JEE MAIN - Physics (2020 - 3rd September Morning Slot - No. 24)
Using screw gauge of pitch 0.1 cm and
50 divisions on its circular scale, the thickness
of an object is measured. It should correctly be
recorded as
2.123 cm
2.124 cm
2.125 cm
2.121 cm
Explanation
Using a screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of an object is measured as:
Measurement = (Main scale reading) + (Circular scale reading × Least count)
where the least count is calculated as the pitch of the screw gauge divided by the number of divisions on the circular scale:
Least count = (Pitch of screw gauge) / (Number of circular scale divisions)
Least count = $${{0.1} \over {50}}$$ = 0.002 cm
Now if we multiply division of circular scale with least count then we get 0th digit of fraction part even.
Here only option B has 0th digit of fraction part even.
Measurement = (Main scale reading) + (Circular scale reading × Least count)
where the least count is calculated as the pitch of the screw gauge divided by the number of divisions on the circular scale:
Least count = (Pitch of screw gauge) / (Number of circular scale divisions)
Least count = $${{0.1} \over {50}}$$ = 0.002 cm
Now if we multiply division of circular scale with least count then we get 0th digit of fraction part even.
Here only option B has 0th digit of fraction part even.
Comments (0)
