NEET2024PhysicsUnits and DimensionsActual
The potential energy of a particle moving along x -direction varies as V= A x^2 x +B . The dimensions of A^2 B are:
Options
- A[ M ^ 3 / 2 ~L ^ 1 / 2 ~T ⁻³ ]
- B[M^ 1 / 2 L⁻³ ]
- C[ M ^2 ~L ^ 1 / 2 ~T ⁻⁴ ]
- D[ ML ^2 ~T ⁻⁴ ]
Correct answer
C. [ M ^2 ~L ^ 1 / 2 ~T ⁻⁴ ]
Step-by-step solution
V= A x^2 x +B As per homogeneous rule B= L ML ^2 ~T ⁻²= A ~L ^2 ~L ^ 1 / 2 A= ML ^ 1 / 2 ~T ⁻² A^2 B = M ^2 L T⁻⁴ ~L ^ 1 / 2 = M ^2 ~L ^ 1 / 2 ~T ⁻⁴