In an experiment, brass and steel wires of length 1 m each with areas of cross-section 1 mm^2 are us

Q: Why is the tension the same in both wires?

When massless wires are connected in series, the force applied at one end is transmitted equally throughout the entire length. Think of it like a single rope; the pull you exert on one end is felt equally at every point along the rope.

Q: Why do we add the elongations?

Since the wires are connected end-to-end, the total change in length of the system is simply the sum of how much the brass wire stretches and how much the steel wire stretches. They stretch independently but contribute to the total length.

Q: Why did we convert millimeters to meters?

Young's modulus is given in Newtons per square meter ($N/m^2$). To ensure our units are consistent and our final answer is correct, all lengths must be converted to the standard SI unit, which is meters.

Q: What if the cross-sectional areas were different?

If the areas were different, we wouldn't be able to factor out the $A$ from the equation. We would have to calculate the elongation for each wire separately using their respective areas and then add them up.

Q: Why doesn't the answer match any of the options?

Sometimes, competitive exams contain errors in their options. In this case, the mathematically correct answer is $8 \times 10^6 \text{ N/m}^2$, which isn't listed. Such questions are usually awarded bonus marks to all students.

Question

In an experiment, brass and steel wires of length 1 m each with areas of cross-section $1	ext{ mm}^2$ are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress requires to produce a net elongation of 0.2 mm is [Take, the Young's modulus for steel and brass are respectively $120 	imes 10^9	ext{ N/m}^2$ and $60 	imes 10^9	ext{ N/m}^2$]

Options:

(A) $1.2 	imes 10^6	ext{ N/m}^2$

(B) $0.2 	imes 10^6	ext{ N/m}^2$

(C) $1.8 	imes 10^6	ext{ N/m}^2$

(D) $4.0 	imes 10^6	ext{ N/m}^2$

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Wires

Imagine an experiment where we have two distinct wires—one made of brass and the other of steel. They are connected end-to-end, in a series configuration, to a rigid support. We then apply a pulling force $F$ at the free end.

This physical setup is a classic example of composite materials under stress. The problem states that both wires have an identical length of $l = 1 	ext{ m}$ and an identical cross-sectional area of $A = 1 	ext{ mm}^2$.

### Hooke's Law: The Elastic Heartbeat

When we pull this combined wire, it will inevitably stretch. According to Hooke's Law, the elongation of any elastic wire is governed by its Young's Modulus. The formula for elongation is given by:

$$\Delta l = \frac{Fl}{AY}$$

Here, $F$ is the applied force, $l$ is the original length, $A$ is the cross-sectional area, and $Y$ is the Young's Modulus of the material.

### The Series Connection: Adding the Stretches

Look closely at the setup. Because both wires are connected in series, the total elongation of the entire system will simply be the sum of the individual elongations of the brass and steel wires.

$$\Delta l_{	ext{net}} = \Delta l_{	ext{brass}} + \Delta l_{	ext{steel}}$$

Let's substitute our elongation formula for both wires. Here is a crucial conceptual point: because they are in series and we assume them to be massless, the tension—or the force $F$—is uniform throughout both wires.

$$\Delta l_{	ext{net}} = \frac{F l_{	ext{brass}}}{A_{	ext{brass}} Y_{	ext{brass}}} + \frac{F l_{	ext{steel}}}{A_{	ext{steel}} Y_{	ext{steel}}}$$

### The Mathematical Symphony: Factoring and Substituting

Since the problem states they have the same length ($l$) and cross-sectional area ($A$), this makes our mathematical life much easier! We can factor out the common terms $\frac{Fl}{A}$ from the equation.

$$\Delta l_{	ext{net}} = \frac{F l}{A} \left( \frac{1}{Y_{	ext{brass}}} + \frac{1}{Y_{	ext{steel}}} ight)$$

Notice that the term $\frac{F}{A}$ is exactly the **stress** we are asked to find. Let's isolate it mentally as we plug in the given values. The net elongation is $0.2 	ext{ mm}$, which we must convert to standard SI units as $0.2 	imes 10^{-3} 	ext{ m}$.

$$0.2 	imes 10^{-3} = \left(\frac{F}{A}ight) (1) \left( \frac{1}{60 	imes 10^9} + \frac{1}{120 	imes 10^9} ight)$$

Don't make a silly mistake with the powers of 10 here. Let's carefully add the fractions inside the bracket. The common denominator is $120 	imes 10^9$.

$$\frac{1}{60 	imes 10^9} + \frac{1}{120 	imes 10^9} = \frac{2 + 1}{120 	imes 10^9} = \frac{3}{120 	imes 10^9} = \frac{1}{40 	imes 10^9}$$

### The Final Reveal: A Bonus Surprise

Finally, let's isolate the stress term, $\frac{F}{A}$.

$$0.2 	imes 10^{-3} = \left(\frac{F}{A}ight) \left( \frac{1}{40 	imes 10^9} ight)$$

$$\frac{F}{A} = 0.2 	imes 10^{-3} 	imes 40 	imes 10^9$$

**$$	ext{Stress} = 8 	imes 10^6 	ext{ N/m}^2$$**

Interestingly, if you look at the options provided in the original exam question, this value doesn't match any of them! This means it was likely a flawed question in the exam, and students were awarded a bonus mark. However, the physics and the math we executed are flawlessly correct.

### The Way Forward: Parallel Realities

So, we've successfully solved the series case. But think about this: what if these two wires were connected in parallel and pulled together by a single rigid bar? Would the stress in both wires still be the same? How would you find the equivalent Young's modulus of that parallel system? Ponder over this, as it is a favorite concept for advanced competitive exams!

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Hooke’s Law for Elongation$

$Net Elongation in Series$

$Substituting the Formula$

$Factoring Common Terms$

$Substituting Given Values$

$Calculating the Bracket Term$

$Final Calculation$

$The Way Forward$

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

The Setup

Hooke's Law

The Series Connection

The Mathematical Symphony

The Final Reveal

The Way Forward

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Select Answer:

Visualized Solution

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

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

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

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

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

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

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

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

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