A vernier callipers has a least count of 0.05 mm. The main scale is...

The least count (LC) of a vernier callipers is given by the difference between the length of one main scale division (MSD) and one vernier scale division (VSD).

LC $= 1$ MSD $- 1$ VSD

Given that LC $= 0.05$ mm and $1$ MSD $= 0.5$ mm, we can find the length of $1$ VSD:

$0.05$ mm $= 0.5$ mm $- 1$ VSD

$1$ VSD $= 0.5$ mm $- 0.05$ mm $= 0.45$ mm

The total length of the vernier scale, which has $50$ divisions, is:

Length of $50$ VSD $= 50 \times 0.45$ mm $= 22.5$ mm

To find how many main scale divisions coincide with this length, we divide the total length of the vernier scale by the length of one main scale division:

Number of coinciding MSD $= \dfrac{22.5 \text{ mm}}{0.5 \text{ mm}} = 45$

Thus, $45$ main scale divisions coincide exactly with $50$ vernier scale divisions.

Answer: $45$