Arjuna JEE 2027: Physics - Lecture 09: Units, Measurements, and Errors

Homework Discussion: Units and Dimensions

  • Question 1: Conversion of Gravitational Constant (GG)     * Problem: Find the value of the gravitational constant in the CGS system given its SI value is 6.6×1011kg1m3s26.6 \times 10^{-11} \, kg^{-1}m^3s^{-2}.     * Dimensional Formula: [G]=[M1L3T2][G] = [M^{-1}L^3T^{-2}].     * Conversion Formula: n1u1=n2u2n2=n1[M1M2]a[L1L2]b[T1T2]cn_1 u_1 = n_2 u_2 \Rightarrow n_2 = n_1 \left[ \frac{M_1}{M_2} \right]^{a} \left[ \frac{L_1}{L_2} \right]^{b} \left[ \frac{T_1}{T_2} \right]^{c}.     * Variables:         * n1=6.6×1011n_1 = 6.6 \times 10^{-11}.         * M1=1kgM_1 = 1 \, kg, M2=1gmM_2 = 1 \, gm.         * L1=1mL_1 = 1 \, m, L2=1cmL_2 = 1 \, cm.         * T1=1sT_1 = 1 \, s, T2=1sT_2 = 1 \, s.     * Calculation:         * n2=6.6×1011[1000gm1gm]1[100cm1cm]3[1s1s]2n_2 = 6.6 \times 10^{-11} \left[ \frac{1000 \, gm}{1 \, gm} \right]^{-1} \left[ \frac{100 \, cm}{1 \, cm} \right]^{3} \left[ \frac{1 \, s}{1 \, s} \right]^{-2}.         * n2=6.6×1011×103×106n_2 = 6.6 \times 10^{-11} \times 10^{-3} \times 10^6.         * n2=6.6×108n_2 = 6.6 \times 10^{-8}.     * Result: The value in CGS is 6.6×108gm1cm3s26.6 \times 10^{-8} \, gm^{-1}cm^3s^{-2}.

  • Question 2: Modified Unit System for Planck's Constant (hh)     * Problem: The value of Planck's constant is 6.6×1034kgm2s16.6 \times 10^{-34} \, kg \, m^2s^{-1}. Find its value in a system where length is doubled (L2=2L1L_2 = 2L_1) and time is halved (T2=T12T_2 = \frac{T_1}{2}).     * Dimensional Formula: Based on E=hνE = h \nu, [h]=[M1L2T1][h] = [M^1 L^2 T^{-1}].     * Calculation:         * n2=6.6×1034[M1M1]1[L12L1]2[T1T1/2]1n_2 = 6.6 \times 10^{-34} \left[ \frac{M_1}{M_1} \right]^{1} \left[ \frac{L_1}{2L_1} \right]^{2} \left[ \frac{T_1}{T_1 / 2} \right]^{-1}.         * n2=6.6×1034×1×14×(2)1n_2 = 6.6 \times 10^{-34} \times 1 \times \frac{1}{4} \times (2)^{-1}.         * n2=6.6×1034×18n_2 = 6.6 \times 10^{-34} \times \frac{1}{8}.         * n2=0.825×1034=8.25×1035n_2 = 0.825 \times 10^{-34} = 8.25 \times 10^{-35}.

Advanced Concepts: Permittivity and Permeability

  • Permittivity of Free Space (ϵ0\epsilon_0):     * Derived from Coulomb's Law: F=14πϵ0q2r2F = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}.     * Dimensions: [ϵ0]=[M1L3T4A2][\epsilon_0] = [M^{-1}L^{-3}T^4A^2].
  • Permeability of Free Space (μ0\mu_0):     * Related to magnetic fields and masses (F=Gm2r2F = G \frac{m^2}{r^2} analogy used for explanation).
  • Universal Relation through Speed of Light:     * The speed of light (cc) is defined as: c=1μ0ϵ0c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}.     * Dimensional check: 1μ0ϵ0=(speed)2=[L2T2]\frac{1}{\mu_0 \epsilon_0} = (speed)^2 = [L^2 T^{-2}].
  • Instruction: Students are advised not to memorize extra dimensions beyond these; focusing on core formulas is more effective.

Introduction to Errors and Their Types

  • Core Concepts: Errors are the uncertainties in measurements encountered during experiments.
  • Systematic Errors (Known / Predictable):     * Nature: Causes are known; they tend to be in one direction (either positive or negative).     * Types:         * Instrumental Errors: Caused by faulty or incorrectly calibrated instruments (e.g., a scale always showing an extra +0.5g+0.5 \, g).         * Environmental Errors: Caused by external conditions like temperature, pressure, or humidity (e.g., expansion of a ruler in high heat).     * Fix: Proper calibration and adjusting for known parameters.
  • Random Errors (Unknown / Unpredictable):     * Nature: Causes are unknown; result in small, unpredictable fluctuations.     * Fix: Perform more trials. Take multiple readings (nn) and find the average value.     * Relationship: Random Error is inversely proportional to the number of readings (nn). If readings increase by a factor of kk, the random error reduces to 1/k1/k of the initial error.
  • Gross Errors (Human Errors):     * Nature: Caused by human carelessness (the "Reason is you" category).     * Examples: Recording 25 instead of 52, or errors while taking observations and performing calculations.     * Fix: Be careful and attentive (motto: "Savdhani hati durghatna ghati").

Questions and Discussion on Measuring Errors

  • Interaction:     * Q: When we perform experiments do we make mistakes? A: Yes.     * Q: Can there be a faulty instrument which appears to be accurate? A: Yes.     * Q: Can we make mistakes while taking reading? A: Yes.     * Q: Can there be any kind of error which is unknown to us? A: Yes.
  • Question 3: Scaling Random Error     * Problem: If the random error in an experiment of 10 readings is xx, how many readings must be taken to reduce the random error to x/5x/5?     * Solution: Since Random Error 1No. of Readings\propto \frac{1}{\text{No. of Readings}}, to reach x/5x/5, the number of readings must be 5 times the initial count. Result: 10×5=5010 \times 5 = 50 readings.

Calculation and Representation of Errors

  • Experiment Example: Time Period of a Pendulum     * Readings (S.NoS.No): 1) 1.8s1.8 \, s, 2) 1.9s1.9 \, s, 3) 2.0s2.0 \, s, 4) 2.1s2.1 \, s, 5) 2.2s2.2 \, s.
  • Step 1: Mean Value (Average Value):     * Tˉ=T1+T2+T3+T4+T55\bar{T} = \frac{T_1 + T_2 + T_3 + T_4 + T_5}{5}.     * Calculation: 1.8+1.9+2.0+2.1+2.25=2.0s\frac{1.8+1.9+2.0+2.1+2.2}{5} = 2.0 \, s.
  • Step 2: Absolute Error (ΔT\Delta T):     * Calculated for each reading as ReadingAverage Value\text{Reading} - \text{Average Value}.     * ΔT1=1.82.0=0.2\Delta T_1 = 1.8 - 2.0 = -0.2.     * ΔT2=1.92.0=0.1\Delta T_2 = 1.9 - 2.0 = -0.1.     * ΔT3=2.02.0=0\Delta T_3 = 2.0 - 2.0 = 0.     * ΔT4=2.12.0=+0.1\Delta T_4 = 2.1 - 2.0 = +0.1.     * ΔT5=2.22.0=+0.2\Delta T_5 = 2.2 - 2.0 = +0.2.
  • Step 3: Mean Absolute Error (ΔT\overline{\Delta T}):     * The average of the magnitudes (absolute values) of the errors.     * ΔT=ΔT1+ΔT2+...+ΔTnn\overline{\Delta T} = \frac{|\Delta T_1| + |\Delta T_2| + ... + |\Delta T_n|}{n}.     * Calculation: 0.2+0.1+0+0.1+0.25=0.65=0.12s\frac{0.2+0.1+0+0.1+0.2}{5} = \frac{0.6}{5} = 0.12 \, s.
  • Step 4: Reporting the Result:     * Format: Value=Mean Value±Mean Absolute ErrorValue = \text{Mean Value} \pm \text{Mean Absolute Error}.     * Report: T=2±0.12sT = 2 \pm 0.12 \, s.
  • Step 5: Fractional Error:     * Fractional Error=Mean Absolute ErrorMean Value\text{Fractional Error} = \frac{\text{Mean Absolute Error}}{\text{Mean Value}}.     * Calculation: 0.122=0.06\frac{0.12}{2} = 0.06.
  • Step 6: Percentage Error:     * Percentage Error=Fractional Error×100\text{Percentage Error} = \text{Fractional Error} \times 100.     * Calculation: 0.06×100=6%0.06 \times 100 = 6\%.     * Note: Errors up to 5-7%5\text{-}7\% are generally considered acceptable.

Practical Numerical Applications

  • Question 4: Reporting Length     * Given: Results reported as 5±0.0155 \pm 0.015.     * Analysis:         * Mean Value (LavgL_{avg}): 5.         * Mean Absolute Error (ΔL\Delta L): 0.015.         * Fractional Error: 0.0155=0.003\frac{0.015}{5} = 0.003.         * Percentage Error: 0.3%0.3\%.
  • Question 5: Refractive Index of Glass     * Data: 1.45, 1.56, 1.54, 1.44, 1.54, 1.53.     * Calculation:         * Mean (μavg\mu_{avg}): 1.45+1.56+1.54+1.44+1.54+1.536=1.51\frac{1.45+1.56+1.54+1.44+1.54+1.53}{6} = 1.51.         * Absolute Errors: 0.06-0.06, +0.05+0.05, +0.03+0.03, 0.07-0.07, +0.03+0.03, +0.02+0.02.         * Mean Absolute Error (Δμ\overline{\Delta \mu}): 0.06+0.05+0.03+0.07+0.03+0.026=0.043\frac{0.06+0.05+0.03+0.07+0.03+0.02}{6} = 0.043.         * Report: 1.51±0.0431.51 \pm 0.043.         * Percentage Error: 0.0431.51×1002.8%\frac{0.043}{1.51} \times 100 \approx 2.8\%.
  • Question 6: Pendulum Oscillations and Minimum Division     * Data: 100 oscillations recorded as 90s, 91s, 95s, and 92s.     * Note: The minimum division (resolution) of the clock is 1s, which dictates the precision of the reported mean.

Homework and Resources

  • Module Assignments:     * Module 1 Examples: 1 to 5.     * Check your understanding: 1 to 4.     * Topic-wise questions: Extensive list including 1-8, 12-18, 21-26, 28-36, etc.     * Advanced and PYQ levels involve 50+ specific problem numbers.
  • Schedule: Full homework on Errors to be completed by next week (100+ questions).
  • Promotional Note: Arjuna For JEE Main & Advanced Class 11th PCM Combo set includes 16 books with 100-150 high probability questions per chapter.