You measure two quantities as A = 1.0 m ± 0.2 m, B = 2.0 m ± 0.2 m. We should report correct value for √(AB) as

Sol: Let $\displaystyle x = \sqrt{AB} $

$\displaystyle x = \sqrt{(1.0)(2.0)} = 1.414 $

Rounding off to two significant figures ,

x = 1.4 m

$\displaystyle x = \sqrt{AB} $

$\displaystyle \frac{\Delta x}{x} = \pm( \frac{1}{2}\frac{\Delta A}{A} + \frac{1}{2}\frac{\Delta B}{B})$

$\displaystyle \frac{\Delta x}{x} = \pm \frac{1}{2}( \frac{\Delta A}{A} + \frac{\Delta B}{B})$

$\displaystyle \frac{\Delta x}{x} = \pm \frac{1}{2}( \frac{0.2}{1} + \frac{0.2}{2})$

$\displaystyle \frac{\Delta x}{x} = \pm \frac{1}{2}(0.3)$

$\displaystyle \Delta x = \pm \frac{0.3}{2} x $

$\displaystyle \Delta x = \pm \frac{0.3}{2} \times 1.414 $

$\displaystyle \Delta x = \pm 0.212 m = \pm 0.2 m$ ( Rounding off to one significant figures )

x + Δx = 1.4 m ± 0.2 m