$$10$$ divisions on the main scale of a Vernier calliper coincide with $$11$$ divisions on the Vernier scale. If each division on the main scale is of $$5$$ units, the least count of the instrument is :

Find the least count of a Vernier calliper where 10 MSD = 11 VSD and each MSD = 5 units.

MSD stands for Main Scale Division and VSD for Vernier Scale Division. Since 10 main-scale divisions span the same length as 11 vernier-scale divisions, we have $$ 10 \times 5 = 11 \times \text{(1 VSD)} $$. It follows that $$ 1 \text{ VSD} = \frac{50}{11} \text{ units} $$. The least count (LC) of a Vernier instrument is the difference between one main-scale division and one vernier-scale division, so $$ LC = 1 \text{ MSD} - 1 \text{ VSD} = 5 - \frac{50}{11} = \frac{55 - 50}{11} = \frac{5}{11} $$. Therefore, the correct answer is Option D: $$\frac{5}{11}$$.