The problem asks us to verify and derive the correct form of the equation for linear expansivity $\alpha$, using dimensional analysis. Here’s the step-by-step solution:

Problem Details

1. The equation for $\alpha$ is given as: $\alpha = \frac{\Delta L}{L_0 (\Delta \theta)^x}.$
2. The SI unit of $\alpha$ is $K^{-1}$ (Kelvin inverse).
3. We are tasked to find the values of $x$ and $y$ (dimensional exponents) in the equation using dimensional analysis, and confirm: $x = -1, \quad y = -1.$

Solution Using Dimensional Analysis

Dimensional Formula of Given Quantities:

1. Change in length ($\Delta L$) and initial length ($L_0$): Both have dimensions of length, denoted as $[L]$.
2. Temperature change ($\Delta \theta$): This has the dimension of temperature, denoted as $[\theta]$.
3. Linear expansivity ($\alpha$): Its SI unit is $K^{-1}$, so the dimension is $[\theta^{-1}]$.

Writing the Dimensional Equation:

From the given equation: $\alpha = \frac{\Delta L}{L_0 \cdot (\Delta \theta)^x}.$ Substitute the dimensions of all terms: [ [\alpha] = \frac{[\Delta L]}{[L_0] \cdot [\Delta \theta]^x}. ]

The dimensions become: [ [\theta^{-1}] = \frac{[L]}{[L] \cdot [\theta]^x}. ]

Simplifying:

Cancel $[L]$ (since $\Delta L$ and $L_0$ both have the same dimension): [ [\theta^{-1}] = [\theta]^{-x}. ]

Equating the powers of $[\theta]$ on both sides: $-1 = -x.$

Thus: $x = 1.$

Final Equation:

The correct form of the equation becomes: $\alpha = \frac{\Delta L}{L_0 \cdot \Delta \theta}.$

This matches the condition where $\alpha$ has the unit $K^{-1}$.

Final Answer:

• $x = 1$

Would you like further details or clarification?
Here are 5 related questions to consider:

1. What is the physical significance of linear expansivity $\alpha$?
2. How would the equation change if volumetric expansivity was considered instead of linear?
3. What are the dimensions of thermal expansivity in terms of mass $[M]$, length $[L]$, time $[T]$, and temperature $[\theta]$?
4. Can this formula be applied universally to all materials, or are there exceptions?
5. How does dimensional analysis help in verifying equations in physics?

Tip: Always verify the consistency of units and dimensions when solving physical problems—it ensures the correctness of the formula.