Mathematics in Physics

JEE Main level, single correct, medium

84 of 121

Question 84 of 121

If Z=A2B3C4Z=\dfrac{{A}^{2}{B}^{3}}{{C}^{4}}, then the relative error in ZZ will be

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Answer: (A) (2ΔAA+3ΔBB+4ΔCC)\left(\dfrac{2\Delta A}{A}+\dfrac{3\Delta B}{B}+\dfrac{4\Delta C}{C}\right)

Given: Z=A2B3C4Z=\dfrac{{A}^{2}{B}^{3}}{{C}^{4}}
When quantities are multiplied or divided, the relative error in the result is the sum of the relative errors in the multipliers.
So, ∆ZZ=[∆AA+∆AA+∆BB+∆BB+∆BB+∆CC+∆CC+∆CC+∆CC]\dfrac{∆Z}{Z}=\left[\dfrac{∆A}{A}+\dfrac{∆A}{A}+\dfrac{∆B}{B}+\dfrac{∆B}{B}+\dfrac{∆B}{B}+\dfrac{∆C}{C}+\dfrac{∆C}{C}+\dfrac{∆C}{C}+\dfrac{∆C}{C}\right]
Thus, relative error is
∆ZZ=[2×∆AA+3×∆BB+4×∆CC]\dfrac{∆Z}{Z}=\left[2\times \dfrac{∆A}{A}+3\times \dfrac{∆B}{B}+4\times \dfrac{∆C}{C}\right]

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