Dimension of density is
[d] =
[ML−3] Dimension of radius is
[r] = [L]
Dimension of surface tension is
[S] =
[] =
[MT−2] The time period of the small liquid drop can be represented as
T α
sxrySz where x. y and z have some numerical values, to be evaluated
Thus T =
KdxrySz ... (i)
where, K is a dimensionless constant.
applying the principle of homogeneity of dimensions in eq. (i) we get,
[T] =
[ML−3]x[L]y[MT−2]z ⇒
[M0L0T−1] =
[M]x+z[L]−3x+y[T]−2z Equating the powers of M L and T on both sides of the above equation, we get
x + z = 0
- 3x + y = 0
- 2z = 1
Solving the above equations, we obtain
x =
y =
z =
− Using above values of x, y and z in eq (I), we get
T α
drS− ⇒ T =
K√ Hence, option 'B' is correct