JEE MAIN - Physics (2021 - 27th July Morning Shift - No. 18)
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 = $${{Pitch} \over {Total\,divisions\,on\,circular\,scale}}$$
In the light of the above statements, choose the most appropriate answer from the options given below :
Reason R :
Least Count = $${{Pitch} \over {Total\,divisions\,on\,circular\,scale}}$$
In the light of the above statements, choose the most appropriate answer from the options given below :
A is not correct but R is correct.
Both A and R are correct and R is the correct explanation of A.
A is correct but R is not correct.
Both A and R are correct and R is NOT the correct explanation of A.
Explanation
Least Count = $${{Pitch} \over {Total\,divisions\,on\,circular\,scale}}$$
In 5 revolution, distance travel, 5 mm
In 1 revolution, it will travel 1 mm.
So least count = $${1 \over {50}}$$ = 0.02
In 5 revolution, distance travel, 5 mm
In 1 revolution, it will travel 1 mm.
So least count = $${1 \over {50}}$$ = 0.02
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