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: Identify the given relation.

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.

Find the dimensional formulas for time period, pressure, density, and energy.

$$[T] = M^{0}L^{0}T^{1}$$ $$[p] = \frac{[F]}{[A]}$$ $$[p] = \frac{MLT^{-2}}{L^{2}}$$ $$[p] = ML^{-1}T^{-2}$$ $$[d] = \frac{[m]}{[V]}$$ $$[d] = \frac{M}{L^{3}}$$ $$[d] = ML^{-3}$$ $$[E] = [F][x]$$ $$[E] = (MLT^{-2})(L)$$ $$[E] = ML^{2}T^{-2}$$

Step 3: Set up the dimensional equation.

Substitute the dimensional formulas into the proportionality relation.

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

Step 4: Equate the exponents of $M$, $L$, and $T$.

Compare the powers on both sides of the equation.

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

Step 5: Solve the system of equations.

Simplify the third equation to find a relation between $a$ and $c$.

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

Substitute this relation into the first equation to solve for $b$.

$$-\frac{1}{2} + b = 0$$ $$b = \frac{1}{2}$$

Substitute the value of $b$ into the second equation.

$$-a - \frac{3}{2} + 2c = 0$$ $$-a + 2c = \frac{3}{2}$$

Express $a$ in terms of $c$.

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

Substitute this expression into the modified second equation.

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

Solve for $a$ using the value of $c$.

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

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