Question

The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. A student measures the diameter of a wire using this screw gauge and he gets the main scale reading as 5 mm and circular divisions as 25. If the screw gauge has a positive zero error of 0.03 mm, the correct value of diameter will be:
(a). 5.28 mm
(b). 5.25 mm
(c ). 5.20 mm
(d). 5.22 mm

Answer 1

- Hint: The screw gauge is a measuring device and if it has an error suitable changes should be made to the value obtained from the measurement.

Complete step-by-step solution -

So we are given a screw gauge of pitch 1 mm and it has 100 divisions on the circular scale. The least count of the screw gauge is calculated using the formula,
$\text{least count=}\left( \dfrac{\text{Pitch of the screw gauge}}{\text{Number of divisions on the circular scale}} \right)$
So, our pitch of the screw gauge is given as 1 mm and number of divisions is 100, so substituting in the above formula we get,
$\begin{align}
  & \text{least count=}\left( \dfrac{1}{100} \right) \\
 & \therefore \text{ least count}=0.01\text{ mm} \\
\end{align}$ ….. equation (1)
So the least count is found to be 0.01 mm. We usually take the readings from the main scale and circular scale. The actual measurement that we do with the screw gauge is given by the formula,
$\text{Initial Reading}=\text{ M}\text{.S}\text{.R}+\text{ C}\text{.R}\times \text{L}\text{.C}$
Where,
M.S.R is the main scale reading
C.R is the circular scale reading
L.C is the least count of the screw gauge from equation (1).
Calculating the initial with M.S.R = 5 mm, C. R= 25 and L.C=0.01 mm, we get
$\begin{align}
  & \text{Initial Reading}=5+25\times 0.01 \\
 & \therefore \text{ Initial Reading}=5.25\text{ mm} \\
\end{align}$
We now have calculated the initial reading, but we need the actual reading or the true reading which can be found out considering the zero error. The actual reading is given by the formula,
$\text{Actual Reading}=\text{Initial Reading}-\text{Zero Error}$
According to the question the zero error given is positive zero error of 0.03 mm. Since it is a positive zero error you can directly substitute in the equation. (We have to put a negative sign for negative zero error.)

 $\text{Actual Reading}=(5.25-0.03)\text{mm}$
$\therefore \text{ Actual Reading}=5.22\text{ mm}$

 So the correct answer is option (D) 5.22 mm.

Note: There are two types of zero error.
(1). Positive zero error: If the zero mark on the circular scale does not coincide with the baseline of the main scale and it is below the baseline of the main scale, the error is said to be positive.
(2). Negative Zero error: If the zero mark on the circular scale does not coincide with the baseline of the main scale and it is above the baseline, the error is said to be negative.