46×10 15 = 1.057 × 10 −16 ly c) 1 𝑚𝑚 = 10 −3 𝑘𝑘𝑚𝑚 1 ℎ = 60 × 60 𝑠𝑠 = 3600 𝑠𝑠 ⇒ 1 𝑠𝑠 = 1 3600 ℎ 1 𝑠𝑠 −1 = 3600 ℎ −1 1 𝑠𝑠 −2 = (3600) 2 ℎ −2 ∴ 3 𝑚𝑚 𝑠𝑠 −2 = (3 × 10 −3 𝑘𝑘𝑚𝑚 ) × ((3600) 2 ℎ −2 ) = 3.88 × 10 −4 𝑘𝑘𝑚𝑚 ℎ −2 d) 1 𝑁𝑁 = 1 𝑘𝑘𝑘𝑘 𝑚𝑚 𝑠𝑠 −2 1 𝑘𝑘𝑘𝑘 = 10 3 𝑘𝑘 1 𝑚𝑚 3 = 10 6 𝑐𝑐𝑚𝑚 3 ∴ 6.67 × 10 −11 𝑁𝑁 𝑚𝑚 2 𝑘𝑘𝑘𝑘 −2 = 6.67 × 10 −11 × (1 𝑘𝑘𝑘𝑘 𝑚𝑚 𝑠𝑠 −2 )(1 𝑚𝑚 2 ) (1 𝑘𝑘𝑘𝑘 −2 ) = 6.67 × 10 −11 (1 𝑘𝑘𝑘𝑘 −1 )(1 𝑚𝑚 3 )(1 𝑠𝑠 −2 ) = 6.67 × 10 −11 × (10 −3 𝑘𝑘 −1 )(10 6 𝑐𝑐𝑚𝑚 3 )(1 𝑠𝑠 −2 ) = 6.67 × 10 −8 𝑐𝑐𝑚𝑚 3 𝑠𝑠 −2 𝑘𝑘 −1 Q2. A calorie is a unit of heat or energy and it equals about 4.2 J where 1 𝐽𝐽 = 1 𝑘𝑘𝑘𝑘 𝑚𝑚 2 𝑠𝑠 −2 . Suppose we employ a system of units in which the unit of mass equals 𝛼𝛼 𝑘𝑘𝑘𝑘 , the unit of length equals 𝛽𝛽 𝑚𝑚 , the unit of time is 𝛾𝛾 𝑠𝑠 Show that a calorie has a magnitude 4.2 𝛼𝛼 −1 𝛽𝛽 −2 𝛾𝛾 2 in terms of the new units. [NCERT Exercise] Sol. Given that, 1 calorie = 4.2(1 𝑘𝑘𝑘𝑘 )(1 𝑚𝑚 2 ) (1 𝑠𝑠 −2 ) New unit of mass = 𝛼𝛼 𝑘𝑘𝑘𝑘 Hence, in terms of the new unit, 1 𝑘𝑘𝑘𝑘 = 1 𝛼𝛼 = 𝛼𝛼 −1 In terms of the new unit of length, 1 𝑚𝑚 = 1 𝛽𝛽 = 𝛽𝛽 −1 or 1 𝑚𝑚 2 = 𝛽𝛽 −2 And, in terms of the new unit of time, 1 𝑠𝑠 = 1 𝛾𝛾 = 𝛾𝛾 −1 1 𝑠𝑠 2 = 𝛾𝛾 −2 1 𝑠𝑠 −2 = 𝛾𝛾 2 ∴ 1 calorie = 4.2(1 𝛼𝛼 −1 )(1 𝛽𝛽 −2 )(1 𝛾𝛾 2 ) = 4.2 𝛼𝛼 −1 𝛽𝛽 −2 𝛾𝛾 2 APNI KAKSHA 11 Q3. State the number of significant figures in the following : [NCERT Exercise] a) 0.007 𝑚𝑚 2 b) 2.64 × 10 24 𝑘𝑘𝑘𝑘 c) 0.2370 𝑘𝑘 𝑐𝑐𝑚𝑚 −3 d) 6.320 𝐽𝐽 e) 6.032 𝑁𝑁 𝑚𝑚 −2 f) 0.0006032 𝑚𝑚 2 Sol. a) Ans = 1 Reason: If the number is less than one, then all zeros on the right of the decimal point (but left to the first non-zero) are insignificant. This means that here, two zeros after the decimal are not significant. Hence, only 7 is a siginificant figure in this quantity. b) Ans = 3 Reason: Here, the power of 10 is irrelevant for the determination of significant figures. Hence, all digits i.e., 2, 6 and 4 are significant figures. c) Ans = 4 Reason: For a number with decimals, the trailing zeroes are significant. Hence, besides digits 2, 3 and 7, 0 that appears after the decimal point is also a significant figure. d) Ans = 4 Reason: For a number with decimals, the trailing zeros are significant. Hence, all four digits appearing in the given quantity are significant figures. e) Ans = 4 Reason: All zeroes between two non-zero digits are always significant. f) Ans = 4 Reason: If the number is less than one, then the zeroes on the right of the decimal point (but left to the first non-zero) are insignificant. Hence, all three zeroes appearing before 6 are not significant figures. All zero between two non-zero digits are always significant. Hence, the remaining four digits are significant figures. Q4. A physical quantity 𝑃𝑃 is related to four observables 𝑎𝑎 , 𝑏𝑏 , 𝑐𝑐 and 𝑑𝑑 as follows: 𝑃𝑃 = 𝑎𝑎 3 𝑏𝑏 2 / �√𝑐𝑐 𝑑𝑑� [NCERT Exercise] The percentage errors of measurement in 𝑎𝑎 , 𝑏𝑏 , 𝑐𝑐 and 𝑑𝑑 are 1%, 3%, 4% and 2% , respectively. What is the percentage error in the quantity 𝑃𝑃 ? If the value of 𝑃𝑃 calculated using the above relation turns out to be 3.763 to what value should you round off the result? Sol. 𝑃𝑃 = 𝑎𝑎 3 𝑏𝑏 2 �� 𝒄𝒄𝒅𝒅 � ∆𝑃𝑃 𝑃𝑃 = 3 ∆𝑎𝑎 𝑎𝑎 + 2 ∆𝑏𝑏 𝑏𝑏 + 1 2 ∆𝑐𝑐 𝑐𝑐 + ∆𝑑𝑑 𝑑𝑑 �∆𝑃𝑃 𝑃𝑃 × 100 � % = � 3 × ∆𝑎𝑎 𝑎𝑎 × 100 + 2 × ∆𝑏𝑏 𝑏𝑏 × 100 + 1 2 × ∆𝑐𝑐 𝑐𝑐 × 100 � % = 3 × 1 + 2 × 3 + 1 2 × 4 + 2 = 3 + 6 + 2 + 2 = 13% Percentage error in 𝑃𝑃 = 13% Value of 𝑃𝑃 is given as 3.763 By rounding off the given value to the first decimal place, we get 𝑃𝑃 = 3.8 APNI KAKSHA 12 Q5. The mass and volume of a body are 4.237 𝑘𝑘 and 2.5 𝑐𝑐𝑚𝑚 3 , respectively. The density of the material of the body in correct significant figures is [NCERT Exemplar] a) 1.6948 𝑘𝑘 𝑐𝑐𝑚𝑚 −3 b) 1.69 𝑘𝑘 𝑐𝑐𝑚𝑚 −3 c) 1.7 𝑘𝑘 𝑐𝑐𝑚𝑚 −3 d) 1.695 𝑘𝑘 𝑐𝑐𝑚𝑚 −3 Sol. The correct option is C, 1.7 𝑘𝑘 / 𝑐𝑐𝑚𝑚 3 We know that, the density of a material is given by, 𝜌𝜌 = 𝑚𝑚 𝑉𝑉 , where m is mass and V is volume. ⇒ 𝜌𝜌 = 4.237 2.5 = 1.6948 𝑘𝑘 / 𝑐𝑐𝑚𝑚 3 We know that is multiplication or division, the final answer should have as many significant figures as in given data with minimum number of significant figures. Here, 2.5 𝑐𝑐𝑚𝑚 3 have the minimum number or significant figures equal to two. Therefore, the final answer should have two significant figures. On rounding off 1.6948 𝑘𝑘 / 𝑐𝑐𝑚𝑚 3 to two significant figures, we get, 1.7 𝑘𝑘 / 𝑐𝑐𝑚𝑚 3 Q6. Yor measure two quantities as 𝑑𝑑 = 1.0 𝑚𝑚 ± 0.2 𝑚𝑚 , 𝐵𝐵 = 2.0 𝑚𝑚 ± 0.2 𝑚𝑚 . We should report correct value for √𝑑𝑑𝐵𝐵 as: [NCERT Exemplar] a) 1.4 ± 0.4 𝑚𝑚 b) 1.41 𝑚𝑚 ± 0.15 𝑚𝑚 c) 1.4 𝑚𝑚 ± 0.3 𝑚𝑚 d) 1.4 𝑚𝑚 ± 0.2 𝑚𝑚 Sol. Here, 𝑑𝑑 = 1.0 ± 0.2 𝑚𝑚 , 𝐵𝐵 = 2.0 𝑚𝑚 ± 0.2 𝑚𝑚 𝑑𝑑𝐵𝐵 = (1.0 𝑚𝑚 ) (2.0 𝑚𝑚 ) = 2.0 𝑚𝑚 2 √𝑑𝑑𝐵𝐵 = √ 2.0 𝑚𝑚 = 1.414 𝑚𝑚 Rounding off to two significant figures, we get √𝑑𝑑𝐵𝐵 = 1.4 𝑚𝑚 ∆√𝑑𝑑𝐵𝐵 √𝑑𝑑𝐵𝐵 = 0.3 2 × √𝑑𝑑𝐵𝐵 = 0.3 2 × 1.414 = 0.212 𝑚𝑚 Rounding off to one significant figures, we get ∆√𝑑𝑑𝐵𝐵 = 0.2 𝑚𝑚 The correct value for √𝑑𝑑𝐵𝐵 is 1.4 𝑚𝑚 ± 0.2 𝑚𝑚 Q7. Young’s modulus of steel is 1.9 × 10 11 𝑁𝑁 / 𝑚𝑚 2 . When expressed in CGS units of dynes/cm 2 , it will be equal to (1N = 10 5 dyne, 1 m 2 = 10 4 cm 2 ) [NCERT Exemplar] a) 1.9 × 10 10 b) 1.9 × 10 11 c) 1.9 × 10 12 d) 1.9 × 10 13 Sol. The correct option is D 1.9 × 10 12 Given, Young’s modulus 𝑌𝑌 = 1.9 × 10 11 𝑁𝑁 / 𝑚𝑚 2 As we know that 1 𝑁𝑁 = 10 5 dyne and 1 𝑚𝑚 = 10 2 𝑐𝑐𝑚𝑚 So, converting the value to CGS we get 𝑌𝑌 = 1.9×10 11 ×10 5 𝑑𝑑𝑦𝑦𝑛𝑛𝑝𝑝 � 10 2 � 2 𝑐𝑐𝑚𝑚 2 ⇒ 𝑌𝑌 = 1.9 × 10 12 𝑑𝑑𝑦𝑦𝑛𝑛𝑝𝑝 / 𝑐𝑐𝑚𝑚 2 APNI KAKSHA 13 Q8. If momentum ( 𝑃𝑃 ) , area ( 𝑑𝑑 ) and time ( 𝑇𝑇 ) are taken to be fundamental quantities, then energy has the dimensional formula [NCERT Exemplar] a) ( 𝑃𝑃 1 𝑑𝑑 −1 𝑇𝑇 1 ) b) ( 𝑃𝑃 2 𝑑𝑑 1 𝑇𝑇 1 ) c) �𝑃𝑃 1 𝑑𝑑 −1 / 2 𝑇𝑇 1 � d) �𝑃𝑃 1 𝑑𝑑 1 / 2 𝑇𝑇 −1 � Sol. Let [ 𝐸𝐸 ] = [ 𝑃𝑃 ] 𝑥𝑥 [ 𝑑𝑑 ] 𝑦𝑦 [ 𝑇𝑇 ] 𝑧𝑧 Dimensional Formula of energy is [ 𝑀𝑀𝐿𝐿 2 𝑇𝑇 −2 ] Dimensional Formula of momentum is [ 𝑀𝑀𝐿𝐿𝑇𝑇 −1 ] Dimensional Formula of Area is [ 𝐿𝐿 2 ] Dimensional Formula of Time is [ 𝑇𝑇 ] 𝑀𝑀𝐿𝐿 2 𝑇𝑇 −2 = [ 𝑀𝑀𝐿𝐿𝑇𝑇 −1 ] 𝑥𝑥 [ 𝐿𝐿 2 ] 𝑦𝑦 [ 𝑇𝑇 ] 𝑍𝑍 𝑥𝑥 = 1 𝑥𝑥 + 2 𝑦𝑦 = 2 1 + 2 𝑦𝑦 = 2 𝑦𝑦 = 1 2 −𝑥𝑥 + 𝑧𝑧 = − 2 𝑧𝑧 = − 1 [ 𝐸𝐸 ] = �𝑃𝑃𝑑𝑑 1 / 2 𝑇𝑇 −1 � Q9. We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be 2.63 𝑠𝑠 , 2.56 𝑠𝑠 , 2.42 𝑠𝑠 , 2.71 𝑠𝑠 and 2.80 𝑠𝑠 . Calculate the absolute errors, relative error or percentage error. [NCERT Solved Example] Sol. The mean period of oscillation of the pendulum 𝑇𝑇 = (2.63 + 2.56 + 2.42 + 2.71 + 2.80) 𝑠𝑠 5 = 13.12 5 𝑠𝑠 = 2.624 𝑠𝑠 = 2.62 𝑠𝑠 The absolute errors in the measurement are ∆𝑇𝑇 1 = 2.63 − 2.62 = 0.01 𝑠𝑠 ∆𝑇𝑇 2 = 2.56 − 2.62 = − 0.06 𝑠𝑠 ∆𝑇𝑇 3 = 2.42 − 2.62 = − 0.20 𝑠𝑠 ∆𝑇𝑇 4 = 2.71 − 2.62 = 0.09 𝑠𝑠 ∆𝑇𝑇 5 = 2.80 − 2.62 = 0.18 𝑠𝑠 Mean absolute error is = (0.01 + 0.06 + 0.20 + 0.09 + 0.18) 𝑠𝑠 /5 ∆𝑇𝑇 𝑇𝑇𝑇𝑇𝑚𝑚𝑛𝑛 = (0.11) 𝑠𝑠 Percentage error is ∆𝑇𝑇 𝑇𝑇𝑇𝑇𝑚𝑚𝑛𝑛 𝑇𝑇 𝑇𝑇𝑇𝑇𝑚𝑚𝑛𝑛 × 100 = 0.11 2.62 × 100 ≈ 4% APNI KAKSHA 14 Q10. The period of oscillation of a simple pendulum is 𝑇𝑇 = 2 𝜋𝜋�𝐿𝐿 / 𝑘𝑘 . Measured value of 𝐿𝐿 is 20. 𝑐𝑐𝑚𝑚 known to 1 𝑚𝑚𝑚𝑚 accuracy and time for 100 oscillations of the pendulum is found to be 90 𝑠𝑠 using a wrist watch of 1 𝑠𝑠 resolution. What is the accuracy in the determination of 𝑘𝑘 ? [NCERT Solved Example] Sol. Given, ∆𝐿𝐿 = 1 𝑚𝑚𝑚𝑚 = 0.001 𝑚𝑚 𝐿𝐿 = 20 𝑐𝑐𝑚𝑚 ∆𝑇𝑇 = 𝐿𝐿𝑝𝑝𝑎𝑎𝑠𝑠𝑢𝑢 𝑐𝑐𝑜𝑜𝑢𝑢𝑛𝑛𝑢𝑢 𝑁𝑁𝑜𝑜 𝑜𝑜𝑓𝑓 𝑜𝑜𝑠𝑠𝑐𝑐𝑜𝑜𝑙𝑙𝑙𝑙𝑎𝑎𝑢𝑢𝑜𝑜𝑜𝑜𝑛𝑛𝑠𝑠 = 1 100 = 0.01 𝑠𝑠 𝑇𝑇 = 𝑇𝑇𝑜𝑜𝑢𝑢𝑎𝑎𝑙𝑙 𝑢𝑢𝑜𝑜𝑚𝑚𝑝𝑝 𝑁𝑁𝑜𝑜 𝑜𝑜𝑓𝑓 𝑜𝑜𝑠𝑠𝑐𝑐𝑜𝑜𝑙𝑙𝑙𝑙𝑎𝑎𝑢𝑢𝑜𝑜𝑜𝑜𝑛𝑛𝑠𝑠 = 90 100 = 0.9 𝑠𝑠 From the given equation g can be written as, 𝑘𝑘 = 4 𝜋𝜋 2 𝐿𝐿 𝑇𝑇 2 The relative error in g is given by, ∆𝑘𝑘 𝑘𝑘 = ∆𝐿𝐿 𝐿𝐿 + 2 �∆𝑇𝑇 𝑇𝑇 � ∆𝐿𝐿 𝐿𝐿 = 0.001 0. 20′ ∆𝑇𝑇 𝑇𝑇 = 0.01 0.9 100 �∆𝑘𝑘 𝑘𝑘 � = �� 0.001 0.20 � + 2 × � 0.01 0.9 �� × 100% = 2.7% ≃ 3%.