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A student forgot Newton's formula for speed of sound but he knows there were speed `(upsilon)` pressure (p) and density (d) in the formula. He then starts using dimensional analysis method to find the actual relation. `upsilon = kp^(x)d^(y)`. Where k is a dimensionless constant on the basis of above passage answer the following questions.
The value of x is

A

1

B

`(1)/(2)`

C

`-(1)/(2)`

D

2

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To find the value of \( x \) in the equation \( \upsilon = k p^{x} d^{y} \) using dimensional analysis, we will follow these steps: ### Step 1: Identify the dimensions of each variable 1. **Speed (\( \upsilon \))**: The dimension of speed is \( L T^{-1} \) (length per time). 2. **Pressure (\( p \))**: Pressure is defined as force per unit area. The dimension of force is \( M L T^{-2} \), and area is \( L^2 \). Therefore, the dimension of pressure is: \[ [p] = \frac{[F]}{[A]} = \frac{M L T^{-2}}{L^2} = M L^{-1} T^{-2} \] 3. **Density (\( d \))**: Density is defined as mass per unit volume. The dimension of volume is \( L^3 \). Therefore, the dimension of density is: \[ [d] = \frac{[M]}{[V]} = \frac{M}{L^3} = M L^{-3} \] ### Step 2: Write the dimensional equation We can express the dimensions of \( \upsilon \) in terms of \( p \) and \( d \): \[ [L T^{-1}] = [p^{x}] [d^{y}] \] Substituting the dimensions we found: \[ L T^{-1} = (M L^{-1} T^{-2})^{x} (M L^{-3})^{y} \] ### Step 3: Expand the right-hand side Expanding the right-hand side gives: \[ L T^{-1} = M^{x+y} L^{-x-3y} T^{-2x} \] ### Step 4: Equate the dimensions Now we equate the dimensions on both sides: 1. For mass (\( M \)): \[ 0 = x + y \quad \text{(since there is no mass on the left side)} \] 2. For length (\( L \)): \[ 1 = -x - 3y \] 3. For time (\( T \)): \[ -1 = -2x \] ### Step 5: Solve the equations From the time equation: \[ -1 = -2x \implies x = \frac{1}{2} \] Now substituting \( x = \frac{1}{2} \) into the mass equation: \[ 0 = \frac{1}{2} + y \implies y = -\frac{1}{2} \] ### Conclusion The value of \( x \) is: \[ \boxed{\frac{1}{2}} \]
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