The diameter of a spherical bob is measured using a Vernier callipers. 9 divisions of the main scale

Q: Why do we subtract a positive zero error?

A positive zero error means the instrument reads a value even when the jaws are fully closed. It's like a weighing scale showing 2 kg when empty. To get the true weight, you must subtract that extra 2 kg. Similarly, we subtract the positive zero error from the measured reading.

Q: How did we get 1 VSD = 0.9 mm?

The problem states that 9 main scale divisions (MSD) are equal to 10 vernier scale divisions (VSD). Since 1 MSD is 1 mm, 9 MSD is 9 mm. Therefore, 10 VSD = 9 mm, which means 1 VSD = 9/10 mm = 0.9 mm.

Q: Why is the least count formula 1 MSD - 1 VSD?

The least count is the smallest difference in length that the scale can resolve. It is exactly the difference in length between one main scale division and one vernier scale division. This tiny difference is what allows the vernier scale to measure fractions of a main scale division.

Q: What if the zero error was negative?

If the zero error was negative, it would mean the instrument reads less than the actual value (like a scale starting at -2 kg). In that case, you would subtract a negative value, which mathematically means you add the magnitude of the error to the measured reading.

Q: Why did we divide by 2 at the end?

The Vernier callipers measures the distance between its jaws, which corresponds to the full width or diameter of the spherical bob. The question specifically asks for the radius, which is half of the diameter.

Question

The diameter of a spherical bob is measured using a Vernier callipers. 9 divisions of the main scale, in the vernier calipers, are equal to 10 divisions of vernier scale. One main scale division is $1 	ext{ mm}$. The main scale reading is $10 	ext{ mm}$ and 8th division of vernier scale was found to coincide exactly with one of the main scale division. If the given vernier callipers has positive zero error of $0.04 	ext{ cm}$, then the radius of the bob is .......... $	imes 10^{-2} 	ext{ cm}$.

Sigma Percentile · Accepted Answer

The beauty of physics lies in precision, and the Vernier callipers is a masterpiece of precise measurement. Invented by Pierre Vernier in 1631, this brilliant instrument allows us to measure lengths far smaller than the smallest division on a standard ruler. In this problem, we are tasked with finding the radius of a spherical bob using a Vernier callipers that has a slight imperfection—a zero error.

Let's embark on this journey of precision and uncover the true radius of the bob step by step.

### Analyzing the Setup

Before we can take any measurements, we must understand the instrument itself. The core of a Vernier callipers is the relationship between its main scale and its sliding vernier scale. The problem states that **9 divisions of the main scale (MSD) are equal to 10 divisions of the vernier scale (VSD)**.

We are also given that one main scale division is exactly $1 	ext{ mm}$. This means that 9 main scale divisions cover a distance of $9 	ext{ mm}$. Since 10 vernier scale divisions fit exactly into this $9 	ext{ mm}$ space, the length of a single vernier scale division is:

$$1 	ext{ VSD} = \frac{9}{10} 	ext{ MSD} = 0.9 	ext{ mm}$$

### The Master Equation

The magic of the Vernier callipers comes from its **Least Count (LC)**. The least count is the absolute smallest difference in length that the instrument can reliably resolve. It is defined as the difference between one main scale division and one vernier scale division.

$$	ext{LC} = 1 	ext{ MSD} - 1 	ext{ VSD}$$

Substituting the values we just found:

$$	ext{LC} = 1 	ext{ mm} - 0.9 	ext{ mm} = 0.1 	ext{ mm}$$

To keep our units consistent with the final required answer, let's convert this to centimeters:

$$	ext{LC} = 0.01 	ext{ cm}$$

### Taking the Measurement

Now, we place the spherical bob between the jaws of the callipers. We read the main scale just before the zero mark of the vernier scale. The problem tells us the **Main Scale Reading (MSR)** is $10 	ext{ mm}$, which is $1.0 	ext{ cm}$.

Next, we look for the vernier scale division that perfectly aligns with any mark on the main scale. We are told the **8th division** coincides perfectly. This is our **Vernier Scale Reading (VSR)**.

The formula for the measured diameter is:

$$	ext{Measured Diameter} = 	ext{MSR} + (	ext{VSR} 	imes 	ext{LC})$$

Plugging in our numbers:

$$	ext{Measured Diameter} = 1.0 	ext{ cm} + (8 	imes 0.01 	ext{ cm}) = 1.08 	ext{ cm}$$

### The Zero Error Trap

If the instrument were perfect, $1.08 	ext{ cm}$ would be our final diameter. However, the problem throws a curveball: the callipers has a **positive zero error of $0.04 	ext{ cm}$**.

What does a positive zero error mean? Imagine a weighing scale that reads $2 	ext{ kg}$ even when nothing is on it. If you step on it and it reads $62 	ext{ kg}$, your true weight is actually $60 	ext{ kg}$. The scale is inherently biased upwards. Similarly, our callipers reads $0.04 	ext{ cm}$ when the jaws are completely closed.

To find the true diameter, we must subtract this inherent bias from our measured reading:

$$	ext{True Diameter} = 	ext{Measured Diameter} - 	ext{Zero Error}$$

$$	ext{True Diameter} = 1.08 	ext{ cm} - 0.04 	ext{ cm} = 1.04 	ext{ cm}$$

### Final Calculation

We have successfully found the true diameter of the bob. But wait! A classic trap in physics exams is asking for a different parameter than the one you just calculated. The question asks for the **radius**, not the diameter.

The radius is simply half of the diameter:

$$r = \frac{	ext{True Diameter}}{2} = \frac{1.04 	ext{ cm}}{2} = 0.52 	ext{ cm}$$

The final step is to format our answer to match the requested structure of $.......... 	imes 10^{-2} 	ext{ cm}$.

$$0.52 	ext{ cm} = 52 	imes 10^{-2} 	ext{ cm}$$

Therefore, the numerical value to be filled in the blank is **52**.

Enter Numerical Value:

Visualized Solution

\text{The Setup}

\text{Least Count (LC) Formula}

\text{Calculating Least Count}

\text{Measured Diameter}

\text{Computing Measured Diameter}

\text{True Diameter and Zero Error}

\text{Computing True Diameter}

\text{Finding the Radius}

\text{Conclusion}

The Sigma Insight: Vernier Calipers and Screw Gauge

Analyzing the Setup

The Master Equation

Taking the Measurement

The Zero Error Trap

Final Calculation

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