Question

The thimble of a screw gauge has 50 divisions. The spindle advances 1mm when the screw is turned through two revolutions. What is the pitch of the screw gauge?
$\begin{aligned} & \left( A \right)0.5mm \\\ & \left( B \right)0.1mm \\\ & \left( C \right)0.05mm \\\ & \left( D \right)0.005mm \\\ \end{aligned}$

Answer 1

Screw gauge is a device used to measure very small lengths, that is lengths of the order of the fraction of a millimeter. The device is used to measure the diameter of very thin wires, or the thickness of thin sheets. A Screw gauge is a more accurate device than a Vernier caliper as it has a less value of least count measure.

Complete answer:
In the problem given to us, the thimble of the screw gauge has 50 divisions. The number of divisions on the thimble of a screw gauge is actually the number of circular divisions. Therefore, we can write:
Number of circular divisions on the screw gauge $=50$
Now, since nothing is mentioned about error, we will assume there is no error in the measuring instrument.
Let the linear distance moved in two revolutions be denoted by ‘d’. Then, according to the data given in the problem this is equal to:
$\Rightarrow d=1mm$
Now, the formula for the pitch of the screw gauge can be written as:
$\Rightarrow \text{Pitch}=\dfrac{\text{Distance moved by the spindle (d)}}{\text{Number of revolutions}}$
Putting the value of ‘d’ from above and using the number of revolutions value as ‘2’, we get:
$\begin{aligned} & \Rightarrow \text{Pitch}=\dfrac{1}{2}mm \\\ & \therefore \text{Pitch}=0.5mm \\\ \end{aligned}$
Hence, the Pitch of the Screw gauge comes out to be 0.5mm .

Hence, option (A) is the correct option.

Note:
While solving any problem from length and measurement, we should always re-check our solution at the end of the process. Also, we should take note of errors in the measurement, if any. And in case, nothing is mentioned in the problem, then the error of the measuring instrument is taken to be zero.