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=……

Options

A. `M^(1)L^(5/2)T^(-2)`
B. `M^(1)L^(2)T^(-2)`
C. `M^(3/2)L^(3/2)T^(-2)`
D. `M^(1)L^(7/2)T^(-2)`

Answer

A, B

Explanation

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)`