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The heat generated in a circuit is given by $\mathrm{Q}=\mathrm{I}^{2} \mathrm{Rt},$ where I is current, $\mathrm{R}$ is resistance and tis time. If the percentage errors in measuring I. $\mathrm{R}$ and $\mathrm{t}$ are $2 \%, 1 \%$ and $1 \%$ respectively, then the maximum error in measuring heat will be
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$6 \%$
$\begin{aligned} \frac{\Delta Q}{Q} \times 100 &=\frac{2 \Delta I}{I} \times 100+\frac{\Delta R}{R} \times 100+\frac{\Delta t}{t} \times 100 \\ &=2 \times 2 \%+1 \%+1 \%=6 \% \end{aligned}$
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