Question

Question #b47e0

Answer 1

`a=1/2, b=-1/2`

Explanation:

Given equation is

`PpropL^ag^b`
where `P" has dimension of time"=>T^1`, `L" has dimension of Length"=>L^1` and
`g" acceleration due to gravity has units as "ms^-2=>L^1T^-2`

Now writing given equation in dimensional form we get
`T^1prop(L^1)^a.(L^1T^-2)^b`

To include both dimension on either side of the equation it can be rewritten as
`L^0T^1prop(L^1)^a.(L^1T^-2)^b`
`=>L^0T^1propL^(a+b).T^(-2b)`

Comparing exponents of each dimension on LHS and RHS we get
`0=a+b` ........(1)
`1=-2b` ..........(2)
Solving (2) for `b`, we get
`b=-1/2`
Inserting this values in (1) we get
`a=1/2`