JEE MAIN - Physics (2020 - 9th January Morning Slot - No. 14)
If the screw on a screw-gauge is given six
rotations, it moves by 3 mm on the main scale.
If there are 50 divisions on the circular scale
the least count of the screw gauge is :
0.001 mm
0.01 cm
0.02 mm
0.001 cm
Explicació
Pitch = $${3 \over 6}$$ mm = 0.5 mm
Least count = $${{0.5} \over {50}}$$ mm
= $${1 \over {100}}$$ = 0.01 mm = 0.001 cm
Least count = $${{0.5} \over {50}}$$ mm
= $${1 \over {100}}$$ = 0.01 mm = 0.001 cm