Question

) The period of vibration, T of a liquid drop is given by. T =kapy,
where K is a dimension constant, a'is the radius of the drop, S is
the density of the liquid andy is the surface tension. By
dimension, find the values of the indices x,y,z and the relationship
for T.​

Answer 1

Answer:

The indices x,y, and z not showing but lets assume T =ka^x S^y y^z

Dimension of a the radius is L

dimension of density is ML^-3

dimension of surface tension(F/L) is MT^-2

dimension of period is T

T =ka^x S^y y^z

T = L^x (ML^-3)^y (MT^-2)^z

T = L^x M^yL^-3y M^zT^-2z

M^0 L^0 T^1 = L^(x-3y) M^(y+z) T^(-2z)

T^1 = T^(-2z)

1 = -2z

==> z = -1/2

M^0 = M^(y+z)

y + z =0

y = -z

y = -(-1/2)

y = 1/2

L^0 = L^(x-3y)

x - 3y = 0

x = 3y

x = 3(1/2)

x = 3/2

Thus, x = 3/2, y = 1/2 and z = -1/2

The relationship for T

T = ka^(3/2) S^(1/2) y^(-1/2)