A gas bubble, from an explosion under water, oscillates with a period $T$ proportional to $p^{a}d^{b}E^{c}$. Where '$P$' is the static pressure, '$d$' is the density of water and '$E$' is the total energy of the explosion. Find the values of $a$, $b$ and $c$.

Step 1: State the proportional relationship.

The period of oscillation $T$ is proportional to the static pressure $p$, density of water $d$, and total energy $E$.

$$T \propto p^{a} d^{b} E^{c}$$

Step 2: Determine the dimensions of each physical quantity.

Write the dimensional formulas for pressure, density, and energy in terms of mass ($M$), length ($L$), and time ($T$).

$$[p] = M L^{-1} T^{-2}$$ $$[d] = M L^{-3}$$ $$[E] = M L^{2} T^{-2}$$

Step 3: Set up the dimensional equation.

Equate the dimensions of both sides of the relation.

$$[T] = [p]^{a} [d]^{b} [E]^{c}$$ $$M^{0} L^{0} T^{1} = (M L^{-1} T^{-2})^{a} (M L^{-3})^{b} (M L^{2} T^{-2})^{c}$$ $$M^{0} L^{0} T^{1} = M^{a+b+c} L^{-a-3b+2c} T^{-2a-2c}$$

Step 4: Form a system of linear equations.

Equate the exponents of $M$, $L$, and $T$ on both sides.

$$a + b + c = 0$$ $$-a - 3b + 2c = 0$$ $$-2a - 2c = 1$$

Step 5: Solve for the exponents $a$, $b$, and $c$.

Simplify the equation for $T$.

$$a + c = -\frac{1}{2}$$

Substitute this into the equation for $M$ to find $b$.

$$\left(-\frac{1}{2}\right) + b = 0$$ $$b = \frac{1}{2}$$

Substitute $b = \frac{1}{2}$ into the equation for $L$.

$$-a - 3\left(\frac{1}{2}\right) + 2c = 0$$ $$-a + 2c = \frac{3}{2}$$

Add this equation to $a + c = -\frac{1}{2}$ to solve for $c$.

$$3c = 1$$ $$c = \frac{1}{3}$$

Solve for $a$ using $a + c = -\frac{1}{2}$.

$$a = -\frac{1}{2} - \frac{1}{3}$$ $$a = -\frac{5}{6}$$

Answer: $a = -\dfrac{5}{6}$, $b = \dfrac{1}{2}$, $c = \dfrac{1}{3}$