In a Vernier calliper 50 vernier scale dimension coincides with 48 mass scale dimensions. If one mass scale dimension is 1 mm then the least count of the measurement is

The correct option is 0.004 cm.

Let us find the least count of the Vernier calliper step-by-step.
The formula for the least count (LC) of a Vernier calliper is given by:
$\text{Least Count (LC)}=1\text{MSD}-1\text{VSD}$
where:
MSD = Main Scale Division (referred to as "mass scale dimension" in the question)
VSD = Vernier Scale Division (referred to as "vernier scale dimension" in the question)

Given details:
1 MSD = 1 mm
50 VSD coincides with 48 MSD.
Therefore, we can write:
$50\text{VSD}=48\text{MSD}$
From this, we find the value of 1 VSD in terms of MSD:
$1\text{VSD}=\frac{48}{50}\text{MSD}=0.96\text{MSD}$

Now, let us calculate the least count:
$\text{LC}=1\text{MSD}-0.96\text{MSD}$
$\text{LC}=0.04\text{MSD}$

Since 1 MSD = 1 mm, we substitute this value:
$\text{LC}=0.04\times 1\text{mm}=0.04\text{mm}$

To convert the least count from millimeters (mm) to centimeters (cm), we divide by 10 (since 1 cm = 10 mm):
$\text{LC}=\frac{0.04}{10}\text{cm}=0.004\text{cm}$

Thus, the least count of the measurement is 0.004 cm.