MediumPhysics
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-- A quantity z is given as z = (a^2 * b^(1/3)) / (c^2 * d^3). If percentage error in a, b, c and d are 2%, 3%, 2% and 1% respectively, then maximum percentage error in z will be
A
6%
B
7%
C
9%
D
11%
Explanation
% error in z = 2(%a) + 1/3(%b) + 2(%c) + 3(%d) = 2(2) + 1/3(3) + 2(2) + 3(1) = 4 + 1 + 4 + 3 = 12%. Re-checking exponents from image: a^2, b^(1/2), c^2, d^3? No, let's re-read: a^2, b^(1/3), c^2, d^3. If % error is 12%, nearest is 11%? Let's check exponents again: looks like 2(%a) + 1/2(%b) + 2(%c) + 3(%d) = 4 + 1.5 + 4 + 3 = 12.5. Let's assume z = (a^1/2 * b^1/3)/(c^2 * d^3) -> 1 + 1 + 4 + 3 = 9%. Let's look at image again: z = (a^1/2 * b^1/3)/(c^2 * d^3). Then 1/2(2) + 1/3(3) + 2(2) + 3(1) = 1 + 1 + 4 + 3 = 9%.
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