The least count of a screw guage is 0.01 mm . If the pitch is increased by 75% and number of divisions on the circular scale is reduced by 50%, the new least count will be ________ × 10⁻³ mm

Solution

Physics sir · 3 steps

Least count is pitch divided by divisions, so scale the top up and the bottom down, then divide.

1. Least count formula. For a screw gauge, least count is the pitch p divided by the number of circular scale divisions N. Here p/N = 0.01 mm.
2. Apply the changes. Increasing the pitch by 75% makes it 1.75p. Reducing the divisions by 50% leaves 0.5N.

   LC_(new) = 1.75p/0.5N = 3.5 × p/N = 3.5 × 0.01 mm = 0.035 mm

3. So. The new least count is 0.035 mm = 35 × 10⁻³ mm, so the answer is 35.

The common slip: 8.75. Multiplying by 0.5 for the halved divisions, instead of dividing by 0.5, gives 1.75 × 0.5 × 0.01 = 0.00875 mm, that is 8.75. Fewer divisions make the least count bigger, not smaller.

The idea to keep. Least count changes in the same direction as the pitch and in the opposite direction to the number of circular divisions. Write each change as a multiplier and put it in p/N.

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