
While measuring the thickness of the glass plate using a screw gauge, the main scale reads $2\;mm$ and head scale coinciding division is $57\;$. If the least count of the screw gauge is $0.01\;mm$ find out the thickness of the glass plate. (Zero error of the screw gauge is zero).
A. $2.57\;mm$
B. $6.25\;mm$
C. $3.25\;mm$
D. $16.3\;mm$
Answer
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Hint: A screw gauge is used to measure small diameters (of wires) and thickness (of sheets and plates). It has two divisions, the main scale, which is generally in millimetres, and a circular scale division which has $50\;$ or $100\;$ equal parts. A full rotation of this circular scale, moves the main scale forward by $1\;$ division (millimetres).
Complete step by step answer:
In a screw gauge, there are two scales which are used to make a precise measurement.
The main scale- This is the linear scale where each division is $1\;mm$ long. This is also called the pitch scale. As the ratchet of a screw jack is rotated, the main scale has to fit the object between the stud and the screw. One full rotation of the ratchet moves the main scale by $1\;$ division.
The circular scale- This is a circular scale where a circle is divided into $50\;$ or $100\;$ equal parts. The number on the circular scale which coincides with the division line of the linear scale gives the other significant digits of the thickness of the object. The number of divisions on it tells the relation between the pitch scale and the circular scale.
Given that the screw jack has no zero error,
The thickness of an object is given by-
$T = MSR + LC \times n$
Where $MSR\;$ is the main scale reading, which is given equal to, $2\;mm$.
$LC\;$ is the least count of the scale, which is given equal to, $0.01\;mm$
And $n\;$ is the reading of the division of circular scale that coincides with the main scale, this is, $57\;$.
Putting these values,
$T = 2 + \left( {0.01 \times 57} \right)$
$T = 2.57mm$
Therefore option (A) is correct.
Note: A screw jack is also known as a micrometer because it is capable of measuring the thickness of objects which are precise to a scale of a micrometre or ${10^{ - 6}}m$. It consists of a carefully built screw inside the apparatus that rotates as well as moves forward when rotated.
Complete step by step answer:
In a screw gauge, there are two scales which are used to make a precise measurement.
The main scale- This is the linear scale where each division is $1\;mm$ long. This is also called the pitch scale. As the ratchet of a screw jack is rotated, the main scale has to fit the object between the stud and the screw. One full rotation of the ratchet moves the main scale by $1\;$ division.
The circular scale- This is a circular scale where a circle is divided into $50\;$ or $100\;$ equal parts. The number on the circular scale which coincides with the division line of the linear scale gives the other significant digits of the thickness of the object. The number of divisions on it tells the relation between the pitch scale and the circular scale.
Given that the screw jack has no zero error,
The thickness of an object is given by-
$T = MSR + LC \times n$
Where $MSR\;$ is the main scale reading, which is given equal to, $2\;mm$.
$LC\;$ is the least count of the scale, which is given equal to, $0.01\;mm$
And $n\;$ is the reading of the division of circular scale that coincides with the main scale, this is, $57\;$.
Putting these values,
$T = 2 + \left( {0.01 \times 57} \right)$
$T = 2.57mm$
Therefore option (A) is correct.
Note: A screw jack is also known as a micrometer because it is capable of measuring the thickness of objects which are precise to a scale of a micrometre or ${10^{ - 6}}m$. It consists of a carefully built screw inside the apparatus that rotates as well as moves forward when rotated.
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