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# The least count of the main scale of the screw gauge is 1mm. The minimum number of divisions on its circular scale required to, measure 5μm diameter of the wire is(A) 50(B) 100(C) 200(D) 500

Last updated date: 13th Jun 2024
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Hint: The least count of an instrument is the distance which you can measure effectively. Quantitatively, least count is defined as the pitch of the instrument divided by the number of divisions on the instruments. In this question we are given the least count as 1mm. we are also given the pitch of the scale which is equal to $5\mu m$ . Substitute these 2 values in the equation to get the number of divisions.

Complete step-by-step solution
Screw gauge is a measuring instrument used to measure very small units like diameter of a thin wire, thickness of a metal sheet and other smaller values. It has two scales: the head scale and the pitch scale. The ratchet at the end is made to rotate to fix the object in place and find the measurement.
The least count (LC) of the instrument means the smallest value up to which you can measure.
Least count of screw gauge is $1mm = 1 \times {10^{ - 3}}m$
The pitch is given = $5\mu m = 5 \times {10^6}m$
To find the number of divisions (N) we use the formula,
$LC = \dfrac{{Pitch}}{N}A$
After substituting the values we get,
$N = \dfrac{{LC}}{{Pitch}} \\ \Rightarrow N = \dfrac{{1 \times {{10}^{ - 3}}}}{{5 \times {{10}^{ - 6}}}} \\ \Rightarrow N = 200 \\$

Hence, the number of divisions on the circular scale needed is 200 and the option is C.

Note
Make sure that you convert each variable on the same scale. In this equation, we were given the value of least count and pitch in millimeters and nanometers or microns. We need to first convert it into the same unit and then proceed.
Other than screw gauge, vernier calipers, traveling microscopes are some instruments to measure small units which have their respective least counts.