Question

An explosion at time \(t = 0\) releases energy \(E\) at the origin in a space filled with a gas of density \(\rho\). Subsequently, a hemispherical blast wave propagates radially outwards as shown in the figure. 
Let \(R\) denote the radius of the front of the hemispherical blast wave. The radius \(R\) follows the relationship \(R = k t^a E^b \rho^c\), where \(k\) is a dimensionless constant. The value of exponent \(a\) is ................ 
(Rounded off to one decimal place) 

 

Hint:

Dimensional analysis is a powerful tool for checking equations and deriving relationships between physical quantities. Memorize the fundamental dimensions of common physical quantities like Force ([MLT⁻²]), Energy/Work ([ML²T⁻²]), Power ([ML²T⁻³]), Pressure ([ML⁻¹T⁻²]), and Viscosity ([ML⁻¹T⁻¹]). This will save you significant time in the exam.

Answer 1

Step 1: Understanding the Concept:
This problem uses the principle of dimensional analysis. It states that any physically meaningful equation must be dimensionally homogeneous, meaning the dimensions on both sides of the equation must be the same. By expressing each variable in the given relationship in terms of fundamental dimensions (Mass [M], Length [L], Time [T]), we can solve for the unknown exponents.
Step 2: Key Formula or Approach:
The relationship is given as \(R = k t^a E^b \rho^c\). We need to write down the fundamental dimensions for each quantity:
- Radius, \(R\): [L]
- Time, \(t\): [T]
- Energy, \(E\): Energy is work (Force × Distance), so [F][L] = [\(MLT^{-2}\)][L] = [\(ML^{2}T^{-2}\)].
- Density, \(\rho\): Mass/Volume = [M]/[\(L^3\)] = [\(ML^{-3}\)].
- Constant, \(k\): Dimensionless, [\(M^{0}L^{0}T^{0}\)].
Step 3: Detailed Explanation or Calculation:
Substitute the dimensions into the equation:
\[ [L] = [T]^a [ML^2T^{-2}]^b [ML^{-3}]^c \] Now, expand the exponents on the right side:
\[ [M^0 L^1 T^0] = [T^a] [M^b L^{2b} T^{-2b}] [M^c L^{-3c}] \] Combine terms with the same base dimension by adding their exponents:
\[ [M^0 L^1 T^0] = [M^{b+c}] [L^{2b-3c}] [T^{a-2b}] \] For the equation to be dimensionally homogeneous, the exponents of M, L, and T on both sides must be equal. This gives us a system of three linear equations:
1. For Mass (M): \( b + c = 0 \)
2. For Length (L): \( 2b - 3c = 1 \)
3. For Time (T): \( a - 2b = 0 \)
Let's solve this system of equations.
From equation (1), we get \(c = -b\).
Substitute this into equation (2):
\[ 2b - 3(-b) = 1 \] \[ 2b + 3b = 1 \] \[ 5b = 1 \implies b = \frac{1}{5} = 0.2 \] Now use equation (3) to find the value of \(a\):
\[ a - 2b = 0 \implies a = 2b \] \[ a = 2 \times \frac{1}{5} = \frac{2}{5} = 0.4 \] (For completeness, we can also find c: \(c = -b = -1/5 = -0.2\)).
The question asks for the value of exponent \(a\).
Step 4: Final Answer:
The value of exponent \(a\) is 0.4.
Step 5: Why This is Correct:
The solution is based on the fundamental principle of dimensional homogeneity. By correctly identifying the dimensions of energy, density, time, and length, we established a system of linear equations for the unknown exponents. Solving this system yields \(a=0.4\). This result, known as part of the Taylor-von Neumann-Sedov blast wave solution, is a classic result from dimensional analysis, and the calculated value falls exactly within the provided answer range of 0.39 to 0.41.