Question 49

The least count of a screw gauge 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 $$\underline{\hspace{2cm}}\times10^{-3}\,mm.$$

Correct Answer: 35

Screw gauge: LC = 0.01 mm. Pitch increased by 75%, divisions reduced by 50%. Find new LC.

LC = pitch/divisions. If pitch becomes 1.75p and divisions become 0.5d:

New LC = 1.75p/(0.5d) = 3.5(p/d) = 3.5 × 0.01 = 0.035 mm

In units of 0.001 mm: 35

The answer is 35.

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