A physical quantity P is related to four observables a, b, c and d as follows:

P = $$\dfrac{a^{3}b^{2}}{\sqrt{cd}}​$$

The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result?

Asked by Abhisek | 1 year ago |  86

Solution :-

$$\dfrac{a^3b^2}{\sqrt{cd}}​ \dfrac{\Delta P}{P}$$​ = $$\dfrac{3\Delta a}{a}​ + \dfrac{2\Delta b}{b}$$$$+ \dfrac{1}{2}​ \dfrac{\Delta c}{c}​ + \dfrac{\Delta d}{d}$$

$$( \dfrac{\Delta P}{P}​ \times 100 ) %$$ = $$( 3 \times \dfrac{\Delta a}{a}​ \times 100 + 2 \times \dfrac{\Delta b}{b}​ \times 100)$$

= 3 x 1 + 2 x 3 + $$\dfrac{1}{2}$$x 4 + 2

= 3 + 6 + 2 + 2 = 13 %

P = 4.235

ΔP = 13 % of P

13P​ = $$\dfrac{13P}{100}$$

=$$\dfrac{13\times 4.235}{100}$$

= 0.55

The error lies in the first decimal point, so the value of p = 4.3

Answered by Pragya Singh | 1 year ago

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