In the given figure, chord AB is larger than chord CD. The relation between OM and ON is :

1. OM = ON

2. OM < ON

3. OM > ON

4. OM + ON = AB

Join OC and OB.

Given,

⇒ AB > CD

⇒ $\dfrac{AB}{2} \gt \dfrac{CD}{2}$

⇒ BM > CN

From figure,

OC = OB = radius = r.

In right angle triangle ONC,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = ON² + CN²

⇒ r² = ON² + CN²

⇒ ON² = r² - CN²

⇒ ON = $\sqrt{r^2 - CN^2}$

In right angle triangle OMB,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OB² = OM² + BM²

⇒ r² = OM² + BM²

⇒ OM² = r² - BM²

⇒ OM = $\sqrt{r^2 - BM^2}$

Since, BM > CN

∴ r² - BM² < r² - CN²

⇒ $\sqrt{r^2 - BM^2} \lt \sqrt{r^2 - CN^2}$

⇒ OM < ON.

Hence, Option 2 is the correct option.