Two wires are made of the same material and have the same volume. However, wire 1 has cross-sectiona

Q: Why is Young's Modulus ($Y$) the same for both wires?

Young's Modulus is an intrinsic property of the material. Since the problem states that both wires are made of the "same material", their $Y$ values are identical, regardless of their shape or size.

Q: How did we know that the volume is $A \cdot l$?

A wire is essentially a long cylinder. The volume of any uniform cylinder is the area of its base (cross-sectional area $A$) multiplied by its height (length $l$).

Q: Why does a thicker wire need more force to stretch?

A thicker wire has more atomic bonds in its cross-section that need to be pulled apart simultaneously. It's like trying to stretch a thick rubber band versus a thin one; the thick one is much stiffer.

Q: Can I solve this using the concept of a spring constant?

Yes! A wire acts like a spring with constant $k = \frac{YA}{l}$. For the second wire, $A$ is 3 times larger and $l$ is 3 times smaller, making $k_2 = \frac{Y(3A)}{l/3} = 9k_1$. So, it takes 9 times the force for the same extension.

Q: What if the wires had the same length instead of the same volume?

If they had the same length, then $l_1 = l_2$. Equating the extensions would give $\frac{F}{A} = \frac{F'}{3A}$, which means $F' = 3F$. The "same volume" constraint is what makes the second wire shorter, adding another factor of 3.

Question

Two wires are made of the same material and have the same volume. However, wire 1 has cross-sectional area $A$ and wire 2 has cross-sectional area $3A$. If the length of wire 1 increases by $\Delta x$ on applying force $F$, how much force is needed to stretch wire 2 by the same amount? [AIEEE 2009]

Options:

(A) $F$

(B) $4F$

(C) $6F$

(D) $9F$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are in a physics lab, and you have two wires hanging from the ceiling. They are made of the exact same material—let's say steel—and they contain the exact same amount of metal. This means their **volumes are identical**.

However, they look completely different. Wire 1 is thin, with a cross-sectional area of $A$. Wire 2 is much thicker, with a cross-sectional area of $3A$. Because they have the same volume, the thicker wire must be shorter.

We know that the volume of a cylinder is given by the product of its area and length:
$$V = A \cdot l$$

Since $V_1 = V_2$, we can write:
$$A_1 l_1 = A_2 l_2$$

Substituting the given areas, we get:
$$A \cdot l_1 = 3A \cdot l_2$$

The $A$ cancels out beautifully, leaving us with:
$$l_1 = 3 l_2$$

This tells us that Wire 1 is three times as long as Wire 2. This makes perfect physical sense! If you take a piece of clay and make it three times thicker, it will become one-third of its original length.

### The Master Equation
Now, let's talk about stretching these wires. When we apply a force to a wire, it acts like a spring. The relationship between the applied force $F$, the original length $l$, the cross-sectional area $A$, and the extension $\Delta x$ is governed by Young's Modulus $Y$:
$$Y = \frac{F/A}{\Delta x/l}$$

Rearranging this to solve for the extension $\Delta x$, we get our master equation:
$$\Delta x = \frac{F l}{A Y}$$

The problem states that both wires are stretched by the **exact same amount**, meaning $\Delta x_1 = \Delta x_2$. Let's set up the equations for both wires.

For Wire 1, a force $F$ is applied:
$$\Delta x_1 = \frac{F l_1}{A Y}$$

For Wire 2, an unknown force $F'$ is applied:
$$\Delta x_2 = \frac{F' l_2}{3A Y}$$

### Final Calculation
Since the extensions are equal, we can equate the two expressions:
$$\frac{F l_1}{A Y} = \frac{F' l_2}{3A Y}$$

Look at how elegant this gets! The material is the same, so Young's Modulus $Y$ is the same for both. We can immediately cancel out $A$ and $Y$ from both sides:
$$F \cdot l_1 = \frac{F' \cdot l_2}{3}$$

Now, remember that length relationship we found earlier? We know that $l_1 = 3 l_2$. Let's substitute this into our equation:
$$F \cdot (3 l_2) = \frac{F' \cdot l_2}{3}$$

The $l_2$ terms cancel out on both sides:
$$3F = \frac{F'}{3}$$

Multiplying both sides by 3, we find our final answer:
$$F' = 9F$$

It takes **nine times the force** to stretch the thicker wire by the same amount! Why is this the case? You can think of a wire as a spring with stiffness $k = \frac{YA}{l}$. By making the wire three times thicker (increasing $A$ by a factor of 3) and one-third as long (decreasing $l$ by a factor of 3), its overall stiffness increases by a factor of $3 	imes 3 = 9$. It becomes a incredibly stiff spring!

Select Answer:

Visualized Solution

$V_{1} = V_{2}$

$A_{1} l_{1} = A_{2} l_{2}$

$\frac{l _{1}}{l _{2}} = 3$

$Δ x = \frac{F l}{A Y}$

$Δ x_{1} = Δ x_{2}$

$\frac{F l _{1}}{A Y} = \frac{F ^{'} l _{2}}{3 A Y}$

$F^{'} = 9 F$

$k = \frac{Y A}{l}$

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

Analyzing the Setup Imagine you are in a physics lab, and you have two wires hanging from the ceiling

The Master Equation Now, let's talk about stretching these wires

Final Calculation

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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; } V1​=V2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​l1​=A2​l2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l2​l1​​=3

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δx1​=Δx2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AYFl1​​=3AYF′l2​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F′=9F

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