A U - tube of uniform cross section contains mercury upto a height h in either limb. The mercury in one limbe is depressed a little and then relased. Obtain an expression for time period (T) of oscillation, assuming that T depends on h,`rho` and g, where `rho` is density of mercury.

Answer

`T =Ksqrt((h)/(g))`

Explanation

Let `T = Kh^a rho^b g^c`
where a,b,c are the dimensions and k is
dimensionless constant.
Writing the dimensions in (i) we get
`[M^0L^0T^1] = L^a(ML^(-3))^b (LT^(-2))^c`
`=M^b L^(a -3b +c T^(-2c)`
Applying the principle of homogeneity of
dimensions, we get
`b= 0, a-3b +c =0, -2c =1, c = (1)/(2).`
From `a-3b + c =0, a-3xx0 - (1)/(2) =0, a =(1)/(2)`1
Putting these value in (i) we get
`T = K h^(1//2) rho^0 g^(-1//2)`
`T = K(sqrth/sqrtg)`