Answer
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Hint: Least count of a micrometer screw gauge is the smallest division on the main scale divided by the total number of divisions.
Complete step by step solution:
Micrometer screw gauge is a device having a calibrated screw. Micrometer screw gauge has two scales, one rotating that can be found on its rotating cylindrical part called circular scale. The other part is its stationery sleeve which is called the main scale or sleeve scale. The circular scale is divided into 50 or 100 equal parts.
The number of divisions on the main scale of a micrometer screw gauge is 10 to a centimeter. Therefore the smallest division on the main scale will be \[1mm\].
The smallest division on the main scale of a micrometer screw gauge is known as the pitch (p) of the screw gauge. So, $p = 100$.
The number of circular divisions is(n) = 100
The least count of a screw gauge = $\dfrac{p}{n} = \dfrac{{1\,mm}}{{100}} = 0.01\,mm$
Hence option (A) is correct.
Note: When both ends of a screw gauge coincides with each other in such a way the zeros of both circular scale and main scale exactly overlap each other, then there is no error. If it is not so, the screw gauge has an error called zero error. Zero error can be negative or positive. The positive or negative error of the screw gauge is equal to the product of the number of divisions on circular scale coinciding with the main scale and the least count of the micrometer screw gauge.
Complete step by step solution:
Micrometer screw gauge is a device having a calibrated screw. Micrometer screw gauge has two scales, one rotating that can be found on its rotating cylindrical part called circular scale. The other part is its stationery sleeve which is called the main scale or sleeve scale. The circular scale is divided into 50 or 100 equal parts.
The number of divisions on the main scale of a micrometer screw gauge is 10 to a centimeter. Therefore the smallest division on the main scale will be \[1mm\].
The smallest division on the main scale of a micrometer screw gauge is known as the pitch (p) of the screw gauge. So, $p = 100$.
The number of circular divisions is(n) = 100
The least count of a screw gauge = $\dfrac{p}{n} = \dfrac{{1\,mm}}{{100}} = 0.01\,mm$
Hence option (A) is correct.
Note: When both ends of a screw gauge coincides with each other in such a way the zeros of both circular scale and main scale exactly overlap each other, then there is no error. If it is not so, the screw gauge has an error called zero error. Zero error can be negative or positive. The positive or negative error of the screw gauge is equal to the product of the number of divisions on circular scale coinciding with the main scale and the least count of the micrometer screw gauge.
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