Correct Answer: $\frac{4 \sqrt{2}-3 \sqrt{3}}{2}$ cm

Solution :
Let $OF$ and $OE$ be perpendiculars from centre $O$ to chords $AB$ and $CD$.
We know perpendiculars from the centre to the chord bisect the chord.
Given, the radius of the circle = 3 cm
AB = 2 cm
CD = 3 cm
So, FB = FA = 1 cm
And, EC = ED = $\frac{3}2$ cm
Now from $\triangle$OFB using Pythagoras theorem,
$OF^2 = 3^2 - 1^2$
⇒ $OF^2 = 9-1$
⇒ $OF^2 = 8$
⇒ $OF = 2\sqrt2$
From $\triangle$OEC using Pythagoras theorem,
$OE^2 = 3^2 - (\frac{3}2)^2$
⇒ $OE^2 = 9-(\frac{9}4)$
⇒ $OE^2 = \frac{27}4$
⇒ $OE=\frac{3\sqrt3}2$
So, the distance between AB and CD,
⇒ FE $=$ OF – OE
⇒ FE $= 2\sqrt2 - \frac{3\sqrt3}{2}$
⇒ FE $= \frac{4\sqrt2-3\sqrt3}{2}$ cm
So, the perpendicular distance between the two chords is $\frac{4\sqrt2-3\sqrt3}{2}$ cm.
Hence, the correct answer is $\frac{4\sqrt2-3\sqrt3}{2}$ cm.