Using dimension derive the formular for the period of vibration of a drum of liquid at the end of a tube of radius r assuming that the period t depends on its surface tension, density and radius r of the drum

T = R^(3/2) ¥^(-1/2) d^(1/2)

Explanation:

T = R^a ¥^b d^c 1

Where,

R is the radius of drop,

¥ is the surface tension and

d is the density of liquid .

Dimension of period(T) = [ T ]

dimension of radius (R) = [ L ]

dimension of surface tensn ¥ (F/l) = [MT⁻²]

dimension of density (M/V)= [ ML⁻³]

[ T ] = [L]^a [MT⁻²]^b [ML⁻³]^c

[T] = L^(a -3c) M^(b + c) T^(-2b)

Compare both sides,

-2b = 1

===> b = -1/2 ,

b + c = 0

-1/2 + c = 0

> c = 1/2 and

a -3c = 0

a = 3c =3(1/2)

a = 3/2

Put the value of a, b and c in equation 1 above

Thus,

T = R^(3/2) ¥^(-1/2) d^(1/2)

Goodluck

magnets not strong enough

explanation: