How to Calculate Error in Measurement ?

Error in Measurement : The difference between the true value and the measured values of a physical quantity is called error in measurement .
(a) Absolute error in the measurement of a physical quantity is the magnitude of the difference between the true value and the individual measured value of the quantity.
By definition, absolute errors in the individual measured values of the quantity are

∆a1 = am – a1

∆a2 = am – a2
………
∆an = am – an
(b) Mean absolute error : it is the arithmetic mean of the magnitudes of absolute errors in all the measurements of the quantity. It is represented by ∆amean. thus,

$\large \Delta a_{mean} = \frac{|\Delta a_1| +|\Delta a_2| + ….+|\Delta a_n|}{n} $

(c)Relative error or Fractional error
The relative error or fractional error of measurement  is defined as the ratio of mean absolute error to the mean value of the quantity measured. Thus,

$\large Relative \;Error = \frac{\Delta a_{mean}}{a_m}$

(d) Percentage error : When the relative/fractional error is expressed in percentage, we call it percentage error.

$\large Percentage \;Error = \frac{\Delta a_{mean}}{a_m} \times 100 $%

Propagation of Errors :

(a) Error in Sum :
x = a + b

Absolute error in x is

Δx = ±(Δa + Δb)

(b) Error in Difference :
x = a – b

Absolute error in x is

Δx = ±(Δa + Δb)

(c) Error in Product

x = a × b

$ \displaystyle \frac{\Delta x}{x} = \pm (\frac{\Delta a}{a} + \frac{\Delta b}{b} ) $

(d) Error in Quotient

$\large x = \frac{a}{b}$

$ \large \frac{\Delta x}{x} = \pm (\frac{\Delta a}{a} + \frac{\Delta b}{b} ) $

(e) Errors in powers of quantities :

$ \displaystyle x = \frac{a^n}{b^m} $

Taking log ,

$\large ln x = ln a^n – ln b^m $

$\large ln x = n ln a – m ln b  $

Differentiating ,

$ \displaystyle \frac{dx}{x} = n\frac{da}{a} – m\frac{db}{b} $

Maximum fractional error in x,

$ \displaystyle \frac{\Delta x}{x} = \pm ( n\frac{\Delta a}{a} + m\frac{\Delta b}{b} )$

Solved Example : The initial and final temperature of water as recorded by an observer are
(40.6 ± 0.2)°C and (78.3 ± 0.3 )°C. Calculate the rise in temperature with proper error limit.

Solution: Let θ1 = 40.6 °C,  Δθ1 = ± 0.2 °C

θ2 = 78.3 °C,  Δθ2 = ± 0.3  °C

⇒ θ = θ2 – θ1 = 78.3 – 40.6 = 37.7 °C

& Δθ = ± (Δθ1 + Δθ2)= ± (0.2 + 0.3) = ± 0.5°C

Hence rise in temperature

= (37.7 ± 0.5)°C

Solved Example : The sides of a rectangle are (10.5 ± 0.2) cm and (5.2 ± 0.1)cm. Calculate its perimeter with error limit.

Solution: Here, l = (10.5 ± 0.2) cm

b = (5.2 ± 0.1)cm

P = 2(l + b) = 2 (10.5 + 5.2) = 31.4cm

ΔP = ± 2 (Δl + Δb) = ±0.6

Hence perimeter = (31.4 ± 0.6) cm.

Solved Example : A potential difference of V = (10 ±1) volt is applied across a resistance of (5± 2) Ohm. Calculate the current with error limits.

Solution.      Form Ohm’s law, V = IR

$ \displaystyle I = \frac{V}{R} = \frac{10}{5} = 2 A $

$ \displaystyle \frac{\Delta I}{I} = \pm (\frac{\Delta V}{V} + \frac{\Delta R}{R} ) $

$ \displaystyle \frac{\Delta I}{I} = \pm (\frac{1}{10} + \frac{2}{5} ) $

$ \displaystyle \frac{\Delta I}{I} = \pm (0.1 + 0.4) = \pm 0.5 $

$ \displaystyle \frac{\Delta I}{2} = \pm 0.5 $

ΔI =± 0.5 × 2 = ±1

Current with error limits  = (2± 1)A.

Solved Example : The lengths of two cylinders are measured to be l1 = (5.62 ± 0.01) cm and l2 = (4.34 ± 0.02) cm. calculate difference in lengths with error limits.

Sol. Here, l1 = (5.62 ± 0.01) cm

l2 = (4.34 ± 0.02) cm

l’ = l1 – l2 = 5.62 – 4.34 = 1.28 cm.

∆ l’ = ± (∆l1 + ∆l2)

= ±(0.01 + 0.02) = ± 0.03

Hence, difference in lengths = (1.28 ± 0.03)cm

Solved Example : The length and breadth of a rectangular lamina are measured to be (2.3 ± 0.2)cm and (1.6 ± 0.1)cm. calculate area of the lamina with error limits.

Sol. Here, l = (2.3 ± 0.2) cm and b = (1.6 ± 0.1) cm,

A = l × b = 2.3 × 1.6 = 3.68 cm2

$\large \frac{\Delta A}{A} = \pm (\frac{\Delta l}{l} + \frac{\Delta b}{b}) $

$\large \frac{\Delta A}{A} = \pm (\frac{0.2}{2.3} + \frac{0.1}{1.6}) $

$\large = \pm \frac{0.55}{3.68}$

$\large \Delta A = \pm \frac{0.55}{3.68} \times A $

$\large \Delta A = \pm \frac{0.55}{3.68} \times 3.68 = \pm 0.55 $

$\large A = (3.68 \pm 0.55) cm^2 $

Exercise : A force F is applied on a square plate of side L. If percentage error in determination of L is 2 % and that in F is 4 %, what is permissible error in pressure ?

(a)  2 %       (b)  4 %         (c)  6 %      (d)  8 %

Exercise: The density of a cube is measured by measuring its mass and length of its sides. If maximum error in measurement of mass and lengths are 4 % and 3 % respectively, the maximum error in the measurement of density would be

(a)   9 %      (b)   13 %        (c)   12 %     (d)   7 %

Exercise: A thin copper wire of length l metre increases in length by 2 % when heated through 10ºC. What is the percentage increase in area when a square copper sheet of length l metre is heated through 10ºC ?

(a)  4 %    (b)  8 %   (c)  16 %  (d)  none of the above

Also Read :

→  System of Units : Fundamental Units & Derived Units
→  Dimensional Analysis & Uses of Dimensional equations
→  Significant figures , Rules of counting significant figure
→  Vernier Callipers & Screw Gauge
→  Solved Problems : Physical World & Measurement

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