The potential energy of a particle changes with distance x from a fixed origin as V= A xx+B, where A and B are constants with appropriate dimensions. The dimensions of AB are .
- [ M1L5/2T-2]
- [ M3/2L5/2T-2]
- [ M1L2T-2]
- [ M1L7/2T-2]
Answer
(4) [ M1L7/2T-2]
Solution
∵ x+B is a sum, B has the dimensions of x:
[B]=[L]
V= A xx+B ⇒ [V]= [A][ L1/2][L]=[A][ L-1/2]
∵ V is an energy, [V]=[ ML2T-2]
⇒ [A]=[ ML2T-2][ L1/2]=[ ML5/2T-2]
∴ [AB]=[ ML5/2T-2][L]=[ M1L7/2T-2]
Answer: (4)