Question
If $\mathrm{P}=\left[\left(\mathrm{A}^{2} \mathrm{~B}\right) /\left(\mathrm{C}^{3}\right)\right]$ where percentage error in $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$are respectively $\pm 2 \% \pm 3 \%$ and $\pm 5 \%$ then total percentage error in measurement of $\mathrm{p}$(a) $18 \%$(b) $14 \%$(c) $21 \%$(d) $12 \%$
Step 1
Step 1: Given that $P=\left[\left(A^{2} B\right) /\left(C^{3}\right)\right]$, the percentage error in $P$ is given by the sum of the percentage errors in $A$, $B$, and $C$ multiplied by their respective powers. Show more…
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Units and Measurements
Round 1
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