QNEET-2024-Ph-169: NEET 2024 Physics - Class 11 NEET-2024-Ph-169 Solution: The potential energy of a particle moving along x-direction…

To determine the dimensions of A 2 B , we start with the given potential energy expression: V = A x 2 x + B The dimensions of potential energy ( V ) are [ M L 2 T _ 2 ] . So, we need to consider the dimensions of the other terms in the expression to match these dimensions. Let's break it down step-by-step. For the term x , where x represents distance: x has dimensions [ L 1 / 2 ] Since both terms in the denominator x + B have to be of the same dimensions for them to be added together, the dimensions of B must also be [ L 1 / 2 ] . Now the expression becomes: V = A x 2 L 1 / 2 Thus, the dimensions must be: [ V ] = [ A x 2 L 1 / 2 ] Substitute the dimensions of V , x , and L : [ M L 2 T _ 2 ] = [ A L 2 L 1 / 2 ] Simplifying the right-hand side: [ M L 2 T _ 2 ] = [ A L 3 / 2 ] Therefore, the dimensions of A must be: [ A ] = [ M L 2 T _ 2 ] _ [ L _ 3 / 2 ] [ A ] = [ M L 1 / 2 T _ 2 ] We have dimensions of A and B . Now we need to find the dimensions of A 2 B : [ A 2 ] = [ M L 1 / 2 T _ 2 ] 2 [ A 2 ] = [ M 2 L T _ 4 ] Divide by the dimensions of B : [ A 2 B ] = [ M 2 L T _ 4 L 1 / 2 ] [ A 2 B ] = [ M 2 L 1 / 2 T _ 4 ] Thus, the dimensions of A 2 B are: Option C: [ M 2 L 1 / 2 T _ 4 ]