If Z=\frac{A^2 B^3}{C^4}, then the relative error in Z will be : [25-Jun-2022-Shift-1]

Correct Answer is : 2∆AA+3∆BB+4∆CC

\frac{∆ A}{A}+\frac{∆ B}{B}+\frac{∆ C}{C}

\frac{2 ∆ A}{A}+\frac{3 ∆ B}{B}-\frac{4 ∆ C}{C}

\frac{2 ∆ A}{A}+\frac{3 ∆ B}{B}+\frac{4 ∆ C}{C}

\frac{∆ A}{A}+\frac{∆ B}{B}-\frac{∆ C}{C}

JEE Main 25 June 2022 Shift 1 Solved Paper

Section: Physics

Question : 31 of 90

Marks: +1, -0

If Z=‌

A2B3

, then the relative error in Z will be :

[25-Jun-2022-Shift-1]

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