CBSE Class 11 Physics

Laws of Motion Class 11 Physics - The Law Of Inertia

Laws of Motion Class 11 Physics - The Law Of Inertia

Motion in a Straight Line Class 11 Physics - Motion In A Plane - Acceleration

Motion in a Straight Line Class 11 Physics - Motion In A Plane - Acceleration

Units and Measurements - Significant Figures Measurement of any physical quantity gives out a number. Suppose, measurement of time period of simple pendulum is 02.71 s. If you carefully look at this number then you will come to know that the information is reliably conveyed by last three digits only and the first digit is unnecessary. So, we can say that, there are three significant figures in this number. In simple words, The number of figures, present in any value, which are known with some degree of reliability known as the significant figures. Rules for determining significant figures, The change of unit, does not change the number of significant figures. e.g. 3.561 m has four significant figures. Even if we change the unit i.e. 356.1cm or 3561 mm.still the number of significant figures will be four. All non-zero digits are significant. e.g. the number 4587 has four significant figures. All zeros between the non zero digits are significant. e.g. 4.05 has three significant fig

Units and Measurements - Significant Figures Measurement of any physical quantity gives out a number. Suppose, measurement of time period of simple pendulum is 02.71 s. If you carefully look at this number then you will come to know that the information is reliably conveyed by last three digits only and the first digit is unnecessary. So, we can say that, there are three significant figures in this number. In simple words, The number of figures, present in any value, which are known with some degree of reliability known as the significant figures. Rules for determining significant figures, The change of unit, does not change the number of significant figures. e.g. 3.561 m has four significant figures. Even if we change the unit i.e. 356.1cm or 3561 mm.still the number of significant figures will be four. All non-zero digits are significant. e.g. the number 4587 has four significant figures. All zeros between the non zero digits are significant. e.g. 4.05 has three significant fig

Units and Measurement Class 11 Physics - Combination Of Errors Error Of A Sum Or Difference Consider two physical quantities A and B. The values of these quantities are given as A △A, B△B. where △A and △B are the absolute errors. Let us find Error of a Sum: take the third quantity Z such that Z = A + B for which △Z is the resultant error due to sum of errors. Adding the two physical quantities we get, Z△Z = (A △A) + (B△B) i.e. Z△Z = (A+B) (△A + △B) The maximum possible error △Z = △A + △B … (1) in case of error of a difference: we will subtract the two physical quantities. i.e. Z △Z = (A △A) - (B△B) = (A-B) -( △A △B) = (A-B) - (△A △B) since △B = △B = (A-B) (△A + △B) the maximum possible error in Z will be △Z = △A + △B … (2) Conclusion: From (1) and (2) we can conclude that, when two quantities are added or subtracted, the resultant absolute error in the final result will be the sum of individual absolute errors of different quantities. Example: The temperature is measu

Units and Measurement Class 11 Physics - Combination Of Errors Error Of A Sum Or Difference Consider two physical quantities A and B. The values of these quantities are given as A △A, B△B. where △A and △B are the absolute errors. Let us find Error of a Sum: take the third quantity Z such that Z = A + B for which △Z is the resultant error due to sum of errors. Adding the two physical quantities we get, Z△Z = (A △A) + (B△B) i.e. Z△Z = (A+B) (△A + △B) The maximum possible error △Z = △A + △B … (1) in case of error of a difference: we will subtract the two physical quantities. i.e. Z △Z = (A △A) - (B△B) = (A-B) -( △A △B) = (A-B) - (△A △B) since △B = △B = (A-B) (△A + △B) the maximum possible error in Z will be △Z = △A + △B … (2) Conclusion: From (1) and (2) we can conclude that, when two quantities are added or subtracted, the resultant absolute error in the final result will be the sum of individual absolute errors of different quantities. Example: The temperature is measu

Units and Measurement Class 11 Physics - Absolute Error, Relative Error, Percentage Error Numerical Question: From the experiment of measuring the time period of simple pendulum, the successive measurements for time period are 2.54s, 2.36s, 2.60s. Find absolute errors, relative error and percentage error. Answer: amean=2.54+2.36+2.603 s = 7.50/3 = 2.50 s absolute errors △a1=a1-amean = △a1=2.54-2.50 s = △a1=0.04 s similarly, △a2=2.36-2.50 s= (-0.14) s △a3=2.60-2.50 s= 0.10 s mean absolute error, Δamean= |Δa1|+|Δa2|+ ... + |Δan|n Δamean= 0.04+ 0.14 + 0.103s= 0.093s = 0.09s Relative error = Δameanamean=0.092.50=0.03 Percentage Error, δa= Δameanamean 100 % = 0.092.50100=3%

Units and Measurement Class 11 Physics - Absolute Error, Relative Error, Percentage Error Numerical Question: From the experiment of measuring the time period of simple pendulum, the successive measurements for time period are 2.54s, 2.36s, 2.60s. Find absolute errors, relative error and percentage error. Answer: amean=2.54+2.36+2.603 s = 7.50/3 = 2.50 s absolute errors △a1=a1-amean = △a1=2.54-2.50 s = △a1=0.04 s similarly, △a2=2.36-2.50 s= (-0.14) s △a3=2.60-2.50 s= 0.10 s mean absolute error, Δamean= |Δa1|+|Δa2|+ ... + |Δan|n Δamean= 0.04+ 0.14 + 0.103s= 0.093s = 0.09s Relative error = Δameanamean=0.092.50=0.03 Percentage Error, δa= Δameanamean 100 % = 0.092.50100=3%