If v stands for velocity of sound, E is elasticity and d the density, then find x in the equation `v=((d)/(E))^(x)`

Options

A. 1
B. `1//2`
C. 2
D. `-1//2`

Answer

To solve the equation \( v = \left( \frac{d}{E} \right)^{x} \) for \( x \), we will use dimensional analysis. Here are the steps:

### Step 1: Identify the dimensions of each variable
- **Velocity (v)**: The dimension of velocity is given by \( [v] = L T^{-1} \), where \( L \) is length and \( T \) is time.
- **Density (d)**: Density is defined as mass per unit volume. Therefore, the dimension of density is:
\[ [d] = \frac{M}{L^3} = M L^{-3} \]
- **Elasticity (E)**: Elasticity is defined as stress divided by strain. Stress has the dimension of force per area. The dimension of force is \( [F] = M L T^{-2} \), and area is \( L^2 \). Thus:
\[ [E] = \frac{[F]}{[A]} = \frac{M L T^{-2}}{L^2} = M L^{-1} T^{-2} \]

### Step 2: Substitute the dimensions into the equation
Now, substituting the dimensions into the equation \( v = \left( \frac{d}{E} \right)^{x} \):
\[ [L T^{-1}] = \left( \frac{M L^{-3}}{M L^{-1} T^{-2}} \right)^{x} \]

### Step 3: Simplify the right side
Simplifying the expression inside the parentheses:
\[ \frac{d}{E} = \frac{M L^{-3}}{M L^{-1} T^{-2}} = \frac{L^{-3}}{L^{-1} T^{-2}} = L^{-3 + 1} T^{2} = L^{-2} T^{2} \]
Thus, we have:
\[ v = \left( L^{-2} T^{2} \right)^{x} \]

### Step 4: Expand the right side
Expanding the right side gives:
\[ v = L^{-2x} T^{2x} \]

### Step 5: Set dimensions equal
Now, we equate the dimensions from both sides:
\[ L T^{-1} = L^{-2x} T^{2x} \]

### Step 6: Equate the coefficients
From the dimensions of length \( L \):
\[ 1 = -2x \quad \Rightarrow \quad x = -\frac{1}{2} \]

From the dimensions of time \( T \):
\[ -1 = 2x \quad \Rightarrow \quad x = -\frac{1}{2} \]

Both equations give the same value for \( x \).

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{-\frac{1}{2}} \]