UPSEE Solved Model Paper 7

Dimension of density is
[d] = [ML−3]
Dimension of radius is
[r] = [L]
Dimension of surface tension is
[S] = [

force

length

] = [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 =

1

2

y =

3

2

z = −

1

2

Using above values of x, y and z in eq (I), we get
T α d

1

2

r

3

2

S−

1

2

⇒ T = K√

dr3

S

Hence, option 'B' is correct