First slide

Accuracy; precision of instruments and errors in measurements

Question

The volumes of two bodies are measured to be $V_{1} = (10.2 \pm 0.02) c m^{3} a n d V_{2} = (6.4 \pm 0.01) c m^{3} .$ The sum and difference in volumes with error limits is

Moderate

A

$(16.6 \pm 0.03) c m^{3} a n d (3.8 \pm 0.03) c m^{3}$
B

$(16.6 \pm 0.01) c m^{3} a n d (3.8 \pm 0.01) c m^{3}$
C

$(16.2 \pm 0.03) c m^{3} a n d (3.6 \pm 0.03) c m^{3}$
D

$(16.2 \pm 0.01) c m^{3} a n d (3.6 \pm 0.01) c m^{3}$

Solution

Given, $V_{1} = (10.2 \pm 0.02) c m^{3}$
and $V_{2} = (6.4 \pm 0.01) c m^{3}$
$Δ V = \pm (Δ V_{1} + Δ V_{2})$
$= \pm (0.02 + 0.01) c m^{3} = \pm 0.03 c m^{3}$
$V_{1} + V_{2} = (10.2 + 6.4) c m^{3} = 16.6 c m^{3}$
and $V_{1} - V_{2} = (10.2 - 6.4) c m^{3} = 3.8 c m^{3}$
Hence, sum of volume $= (16.6 \pm 0.03) c m^{3}$
and difference of volume $= (3.8 \pm 0.03) c m^{3}$
In product If Z = AB, then maximum fractional error is $\frac{Δ Z}{Z} = \pm (\frac{Δ A}{A} + \frac{Δ B}{B})$
Therefore, maximum fractional error in product of two (or more) quantities is equal to sum of fractional errors in the individual quantities.
In division If Z = A /B, then maximum fractional error is $\frac{Δ Z}{Z} = \pm (\frac{Δ A}{A} + \frac{Δ B}{B})$
Therefore, maximum fractional error in product of two ( or more) quantities is equal to sum of fractional errors in the individual quantities.

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