JEE MAIN - Physics (2022 - 25th June Morning Shift - No. 1)
If $$Z = {{{A^2}{B^3}} \over {{C^4}}}$$, then the relative error in Z will be :
$${{\Delta A} \over A} + {{\Delta B} \over B} + {{\Delta C} \over C}$$
$${{2\Delta A} \over A} + {{3\Delta B} \over B} - {{4\Delta C} \over C}$$
$${{2\Delta A} \over A} + {{3\Delta B} \over B} + {{4\Delta C} \over C}$$
$${{\Delta A} \over A} + {{\Delta B} \over B} - {{\Delta C} \over C}$$
Explanation
$$Z = {{{A^2}{B^3}} \over {{C^4}}}$$
$$\therefore$$ $${{\Delta Z} \over Z} = {{2\Delta A} \over A} + 3 \times {{\Delta B} \over B} + {{4\Delta C} \over C}$$
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