If pressure [P] , velocity [V] and time [T] are chosen as the...

Let $[Force] = [P]^a [V]^b [T]^c$

Substituting the standard dimensional formulas for each quantity:
$[MLT^{-2}] = [ML^{-1}T^{-2}]^a [LT^{-1}]^b [T]^c$
$[MLT^{-2}] = [M^a L^{-a+b} T^{-2a-b+c}]$

Comparing the powers of $M$, $L$, and $T$ on both sides:
For $M$: $a = 1$
For $L$: $-a + b = 1 \Rightarrow -1 + b = 1 \Rightarrow b = 2$
For $T$: $-2a - b + c = -2 \Rightarrow -2(1) - 2 + c = -2 \Rightarrow -4 + c = -2 \Rightarrow c = 2$

Therefore, the dimensional formula of force in this new system is $[P][V]^2[T]^2$.

Answer: $[P][V]^2[T]^2$