Question

measure of two quantities as A=1.0m ±0.2m B=2.0m±0.2m.we should report correct value for √AB as​

Answer 1

There are two significant figures in 1.0 & 2.0

\sqrt{AB}=\sqrt{1.0\times 2.0}=\sqrt{2}

=1.414\; m

After roundig up 1.0 &2.0 we het two sigificat features, i.e.

X=\sqrt{AB}

=1.4

Now, \frac{\Delta X}{X}=\frac{1}{2}[\frac{\Delta A}{A}+\frac{\Delta B}{B}]

\frac{\Delta X}{1.4}=0.1\left [ \frac{2.0+1.0}{1.0\times 2.0} \right ]

\Delta X=0.1(\frac{3}{2})((1.4)

=0.21\ m after rounding up

Explanation:

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