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 ______ $ imes10^{-3}$ mm.

Question

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 ______ $	imes10^{-3}$ mm.

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Least count $=\frac{	ext{pitch}}{	ext{no. of divisions}}$. New value $=\frac{1.75}{0.5}	imes0.01=0.035$ mm $=35	imes10^{-3}$ mm. Answer: 35.

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 ______ $\times10^{-3}$ mm.

Explanation

Reviewed by Dr. Abubacker

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