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Physics · 2022

JEE Main · 25 June 2022, Shift 1 · Q31

If Z=(A^2 B^3)/C^4, then the relative error in Z will be:

If $\displaystyle Z=\frac{A^2 B^3}{C^4}$, then the relative error in $\displaystyle Z$ will be :
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JEE Main 2022 Physics question, with the answer from NTA’s final answer key. Where our answers come from.