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Question: A physical quantity P is related to four observation a, b, c and d as follows\[P=\dfrac{{{a}^{3}}{{b...

A physical quantity P is related to four observation a, b, c and d as followsP=a3b3cdP=\dfrac{{{a}^{3}}{{b}^{3}}}{\sqrt{c}d}.The percentage errors of measurement in a, b, c and d are 1%,3%,4%1\%,3\%,4\% and 2%2\%respectively. What is the percentage error in the quantity P? If the value of P. Calculated using the above relation terms out to be 3.7633.763 to what value you should round off the result.

Explanation

Solution

In this type of question of percentage error it should be needed to convert the given relation into the percentage error and by calculating the total percentage error can be calculated.

Complete step by step answer:
Basically the error is defined as a difference of the desired value to the actual value of the measurement. Errors are normally like systematic error, random error and blunders.
The systematic error can be eliminated many times.

Given, Percentage error of measurement in a, b, c and d are 1%,3%,4%1\%,3\%,4\% and 2%2\%
Also given relation
P=a3b2cd=a3b3c12dP=\dfrac{{{a}^{3}}{{b}^{2}}}{\sqrt{c}d}=\dfrac{{{a}^{3}}{{b}^{3}}}{{{c}^{\dfrac{1}{2}}}d}
So, Maximum fraction error in P is,
ΔPP=±[3Δaa+2Δbb+12Δcc+Δdd]\Rightarrow \dfrac{\Delta P}{P}=\pm \left[ \dfrac{3\Delta a}{a}+\dfrac{2\Delta b}{b}+\dfrac{1}{2}\dfrac{\Delta c}{c}+\dfrac{\Delta d}{d} \right]
ΔPP×100=±(3Δaa+2Δbb+12Δcc+Δdd)×100\Rightarrow \dfrac{\Delta P}{P}\times 100=\pm \left( \dfrac{3\Delta a}{a}+\dfrac{2\Delta b}{b}+\dfrac{1}{2}\dfrac{\Delta c}{c}+\dfrac{\Delta d}{d} \right)\times 100
ΔPP×100=±[3×1100+2×3100+12×4100+2100]×100\Rightarrow \dfrac{\Delta P}{P}\times 100=\pm \left[ 3\times \dfrac{1}{100}+2\times \dfrac{3}{100}+\dfrac{1}{2}\times \dfrac{4}{100}+\dfrac{2}{100} \right]\times 100
ΔPP×100=±[3+6+2+2100]×100\Rightarrow \dfrac{\Delta P}{P}\times 100=\pm \left[ \dfrac{3+6+2+2}{100} \right]\times 100
ΔPP×100=13%\Rightarrow \dfrac{\Delta P}{P}\times 100=13\%
Percentage error in P=13%P=13\%
Value of P is given as 3.7633.763
By rounding off the given value to the first decimal place we get; P=3.8P=3.8

Note:
Here we can calculate the standard error. A small standard error is a good thing for any measurement.
For calculating total error in any function at First it is advised to calculate the fractional error and substitute the given percentage error
Here given P=a3b2cdP=\dfrac{{{a}^{3}}{{b}^{2}}}{\sqrt{c}d}
ΔPP=±[3Δaa+2Δbb+12Δcc+Δdd]\Rightarrow \dfrac{\Delta P}{P}=\pm \left[ \dfrac{3\Delta a}{a}+\dfrac{2\Delta b}{b}+\dfrac{1}{2}\dfrac{\Delta c}{c}+\dfrac{\Delta d}{d} \right]