Mathematics in Physics

JEE Main level, single correct, medium

72 of 121

Question 72 of 121

A physical quantity PP is given as P=a2b3cdP=\dfrac{{a}^{2}{b}^{3}}{c\sqrt{d}}. The percentage error in the measurement of a,b,ca,b,c and dd are 11%,2%,3% and 44% respectively. The percentage error in the measurement of quantity PP will be

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Answer: (A) 1313%

The given quantity is P=a2b3cdP=\dfrac{{a}^{2}{b}^{3}}{c\sqrt{d}}.
The percentage error for a quantity can be given by the formula ∆xx×100\dfrac{∆x}{x}\times 100.
Relation of the relative errors will be,
(ΔPP)=[(2Δaa)+(3Δbb)+(Δcc)+(12Δdd)]\left(\dfrac{\Delta P}{P}\right)=\left[\left(\dfrac{2\Delta a}{a}\right)+\left(\dfrac{3\Delta b}{b}\right)+\left(\dfrac{\Delta c}{c}\right)+\left(\dfrac{1}{2}\dfrac{\Delta d}{d}\right)\right]
Hence, the percentage error is calculated by substituting the values,
(ΔPP)×100=(2×1)+(3×2)+3+(12×4)=13\left(\dfrac{\Delta P}{P}\right)\times 100=\left(2\times 1\right)+\left(3\times 2\right)+3+\left(\dfrac{1}{2}\times 4\right)=13%.

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