if energy velocity and time are fundamental units what will be the dimensions of surface tension

Open in App

Solution

Let the relation be
$S = E^{a} V^{b} T^{c} - - - (1)$
where
S = Surface Tension
E = Energy
V = Velocity
T = Time
Their dimensions are
$[S] = M T^{- 2}, [E] = M L^{2} T^{- 2}, [V] = L T^{- 1}, [T] = T$
Putting in Eq.(1) we get
$M T^{- 2} = {(M L^{2} T^{- 2})}^{a} {(L T^{- 1})}^{b} {(T)}^{c} \Rightarrow M T^{- 2} = M^{a} L^{2 a + b} T^{- 2 a - b + c}$
Equating the coefficient we get
$a = 1 2 a + b = 0 \Rightarrow b = - 2 - 2 a - b + c = - 2 \Rightarrow c = - 2$
Therefore, dimension of surface tension is $[S] = E V^{- 2} T^{- 2}$

Suggest Corrections

Similar questions

(E)

(v)

(T)

If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, thendimensional formula of surface tension will be :

E

V

T

^{'} K^{'},

^{'} v^{'}

^{'} T^{'}

S =

V

k

T

Join BYJU'S Learning Program