A steel wire of diameter 0.5 mm and Young's modulus 2 x 10^11 N m^-2 carries a load of mass m. The l

Q: Why do we need a reference wire?

The reference wire compensates for any yielding of the support or thermal expansion of the environment. If the ceiling sags slightly under the weight, both wires sag together, ensuring the relative measurement between the main scale and vernier scale remains perfectly accurate.

Q: Why is $\pi$ given as 3.2?

In competitive exams like JEE, examiners often provide slightly modified constants to make the final arithmetic clean and perfectly cancel out, saving you time and reducing calculation errors. Always use the given constants!

Q: What does '10 divisions of vernier scale correspond to 9 divisions of main scale' mean?

It means the total length of 10 vernier divisions is exactly equal to the length of 9 main scale divisions ($9\text{ mm}$). Therefore, one single vernier division is slightly smaller ($0.9\text{ mm}$) than a main scale division ($1.0\text{ mm}$).

Q: Why did we only use the additional mass ($1.2\text{ kg}$) and not the initial mass $m$?

The initial mass $m$ was already hanging there when the zeros of the scales were perfectly aligned. We are only interested in the *change* in length (extension) caused by the *change* in force (the additional $1.2\text{ kg}$).

Q: What if the extension was $0.35\text{ mm}$? Which division would coincide?

If the extension was $0.35\text{ mm}$, it would mean the zero mark is exactly halfway between the 3rd and 4th vernier coincidence. However, since the least count is $0.1\text{ mm}$, a standard vernier scale cannot precisely resolve $0.05\text{ mm}$ increments. The question is designed to yield an exact multiple of the least count.

Question

A steel wire of diameter $0.5	ext{ mm}$ and Young's modulus $2 	imes 10^{11}	ext{ N m}^{-2}$ carries a load of mass $m$. The length of the wire with the load is $1.0	ext{ m}$. A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count $1.0	ext{ mm}$, is attached. The 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by $1.2	ext{ kg}$, the vernier scale division which coincides with a main scale division is ......... .
(Take, $g = 10	ext{ m s}^{-2}$ and $\pi = 3.2$).

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a laboratory, looking at a classic Searle's apparatus setup. You see two wires hanging side by side from a rigid ceiling.

The first wire is the **reference wire**. It hangs straight down and holds the main scale. Why do we need a reference wire? Because if the ceiling sags slightly under the weight, both wires will sag together. This ensures that our relative measurement between the two scales remains perfectly accurate.

The second wire is our **experimental steel wire**. It holds the vernier scale and carries an initial mass $m$. At this starting point, the zero mark of the vernier scale perfectly aligns with the zero mark of the main scale.

### Understanding the Vernier Scale

Before we add any extra weight, we must understand our measuring tool. The problem states that 10 divisions of the vernier scale correspond to 9 divisions of the main scale.

Since one main scale division (MSD) is exactly $1.0 	ext{ mm}$, the length of 9 MSD is $9.0 	ext{ mm}$. This means 10 vernier scale divisions (VSD) span $9.0 	ext{ mm}$, making a single VSD equal to $0.9 	ext{ mm}$.

The **least count (LC)** is the smallest measurement the scale can resolve, defined as the difference between one MSD and one VSD:

$$ 	ext{LC} = 1 	ext{ MSD} - 1 	ext{ VSD} = 1.0 	ext{ mm} - 0.9 	ext{ mm} = 0.1 	ext{ mm} $$

This tells us that for every division we move down the vernier scale, it lags behind the main scale by exactly $0.1 	ext{ mm}$.

### The Master Equation

Now, we disrupt the equilibrium. We add an additional load of $\Delta m = 1.2 	ext{ kg}$ to the steel wire. This extra mass exerts a downward gravitational force, stretching the atomic bonds within the steel.

To find out exactly how much the wire stretches, we call upon **Hooke's Law**, expressed in terms of Young's Modulus ($Y$):

$$ \Delta l = \frac{F l}{A Y} $$

Here, the force $F$ is the weight of the additional mass ($\Delta m \cdot g$), $l$ is the original length, and $A$ is the cross-sectional area of the wire.

### Crunching the Numbers

Let's calculate the cross-sectional area first. The diameter $d$ is given as $0.5 	ext{ mm}$, which is $5 	imes 10^{-4} 	ext{ m}$.

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

Notice that the problem generously provides $\pi = 3.2$. This is a classic exam trick to make the arithmetic clean!

$$ A = \frac{3.2 	imes (5 	imes 10^{-4})^2}{4} = \frac{3.2 	imes 25 	imes 10^{-8}}{4} = 0.8 	imes 25 	imes 10^{-8} = 2 	imes 10^{-7} 	ext{ m}^2 $$

Now, we substitute all our known values into the extension formula:

$$ \Delta l = \frac{1.2 	imes 10 	imes 1.0}{2 	imes 10^{-7} 	imes 2 	imes 10^{11}} $$

$$ \Delta l = \frac{12}{4 	imes 10^4} = 3 	imes 10^{-4} 	ext{ m} $$

Converting this back to millimeters, we get a precise extension of **$0.3 	ext{ mm}$**.

### Final Calculation

The steel wire has stretched by $0.3 	ext{ mm}$. This means the zero mark of the vernier scale has shifted downwards by exactly $0.3 	ext{ mm}$ relative to the main scale.

To find which vernier division coincides with a main scale mark, we divide the total extension by the least count:

$$ n = \frac{\Delta l}{	ext{LC}} = \frac{0.3 	ext{ mm}}{0.1 	ext{ mm}} = 3 $$

Because each vernier division covers a lag of $0.1 	ext{ mm}$, it takes exactly 3 divisions for the vernier mark to catch up and perfectly align with a main scale mark.

Therefore, the **3rd division** of the vernier scale coincides with the main scale!

Enter Numerical Value:

Visualized Solution

$Initial Setup$

$Least Count (LC)$

$Hooke’s Law$

$Raw Substitution$

$Area Calculation$

$Extension Calculation$

$Vernier Coincidence$

$Conclusion$

The Sigma Insight: Young's Modulus, Bulk Modulus and Modulus of Rigidity

Analyzing the Setup

Understanding the Vernier Scale

The Master Equation

Crunching the Numbers

Final Calculation

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Enter Numerical Value:

Visualized Solution

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

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

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

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

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

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

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

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