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Physics Question on Units and measurement

If Z=A2B3C4,Z=\frac {A^2B^3}{C^4}, then the relative error in Z will be

A

△AA+△BB+△CC\frac {△A}{A}+\frac {△B}{B}+\frac {△C}{C}

B

2△AA+3△BB−4△CC\frac {2△A}{A}+\frac {3△B}{B}-\frac {4△C}{C}

C

2△AA+3△BB+4△CC\frac {2△A}{A}+\frac {3△B}{B}+\frac {4△C}{C}

D

△AA+△BB−△CC\frac {△A}{A}+\frac {△B}{B}-\frac {△C}{C}

Answer

2△AA+3△BB+4△CC\frac {2△A}{A}+\frac {3△B}{B}+\frac {4△C}{C}

Explanation

Solution

Z=A2B3C4Z=\frac {A^2B^3}{C^4}

∴ △ZZ=2△AA+3△BB+4△CC\frac {△Z}{Z} = \frac {2△A}{A}+\frac {3△B}{B}+\frac {4△C}{C}

So, the correct option is (C): 2△AA+3△BB+4△CC\frac {2△A}{A}+\frac {3△B}{B}+\frac {4△C}{C}