Frequency of sound that can be produced by a pipe depends on length (l) of the pipe, atmospheric pressure (p) and density (d) of air, according to relation `v=(p^b d^c)/t^a` . Find the value of (a +b + c).

Answer

To solve the problem, we need to analyze the given relation for frequency \( v \) in terms of atmospheric pressure \( p \), density \( d \), and length \( l \) of the pipe. The relation is given as:

\[ v = \frac{p^b d^c}{l^a} \]

We need to find the value of \( a + b + c \).

### Step 1: Identify the dimensions of each variable
1. **Pressure (\( p \))**: The dimension of pressure is given by force per unit area.
- Force has the dimension \( [M L T^{-2}] \) (mass times acceleration).
- Area has the dimension \( [L^2] \).
- Therefore, the dimension of pressure is:
\[ [p] = \frac{[M L T^{-2}]}{[L^2]} = [M L^{-1} T^{-2}] \]

2. **Density (\( d \))**: The dimension of density is mass per unit volume.
- Volume has the dimension \( [L^3] \).
- Therefore, the dimension of density is:
\[ [d] = \frac{[M]}{[L^3]} = [M L^{-3}] \]

3. **Length (\( l \))**: The dimension of length is simply:
\[ [l] = [L] \]

4. **Frequency (\( v \))**: The dimension of frequency is the inverse of time.
\[ [v] = [T^{-1}] \]

### Step 2: Substitute the dimensions into the equation
Substituting the dimensions into the equation \( v = \frac{p^b d^c}{l^a} \):

\[ [T^{-1}] = \frac{([M L^{-1} T^{-2}])^b \cdot ([M L^{-3}])^c}{([L])^a} \]

### Step 3: Expand the right-hand side
Expanding the right-hand side gives:

\[ [T^{-1}] = \frac{[M^b L^{-b} T^{-2b}] \cdot [M^c L^{-3c}]}{[L^a]} \]

Combining the terms:

\[ [T^{-1}] = \frac{[M^{b+c} L^{-b-3c} T^{-2b}]}{[L^a]} = [M^{b+c} L^{-b-3c-a} T^{-2b}] \]

### Step 4: Equate the dimensions
Now we equate the dimensions on both sides:

1. For mass \( M \):
\[ b + c = 0 \quad \text{(1)} \]

2. For length \( L \):
\[ -b - 3c - a = 0 \quad \text{(2)} \]

3. For time \( T \):
\[ -2b = -1 \quad \Rightarrow \quad 2b = 1 \quad \Rightarrow \quad b = \frac{1}{2} \quad \text{(3)} \]

### Step 5: Solve for \( c \) and \( a \)
Substituting \( b = \frac{1}{2} \) into equation (1):

\[ \frac{1}{2} + c = 0 \quad \Rightarrow \quad c = -\frac{1}{2} \quad \text{(4)} \]

Now substituting \( b = \frac{1}{2} \) and \( c = -\frac{1}{2} \) into equation (2):

\[ -\frac{1}{2} - 3(-\frac{1}{2}) - a = 0 \]

This simplifies to:

\[ -\frac{1}{2} + \frac{3}{2} - a = 0 \quad \Rightarrow \quad 2 - a = 0 \quad \Rightarrow \quad a = 1 \quad \text{(5)} \]

### Step 6: Calculate \( a + b + c \)
Now we can find \( a + b + c \):

\[ a + b + c = 1 + \frac{1}{2} - \frac{1}{2} = 1 \]

### Final Answer
Thus, the value of \( a + b + c \) is:

\[ \boxed{1} \]