Question

The potential energy u of a particle varies with distance x from a fixed origin as u=A x^1/2/x+B where A&B are constants. what will be the dimensions of A&B

Answer 1

According to the principle of homogeneity kf dimensions
Now dimension of B b dimension of x
[B] = [L]

For constant 'A'
The above is rearrange as. U= (A√x)/(x+B)

A= U(x+B)/(√x)
Now in dimension
[A] = [U(x+B)/(√x)]
= [ML²T^(-2)][L]/√[L]
=[ML^(5/2)T^(-2)].

These are the required dimension of constant'A'and 'B'.
Solution by MOHAMMAD M. H. RAZA

Answer 2

hope it helps you.......