If the time period (t) of vibration of a liquid drop depends on density (rho) of the liquid, radius (r) of the drop and surface tension (S), then the expression of t is where k is a dimensionless constant.

If the time period (t) of vibration of a liquid drop depends on densit

1.	If the time period (t) of vibration of a liquid drop depends on density (rho) of the liquid, radius (r) of the drop and surface tension (S), then the expression of t is where k is a dimensionless constant.
Answer» `t= ksqrt((rhor^(3))/(S))` surface tension `t= ksqrt((rhor^(3))/(S^(1//2)))` `t= ksqrt((rhor)/(S))` Solution :Let `t= krho^(x)r^(y)S^(z)` where k is a dimensionless constant Equating dimensions on both sides, we get `[T]= [ML^(-3)]^(x)[L]^(y)[ML^(0)T^(-2)]^(z)` or `[M^(0)L^(0)T]= [M^(x+z)L^(-3x+z)T^(-2Z)]` According to PRINCIPLE of homogeneity of dimensions we get x+z= 0...(i) -3x+y=0 ..(ii) -2z = 1 ...(iii) SOLVING equations (i), (ii), and (iii), we get `x= (1)/(2), y= (3)/(2), z= (-1)/(2) :. t= krho^(1//2)r^(3//2)S^(-1//2)` or `t= ksqrt((rhor^(3))/(S))`