Consider a vernier caliper in which each 1 cm on the main scale is divided into 8 equal divisions an

Q: Why do we subtract 1 VSD from 1 MSD to find the least count?

The least count is the smallest measurement possible. In a Vernier caliper, this is exactly the difference in length between one main scale division and one Vernier scale division.

Q: How did we get 1 VSD = 4/5 MSD?

The problem states that 5 divisions of the Vernier scale coincide with 4 divisions of the main scale. Equating their lengths gives $5 \text{ VSD} = 4 \text{ MSD}$, which simplifies to $1 \text{ VSD} = \frac{4}{5} \text{ MSD}$.

Q: Why is the pitch of the screw gauge equal to 2 linear divisions?

Pitch is defined as the linear distance moved by the screw in one complete rotation. The problem explicitly states that one complete rotation moves it by two divisions on the linear scale.

Q: Why do we divide by 100 to find the least count of the screw gauge?

The circular scale has 100 divisions. The least count is the distance moved when the circular scale is rotated by just one division, which is $\frac{\text{Pitch}}{\text{Total Circular Divisions}}$.

Q: Why did we convert the final answers to millimeters?

All the options are given in millimeters, so we must convert our calculated values from centimeters to millimeters to match and verify the correct options.

Question

Consider a vernier caliper in which each $1	ext{ cm}$ on the main scale is divided into $8$ equal divisions and a screw gauge with $100$ divisions on its circular scale. In the vernier callipers, $5$ divisions of the vernier scale coincide with $4$ divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then

Options:

(A) if the pitch of the screw gauge is twice the least count of the vernier caliper, the least count of the screw gauge is $0.01	ext{ mm}$

(B) if the pitch of the screw gauge is twice the least count of the Vernier caliper, the least count of the screw gauge is $0.005	ext{ mm}$

(C) if the least count of the linear scale of the screw gauge is twice the least count of the Vernier calipers, the least count of the screw gauge is $0.01	ext{ mm}$

(D) if the least count of the linear scale of the screw gauge is twice the least count of the vernier caliper, the least count of the screw gauge is $0.005	ext{ mm}$

Sigma Percentile · Accepted Answer

This problem is a beautiful exercise in understanding the fundamental principles of measuring instruments: the Vernier Caliper and the Screw Gauge. It tests your ability to derive the least count from basic definitions rather than just memorizing formulas.

### Decoding the Vernier Caliper

Let's start by breaking down the Vernier Caliper. The problem states that $1	ext{ cm}$ on the main scale is divided into $8$ equal divisions. This gives us the value of one Main Scale Division (MSD):

$$1	ext{ MSD} = \frac{1}{8}	ext{ cm} = 0.125	ext{ cm}$$

Next, we look at the Vernier scale. We are told that $5$ divisions of the Vernier scale coincide exactly with $4$ divisions on the main scale. This relationship allows us to find the length of one Vernier Scale Division (VSD):

$$5	ext{ VSD} = 4	ext{ MSD}$$

$$1	ext{ VSD} = \frac{4}{5}	ext{ MSD} = \frac{4}{5} 	imes 0.125	ext{ cm} = 0.1	ext{ cm}$$

The least count of a Vernier Caliper is the smallest difference it can measure, which is the difference between one main scale division and one Vernier scale division:

$$LC_{VC} = 1	ext{ MSD} - 1	ext{ VSD}$$

$$LC_{VC} = 0.125	ext{ cm} - 0.1	ext{ cm} = 0.025	ext{ cm}$$

### Unraveling the Screw Gauge

Now, let's analyze the Screw Gauge. It has $100$ divisions on its circular scale. The crucial piece of information is that one complete rotation of the circular scale moves it by two divisions on the linear scale.

The pitch of a screw gauge is defined as the linear distance moved in one complete rotation. Therefore:

$$	ext{Pitch} = 2 	imes 	ext{MSD}_{SG}$$

The least count of the screw gauge is the pitch divided by the total number of circular divisions:

$$LC_{SG} = \frac{	ext{Pitch}}{100}$$

### Testing the Hypotheses: Options (a) and (b)

Options (a) and (b) propose a scenario where the pitch of the screw gauge is twice the least count of the Vernier caliper. Let's test this condition:

$$	ext{Pitch} = 2 	imes LC_{VC} = 2 	imes 0.025	ext{ cm} = 0.05	ext{ cm}$$

Using this pitch, we can calculate the least count of the screw gauge:

$$LC_{SG} = \frac{0.05	ext{ cm}}{100} = 0.0005	ext{ cm}$$

Since the options are in millimeters, we convert our result:

$$LC_{SG} = 0.005	ext{ mm}$$

Comparing this with the options, we see that **Option (b) is correct** and Option (a) is incorrect.

### Testing the Hypotheses: Options (c) and (d)

Options (c) and (d) propose a different scenario: the least count of the linear scale of the screw gauge (which is $1	ext{ MSD}_{SG}$) is twice the least count of the Vernier caliper. Let's test this new condition:

$$	ext{MSD}_{SG} = 2 	imes LC_{VC} = 2 	imes 0.025	ext{ cm} = 0.05	ext{ cm}$$

Remembering our earlier deduction that the pitch is twice the linear scale division:

$$	ext{Pitch} = 2 	imes 	ext{MSD}_{SG} = 2 	imes 0.05	ext{ cm} = 0.1	ext{ cm}$$

Now, we calculate the least count for this scenario:

$$LC_{SG} = \frac{0.1	ext{ cm}}{100} = 0.001	ext{ cm}$$

Converting to millimeters:

$$LC_{SG} = 0.01	ext{ mm}$$

Comparing this with the options, we find that **Option (c) is correct** and Option (d) is incorrect.

Thus, the correct choices are (b) and (c).

Select Answer:

Visualized Solution

$Main Scale of Vernier Caliper$

$Vernier Scale of Vernier Caliper$

$Least Count of Vernier Caliper$

$Screw Gauge Properties$

$Evaluating Options (a) and (b)$

$Least Count for (a) and (b)$

$Evaluating Options (c) and (d)$

$Least Count for (c) and (d)$

The Sigma Insight: Vernier Calipers and Screw Gauge

Decoding the Vernier Caliper

Unraveling the Screw Gauge

Testing the Hypotheses

Testing the Hypotheses

Similar Questions

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Main Scale of Vernier Caliper

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vernier Scale of Vernier Caliper

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Least Count of Vernier Caliper

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Screw Gauge Properties

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Options (a) and (b)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Least Count for (a) and (b)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Options (c) and (d)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Least Count for (c) and (d)

The Sigma Insight: Vernier Calipers and Screw Gauge

Decoding the Vernier Caliper

Unraveling the Screw Gauge

Testing the Hypotheses

Testing the Hypotheses

Similar Questions

If the screw on a screw gauge is given six rotations, it moves by 3 mm on the main scale. If there are 50 divisions on the circular scale, the least count of the screw gauge is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5μm diameter of a wire is

$Main Scale of Vernier Caliper$

$Vernier Scale of Vernier Caliper$

$Least Count of Vernier Caliper$

$Screw Gauge Properties$

$Evaluating Options (a) and (b)$

$Least Count for (a) and (b)$

$Evaluating Options (c) and (d)$

$Least Count for (c) and (d)$

The least count of the main scale of a screw gauge is $1 mm$ . The minimum number of divisions on its circular scale required to measure $5 μ m$ diameter of a wire is