When the temperature of a metal wire is increased from 0°C to 10°C, its length increases by 0.02%. T

Q: Why did we ignore the change in mass?

Mass is an intrinsic property of the object and depends only on the amount of matter, which does not change upon heating.

Q: Why is the percentage change in density positive when density actually decreases?

The question asks for the "percentage change", which typically refers to the magnitude of the change unless a sign convention is explicitly required by the options.

Q: Is $\gamma = 3\alpha$ always true?

It is true for isotropic materials (materials that have the same properties in all directions). For anisotropic materials, $\gamma = \alpha_x + \alpha_y + \alpha_z$.

Q: Why can we just multiply the percentage change in length by 3?

For small changes, the binomial expansion $(1 + x)^3 \approx 1 + 3x$ applies. Here, $x$ is the fractional change in length, making the fractional change in volume exactly 3 times that of length.

Q: Does the initial temperature of $0^\circ\text{C}$ matter?

The specific initial temperature doesn't change the formula, but the temperature difference $\Delta T = 10^\circ\text{C}$ is what drives the expansion. However, since the percentage change in length is already given, we don't even need to use $\Delta T$ in our final calculation!

Question

When the temperature of a metal wire is increased from $0^\circ	ext{C}$ to $10^\circ	ext{C}$, its length increases by $0.02\%$. The percentage change in its mass density will be closest to

Options:

(A) 0.06

(B) 2.3

(C) 0.008

(D) 0.8

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a solid metal wire resting at a cool $0^\circ	ext{C}$. It has a specific mass $M$, an initial length $L_0$, and an initial volume $V_0$. Because density is simply how tightly packed this mass is, we can define its initial density as $ho_0 = \frac{M}{V_0}$.

Now, we turn up the heat, bringing the wire to $10^\circ	ext{C}$. As thermal energy flows into the metal, the atoms vibrate more vigorously and push further apart. This causes the wire to expand in all directions—its length increases by $\Delta L$ and its volume increases by $\Delta V$.

However, the total number of atoms hasn't changed. The mass $M$ remains absolutely constant.

### The Master Equation

Since the mass is constant, the expanding volume means the density must decrease. Let's look at the mathematical relationship. We know that:
$$ ho = \frac{M}{V} $$

If we take the natural logarithm on both sides and differentiate to find the fractional changes, we get:
$$ \frac{\Delta ho}{ho} = \frac{\Delta M}{M} - \frac{\Delta V}{V} $$

Because the mass doesn't change, $\Delta M = 0$. This simplifies our equation to:
$$ \frac{\Delta ho}{ho} = - \frac{\Delta V}{V} $$

The negative sign perfectly captures our physical intuition: as volume goes up, density goes down. Since the question asks for the "percentage change" (which implies the magnitude of the change), we can focus on the absolute values:
$$ \left| \frac{\Delta ho}{ho} ight| = \frac{\Delta V}{V} $$

### Connecting Volume to Length

We need the fractional change in volume, but the problem only gives us the fractional change in length ($0.02\%$). How do we bridge this gap?

For any isotropic material (a material that expands equally in all directions), the coefficient of volume expansion ($\gamma$) is exactly three times the coefficient of linear expansion ($\alpha$). 
$$ \gamma = 3\alpha $$

We know the fundamental thermal expansion formulas:
$$ \frac{\Delta V}{V} = \gamma \Delta T $$
$$ \frac{\Delta L}{L} = \alpha \Delta T $$

By substituting $\gamma = 3\alpha$ into the volume equation, we reveal the hidden link:
$$ \frac{\Delta V}{V} = 3\alpha \Delta T = 3 \left( \frac{\Delta L}{L} ight) $$

### Final Calculation

We have successfully connected the change in density directly to the change in length! 
$$ \left| \frac{\Delta ho}{ho} ight| = 3 \left( \frac{\Delta L}{L} ight) $$

To find the percentage change, we simply multiply both sides by 100:
$$ 	ext{Percentage change in density} = 3 	imes (	ext{Percentage change in length}) $$

The problem states that the percentage change in length is $0.02\%$. Substituting this value in:
$$ 	ext{Percentage change in density} = 3 	imes 0.02\% = \mathbf{0.06\%} $$

The density changes by $0.06\%$. It's a small, subtle shift, but mathematically rigorous and physically profound!

Animated Solution for Physics - Properties of Solids and Liquids: When the temperature of a metal wire is increased from $0^{\circ} C$ to $1 0^{\circ} C$ , its length increases by $0.02%$ . The percentage change in its mass density will be closest to

Select Answer:

Visualized Solution

Initial State at $T = 0^{\circ} C$

Heating and Expansion

Density and Volume Relation

Volume and Linear Expansion

Connecting Density to Length

Substituting the Values

Final Answer

The Way Forward

The Sigma Insight: Thermal Expansion

Analyzing the Setup

The Master Equation

Connecting Volume to Length

Final Calculation

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