If in a verneir callipers 10 VSD coincides with 8 MSD then the least count of verneir callipers is (given 1 MSD=1mm)
Answers
A vernier callipers has a Least Count, and not a pitch. We talk of pitch in a screw gauge or a spherometer. These two instruments have a threaded screw, which moves forward or backward when you rotate the head. The distance advanced on one full rotation of the head is called pitch of the instrument.
There are two methods by which we can find the least count of a vernier callipers. The significance of least count is that it gives the least thickness/distance that can be measured with the instrument.
Method 1:
Number of divisions in which 1 cm is divided on the main scale = 20
Value of the smallest division on the main scale = 1/20 cm= 0.05 cm
Number of divisions on vernier scale =20
Least Count of the vernier callipers= (value of smallest division on the main scale)/(number of divisions on the vernier scale) = 0.05 cm/20= 0.0025 cm
The least count of the vernier callipers= 0.0025 cm.
Method 2:
Smallest division on the main scale = 0.05 cm
20 vernier divisions = 19 main scale divisions
1 vernier division = 19/20 main scale divisions
Difference between 1main scale division and 1 vernier division= 1 msd - 19/20 msd =1/20 msd= 1/20×0.05cm= 0.0025 cm.
The least count of the vernier callipers = 0.0025cm
Both methods give the same value of the vernier constant, or least count as it should.
I personally prefer the first method as it is applicable for vernier callipers, screw gauge and spherometer. It is also simpler.
HOPE IT HELPS
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⚫10 VSD = 8 MSD
⚫1 VSD = 0.8 MSD
⚫Least count = 1 MSD - 1 VSD
⚫1 MSD - 0.8 MSD
⚫0.2 MSD
⚫0.2 × 1mm
▶️▶️0.2mm