JEE MAIN - Physics (2022 - 25th July Evening Shift - No. 16)
The maximum error in the measurement of resistance, current and time for which current flows in an electrical circuit are $$1 \%, 2 \%$$ and $$3 \%$$ respectively. The
maximum percentage error in the detection of the dissipated heat will be :
2
4
6
8
Explanation
Given, $\frac{\Delta R}{R} \times 100=1 \%$
$$ \frac{\Delta F}{F} \times 100=2 \% \text { and } \frac{\Delta t}{t} \times 100=3 \% $$
We know that, heat produced due to current flowing through a resistor $R$.
$ H =I^{2} R t $
$ \Rightarrow \frac{\Delta H}{H} =\frac{2 \Delta I}{I}+\frac{\Delta R}{R}+\frac{\Delta t}{t}$
$ \Rightarrow \frac{\Delta H}{H} \times 100 =2\left(\frac{\Delta I}{I} \times 100\right)+\frac{\Delta R}{R} \times 100+\frac{\Delta t}{t} \times 100$
$ =2 \times 2 \%+1 \%+3 \%=8 \% $
$$ \frac{\Delta F}{F} \times 100=2 \% \text { and } \frac{\Delta t}{t} \times 100=3 \% $$
We know that, heat produced due to current flowing through a resistor $R$.
$ H =I^{2} R t $
$ \Rightarrow \frac{\Delta H}{H} =\frac{2 \Delta I}{I}+\frac{\Delta R}{R}+\frac{\Delta t}{t}$
$ \Rightarrow \frac{\Delta H}{H} \times 100 =2\left(\frac{\Delta I}{I} \times 100\right)+\frac{\Delta R}{R} \times 100+\frac{\Delta t}{t} \times 100$
$ =2 \times 2 \%+1 \%+3 \%=8 \% $
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