JEE Main 2022 · 25 Jun · Shift 1 · Physics Q31

← JEE Main 2022 · 25 Jun · Shift 1

PhysicsSingle correct+4 / -1Question 31 of 90

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

\dfrac{\Delta A}A+\dfrac{\Delta B}B+\dfrac{\Delta C}C

\dfrac{2\Delta A}A+\dfrac{3\Delta B}B-\dfrac{4\Delta C}C

\dfrac{2\Delta A}A+\dfrac{3\Delta B}B+\dfrac{4\Delta C}C

\dfrac{\Delta A}A+\dfrac{\Delta B}B-\dfrac{\Delta C}C

Answer and worked solution

Official answer: C

Relative errors add, each multiplied by the power: $\dfrac{\Delta Z}Z=\dfrac{2\Delta A}A+\dfrac{3\Delta B}B+\dfrac{4\Delta C}C$ (errors never subtract).

(C)

Topics:Errors in measurement

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JEE Main 2022 · 25 Jun · Shift 1, Physics question 31