NTA Abhyas JEE Main2020PhysicsUnits and DimensionsPractice
The potential energy of a particle varies with distance x from a fixed origin as U = A x x 2 + B where A and B are constants. The dimensions of A are -
Options
- AM L 5 2 T - 2
- BM 1 L 2 T - 2
- CM 3 2 L 5 2 T - 2
- DM 1 L 7 2 T - 2
Correct answer
D. M 1 L 7 2 T - 2
Step-by-step solution
U = A x x 2 + B .....(i) Where U = Potential energy x = distance Dimensional formula of potential energy, U = M 1   L 2   T - 2 Dimensional formula of distance, x = M 0   L 1   T 0 From equation (i), we can write [ x 2 ] = [ B ] [ B ] =   M 0   L 1   T 0 2 [ B ] = M 0   L 2   T 0 Dimensional formula of B = M 0   L 2   T 0 We can write equation (i) as, U = A x x 2 U = A x 3 2 A = U x 3 2 A = M 1   L 2   T - 2 × M 0   L 1   T 0 3 2 A = M