The potential energy of a particle changes with distance x from a fixed origin as V = A x x + B , where A and B are constant with appropriate dimensions. The dimensions of AB are _______.

Step-by-step solution

By the principle of dimensional homogeneity, quantities added or subtracted must have the same dimensions. In the denominator, B is added to x (distance). [B] = [x] = [L] The dimensions of potential energy V are [M^1 L^2 T⁻²] . From the given equation V = A x x + B , we can write the dimensional formula as: [V] = [A][x]^ 1/2 [x + B] [M^1 L^2 T⁻²] = [A][L]^ 1/2 [L] [M^1 L^2 T⁻²] = [A][L]^ -1/2 [A] = [M^1 L^2 T⁻²][L]^ 1/2 = [M^1 L^ 5/2 T⁻²] Now, the dimensions of AB are: [AB] = [A][B] = [M^1 L^ 5/2 T⁻²][L] = [M^1 L^