If $Z=\frac{A^{2} B^{3}}{C^{4}}$, then the relative error in $Z$ will be :

(C) $\frac{2 \Delta \mathrm{~A}}{\mathrm{~A}}+\frac{3 \Delta \mathrm{~B}}{\mathrm{~B}}+\frac{4 \Delta \mathrm{C}}{\mathrm{C}}$

(A) $\frac{\Delta \mathrm{A}}{\mathrm{A}}+\frac{\Delta \mathrm{B}}{\mathrm{B}}+\frac{\Delta \mathrm{C}}{\mathrm{C}}$

(B) $\frac{2 \Delta \mathrm{~A}}{\mathrm{~A}}+\frac{3 \Delta \mathrm{~B}}{\mathrm{~B}}-\frac{4 \Delta \mathrm{C}}{\mathrm{C}}$

(D) $\frac{\Delta \mathrm{A}}{\mathrm{A}}+\frac{\Delta \mathrm{B}}{\mathrm{B}}-\frac{\Delta \mathrm{C}}{\mathrm{C}}$

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Units & Measurement

Z = \frac{A ^{2} B ^{3}}{C ^{4}}

, then the relative error in

Z

will be :

Answer: (C)

For a quantity

Z = \frac{A ^{2} B ^{3}}{C ^{4}}

, the relative error is found by taking the differential and dividing by

Z

. Using the rule that for products and quotients, relative errors add with coefficients equal to the powers:

\frac{Δ Z}{Z} = 2 \frac{Δ A}{A} + 3 \frac{Δ B}{B} + 4 \frac{Δ C}{C}

Note that the term for

C

is positive because

C

appears in the denominator with power 4, and the relative error formula uses absolute values of the exponents.

Answer: (C)

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Single correct option