Question 242: Units , Dimensions and Error Analysis

Question

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 ________ $	imes 10^{-3} \mathrm{~mm}$

Accepted Answer

The least count (L.C.) of a screw gauge is given by:
$ 	ext{L.C.} = \frac{	ext{Pitch}}{	ext{Number of Divisions on Circular Scale}} $
Let the original pitch be $p$ and the original number of divisions be $n$. Then, the original least count is:
$ \frac{p}{n} = 0.01 \, 	ext{mm} $
After modifications:
The pitch is increased by 75%, so the new pitch is:
$    p_{	ext{new}} = p + 0.75p = 1.75p    $
The number of divisions is reduced by 50%, so the new number of divisions is:
$    n_{	ext{new}} = 0.5n    $
The new least count is then:
$ 	ext{L.C.}_{	ext{new}} = \frac{p_{	ext{new}}}{n_{	ext{new}}} = \frac{1.75p}{0.5n} = \frac{1.75}{0.5} 	imes \frac{p}{n} = 3.5 	imes \frac{p}{n} $
Substitute the original least count:
$ 	ext{L.C.}_{	ext{new}} = 3.5 	imes 0.01 \, 	ext{mm} = 0.035 \, 	ext{mm} $
Expressed in the form $	imes 10^{-3} \, 	ext{mm}$, this becomes:
$ 0.035 \, 	ext{mm} = 35 	imes 10^{-3} \, 	ext{mm} $
Thus, the new least count is:
$ 35 	imes 10^{-3} \, 	ext{mm} $

Units , Dimensions and Error Analysis

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 ________ \times 10^{-3} \mathrm{~mm}

Integer / Numeric Answer Type