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Standard XI

Physics

Multiplication & Division of Errors

According to ...

Question

According to Joule's law of heating, heat produced $H = I^{2} R t,$ where $I$ is current, $R$ is resistance and $t$ is time. If the errors in the measurements of $I$ , $R$ and $t$ are 3%, 4% and 6% respectively, then error in the measurement of H is

\pm 17 %

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\pm 16 %

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\pm 19 %

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\pm 25 %

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Solution

The correct option is B $\pm 16 %$
$H = I^{2} R t$
$∴ \frac{Δ H}{H} \times 100 = (\frac{2 Δ I}{I} + \frac{Δ R}{R} + \frac{Δ t}{t}) \times 100$
$= (2 \times 3 + 4 + 6) % = 16 %$

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According to Joule's law of heating, heat produced H=I2 Rt, where I is current, R is resistance and t is time. If the errors in the measurements of I, R and t are 3%, 4% and 6% respectively, then error in the measurement of H is

The correct option is B ± 16%H=I2Rt ∴Δ HH×100=(2Δ II+Δ RR+Δ tt)×100 =(2×3+4+6)%=16%

According to Joule's law of heating, heat produced $H = I^{2} R t,$ where $I$ is current, $R$ is resistance and $t$ is time. If the errors in the measurements of $I$ , $R$ and $t$ are 3%, 4% and 6% respectively, then error in the measurement of H is

The correct option is B $\pm 16 %$
$H = I^{2} R t$
$∴ \frac{Δ H}{H} \times 100 = (\frac{2 Δ I}{I} + \frac{Δ R}{R} + \frac{Δ t}{t}) \times 100$
$= (2 \times 3 + 4 + 6) % = 16 %$