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 ____ mm. [2025]

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Solution

Q.
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 ____ $\times 10^{- 3}$ mm. [2025]

Ans.
(35)

Given least count of Screw Gauge = 0.01 mm

$L . C = \frac{pitch}{No. of circular turns} = \frac{P}{N} = 0.01 mm$

$New pitch = \frac{P (1 + 0.75)}{N (1 - 0.5)} = \frac{P}{N} [\frac{1.75}{0.5}]$

$= (0.01) 3.5 = 0.035 mm = 35 \times 10^{- 3} mm$

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