JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement
If \(Z=\frac{A^{2} B^{3}}{C^{4}}\), then the relative error in \(Z\) will be
- A \(\frac{\Delta A }{ A }+\frac{\Delta B }{ B }+\frac{\Delta C }{ C }\)
- B \(\frac{2 \Delta A }{ A }+\frac{3 \Delta B }{ B }-\frac{4 \Delta C }{ C }\)
- C \(\frac{2 \Delta A }{ A }+\frac{3 \Delta B }{ B }+\frac{4 \Delta C }{ C }\)
- D \(\frac{\Delta A }{ A }+\frac{\Delta B }{ B }-\frac{\Delta C }{ C }\)
Answer & Solution
Correct Answer
(C) \(\frac{2 \Delta A }{ A }+\frac{3 \Delta B }{ B }+\frac{4 \Delta C }{ C }\)
Step-by-step Solution
Detailed explanation
\(Z=\frac{A^{2} B^{3}}{C^{4}}\) In case of error \(\frac{ dZ }{ Z }=\frac{2 d A}{ A }+\frac{3 dB }{ B }+\frac{4 dC }{ C }\) \(\frac{\Delta Z }{ Z }=\frac{2 \Delta A }{ A }+\frac{3 \Delta B }{ B }+\frac{4 \Delta C }{ C }\)
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