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 \( x \), where \( v \) is the velocity of sound, \( E \) is elasticity, and \( d \) is density, we will compare the dimensions of both sides of the equation.

### Step-by-Step Solution:

1. **Identify the dimensions of velocity (v)**:
- The velocity of sound \( v \) has dimensions of length per time, which can be expressed as:
\[ [v] = LT^{-1} \]

2. **Identify the dimensions of density (d)**:
- Density \( d \) is defined as mass per unit volume. Thus, its dimensions are:
\[ [d] = \frac{M}{L^3} = ML^{-3} \]

3. **Identify the dimensions of elasticity (E)**:
- Elasticity \( E \) has dimensions of pressure, which is force per unit area. The dimensions of force are \( [F] = MLT^{-2} \) and area is \( L^2 \). Therefore, the dimensions of elasticity are:
\[ [E] = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} \]

4. **Substitute the dimensions into the equation**:
- Now, substituting the dimensions of \( d \) and \( E \) into the right-hand side of the equation:
\[ \frac{d}{E} = \frac{ML^{-3}}{ML^{-1}T^{-2}} = \frac{1}{L^{-3}} \cdot \frac{L^1T^2}{1} = L^{-2}T^2 \]

5. **Raise the right-hand side to the power of x**:
- The equation now becomes:
\[ v = \left( L^{-2}T^2 \right)^x \]
- This expands to:
\[ v = L^{-2x}T^{2x} \]

6. **Set the dimensions equal to each other**:
- Now we equate the dimensions from both sides:
\[ LT^{-1} = L^{-2x}T^{2x} \]

7. **Compare the powers of L and T**:
- For the dimensions of \( L \):
\[ 1 = -2x \quad \Rightarrow \quad x = -\frac{1}{2} \]
- For the dimensions of \( T \):
\[ -1 = 2x \quad \Rightarrow \quad x = -\frac{1}{2} \]

8. **Conclusion**:
- In both cases, we find that:
\[ x = -\frac{1}{2} \]
- Therefore, the value of \( x \) is \( -\frac{1}{2} \).

### Final Answer:
\[ x = -\frac{1}{2} \]