Question

# Screw gauge is an instrument used to measure the dimensions of very small objects up to ……….A) 0.1cmB) 0.01cmC) 0.1mmD) 0.01mm

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Hint: The statement ‘dimensions of very small objects up to’ basically means that they are asking for the least count of the screw gauge and least count of screw gauge can be calculated by-
$L.C = \dfrac{p}{n}$
Where, $p$ is the pitch and $n$ is the number of divisions on circular scale.

The least count of a screw gauge is defined as the ratio between the pitch and the number of divisions on the circular scale i.e.
$L.C = \dfrac{p}{n}$
Where p= pitch and n= number of divisions on circular scale.
For a standard screw gauge the number of division on the circular scale is always fixed and equals to 100
Therefore, n = 100
Pitch is defined as the linear distance moved by the spindle on giving one complete revolution.
For a standard screw gauge, pitch (p) = 1mm
Therefore,
$L.C = \dfrac{1}{{100}}mm \\$
We can write this as,
$L.C = {10^{ - 2}}mm \\$
On simplifying it, we get
$\Rightarrow L.C = 0.01mm \\$

$\therefore$ Screw gauge can measure upto 0.01mm. So, Option D is the correct answer.

Note:
Although screw gauge is the more accurate device as compared to vernier callipers, when we need to find the internal diameter of a hollow body then it can’t be measured with the help of screw gauge, in such cases only Vernier callipers are used. This is the limitation of a screw gauge.
Least count of any instrument or device is the least possible measurement that we can take with the help of that particular instrument or device.