NEET2010PhysicsUnits and DimensionsActual
The potential energy of a particle varies with distance x from fixed a origin as v= ( A x x+B ) ; where, A and B are constants. The dimensions of A B are
Options
- A[ ML ^ 5 / 2 ~T ⁻² ]
- B[ ML ^2 ~T ⁻² ]
- C[ M ^ 3 / 2 ~L ^ 3 / 2 ~T ⁻² ]
- D[ ML ^ 7 / 2 ~T ⁻² ]
Correct answer
D. [ ML ^ 7 / 2 ~T ⁻² ]
Step-by-step solution
Given, v= A x x+B ....(i) Dimensions of v= dimensions of potential energy = [ ML ^2 ~T ⁻² ] From Eq. (i) Dimensions of B= dimensions of x= [ M ^0 LT ^0 ] Dimensions of A aligned & = dimension of v dimensions of (x+B) dimensions of x & = [ ML ^2 ~T ⁻² ] [ M ^0 LT ^0 ] [ M ^0 ~L ^ 1 / 2 ~T ^0 ] = [ ML ^ 5 / 2 ~T ⁻² ] aligned Hence, dimensions of A B= [ ML ^ 5 / 2 ~T ⁻² ] [ M ^0 LT ^0 ]= [ ML ^ 7 / 2 ~T ⁻² ]