Assertion (A) If in five complete rotations of the circular scale, the distance travelled on main sc

Q: Why do we divide the total distance by the number of rotations to find the pitch?

Pitch is defined as the distance moved in *one* single rotation. If $5$ rotations move the screw by $5 \text{ mm}$, dividing by $5$ gives the distance for $1$ rotation, which is $1 \text{ mm}$.

Q: What exactly does the 'Least Count' represent physically?

It represents the absolute smallest length you can measure with the instrument. For this screw gauge, turning the circular scale by just one single mark out of the $50$ moves the jaw by $0.02 \text{ mm}$.

Q: I got $0.02$ as my answer, why is the assertion wrong?

You likely forgot to check the units! Your calculation of $1/50$ gives $0.02 \text{ mm}$. The assertion states the value is $0.001 \text{ cm}$. You must convert $0.02 \text{ mm}$ to $0.002 \text{ cm}$ to see that the assertion's value is incorrect.

Q: Is the formula in the Reason statement always true for every screw gauge?

Yes! The formula $\text{Least Count} = \frac{\text{Pitch}}{\text{Total divisions}}$ is the universal defining formula for the least count of any standard screw gauge.

Q: What if the screw gauge had $100$ divisions instead of $50$?

Then the least count would be even smaller, making the instrument more precise! It would be $1 \text{ mm} / 100 = 0.01 \text{ mm}$, or $0.001 \text{ cm}$. In that specific case, the assertion would have been correct!

Question

Assertion (A) If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is $5 	ext{ mm}$ and there are $50$ total divisions on circular scale, then least count is $0.001 	ext{ cm}$.
Reason (R) Least count = $\frac{	ext{Pitch}}{	ext{Total divisions on circular scale}}$
In the light of the above statements, choose the most appropriate answer from the options given below.

Options:

(A) Both A and R are correct and R is the correct explanation of A.

(B) Both A and R are correct and R is not the correct explanation of A.

(C) A is correct but R is not correct.

(D) A is not correct but R is correct.

Sigma Percentile · Accepted Answer

Have you ever wondered how we measure the thickness of a single sheet of paper or the diameter of a thin wire? Enter the **screw gauge**, a marvel of mechanical precision. In this problem, we are tasked with evaluating an assertion and a reason regarding the least count of a screw gauge. Let's break it down step-by-step and uncover the physics behind the measurement!

### Understanding the Pitch

Imagine turning a screw into a piece of wood. With every full rotation, the screw advances by a specific distance. In a screw gauge, this linear distance moved by the main scale during one complete rotation of the circular scale is called the **pitch**.

The problem states that in $5$ complete rotations, the circular scale travels a distance of $5 	ext{ mm}$ on the main scale.

To find the pitch, we simply need to determine the distance traveled in just *one* rotation. It's a straightforward unitary method:

$$ 	ext{Pitch} = \frac{	ext{Total Distance}}{	ext{Number of Rotations}} $$

$$ 	ext{Pitch} = \frac{5 	ext{ mm}}{5} = 1 	ext{ mm} $$

So, our screw gauge has a pitch of $1 	ext{ mm}$. Every time you turn the thimble a full $360^\circ$, the jaws open or close by exactly one millimeter.

### The Master Equation: Least Count

Now, let's look at the Reason (R) provided in the question. It states:

$$ 	ext{Least Count} = \frac{	ext{Pitch}}{	ext{Total divisions on circular scale}} $$

Is this correct? Absolutely! The least count is the smallest measurement an instrument can accurately make. If one full rotation (which is $1 	ext{ mm}$) is divided into $50$ equal parts on the circular scale, then turning the scale by just *one* division moves the screw by a tiny fraction of that pitch.

This formula is the fundamental principle of a screw gauge. Therefore, **Reason (R) is perfectly correct**.

### The Final Calculation and The Unit Trap

Let's put our formula to the test and calculate the actual least count of our screw gauge. We know the pitch is $1 	ext{ mm}$ and there are $50$ divisions on the circular scale.

$$ 	ext{Least Count} = \frac{1 	ext{ mm}}{50} = 0.02 	ext{ mm} $$

We have our least count: $0.02 	ext{ mm}$. But wait! Before we declare the Assertion (A) correct or incorrect, we must look closely at the units. The assertion claims the least count is $0.001 	ext{ cm}$.

This is a classic trap! We need to convert our answer from millimeters to centimeters. Since $1 	ext{ cm} = 10 	ext{ mm}$, we divide by $10$:

$$ 0.02 	ext{ mm} = \frac{0.02}{10} 	ext{ cm} = 0.002 	ext{ cm} $$

Our calculated least count is $0.002 	ext{ cm}$, but the Assertion (A) states it is $0.001 	ext{ cm}$.

### Conclusion

Because our calculated value does not match the value given in the assertion, we can confidently say that **Assertion (A) is incorrect**.

However, the formula provided in Reason (R) is the exact, correct formula we used to find the true least count. Thus, **Reason (R) is correct**.

This leads us directly to our final answer: Assertion is not correct, but Reason is correct. Always remember to double-check your units—they are the silent guardians of physics!

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Visualized Solution

$Analyzing the Setup$

$Calculating Pitch$

$Evaluating Reason (R)$

$Calculating Least Count$

$Evaluating Assertion (A)$

$Final Conclusion$

The Sigma Insight: Vernier Calipers and Screw Gauge

Understanding the Pitch

The Master Equation

The Final Calculation and The Unit Trap

Conclusion

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$N$ divisions on the main scale of a vernier calipers coincide with $(N + 1)$ divisions on the vernier scale. If each division on the main scale is of $a$ units, determine the least count of instrument.

One main scale division of a vernier callipers is $a$ cm and $n$ th division of the vernier scale coincide with $(n - 1)$ th division of the main scale. The least count of the callipers (in mm) is

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Setup

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

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

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

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

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