The potential energy of a particle from a distance x from an origin, changes according to the formula U=(Asqrtx)/(x+B) where A and B are constant so the dimension of AB=……

The potential energy of a particle from a distance x from an origin, c

1.	The potential energy of a particle from a distance x from an origin, changes according to the formula U=(Asqrtx)/(x+B) where A and B are constant so the dimension of AB=……
Answer» `M^(1)L^(5/2)T^(-2)` `M^(1)L^(2)T^(-2)` `M^(3/2)L^(3/2)T^(-2)` `M^(1)L^(7/2)T^(-2)` Solution :In `U=(Asqrtx)/(x+B)`, the DIMENSION of B `=M^(0)L^(1)T^(0)""....(i)` ( `:.x+B & x` is distance) The dimension of `Axx"(distance")^(1/2)` = The dimension of U `xx` dimension of distance `("":.Asqrtx=U(x+B))` `:.` The dimension of A `=(M^(1)L^(2)T^(-)xxL^(1))/(L^(1/2))` `:.[A]=M^(1)L^(5/2)T^(-2)""......(ii)` `:.` The dimension of AB `=M^(1)L^(5/2)T^(-2)xxL^(1)` `:.[AB]=M^(1)L^(7/2)T^(-2)`