If Energy (E), velcity (v) and time (T) are fundamental units. What will be the dimension of surface tension ?

Options

A. `[EV^(-1) T^(-2)]`
B. `[EV^(-2)T^(-2)]`
C. `[E^(-2) V^(-1) T^(-3)]`
D. `[EV ^(-2)T^(-1)]`

Answer

B

Explanation

Applying dimensional analysis
`S prop E^(a)V^(b) T ^(c )`
`[M^(1)L^(0) T^(-2)] = k [M^(1)L^(2)T^(-2)]^(a) [L^(1)T^(-1)]^(b)[ T^(1)]^(c )`
`[M^(1)L^(0)T^(-2)] = k [M^(a)L^(2a) T^(-2a)] [L^(b)T^(-b)] [T^(c )]`
`[M^(1)L^(0)T^(2)] = k [M]^(a) [L]^(2a +b) [T]^(-2a-b+c)`
Comparision
`a =1 " "2a+b=0 " " -2= -2a -b +c`
`" " b=-2 " " -2 = -2(1)+ 2+c`
`" " c =-2`
So the dimensional formula for surface tension will be `[E^(1)V^(-2)T^(-2)]`
Alternate solution :
Surface Tension = `("Surface energy")/("Area")`
[Surface tension] = `([E])/([V. T]^(2)) = [E V^(-2)T^(-2)]`