Question

The pitch of the screw gauge is $1mm$ and there are $100$ divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lines $8$ divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while the ${72}^{nd}$ division on the circular scale coincides with the reference line. The radius of the wire is

• A. $1.64mm$
• B. $1.80mm$
• C. $0.82mm$
• D. $0.90mm$

Answer 1

The correct option is C

$0.82mm$

Step 1: Given Data

Pitch $p=1mm$

Number of divisions $d=100$

Zero error $z=8$ divisions below the reference line

Reading of the division on the circular scale $r={72}^{nd}$ division

Step 2: Formula Used

Measured Reading = Main Scale Reading + Reading of the division on the circular scale x Least Count

Step 3: Calculate the Least Count

We know that the least count is given as,

$$\begin{gathered} LC=\frac{p}{d} \\ =\frac{1mm}{100} \\ =0.01mm \end{gathered}$$

Step 4: Calculate the Zero Error

Since the zero error is below the reference line, it is a positive zero error, which can be given as

$$\begin{gathered} ZE=z\times LC \\ =8\times 0.01 \\ =0.08mm \end{gathered}$$

Step 5: Calculate the Measured Reading

We know that the measured reading can be given as,

$$\begin{gathered} MR=1mm+r\times LC \\ =1mm+72\times 0.01 \\ =1mm+0.72 \\ =1.72mm \end{gathered}$$

Step 6: Calculate the True Reading

The true reading can be given as,

$$\begin{gathered} TR=MR-ZE \\ =1.72-0.08 \\ =1.64mm \end{gathered}$$

Step 7: Calculate the Radius

Using the screw gauge we measured the diameter of the wire.

Therefore the radius of the wire is,

$$\begin{gathered} Ra=\frac{TR}{2} \\ =\frac{1.64}{2} \\ =0.82mm \end{gathered}$$

Hence, the radius of the wire is $0.82mm$ and therefore the correct answer is option (C).