Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is

Q: Why is the pitch of the screw 1.0 mm and not 0.5 mm?

The problem states that one full rotation of the circular scale shifts the main scale by *two* divisions. Since one main scale division is 0.5 mm, two divisions equal 1.0 mm. The pitch of the screw is defined as the distance moved in one full rotation, which is 1.0 mm.

Q: Why do we subtract the zero error?

A positive zero error means the gauge reads a value greater than zero even when nothing is between the jaws. To get the true measurement, we must subtract this 'extra' reading from our final measured value.

Q: How did we get the mean absolute error of 0.02 mm?

We took the absolute difference between each individual reading and the mean diameter ($|2.16 - 2.14| = 0.02$ and $|2.12 - 2.14| = 0.02$). The average of these absolute differences gives the mean absolute error.

Q: Why was this question dropped in the official JEE Advanced exam?

The phrasing 'pitch of the main scale is 0.5 mm' is technically ambiguous. 'Pitch' usually refers to the distance between threads on the screw itself, not the markings on the main scale. This non-standard terminology likely led to the question being dropped, even though the mathematical intent points clearly to option (C).

Q: How do we calculate the error in the area?

Since Area $A = \frac{\pi d^2}{4}$, the relative error follows the power rule: $\frac{\Delta A}{A} = 2 \frac{\Delta d}{d}$. We substitute the values of $A$, $d$, and $\Delta d$ to solve for the absolute error $\Delta A$.

Question

Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is $0.5	ext{ mm}$. The circular scale has $100$ divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below.

\begin{array}{|l|c|c|}
\hline
	extbf{Measurement condition} & 	extbf{Main scale reading} & 	extbf{Circular scale reading} \
\hline
	ext{Two arms of gauge touching} & & \
	ext{each other without wire} & 0	ext{ division} & 4	ext{ division} \
\hline
	ext{Attempt-1: With wire} & 4	ext{ divisions} & 20	ext{ divisions} \
\hline
	ext{Attempt-2: With wire} & 4	ext{ divisions} & 16	ext{ divisions} \
\hline
\end{array}

What are the diameter and cross-sectional area of the wire measured using the screw gauge?

Options:

(A) $2.22 \pm 0.02 	ext{ mm}, \pi(1.23 \pm 0.02) 	ext{ mm}^2$

(B) $2.22 \pm 0.01 	ext{ mm}, \pi(1.23 \pm 0.01) 	ext{ mm}^2$

(C) $2.14 \pm 0.02 	ext{ mm}, \pi(1.14 \pm 0.02) 	ext{ mm}^2$

(D) $2.14 \pm 0.01 	ext{ mm}, \pi(1.14 \pm 0.01) 	ext{ mm}^2$

Sigma Percentile · Accepted Answer

### Decoding the Screw Gauge

When we first look at this problem, the terminology might seem a bit unusual. The problem states that the "pitch of the main scale" is $0.5 	ext{ mm}$. In standard terminology, this simply means that one Main Scale Division (MSD) is $0.5 	ext{ mm}$.

However, the real catch lies in the next sentence: "for one full rotation of the circular scale, the main scale shifts by two divisions." This is the critical piece of information we need to find the true pitch of the screw.

### The Hidden Trap: Calculating the Least Count

The pitch of a screw gauge is defined as the linear distance moved by the spindle in one complete rotation. Since one rotation shifts the main scale by two divisions, and each division is $0.5 	ext{ mm}$, the actual pitch of the screw is:

$$ 	ext{Pitch} = 2 	imes 0.5 	ext{ mm} = 1.0 	ext{ mm} $$

Now, to find the Least Count (LC), we divide the pitch by the total number of divisions on the circular scale ($100$ divisions):

$$ 	ext{LC} = \frac{1.0 	ext{ mm}}{100} = 0.01 	ext{ mm} $$

### Accounting for the Zero Error

Before taking any measurements, we must calibrate our instrument. When the two arms of the gauge touch without any wire, the main scale reads $0$, but the circular scale reads $4$ divisions. This indicates a positive zero error.

$$ 	ext{Zero Error} = 	ext{MSR} + (	ext{CSR} 	imes 	ext{LC}) $$
$$ 	ext{Zero Error} = 0 + (4 	imes 0.01) = +0.04 	ext{ mm} $$

We must subtract this zero error from all our subsequent readings to get the true values.

### Measuring the Diameter

Let's calculate the diameter for both attempts. For Attempt 1, the main scale reading is $4$ divisions. Since $1 	ext{ MSD} = 0.5 	ext{ mm}$, the main scale reading in millimeters is $4 	imes 0.5 = 2.0 	ext{ mm}$. The circular scale reading is $20$.

$$ d_1 = 2.0 + (20 	imes 0.01) - 0.04 = 2.16 	ext{ mm} $$

For Attempt 2, the main scale reading is still $2.0 	ext{ mm}$, but the circular scale reading is $16$.

$$ d_2 = 2.0 + (16 	imes 0.01) - 0.04 = 2.12 	ext{ mm} $$

To find the most accurate diameter, we take the mean of these two readings:

$$ d_{	ext{mean}} = \frac{2.16 + 2.12}{2} = 2.14 	ext{ mm} $$

The absolute error for each reading is the difference from the mean, which is $|2.14 - 2.16| = 0.02 	ext{ mm}$. Thus, the reported diameter is **$2.14 \pm 0.02 	ext{ mm}$**.

### Error Propagation in Area

Now, we need to find the cross-sectional area and its associated error. The formula for the area of a circle is:

$$ A = \frac{\pi d^2}{4} $$

Substituting our mean diameter:

$$ A = \frac{\pi (2.14)^2}{4} = \pi 	imes \frac{4.5796}{4} \approx \pi(1.14) 	ext{ mm}^2 $$

For the error in area, we use the rules of error propagation. Since area is proportional to the square of the diameter, the relative error in area is twice the relative error in diameter:

$$ \frac{\Delta A}{A} = 2 \frac{\Delta d}{d} $$

Solving for the absolute error $\Delta A$:

$$ \Delta A = 2 	imes \frac{0.02}{2.14} 	imes \pi(1.1449) = \pi 	imes 0.0214 \approx \pi(0.02) 	ext{ mm}^2 $$

So, the final cross-sectional area is **$\pi(1.14 \pm 0.02) 	ext{ mm}^2$**.

### The Final Verdict

Matching our results with the given options, we find that Option (C) is the perfect fit. It is worth noting that JEE Advanced officially dropped this question, likely due to the non-standard phrasing of "pitch of the main scale," which caused confusion among students. However, mathematically and logically, the intended path leads directly to Option (C).

Select Answer:

Visualized Solution

$Understanding the Screw Gauge$

$Calculating Least Count$

$Zero Error Calculation$

$First Measurement$

$Second Measurement$

$Mean Diameter and Absolute Error$

$Cross-Sectional Area$

$Conclusion$

The Sigma Insight: Vernier Calipers and Screw Gauge

Decoding the Screw Gauge

The Hidden Trap

Accounting for the Zero Error

Measuring the Diameter

Error Propagation in Area

The Final Verdict

Similar Questions

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 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

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.

Select Answer:

Visualized Solution

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

.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; } Zero Error Calculation

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mean Diameter and Absolute Error

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

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