Question

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Answer 1

Answer:

Explanation:

Here, `P =(a^3 b^2)/(sqrtc d)`

Maximum fractional error in P is given by

`(Delta P)/(P) = +-[3(Delta a)/(a) +(Deltab)/(b) +(1)/(2) (Deltac)/(c ) +(Deltad)/(d)] = +-[ 2((1)/(100))+2((3)/(100))+(1)/(2)((4)/(100))+(2)/(100)] = +- (13)/(100) =+-0.13`

`"Percentage error in" P = (DeltaP)/(P)xx100 = +- 0.13xx100 = +- 13%`

As the result (13%error) has two significant figures, therefore, if P turns out to be 3.763, the result would

be rounded off to 3.8.

Answer 2

QUESTION:

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

$\huge{ \red{ \boxed{ \boxed{P= \dfrac{a^3b^2}{\sqrt{c}d}}}}}$

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.

ANSWER:

$\Delta p\% =13\%$

GIVEN:

• The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively.

$\bull P= \dfrac{a^3b^2}{\sqrt{c}d}$

TO FIND:

• The percentage error in the quantity P.

EXPLANATION:

${ \blue{ \boxed{ \boxed{If \: \: P= y^3x^2,\ then \: \: \frac{ \Delta p}{p} =3\frac{ \Delta y}{y} + 2\frac{ \Delta x}{x} }}}}$

$\dfrac{ \Delta p}{p} =3\dfrac{ \Delta a}{a} + 2\dfrac{ \Delta b}{b} + \dfrac{1}{2} \dfrac{ \Delta c}{c} + \dfrac{ \Delta d}{d}$

It is given in percentage so multiply 100 on both sides.

$Let \ \dfrac{\Delta p}{p} \times 100 = \Delta p\%$

$Let \ \dfrac{\Delta a}{a} \times 100 = \Delta a\%$

$Let \ \dfrac{\Delta b}{b} \times 100 = \Delta b\%$

$Let \ \dfrac{\Delta c}{c} \times 100 = \Delta c\%$

$Let \ \dfrac{\Delta d}{d} \times 100 = \Delta d\%$

$\Delta p\% =3 \Delta a\% + 2\Delta b\% + \dfrac{1}{2} \Delta c\% + \Delta d\%$

$\Delta p\% =3 \times 1\% + 2 \times 3\% + \dfrac{1}{2} \times 4\% + 2\%$

$\Delta p\% =3\% + 6\% + 2\% + 2\%$

$\Delta p\% =13\%$

Hence he percentage error in the quantity P = 13%