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 :

Solution

Physics sir · 3 steps

Find the pitch first, then share it among the 50 circular divisions, and watch the unit at the end.

1. Find the pitch. The pitch is how far the screw moves in one full rotation. Six rotations move it 3 mm, so one rotation moves it 3/6 = 0.5 mm.
2. Least count. One full rotation turns all 50 circular divisions past the line. So one division moves the screw by the pitch divided by 50.

   LC = pitch/(number of circular divisions) = (0.5 mm)/50 = 0.01 mm

3. So. Since 1 cm = 10 mm, 0.01 mm = 0.001 cm. That is option (d).

The idea to keep. Least count of a screw gauge is pitch divided by circular divisions. Get the pitch from distance moved over number of rotations, then match the unit used in the options.

Answer from the NTA final answer key. Solution by PrepWiser.