Assertion A : If in five complete rotations of the circular scale,
Question:

Assertion A : If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is 5 mm and there are 50 total divisions on circular scale, then least count is 0.001 cm.

Reason R :

Least Count $=\frac{\text { Pitch }}{\text { Totaldivisionson circular scale }}$

In the light of the above statements, choose the most appropriate answer from the options given below : 

  1. A is not correct but R is correct. 

  2. Both A and R are correct and R is the correct explanation of A. 

  3. A is correct but R is not correct

  4. Both A and R are correct and R is NOT the correct explanation of A.


Correct Option: 1,

Solution:

Least count $=\frac{\text { Pitch }}{\text { total division on circular scale }}$

In 5 revolution, distance travel, 5 mm

In 1 revolution, it will travel 1 mm.

So least count $=\frac{1}{50}=0.02$

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