Name: Accuracy and Precision
Uploaded: 2019-07-19
Duration: 6 min 1 s
Description: Precision and accuracy are two terms that have been twisting many brains into thinking they are synonymous with each other. In reality, they are two very dif...

Question

A physical quantity X is related to four measurable quantities

a, b, c and d

as follows:

X = a^{2} b^{3} c^{5 / 2} d^{- 2}

.
The percentage error in the measurement of

a, b, c and d

are

1 %, 2 %, 3 %

and

4 %

respectively. What is the percentage error in quantity

X

? If the value of

X

calculated on the basis of the above relation is

2.763

, to what value should you round off the result?

Accepted Answer

Percentage error in quantities are the following:

$For X: \frac{△ X}{X} \times 100$

$For a : \frac{△ a}{a} \times 100 = 1 %$

$For b : \frac{△ b}{b} \times 100 = 2 %$

$1 %, 2 %, 3 %$

$For d : \frac{△ d}{d} \times 100 = 4 %$

So, for maximum error: $\frac{△ X}{X} \times 100 = \pm [2 \frac{△ a}{a} + 3 \frac{△ b}{b} + \frac{5}{2} \frac{△ c}{c} + 2 \frac{△ d}{d}] \times 100$

$\Rightarrow \frac{△ X}{X} \times 100 = \pm [\frac{2 \times 1}{100} + \frac{3 \times 2}{100} + \frac{5}{2} \times \frac{3}{100} + \frac{2 \times 4}{100}] 100$

$= \pm \frac{100}{100} [2 + 6 + \frac{15}{2} + 8]$

$\frac{Δ X}{X} \times 100 = \pm [16 + \frac{15}{2}] = \pm [\frac{32 + 15}{2}] = \pm \frac{47}{2} = \pm 23.5 %$

Mean absolute error $= \pm \frac{235}{100} = \pm 0.235$

So, by rounding off the mean absolute error will be $\pm 0.24$ .

Since, the mean absolute error is having two significant figures then rounding off $X = 2.763$ in two significant figure i.e. $X = 2.8$ .

Important Questions on Units and Measurement

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