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(C) The period $T$ is given by $T \propto P^{a} d^{b} E^{c}$.Dimensional formulas are: $[T] = [T^1]$,$[P] = [M^1 L^{-1} T^{-2}]$,$[d] = [M^1 L^{-3}]$,

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$\frac{-5}{6}, \frac{1}{2}, \frac{1}{3}$

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(C) The period $T$ is given by $T \propto P^{a} d^{b} E^{c}$.Dimensional formulas are: $[T] = [T^1]$,$[P] = [M^1 L^{-1} T^{-2}]$,$[d] = [M^1 L^{-3}]$,and $[E] = [M^1 L^2 T^{-2}]$.Substituting these into the equation: $[T^1] = [M^1 L^{-1} T^{-2}]^{a} [M^1 L^{-3}]^{b} [M^1 L^2 T^{-2}]^{c}$.Equating the powers of $M, L,$ and $T$ on both sides:For $M: a + b + c = 0$ (Eq. $1$)For $L: -a - 3b + 2c = 0$ (Eq. $2$)For $T: -2a - 2c = 1$ (Eq. $3$)From Eq. $3$,$a + c = -1/2$,so $c = -1/2 - a$.Substitute $c$ into Eq. $1$: $a + b + (-1/2 - a) = 0 \implies b = 1/2$.Substitute $b = 1/2$ into Eq. $2$: $-a - 3(1/2) + 2c = 0 \implies -a + 2c = 3/2$.Substitute $c = -1/2 - a$ into this: $-a + 2(-1/2 - a) = 3/2 \implies -a - 1 - 2a = 3/2 \implies -3a = 5/2 \implies a = -5/6$.Finally,$c = -1/2 - (-5/6) = -3/6 + 5/6 = 2/6 = 1/3$.Thus,$a = -5/6, b = 1/2, c = 1/3$.

$A$ gas bubble formed 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 energy of explosion. Then $a, b, c$ are,respectively:

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