JEE MAIN - Physics (2022 - 25th June Evening Shift - No. 19)
For $$z = {a^2}{x^3}{y^{{1 \over 2}}}$$, where 'a' is a constant. If percentage error in measurement of 'x' and 'y' are 4% and 12% respectively, then the percentage error for 'z' will be _______________%.
Answer
18
Explanation
% error in $$z = 3 \times 4 + {1 \over 2} \times 12$$
$$ = 12 + 6 = 18\% $$
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