A gas bubble, from an explosion under water, oscillates with a period proportional to `P^a d^b E^c`, where P is the static pressure , d is the density and E is the total energy of the explosion. Find the values of a,b and c.

Answer

To solve the problem, we need to find the values of \( a \), \( b \), and \( c \) in the relationship \( T \propto P^a d^b E^c \), where \( T \) is the period of oscillation, \( P \) is the static pressure, \( d \) is the density, and \( E \) is the total energy of the explosion. We will use dimensional analysis to find these values.

### Step 1: Identify the dimensions of each variable

1. **Time period \( T \)**: The dimension of time is given as:
\[ [T] = T^1 \]

2. **Pressure \( P \)**: Pressure is defined as force per unit area. The dimensions of force are \( [M L T^{-2}] \) and area is \( [L^2] \). Therefore, the dimension of pressure is:
\[ [P] = \frac{[M L T^{-2}]}{[L^2]} = [M L^{-1} T^{-2}] \]

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

4. **Energy \( E \)**: The dimension of energy can be derived from kinetic energy, which is \( \frac{1}{2}mv^2 \). The dimension of velocity \( v \) is \( [L T^{-1}] \), so:
\[ [E] = [M][L^2 T^{-2}] = [M L^2 T^{-2}] \]

### Step 2: Set up the dimensional equation

The relationship given is:
\[ [T] = k [P]^a [d]^b [E]^c \]
Substituting the dimensions we found:
\[ [T^1] = [M L^{-1} T^{-2}]^a [M L^{-3}]^b [M L^2 T^{-2}]^c \]

### Step 3: Expand the dimensions

Now we expand the right-hand side:
\[ [T^1] = [M^{a+b+c} L^{-a-3b+2c} T^{-2a-2c}] \]

### Step 4: Equate the dimensions

We equate the dimensions on both sides:
1. For mass \( M \):
\[ a + b + c = 0 \quad \text{(1)} \]

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

3. For time \( T \):
\[ -2a - 2c = 1 \quad \text{(3)} \]

### Step 5: Solve the equations

From equation (3):
\[ -2a - 2c = 1 \implies a + c = -\frac{1}{2} \quad \text{(4)} \]

Substituting equation (4) into equation (1):
\[ a + b + c = 0 \implies b = -a - c = -(-\frac{1}{2}) = \frac{1}{2} \]

Now substituting \( b \) into equation (1):
\[ a + \frac{1}{2} + c = 0 \implies c = -a - \frac{1}{2} \quad \text{(5)} \]

Substituting equation (5) into equation (4):
\[ a - a - \frac{1}{2} = -\frac{1}{2} \implies a = -\frac{5}{6} \]

Now substituting \( a \) back into equation (5) to find \( c \):
\[ c = -(-\frac{5}{6}) - \frac{1}{2} = \frac{5}{6} - \frac{3}{6} = \frac{1}{3} \]

Finally, substituting \( a \) into the equation for \( b \):
\[ b = -(-\frac{5}{6}) - \frac{1}{3} = \frac{5}{6} - \frac{2}{6} = \frac{1}{2} \]

### Final Values
Thus, we have:
\[ a = -\frac{5}{6}, \quad b = \frac{1}{2}, \quad c = \frac{1}{3} \]