We use the method of dimensions to solve for the exponents $a$, $b$, and $c$.

1. Identify Dimensions: Period ($T$): $[T]$ Pressure ($P$): Force/Area = $[ML^{-1}T^{-2}]$ . Density ($D$): Mass/Volume = $[ML^{-3}]$ . Surface Tension ($S$): Force/Length = $[MT^{-2}]$ .

2. Equate Dimensions: $[T] = [P]^a [D]^b [S]^c$ $[M^0 L^0 T^1] = [ML^{-1}T^{-2}]^a [ML^{-3}]^b [MT^{-2}]^c$ $[M^0 L^0 T^1] = [M^{a+b+c} L^{-a-3b} T^{-2a-2c}]$

3. Solve System of Equations: For Mass ($M$): $a + b + c = 0$ (Eq. 1) For Length ($L$): $-a - 3b = 0 \Rightarrow a = -3b$ (Eq. 2)

• For Time ($T$): $-2a - 2c = 1$ (Eq. 3)

4. Calculation: Substitute $a = -3b$ into Eq. 1: $(-3b) + b + c = 0 \Rightarrow c = 2b$. Substitute $a = -3b$ and $c = 2b$ into Eq. 3: $-2(-3b) - 2(2b) = 1$ $6b - 4b = 1 \Rightarrow 2b = 1 \Rightarrow b = 1/2$ Find $a$: $a = -3(1/2) = -3/2$. Find $c$: $c = 2(1/2) = 1$.

Therefore, $a = -3/2, b = 1/2, c = 1$.