If the screw on a screw gauge is given six rotations, it moves by 3 mm on the main scale. If there a

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

The pitch is defined as the linear distance moved in exactly *one* complete rotation. Since we are given the distance for 6 rotations, dividing by 6 gives us the distance for 1 rotation.

Q: What does the least count actually represent physically?

It represents the linear distance the screw moves when the circular scale is rotated by just *one* single division. It's the absolute smallest length the instrument can measure accurately.

Q: Why did we convert mm to cm at the end?

Always let the options guide your final steps! The calculated least count was 0.01 mm, but all the multiple-choice options were provided in centimeters.

Q: What if the screw gauge had a zero error?

Zero error doesn't affect the least count or the pitch; it only affects the final measurement reading. The least count is an intrinsic property of the instrument's scales.

Q: Can the least count be negative?

No, the least count is a magnitude of length and is always positive. However, zero error can be positive or negative depending on where the zero mark rests.

Question

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

Options:

(A) 0.001 cm

(B) 0.01 cm

(C) 0.02 cm

(D) 0.001 cm

Sigma Percentile · Accepted Answer

The problem of finding the least count of a screw gauge is a classic application of understanding how rotational motion translates into linear measurement. Let's break down the mechanics of this elegant instrument and solve the problem step-by-step.

### Understanding the Screw Gauge

Imagine you are holding a screw gauge. It consists of two primary scales: the **main scale** (which is fixed and linear) and the **circular scale** (which rotates). When you turn the circular scale, the internal screw mechanism causes it to move forward or backward along the main scale.

The fundamental principle here is that a certain number of full rotations of the circular scale corresponds to a specific linear distance moved on the main scale.

### Finding the Pitch

The first crucial parameter we need to determine is the **pitch** of the screw gauge. The pitch is defined as the linear distance the screw moves on the main scale during exactly one complete rotation.

The problem states that when the screw is given $6$ rotations, it moves by $3 	ext{ mm}$ on the main scale. We can set up a simple ratio to find the pitch:

$$ 	ext{Pitch} = \frac{	ext{Distance moved}}{	ext{Number of rotations}} $$

Substituting the given values:

$$ 	ext{Pitch} = \frac{3 	ext{ mm}}{6} = 0.5 	ext{ mm} $$

This tells us that every single full rotation of the circular scale advances the screw by exactly $0.5 	ext{ mm}$.

### Calculating the Least Count

Now that we have the pitch, we can find the **least count**. The least count is the absolute smallest measurement the instrument can accurately record. It represents the linear distance moved when the circular scale is rotated by just one single division.

The formula for the least count is:

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

We know the pitch is $0.5 	ext{ mm}$, and the problem tells us there are $50$ divisions on the circular scale. Let's plug these in:

$$ 	ext{LC} = \frac{0.5 	ext{ mm}}{50} $$

$$ 	ext{LC} = 0.01 	ext{ mm} $$

So, the smallest distance this screw gauge can measure is $0.01 	ext{ mm}$.

### The Final Catch: Unit Conversion

We have our answer, but there is a catch! If you look closely at the options provided in the question, they are all in **centimeters (cm)**, not millimeters (mm). This is a classic trap in physics exams.

To convert millimeters to centimeters, we must divide by $10$:

$$ 	ext{LC} = \frac{0.01}{10} 	ext{ cm} $$

$$ 	ext{LC} = 0.001 	ext{ cm} $$

This perfectly matches option (a). By carefully understanding the physical meaning of pitch and least count, and keeping a sharp eye on our units, we've successfully navigated the problem!

Animated Solution for Physics - Physics and Measurement: 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

Select Answer:

Visualized Solution

\text{Screw Gauge Setup}

\text{Pitch Formula}

\text{Substituting Values}

\text{Calculating Pitch}

\text{Least Count Formula}

\text{Calculating Least Count}

\text{Unit Conversion}

\text{The Way Forward}

The Sigma Insight: Vernier Calipers and Screw Gauge

Understanding the Screw Gauge

Finding the Pitch

Calculating the Least Count

The Final Catch

Similar Questions

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

A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading $0 mm$ Circular scale reading $52$ divisions Given that $1 mm$ on main scale corresponds to $100$ divisions of the circular scale. The diameter of wire from the above data is

The circular scale of a screw gauge has 50 divisions and pitch of 0.5 mm. Find the diameter of sphere. Main scale reading is 2.

A vernier calipers has $1 mm$ marks on the main scale. It has $20$ equal divisions on the vernier scale which match with $16$ main scale divisions. For this vernier calipers, the least count is