We know,

surface tension (S) = $${F \over L}$$ = $${{\left[ {ML{T^{ - 2}}} \right]} \over {\left[ L \right]}}$$

$$ \therefore $$ [S] = $${\left[ {M{T^{ - 2}}} \right]}$$

Moment of inertia (I) = mr²

$$ \therefore $$ [I] = $${\left[ {M{L^2}} \right]}$$

Planck's constant (h) = $${E \over f}$$ = Et

$$ \therefore $$ [h] = $${\left[ {M{L^2}{T^{ - 1}}} \right]}$$

Also linear momentum (p) = mv = $${\left[ {ML{T^{ - 1}}} \right]}$$

Now we have to express p in terms of s, I and h.

$$ \therefore $$ Let, [P] = [S^a I^b h^c]

$$ \Rightarrow $$ $${\left[ {ML{T^{ - 1}}} \right]}$$ = $${\left[ {M{T^{ - 2}}} \right]}$$^a $${\left[ {M{L^2}} \right]}$$^b $${\left[ {M{L^2}{T^{ - 1}}} \right]}$$^c

$$ \Rightarrow $$ $${\left[ {ML{T^{ - 1}}} \right]}$$ = [ M^{a + b + c} L^{2b +2c} T^{- 2a - c} ]

By comparing the dimensions of both sides, we get

a + b + c = 1 .........(1)

2b +2c = 1 ..............(2)

- 2a - c = -1 ...................(3)

By solving those three equations we get,

a = $${1 \over 2}$$

b = $${1 \over 2}$$

c = 0

$$ \therefore $$ linear momentum [p] = [S^1/2I^1/2h⁰]