Question

When a piece of wire is held diametrically in a screw gauge [pitch =1mm, number of divisions on circular scale =100], the readings obtained are as shown. Now, if we measure the same with the help of vernier calipers [1MSD=1mm,10 divisions of vernier coinciding with 9 divisions of the main scale] having a negative zero error of 0.5mm, then which of the following figures correctly represents the reading?

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The correct option is A

Vernier Calliper:

1. The length of a rod, the diameter of a sphere, or the inner and outer radius of a hollow cylinder are all calculated using a vernier caliper.
2. The main scale and the sliding vernier scale make up a vernier caliper.
3. The vernier scale ensures that readings are accurate to $0.02\mathrm{mm}$.
4. The measuring object is held in the vernier caliper's jaws, and the reading is taken.

Given data:

1. In this question, we were given information from two distinct instruments that measured the diameter of a wire.
2. In addition, we were given certain reading figures on vernier calipers as possibilities.
3. As a result, we'll calculate the screw gauge readings first, then choose the figure that best indicates a similar reading on a vernier caliper from the options.

Formula Used:

$$\begin{gathered} \mathrm{R}=\mathrm{LSR}+(\mathrm{CSR}\times \mathrm{LC}) \\ \mathrm{R}=[\mathrm{MSR}+(\mathrm{VSR}\times \mathrm{LC}\left)\right] \end{gathered}$$

A positive error occurs when zero is on the left side and coincides with the reading on the circular scale.

As a result of the figure, the screw gauge reading has a positive zero inaccuracy. We also know that the final measurement is subtracted from the positive zero error.

The least count of screw gauge $LC=1mm/100=0.01mm$

As zero on a linear scale (left image) coincides with 20 on a circular scale so it has a

$$\begin{gathered} \mathrm{Positive}\mathrm{zero}\mathrm{error}=20\times 0.01 \\ =0.2\mathrm{mm} \\ =0.02\mathrm{cm} \end{gathered}$$ and it should be subtracted from the final reading.

From the right image, the final reading

$$\begin{gathered} \mathrm{R}=\mathrm{LSR}+(\mathrm{CSR}\times \mathrm{LC}) \\ =3+(50\times 0.01) \\ =3.5\mathrm{mm} \\ =0.35\mathrm{cm} \end{gathered}$$

The final reading with zero error is

$$\begin{gathered} \mathrm{R}=0.35-0.02 \\ =0.33\mathrm{cm}...\left(1\right) \end{gathered}$$

The least count of calipers is

$$\begin{gathered} \mathrm{LC}=\frac{1}{10}\mathrm{mm} \\ =0.1\mathrm{mm} \\ =0.01\mathrm{cm}...\left(2\right) \end{gathered}$$

Final reading R with zero error for calipers is

For A:

$$\begin{gathered} \mathrm{R}=[\mathrm{MSR}+(\mathrm{VSR}\times \mathrm{LC}\left)\right]+0.05\mathrm{cm} \\ =[0.2+(8\times 0.01\left)\right]+0.05 \\ =0.33\mathrm{cm}...\left(\mathrm{A}\right) \end{gathered}$$

For B:

$$\begin{gathered} \mathrm{R}=[\mathrm{MSR}+(\mathrm{VSR}\times \mathrm{LC}\left)\right]+0.05\mathrm{cm} \\ =[0.3+(8\times 0.01\left)\right]+0.05 \\ =0.43\mathrm{cm}...\left(\mathrm{B}\right) \end{gathered}$$

For C:

$$\begin{gathered} \mathrm{R}=[\mathrm{MSR}+(\mathrm{VSR}\times \mathrm{LC}\left)\right]+0.05\mathrm{cm} \\ =[0.2+(10\times 0.01\left)\right]+0.05 \\ =0.35\mathrm{cm}...\left(\mathrm{C}\right) \end{gathered}$$

For D:

$$\begin{gathered} \mathrm{R}=[\mathrm{MSR}+(\mathrm{VSR}\times \mathrm{LC}\left)\right]+0.05\mathrm{cm} \\ =[0.3+(6\times 0.01\left)\right]+0.05 \\ =0.41\mathrm{cm}...\left(\mathrm{D}\right) \end{gathered}$$

The equations ($1$) and ($\mathrm{A}$) are the same. Thus, option A is matching with the value of the screw gauge reading.

Therefore, the correct option is A.