JEE Physics Errors in Measurement Notes Set 01

Types of Errors

Every measured value differs a little from the true value. The difference is the error, and it is of three kinds.
1. Systematic errors always push the result the same way. Their causes are instrument faults (zero error, worn scale), a faulty method or approximation (for example ignoring the end correction in a resonance tube) and personal habits such as consistently reading from one side. They can be traced and removed by correction.
2. Random errors change unpredictably from reading to reading, because of small changes in conditions or in the observer. They are reduced by taking many readings and using the mean.
3. Gross errors are plain mistakes such as misreading a scale or writing a wrong digit. Such readings are rejected.

Two further words matter. Accuracy tells how close a result is to the true value, and precision tells how close repeated readings are to each other. A set of readings can be precise but not accurate if a systematic error is present.

Absolute, Mean Absolute, Relative and Percentage Error

If the readings are \( a_1, a_2, \dots, a_n \), the best value is the arithmetic mean \( a_m = \frac{a_1 + a_2 + \dots + a_n}{n} \).
Absolute error of one reading: \( \Delta a_i = |a_m - a_i| \).
Mean absolute error: \( \Delta a = \frac{|\Delta a_1| + |\Delta a_2| + \dots + |\Delta a_n|}{n} \). The result is written as \( a = a_m \pm \Delta a \).
Relative (fractional) error \( = \frac{\Delta a}{a_m} \), and percentage error \( = \frac{\Delta a}{a_m} \times 100\% \). Relative and percentage errors are pure numbers, with no unit.
When only one reading is taken, the least count of the instrument is taken as the absolute error.

Combination of Errors

Let the errors in A and B be \( \Delta A \) and \( \Delta B \), and let Z be the result. The worst case (maximum error) is as follows.

The three rules come from taking logarithms and differentiating. Signs of the individual terms are made positive because in the worst case the errors do not cancel. A quantity raised to a high power, or one that is measured with a coarse instrument, dominates the final error. For a result that depends on the difference of two nearly equal readings the relative error becomes very large.
Parallel resistors cannot use the quotient rule directly. From \( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \) we get \( \frac{\Delta R}{R^2} = \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2} \).
For a function of an angle, \( \Delta(\cos\theta) = \sin\theta\,\Delta\theta \) with \( \Delta\theta \) in radian.
The mean of N readings has an error that falls roughly as \( 1/\sqrt{N} \) for random errors.

Worked Example 1: Combination of Errors

The period of a pendulum is measured as \( T = 2.00 \pm 0.02 \) s and its length as \( l = 1.00 \pm 0.01 \) m. Find g and its percentage error.
Since \( g = \frac{4\pi^2 l}{T^2} \), \( \frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T} = 1\% + 2(1\%) = 3\% \).
\( g = \frac{4\pi^2 (1.00)}{(2.00)^2} = 9.87 \) m/s², so \( g = 9.87 \pm 0.30 \) m/s².

Significant Figures

The significant figures of a number are the digits whose values are reliably known, plus the first doubtful digit.
1. All non-zero digits are significant.
2. Zeros between non-zero digits are significant (1002 has four).
3. Zeros to the left of the first non-zero digit are not significant (0.0043 has two).
4. Trailing zeros after a decimal point are significant (2.500 has four).
5. Trailing zeros in a whole number without a decimal point are ambiguous. Writing in scientific notation removes the doubt: \( 1.00 \times 10^6 \) has three, \( 1.0000 \times 10^6 \) has five.
6. Powers of ten and a change of unit do not alter the number of significant figures: 0.000010 km = 1.0 cm = 1.0 × 10⁻⁵ km, all with two.
Exact numbers (such as 2 in 2πr, or a count of 10 oscillations) have unlimited significant figures.

Rules for Calculation and Rounding

Multiplication and division: the result has as many significant figures as the factor with the fewest. Example: 3.0 × 800.0 = 2.4 × 10³.
Addition and subtraction: the result keeps as many decimal places as the term with the fewest decimal places. Example: 160.45 + 6.732 = 167.18, and 478 km + 0.397 km = 478 km.
Rounding off: if the first digit to be dropped is less than 5, leave the last kept digit unchanged; if it is more than 5, increase it by 1; if it is exactly 5 followed by nothing, make the last kept digit even (4.565 to three figures gives 4.56, and 4.575 gives 4.58).
Do the rounding only at the end of a multi-step calculation.

Least Count of Instruments

The least count (L.C.) is the smallest value an instrument can measure directly.
Vernier callipers: \( \text{L.C.} = 1\ \text{MSD} - 1\ \text{VSD} = \frac{\text{value of 1 MSD}}{\text{number of vernier divisions}} \). If N vernier divisions equal (N − 1) main scale divisions, 1 VSD = \( \frac{N-1}{N} \) MSD. For 10 vernier divisions and 1 MSD = 1 mm, L.C. = 0.1 mm.
Reading = main scale reading + (coinciding vernier division × L.C.).
Screw gauge: \( \text{L.C.} = \frac{\text{pitch}}{\text{number of circular scale divisions}} \), where pitch is the distance the screw advances in one full turn. Reading = main scale reading + (circular scale reading × L.C.).
Spherometer: same formula as the screw gauge. For a stop watch the least count is its smallest division, and for a metre rule it is usually 1 mm.
Angles: if 29 main divisions of 0.5° equal 30 vernier divisions, L.C. = 0.5°/30 = 1 minute.

Zero Error

If the instrument does not read zero when the jaws are closed, the reading must be corrected. True reading = observed reading − zero error, with the sign of the zero error included.
Vernier callipers: if the vernier zero lies to the right of the main zero, the zero error is positive and equals (coinciding vernier division × L.C.). If it lies to the left, the error is negative and equals − (total vernier divisions − coinciding division) × L.C.
Screw gauge: if the circular scale reads k divisions with the reference line when closed and k is below about half the divisions, the error is + k × L.C. If k is high (for example 45 out of 50), the error is − (50 − k) × L.C.
A negative zero error is therefore added to the observed reading, and a positive zero error is subtracted.

Worked Example 2: Screw Gauge with Zero Error

A screw gauge has pitch 0.5 mm and 50 circular divisions. With the jaws closed the 46th division is on the reference line. A wire is then placed and the main scale shows 2 mm with the 30th division on the line.
L.C. = 0.5/50 = 0.01 mm. Zero error = − (50 − 46) × 0.01 = − 0.04 mm.
Observed = 2 + 30 × 0.01 = 2.30 mm. Corrected diameter = 2.30 + 0.04 = 2.34 mm.

Standard Experiments and Their Error Formulae

Simple pendulum: \( g = \frac{4\pi^2 l}{T^2} \) and \( \frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T} \). The time of n oscillations is measured, so the error in one period is the error of the stop watch divided by n. Use a small angle of oscillation and measure l to the centre of the bob.
Young's modulus by Searle's method: \( Y = \frac{4MgL}{\pi d^2 l} \) and \( \frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta L}{L} + 2\frac{\Delta d}{d} + \frac{\Delta l}{l} \). The diameter is measured with a screw gauge, the extension l with the micrometer screw. Wait for a short time after each load so that the wire reaches its full extension, and take readings while loading and unloading to remove the effect of hysteresis.
Resonance column: for two successive resonances, \( \lambda = 2(l_2 - l_1) \) and \( v = f\lambda \), so \( \Delta v = 2f(\Delta l_1 + \Delta l_2) \) when f is exact. A single resonance obeys \( l_1 + 0.3d = \frac{\lambda}{4} \), where 0.3d is the end correction for a tube of diameter d.
Metre bridge: \( \frac{R}{S} = \frac{l}{100 - l} \), so \( \frac{\Delta S}{S} = \frac{\Delta l}{l} + \frac{\Delta l}{100 - l} \). The error is least when the balance point is near the middle (50 cm). The bridge wire should have high resistivity and a low temperature coefficient, and the resistances in the two gaps can be interchanged to reduce the effect of a non-uniform wire.
u–v method for a mirror or lens: \( \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \) (magnitudes for a real image). The image pin is located without parallax.
Prism: \( \mu = \frac{\sin\frac{A + \delta_m}{2}}{\sin\frac{A}{2}} \); if A is exact, \( \frac{\Delta\mu}{\mu} = \frac12\cot\frac{A + \delta_m}{2}\,\Delta\delta_m \).
Ohm's law and resistance: \( R = \frac{V}{I} \) and \( \frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I} \).

Parallax

When two objects lie on the same line of sight, the nearer one hides the farther one. If the eye is moved sideways, the farther object appears to move in the direction opposite to the eye relative to the nearer one. When the apparent relative shift vanishes, the two objects are at the same position. This test is used to place a pin exactly at the position of an image in the u–v method.

Worked Example 3: Reporting a Measurement

Five readings of a time are 90 s, 91 s, 95 s, 92 s and 92 s. The mean is 92 s. The absolute deviations are 2, 1, 3, 0 and 0 s, with mean 1.2 s. Since the instrument has a least count of 1 s, the error is written with the same decimal places as the readings: T = 92 ± 1 s.

Key Points to Remember

1. Maximum error: absolute errors add in sum and difference; relative errors add (multiplied by the powers) in products, quotients and powers.
2. Percentage error has no unit, and only the least count (or the mean absolute error) of the instrument is used as the error of a reading.
3. Least count of vernier = 1 MSD − 1 VSD; of screw gauge = pitch / number of circular divisions.
4. Corrected reading = observed reading − zero error (with sign).
5. Quote results with the same number of significant figures as the least precise data, and give the error to one or two significant figures.